# Some ‘interest’ing problems for police job aspirants

Hyderabad: This article is in continuation to the last article on preparation for the Sub-Inspector of Police recruitment exam. Here are some previous year questions and answers along with explanations on the Simple Interest and Compound Interest topic. 1. A bank lends Rs.30,00,000 to a person P at 12.5% simple interest and Rs.10,00,000 to another […]

Published Date - 11:15 PM, Fri - 17 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 previous year questions and answers along with explanations on the Simple Interest and Compound Interest topic.

**1. A bank lends Rs.30,00,000 to a person P at 12.5% simple interest and Rs.10,00,000 to another person Q at 10.5% simple interest, for t years. If the bank received an interest of Rs.16,80,000 at the end of the period t, then t = (SI MAINS 2018)**

a) 2 1/2 b) 3 c) 3 1/2 d) 4

**Ans: c**

**Explanation:**

Simple Interest = P × T × R/ 100

Time period = t (both cases)

Given that

1680000 = (3000000 × 12.5 × t)/100 (1000000 × 10.5 × t)/100

1680000 = 375000t 105000t

1680000 = 480000t

t = 168/48 = 7/2 or 3 1/2

**2. A certain sum is given on compound interest at some rate. The sum amounts to Rs 7350 and Rs 8575 at the end of 2 years and 3 years respectively. The sum given on compound interest is: (SI PRE 2018)**

a) Rs 5,000 b) Rs 5,400 c) Rs 6,000 d) Rs 6,400

**Ans: b**

**Explanation:**

In compound interest the amount after this year is principle for the next year. We can consider it as simple interest for year.

SI for on Rs 7350 for 1 yr = 8575 – 7350 = Rs 1225

S.I = P × T × R/100

1225 = 7350 × 1 × R/100

R = 50/3

In compound interest A = P (1 R/100)^2

7350 = P (1 50/(3 × 100))^2

7350 = P × 7/6 × 7/6

P = 7350 × 36/49 = 5400

**3. A lent Rs 5000 to B for 2 years and Rs 3,000 to C for 4 years on simple interest at the same rate of interest and received 2,200 in all form both of them as interest. The rate of interest per annum is (SI PRE 2018)**

a) 10% b) 8% c) 7% d) 6%

**Ans: a**

**Explanation:**

Simple Interest = P × T × R/100

Interest rate = R% in both cases

Given that

2200 = (5000 × 2 × R)/100 (3000 × 4 × R)/100

2200 = 100R 120R

2200 = 220R

R = 10%

**4. A sum of Rs 7,200 amounts to Rs 9,360 in 6 years at a certain rate of simple interest. The amount, if the rate of interest is increased by 4% is (in rupees) (SI 2016)**

a) 10,080 b) 10,880 c) 11,088 d) 12,880

**Ans: c**

**Explanation:**

Simple Interest = P × T × R/100

2160 = (7200 × 6 × R)/100

R = 2160/(72 × 6) = 5

In the second case

P = 7200; R = 5 4 = 9 ; T = 6

SI = (7200 × 6 × 9)/100 = 3888

Amount = 7200 3888 = 11088

**5. The present worth of Rs 132 due in two years at 5% simple interest per annum (in rupees) is (SI 2016)**

a) 120 b) 122 c) 112 d) 118.80

**Ans: a**

**Explanation:**

Assume that the principle = P

S.I= P × T × R/100

Given A = 132

S.I = Amount – Principle = 132 – P

132 – P = (P × 5 × 2)/100

1320 – 10P = P

11P = 1320

P = 1320/11 = 120

**6. A person borrowed a sum of money with simple interest as given below:**

**6% per annum for the first three years**

**8% per annum for the next five years**

**12% per annum for the next four years**

**If he paid a total interest of Rs 4,240, the money he borrowed (in rupees) is (SI 2016)**

a) 3,000 b) 3,500 c) 4,000 d) 4,050

**Ans: c**

**Explanation:**

Let the principle = P

(P × 6 × 3/100) (P × 8 × 5/100) (P × 12 × 4/100) = 4240

18P 40P 48P = 424000

106P = 424000

P = 4000

**To be continued…**

**By**