14 May 2016

SBI Clerk Prelims 2016- Practice Aptitude Questions (Simple Interest Problems)

SBI Clerk Prelims 2016- Practice Aptitude Questions (Simple & Compound Interest)

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)

More Practice Aptitude Questions for SBI Clerk -Click Here



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