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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 […]

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

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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…