# Exercise : 5

1. An automobile financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes :
(a) 10%
(b) 10.25%
(c) 10.5%
(d) None of these

#### View Ans & Explanation

Ans.b

Let the sum be Rs 100. Then,

S.I. for first 6 months = $\left ( \frac{100 \times 10 \times 1}{100 \times 2} \right )$ = 5

S.I. for last 6 months = $\left ( \frac{105 \times 10 \times 1}{100 \times 2} \right )$ = 5.25

So, amount at the end of 1 year = (100 + 5 + 5.25) = 110.25.

∴ Effective rate = (110.25 – 100) = 10.25%.

2. 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 of them as interest. The rate of interest per annum is:
(a) 5%
(b) 7%
(c) 718%
(d) 10%
(e) None of these

#### View Ans & Explanation

Ans.d

Let the rate be R% p.a. Then,

$\left ( \frac{5000 \times R \times 2}{100} \right ) + \left ( \frac{3000 \times R \times 4}{100} \right ) = 2200$

⇒ 100R + 120R = 2200 ⇒ R = (2200220) = 10.

3. A sum of 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of 362.50 more is lent but at the rate twice the former. At the end of the year, 33.50 is earned as interest from both the loans. What was the original rate of interest?
(a) 3.6%
(b) 4.5%
(c) 5%
(d) 3.46%
(e) None of these

#### View Ans & Explanation

Ans.d

Let the original rate be R%. Then, new rate = (2R)%

$\left ( \frac{725 \times R \times 1}{100} \right ) + \left ( \frac{362.50 \times 2R \times 1}{100 \times 3} \right ) = 33.50$

⇒ (2175 + 725)R = 33.50 × 100 × 3 = 10050

⇒ R = 100502900 = 3.46%

4. 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:
(a) 0.1%
(b) 0.2%
(c) 0.3%
(d) 0.4%
(e) None of these

#### View Ans & Explanation

Ans.c

$\left ( \frac{1500 \times R_{1} \times 3}{100} \right ) - \left ( \frac{1500 \times R_{2} \times 3}{100} \right ) = 13.50$

⇒ 4500(R1 - R2) = 1350 ⇒ R1 - R2 = 13504500 = 0.3%

5. The rates of simple interest in two banks A and B are in the ratio 5 : 4. A person wants to deposit his total savings in two banks in such a way that he received equal half-yearly interest from both. He should deposit the savings in banks A and B in the ratio.
(a) 2 : 5
(b) 4 : 5
(c) 5 : 2
(d) 5 : 4
(e) None of these

#### View Ans & Explanation

Ans.b

Let the savings be X and Y and the rates of simple interest be 5x and 4x respectively.

Then, X × 5x × 12 × 1100 = Y × 4x × 12 × 1100 or XY = 45

i.e., X : Y = 4 : 5.

6. The price of a T.V. set worth 20,000 is to paid in 20 instalments of 1000 each. If the rate of interest be 6% per annum, and the first instalment be paid at the time of purchase, then the value of the last instalment covering the interest as well will be :
(a) 1050
(b) 2050
(c) 3000
(d) None of these
(e) 2020

#### View Ans & Explanation

Ans.d

Money paid in cash = 1000.

Balance payment = (20000 – 1000) = 19000.

7. Mr. Thomas invested an amount of 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be 3508, what was the amount invested in Scheme B?
(a) 6400
(b) 6500
(c) 7200
(d) 7500
(e) None of these

#### View Ans & Explanation

Ans.a

Let the sum invested in Scheme A be x and that in Scheme B be (13900 – x).

Then, $\left ( \frac{x \times 14 \times 2}{100} \right ) + \left [ \frac{\left ( 13900 - x \right ) \times 11 \times 2}{100} \right ] = 3508$

⇒ 28x – 22x = 350800 – (13900 × 22) ⇒ 6x = 45000

⇒ x = 7500.

So, sum invested in Scheme B = (13900 – 7500) = 6400.

