Interest Calculations

1) What will be the simple interest earned on an amount of Rs.17600 in 9 months at the rate of 4\({1 \over {2}}\)% per annum?

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SI =  \({PNR \over {100}}\)

=    \({17600*({3 \over {4}})*({9 \over {2}}) \over {100}}\) = Rs.594 .  Hence (a).

2) A sum of money at simple interest amounts to Rs.876 in 3 years and to Rs.918 in 4 years. Find the principal.

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SI for 1 st year = (918 – 876) = Rs.42

SI for 3 years = 42 * 3 = Rs.126

Principal = (876 – 126 ) = Rs.750   Hence (b).

3) A sum of money becomes 5 times its value under 8 years. In how many years will it become 17

times itself?

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If a sum becomes 5 times itself in 8 years , the interest will be 4 times that sum.

For the sum to become 17 times itself, the interest has to become 16 times the sum. 

SI for n consecutive years follows AP. It will take (8 * 4) = 32 years.    Hence (d).

4) A man borrows Rs.40000 from a bank at 8% per annum. Find the amount after for 3\({1 \over {2}}\) years

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  SI =  \({PNR \over {100}}\)
= \({ 40000*({7 \over {2}})* 8 \over {100}}\) = Rs.11200

Amount = P + SI = 40000 + 11200 = Rs.51200 .   Hence (c).

5) The difference between the simple and compound interest after 2 years at 7% per annum is Rs.98. What is the simple interest after 4 years at the same interest?

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Difference between SI and CI after 2 years = Rs.98 =  \({PR{}{}{}{}{}{}{}{}{}{}{}{} \over {100}}\)  

P = Rs.20000

SI after 4 years at 7% per annum SI = \({ 20000*4* 7 \over {100}}\) = Rs.5600 .  Hence (b).

6)

The simple interest for 2 years is Rs.4200 and the compound interest for the second year is Rs.2415 for the same principal at the same rate. Find the principal amount.

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SI for 2 years = Rs.4200 SI for 1 year = Rs.2100 = PR …..(i) CI for 2nd year = Rs.2415 = PR + PR2 => PR2 = 2415 – 2100 à Rs.315 …..(ii) Divide equation ii and i R = \({315 \over {2100}}\) à R = 15% Substitute R value in equation i P * \({15 \over {100}}\) = 2100 = Rs.14000. Hence (c).

7)

The ratio of compound interest to simple interest on a certain sum at 16% per annum for 2 years is:

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SI = \({PNR \over {100}}\) = \({P*2*16 \over {100}}\) = 0.32P CI = P [(1+ \({R \over {100}}\) ) n - 1] = P [(1+ \({16 \over {100}}\) ) 2 - 1] = 0.3456P Ratio of CI to SI = \({0.3456P \over {0.32P}}\) = 27: 25 Hence (b).

8)

A invested an amount of Rs.15200 divided in two different banks A and B at the simple interest rate of 16% p.a. and 10% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs.3424, what was the amount invested in bank B?

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Let the invested amount in bank A and B be x and 15200-x respectively. SI for 2 years at 16% p.a. = \({x*2*16 \over {100}}\) = \({32x \over {100}}\) SI for 2 years at 10% p.a. = \({(15200-x)*2*10 \over {100}}\) = \({(15200-x)*20 \over {100}}\) Total SI earned in 2 years = Rs.3424 \({32x \over {100}}\) + \({(15200-x)*20 \over {100}}\) = 3424 à x = Rs.3200 Amount invested in bank B = 15200 – x = 15200 – 3200 = Rs.12000 Hence (c).

9) The difference between the simple interest and the compound interest for a certain sum after 3 years is Rs.2736 at 4% per annum. Find the principal.

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CI – SI = Rs.2736 (after 3 years)

Difference between SI and CI after 3 years = PR 3 + 3PR 2 = Rs.2736

R = 4%

P =  \({2736*100*100*100 \over {R+300R}}\)

= \({2736*100*100*100 \over {16(300+4)}}\) = Rs.562500 .  Hence (d).

