# SBI Clerk Prelims 2016- Practice Aptitude Questions (Percentage)

SBI Clerk Prelims 2016- Practice Aptitude Questions (Percentage) Set-32:
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).Ratan bought a car at a discountof 16.67% on listed price from an automobile company. If the listed price of the car is Rs. 6, 60, 000, then find the amount of discount.
a)  Rs. 1,10,000
b)  Rs. 1,00,000
c)  Rs. 1,01,000
d)  Rs. 1,25,000
e)  None of these
2).Two candidates contested in a election. If one got 520 votes which was 65% of votes, then what was the total number of votes?
a)  858
b)  702
c)   780
d)   754
e)  None of these
3).In an election between two candidates, one got 52% of the valid votes. 25% of the total votes were invalid. The total number of votes were 8400. How many valid votes did the other person get?
a)  3276
b)  3196
c)  3024
d)  Cannot be determined
e)  None of these
4).A student scores 55% marks in 8 papers of 100 marks each. He scores 15% of his total marks in English. How much does he score in English?
a)   55
b)  66
c)  77
d)  44
e)  None of these
5).A student has to secure a minimum 35% marks to pass in an examination. If he gets 200 marks and fails by 10 marks, then the maximum marks are:
a)  300
b)  400
c)  500
d)  600
e)  None of these
6).A’s salary is 50% more than that of B. How much % is B’s salary less than A?
a)  50
b)  33(1 / 3)
c)  45
d)  66(2 / 3)
e)  None of these
7).What percent decrease in salaries would exactly cancel out 20% increase?
a)    20
b)  16(2 / 3)
c)  33(1 / 3)
d)  18
e)  None of these
8).The price of a car is first increased by 10%& then decreased by 15%. What is the net percentage change in price of the car ? Also find the new price if the price was initially Rs. 2,40,000.
a)  Rs. 2,48,000
b)  Rs. 2,24,000
c)  Rs. 2,42,000
d)  Cannot be determined
e)  None of these
9).What will be the net% change in salary of Abha it is first decreased by 25% and then increased by 30%.
a)  -5/2 %
b)  5/2%
c)  2/5%
d)  6/5%
e)  None of these
10).The cost of manufacturing one unit of an article goes up by 2% in 1992 and 5% in 1993. Find the net % increase  in the cost till 1993. Also find the cost in 1993, if it was Rs. 500, in 1992 (before increase).
a)  Rs. 535. 5
b)  Rs. 535
c)  Rs. 500
d)  Cannot be determined
e)  None of these
1). a) 2). e) 3). c) 4). b) 5). d) 6). b) 7). b) 8). b) 9). a) 10). a)
Solution:

1).16.67 % = 1 / 6
Hence discount = (1 / 6) × 6,60,000
= Rs. 1,10,000

2).We are given 65% of votes = 520,
Let the total number of votes be x, this means (65 / 100)x = 520 and so
x = 520 × (100 / 65) = 800

3).Total votes (valid + invalid) = 8400
Total invalid votes = 25% of 8400 or (1 / 4) of 8400 = 2100
Total valid votes = 8400 – 2100 = 6300
[or directly calculate valid votes as (100% – 25% = 75%) of the total votes i.e. 8400]
One got 52% of the valid votes, other will get (100% – 52% = 48%) of the valid votes
i.e. (48 / 100) × 6300 = 48 × 63 = 3024

4).Maximum marks = 8 × 100 = 800
Total marks scored = 55% of 800
= (55 / 100) × 800 = 55 × 8 = 440
= 440 and marks scored in English = (15 / 100) × 440 = 66

5).Let the maximum marks be x, then 200 + 10 = 210 ⇒ (35 / 100)x
And x = 210 × (100 / 35) = 600 marks
Or
[ you can also solve by checking options]
Checking option a) 35% of 300 = 105 hence ruled out,
Checking option b) 35% of 400 = 140 hence ruled out,
Checking option c) 35% of 500 = 175 hence ruled out,
Checking option d) 35% of 600 = 210 valid option,

6).Method 1: A’s salary is 50% more than B.
Hence B’s salary must be decreased by
[ x / (100 + x) × 100 ] % where x is % increase
= (50 / 150) × 100 = (100 / 3)% = 33(1 / 3)%
B’s salary is 33(1 / 3)% less than A
Method 2 :
Step 1 : 50% = 1 / 2
Step 2 : 1 / (2 + 1) = 1 / 3 = 33(1 / 3)% (mentally add numerator to the denominator)

7).Doing it in the fraction way
Step 1 : 20% = 1 / 5
Step 2 = 1 / (5 + 1) = 1 / 6 which is 16.67%

8).Using formula, % change = [ a + b + (ab / 100) ] %
Here a = 10, b = -15
% change = [ 10 + (- 15) + (10 (-15) / 100) ] %
= [ -5 -(3 / 2) ]% = -13 / 2%
There is net decrese by 6.5% in price of the car.
New Price = 2,40,000 × [(100 + 10) / 100] × [(100 – 15) / 100]
= 2,40,000 × (11 / 10) × (85 / 100) = Rs. 2,24,400.

9).Net % change = [ a + b + (ab / 100) ] %
Here a = -25, b = 30
= [ -25 + 30 + ((-25)(30)) / 100)%
= 5 – (15 / 2)%
= – 5 / 2% = decreased by 2.5%