In what time will the simple interest on a certain sum of money at 25 2 )% per annum be 3/8 of itself?

Answer

In what time will the simple interest on a certain sum of money at 25 2 )% per annum be 3/8 of itself?
Verified

Hint:- In 8 years money from Interest will be come equal to the principal
amount invested. So, money had been doubled in 8 years.

Let the initial amount of money invested will be Rs. x.
Then after 8 years money had become 2x.
Out of Rs. 2x, money from interest will be 2x – initial amount invested = 2x – x = x.
Let the rate of interest be r.

So, now we will use a simple interest formula.
According to Simple Interest (S.I) formula.
\[ \Rightarrow S.I. = \dfrac{{PRT}}{{100}}\]. Where P is principal amount, R is rate of interest and T will be time period.

So, putting the values in the above formula. We will get,
\[ \Rightarrow x = \dfrac{{xr(8)}}{{100}}\]
On solving the above equation. We will get,
\[ \Rightarrow {\text{ }}r{\text{ }} = {\text{ }}\dfrac{{100}}{8}{\text{ }} = {\text{ }}12.5\]

Hence, the rate of interest to double a money in 8 years will be 12.5% per annum.

Note:- Whenever we came up with this type of problem where we are asked to
find rate of interest then first, we will find the interest on principal amount by
subtracting principal amount from the money after 8 years and then we will
assume rate of interest to be r and then apply, Simple Interest formula and
find the required value of rate of interest.

Important Points :

  • Borrowed money is called Principal and it is denoted by ‘P’.

  • Money is borrowed for certain time period, that time is called interest time and it is denoted by ‘T’ or ‘t’.

  • The principal becomes Amount when interest is added to it Amount is represented as A.

  • So, Amount = Principal + Interest => A = P + S. I.
    OR
    Interest = Amount – Principal => S. I. = A – P

  • When Interest is payable half – yearly Rate will be half and time will be twice

  • When Interest is payable quarterly Rate will be one-fourth and time will be four times.

  • Rules:

    RULE 1:

    Simple Interest (S.I.)= \( \frac{Principal * Rate * Time}{100} \)
    or,
    S.I.= \( \frac{P * R * T}{100} \)
    P= \( \frac{S.I. * 100}{R * T} \), R= \( \frac{S.I. * 100}{P * T} \), T= \( \frac{S.I. * 100}{P * R} \),

    RULE 2:

    If there are distinct rates of interest for distincttime periods i.e.
    Rate for 1st t1 years -> R1%
    Rate for 2nd t2 years -> R2%
    Rate for 3rd t3 years -> R3%
    or,
    Then, Total S.I. for 3 years = \( \frac{P(R1T1 * R2T2 * R3T3)}{100} \)

    RULE 3:

    If a certain sum becomes ‘n’ times of itself in T years on Simple Interest, then the rate per cent per annum is,
    R% = \( \frac{(n-1)}{T} \)x100% and
    T = \( \frac{(n-1)}{R} \)x100%

    RULE 4:

    If a certain sum becomes n1 times of itself at R1% rate and n2 times of itself at R2% rate, then,
    R2 = \( \frac{(n2-1)}{n1-1} \)xR1 and T2 = \( \frac{(n2-1)}{n1-1} \)xT1

    RULE 5:

    If Simple Interest (S.I.) becomes ‘n’ times of principal i.e.
    S.I. = P × n then.
    RT = n × 100

    RULE 6:

    If an Amount (A) becomes ‘n’ times of certain sum (P) i.e.
    A = Pn then,
    RT = (n – 1) × 100

    RULE 7:

    If the difference between two simple interests is ‘x’ calculated at different annual rates and times, then principal (P) is
    P = \( \frac{(x*100)}{dr * dt} \)
    where dr = Difference in rate & dt = Difference in time

    RULE 8:

    If a sum amounts to x1 in t years and then this sum amounts to x2 in t yrs. Then the sum is given by
    P = \( \frac{((da)*100)}{ (ci) * time} \)
    where da = Difference in amount & ci = Change in time

    RULE 9:

    If a sum with simple interest rate, amounts to ‘A’ in t1 years and ‘B’ in same t2 years, then,
    R% = \( \frac{(B-A)*100}{A.t2-B.t1} \) and
    P = \( \frac{A.t2-B.t1}{t2-t1} \)

    RULE 10:

    If a sum is to be deposited in equal instalments, then,
    Equal instalment = \( \frac{A*200}{T[200+(T-1)r]} \)
    where T = no. of years, A = amount, r = Rate of Interest.

    RULE 11:

    To find the rate of interest under current deposit plan,
    r = \( \frac{S.I. * 2400}{n(n+1)*dA00} \)
    where n = no. of months & dA = deposited ammount

    RULE 12:

    If certain sum P amounts to Rs. A1 in t1 years at rate of R% and the same sum amounts to Rs. A2 in t2 years at same rate of interest R%. Then,
    (i) R = \( \frac{A1-A2}{A2T1-A1T2} \)X100
    (ii) P = \( \frac{A2T1-A1T2}{T1-T2} \)

    RULE 13:

    The difference between the S.I. for a certain sum P1 deposited for time T1 at R1 rate of interest and another sum P2 deposited for time T2 at R2 rate of interest is
    S.I. = \( \frac{P2R2T2-P1R1T1}{100} \)

    TYPE-I

    Q.1

    What sum of money must be given as simple interest for six months at 4% per annum in order to earn ₹150 interest?

    ₹5000

    ₹7500

    ₹10000

    ₹15000

    Ans .

    2


    1. Explanation :

      (2) Using Rule 1,

      \( \frac{150*100}{4} * \frac{2}{1} \) = ₹7500

    Q.2

    A sum of ₹1600 gives a simple interest of ₹252 in 2 years and 3 months. The rate of interest per annum is:

    5\( \frac{1}{2}\)

    8%

    7%

    6%

    Ans .

    2


    1. Explanation :

      (3) Using Rule 1,
      Principal (P) = ₹1600
      T = 2 years 3 months = \( (2+\frac{3}{12})yrs. = (2+\frac{1}{3})yrs. = \frac{9}{4}yrs. \)
      S.I = ₹252
      R = % rate of interest per annum
      => R = \( \frac{100*S.I.}{P*t} \)
      = \( \frac{100*252}{1600*\frac{9}{4}} \)
      Rate of interest = 7% per annum.

    Q.3

    A sum of money lent at simple interest amounts to ₹880 in 2 years and to ₹920 in 3 years. The sum of money (in rupees) is

    700

    760

    784

    800

    Ans .

    4


    1. Explanation :

      (4) If the principal be x and rate of interest be r% per annum,
      then
      SI after 1 year = 920 – 880 = ₹40
      Therefore, SI after 2 years = ₹80
      => 880 = x + 80
      => x = ₹(880 – 80) = ₹800
      Aliter : Using Rule 12,
      P = \( \frac{A2T1-A1T2}{T1-T2} \)
      = \( \frac{920*2-880*3}{2-3} \)
      = \( \frac{1840-2640}{-1} \)
      = \( \frac{-800}{-1} \)
      = ₹800

    Q.4

    At some rate of simple interest, A lent ₹6,000 to B for 2 years and ₹1,500 to C for 4 years and received ₹9,00 as interest from both of them together. The rate of interest per annum was

    5%

    6%

    8%

    10%

    Ans .

    1


    1. Explanation :

      (1) Using Rule 1,
      If rate of interest be R% p.a. then,
      S.I.= \( \frac{Principal * Rate * Time}{100} \)
      Therefore, \( \frac{600*2*R}{100} + \frac{1500*4*R}{100}\)
      = 900
      => 120R + 60R = 900
      => 180R = 900
      => R = \( \frac{900}{180} = 5%\)

    Q.5

    A lent ₹5000 to B for 2 years and ₹3000 to C for 4 years on simple interest at the same rate of interest and received ₹2200 in all from both as interest. The rate of interest per annum is

    7%

    5%

    7\( \frac{1}{8}% \)

    10%

    Ans .

    4


    1. Explanation :

      (4) Using Rule 1,
      Let the rate of interest per annumbe r%
      According to the question,
      \( \frac{5000*2*R}{100} +\frac{300*2*R}{100} = 2000 \)
      => 100R + 120R = 2200
      => 220R = 2200
      => R = \( \frac{2200}{220} =10% \)

    Q.6

    What sum of money will amount to ₹520 in 5 years and to ₹568 in 7 years at simple interest ?

    ₹410

    ₹120

    ₹510

    ₹220

    Ans .

    1


    1. Explanation :

      (1) Simple interest for 2 years
      = ₹(568 – 520) = ₹48
      Therefore, Interest for 5 years
      =₹\( \frac{48}{2} \)*5 = ₹120
      Principal = ₹(520 – 120) = ₹400
      Aliter : Using Rule 12,
      P = \( \frac{A2T1-A1T2}{T1-T2} \)
      = \( \frac{568*5-520*7}{5-7} \)
      = \( \frac{2840-3640}{-2} \)
      = \( \frac{-800}{-2} \)
      =₹ 400

    Q.7

    ₹500 was invested at 12% per annum simple interest and a certain sum of money invested at 10% per annum simple interest. If the sum of the interest on both the sum after 4 years is ₹480, the latter sum of money is :

    ₹450

    ₹750

    ₹600

    ₹550

    Ans .

    3


    1. Explanation :

      (3) Using Rule 1,
      Simple interest gained from ₹500
      = \( \frac{500*12*4}{100} \) = ₹240
      Let the other Principal be x.
      S.I. gained = ₹(480 – 240)= ₹240
      Therefore, \( \frac{x*10*4}{100} \)=240
      => x=\( \frac{240*100}{40} \) = ₹600

    Q.8

    A money lender finds that due to fall in the annual rate of interest 8% to 7\( \frac{3}{4} \)%, his yearly income diminishes by ₹61.50. His capital is

    ₹22400

    ₹23800

    ₹24600

    ₹26000

    Ans .

    3


    1. Explanation :

      Difference in rate
      = \( (8-7\frac{3}{4}) \) % = \( \frac{1}{4} \)%
      Let the capital be ₹x.
      Therefore, \( \frac{1}{4} \)% of x = 61.50
      => x = 61.50 × 100 × 4
      = ₹24600

    Q.9

    A lends ₹2500 to B and a certain sum to C at the same time at 7% annual simple interest. If after 4 years, A altogether receives ₹1120 as interest from B and C, the sumlent to C is

    ₹700

    ₹6500

    ₹4000

    ₹1500

    Ans .

    4


    1. Explanation :

      (4) Using Rule 1,
      Let the sum lent to C be x
      According to the question,
      \( \frac{2500*7*4}{100} + \frac{x*7*4}{100} = 1120 \)
      or 2500 × 28 + 28x = 112000
      or 2500 + x = 4000
      or x = 4000 – 2500 = 1500

    Q.10

    A certain sum of money amounts to ₹756 in 2 years and to ₹873 in 3\( \frac{1}{2} \) years at a certain rate of simple interest. The rate of interest per annum is

    10%

    11%

    12%

    13%

    Ans .

    4


    1. Explanation :

      (4) S.I. for 1\( \frac{1}{2} \)years
      = (873 – 756) = 117
      S.I. for 2 years
      = ₹\( (117*\frac{2}{3}*2) = ₹156 \)
      Therefore, Principal = 756 – 156 = ₹600
      Now, P = 600, T = 2,
      S.I. = 156
      Therefore, R= \( \frac{100*S.I.}{P*T} \)
      \( \frac{100*156}{600*2} =13 \)%
      Aliter : Using Rule 12,
      Rate of interest = \( \frac{A1-A2}{A2T1-A1T2} \)X100
      = \( \frac{756-873}{873*2-756\frac{7}{2}} \)X100
      = \( \frac{-117}{1746-2646} \)X100
      = \( \frac{-117}{-900} \)X100
      =13%

    Q.11

    What sum will amount to ₹7000 in 5 years at 3\(\frac{1}{3}\)% simple interest ?

    ₹6300

    ₹6500

    ₹6000

    ₹5000

    Ans .

    3


    1. Explanation :

      (3) Using Rule 1,
      P = \( \frac{A * 100}{100 + R * T} \)
      =\( \frac{7000 * 100}{1000+\frac{10}{3}*5} \)
      =\( \frac{7000*100*3}{350} \)
      = ₹6000

    Q.12

    A man took a loan from a bank at the rate of 12% per annum at simple interest. After 3 years he had to pay ₹5,400 as interest only for the period. The principal amount borrowed by him was :

    ₹2000

    ₹10000

    ₹20000

    ₹15000

    Ans .

    4


    1. Explanation :

      (4) Using Rule 1,
      Let the principal be x.
      S.I. = \( \frac{Principal * Rate * Time}{100} \)
      => 5400 = \( \frac{x * 12 * 3}{100} \)
      => \( x = \frac{5400 * 100}{12 * 3} \) = ₹15000

    Q.13

    A sum of money at simple interest amounts to ₹1,012 in 2\( \frac{1}{2} \)years and to ₹1,067.20 in 4 years. The rate of interest per annum is :

    2.5%

    3%

    4%

    5%

    Ans .

    3


    1. Explanation :

      (3) Principal + S.I. for \(\frac{5}{2}\)years = ₹1012 ...(i)
      Principal + S.I. for 4 years = ₹1067.20 ...(ii)
      Subtracting equation (i) from (ii)
      S.I. for \(\frac{3}{2}\)years = ₹55.20
      Therefore, S.I. for \(\frac{5}{2}\)years = 55.20 x \(\frac{2}{3}\) x \(\frac{5}{2}\) = ₹92
      Therefore, Principal
      = ₹(1012 – 92) = ₹920
      Therefore, Rate = \(\frac{92*100}{920*\frac{5}{2}}\)
      = \(\frac{2*92*100}{920*5}\) = 4%
      Aliter : Using Rule 12,
      Rate of interest = \( \frac{A1-A2}{A2T1-A1T2} \)X100
      = \( \frac{1012-1067.50}{1067.50*\frac{5}{2}-1012*4} \)X100
      = \( \frac{-55.2}{2668-4048} \)X100
      = \( \frac{-55.2}{-1380} \)X100
      =4%

    Q.14

    A sum of money lent out at simple interest amounts to ₹720 after 2 years and to ₹1020 after a further period of 5 years. The sum is :

    ₹500

    ₹600

    ₹700

    ₹710

    Ans .

    2


    1. Explanation :

      (2) Principal + SI for 2 years = ₹720 .... (i)
      Principal + SI for 7 years = ₹1020 .....(ii)
      Subtracting equation (i) from (ii) get,
      SI for 5 years
      = ₹(1020 – 720) = ₹300
      Therefore, SI for 2 years
      = ₹300 × \(\frac{2}{5}\) = ₹120
      Therefore, Principal
      = ₹(720 – 120) = ₹600
      Aliter : Using Rule 12,
      P = \( \frac{1020*2-720*7}{2-7} \)
      = \( \frac{2040-5040}{-5} \)
      = \( \frac{-3000}{-5} \)
      = ₹600

    Q.15

    The sum of money, that will give ₹1 as interest per day at the rate of 5% per annum simple interest is

    ₹3650

    ₹36500

    ₹730

    ₹7300

    Ans .

    4


    1. Explanation :

      (4) Using Rule 1,
      The sum of money will give ₹365
      as simple interest in a year.
      => S.I. = \(\frac{PRT}{100}\)
      => 365 = \(\frac{P*5*1}{100}\)
      => P = \(\frac{365*100}{5}\) = ₹7300

    Q.16

    If the simple interest on a certain sum of money for 15 months at 7\(\frac{1}{2}\)% per annum exceeds the simple interest on the same sum for 8 months at 12\(\frac{1}{2}\)% per annum by ₹32.50, then the sum of money (in ₹) is :

    312

    312.50

    3120

    3120.50

    Ans .