8. An amount of 1,00,000 is invested in two types of shares. The first yields an interest of 9% p.a. and the second, 11% p.a. If the total interest at the end of one year is 934% , then the amount invested in each share was:
(a) 52,500; 47,500
(b) 62,500; 37,500
(c) 72,500: 27,500
(d) 82,500; 17,500
(e) None of these

#### View Ans & Explanation

Ans.b

Let the sum invested at 9% be x and that invested at 11% be (100000 – x).

Then,

$\left ( \frac{x \times 9 \times 1}{100} \right ) + \left [ \frac{\left ( 100000 - x \right ) \times 11 \times 1}{100} \right ]$

= $\left ( 100000 \times \frac{39}{4} \times \frac{1}{100} \right )$

$\frac{9x + 1100000 - 11x}{100} = \frac{39000}{4} = 9750$

⇒ 2x = (1100000 – 975000) = 125000 ⇒ x = 62500.

∴ Sum invested at 9% = 62500.

Sum invested at 11% = (100000 – 62500) = 37500.

9. David invested certain amount in three different schemes A, B and C with the rate of interest 10% p.a., 12% p.a. and 15% p.a. respectively. If the the total interest accrued in one year was 3200 and the amount invested in Scheme C was 150% of the amount invested in Scheme A and 240% of the amount invested in Scheme B, what was the amount invested in Scheme B?
(a) 5000
(b) 6500
(c) 8000
(d) cannot be determined
(e) None of these

#### View Ans & Explanation

Ans.a

Let x, y and z be the amounts invested in schemes A, B and C respectively. Then,

$\left ( \frac{x \times 10 \times 1}{100} \right ) + \left ( \frac{y \times 12 \times 1}{100} \right ) + \left ( \frac{z \times 15 \times 1}{100} \right ) = 3200$

⇒ 10x + 12y + 15z = 320000 ..... (i)

Now, z = 240% of y = 125y..... (ii)

And, z = 150% of x = 32x

⇒ x = 23z = (23 × 125)y = 85y......(iii)

From (i),(ii) and (iii), we have :

16y + 12y + 36y = 320000 ⇒ 64y = 320000 ⇒ y = 5000.

∴ Sum invested in Scheme B = 5000.

10. A person invested in all 2600 at 4%, 6% and 8% per annum simple interest. At the end of the year, he got the same interest in all the three cases. The money invested at 4% is:
(a) 200
(b) 600
(c) 800
(d) 1200
(e) None of these

#### View Ans & Explanation

Ans.d

Let the parts be x, y and [2600 – (x + y)].Then,

$\frac{x \times 4 \times 1}{100} = \frac{y \times 6 \times 1}{100} = \frac{\left [ 2600 - \left ( x + y \right ) \right ] \times 8 \times 1}{100}$

yx = 46 = 23 or y = 23x.

So, $\frac{x \times 4 \times 1}{100} = \frac{\left [ 2600 - \frac{5}{3}x \right ] \times 8}{100}$

⇒ 4x = $\frac{\left ( 7800 - 5x \right ) \times 8}{3} \Rightarrow 52x = \left ( 7800 \times 8 \right )$

⇒ x = $\left ( \frac{7800 \times 8}{52} \right ) = 1200.$

∴ Money invested at 4% = 1200.

11. Divide 2379 into 3 parts so that their amounts after 2, 3 and 4 years respectively may be equal, the rate of interest being 5% per annum at simple interest. The first part is:
(a) 759
(b) 792
(c) 818
(d) 828
(e) None of these

#### View Ans & Explanation

Ans.d

Let the parts be x, y and [2379 – (x + y)].

x + (x × 2 × 5100) = y + (y × 3 × 5100)

= z + (z × 4 × 5100)

11x10 = 23y20 = 6z5 = k ⇒ x = 10k11, y = 20k23, z = 5k6

But x + y + z = 2379.

10k11 + 20k23 + 5k6 = 2379

⇒ 1380 k + 1320 k + 1256 k = 2379 × 11 × 23 × 6

⇒ k = $\frac{2379 \times 11 \times 23 \times 6}{3965} = \frac{3 \times 11 \times 23 \times 6}{5}$

∴ x = $\left (\frac{10}{11} \times \frac{3 \times 11 \times 23 \times 6}{5} \right ) = 828$

Hence, the first part is 828.