Quantitative Aptitude Practice Test for SSC CGL (TIER-I)

QUANTITATIVE APTITUDE PRACTICE TEST

1.The value of 
1/15 + 1/35 + 1/63 + 1/99 + 1/143 is
a)8/39
b)5/39
c)1/39
d)11/39

2.A can do a piece of work in 70 days and B is 40% more efficient than A. The number of days taken by B to do the same work is:
a)50 days
b)20 days
c)30 days
d)35 days

3.Area of the base of a pyramid is 57 sq.cm and height is 10cm. What is the volume of the pyramid?
a)190 cu.cm
b)290 cu.cm
c)390 cu.cm
d)490 cu.cm

4.For a certain article, if discount is 25% the profit is 25%. If the discount is 10%, then the profit is:
a)60%
b)20%
c)50%
d)10%

5.If y : x = 4 : 15, then the value of [(x - y)/(x + y)] is:
a)7/19
b)5/17
c)13/15
d)11/19

6.Some carpentars promised to do a job in 9 days but 5 of them were absent and remaining men did the job in 12 days. The original number of the carpentars was:
a)22
b)16
c)20
d)17

7.The average of two numbers is 62. If 2 is added to the smallest number, the ratio between them becomes 1:2. What is the difference of the numbers?
a)55
b)44
c)33
d)22

8.A shopkeeper makes a profit of 20% even after giving a discount of 10% on the marked price. If the marked price is Rs.500, then the CP of the article is:
a)Rs.375
b)Rs.525
c)Rs.456
d)Rs.395

9.If 40% of (A+B) = 60% of (A-B), then the value of (2A-3B)/(A+B) is:
a)1/7
b)5/7
c)7/6
d)1/4

10.A number is first increased by 10% and then it is further increased by 20%. The original number is increased altogether by?
a)25%
b)32%
c)20%
d)30%

Answer key

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

Solution

1.1/15 + 1/35 + 1/63 + 1/99 + 1/143
= 1/(3×5) + 1/(5×7) + 1/(7×9) + 1/(9×11) + 1/(11×13)
= 1/2(1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13)
= 1/2 (1/3 - 1/13)
= 5/39

2.A : B = D2 : D1
or, 100 : 140 = D2 : 70
or, D2 = (100×70)/140 = 50

3.Volume of the pyramid = 1/3 × height × area of the base = 1/3 × 10 × 57 = 190 cu.cm

4.If the marked price be Rs.x and CP be Rs.100,
then, (x×75)/100 = 125
or, x = 500/3
SP after a discount of 10% = 500/3 × 90/100 = Rs.150
Therefore, profit = 50%

5.y : x = 4 : 15
or, x : y = 15 : 4
By componendo and dividendo,
(x - y)/(x + y) = (15 - 4)/(15 + 4) = 11/19

6.Let the original number of carpentars be x.
M1D1 = M2D2
or, x × 9 = (x - 5) × 12
or, 9x = 12x - 60
or, x = 20

7.Sum of the numbers = 124
If the larger number be x, then smaller number = (124 - x)
so, (124 - x + 2)/x = 1/2
or, 252 - 2x = x
or, x = 84
Therefore, smaller number = 124 - 84 = 40
Difference = 84 - 40 = 44

8.If the CP of the article be Rs.x,
then, (500×90)/100 = (x×120)/100
or, x = (5×450)/6 = 375

9.(A+B)×40/100 = (A-B)×60/100
or, 2A + 2B = 3A - 3B
or, A = 5B
Therefore, (2A-3B)/(A+B) = (10B-3B)/(5B+B) = 7B/6B = 7/6

10.Required% increase = [10 + 20 + (10×20)/100] = 32%