    3


    1. Explanation :

      (3) Using Rule 1,
      Let the sum be x.
      Using formula, I = \(\frac{PRT}{100}\) we have
      \( \frac{x*\frac{15}{12}*\frac{15}{2}}{100} \) - \( \frac{x*\frac{8}{12}*\frac{25}{2}}{100} \)
      = 32.50
      => \( \frac{25x}{2400} \) = 32.50
      => x = \( \frac{32.50*2400}{25} \) = 3120
      Therefore, Required sum = ₹3120

    Q.17

    What annual instalment will discharge a debt of ₹6450 due in 4 years at 5% simple interest ?

    ₹1500

    ₹1835

    ₹1935

    ₹1950

    Ans .

    1


    1. Explanation :

      (1) Let each instalment be x
      Then,
      \((x+\frac{x*5*1}{100})+(x+\frac{x*5*2}{100})+(x+\frac{x*5*3}{100})+x\)=6450
      => \((x+\frac{x}{20})+(x+\frac{x}{10})+(x+\frac{3x}{20})+x\)=6450
      => \((\frac{21x}{20})+(\frac{11x}{10})+(\frac{23x}{20})+x\)=6450
      => \( \frac{21x+22x+23x+20x}{20} \)=6450
      => \( \frac{86x}{20} \) = 6450
      => x = \( \frac{6450*20}{86} \) = ₹1500
      Aliter : Using Rule 10,
      Equal instalment
      = \( \frac{6450*200}{4[200+(4-1)*5]} \)
      = \(\frac{6450*200}{4(215)}\)
      = \(\frac{6450*50}{215}\)
      = ₹1500

    Q.18

    In what time will ₹72 become 81 at 6\(\frac{1}{4}\)% per annum simple interest ?

    2 years

    3 years

    2 years 6 months

    None of these

    Ans .

    1


    1. Explanation :

      (1) Using Rule 1,
      Interest = ₹(81–72)= ₹9
      Let the time be t years.
      Then, 9 = \( \frac{72*25*t}{4*100} \)
      => t = \(\frac{9*400}{72*25}\)=2 years

    Q.19

    The simple interest on ₹7,300 from 11 May, 1987 to 10 September, 1987 (both days included) at 5% per annum is

    ₹123

    ₹103

    ₹200

    ₹223

    Ans .

    1


    1. Explanation :

      Using Rule 1,
      Time from 11 May to 10 September,
      1987
      = 21 + 30 + 31 + 31 + 10
      = 123 days
      Therefore, 123 days = \( \frac{123}{365} \) year Therefore, S.I. = \( \frac{7300*123*5}{365*100}\) = ₹123

    Q.20

    A person borrows ₹5,000 for 2 years at 4% per annum simple interest. He immediately lends it to another person at 6\(\frac{1}{4}\)% per annum simple interest for 2 years. His gain in the transaction is

    ₹112.50

    ₹450

    ₹225

    ₹150

    Ans .

    3


    1. Explanation :

      (3) Using Rule 1,
      Case I :
      S.I. = \( \frac{5000*2*4}{100}\) = ₹400
      Case II :
      S.I. = \( \frac{5000*25*2}{100*4}\) = ₹625
      Therefore, ₹(625-400) = ₹225

    Q.21

    A man had ₹16,000, part of which he lent at 4% and the rest at 5% per annum simple interest. If the total interest received was ₹700 in one year, the money lent at 4% per annum was

    ₹12000

    ₹8000

    ₹10000

    ₹6000

    Ans .

    3


    1. Explanation :

      (3) Using Rule 1,
      Let the sum lent at 4% = Rs.x
      Therefore, Amount at 5%= (16000 – x )
      According to the question,
      \( \frac{x*4*1}{100} + \frac{(16000-x)*5*1}{100}\)
      = 700
      => 4x + 80000 – 5x = 70000
      => x = 80000 – 70000
      = Rs. 10000

    Q.22

    ₹1,000 is invested at 5% per annum simple interest. If the interest is added to the principal after every 10 years, the amount will become ₹2,000 after

    15 years

    18 years

    20 years

    16\(\frac{2}{3}\) years

    Ans .

    4


    1. Explanation :

      (4) Using Rule 1,
      After 10 years,
      SI = \( \frac{1000*5*10}{100} \)= ₹ 500
      Principal for 11th year
      = 1000 + 500 = 1500
      SI = ₹(2000 – 1500) = ₹500
      Therefore, T = \( \frac{SI*100}{P*R} = \frac{500*100}{1500*5} = \frac{20}{3}\) yeras =6\(\frac{2}{3}\) years
      Therefore, Total time = 10 + 6\( \frac{2}{3}\) = 16\( \frac{2}{3}\) years

    Q.23

    A sum of money amounts to ₹5,200 in 5 years and to ₹5,680 in 7 years at simple interest. The rate of interest per annum is

    3%

    4%

    5%

    6%

    Ans .

    4


    1. Explanation :

      (4)
      P + S.I. for 5 years = 5200 ..(i)
      P + SI for 7 years = 5680 ...(ii)
      On subtracting equation (i) from (ii
      SI for 2 years = 480
      Therefore, SI for 1 year = ₹240
      Therefore, From equation (i),
      P + 5 × 240 = 5200
      => P = 5200 – 1200 = ₹4000
      Therefore, R = \( \frac{SI*100}{T*P} \)= \( \frac{240*100}{1*4000} \) = 6%
      Aliter: Using Rule 12,
      R = \( \frac{A1-A2}{A2T1-A1T2} \)X100
      = \( \frac{5200-5680}{5680*5-5200*7} \)X100
      = \( \frac{-480}{28400-36400} \)X100
      = \( \frac{-480}{-8000} \)X100
      =6%

    Q.24

    ₹800 becomes ₹956 in 3 years at a certain rate of simple interest. If the rate of interest is increased by 4%, what amount will ₹800 become in 3 years ?

    ₹1020.80

    ₹1025

    ₹1052

    ₹1050

    Ans .

    3


    1. Explanation :

      (3) Using Rule 1,
      S.I. = 956 – 800 = Rs. 156
      Therefore, Rate = \(\frac{SI*100}{Principal*Time}\) = \(\frac{156*100}{800*3}\) = 6.5% per annum
      Therefore, New rate = 10.5%
      Therefore, S.I. = \(\frac{Principal*Time*Rate}{100}\) = \(\frac{800*3*10.5}{100}\) = ₹252
      Therefore, ammount = 800+252 = ₹1052

    Q.25

    A person deposited ₹400 for 2 years, ₹550 for 4 years and ₹1,200 for 6 years. He received the total simple interest of ₹1,020. The rate of interest per annum is

    10%

    5%

    15%

    20%

    Ans .

    1


    1. Explanation :

      (1) Using Rule 1,
      Let the rate of interest be R per
      cent per annum.
      Therefore, \( \frac{400*2*R}{100} + \frac{550*4*R}{100} + \frac{1200*6*R}{100} \) = 1020
      => 8R+22R+72R = 1020
      => 102R = 1020
      => R=\(\frac{1020}{102}\)=10%

    Q.26

    Manoj deposited ₹29400 for 6 years at a simple interest. He got ₹4200 as interest after 6 years. The annual rate of interest was

    2\(\frac{8}{21}\)%

    2\(\frac{7}{20}\)%

    3\(\frac{8}{21}\)%

    4\(\frac{8}{21}\)%

    Ans .

    1


    1. Explanation :

      (1) Using Rule 1,
      4200 = \( \frac{29400*6*R}{100} \)
      => R = \(\frac{4200}{294*6}\) = \(\frac{50}{21}\) = 2\(\frac{8}{21}\)%

    Q.27

    A man lent ₹60,000, partly at 5% and the rest at 4% simple interest. If the total annual interest is ₹2560, the money lent at 4% was

    ₹40000

    ₹44000

    ₹30000

    ₹45000

    Ans .

    2


    1. Explanation :

      (2) Using Rule 1,
      Let the amount lent at 4% be x
      \ Amount lent at 5% = (60000 – x )
      According to the question,
      \(\frac{(60000-x)*5*1}{100}\) + \(\frac{x*4*1}{100}\)
      => 2560
      => 300000-5x + 4x = 256000
      => x = 300000-256000=₹44000

    Q.28

    A sum of money at some rate of simple interest amounts to ₹2,900 in 8 years and to ₹3,000 in 10 years. The rate of interest per annum is

    4%

    2\(\frac{1}{2}\)%

    3%

    2%

    Ans .

    4


    1. Explanation :

      (4) Principal + interest for 8 years= ₹2900... (i)
      Principal + interest for 10 years = ₹3000 ... (ii)
      Subtracting equation (i) from (ii)
      Interest for 2 years = 100
      Therefore, Interest for 8 years = \(\frac{100}{2}\)*8 = ₹400
      From equation (i),
      Principal = ₹(2900 – 400) = ₹2500
      Therefore, Rate = \( \frac{SI*100}{Time*Principal} \) = \( \frac{400*100}{8*2500} \)=2%
      Aliter : Using Rule 12,
      R = \( \frac{A1-A2}{A2T1-A1T2} \)X100
      = \( \frac{2900-3000}{3000*8-2900*10} \)X100
      R = \( \frac{A1-A2}{A2T1-A1T2} \)X100
      = \( \frac{-100}{24000-29000} \)X100
      = \( \frac{-100}{-5000} \)X100
      = 2%

    Q.29

    herIn how many years will a sum of ₹3,000 yield a simple interest of ₹1,080 at 12% per annum ?e

    3 years

    2\(\frac{1}{2}\)years

    2 years

    3\(\frac{1}{2}\)years

    Ans .

    1


    1. Explanation :

      Using Rule 1, Time = \( \frac{S.I. * 100}{P * R} \) =\( \frac{1080*100}{3000*12} \) = 3 years

    Q.30

    A sum of money amounts to ₹850 in 3 years and to ₹925 in 4 years at some rate of simple interest. The sum is :

    ₹550

    ₹600

    ₹625

    ₹700

    Ans .

    3


    1. Explanation :

      (3) Interest for 1 year
      = ₹(925 – 850) = ₹75
      Therefore, If a sum becomes a1 in t1 years
      and a2 in t2 years then rate of
      interest = \( \frac{100(a2-a1)}{(a1t2-a2t1)} \)%
      =\( \frac{100(925-850)}{850*4-3*925} \) = \(\frac{7500}{625}\) = 12%
      Therefore, Principal = \( \frac{SI*100}{Time*Rate} \) = \( \frac{75*100}{1*12} \) = ₹625
      Aliter : Using Rule 12, P = \( \frac{A2T1-A1T2}{T1-T2} \)
      = \( \frac{925*3-850*4}{3-4} \)
      = \( \frac{2775-3400}{-1} \)
      = \( \frac{-625}{-1} \)
      = ₹625

    Q.31

    In what time will 1,860 amount to ₹2,641.20 at simple interest 12% per annum ?

    3 years

    3\(\frac{1}{2}\)years

    4 years

    4\(\frac{1}{2}\) years

    Ans .

    2


    1. Explanation :

      (2) Using Rule 1,
      S.I. = 2641.20 – 1860
      = 781.2
      Time = \( \frac{S.I. * 100}{P * R} \) =\( \frac{781.2*100}{1860*12} \) = 3.5 = 3\(\frac{1}{2}\) years

    Q.32

    The population of a village decreases at the rate of 20% per annum. If its population 2 years ago was 10,000, the present population is

    4600

    6400

    7600

    6000

    Ans .

    2


    1. Explanation :

      (2) Using Rule 18 of ‘percentage’ chapter,
      Present population = 10000 \( (1-\frac{20}{100}) \)2
      = 10000 \( * \frac{4}{5} * \frac{4}{5} \) = 6400

    Q.33

    The sum lent at 5% per annum (i.e. 365 days) simple interest, that produces interest, of ₹2.00 a day, is

    ₹1400

    ₹14700

    ₹14600

    ₹7300

    Ans .

    3


    1. Explanation :

      (3) Using Rule 1,
      Annual interest
      = 365 × 2 = ₹730
      Principal = \( \frac{SI*100}{Time*Rate} \)
      = \( \frac{730*100}{1*5} \) = ₹14600

    Q.34

    A certain sum of money lent out at simple interest amounts to ₹1380 in 3 years and ₹1500 in 5 years. Find the rate per cent per annum.

    3%

    3.5%

    4%

    5%

    Ans .

    3


    1. Explanation :

      (4) If principal = x and rate = r%
      per annum, then
      1380 = x + \(\frac{x*3*r}{100}\) .....(i)
      1500 = x + \(\frac{x*5*r}{100}\) ....(ii)
      S.T. for to years = 1500-1380 = ₹120
      Therefore, \(\frac{x*2*r}{100}\) =120
      Therefore, \(\frac{xr}{100}\) = 60
      Therefore, From equation(i)
      1380 = x + 60 × 3
      => x = 1380 – 180 = ₹1200
      From equation (iii)
      \( \frac{1200*r}{100} \) = 60
      => r = \( \frac{6000}{1200} \) = 5% per annum
      Aliter : Using rule 12
      R = \( \frac{A1-A2}{A2T1-A1T2} \)X100%
      = \( \frac{1380-1500}{1500*3-1380*5} \)X100
      = \( \frac{-120}{4500-6900} \)X100%
      = \( \frac{-120}{-2400} \)X100
      = 5%

    Q.35

    If a sum of money amounts to ₹12,900 and ₹14,250 at the end of 4th year and 5th year respectively at a certain rate of simple interest, then the rate of interest is

    10%

    12%

    18%

    20%

    Ans .

    3


    1. Explanation :

      S.I. for 1 year = 14250 – 12900 = Rs. 1350
      S.I. for 4 years = 1350 × 4 =Rs. 5400
      Therefore, Principal = 12900 – 5400 =Rs. 7500
      Therefore, Rate = \( \frac{SI*100}{Principal*Time} \)
      = \( \frac{5400*100}{7500*4}\)
      =18% per annum
      Aliter : Using Rule 12,
      R = \( \frac{A1-A2}{A2T1-A1T2} \)X100
      = \( \frac{12900-14250}{14250*4-12900*5} \)X100
      = \( \frac{-1350}{57000-64500} \)X100
      = \( \frac{1350}{7500} \)X100
      = 18%

    Q.36

    In what time will ₹8,000, at 3% per annum, produce the same interest as ₹6, 000 does in 5 years at 4 % simple interest ?

    5 years

    6 years

    3 years

    4 years

    Ans .

    1


    1. Explanation :

      (1) Using Rule 1,
      Required time = t years
      S.I. = \( \frac{Principal * Rate * Time}{100} \)
      Therefore, \( \frac{6000*5*4}{1000} \) = \( \frac{8000*3*t}{100} \)
      => 6000*4*5=8000*3*t
      Therefore, t = \( \frac{6000*4*5}{8000*3} \)=5 years

    Q.37

    The principal which gives ₹1 interest per day at a rate of 5% simple interest per annum is

    ₹5000

    ₹7300

    ₹36500

    ₹3650

    Ans .

    2


    1. Explanation :

      (2) Using Rule 1,
      Principal = \( \frac{S.I. * 100}{R * T} \)
      = \( \frac{1*100}{\frac{1}{365}*5} \) = \( \frac{365*100}{5} \)
      = Rs. 7300

    Q.38

    A sum of money lent out at simple interest amounts to ₹720 after 2 years and Rs. 1020 after a further period of 5 years. Find the principal.

    ₹600

    ₹1740

    ₹6000

    ₹120

    Ans .

    1


    1. Explanation :

      S.I. for 5 years
      = Rs. (1020 – 720) = Rs. 300
      Therefore, S.I. for 2 years
      = \( \frac{300}{5} * 2\) = Rs.120
      Therefore, Principal = Rs. (720 – 120)
      = Rs. 600

    Q.39

    The simple interest on Rs. 36,000 for the period from 5th January to 31st May, 2013 at 9.5% per annum is

    Rs. 1,338

    Rs. 1425

    Rs. 1400

    Rs. 1368

    Ans .

    4


    1. Explanation :

      (4) Using Rule 1,
      Number of days from 5th January to 31st May = 26 + 28 + 31 + 30 + 31 = 146
      Therefore, S.I. = \( \frac{P*T*R}{100} \) = \( \frac{36000*146*9.5}{365*100} \) = Rs.1368

    Q.40

    Alipta got some amount of money from her father. In how many years will the ratio of the money and the interest obtained from it be 10:3 at the rate of 6% simple interest per annum?

