# SSC CGL 2018 Practice Test Papers | Quantitative Aptitude (Day-3)

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SSC CGL 2018 | Quantitative Aptitude (Day-3)

maximum of 11 points

Click โStart Quizโ to attend these Questions and view Explanation

1. If a/b+b/a=1,aโ 0, bโ 0, the value of a3+b3 is ______

(a) 0

(b) 1

(c)ย  -1

(d) 2

1. ABCD is a cyclic parallelogram. The angle c is equal to____

(a) 30

(b) 60

(c) 90

(d) 45

1. If sinัฒ+ cosecัฒ= 2 then sin100ัฒ+ cosec100ัฒ is equal to

(a) 1

(b) 2

(c) 200

(d) 100

4 . A man walks a distance of 40 km from home to his office at 8 km/hr, he was late by 80 minutes.

If he walks at 10 km/hr He will reach the office____

(a) 20 minutes late

(b) 30 minutes late

(c) 30 minutes earlier

(d) 20 minutes earlier

1. If 2x+8=82x+1 then what is the value of x?

(a) 3

(b) 5

(c) 1

(d) 2

1. Fraction between 2/5 and 4/9 is

(a) 3/7

(b) 1/2

(c) 2/3

(d) 4/5

1. If 17200 is divided by 18, the remainder is______

(a) 17

(b) 1

(c) 16

(d) 2

8. Three persons A, B , C invest in a business in the ratio 8:5:7. After 3 months A left the business. If the total profit is 80000 what is difference of profit between B and C?

1. 12000
2. 20000
3. 30000
4. 15000

9). The least positive number when subtracted from 5 equals to 6 times the reciprocal of its number. What is the number?

a.6

b.2

c.3

d.10

ยญย 10). Ifย cos4A-sin4Aย =ย X,ย thenย valueย ofย Xย is?

a.cos2A+1

b. cos2A -1

c.2cos2A+1

d. 2cos2A-1

11. Five bells A, B, C, D,E ring at intervals of 3,6,9,18 and 36 minutes and another bell F rings at an interval of X minutes. All the six bells ring together at every 180 minutes. How many X values are possible?

1. 10
2. 5
3. 4
4. 6

a/b+b/a=1

=>ย  (a2+b2)/ab=1

=> a2+b2=ab

=> a2+b2-ab=0

Therefore a3+b3=(a+b)(a2-ab+b2)=0

ABCD is acyclic parallelogram.

โ A+โ C=180

2โ C=180

โ C=90

sinัฒ+ cosecัฒ= 2

sinัฒ+1/ sinัฒ=2

sin2 ัฒ-2 sinัฒ+1=0

(sinัฒ-1)2=0

sinัฒ=1=> cosecัฒ=1

sin100ัฒ+ cosec100ัฒ=1+1=2

Time taken to cover 40 km at the speed of 5 km/hr =40/8=5 hrs

Fixed time= 5 hrsย  – 40 minutes= 4 hrs 20 minutes

Time taken to cover 40 km at the speed of 10 km/hr= 40/10=4 hrs

Required time= 4 hrs 20 minutes-4 hrs = 20 minutes earlier

2x+8=23( 2x+1)

X+8=3(2x+1)

X+8=6x+3

5x=5

X=1

2/5 = 0.40 andย ย  4/9 =0.44

3/7=0.42

1/2=0.5

2/3=0.66

4/5=0.8

17200 = (18=1)200

17200/18 = (18=1)200/18

Remainder=1

8x*2: 5x*12: 7x*12= 4:15:21

80000/40x*(21-15)x=12000

(5-x)=6/x

5x-x2-6=0

x= 2,3

Least number is 2

cos4Aโ sin4Aย =ย (cos2A+sin 2 A) (cos 2A-sin 2A)

=1(cos2A-(1-cos2A))ย ย ย ย ย  โด (cos2x+sin 2x =1)

=2cos2A-1