### Practice Aptitude Questions for SSC CGL,CHSL,CPO Exam

Practice Aptitude Questions With Solutions SSC CGL, CHSL Exams Set-28:
Dear Readers, Important Aptitude Questions for SSC CGL,CHSL Exam was given here with solutions, candidates those who are preparing for those exams can use this material.

1).If x – y = (x + y) / 7 = xy / 4, the numerical value of xy is
a)    4 / 3
b)    3 / 4
c)    1 / 4
d)    1 / 3

2).The area of a field in the shape of trapezium measures 1400 m2. The perpendicular distance between its parallel side is 24 m. If the ratio of the parallel sides is 5 : 3, the length of the longer parallel side is
a)    75 m
b)    45 m
c)    120 m
d)    60 m

3).The area of the greatest circle, which can be inscribed in a square whose perimeter is 120 cm, is
a)    (22 / 7) × (15)2 cm2.
b)    (22 / 7) × (7 / 2)2 cm2
c)    (22 / 7) × (15 / 2)2 cm2
d)    (22 / 7) × (9 / 2)2 cm2

4).If (x + y – z)2 + (y + z – x)2 + (z + x – y)2 = 0, then the value of x + y – z is
a)    √3
b)    3√3
c)    3
d)    0

5).Starting from his house one day, a student walks at a speed of 2(1 / 2) km / h and reaches his school 6 min late. Next day at the same time, he increases his speed by 1 km/h and reaches the school 6 min early. How far is the school from his house?
a)    2 km
b)    1(1 / 2) km
c)    1 km
d)    1(3 / 4)km

6).The average of a marks in Mathematics for 5 students was found to be 50. Later, it was discovered that in the case of 1 student, the marks 48 were misread as 84. The correct average is
a)    40.2
b)    40.8
c)    42.8
d)    48.2

7).D and E are the points in the sides AB and AC respectively in ∆ ABC, AD = 8 cm, DB = 12 cm, AE = 6 cm and EC = 9cm, then BC is equal to
a)    2 / 5 DE
b)    5 / 2 DE
c)    3 / 2 DE
d)    2 / 3 DE

8).A vertical stick 15 m long casts a shadow 12 m long on the ground. At the same time, a tower casts a shadow 50 m long on the ground. The height of the tower is
a)    60 m
b)    62 m
c)    62.5 m
d)    63 m

9).A man gets Rs. 13 more by selling an article at a profit of 12(1 / 2)%, then selling it at a loss of 12(1 / 2)%. The cost price of the article is
a)    Rs. 25.50
b)    Rs. 38
c)    Rs. 52
d)    Rs. 65

10).A walks at a uniform rate of 4km an hour and 4h after his start, B bicycles after him at the uniform rate of 10 km an hour. How far from the starting point will B catch A?
a)    16.7 km
b)    18.6 km
c)    21.5 km
d)    26.7 km

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

Solution:
1). x – y = (x + y) / 7 = xy / 4 = k
x – y = k; x + y = 7k; xy = 4k
(x + y)2 – (x – y)2 = 49k2 – k2
k = 1 / 3
xy = 4k = 4 × (1 / 3) = 4 / 3
2). Let the parallel sides be 5x m and 3x m.
Area of trapezium = 1 / 2 (Sum of parallel sides) × Distance between them
1440 = 1 / 2(5x + 3x) × 24
12 × 8x = 1440
x = 1440 / (12 × 8) = 15
The longer parallel side = 5x = 5 × 15 = 75 m
3). Side of the square = 120 / 4 = 30 cm

Clearly, diameter of the greatest circle = side of the square = 30 cm

Radius = 30 / 2 = 15 cm
Required area = π × (Radius)2
= (22 / 7) × (15)2 cm2
4). (x + y – z)2 + (y + z – x)2 + (z + x – y)2 = 0
(x + y – z) = 0
[∵ (a – b)2 + (b – c)2 + (c – a)2 = 0 provides a – b = 0]
5). Let the required distance be x km.
Difference of time = 6 + 6 = 12 min
= 12 / 60 = 1 / 5 h
According to the question,
[x / (5 / 2)] – [x / (7 / 2) = 1 / 5
(2x / 5) – (2x / 7) = 1 / 5
(14x – 10x) / 35 = 1 / 5
4x / 35 = 1 / 5
x = 35 / (4 × 5) = 7 / 4 = 1(3 / 4)km
6). Total marks obtained by 5 students = 50 × 5 = 250
Now, In this total marks, 84 is included Instead of 48,
Correct total marks = 250 – 84 + 48 = 214
Correct average = 214 / 5 = 42.8
7). In ∆ ADE and ∆ABC

AD / AB = 8 / 20 = 2 / 5

AE / EC = 6 / 15 =2 / 5
So, AD / AB = AE / AC
And ∠A = ∠A (common)
DE / BC = AD / AB   ⇒ DE / BC = 2 / 5
BC = 5 / 2 DE
8). Let AB be a vertical stick and AC be its shadow
Also, let PQ be a tower having shadow PR.

As, ∆ABC ~ ∆ PQR
AB / PQ = AC / PR
15 / x = 12 / 50
x = (15 × 50) / 12 = 62.5m
Hence, height of the tower is 62.5 m.
9). Let the CP of article be Rs.x
According to the question,
[100 + (25 / 2)] % of x - [100 - (25 / 2)] % of x = 13
⇒ (x / 100) [100 + (25 / 2)] - [100 - (25 / 2)] = 13
(x / 100) × 25 = 13
x = 13 × 4 = Rs. 52