    7 years

    3 years

    5 years

    4 years

    Ans .

    3


    1. Explanation :

      (3)\( \frac{Principal}{Interest} \) = \( \frac{10}{3} \)
      \( \frac{Interest}{Principal} \) = \( \frac{3}{10} \)
      Time = \( \frac{SI*100}{P*R} \)
      \( \frac{3}{10} * \frac{100}{6} \) = 5 years

    Q.41

    The sum of money that will yield Rs. 60 as simple interest at the rate of 6% per annum in 5 years

    200

    225

    175

    300

    Ans .

    1


    1. Explanation :

      (1) Principal = \( \frac{S.I. * 100}{R * T} \)
      =\( \frac{60*100}{5*6} \) = Rs.200

    Q.42

    If a sum of money becomes Rs. 4000 in 2 years and Rs. 5500 in 4 years 6 months at the same rate of simple interest per annum, then the rate of simple interest is

    21\(\frac{3}{7}\)%

    21\(\frac{2}{7}\)%

    21\(\frac{1}{7}\)%

    21\(\frac{5}{7}\)%

    Ans .

    1


    1. Explanation :

      (1) According to the question,
      S.I. for 2 years 6 months
      = Rs. (5500 – 4000)
      => S.I. for \( \frac{5}{2} \) years = Rs.1500
      Therefore, S.I. for 1 year = \( \frac{1500*2}{5} \)
      = Rs.600
      Rate = \( \frac{S.I.*100}{Principal*Time} \)
      = \( \frac{1200*100}{2800*2} \) = \( \frac{150}{7} \)
      = 21 \( \frac{3}{7}% \)per annum

    Q.43

    The simple interest on a certain sum of money at the rate of 5% per annum for 8 years is Rs. 840. Rate of interest for which the same amount of interest can be received on the same sum after 5 years is :

    7% per annum

    8% per annum

    9% per annum

    10% per annum

    Ans .

    2


    1. Explanation :

      (2) Principal = \( \frac{S.I. * 100}{R * T} \)
      = \( \frac{840*100}{8*5} \)
      = Rs.2100
      Case II,
      S.I. = Rs. 840
      Principal = Rs. 2100
      Time = 5 years
      Rate = \( \frac{S.I. * 100}{P * T} \)
      =\( \frac{840 * 100}{2100*5} \)
      = 8 % per annum

    Q.44

    A sum of Rs. 2800 is divided into two parts in such a way that the interest on both the parts is equal. If the first part is lent at 9% p.a. for 5 years and second part is for 6 years at 10% p.a., find the two sums.

    Rs. 1800, Rs. 1000

    Rs. 1600, Rs. 1200

    Rs. 1400, Rs. 1400

    Rs. 1300, Rs. 1500

    Ans .

    2


    1. Explanation :

      (2) Let first part be x.
      Therefore, Second part
      = Rs. (2800 – x)
      According to the question,
      (S.I.)= \( \frac{Principal * Rate * Time}{100} \)
      Therefore, \( \frac{x*5*9}{100}\) = \( \frac{(2800-x)*6*10}{100} \)
      3x = 4*2800-4x
      7x = 4*2800
      x = \( \frac{4-2800}{7} \)
      =Rs.1600

    Q.45

    The simple interest on a sum for 5 years is two-fifth of the sum. The rate of interest per annum is

    0.1

    0.08

    0.06

    0.04

    Ans .

    2


    1. Explanation :

      (2) According to the question,
      \( \frac{S.I.}{Principal} = \frac{2}{5} \)
      Rate = \( \frac{S.I. * 100}{P * T} \)
      = \( \frac{2}{5} * \frac{100}{5} \) = 8% per annum

    2. = 0.08 per annum

    Q.46

    If the simple interest on Rs. 400 for 10 years is Rs. 280, the rate of interest per annum is

    7%

    7\(\frac{1}{2}\)%

    7\(\frac{1}{4}\)%

    8\(\frac{1}{2}\)%

    Ans .

    1


    1. Explanation :

      Rate = \( \frac{S.I. * 100}{P * T} \)
      \( \frac{280*100}{400*10} \)
      = 7% per annum

    Q.47

    If the simple interest on Re. 1 for 1 month is 1 paisa, then the rate per cent per annum will be

    10%

    8%

    12%

    6%

    Ans .

    3


    1. Explanation :

      Rate = \( \frac{S.I. * 100}{P * T} \)
      = \( \frac{\frac{1}{100}*100}{1*\frac{1}{12}} \)= 12% per annum

    Q.48

    How much simple interest will Rs. 4000 earn in 18 months at 12% per annum?

    Rs. 216

    Rs. 360

    Rs. 720

    Rs. 960

    Ans .

    3


    1. Explanation :

      (3)S.I.= \( \frac{Principal * Rate * Time}{100} \)
      =Rs. \( (4000*\frac{18}{12}*\frac{12}{100}) \)
      = Rs.720

    Q.49

    In how many years a sum of Rs. 3000 will yield an interest of Rs. 1080 at 12% per annum simple interest ?

    4 years

    3 years

    5 years

    2\(\frac{1}{2}\) years

    Ans .

    2


    1. Explanation :

      (2)T= \( \frac{S.I. * 100}{P * R} \)
      =\( \frac{1080*100}{3000*12} \)
      = 3 yers

    Q.50

    In simple interest rate per annum a certain sum amounts to Rs. 5,182 in 2 years and Rs. 5,832 in 3 years. The principal in rupees is

    Rs.2882

    Rs.5000

    Rs.3882

    Rs.4000

    Ans .

    3


    1. Explanation :

      (3) Let the principal be Rs. x.
      According to the question,
      x + S.I. for 2 years = Rs. 5182 ...(i)
      x + S.I. for 3 years = Rs. 5832 ...(ii)
      By equation (ii) – (i),
      S.I. for 1 year
      = Rs. (5832 – 5182)
      = Rs. 650
      Therefore, S.I. for 2 years = Rs. (2 × 650) = Rs. 1300
      Therefore, Principal = Rs. (5182 – 1300) = Rs. 3882

    Q.51

    For what sum will the simple interest at R% per annum for 2 years will be R ?

    Rs.\(\frac{100}{2R}\)

    Rs.50

    Rs.\(\frac{100}{R}\)

    Rs.\(\frac{200}{R}\)

    Ans .

    2


    1. Explanation :

      P= \( \frac{S.I. * 100}{R * T} \)
      = \( \frac{R * 100}{R * 2} \)
      = Rs.50

    Q.52

    The amount to be paid, when principal = Rs. 2000, rate of simple interest (R) = 5%, T = 2 years, is :

    Rs.3200

    Rs.2400

    Rs.2200

    Rs.3400

    Ans .

    3


    1. Explanation :

      (3)S.I.= \( \frac{Principal * Rate * Time}{100} \)
      = \( \frac{2000*2*5}{100} \) = Rs.250
      Therefore, Required amount = Rs. (2000 + 200) = Rs. 2200

    Q.53

    The rate of simple interest for which Rs. 6,000 will amount to Rs. 6,900 in 3 years is

    5%

    7%

    2%

    4%

    Ans .

    1


    1. Explanation :

      (1) S.I. = Amount – Principal = Rs. (6900 – 6000) = Rs. 900
      Therefore, Rate =\( \frac{S.I. * 100}{P * T} \)
      = \( \frac{900 * 100}{6000*3} \)= 5% per annum

    TYPE - II

    Q.1

    A sum of money becomes \( \frac{7}{6} \) of itself in 3 years at a certain rate of simple interest. The rate per annum is :

    \( 5\frac{5}{9} \)%

    \( 6\frac{5}{9}\)%

    18%

    25%

    Ans .

    1


    1. Explanation :

      (1)Using Rule 3,
      R% = \( \frac{(\frac{7}{6}-1)*100}{3} \)%
      = \( \frac{1}{18} *100\)%
      = \(\frac{50}{9}\)%
      = \( 5\frac{5}{9} \)%

    Q.2

    A sum of money becomes \( \frac{41}{40} \) of itself in \( \frac{1}{4} \)years at a certain rate of simple interest. The rate of interest per annum is

    10%

    1%

    2.5%

    5%

    Ans .

    1


    1. Explanation :

      (1)Using Rule 3,
      R% = \( \frac{(\frac{41}{40}-1)*100}{\frac{1}{4}} \)%
      = \( \frac{1}{40}*4 *100 \)%
      = 10%

    Q.3

    A certain sum of money becomes three times of itself in 20 years at simple interest. In how many years does it become double of itself at the same rate of simple interest ?

    8 years

    10 years

    12 years

    14 years

    Ans .

    2


    1. Explanation :

      (1)Using Rule 3,
      R% = \( \frac{(3-1)*100}{20} \)%
      = 10%
      Now , T = \( \frac{(n-1)}{R} \)x100%
      = \(\frac{2-1}{10}*100\)%
      = 10 years

    Q.4

    At what rate per cent per annum will the simple interest on a sum of money be \( \frac{2}{5} \) of the amount in 10 years ?

    4%

    6%

    5\(\frac{2}{3}\)%

    6\(\frac{2}{3}\)%

    Ans .

    4


    1. Explanation :

      (4) Using Rule 1,
      Let P be the principal and R%
      rate of interest.
      Therefore, S.I. = \( \frac{Principal * Rate * 10}{100} \) = \( \frac{Principal * Rate}{10} \)
      According to the question,
      \( \frac{PR}{10}\) = \( (P+\frac{PR}{10}) * \frac{2}{5}\)
      \( \frac{R}{10}\) = \( (1+\frac{R}{10}) * \frac{2}{5}\)
      \( \frac{R}{10}\) = \( \frac{R}{25} + \frac{2}{5}\)
      \( \frac{R}{10}\) - \( \frac{R}{25} = \frac{2}{5}\)
      \( \frac{5R-2R}{50}\) = \frac{2}{5}\)
      \( \frac{3R}{50}\) = \frac{2}{5}\)
      R = \( \frac{50*2}{3*5} = \frac{20}{3} = 6\frac{2}{3}%

    Q.5

    ₹6,000 becomes ₹7,200 in 4 years at a certain rate of simple interest. If the rate becomes 1.5 times of itself, the amount of the same principal in 5 years will be

    ₹8000

    ₹8250

    ₹9250

    ₹9000

    Ans .

    2


    1. Explanation :

      (2) Using Rule 1,
      SI = (7200–6000) = 1200
      Therefore, SI = \( \frac{PRT}{100} \)
      1200 = \( \frac{600*R*4}{100} \)
      R = \( \frac{1200*100}{6000*4} \)= 5%
      New rate of R = 5×1.5 = 7.5%
      Then, SI = \( \frac{6000*7.5*5}{100} \) = ₹2250
      Therefore, Amount = (6000 + 2250) = ₹8250

    Q.6

    A sum of money at simple interest trebles itself in 15 years. It will become 5 times of itself in

    40 years

    36 years

    30 years

    25 years

    Ans .

    3


    1. Explanation :

      (3)Using Rule 3,
      R% = \( \frac{(3-1)*100}{15} \)%
      = \(\frac{40}{3}\)%
      Now , T = \( \frac{(n-1)}{R} \) years
      = \(\frac{2-1}{\frac{40}{3}}*100\)
      = 30 years

    Q.7

    If a sum of money at simple interest doubles in 12 years, the rate of interest per annum is

    16\(\frac{2}{3}\)%

    7.5%

    8\(\frac{1}{3}\)%

    10%

    Ans .

    3


    1. Explanation :

      (3)Using Rule 3,
      R% = \( \frac{(2-1)*100}{12} \)%
      = \(\frac{25}{3}\)%
      = \(8\frac{1}{3}\)%

    Q.8

    At what rate of simple interest per annum will a sum become \( \frac{7}{4} \) of itself in 4 years ?

    18%

    18 \( \frac{1}{4} \)

    18 \( \frac{3}{4} \)%

    18 \( \frac{1}{2} \)%

    Ans .

    3


    1. Explanation :

      (3)Using Rule 3,
      R% = \( \frac{(\frac{7}{4}-1)*100}{12} \)%
      = \(\frac{75}{4}\)%
      = \(18\frac{3}{4}\)%

    Q.9

    A sum of money at a certain rate per annum of simple interest doubles in the 5 years and at a different rate becomes three times in 12 years. The lower rate of interest per annum is

    15%

    20%

    15\(\frac{3}{4}\)%

    16\( \frac{2}{3} \)%

    Ans .

    4


    1. Explanation :

      (4)Using Rule 3,
      R1 = \( \frac{(2-1)*100}{5} \)%
      = 20%
      R2 = \( \frac{(3-1)*100}{12} \)%
      = \(16\frac{2}{3}\)%
      Lowest rate of interest = = \(16\frac{2}{3}\)%

    Q.10

    In how many years will a sum of money double itself at 6\(\frac{1}{4}\)% simple interest per annum ?

    24 years

    20 years

    16 years

    12 years

    Ans .

    3


    1. Explanation :

      (3)Using Rule 3,
      T = \( \frac{(n-1)}{R} \)% years
      = \(\frac{2-1}{\frac{25}{4}}*100\) years
      = 16 years

    Q.11

    At a certain rate of simple interest, a certain sum of money becomes double of itself in 10 years. It will become treble of itself in

    15 years

    18 years

    20 years

    30 years

    Ans .

    3


    1. Explanation :

      (3)Using Rule 3,
      R% = \( \frac{(2-1)*100}{10} \)%
      = 10%
      Now , T = \( \frac{(n-1)}{R} \)*100 years
      = \(\frac{3-1}{10}*100\)
      = 20 years

    Q.12

    In how much time, will a sum of money become double of itself at 15% per annum simple interest?

    6\(\frac{1}{4}\) years

    6\(\frac{1}{2}\) years

    6\(\frac{1}{3}\) years

    6\(\frac{2}{3}\) years

    Ans .

    4


    1. Explanation :

      (4)Using Rule 3,
      T = \( \frac{(n-1)}{R} \)% years
      = \(\frac{2-1}{15}*100\)
      = \(\frac{20}{3}\) years
      = \(6\frac{2}{3}\) years

    Q.13

    In how many years will a sum of money double itself at 12% per annum?

    8 yrs. 6 months

    6 yrs. 9 months

    8 yrs. 4 months

    7 yrs. 4 months

    Ans .

    3


    1. Explanation :

      (3)Using Rule 3,
      T = \( \frac{(n-1)}{R} \)% years
      = \(\frac{2-1}{12}*100\)
      = \(\frac{25}{3}\) years
      = \(8\frac{1}{3}\) years
      = 8 years, 4 months.

    Q.14

    A sum amounts to double in 8 years by simple interest. Then the rate of simple interest per annum is

    10%

    12.5%

    15%

    20%

    Ans .

    2


    1. Explanation :

      (2)Using Rule 3,
      R% = \( \frac{(2-1)*100}{8} \)%
      = 12.5%

    Q.15

    A sum doubles itself in 16 years, then in how many years will it triple itself; rate of interest being simple

    25 years

    32 years

    48 years

    64 years

    Ans .

    2


    1. Explanation :

      (2)Using Rule 3,
      R% = \(\frac{n-1}{T}*100\) %
      =( \frac{(2-1)*100}{16} \)%
      = \(6\frac{1}{4}\)%
      Now , T = \( \frac{(n-1)}{R} \)*100
      = \(\frac{3-1}{\frac{25}{4}}*100\)
      = 32 years

    Q.16

    In certain years a sum of money is doubled to itself at 6\(\frac{1}{4}\)% simple interest per annum, then the required time will be

    16 years

    12\(\frac{1}{2}\) years

    8 years

    10\(\frac{2}{3}\) years

    Ans .

