# SSC CGL EXAMS 2018 | Quantitative Aptitude Practice Questions (Day-13)

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SSC CGL EXAMS 2018 | Quantitative Aptitude Questions (Day-13)

maximum of 10 points

1) sin ฮธ โ cos ฮธ = โ2cos ฮธ, then find the value of sin ฮธ + cos ฮธ?

a) โ2 sin ฮธ

b) 2 sin ฮธ cos ฮธ

c) โ2 sin2 ฮธ

d) โ2 sin ฮธ cos ฮธ

2) What percentage of numbers is there from 1 to 75 has 1 or 6 in the unitโs digits?

a) 25%

b) 20%

c) 50%

d) 33.33%

3) A Manโs age is thrice as much of his son and twice as much of his wife. Average age of this family is 22. What is the sum of ages of his wife and son?

a) 36 years

b) 66 years

c) 30 years

d) 25 years

4) Find the number of diagonals of 8 side closed figure?

a) 22

b) 16

c) 20

d) 25

5) In a triangle PQR, O is an in center of โPQR. If โ P = 600. Find the โ QOR.

a) 1100

b) 1000

c) 1200

d) 900

6) P is completed a job in 8 hrs in a day. The same job is completed by R in 6 hrs a day. They are working together, how long the work will be completed?

a) 3 (1/7) hours

b) 3 (4/7) hours

c) 3 (3/7) hours

d) 3 (2/5) hours

7) The lengths of two parallel sides of a trapezium are 28 cm and 12 cm. If its height is 15 cm, then what is the area of trapezium?

a) 540 cm2

b) 300 cm2

c) 250 cm2

d) 350 cm2

8) Find m if the ordinate and abscissa of the point (6m, m2 + 5m โ 12) are equal

a) โ 2 or 3

b) 2 or 3

c) โ 3 or 4

d) 3 or โ 4

9) The curved surface area of a cylindrical pillar is 660 m2 and its volume 2310 m3. Find the height of the cylinder

a) 11 m

b) 15 m

c) 7 m

d) 8 m

10) When 15% of fruits were rotten, a seller remains some fruits unsold after 60 fruits were sold. On another day if only 10% of fruits were rotten, he can sale 65 fruits with same number of unsold fruits on previous day. How many fruits actually he has daily if he has same number of fruits on each day?

a) 100

b) 150

c) 300

d) 250

Sin ฮธ โ cos ฮธ = โ2cos ฮธ

Squaring both side,

(sin ฮธ โ cos ฮธ)2 = 2cos2 ฮธ

sin2 ฮธ +cos2 ฮธ โ 2 sin ฮธ cos ฮธ = 2 cos2ฮธ

sin2 ฮธ = cos2 ฮธ + 2 sin ฮธ.cos ฮธ

Add Sin2 ฮธ on both sides

2 sin2 ฮธ = sin2 ฮธ +cos2 + 2sin ฮธ cos ฮธ

2 sin2 ฮธ = (sin ฮธ + cos ฮธ) 2

sin ฮธ + cos ฮธ = โ2 sin ฮธ

The Number which have 1 or 6 in unitโs digits are 1, 11, 21, 31, 41, 51, 61, 71, 6, 16, 26, 36, 46, 56, 66.

Total number with unitโs digit as 1 or 6 = 15

Total number = 75

Required percentage = (15*100)/ 75 = 20%

Manย  ย  ย  ย : ย ย ย ย ย ย  His son

3ย  ย  ย  ย  ย  :ย ย ย ย ย  ย  ย ย ย ย 1

Manย  ย  ย  ย :ย ย ย ย ย ย ย  His wife

2ย  ย  ย  ย  ย  ย :ย  ย  ย  ย  ย  ย 1

Manย  ย  ย  ย : ย ย ย ย ย ย  Wifeย  ย  ย  : ย ย ย ย ย  Son

6ย  ย  ย  ย  ย  ย :ย  ย  ย  ย  ย 3ย  ย  ย  ย  :ย  ย  ย  ย  2

Average age of this family = 22

Total age of this family = 22*3 = 66

6x + 3x+ 2x = 66 => 11x = 66

X = 6

Sum of ages of wife and son = 3x + 2x = 5x = 5 * 6 = 30 years

No. of diagonals of polygon = n (n โ 3)/2 = 8 (8 โ 3) / 2 = (8 * 5) /2 = 20

The angle made by any side with the in centre = 90ยฐ + half the opposite angle.

= 90ยฐ + โ P/2

= 900 + ยฝ *600 = 90ยฐ + 30ยฐ = 1200

P = 1/8 hours

R = 1/6 Hours

P + R = 1/8 + 1/6 = 7/24

= 3 (3/7) hours

Area of Trapezium = Half of sum of two parallel Sides * Heights

= (ยฝ)* [(s1+s2) * H]

= (28 + 12) /2 * 15 = 300 cm2

The ordinate of the point is m2 + 5m โ 12

Abscissa = 6m

6m = m2 + 5m โ 12

m2 + 5m โ 6m โ 12 = 0

m2 – m โ 12 = 0

m2 โ 4m + 3m โ 12 = 0

m (m โ 4) + 3 (m โ 4) = 0

m = – 3

m = 4

Curved surface area of the cylinder= 2ฯrh = 660 m2

Volume of the cylinder = ฯr2h = 2310 m3

(ฯr2 h)/(2ฯrh )= (2310 )/(660 )

R = 7

Curved surface area of the cylinder= 2ฯrh = 660 m2

= 2* 22/7 * 7 * h = 660 m2

= > h = 15 m

Total Fruits = x

According to the question, unsold fruits were remain same

(x – 15x/100 – 60) = (x โ 10x/100 – 65)

15x/100 + 60 = 10x/100 + 65

15% of x โ 10 % of x = 65 – 60

5x/100 = 5

5x = 500

X = 100

Total number of fruits = 100