# Get busy solving these arithmetic problems

Published: Published Date - 11:15 PM, Sun - 12 June 22

**Hyderabad:** This article is in continuation to the last article on preparation for the Sub-Inspector of Police recruitment exam. Here are some practice questions and answers along with explanations on the Simple Interest and Compound Interest topic.

**1.** **Arun lent Rs 600 to Varun for 2 years and Rs 150 to Tarun for 4 years for the same rate of interest and received altogether from both Rs 90 as interest. The rate of interest when simple interest being calculated is**

A. 5% B. 10% C. 12% D. 15%

**Ans:A**

**Explanation:**

Simple interest = (PTR/100)

S.I1 = 600 × 2 × R/100 = 12R

From the question it is clear that we have to find out the rate of interest i.e.

S.I2 = 150 × 4 ×R/100 = 6R

Interest rate given as common, then

S.I1 S.I2 = 90

18R = 90

R = 5%

**2. A man invested 1/3 of his capital at 7%; 1/4 at 8% and the remainder at 10%. If his annual income is Rs 561 then the capital is**

A. Rs 7200 B. Rs 6600 C. Rs 6200 D. Rs 8600

**Ans: B**

**Explanation:**

Let capital be x

Given that annual income = Rs 561

Then, (x × 7 × 1/ 3 × 100) (x × 8 × 1/ 4 ×100 ) ( x × 10 × 5/ 12 × 100) = 561

7x/3 2x 25x/6 = 561 × 100

51x = 336600

x = 6600

**3. Divide Rs 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. Then the first part is**

A. Rs 818 B. Rs 828 C. Rs 759 D. Rs 792

**Ans: B**

**Explanation:**

Let parts be x, y, z

Given that R = 5%

Simple interest = (PTR/100)

x (x × 5 × 2/100) = y ( y × 5 × 3/100) = z ( z × 5 × 4/100)

Let the three fractions be equals to k. Then,

11x/10 = 23y/20 = 6z/5 = k

We know that x y z = 2379

10k/11 20k/23 5k/6 = 2379

1380k 1320k 1265k = 3611322

3965k = 3611322 =>; Therefore, k = 3611322/ 3965 = 910.8

First part = 10k/11 = Rs 828

**4. An amount of Rs 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 9 3/4% then the amount invested in each share was**

A. Rs 52,500; Rs 47,500 B. Rs 62,500; Rs 37,500

C. Rs 72,500; Rs 27,500 D. Rs 82,500; Rs 17,500

**Ans: B**

**Explanation:**

Let the sum invested at 9% be x and at 11% be (100000-x)

Simple interest = (PTR/100)

(x × 9 × 1/100) ((100000 – x) × 11 × 1/ 100) = ( 39000/4)

x = 62500

Sum invested at 9% = Rs 62500

Sum invested at 11% = Rs 37500

**5. What will be the ratio of simple interest earned by certain amount at the same rate of interest for 3 years and that for 9 years?**

A. 2:3 B.3:2 C. 1:3 D. 1:4

**Ans: C**

**Explanation:**

Simple interest = (PTR/100)

Required ratio = (P3R/100) : (P9R/100) = 1:3

**6. The simple interest on a sum of money will be Rs 600 after 10 years. If the principal is tripled after 5 years, what will be the total interest at the end of tenth year?**

A. Rs 600 B. Rs 900 C. Rs 1200 D. Rs 1500

**Ans: C**

**Explanation:**

Simple interest = (PTR/100)

R = (100 × S.I/ PT) = (100 × 600/ 10x)

= (6000/ x)

For first 5 years Simple interest on amount x

= (x × 6000 × 5/ 100x) = Rs 300

Simple interest on amount 3x for next 5 years

= (3x × 6000 × 5/100x) = Rs 900

Total amount = (300 900)

= Rs 1200

**7. Rs 7,500 is borrowed at compound interest at the rate of 4% per anum. What will be the amount to be paid after one year, if the interest is compounded half-yearly?**

A. Rs 6,600 B. Rs 7,803 C. Rs 7,603 D. Rs 7,853

**Ans: B**

**Explanation:**

Given that P = 7500, R = 4, n = 1

A = P (1 R/200)2n

= 7500 × 51/50 × 51/50

A = Rs 7,803

**8. A certain sum grows at compound interest of 10% p.a. for 3 years. What is the overall effective simple interest rate p.a. for the same period?**

A. 11% B. 11 1/30 % C. 11 1/3 % D. 12%

**Ans: B**

**Explanation:**

Compound interest = P [(1 r/100) n – 1]

P[ (1 r/100)n – 1] = PTR/100

[(1 10/100)3 – 1] = 3R/100

331/30 = R

Simple interest effected = R = 11 1/30%

**To be continued…**