    1


    1. Explanation :

      (1)Using Rule 3,
      T = \( (\frac{(n-1)}{R}) \)*100 %
      = \(\frac{3-1}{\frac{25}{4}}*100\)
      = 16 years

    Q.17

    The simple interest on a sum of money is \( \frac{8}{25} \) of the sum. If the number of years is numerically half the rate percent per annum, then the rate percent per annumis

    5

    8

    6\(\frac{1}{4}\)

    4

    Ans .

    2


    1. Explanation :

      (2) Using Rule 1,
      Rate = R% per annum
      Therefore, Time = \(\frac{R}{2}\) years
      Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
      R = \( \frac{8}{25} * \frac{100}{\frac{R}{2}}\)
      R2 = \( \frac{8*200}{25} \) = 64
      R = 8% per annum

    Q.18

    A certain sum doubles in 7 years at simple interest. The same sum under the same interest rate will become 4 times in how many years.

    14

    28

    21

    10

    Ans .

    3


    1. Explanation :

      (3) Case I,
      Interest = Principal
      Rate = \( \frac{S.I. * 100}{P * T} \)
      = \( \frac{100}{7} \)% per annum
      Case II,
      Interest = 3 × Principal
      Time = \( \frac{S.I. * 100}{P * R} \)
      = \( \frac{3*100}{\frac{100}{7}} \)
      = 3 * 7 = 21 years

    Q.19

    A certain sum of money amounts to Rs. 2200 at 5% p.a. rate of interest, Rs. 2320 at 8% interest in the same period of time. The period of time is

    3 years

    4 years

    5 years

    2 years

    Ans .

    4


    1. Explanation :

      (4)Difference in rates = 8 – 5 = 3%
      Since, 3% => 2320 – 2200 = 120
      Therefore, 5% => \( \frac{120}{3} \) *5 = 200
      Therefore, Principal = Rs. (2200 – 200) = Rs. 2000
      Therefore, Time = \( \frac{S.I. * 100}{P * R} \)
      = \( \frac{200 * 100}{2000*5} \) = 2 years

    Q.20

    At what per cent of simple interest will a sum of money double itself in 15 years?

    6\( \frac{1}{3}\)%

    6\( \frac{2}{3}\)%

    6\( \frac{1}{2}\)%

    6%

    Ans .

    2


    1. Explanation :

      (2) Let principal be Rs. x.
      Therefore, Amount = Rs. 2x
      Therefore, Interest = Rs. (2x – x) = Rs. x
      Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
      = \( \frac{x*100}{x*15} \) = \( \frac{20}{3} \)
      = \(6\frac{2}{3}\)% per annum

    Q.21

    If a sum of money deposited in a bank at simple interest is doubled in 6 years, then after 12 years, the amount will be

    \( \frac{5}{2} \) times the original ammount

    3 times the original ammount

    \( \frac{7}{2} \) times the original ammount

    4 times the original ammount

    Ans .

    2


    1. Explanation :

      (2) Principal = Rs. x
      Interest = Rs. x
      Time = 6 years
      Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
      = \( \frac{x*100}{x*16} \) = \( \frac{50}{3} \)
      Case II,
      Interest = \(\frac{x*12*50}{100*3}\) = Rs.2x
      i.e., Amount is thrice the principal.

    Q.22

    The rate of simple interest for which a sum of money becomes 5 times of itself in 8 years is :

    30%

    40%

    50%

    55%

    Ans .

    3


    1. Explanation :

      (3) Principal = Rs. x (let)
      Therefore, Amount = Rs. 5x
      Interest = Rs. (5x – x) = Rs. 4x
      Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
      = \( \frac{4x * 100}{x*8} \)= 5% per annum

    Q.23

    If a sum of money doubles itself in 8 years, then the interest rate in percentage is

    8\(\frac{1}{2}\)%

    10%

    10\(\frac{1}{2}\)%

    12\(\frac{1}{2}\)%

    Ans .

    4


    1. Explanation :

      (4) Let principal be Rs. x.
      Therefore, Amount = Rs. 2x
      Interest = Rs. (2x – x) = Rs. x
      Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
      = \( \frac{x * 100}{x*8} \) = \(12\frac{1}{2}\)% per annum

    Q.24

    The rate of simple intereset per annum at which a sum of money doubles itself in 16\(\frac{2}{3}\)years is

    4%

    5%

    6%

    6\(\frac{2}{3}\)%

    Ans .

    3


    1. Explanation :

      (3) According to the question,
      Principal = Rs. x.
      Interest = Rs. x.
      Time = \(\frac{50}{3}\)years
      Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
      = \( \frac{x * 100}{x*\frac{50}{3}} \) = 6% per annum

    TYPE - III

    Q.1

    In what time will the simple interest be \(\frac{2}{5}\) of the principal at 8 per cent per annum?

    8 years

    7 years

    5 years

    6 years

    Ans .

    3


    1. Explanation :

      (3)Using Rule 5,
      Here, n = \( \frac{2}{5} \)and R = 8%
      RT = n*100
      T = \(\frac{n*100}{R}\)
      T = \(\frac{2}{5}*\frac{100}{8}\)
      T = 5 years

    Q.2

    The simple interest on a sum after 4 years is \(\frac{1}{5}\) of the sum. The rate of interest per annum is

    4%

    5%

    6%

    8%

    Ans .

    2


    1. Explanation :

      (2)Using Rule 5,
      Here, n = \( \frac{1}{5} \)and T = 4 years
      R = \(\frac{n*100}{T}\)
      R = \(\frac{1}{5}*\frac{100}{4}\)
      R = 5 %

    Q.3

    Simple interest on a certain sum for 6 years is \( \frac{9}{25} \) of the sum. The rate of interest is

    6%

    6\(\frac{1}{2}\)%

    8%

    8\(\frac{1}{2}\)%

    Ans .

    1


    1. Explanation :

      (1)Using Rule 5,
      Here, n = \( \frac{9}{25} \)and T = 6 years
      R = \(\frac{n*100}{T}\)
      R = \(\frac{9}{25}*\frac{100}{6}\)
      R = 6%

    Q.4

    The simple interest on a sum for 5 years is one fourth of the sum. The rate of interest per annum is

    5%

    6%

    4%

    8%

    Ans .

    1


    1. Explanation :

      (1)Using Rule 5,
      Here, n = \( \frac{1}{4} \)and T = 5 years
      R = \(\frac{n*100}{T}\)
      R = \(\frac{1}{4}*\frac{100}{5}\)
      R = 5%

    Q.5

    On a certain sum, the simple interest at the end of 6\(\frac{1}{4}\) years becomes \(\frac{3}{8}\) of the sum. The rate of interest is

    5%

    6%

    7%

    8%

    Ans .

    2


    1. Explanation :

      (2)Using Rule 5,
      Here, n = \( \frac{3}{8} \)and T = \(\frac{25}{4}\) years
      R = \(\frac{n*100}{T}\)
      R = \(\frac{3}{8}*\frac{100}{\frac{25}{4}}\)
      R = 6%

    Q.6

    The present worth of a bill due 7 months hence is Rs.1200 and if the bill were due at the end of 2\(\frac{1}{2}\) years its present worth would be Rs.1016. The rate per cent is

    5%

    10%

    15%

    20%

    Ans .

    2


    1. Explanation :

      (2) Using Rule 1,
      S.I. = \( \frac{Principal * Rate * Time}{100} \)
      Therefore, 1200 + \( \frac{1200*7*r}{12*100} \) = Amount (A)
      => 1200 + 7r = A .........(i)
      and, 1016 + \( \frac{1016*5*r}{2*100} \) = A
      Therefore, 1016 + 25.4r = A ...(ii)
      \ 1016 + 25.4r = 1200 + 7r
      25.4r – 7r = 1200 – 1016
      18.4r = 184
      r = \(\frac{184}{18.4}\) = 10% per annum

    Q.7

    At the rate of simple interest per annum, the interest on a certain sum of money for 10 years will be \(\frac{2}{5}\)th part of the amount, then the rate of simple interest is

    5%

    6\(\frac{2}{3}\)%

    7%

    4\(\frac{1}{2}\)%

    Ans .

    2


    1. Explanation :

      Using Rule 5, Here, S.I. =\(\frac{2}{5}\)amount
      S.I. =\(\frac{2}{5}\) (P+S.I.)
      S.I. =\(\frac{2}{5}\)S.I. +\(\frac{2}{5}\) P
      \(\frac{3}{5}\) = \(\frac{2}{5}\)P
      S.I. =\(\frac{2}{3}\)P
      Here, n = \( \frac{2}{3} \)and T = 10 years
      R = \(\frac{n*100}{T}\)
      R = \(\frac{2}{3}*\frac{100}{10}\)
      R = \(6\frac{2}{2}\)%

    Q.8

    A and B borrowed Rs. 3000 and Rs. 3200 respectively at the same rate of interest for 2\(\frac{1}{2}\) years. If B paid Rs. 40 more interest than A, find the rate of interest.

    5%

    7%

    8%

    6%

    Ans .

    3


    1. Explanation :

      (3) Using Rule 13,
      Here, P1 = Rs. 3000,
      R1 = R,
      T1 =\(\frac{5}{2}\) years,
      P2 = Rs. 3200,
      R2 = R,
      T2 =\(\frac{5}{2}\) years,
      Difference S.I. = Rs. 40
      40 = \( \frac{3200*R*\frac{5}{2}-3000R*\frac{5}{2}}{100} \)
      4000 = 8000R - 7500R
      R = 8%

    TYPE-IV

    Q.1

    The simple interest on a certain sum at 5% per annum for 3 years and 4 years differ by ₹42. The sum is :

    ₹210

    ₹280

    ₹750

    ₹840

    Ans .

    4


    1. Explanation :

      (4)Using Rule 13,
      P1 = P, R1 = 5%, T1 = 3years.
      P2 = P, R2 = 5%, T2 = 4 years.
      S.I.= 42
      42 = \(\frac{20P-15P}{100}\)
      P = 42 × 20
      P = ₹840

    Q.2

    The difference between the simple interest received from two different sources on ₹1500 for 3 years is ₹13.50. The difference between their rates of interest is:

    0.1%

    0.2%

    0.3%

    0.4%

    Ans .

    3


    1. Explanation :

      (3)Using Rule 13,
      P1 = Rs. 1500, R1, T1 = 3 years.
      P2 = Rs. 1500, R2, T2 = 3 years.
      S.I. = Rs. 13.50
      13.50 = \(\frac{1500*R2*3-1500*R1*3}{100}\)
      \(\frac{1350}{100}\) = \(\frac{4500(R2-R1)}{100}\)
      R2-R1=\(\frac{1350}{4500}\)=\(\frac{27}{90}\)=\(\frac{3}{10}\)=0.3%

    Q.3

    The simple interest on a sum of money is \(\frac{4}{9}\) of the principal and the number of years is equal to the rate percent per annum. The rate per annum is :

    5%

    \(6\frac{2}{3}\)%

    6%

    \(7\frac{1}{5}\)%

    Ans .

    2


    1. Explanation :

      (2) Using Rule 1,
      We know that
      S.I. = \(\frac{PRT}{100}\)
      According to question,
      S.I.= \(\frac{4}{9}\)P
      & R = T (numerically)
      Therefore, \(\frac{4}{9}\)P = \(\frac{P*R*R}{100}\)
      Therefore, R2 = \(\frac{400}{9}\)
      R= \(\frac{20}{3}\)=\(6\frac{2}{3}\)%

    Q.4

    The simple interest on a certain sum for 8 months at 4% per annum is ₹129 less than the simple interest on the same sum for 15 months at 5% per annum. The sum is :

    ₹2580

    ₹2400

    ₹2529

    ₹3600

    Ans .

    4


    1. Explanation :

      (4)Using Rule 13,
      P1 = P, R1 = 4%, T1
      = 8 months = \(\frac{8}{12}\)years
      P2 = P, R2 = 5%, T2 = 15 month = \(\frac{15}{12}\) years
      S.I. = 129
      129 = \(\frac{P*5*\frac{15}{12}-P*4*\frac{8}{12}}{100}\)
      12900 = \( \frac{75P-32P}{12} \)
      12900 = \(\frac{43P}{12}\)
      P = ₹3600

    Q.5

    Mohan lent some amount of money at 9% simple interest and an equal amount of money at 10% simple interest each for two years. If his total interest was Rs. 760, what amount was lent in each case ?

    ₹1700

    ₹1800

    ₹1900

    ₹2000

    Ans .

    4


    1. Explanation :

      (4) Using Rule 1,
      Let the sum lent in each case be x.
      Then, \( \frac{x*9*2}{100} + \frac{x*10*2}{100} \) = 760
      \( \frac{x*2}{100}(9+10) \) = 760
      \( \frac{2*19x}{100} \)
      x = \(\frac{760*100}{2*19}\) = ₹2000

    Q.6

    Simple interest on a certain sum at a certain annual rate of interest is \( \frac{16}{25} \) of the sum. If the number representing rate per cent and time in years be equal, then the rate of interest is

    8%

    \(11\frac{1}{2}\)%

    \(12\frac{1}{2}\)%

    \(12\frac{1}{4}\)%

    Ans .

    1


    1. Explanation :

      (1) Using Rule 5,
      Here , n = \(\frac{16}{25}\) , R=T
      Now , R*R = \(\frac{16}{25}*100\)
      R2 = \(\frac{1600}{25}\)
      R= \(\frac{40}{5}\)=8%

    Q.7

    A sum of ₹1500 is lent out in two parts in such a way that the simple interest on one part at 10% per annum for 5 years is equal to that on another part at 12.5% per annum for 4 years. The sum lent out at 12.5% is :

    ₹500

    ₹1000

    ₹750

    ₹1250

    Ans .

    3


    1. Explanation :

      (3) Using Rule 1,
      Let the sum lent out at 12.5% be x
      Therefore, Sum lent out at 10% = 1500 – x
      Now, \( \frac{(1500-x)*10*5}{100} \) = \( \frac{x*12.5*4}{100} \)
      50(1500-x)=50x
      2x=1500
      x=\( \frac{1500}{2} \) = ₹750

    Q.8

    If the simple interest for 6 years be equal to 30% of the principal, it will be equal to the principal after

    20 years

    30 years

    10 years

    22 years

    Ans .

    1


    1. Explanation :

      Using Rule 5,
      Here, n = \(\frac{30}{100} = \frac{3}{10} \), T = 6 years.
      RT = n × 100
      R × 6 = \(\frac{3}{10}\)*100
      R = 5%
      As,S.I. = P
      S.I. = \( \frac{P * R * T}{100} \)
      100 = RT
      100 = 5 × T
      This is possible only when T = 20.

    Q.9

    Simple interest on ₹500 for 4 years at 6.25% per annum is equal to the simple interest on ₹400 at 5% per annum for a certain period of time. The period of time is

    4 years

    5 years

    \(6\frac{1}{4}\) years

    \(8\frac{2}{3}\) years

    Ans .

    3


    1. Explanation :

      (3) Using Rule 1,
      Let the period of time be T years.
      Then, \( \frac{400*5*T}{100} = \frac{500*4*6.25}{100} \)
      T = \( \frac{500*4*6.25}{400*5} \) = \(\frac{25}{4}\)
      = 6\(\frac{1}{4}\) years

    Q.10

    The simple interest on a sum of money is \( \frac{1}{16} \)of the principal and the number of years is equal to the rate per cent per annum. The rate per annum is

    \(1\frac{1}{2} \)%

    \(2\frac{1}{2} \)%

    \(3\frac{1}{2} \)%

    \(4\frac{1}{2} \)%

    Ans .

    2


    1. Explanation :

      (2)Using Rule 5,
      Here, n = \(\frac{1}{16}\), R = T
      RT = n × 100
      R2 = \(\frac{100}{16}\)
      R = \(\frac{10}{4}\)
      R = \(2\frac{1}{2}\)%

    Q.11

    If ₹12,000 is divided into two parts such that the simple interest on the first part for 3 years at 12% per annum is equal to the simple interest on the second part for 4\(\frac{1}{2}\) years at 16% per annum, the greater part is

    ₹8000

    ₹6000

    ₹7000

    ₹7500

    Ans .