10)

Divide Rs.5046 between P and Q such that P’s share at the end of 11 years is equal to Q’s share at the end of 13 years at 5% rate of compound interest. Find the share of P.

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Let the shares of P and Q be Rs.x and Rs.y respectively.    

Both P and Q get the same amount

x *(1 +  \({5 \over {100}})\) 11 =  y *(1 +  \({5 \over {100}})\) 13

  \({x \over {y}}\)   =  (1 +  \({5 \over {100}})\) 2

=  \({441 \over {400}}\)  

Share of P =   \({441 \over {841}}\) * 5046 = Rs.2646   Hence (a).

11) A sum of money was lent at 4% per annum simple interest. After a year, Rs.8400 is repaid and the rest is financed at 8% per annum. If the second year’s interest is \({46 \over {125}}\) of the first year’s interest, find the sum of money that was lent out.

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  Let the sum of money lent out be Rs.x
  SI for 1 st year =  \({PNR \over {100}}\) =  \({x*4 \over {100}}\) =  \({x \over {25}}\)
  Amount = x +  \({x \over {25}}\) =  \({26x \over {25}}\)
Returned amount after first year = Rs.8400
  Balance amount = ( \({26x \over {25}}\) – 8400)
Second year’s interest =   ( ( \({26x \over {25}}\) – 8400) * 8)/100 =  \({46 \over {125}}\) of the first year’s interest
  (( \({26x \over {25}}\) – 8400) * 8)/100 =  \({46 \over {125}}\) *  \({x \over {25}}\) = Rs.12500 .  Hence (d).

12)

Rs.5720 is divided into three parts such that their interest after 1, 2 and 3 years respectively are equal, the rate of interest being 12% per annum in all cases. Find the difference between largest two parts and smallest two parts.

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Let a, b and c be the 3 parts invested for 1, 2 and 3 years respectively. Let the interest be d \({a*1*12 \over {100}}\) = \({b*2*12 \over {100}}\) = \({c*3*12 \over {100}}\) = d a = \({25d \over {3}}\), b = \({25d \over {6}}\), c = \({25d \over {9}}\) a: b: c = \({25d \over {3}}\): \({25d \over {6}}\): \({25d \over {9}}\) a: b: c = 6: 3: 2 => 5720 6x + 3x + 2x = 5720 à x = 520 Largest two parts are 6x and 3x. and smallest two parts are 3x and 2x. Difference between largest and smallest two parts = 9x – 5x = 4x = 4*520 = Rs.2080. Hence (c).

13) What will be the interest earned on an amount of Rs.50000 in 2 years at the rate of 10% per annum, if it is compounded half-yearly for the first year and then annually for the second year?

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CI is compounded half yearly = p((1+ \({r \over {200}}\) ) 2n – 1)

Amount   after the first year = 50000(1+ \({10 \over {200}})\) 2 = Rs.55125

Amount after second year = 55125 (1+  \({10 \over {100}}\) ) = Rs.60637.5

  Total interest = 60637.5 – 50000 = Rs.10637. 5 .  Hence (a).

14) A bought an Air-cooler Rs.7800. he would pay back in two equal instalments. If the rate of interest be 6% per annum compounded annually, find the amount of each instalment.

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Let the instalment be Rs.x

Amount at the end of 1 st year = 7800 *  \({106 \over {100}}\)

x = \({7800 \over {{1 \over {1+({6 \over {100}})}}+{1 \over {1+({6 \over {100}})}}}}\) = Rs.4255.32 .  Hence (b).

15) A man took a loan of Rs.45000 at 9% simple interest for 2 years and deposited it at 8% compound interest for the same period. What was the gain/loss?

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SI for 2 years =  \({45000*2*9 \over {100}}\) =  Rs.8100

Amount he has to return = 45000 + 8100 = Rs.53100

He invested Rs.45000 at 8% CI

Amount = 45000(1+\({8 \over {100}}\) ) 2 = Rs.52488

Loss = 53100 – 52488 = Rs.612   Hence (d).