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__SBI Clerk Prelims 2016- Practice Aptitude Questions (Simple Interest) Set-39__Dear Readers, Important Practice Aptitude Questions with solution for Upcoming SBI Clerk Exam, candidates those who are preparing for those exams can use this practice questions.

**1).**Find the amount paid by a borrower after 3 years, if he has taken a loan of Rs. 5000 at 10% per annum, simple interest.

a) Rs. 6000

b) Rs. 6500

c) Rs. 5500

d) Rs. 6750

e) None of these

**2).**Find the amount of money that Abha should lend at 15% p.a If she wants to earn an interest of Rs. 675 in 2(1 / 2) years.

a) Rs. 2000

b) Rs. 1850

c) Rs. 1800

d) Rs. 2010

e) None of these

**3).**A sum of money amounts to Rs. 1650 in two years and to Rs. 1875 in five years. Find the principal and rate of simple interest.

a) P=1500, r=5%

b) P=2000, r=5%

c) P=1000, r=4%

d) Cannot be determined

e) None of these

**4).**A certain sum of money amounts to Rs. 15,500 in 2 years at simple rate of interest of 12% p.a. Find the principal.

a) Rs. 13000

b) Rs. 12500

c) Rs. 14000

d) Rs. 15000

e) None of these

**5).**A sum of money doubles itself in 6 years. In how many years will it becomes 6 times at the same rate of simple interest?

a) 20 years

b) 25 years

c) 15 years

d) 30 years

e) None of these

**6).**Abhishek lent Rs. 10500, partly to Shahid at 10% p.a. S.I and partly to John, at 15% p.a., S.I. If at the end of 5 years, total amount received by Abhishek, from both, was equal to Rs. 17375, then find the amount of money lent to Shahid.

a) Rs. 6000

b) Rs. 6500

c) Rs. 5500

d) Rs. 4000

e) None of these

**7).**Sachin would have paid Rs. 5280, at the end of 4 years, for a sum of money borrowed, at rate of 8% p.a. S.I. If he wants to repay his loan an year before it was due, then what is the amount paid by him?

a) Rs. 5060

b) Rs. 4960

c) Rs. 4760

d) Rs. 4670

e) None of these

**8)**.Vishwas borrowed a total amount of Rs. 30000 part of it on simple interest rate of 12 p.c.p.a. and remaining on simple interest rate of 10 p.c.p.c. If at the end of 2 years he paid in all Rs. 36480 to settle the loan amount, what was the amount borrowed at 12 p.c.p.a?

a) Rs. 16000

b) Rs. 18000

c) Rs. 17500

d) Rs. 12000

e) None of these

**9).**Veena obtained an amount of Rs. 8376 as simple interest on a certain amount at 8 p.c.p.a. after 6 years. What is the amount invested by Veena?

a) Rs. 17180

b) Rs. 18110

c) Rs. 16660

d) Rs. 17450

e) None of these

**10).**If the simple interest for 6 years be equal to 30% of the principal, then it will be equal to the principal after

a) 20 years

b) 30 years

c) 10 years

d) 22 years

e) None of these

__Answers__**:**

**1). b) 2). c) 3). a) 4). b) 5). d) 6). d) 7). b) 8). d) 9). d) 10). a)**

__Solution:__

**1).**S.I = (P × r × t) / 100

= (5000 × 10 × 3) / 100 = Rs. 1500

Amount payable after 3 years = 5000 + 1500 = Rs. 6500

**Answer: b)**

**2).**S.I = 675 = P × (15 / 100) × (5 / 2)

P = (675 × 100 × 2) / (15 × 5)

P = Rs. 1800

**Answer: c)**

**3).**Interest for a period of 3 years = 1875 – 1650 = Rs. 225

Interest for 1 year = 225 / 3 = Rs. 75

Interest for 2 years = 75 × 2 = Rs. 150

Amount after 2 years = 1650

Principal = 1650 – 150 = Rs. 1500

And rate of interest = (75 / 1500) × 100 = 5% per annum

[ S.I for 1 year = (P × r × t) / 100

75 = (1500 × r × t) / 100 ]

**Answer: a)**

**4).**Amount = P + Interest

15500 = P + (P × 12 × 2) / 100

15500 = (124 / 100)P

P = 12500

**Answer: b)**

**5).**Method 1 :

When the sum doubles itself, means interest equal to amount of principal is earned in 6 years.

This means

SI = P in 6 years [then only Amount = P + P = 2P]

Now, SI should be equal to 5P such that amount becomes 6P

So, SI is P in 6 years, it would be 5P in 6 × 5 = 30 years

Method 2:

Let the principal amount = P

2P = P + (P × 6 × r) / 100

P = (6 × P × r) / 100

r = (100 / 6)% per annum

Finding when amount becomes 6 times:

6P = P + [(P × 100 × t) / 600 ]

5P = (P × t) / 6

t = 30 years

**Answer: d)**

**6).**Let the amount lent to Shahid =x, then

[ x + (x × 10 × 5) / 100 ] + [ (10500 – x) + (10500 – x)15 × 5 ) / 100 ]

= Rs. 17,375

(3 / 2)x + 18375 – (7 / 4)x = 17375

-(1 / 4)x + 18375 = 17375

x = Rs. 4000

**Answer: d)**

**7).**P + (P × (8 / 100) × 4) = 5280

(33 / 25)P = 5280

P = Rs. 4000

Amount payable after 3 years

= 4000 + [ 4000 × (8 / 100) × 3] = Rs. 4960

**Answer: b)**

**8).**Let the sum borrowed at the rate of 12 p.c.p.a be Rs. x

Sum borrowed at 10 p.c.p.a = Rs. (30000 – x)

Simple interest = Rs. (36480 – 30000) = Rs. 6480

According to the question,

[(x × 2 × 12) / 100] + [((30000 – x) × 2 × 10) / 100] = 6480

24x + 600000 – 20x = 648000

4x = 648000 – 600000 = 48000

x = 48000 / 4 = Rs. 12000

**Answer: d)**

**9).**S.I = (P × r × t) / 100

8736 = (P × 6 × 8) / 100 = (8736 × 100) / (6 × 8)

= Rs. 17450

**Answer: d)**

**10).**Method 1:

We are given S.I = (30 / 100) P in 6 years, so

To make S.I equal to P : multiply (30 / 100) P by (100 / 30)

S.I = (30 / 100)P × (100 / 30) in (6 × 100) / 30 years

S.I = P in 20 years

Method 2 :

30 / 100 P = (P × 6 × r) / 100

r = 5 %

finding time when S.I = P

P = P × (5 / 100) × t

t = 20 years

**Answer: a)**

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