    1


    1. Explanation :

      (1) Using Rule 1,
      Let the larger part of the sum be x
      Therefore, Smaller part = (12000 – x)
      According to the question,
      \( \frac{x*3*12}{100} = \frac{(12000-x)*9*16}{2*100} \)
      36 x = (12000 – x ) 72
      x = (12000 – x ) × 2
      x + 2x = 24000
      3x = 24000
      x = \(\frac{24000}{3}\)
      = ₹8000

    Q.12

    The simple interest on a sum of money is \( \frac{1}{4} \) th of the principal and the number of years is equal to rate per cent per annum. The rate per cent is

    2.5%

    5%

    7.5%

    10%

    Ans .

    2


    1. Explanation :

      Using Rule 5,
      n = \(\frac{1}{5}\), R = T
      RT = n × 100
      R2 = \(\frac{1}{4}*100\)
      R2 = 25
      R = 5%

    Q.13

    Equal sum of money are lent to X and Y at 7.5% per annum for a period of 4 years and 5 years respectively. If the difference in interest, paid by them was ₹150, the sum lent to each was

    ₹500

    ₹1000

    ₹2000

    ₹3000

    Ans .

    3


    1. Explanation :

      Using Rule 13,
      Here, P1 = P, R1 = 7.5%, T1 = 4 years.
      P2 = P, R2 = 7.5%, T2 = 5 years.
      S.I. = Rs. 150
      S.I. = \( \frac{P2R2T2-P1R1T1}{100} \)
      150 = \( \frac{P*7.5*5-P*7.5*4}{100} \)
      15000 = 7.5P
      P = \(\frac{15000}{7.5}\)
      P = ₹2000

    Q.14

    A sum of ₹1750 is divided into two parts such that the interests on the first part at 8% simple interest per annum and that on the other part at 6% simple interest per annum are equal. The interest on each part (In rupees) is

    60

    65

    70

    40

    Ans .

    1


    1. Explanation :

      (1) Using Rule 1,
      Let first part be x and second
      part be(1750 –x )
      According to the question,
      x × \(\frac{8}{100}\) = (1750 – x ) * \(\frac{6}{100}\)
      8x + 6x = 1750 × 6
      14x = 1750 × 6
      x = \(\frac{1750*6}{14}\)= 750
      Therefore, Interest = 8% of 750
      = 750 * \( \frac{8}{100} \) = 60

    Q.15

    A borrows ₹800 at the rate of 12% per annum simple interest and B borrows ₹910 at the rate of 10% per annum, simple interest. In how many years will their amounts of debt be equal ?

    18 years

    20 years

    22 years

    24 years

    Ans .

    3


    1. Explanation :

      (3) Using Rule 1,
      Let the period of time be T years.
      Therefore, 800 + \( \frac{800*12*T}{100} \) = 910 + \(\frac{910*10*T}{100}\)
      800 + 96 T = 910 + 91T
      96 T – 91 T = 910 – 800
      5T = 110
      T = \(\frac{110}{5}\) = 22 years.

    Q.16

    The simple interest on a sum of money is \(\frac{1}{9}\) of the principal and the number of years is equal to rate per cent per annum. The rate per annum is

    3%

    \(\frac{1}{3}\)%

    3\(\frac{1}{3}\)%

    \(\frac{3}{10}\)%

    Ans .

    3


    1. Explanation :

      (3)Using Rule 5,
      Here, n = \(\frac{1}{9}\) , R = T
      RT = n × 100
      R2 =\(\frac{1}{9}\)*100
      R2 = \(\frac{100}{9}\)
      R =\(\frac{10}{3}\)
      R = 3\(\frac{1}{3}\)%

    Q.17

    A person deposited ₹500 for 4 years and ₹600 for 3 years at the same rate of simple interest in a bank. Altogether he received ₹190 as interest. The rate of simple interest per annum was

    4%

    5%

    2%

    3%

    Ans .

    2


    1. Explanation :

      (2) 411, Using Rule 1,
      Let 'r' be the rate of interest
      190 = \(\frac{500*4*r}{100}+\frac{600*3*r}{100}\)
      20r + 18r = 190 v
      38r = 190
      r = \( \frac{190}{38} \) = 5%

    Q.18

    The difference between the simple interest received from two different banks on ₹500 for 2 years is ₹2.50. The difference between their (per annum) rate of interest is :

    0.10%

    0.25%

    0.50%

    1.00%

    Ans .

    2


    1. Explanation :

      (2)Using Rule 7,
      Here, P = Rs. 500, x = Rs. 2.50,
      Difference in time = 2 years.
      Difference in rate = ?
      500 = \(\frac{2.50*100}{(diff. in rate)*2}\)
      Different in rate = 0.25%

    Q.19

    In how many years will the simple interest on a sum of money be equal to the principal at the rate of 16\(\frac{2}{3}\)% per annum ?

    4 years

    5 years

    6 years

    8 yearss

    Ans .

    3


    1. Explanation :

      (3) Using Rule 1,
      Let the principal be x.
      Time = \( \frac{S.I. * 100}{P * R} \)
      = \(\frac{x*100*3}{x*50}\) = 6 years

    Q.20

    The rate of interest per annum at which the total simple interest of a certain capital for 1 year is equal to the total simple interest of the same capital at the rate of 5% per annum for 2 years, is

    \(\frac{5}{2}\)%

    10%

    25%

    12.5%

    Ans .

    2


    1. Explanation :

      (2) Using Rule 1,
      \( \frac{P*R*1}{100} = \frac{P*5*2}{100} \)
      [Since, Capital is same in both cases]
      r × 1 = 5 × 2
      r = 10%

    Q.21

    The simple interest on ₹4,000 in 3 years at the rate of x% per annum equals the simple interest on ₹5,000 at the rate of 12% per annum in 2 years. The value of x is

    10%

    6%

    8%

    9%

    Ans .

    1


    1. Explanation :

      (1) Using Rule 1,
      S.I. = \( \frac{Principal * Rate * Time}{100} \)
      Therefore, \( \frac{4000*3*x}{100} \) = \( \frac{5000*2*12}{100} \)
      x = \( \frac{5*2*12}{4*3} \).
      = 10% per annum

    Q.22

    If x, y, z are three sum of money such that y is the simple interest on x and z is the simple interset on y for the same time and at the same rate of interest, then we have

    z2 = xy

    xyz = 1

    x2 = yz

    y2= zx

    Ans .

    4


    1. Explanation :

      (4) Using Rule 1,
      S.I. = \( \frac{P * R * T}{100} \)
      Therefore, y = \( \frac{x*T*R}{100} \)
      and z = \( \frac{y*T*R}{100} \)
      So, \(\frac{y}{z} = \frac{x}{y}\)
      y2= zx

    Q.23

    Prakash lends a part of ₹20,000 at 8% simple interest and remaining at \(\frac{4}{3}\)% simple interest. His total income after a year was ₹800. Find the sum lent at 8%.

    ₹8000

    ₹12000

    ₹6000

    ₹10000

    Ans .

    1


    1. Explanation :

      (1) Using Rule 1,
      Amount lent at 8% rate of interest = ₹x
      Therefore, Amount lent at \(\frac{4}{3}\)% rate of interest = (20,000 – x)
      Therefore, S.I. = \( \frac{Principal * Rate * Time}{100} \)
      \( \frac{x*8*1}{100} + \frac{(20000-x)*\frac{4}{3}*1}{100} \) = 800
      \( \frac{2x}{25} + \frac{20000-x}{75} \) = 800
      \( \frac{6x+20000-x}{75} \) = 800
      5x + 20,000 = 75 × 800 = 60,000
      5x = 60,000-20,000 = 40,000
      x = \(\frac{40000}{5}\)= 8000

    Q.24

    Ram deposited a certain sum of money in a company at 12% per annum simple interest for 4 years and deposited equal amount in fixed deposit in a bank for 5 years at 15% per annum simple interest. If the difference in the interest from two sources is ₹1350, then the sum deposited in each case is :

    ₹3000

    ₹4000

    ₹5000

    ₹6500

    Ans .

    3


    1. Explanation :

      (3)Using Rule 13.
      Here, P1 = Rs. P, R1 = 12%, T1 = 4 years
      P2 = Rs. P, R2 = 15%, T2 = 5 years
      S.I. = Rs. 1350
      S.I.= \( \frac{P2R2T2-P1R1T1}{100} \)
      1350 = \( \frac{P*15*5-P*12*4}{100} \)
      135000 = 75 P – 48P
      135000 = 75 P
      P = Rs. 5000

    Q.25

    The difference between simple interest and the true discount on Rs. 2400 due 4 years hence at 5% per annum simple interest is

    Rs.30

    Rs.70

    Rs.80

    Rs.50

    Ans .

    3


    1. Explanation :

      (3) Using Rule 1,
      True discount = \( \frac{Amount*Rate*Time}{100+(Rate*Time)} \)
      \(\frac{2400*5*4}{100+(5*4)}\)
      = \(\frac{2400*5*4}{120}\) = Rs.400
      S.I. = \(\frac{2400*5*4}{100}\) = Rs.480
      Required difference = Rs. (480 – 400) = Rs. 80

    TYPE-V

    Q.1

    A sum of ₹1550 was lent partly at 5% and partly at 8% simple interest. The total interest received after 3 years is ₹300. The ratio of money lent at 5% to that at 8% is :

    5 : 8

    8 : 5

    31 : 6

    16 : 15

    Ans .

    4


    1. Explanation :

      (4) Using Rule 1,
      Let the sum lent at the rate of interest 5% per annum is x and at the rate of interest 8% per annum is (1550 – x)
      According to the question,
      \( \frac{x*5*3}{100} + \frac{(1500-x)*8*3}{100} \) = 300
      \( \frac{15x}{100} + \frac{37200-24x}{100} \) = 300
      15x + 37200 – 24x = 300 × 100
      9x = 7200
      Therefore, x = ₹800 and,
      1550 – x = 1550 – 800 = ₹750
      Therefore, Ratio of money lent at 5% to that at 8% = 800 : 750 = 16 : 15

    Q.2

    A person lent ₹5,000 partly at the rate of 4 per cent and partly at the rate of 5 per cent per annum simple interest. The total interest after 2 years is ₹440. To find the sum of money lent at each of the above rates, ₹5,000 is to be divided in the ratio :

    4 : 5

    3 : 2

    5 : 4

    2 : 3

    Ans .

    2


    1. Explanation :

      (2) Using Rule 1,
      Let the sum of x be lent at the rate of 4% and (5000 – x) at the rate of 5%
      \( \frac{x*4*2}{100} + \frac{(5000-x)*5*2}{100} \) = 440
      8x + 50000 – 10x = 44000
      2x = 50000 – 44000 = 6000
      x = ₹3000
      Therefore, ₹(5000 – x)
      = ₹(5000 – 3000) = ₹2000
      Now, Required ratio
      = 3000 : 2000 = 3 : 2

    Q.3

    A person borrows some money for 5 years and loan amount : total interest amount is 5 : 2. The ratio of loan amount : interest rate is equal to :

    2 : 25

    2:1

    5:2

    25:2

    Ans .

    4


    1. Explanation :

      (4) Required ratio = 5 : \(\frac{2}{5}\) = 25 : 2
      \( \frac{loan amount }{Interest Amount} = \frac{5}{2} \)
      Interest Rate = \(\frac{2}{5}\)
      Since, \( [ \frac{P+I}{I} = \frac{5}{2}=> \frac{P}{I}+1 = \frac{5}{2} => \frac{P}{I} = \frac{3}{2} => I = \frac{2}{5} ] \)
      \( \frac{loan amount }{Interest Rate} = \frac{5}{\frac{2}{5}} \)
      \( \frac{25}{2} \) or 25:2

    Q.4

    A person invests money in three different schemes for 6 years, 10 years and 12 years at 10 per cent, 12 per cent and 15 per cent simple interest respectively. At the completion of each scheme, he gets the same interest. The ratio of his investment is

    6:3:2

    2:3:4

    3:4:6

    3:4:2

    Ans .

    1


    1. Explanation :

      (1) Using Rule 1,
      P1 : P2 : P3 = \( \frac{1}{r1t1} : \frac{1}{r2t2} : \frac{1}{r3t3} \)
      = \( \frac{1}{6*10} : \frac{1}{10*12} : \frac{1}{12*15} \)
      \( \frac{1}{60} : \frac{1}{120} : \frac{1}{180} \)
      = 6:3:2

    Q.5

    With a given rate of simple inerest, the ratio of principal and amount for a certain period of time is 4 : 5. After 3 years, with the same rate of interest, the ratio of the principal and amount becomes 5 : 7. The rate of interest is

    4%

    6%

    5%

    7%

    Ans .

    3


    1. Explanation :

      (3) Using Rule 1,
      Case-I,
      Interest = 5x – 4x = x
      Therefore, x = \(\frac{4x*R*T}{100}\)
      T =\( \frac{25}{R} \) years
      Case-II,
      T = \( \frac{25}{R} \)+ 3 = \( \frac{25+3R}{R} \) years
      SI = 7 y – 5y = 2y
      Therefore, 2y = \( \frac{5y*R*(25+3R)}{R*100} \)
      40 = 25 + 3R
      3R = 40 –25 = 15 %
      R = \( \frac{15}{3} \) =5%

    Q.6

    Ratio of the principal and the amount after 1 year is 10:12. Then the rate of interest per annum is :

    12%

    16%

    18%

    20%

    Ans .

    4


    1. Explanation :

      (4) Using Rule 1,
      \( \frac{Principal}{Amount} = \frac{10}{12} \)
      \( \frac{Amount}{Principal} = \frac{Principal+Interest}{Principal} \)
      = \(\frac{12}{10}\)
      1 + \(\frac{Interest}{Principal} = \frac{12}{10} \)
      \(\frac{Interest}{Principal} = \frac{2}{10} = \frac{1}{5} \)
      Therefore, Rate = \(\frac{1}{5}\)*100 = 20%

    Q.7

    In a certain time, the ratio of a certain principal and the simple interest obtained from it are in the ratio 10 : 3 at 10% interest per annum. The number of years the money was invested is

    1 year

    3 years

    5 years

    7 years

    Ans .

    2


    1. Explanation :

      (2) Using Rule 1,
      Time = \( \frac{S.I. * 100}{P * R} \)
      = \( \frac{3}{10} * \frac{100}{10} \) = 3 years

    Q.8

    ₹12,000 is divided into two parts so that the simple interest on the first part for 3 years at 12% per annum may be equal to the simple interest on the second part for 4\(\frac{1}{2}\) years at 16% per annum. The ratio of the first part to the second part is

    2:1

    1:2

    2:3

    3:2

    Ans .

    1


    1. Explanation :

      (1) Using Rule 1,
      First part = Rs. x and second part = (12000 – x )
      Therefore, \( \frac{x*3*12}{100} = \frac{(12000-x)*9*16}{200} \)
      \(\frac{x}{12000-x} = \frac{9*16*100}{3*12*200} \)
      \( \frac{2}{1} \)= 2 : 1

    Q.9

    If ratio of principal and simple interest for 1 year is 25 : 1, then the rate of interest is

    4%

    25%

    5%

    20%

    Ans .

    1


    1. Explanation :

      (1) Using Rule 1,
      Principal : Interest = 25 : 1
      Interest : Principal = 1 : 25
      Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
      = \(\frac{1}{25}\) × 100 = 4% per annum

    Q.10

    If the ratio of principal and the simple interest for 5 years is 10 : 3, then the rate of interest is :

    5%

    6%

    8%

    3%

    Ans .

    2


    1. Explanation :

      (2) Using Rule 1,
      \( \frac{Principal}{Interest} = \frac{10}{3} \)
      \( \frac{Interest}{Principal} = \frac{3}{10} \)
      Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
      = \(\frac{3}{10}*\frac{100}{5}\) = 6% per annum

    Q.11

    A sum of Rs. 4000 is lent out in two parts, one at 8% simple interest and the other at 10% simple interest. If the annual interest is Rs. 352, the sum lent at 8% is

    Rs.2900

    Rs.2200

    Rs.2400

    Rs.3100

    Ans .

