Exercise : 4


1. If there are three sum of money P, Q and R so that P is the simple interest on Q and Q is the simple interest of R, rate % and time are same in each case, then the relation of P, Q and R is given by
(a) P2 = QR
(b) Q2 = PR
(c) R2 = PQ
(d) PQR = 100
(e) None of these
Ans.b

P = \(\frac{Q \times r \times t}{100}\) and Q = \(\frac{R \times r \times t}{100}\)

\(\frac{P}{Q} = \frac{Q}{R} = \frac{r \times t}{100}\)

∴ Q2 = PR.

2. In how many minimum number of complete years, the interest on 212.50 P at 3% per annum will be in exact number of rupees?
(a) 6
(b) 8
(c) 9
(d) 7
(e) None of these
Ans.b

Interest for one year = 212.50 × 3100 × 1 = 518

Thus in 8 years, the interest is 51.

3. A milk man borrowed 2,500 from two money lenders. For one loan, he paid 5% p.a. and for the other, he paid 7% p.a.The total interest paid for two years was 275. How much did he borrow at 7% rate?
(a) 600
(b) 625
(c) 650
(d) 675
(e) None of these
Ans.b

Let he borrowed at 5% = x

∴ he borrowed at 7% = (2500 – x)

Now I1 + I2 = 275

\(\frac{x \times 5 \times 2}{100} + \frac{\left(2500 - x \right) \times 7 \times 2}{100} = 275\)

⇒ 10x + 14(2500 – x) = 27500

⇒ 4x = 35000 – 27500 = 7500

⇒ x = Rs 1875

∴ Sum borrowed at 7% rate = 2500 – 1875 = 625

4. What annual instalment will discharge a debt of 4,200 due in 5 years at 10% simple interest?
(a) 500 per year
(b) 600 per year
(c) 700 per year
(d) 800 per year
(e) None of these
Ans.c

Shortcut method :

If borrowed amount be M and it is to be paid in equal instalments, then

\(M = na + \frac{ra}{100 \times Y} \times \frac{n\left(n - 1 \right)}{2}\)

where Y = no. of instalments per annum

a = annual instalment

Here, M = 4200, y = 1, r = 10, n = 5, a = ?

4200 = 5a + 10a100 × 5(5 - 1)2

⇒ 4200 = a[5 + 1] ⇒ 6a = 4200

⇒ a = 700

5. Adam borrowed some money at the rate of 6% p.a. for the first two years, at the rate of 9% p.a. for the next three years, and at the rate of 14% p.a. for the period beyond five years. If he pays a total interest of 11,400 at the end of nine years, how much money did he borrow?
(a) 10,000
(b) 12,000
(c) 14,000
(d) 16,000
(e) None of these
Ans.b

Let the sum borrowed be x. Then,

\(\left(\frac{x \times 6 \times 2}{100} \right) + \left(\frac{x \times 9 \times 3}{100} \right) + \left(\frac{x \times 14 \times 4}{100} \right)\) = 11400

⇒ (325x + 27100x + 1425x) = 11400 ⇒ 95100x = 11400

⇒ x = \(\left(\frac{11400 \times 100}{95} \right)\) = 12000

Hence, sum borrowed = 12,000.

6. A person borrows 5000 for 2 years at 4% p.a. simple interest. He immediately lends it to another person at 614 % p.a. for 2 years. Find his gain in the transaction per year.
(a) 112.50
(b) 125
(c) 150
(d) 167.50
(e) None of these
Ans.a

Gain in 2 years

= \(\left [ \left ( 5000 \times \frac{25}{4} \times \frac{2}{100} \right ) - \left ( \frac{5000 \times 4 \times 2}{100} \right )\right ]\)

= (625 – 400) = 225.

∴ Gain in 1 year = (2252) = 112.50

7. A certain amount earns simple interest of 1750 after 7 years Had the interest been 2% more, how much more interest would it have earned?
(a) 35
(b) 245
(c) 350
(d) Cannot be determined
(e) None of these
Ans.d

We need to know the S.I., principal and time to find the rate. Since the principal is not given, so data is inadequate.

8. What will be the ratio of simple interest earned by certain amount at the same rate of interest for 6 years and that for 9 years?
(a) 1 : 3
(b) 1 : 4
(c) 2 : 3
(d) Data inadequate
(e) None of these
Ans.c

Let the principal be P and rate of interest be R%.

∴ Required Ratio = \(\left [ \frac{\left ( \frac{P \times R \times 6}{100} \right )}{\left ( \frac{P \times R \times 9}{100} \right )} \right ] = \frac{6PR}{9PR} = \frac{6}{9} = 2 : 3\)

9. Two equal sums of money were invested, one at 4% and the other at 4.5%. At the end of 7 years, the simple interest received from the latter exceeded to that received from the former by 31.50. Each sum was :
(a) 1,200
(b) 600
(c) 750
(d) 900
(e) None of these
Ans.d

Difference of S.I. = \(\sqrt{31.50}\)

Let each sum be x. Then

\(\frac{x \times 4\tfrac{1}{2} \times 7}{100} - \frac{x \times 4 \times 7}{100} = 31.50\)

or 7100x × 12 = 632

or x = 900

10. 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 8160, how much money did he borrow?
(a) 8000
(b) 10,000
(c) 12,000
(d) Data inadequate
(e) None of these
Ans.a

Let the sum be x. Then,

\(\left ( \frac{x \times 6 \times 3}{100} \right ) + \left ( \frac{x \times 9 \times 5}{100} \right ) + \left ( \frac{x \times 13 \times 3}{100} \right ) = 8160\)

⇒ 18x + 45x + 39x = (8160 × 100)

⇒ 102x = 816000

⇒ x = 8000.