Dear Aspirants, Here we have given the Important SSC CGL Exam 2018 Practice Test Papers. Candidates those who are preparing for SSC CGL 2018 can practice these questions to get more confidence to Crack SSC CGL 2018 Examination.
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1) The marked price of a doll is 25% higher than the cost price. The Shopkeeper gave 20% discount to the customer. In this transaction, the seller loses
a) 5%
b) Gain 5%
c) No loss no gain
d) Loss 2.25%
2) ( x+a): (x+b): (x+c) =11:9: 12 where x = a + b + c then find the ratio of a:(c-b)
a) 1:1
b) 1:3
c) 3:1
d) 1:2
3) After selling 8/9 of total products at a profit of 25% and remaining products at a loss of 10%, the shopkeeper gains Rs. 190 as profit. Find the total number of products.
a) 870
b) 720
c) 900
d) 950
4) Two liquids are mixed in the ratio of 3: 2 and it is sold at the price of Rs. 11 per liter with profit of 10%. If the second liquid costs Rs.2 less than the first, then find the cost price of first liquid?
a) 8.8
b) 10.8
c) 7
d) 6.3
5) A boat takes 3 hours to travel from place A to place B downstream and back from B to A upstream. If the speed of the boat in still water is 4 kmph what is the distance between the two places?
a) 100 km
b) 120 km
c) 150 km
d) Cannot be determined
6) Number of diagonals in a pentagon is
a) 10
b) 7
c) 4
d) 5
7) The area of rectangle, whose length is 10 more than half of the breath, is 192 sq. units. What is length of diagonal of the rectangle?
a) 30
b) 20
c) 25
d) 15
8) A student obtained an average of 84 marks in 8 subjects. His highest mark in these subjects was 4 more than second highest. His average marks in the remaining 6 subjects were 82. Find the highest mark obtained by him?
a) 90
b) 91
c) 92
d) 93
9) A water purifier is available for Rs. 500 down payment followed by two installments of Rs. 2000 each. If the interest paid is Rs. 500, find the price of water purifier.
a) 4500
b) 5000
c) 4000
d) 3500
10) Find the solution set of (x-7) (x+8) ≤ 0
a) [−8, 7]
b) (8,−7)
c) (7, 1)
d) (1, 8]
Answers :
1). Answer: c)
Let CP=100
MP=100×125/100=125
SP= 125×80/100=100
CP+SP
He has no loss no gain
2). Answer: a)
3). Answer: c)
Let total products = 9 and CP be Rs.1 per article
SP = 8×125/100+ 1×90/100 = 10.9
CP = 9
Profit = (10.9 – 9) = 1.9
9—— 1.9
?——–190
Total number of products = 9×190/1.9 = 900
4). Answer: b)
CP of the mixture= 11×100/110=10
x x-2
10
3 2
(x – 10) /(12 – x ) = 2/3
3x – 30 = 24 − 2x
5x = 54
x = 10.8
5). Answer: d)
Let the distance from place A to B be x km and the speed of current being y km/hr.
Now, [x/(4+y)] + [x/(4-y)] = 3
(4x – xy + xy + 4x) / [(4 – y) (4 + y)] = 3
3 (16 – y2) = 8x
48 – y2 = 8x
So, we can’t find the distance.
6). Answer: d)
Number of diagonals in a pentagon is 5
7). Answer: b)
8). Answer: c)
Let x and y be the highest and second highest mark respectively.
Total marks = 84×8 = 672
Average of remaining 6 subjects = 82
Marks of remaining 6 subjects = 82×6 = 492
Total mark of remaining two subjects
- x + y = 672 – 492 = 180 —– (i)
x – y = 4 ——– (ii)
From (i) and (ii)
x = 92, y = 88
Highest mark = 92
9). Answer: c)
Total Amount paid = 500 + 2000 + 2000 = 4500
Price = 4500 – Interest = 4500 – 500 = 4000
10). Answer: a)
(x – 7) (x+8) ≤ 0
−8 ≤ x ≤ 7
x lies between [ −8,7]