    3


    1. Explanation :

      (3) Principal lent at 8% S.I. = Rs. x.
      Therefore, Principal lent at 10% S.I. = Rs. (4000 – x)
      S.I. = \( \frac{Principal * Rate * Time}{100} \)
      \(\frac{x*8}{100}+\frac{(4000-x)*10}{100}\) = 352
      8x + 40000 – 10x = 35200
      2x = 40000 – 35200 = 4800
      x = \( \frac{4800}{2} \) = Rs. 2400

    TYPE-VI

    Q.1

    A sum of ₹400 amounts to ₹480 in 4 years. What will it amount to if the rate of interest is increased by 2%?

    ₹484

    ₹560

    ₹512

    None of these

    Ans .

    3


    1. Explanation :

      (3) Using Rule 1,
      Interest = ₹. (480–400) = ₹80
      Therefore, 80 = \(\frac{400*r*4}{100}\)
      r = 5
      Now, r = 7% (2% increase)
      Therefore, S.I. = \( \frac{400*7*4}{100} \) = 112
      Therefore, Amount = ₹(400+112) = ₹512

    Q.2

    A man loses ₹55.50 yearly when the annual rate of interest falls from 11.5% to 10%. His capital (in rupees) is

    3700

    7400

    3325

    11100

    Ans .

    1


    1. Explanation :

      1) Using Rule 1,
      Let his capital be x.
      According to the question,
      \( \frac{x*11.5}{100} - \frac{x*10}{100} = 55.50 \)
      or (11.5 – 10)x = 5550
      or 1.5x = 5550
      or x = \( \frac{5550}{1.5} \) = ₹3700

    Q.3

    If the annual rate of simple interest increases from 10% to 12 \(\frac{1}{2}\)% , a man’s yearly income increases by ₹1250. His principal (in rupees) is

    50000

    45000

    60000

    65000

    Ans .

    1


    1. Explanation :

      (1) Using Rule 1,
      Change in SI
      = \( (\frac{25}{2}-10) \)% = \(\frac{5}{2}\)%
      Therefore, \(\frac{5}{2}\)% of principal = ₹1250
      Therefore, Principal = ₹ \( \frac{1250*2*100}{5} \) = ₹50000

    Q.4

    A sum was invested on simple interest at a certain rate for 2 years. Had it been put at 3% higher rate, it would have fetched ₹72 more. The sum is

    ₹1200

    ₹1500

    ₹1600

    ₹1800

    Ans .

    1


    1. Explanation :

      (1)Using Rule 13,
      P1 = P, R1 = R, T1 = 2
      P2 = P, R2 = R + 3, T2 = 2
      S.I.= 72
      72 = \( \frac{P*(R+3)*2-P*R*2}{100} \)
      7200 = 6P
      P = 1200

    Q.5

    A sum of money was lent at simple interest at a certain rate for 3 years. Had it been lent at 2.5% per annum higher rate, it would have fetched ₹540 more. The money lent was :

    ₹6400

    ₹6472

    ₹6840

    ₹7200

    Ans .

    4


    1. Explanation :

      (4) If the sum lent be Rs. x, then
      \( \frac{x*2.5*3}{100} \) = 540
      x = \(\frac{540*100}{2.5*3}\) = ₹7200

    Q.6

    A sum of money was invested at a certain rate of simple interest for 2 years . Had it been invested at 1% higher rate, it would have fetched ₹24 more i nterest. The sum of money is :

    ₹1200

    ₹1050

    ₹1000

    ₹9600

    Ans .

    1


    1. Explanation :

      (1) \( \frac{P*1*2}{100} \) = 24
      P = \( \frac{2400}{2} \) = ₹1200

    Q.7

    A person who pays income tax at the rate of 4 paise per rupee, find that a fall of interest rate from 4% to 3.75% diminishes his net yearly income by ₹48. What is his capital ?

    ₹24000

    ₹25000

    ₹20000

    ₹18000

    Ans .

    3


    1. Explanation :

      (3) If the capital after tax deduction
      be x, then
      x × (4 – 3.75) % = 48
      \(\frac{x*0.25}{100}\)= 48
      \(\frac{x*25}{10000}\)= 48
      \(\frac{x}{400}\)= 48
      x = 48 × 400 = ₹19200
      Therefore, Required capital
      = \( \frac{19200*100}{96} \)
      = ₹20000

    Q.8

    A sum was lent at simple interest at a certain rate for 2 years. Had it been lent at 3% higher rate, it would have fetched ₹300 more. The original sum of money was :

    ₹5000

    ₹6000

    ₹7000

    ₹4000

    Ans .

    1


    1. Explanation :

      (1)Using Rule 13.
      P1 = P, R1 = R, T1 = 2.
      P2 = P, R2 = R + 3, T2 = 2.
      S.I.= 300
      300 = \( \frac{P*(R+3)*2-P*R*2}{100} \)
      300 = \( \frac{6P}{100} \)
      p = ₹5000

    Q.9

    A sum of ₹2,400 amounts to ₹3,264 in 4 years at a certain rate of simple interest. If the rate of interest is increased by 1%, the same sum in the same time would amount to

    ₹3288

    ₹3312

    ₹3340

    ₹3360

    Ans .

    4


    1. Explanation :

      (4) Using Rule 1,
      S.I. = 3264 – 2400 = ₹864
      Rate = \( \frac{S.I. * 100}{P * T} \)
      = \( \frac{864*100}{2400*4} \) = 9% per annum
      New rate = 10% per annum
      Therefore, S.I. = \( \frac{2400*10*4}{100} \) = ₹960
      Therefore, Amount = 2400 + 960 = ₹3360

    Q.10

    ₹800 amounts to ₹920 in 3 years at simple interest. If the interest rate is increased by 3%, it would amount to

    ₹1056

    ₹1112

    ₹1182

    ₹992

    Ans .

    4


    1. Explanation :

      (4) Using Rule 1,
      S.I. = ₹(920 – 800) = ₹120
      Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
      = \( \frac{120*100}{800*3} \)= 5% per annum
      New rate = 8% per annum
      Therefore, S.I. = \( \frac{800*3*8}{100} \) = ₹192
      Therefore, Amount = (800 + 192) = ₹992

    Q.11

    A sum of Rs. 800 amounts to Rs. 920 in 3 years at the simple interest rate. If the rate is increased by 3% p.a., what will be the sum amount to in the same period ?

    ₹992

    ₹962

    ₹942

    ₹982

    Ans .

    1


    1. Explanation :

      (1) Using Rule 1,
      Case I,
      S.I. = 920 – 800 = ₹120
      Rate = \( \frac{S.I. * 100}{P * T} \)
      = \(\frac{120*100}{800*3}\) = 5% per annum
      Case II,
      Rate = 8% per annum
      S.I. = \( \frac{800*8*3}{100} \) = ₹192
      Therefore, Amount = Principal + S.I.
      = (800 + 192) = ₹992

    Q.12

    The amount ₹2,100 became ₹2,352 in 2 years at simple interest. If the interest rate is decreased by 1%, what is the new interest ?

    ₹210

    ₹220

    ₹242

    ₹252

    Ans .

    1


    1. Explanation :

      (1) Using Rule 1,
      S.I. = 2352 – 2100 = 252
      Rate = \( \frac{S.I. * 100}{P * T} \).
      = \( \frac{252 * 100}{2100*2} \) = 6% per annum
      New rate = 5%
      Therefore, S.I. = \(\frac{252*5}{6}\) = ₹210

    Q.13

    A sum of Rs. 800 becomes Rs. 956 in 3 years at a certain rate of simple interest. If the rate of interest is increased by 4%, what amount will the same sum become in 3 years ?

    Rs.1025

    Rs.1042

    Rs.1052

    Rs.1024

    Ans .

    3


    1. Explanation :

      (3) Using Rule 1,
      S.I. = 956 – 800 = Rs. 156
      Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
      = \( \frac{156 * 100}{800*3} \)
      = 6.5%
      New rate = (6.5 + 4)%
      = 10.5%
      Thereforel, S.I. = \( \frac{Principal * Rate * Time}{100} \)
      = \( \frac{800*3*10.5}{100} \)
      = Rs. 252
      Therefore, Amount = Rs.(800 + 252)
      = Rs.1052

    Q.14

    The rate of simple interest per annum of bank being decreased from 5% to 3\(\frac{1}{2}\) %, the annual income of a person from interest was less by Rs. 105. The sum deposited at the bank was

    Rs.6000

    Rs.7200

    Rs.6800

    Rs.7000

    Ans .

    4


    1. Explanation :

      (4) Using Rule 1,
      Amount deposited in bank = Rs. x (let)
      Difference of rates = 5 – \( \frac{7}{2} \)= \(\frac{3}{2}\) % per annum
      Therefore, S.I. = \( \frac{Principal * Rate * Time}{100} \)
      \( \frac{x*1*3}{100} \) = 105
      x = \(\frac{105*200}{3}\) = Rs. 7000

    TYPE-VII

    Q.1

    A sum of ₹10,000 is lent partly at 8% and remaining at 10% per annum. If the yearly interest on the average is 9.2%, the two parts are :

    ₹4000,₹6000

    ₹4500,₹5500

    ₹5000,₹5000

    ₹5500,₹4500

    Ans .

    1


    1. Explanation :

      (1) Using Rule 1,
      Let x be lent at 8%, then (10000 – x) is lent at 10%.
      Accordingly,
      \( \frac{10000*9.2*t}{100} = \frac{x*8*t}{100} + \frac{(10000-x)*10*t}{100} \)
      \( \frac{92000t}{100} = \frac{8xt}{100} + \frac{(10000-x)10t}{100} \)
      92000t = 8xt + (10000 – x) 10t
      92000 = 8x + 100000 – 10x
      2x = 8000
      x = 4000
      Therefore, First part = ₹4000
      Second part = ₹6000

    Q.2

    A sum of ₹1000 is lent out partly at 8% and the remaining at 10% per annum. If the yearly income on the average is 9.2%, the two parts respectively are

    ₹400,₹600

    ₹450,₹550

    ₹500,₹500

    ₹550,₹450

    Ans .

    1


    1. Explanation :

      (1) Let x be lent on 8%.
      Therefore, (1000 – x ) is lent on 10%.
      Interest = 9.2% of 1000 = 92
      Therefore, 92 = \( \frac{x*8}{100} + (\frac{1000-x}{100})*10 \)
      8x + 10000 – 10x = 9200
      – 2x = 9200 – 10000
      x = \(\frac{800}{2}\) = ₹400 = first part
      Therefore, Second part = 600

    Q.3

    An old article is available for 12,000 at cash payment or is available for ₹7,000 cash payment and a monthly instalment of ₹630 for 8 months. The rate per cent per annum is

    2.1%

    3%

    3.25%

    3.3%

    Ans .

    1


    1. Explanation :

      (1) Interest = (7000 + 630 × 8) – 12000
      = (7000 + 5040) – 12000
      = 12040 – 12000 = 40
      Total Principal
      = 5000 + 4370 + 3740 + 3110 + 2480 + 1850 + 1220 + 590
      = 22360
      Rate = \( \frac{40*100*12}{22360*1} \) = 2.1 per cent

    Q.4

    The effective annual rate of interest, corresponding to a nominal rate of 6% per annum payable half yearly, is :

    6.06%

    6.07%

    6.08%

    6.09%

    Ans .

    4


    1. Explanation :

      (4) Let the sum be ₹100.
      For initial six months, Interest
      = 100 * \( \frac{6}{100} * \frac{6}{12} \) = 3 %
      Now, sum = 100 + 3 = ₹103
      For another six months,
      Interest = 103 * \( \frac{6}{100} * \frac{6}{12} \) = 3.09
      Therefore, Rate of interest per annum = 3 + 3.09 = 6.09%

    Q.5

    A person lends 40% of his sum of money at 15% per annum, 50% of rest at 10% per annum and the rest at 18% per annum rate of interest. What would be the annual rate of interest, if the interest is calculated on the whole sum ?

    13.4%

    14.33%

    14.4%

    13.33%

    Ans .

    3


    1. Explanation :

      (3) Let the person have 100.
      Then SI for 1 year = ₹\( (\frac{40*15*1}{100} + \frac{30*10*1}{100} + \frac{30*18*1}{100} ) \)
      = ₹(6+3+5.4) = ₹14.4
      Therefore, Rate of interest on whole sum = 14.4%

    Q.6

    Ramesh deposited ₹15600 in a fixed deposit at the rate of 10% per annum simple interest. After every second year, he adds his interest earnings to the principal. The interest at the end of fourth year is

    ₹1716

    ₹1560

    ₹3432

    ₹1872

    Ans .

    4


    1. Explanation :

      (4) SI earned after two years
      = \( \frac{15600*10*2}{100} \) = ₹3120
      Therefore, Principal for next two years
      = ₹(15600 + 3120) = ₹18720
      SI earned at the end of fourth year = \(\frac{18720*10*1}{100}\) = ₹1872

    Q.7

    A part of ₹1500 was lent at 10% per annum and the rest at 7% per annum simple interest. The total interest earned in three years was ₹396. The sum lent at 10% was

    ₹900

    ₹800

    ₹700

    ₹600

    Ans .

    1


    1. Explanation :

      (1) Let x be lent at 10% per annum.
      Therefore, (1500 – x ) is lent at 7% per annum.
      Now,
      \( \frac{x*10*3}{100} + \frac{(1500-x)*7*3}{100} \) = 396
      30x + 31500 – 21x
      = 39600
      9x = 39600 – 31500
      x = \( \frac{8100}{9} \) = 900

    Q.8

    What equal instalment of annual payment will discharge a debt which is due as ₹848 at the end of 4 years at 4% per annum simple interest ?

    ₹212

    ₹200

    ₹250

    ₹225

    Ans .

    2


    1. Explanation :

      Using Rule 10,
      Here, A = 848,
      T = 4 years, r = 4%
      Equal instalment = \( \frac{848*200}{4[200+(4-1)4]} \)
      = \( \frac{848*200}{4*212} \) = ₹200

    Q.9

    Out of ₹50,000, that a man has, he lends ₹8000 at 5\( \frac{1}{2} \)% per annum simple interest and Rs. 24,000 at 6 % per annum simple interest. He lends the remaining money at a certain rate of interest so that he gets total annual interest of ₹3680. The rate of interest per annum, at which the remaining money is lent, is

    5%

    7%

    10%

    12%

    Ans .

    3


    1. Explanation :

      (3) Using Rule 1.
      Remaining amount
      = (50000 – (8000 + 24000)) = 18000
      Let 18000 be lent at the rate of r% p.a.
      According to the question,
      \( \frac{8000*11*1}{2*100} + \frac{24000*6*1}{100} + \frac{18000*r*1}{100} \) = 3680
      440 + 1440 + 180r = 3680
      1880 + 180r = 3680
      180r = 3680 – 1880 = 1800
      r = \(\frac{1800}{180}\)= 10%

    Q.10

    A man invests half his capital at the rate of 10% per annum, onethird at 9% and the rest at 12% per annum. The average rate of interest per annum, which he gets, is

    9%

    10%

    10.5%

    12%

    Ans .

    2


    1. Explanation :

      (2) Using Rule 1.
      Let the principal be x .
      Therefore, I1 = \( \frac{x*10*1}{2*100} = \frac{x}{20} \)
      I2 = \( \frac{x*9*1}{3*100} = \frac{3x}{20} \)
      I3 = \( \frac{x}{6} * \frac{12*1}{100} = \frac{x}{50} \)
      Therefore, I1 + I2 + I3 = \( (\frac{x}{20}+\frac{3x}{100}+\frac{x}{50}) \)
      =\( \frac{5x+3x+2x}{100} = \frac{x}{10}\)
      Therefore, Average annual rate = 10%

    Q.11

    John invested a sum of money at an annual simple interest rate of 10%. At the end of four years the amount invested plus interest earned was ₹770. The amount invested was

    ₹650

    ₹350

    ₹550

    ₹500

    Ans .

