# Solve interest related problems in no time

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. P lends Rs 15,000 to Q for a simple interest 8% per annum and Q lends […]

Published Date - 16 June 2022, 11:10 PM

**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. P lends Rs 15,000 to Q for a simple interest 8% per annum and Q lends the same amount to R for a simple interest 11% per annum. Then the gain of Q in 3 years, in rupees, is (SI MAINS 2018)**

a) 450 b) 900 c) 1350 d) 1800

**Ans: c**

**Explanation:**

Simple Interest from Q to P = P x Tx R/100

=15000 x 3 x 8/100 = 3600

Simple Interest from R to Q = 15000 x 3 x 8/100 = 4950

Gain of Q is= 4950-3600=1350

**2. In 4 years Rs 6000 amount became Rs 8500 for some rate of simple interest. Then the period required for Rs 1200 to become an amount of Rs 1825 at the same rate of interest is (SI MAINS 2018)**

a) 5 years b) 6 years c) 4 years d) 8 years

**Ans: a**

**Explanation:**

Simple Interest = P x Tx R/100

2500= 6000×4×R/100

R= 250/24= 125/12

In the second case,

P=1200; SI= 625; r= 125/12; T=?

T= (SI×100)/(P×R)

T= (625×100×12)/(1200×125)=5

**3. A sum of money becomes Rs 1,08,000 after 3 years and becomes Rs 1,92,000 after 9 years on same rate of compound interest. Then, that sum of money is (in rupees) (SI MAINS 2018)**

a) 96, 000 b) 90,000 c) 85,000 d) 81,000

**Ans: d**

**Explanation:**

Let the rate of interest be R and 1 R/100 = k

Hence, P×k3 = 108000 and P×k9 = 192000

Dividing the second with the first we get k6= 16/9

K3= 4/3

Substituting this value in the first equation P×4/3=108000

P= 81000

**4. A sum of money is doubled at some compound interest in 8 years. The number of years in which the amount becomes 4 times with same rate of interest is (SI MAINS 2018)**

a) 12 b) 14 c) 16 d) 32

**Ans: c**

**Explanation:**

If the principle= x

Given that sum of money doubles in 8 years i.e., 2x

It will become 4 times when it doubles 2 times.

Then 8×2 = 16

**5. An amount of Rs 5,00,000 becomes Rs 5,51,250 on some compound interest compounded half-yearly in one year. Then, the rate of compound interest is (SI MAINS 2018)**

a) 12% b) 10% c) 9% d) 8%

**Ans: b**

**Explanation:**

Let the rate of interest be 2R and 1 R/100 = k

Hence, by the given data: 551250=500000 × k2

k2= 1.1025

k=1.05

1 R/100 = 1.05 =>; R 100=105

Rate of interest- 2R= 10%

**6. A person invested in three different schemes A, B and C at the rates of simple interests 8%, 10% and 12% respectively. The amount invested in the scheme C is equal to 225% of the amount invested in scheme A and is also equal to 90% of the amount invested in scheme B. If the total interest earned by him in one year is Rs 48,000 then the amount the person invested in scheme C is (in rupees) (SI MAINS 2018)**

a) 1,70,000 b) 1,80,000 c) 1,90,000 d) 2,00,000

**Ans: b**

**Explanation:**

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

Then

(X x 8 x1/100) (Y x 10 x 1/100 ) (Zx12x1/100) = 48000

8x 10y 12z=4800000

Given

z=225%of x → z= 9/4 x → x=4/9 z

z=90% of y → z= 9/10 y → y= 10/9 z

By substituting these values in the equation

8×4/9 z 10×10/9 z 12z= 4800000

32z 100z 108z=4800000×9

240z=4800000×9

Z=20000×9=180000

**To be continued…**

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