    3


    1. Explanation :

      (3) Using Rule 1.
      If the principal be x, then
      Simple interest = (770 – x)
      Therefore, Principal = \( \frac{S.I. * 100}{R * T} \)
      x = \( \frac{(700-x)*100}{4*10} \)
      2x =(770 – x) × 5
      2x + 5x = 770 × 5
      7x = 770 × 5
      Therefore, x = \(\frac{770*5}{7}\) = ₹550

    Q.12

    Arun lends ₹20,000 to two of his friends. He gives ₹12,000 to the first at 8% p.a. simple interest. Arun wants to make a profit of 10% on the whole. The simple interest rate at which he should lend the remaining sum of money to the second friend is

    8%

    16%

    12%

    13%

    Ans .

    4


    1. Explanation :

      (4) Using Rule 1.
      S.I. on ₹12000 = \( \frac{12000*8*1}{100} \) = ₹960
      Desired gain on ₹20000
      = 20000 * \(\frac{10}{100}\)= ₹2000
      Therefore, S.I. on ₹8000 = 2000 – 960 = ₹1040
      Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
      =\( \frac{1040*100}{8000} \) = 13% per annum

    Q.13

    A person invests ₹12,000 as fixed deposit at a bank at the rate of 10% per annum simple interest. But due to some pressing needs he has to withdraw the entire money after 3 years, for which the bank allowed him a lower rate of interest. If he gets ₹3,320 less than what he would have got at the end of 5 years, the rate of interest allowed by the bank is

    \( 7\frac{5}{9} \)%

    \( 7\frac{4}{9} \)%

    \( 7\frac{8}{9} \)%

    \( 8\frac{7}{9} \)%

    Ans .

    2


    1. Explanation :

      (2) Using Rule 1.
      S.I. after five years
      = \( \frac{Principal * Rate * Time}{100} \) = \( \frac{12000*5*10}{100} \) = 6000
      Interest earned
      = (6000 – 3320) = 2680
      Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
      = \( \frac{2680 * 100}{12000*3} \) = \( \frac{67}{9} = 7\frac{4}{9}\)%

    Q.14

    A certain scheme of investment in simple interest declares that it trebles the investment in 8 years. If you want to quadruple your money through that scheme, you have to invest it for :

    11 years 6 months

    10 years 8 months

    10 years

    12 years

    Ans .

    4


    1. Explanation :

      (4) Using Rule 1.
      Case I
      Let principal be x then Amount = 3x
      S.I. = 2x
      Therefore, Rate = \( \frac{S.I. * 100}{P * T} \)
      = \(\frac{2x*100}{x*8}\)=25%
      Case II
      Time = \( \frac{S.I. * 100}{P * R} \)
      = \(\frac{3x*100}{x*25}\)= 12 years

    Q.15

    If a man receives on one-fourth of his capital 3% interest, on two third 5% and on the remainder 11%, the percentage he receives on the whole is

    4.5%

    5%

    5.5%

    5.2%

    Ans .

    2


    1. Explanation :

      (2) Using Rule 1.
      Required percent
      = \( \frac{1}{4}*3 + \frac{2}{3}*5 + (1-\frac{1}{4}-\frac{2}{3})*11 \)
      =\( \frac{3}{4}+\frac{10}{3}+\frac{11}{12} =\frac{9+40+11}{12} \) =5%

    Q.16

    At the same rate of simple interest sum of the interest of ₹300 for 4 years and the interest of ₹400 for 3 years is ₹120. The rate of interest is

    5%

    4%

    6%

    10%

    Ans .

    1


    1. Explanation :

      (1) Using Rule 1.
      120 = \( \frac{300*4*r}{100} + \frac{400*3*r}{100} \)
      24r = 120
      r = \frac{120}{24} = 5% per annum

    Q.17

    Nitin borrowed some money at the rate of 6% p.a. for the first three years, 9% p.a. for the next five years and 13% p.a. for the period beyond eight years. If the total interest paid by him at the end of eleven years is ₹8,160, the money borrowed by him (in ₹) was

    12000

    6000

    8000

    10000

    Ans .

    3


    1. Explanation :

      (3) Using Rule 1.
      If the sum of money be x, then
      =\( \frac{x*6*3}{100}+\frac{x*5*9}{100}+\frac{x*3*13}{100} \) = 8160
      18x + 45x + 39x = 816000
      102x = 816000
      x = \(\frac{816000}{102}\)
      = 8000

    Q.18

    Two equal sums were lent out at 7% and 5% S.I. respectively. The interest earned on the two loans add up to ₹960 for 4 years. The total sum lent out in

    ₹3500

    ₹2500

    ₹2000

    ₹3000

    Ans .

    3


    1. Explanation :

      (3) Using Rule 1.
      If each amount lent be x, then
      =\( \frac{x*7*4}{100}+\frac{x*5*4}{100} \) = 960
      \( \frac{48x}{100} \) = 960
      x = \(\frac{960*100}{48}\)
      x = ₹2000

    Q.19

    Mohan lends Rs. 500 to John and a certain sum to Tom at the same time at a simple interest of 8% per annum. If in 4 years, he altogether receives Rs. 210 as interest from the two, then the sum of money he lent to Tom was

    Rs.144.75

    Rs.148

    Rs.156.25

    Rs.165.50

    Ans .

    3


    1. Explanation :

      (3) Using Rule 1.
      Let the money lent to Tom be Rs. x.
      Simple interest = \( \frac{Principal * Rate * Time}{100} \)
      Therefore, \( \frac{500*8*4}{100} + \frac{x*8*4}{100} \) = 210
      160 + \( \frac{32x}{100} \) = 210
      \( \frac{32x}{100} \) = 210 – 160 = 50
      x = \( \frac{50*100}{32} \) = Rs. 156.25

    Q.20

    What should be the least number of years in which the simple interest on Rs. 2600 at 6\(\frac{2}{3}\) % will be an exact number of rupees ?

    3

    2

    5

    4

    Ans .

    1


    1. Explanation :

      (1) Using Rule 1.
      Rate = \( \frac{20}{3} \)% per annum
      Therefore, S.I.= \( \frac{Principal * Rate * Time}{100} \)
      = \( \frac{2600*20*T}{3*100} \)
      Therefore, Required Time = 3 years

    Q.21

    Ram bought a bike for Rs. 60,000. He paid Rs. 10000 cash down and the rest at the end of 2 years at 15% simple interest. How much more did he pay as simple interest ?

    Rs.15000

    Rs.25000

    Rs.35000

    Rs.50000

    Ans .

    1


    1. Explanation :

      (1) Using Rule 1.
      Principal = Rs. (60000 – 10000)
      = Rs. 50000
      Therefore, S.I. = \( \frac{50000*15*2}{100} \)
      = Rs. 15000

    Q.22

    A sum of Rs. 7,930 is divided into three parts and given on loan at 5% simple interest to A, B and C for 2, 3 and 4 years respectively. If the amounts of all three are equal after their respective periods of loan, then A received a loan of

    Rs.3050

    Rs.2760

    Rs.2750

    Rs.2800

    Ans .

    2


    1. Explanation :

      (2) Using Rule 1.
      Let the loans taken by A, B and C be Rs. x, Rs. y and Rs. z respectively.
      Therefore, x + y + z = Rs. 7930
      S.I. = \( \frac{Principal * Rate * Time}{100} \)
      According to the question,
      x+\(\frac{x*2*5}{100}\)= y+\(\frac{y*3*5}{100}\)= z+\(\frac{z*5*4}{100}\)
      \(\frac{100x+10x}{100} = \frac{100y+15y}{100} = \frac{100z+20z}{100} \)
      110x = 115y = 120z
      22x = 23y = 24z
      \(\frac{22x}{6072} = \frac{23y}{6072} = \frac{24z}{6072} \)
      \(\frac{x}{276} = \frac{y}{264} = \frac{z}{253} \)
      x:y:z = 276:264:253
      Sum of terms of ratio = 276 + 264 + 253 = 793
      Therefore, A's Loan = \(\frac{276}{793}\)*7930
      = Rs.2760

    Q.23

    A man buys a TV priced at Rs. 16000. He pays Rs. 4000 at once and the rest after 15 months on which he is charged a simple interest at the rate of 12% per year. The total amount he pays for the TV is

    Rs.18200

    Rs.17800

    Rs.16800

    Rs.17200

    Ans .

    2


    1. Explanation :

      (2) Using Rule 1.
      Remaining amount
      = Rs. (16000 – 4000)
      = Rs. 12000
      \Therefore, S.I. = \( \frac{Principal * Rate * Time}{100} \)
      = \( \frac{12000*15*12}{12*100} \) = Rs.1800
      Therefore, Total amount paid
      = Rs. (16000 + 1800)
      = Rs. 17800

    Q.24

    If Rahim deposited the same amount of Rs. x in a bank at the beginning of successive 3 years and the bank pays simple interest of 5% per annum, then the amount at his credit at the end of 3rd year will be :

    Rs. \( \frac{861x}{400} \)

    Rs. \( \frac{1261x}{400} \)

    Rs. \( \frac{21x}{20} \)

    Rs. \( \frac{25241x}{8000} \)

    Ans .

    4


    1. Explanation :

      (4)Using Rule 1.
      S.I. after 1 year = \( \frac{Principal * Rate * Time}{100} \)
      = \( \frac{x*5}{100} \) = Rs. \( \frac{x}{20} \)
      Principal for 2nd year
      = Rs. \( 2x+\frac{x}{20} \) = Rs. \( \frac{41x}{20} \)
      S.I. after second year
      = Rs. \( \frac{41x}{20}*\frac{5}{100} \)
      = Rs. \( \frac{41x}{400} \)
      Principal for third year
      = Rs. \( 3x+\frac{41x}{400} \)
      = Rs. \( \frac{1200x+41x}{400} \)
      = Rs. \( \frac{1241x}{400} \)
      Therefore, S.I. after 3rd year
      = Rs. \( \frac{1241x}{400} * \frac{5}{100} \)
      = Rs. \( \frac{1241x}{8000} \)
      Therefore, Required amount
      = Rs. \( (3x+\frac{1241x}{8000}) \)
      = Rs. \( \frac{24000x+1241x}{8000} \)
      = Rs. \( \frac{25241x}{8000}\)

    Q.25

    A boy aged 12 years is left with Rs. 100,000 which is under a trust. The trustees invest the money at 6% per annum and pay the minor boy a sum of Rs. 2500, for his pocket money at the end of each year. The expenses of trust come out to be Rs. 500 per annum. Find the amount that will be handed over to the minor boy after he attains the age of 18 years.

    Rs.120000

    Rs.150000

    Rs.118000

    Rs.125000

    Ans .

    3


    1. Explanation :

      (3) Using Rule 1.
      S.I. = \( \frac{Principal * Rate * Time}{100} \)
      = \( \frac{100000*6*6}{100} \)
      = Rs.36000
      Total pocket money = 6 × 2500 = Rs. 15000
      Total expenses of trust = 6 × 500 = Rs. 3000
      Total expenses = Rs. (15000 + 3000) = Rs. 18000
      Therefore, Amount to be received by the boy
      = Rs. (100000 + 36000 – 18000)
      = Rs. 118000

    Q.26

    If A borrowed Rs. P at x% and B borrowed Rs. Q (> P) at y% per annum at simple interest at the same time, then the amount of their debts will be equal after

    100 \( (\frac{Q-P}{Px-Qy}) \) years

    100 \( (\frac{Px-Qy}{Q-P}) \) years

    100 \( (\frac{Px-Qy}{P-Q}) \) years

    100 \( (\frac{P-Q}{Px-Qy}) \) years

    Ans .

    1


    1. Explanation :

      (1) Let amounts be equal in T years.
      S.I. = \( \frac{Principal * Rate * Time}{100} \)
      Therefore, P + \( \frac{P*x*T}{100} \) = Q + \( \frac{Q*y*T}{100} \)
      \(\frac{PxT}{100}-\frac{QyT}{100}\) = Q-P
      T\( \frac{Px-Qy}{100} \)= Q-P
      T=100\( (\frac{Q-P}{Px-Qy}) \)

    Q.27

    A money lender claims to lend money at the rate of 10% per annum simple interest. However, he takes the interest in advance when he lends a sum for one year. At what interest rate does he lend the money actually?

    10%

    10\(\frac{1}{9}\)%

    11%

    11\(\frac{1}{9}\)%

    Ans .

    4


    1. Explanation :

      (4) Let the principal be Rs. 100
      Interest = Rs. 10
      Actual principal = Rs. 90
      Q Interest on Rs. 90 = Rs. 10
      Therefore, Interest on Rs. 100
      = \(\frac{10}{90}\)*100
      = \(\frac{100}{9}\) = 11\(\frac{1}{9}\)%

    Q.28

    Ramesh borrowed a sum at 5 per annum simple interest from Rahul. He returns the amount after 5years. Rahul returns 2 % of the total amount received. How much did Ramesh borrowed if he received Rs. 5?

    Rs.250

    Rs.200

    Rs.150

    Rs.175

    Ans .

    2


    1. Explanation :

      (2) Let the principal be Rs. P.
      S.I. = \( \frac{Principal * Rate * Time}{100} \)
      = \( \frac{P*5*5}{100} \) = Rs. \( \frac{P}{4} \)
      Amount = P + \( \frac{P}{4} \) = Rs. \( \frac{5P}{4} \)
      According to the question,
      \( \frac{5P}{4} * \frac{2}{100} \) = 5
      \( \frac{P}{4} \) = 5
      P = 40*5
      Rs.200

    Q.29

    A man buys a watch for Rs. 1950 in cash and sells it for Rs. 2200 at a credit of 1 year. If the rate of interest be 10% per annum, then how much profit or loss will he have?

    Rs.55 gain

    Rs.30 profit

    Rs.30 loss

    Rs.30 profit

    Ans .

    1


    1. Explanation :

      (1) Principal = Rs. 1950, Rate =10% per annum
      S. I. = \(\frac{Principal*Rate*Time}{100}\)
      =\(\frac{1950*1*10}{100}\)
      = Rs.195
      Therefore, Amount = Rs.(1950+195)= Rs.2145
      Therefore, 2200-2145=55
      Therefore, There is gain of 55 Rs.

    Q.30

    A money lender lends Rs. 400 for 3 years to a person and lends Rs. 500 for 4 years to the other person at the same rate of simple interest. If altogether he receives Rs. 160 as interest, what is the rate of interest per annum ?

    5%

    7%

    9%

    10%

    Ans .

    1


    1. Explanation :

      (1) Using Rule 1.
      160 = \( \frac{500*4*r}{100} + \frac{400*3*r}{100} \)
      32r = 160
      r = \frac{160}{32} = 5% per annum

    TEST YOURSELF

    Q.1

    A Co-operative Bank gives H.B. loans under the condition that if the loan be cleared with interest in five years, the rate of simple interest per year is 5%, otherwise it will be 7%. Mr. Rahim and Mr. Ram take the same amount of H.B. loan and clear the loan with interest in 5 and 8 years respectively. If Ram pays Rs. 62,000 more, what is the amount of loan taken by each of them ?

    Rs.200000

    Rs.180000

    Rs.190000

    Rs.210000

    Ans .

    1


    1. Explanation :

      (1) Let amount of loan per head be Rs. x.
      S.I. = \( \frac{Principal * Rate * Time}{100} \)
      Therefore, \( \frac{x*7*8}{100}- \frac{x*5*5}{100} \) = 62000
      \( \frac{56x}{100}- \frac{25x}{100} \) = 62000
      31x = 6200000
      x = Rs.200000

    Q.2

    A person on retirement gets Rs. 3,20,000 from his gratuity and P.F. He wants to invest this amount in Post Office and Bank in such a way that he earns a total interest of Rs. 27,600 every year. If the annual rate of interest in Post Office and Bank be respectively 9% and 8%, What are the amounts invested in Post Office and Bank respectively ?

    Rs.200000 , Rs.120000

    Rs.180000 and Rs.140000

    Rs.185000 , Rs.135000

    None of these

    Ans .

    1


    1. Explanation :

      (1) Let the amount in Post office be Rs. x.
      Therefore, Amount in Bank = Rs. (320000 – x)
      S.I. = \( \frac{Principal * Rate * Time}{100} \)
      Therefore, \( \frac{x*9}{100} + \frac{(320000-x)*8}{100} \) = 027600
      9x + 2560000 – 8x = 2760000
      x = 2760000 – 2560000
      = Rs. 200000
      Therefore, Amount in bank = Rs. (320000 – 200000)
      = Rs. 120000

    Q.3

    A person on retirement gets Rs. 3,20,000 from his gratuity and P.F. He wants to invest this amount in Post Office and Bank in such a way that he earns a total interest of Rs. 27,600 every year. If the annual rate of interest in Post Office and Bank be respectively 9% and 8%, What are the amounts invested in Post Office and Bank respectively ?

    Rs.2500,Rs.4500

    Rs.4000,Rs.6000

    Rs.5000,Rs.7000

    Rs.6000,Rs.8000

    Ans .

    3


    1. Explanation :

      (3) Sum given to Anil = Rs. x
      Sum given to Sunil =Rs. (x + 2000)
      S.I. = \( \frac{Principal * Rate * Time}{100} \)
      \( \frac{(x+2000)*12*3}{100} - \frac{x*10*3}{100} \) = 1020
      36x + 72000 – 30x = 102000
      6x = 102000 – 72000 = 30000
      x = Rs. 5000
      Therefore, Sum given to Sunil = Rs. 7000

    Q.4

    A person made a fixed deposit of Rs. 30,000 in a bank for 5 years at 10% simple interest per annum. He had to withdraw the whole amount after 3 years to meet the expenses of his daughter’s marriage and he received Rs. 7800 less than what he would have got after 5 years. What is the rate of simple interest per annum paid by the bank for this premature encashment ?

    6%

    7%

    8%

    9%

    Ans .

    3


    1. Explanation :

      (3) S.I.= \( \frac{P * R * T}{100} \)
      Let the required rate of interest
      be R% per annum.
      \( \frac{30000*5*10}{100} - \frac{30000*3*R}{100} \) = 7800
      15000 – 900R = 7800
      900R = 15000 – 7800 = 7200
      R = \(\frac{7200}{900}\) = 8% per annum.

    Q.5

    In 4 years, ₹6000 amounts to ₹8,000. In what time at the same rate will ₹525 amount to ₹700 ?

    5 years

    3 years

    4 years

    None of these

    Ans .

    3


    1. Explanation :

      (3) Case-I,
      Interest = 8000 – 6000
      = Rs. 2000
      Rate = \( \frac{S.I. * 100}{P * T} \) = \( \frac{2000 * 100}{6000*4} \)
      =\(\frac{25}{3}\)%
      Therefore, Case-II,
      Time = \(\frac{175*100}{525*\frac{25}{3}}\) = 4 years

    Q.6

    Find the interest on ₹1460 at 10% from 5th February, 1992 to 25th April, 1992

    ₹32

    ₹36

    ₹40

    ₹34

    Ans .

    1


    1. Explanation :

      (1) P = ₹1460 , R = 10%
      1992 is a leap year
      Therefore, T = (24 + 31 + 25) days = 80 days.
      I = \(\frac{PRD}{36500}\)
      I = \( \frac{1460*10*80}{36500} \)
      I = ₹32.
      Note :We have excluded 5th February but included 25th

    Q.7

    Find the amount Ram will get after 2 years when he invests ₹15000 at 15% interest.

    ₹18500

    ₹19500

    ₹17500

    ₹16500

    Ans .

    2


    1. Explanation :

      (2) Here, P = ₹15000
      R = 15%
      T = 2 years
      A = \(P\frac{100+RT}{100})\)
      = \(15000\frac{100+15*2}{100})\)
      = 15000 * \(\frac{130}{100}\)
      A = ₹19500

    Q.8

    At what rate per annum will a sum of ₹5000 amount to ₹6000 in 4 years?

    6% per annum

    4% per annum

    5% per annum

    4.5% per annum

    Ans .

    3


    1. Explanation :

      (3) Here, P = ₹5000
      A = ₹6000
      T = 4 years
      So, I = A – P
      = ₹(6000 – 5000) = ₹1000
      R = \( \frac{100I}{PT} \)
      R = \( \frac{100*1000}{5000*4} \)
      R = 5%.

    Q.9

    Ram lent ₹1200 to Shyam for 5 years and ₹1500 to Mohan for 2 years received altogether ₹900 as interest. Find the rate per annum.

    8.5%

    8%

    9%

    10%

    Ans .

    4


    1. Explanation :

      (4) I = I1 + I2
      I = \( \frac{P1*R*T1}{100} + \frac{P2*R*T2}{100} \)
      I = \( \frac{R}{100} \)(P1T1 + P2T2)
      or, R = \( \frac{100I}{P1T1+P2T2} \)
      Here, I = 900
      P1 = 1200
      T1 = 5 years
      P2 = 1500
      T2 = 2 years
      R = \( \frac{100*900}{(1200*5)*(1500*2)} \)
      R = \( \frac{90000}{9000} \)
      R = 10%.
      Note : In case of more than two investment, sum the products of principal and time of each case.

    Q.10

    A certain sum of money amounts to Rs. 1680 in 3 years and to ₹1800 in 5 years. Find the sum and the rate of interest.

    ₹1500;4%

    ₹1200;4%

    ₹1600;5%

    ₹1800;5%

    Ans .

    1


    1. Explanation :

      (1) A = P + I
      So, P remains same in both cases. Only amount of interest is different in two cases because the time period are different.
      P + Interest for 5 years = ₹1800
      and P + Interest for 3 years = ₹1680
      On subtraction we get,
      Interest for 2 years = ₹120
      Now, we solve for the case of 3 years .
      Interest for 3 years = ₹120 * \( \frac{3}{2} \) = ₹180
      And amount after 3 years = ₹1680
      Principal (P) = A – I = ₹(1680 – 180) = ₹1500.
      R = \( \frac{100I}{PT} \)
      R = \( \frac{100*180}{1500*3} \)
      R = 4%.
      Note : Alternatively, we could have solved for 5 years too and got the same answer.

    Q.11

    In how many years will a sum of money double itself at 5% rate of interest?

    18 years

    20 years

    22 years

    15 years

    Ans .

    2


    1. Explanation :

      (2) A sum doubles itself when amount of interest becomes equal to the principal.
      So, I = P
      Given, R = 5%
      T = \( \frac{100I}{PR} \)
      On substitution we get,
      T = \( \frac{100*P}{P*5} \)
      T = 20 years.

    Q.12

    A man lends a certain sum of money and gets an interest equal to \( \frac{1}{16} \)th of the principal. The time for which money was lent is equal to the rate of interest. Find the rate of interest per annum.

    4%

    3.5%

    3%

    2.5%

    Ans .

    4


    1. Explanation :

      I.= \( \frac{P * R * T}{100} \)
      Given : I = \( \frac{P}{16} \)
      and T = R
      So, on substitution we get
      \( \frac{P}{16} = \frac{P*R*R}{100} \)
      R2 = \(\frac{100}{16}\)
      R = \( \frac{10}{4} \) % = \( \frac{5}{2} \)% = \(2\frac{1}{2}\)%

    Q.13

    A man borrowed ₹16000 from two persons. He paid 6% interest to one and 10% per annum to the other. In one year he paid total interest 1120. How much did he borrow at each rate?

    ₹10000 ; ₹6000

    ₹12000 ; ₹4000

    ₹11000 ; ₹5000

    ₹12500 ; ₹3500

    Ans .

    2


    1. Explanation :

      (2) Let the sum borrowed at 6% be x = P1
      Then the sum borrowed at 10% = (16000 – x ) = P2
      Time is one year in both cases
      R1 = 6%
      R2 = 10%
      I = I1 + I2
      I = \( \frac{P1*R*T1}{100} + \frac{P2*R*T2}{100} \)
      I = \( \frac{R}{100} \)(P1T1 + P2T2)
      P1T1 + P2T2 = \(\frac{100I}{T}\) On substitution we get,
      (x × 6) + (16000 – x)10 = \(\frac{100*1120}{1} \)
      160000 – 4x =112000
      4x = 48000
      x = 12000

    Q.14

    A borrowed ₹1500 at 4% per annum and ₹1400 at 5% per annum for the same period. He paid ₹390 as total interest. Find the time for which he borrowed the sum.

    3.5 years

    2.5 years

    3 years

    4 years

    Ans .

    3


    1. Explanation :

      I = I1 + I2
      I = \( \frac{P1*R*T1}{100} + \frac{P2*R*T2}{100} \)
      or T = \(\frac{100I}{P1R1*P2R2}\)
      = \(\frac{100*390}{(1500*4)*(1400*5)}\)
      = \( \frac{39000}{13000} \)
      T = 3 years

    Q.15

    Find the annual instalment that will discharge a debt of ₹12900 due in 4 years at 5% per annum simple interest.

    ₹2750

    ₹2150

    ₹2500

    ₹3000

    Ans .

    4


    1. Explanation :

      (4) Let each equal annual instalment be x.
      First instalment is paid after 1 year and hence will remain with the lender for the remaining (4 – 1) = 3 years.
      Similarly, second instalment will remain with the lender for 2 years, third instalment for 1 year and the final fourth instalment remain x as such.
      A = A1 + A2 + A3 + A4
      A = \(P(\frac{100+RT}{100})\)
      A = x\([\frac{100+5*3}{100}+\frac{100+5*2}{100}+\frac{100+5*1}{100}+\frac{100+5*0}{100}]\)
      12900 = x\([\frac{115+110+105+100}{100}]\)
      12900 = \( \frac{430}{100} \)x
      x = \( \frac{12900*100}{430} \)
      x = ₹3000

    Q.16

    A certain sum of money amounts to ₹6780 in 2 years and to ₹7360.50 in 3\(\frac{1}{2}\) years. Find the sum and the rate of interest.

    ₹6006 ; 6.4 % per annum

    ₹8006 ; 6.4 % per annum

    ₹5006 ; 5% per annum

    ₹5506 ; 5% per annum

    Ans .

    1


    1. Explanation :

      (1) Principal + S.I. for 3 \( \frac{1}{2} \)years = ₹7360.50 ....... (i)
      Principal + S.I. for 2 years = ₹6780 ...... (ii)
      On subtracting equation (ii) from (i),
      S.I. for 1 \(\frac{1}{2}\) years = ₹580.50
      Therefore, S.I. for 2 years = ₹\((\frac{580.50*2*2}{3})\) = ₹774
      Therefore Principal = ₹(6780 – 774 ) = ₹6006
      And, rate of interest
      =\( \frac{774*100}{6006*2} \)
      = 6.4% per annum.

    Q.17

    If 5600 amounts to 6678 in 3\(\frac{1}{2}\) years, what will 9600 amount to in 5\(\frac{1}{4}\) years at the same rate of interest ?

    ₹12732

    ₹12372

    ₹12722

    ₹12237

    Ans .

    2


    1. Explanation :

      (2) Interest = ₹(6678 – 5600) = ₹1078
      Rate = \( \frac{S.I. * 100}{P * T} \)
      = \( \frac{1078*100*2}{5600*7} \)
      = \( 5\frac{1}{2} \)% per annum
      Therefore, S.I. on ₹9600 for 5 \( \frac{1}{4} \) years
      = ₹\( (\frac{9600}{100}*\frac{21}{4}*\frac{11}{2}) \) = ₹2772
      therefore, Amount = ₹(9600 + 2772) = ₹12372

    Q.18

    A man promises to his wife a birthday present, given her each year a number of rupees equal to the number of years in her age. If her birthday falls on August 8, what sum must be placed at simple interest at 7% on January 1 before she is 63 (non leap year) in order to raise the required sum ?

    ₹1600

    ₹1550

    ₹1500

    ₹1450

    Ans .

    3


    1. Explanation :

      (3) Let the sum be 100.
      Number of days from January 1 to August 8 = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 7 = 219 days
      = \( \frac{219}{365} \) year = \( \frac{3}{5} \)year
      S.I. on ₹100 for \(\frac{3}{5}\) year at 7% = ₹\((\frac{100*3*7}{100*5})\)= ₹\(\frac{21}{5}\)
      If required money is ₹\( \frac{21}{5} \), sum = ₹100
      If required money is Rs. 63, sum = \((100*\frac{5}{21}*63)\)
      = ₹1500

    Q.19

    It is decided that a loan of ₹10,000 will be paid off at the rate of ₹800 per month in 15 equal instalments. Find out the rate of return on investment.

    17% P.a.

    18% P.a.

    15% P.a.

    16% P.a.

    Ans .

    4


    1. Explanation :

      (4) Number of monthly instalments = 15
      Monthly instalment = 800
      Time (T) = \( \frac{15}{12} = 1\frac{1}{4} \)
      Therefore Total amount paid = (800 × 15) = 12,000
      Interest = (12,000 – 10,000) = 2,000
      Therefore, Rate of return
      = \( \frac{100*2000*1*4}{10000*5} \)= 16%

    Q.20

    A person takes loan of ₹4,000 on the condition that he would pay it in the monthly instalment of ₹500. He has to pay interest @ 6% on the outstanding balances, then find out the average rate of interest received by the creditor.

    \(3\frac{3}{8}\)% per annum

    \(2\frac{3}{8}\)% per annum

    \(4\frac{3}{8}\)% per annum

    \(3\frac{1}{8}\)% per annum

    Ans .

    1


    1. Explanation :

      (1) Monthly instalment = ₹500
      Total loan = ₹4000
      Therefore, Number of instalments = \(\frac{4000}{500}\)=8
      Once the payment starts, outstanding balances will go on diminishing.
      Hence, from point of view of interest, principal = 4000 + 3500 + 3000 + 2,500 + 2000 + 1500 + 1,000 + 500 = ₹18,000
      Therefore, Interest on ₹18,000 for 1 month at 6% P.a.
      = \( \frac{18000*6*1}{100*12} \) = ₹90
      Average rate of interest
      = \( \frac{I*100}{PT} \)
      T= 8 months = \( \frac{8}{12} \) years
      = \( \frac{90*100*13}{4000*8} \) = \(\frac{27}{8}\)% = \( 3\frac{3}{8} \)%

    Q.21

    Divide 6800 into two parts so that S.I. on the first part for 3 \(\frac{1}{3}\) years at 6% may be equal to the interest on the second part for 3\(\frac{1}{2}\) years at 4% Percent Per annum

    ₹2600;₹4200

    ₹2800;₹4000

    ₹2500;₹4300

    ₹2700;₹4100

    Ans .

    2


    1. Explanation :

      (2) Let the first part be x. Then second part = (6800 – x)
      Interest on first part for 3\(\frac{1}{3}\)years at 6%
      = \( \frac{x*6*\frac{10}{3}}{100} \) = \( \frac{x}{5} \)
      Interest on second part for 3\( \frac{1}{2} \)years at 4%
      = \( \frac{(6800-x)*4*\frac{7}{2}}{100} \)
      = ₹\( \frac{(6800-x)*7}{50} \)
      According to the problem,
      \( \frac{x}{5} = \frac{(6800-x)*7}{50} \)
      10x = (6800 – x) 7
      10x = 47600 – 7x
      17x = 47600
      x = 2800
      Hence, first part = ₹2800 and second part
      = ₹(6800 – 2800) = ₹4000.





    At what rate of simple interest will a sum of money doubles itself in 25 2 years?

    Detailed Solution The Sum of money doubles itself in 25 years. Concept: Simple interest is the interest calculated on the principal portion of the loan or the original contribution to the saving account. ∴ The rate of interest per annum is 4%.

    In what time will the simple interest on a certain sum of money at 7.5% per annum be 3/8 of itself?

    ⇒ T = 10 years.

    In what time will the simple interest on a certain sum of money at 61 4 Pa be 3/8 of itself?

    ∴Rate(R)=100×IP×T=100×3x8x×254=100×32×25=6% Q. The simple interest on a certain sum of money is 38th of the sum in 614 years.

    In what time will the simple interest on a certain sum of money at 6 1 2 per annum be 3/8th of itself?

    ⇒T=6 years.