# Quantitative Aptitude Questions Daily Quiz Day – 30

0
142 Dear Aspirants, Our SSC Crackers team is providing a new series of Quantitative Aptitude Questions for Upcoming Exam so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.

Start Quiz

1) Find the radius of the circle circumscribing the equilateral triangle of area 48√3 cm2.

a) 6√2 cm

b) 12 cm

c) 8 cm

d) 8√3 cm

2) In a triangle ABC, AB is extended to E such that ∠CBE = 1300. If 6∠A = 7∠C, find the value of ∠A.

a) 900

b) 600

c) 700

d) 450

3) O is the incentre of triangle PQR such that angle PQR = 70o and angle PRQ = 60o.Find angle ROQ.

a) 75o

b) 95o

c) 115o

d) 140o

4) Two cylindrical containers are filled with water. The radius and height of the first are 12 cm and 26 cm and of the second are 6 cm and 58 cm respectively. Find the radius of the cylindrical vessel 32 cm in height, which can just hold the water of the two given containers?

a) 13.5 cm

b) 14.5 cm

c) 15.5 cm

d) 16.5 cm

5) O is the centre of a circle and DE and PQ are two tangents to it touching at E and Q respectively. If the radius of the circle is 4cm, DE = 12cm and PQ = 3cm, find OD – OP.

a) (2√10 – 5) cm

b) 8 cm

c) (4√10 – 5) cm

d) 6 cm

6) The sum of the surface area of a cube and area of a rectangle is 2820 cm2. If the length and breadth of rectangle is 40% more and 25% less than the side of cube, find the perimeter of rectangle.

a) 108 cm

b) 86 cm

c) 72 cm

d) 64 cm

7) The curved surface area and total surface area of a cylinder are 1760 cm2 and 2992 cm2. Find the ratio of radius of cylinder to its height.

a) 2 : 3

b) 7 : 10

c) 1 : 2

d) 7 : 12

Directions (8 – 10): These questions are to be answered on the basis of the pie chart which gives marks scored by a student in different subjects – English, Hindi, Mathematics, Science and Social Science in an examination. The total marks obtained in the examination by the student are 540. 8) The subject in which the student scored 22.2% marks is ___.

a) Hindi

b) Science

c) Social Science

d) English

9) In which subject did the student score 105 marks?

a) Mathematics

b) Hindi

c) Science

d) English

10) The marks obtained in Mathematics are what percentage of the total marks?

a) 20%

b) 30%

c) 35%

d) 25%

Area of equilateral triangle = 48√3 cm2

(√3/4) × Side2 = 48√3

Side2 = 192

Side = √192 = 8√3 cm

Therefore, circumradius of equilateral triangle = (side)/√3 = (8√3)/√3= 8 cm 6∠A = 7∠C

∠A : ∠C = 1/6 : 1/7 = 7:6

∠A + ∠C = 1300 [An exterior angle of a triangle is equal to the sum of the two opposite interior angles]

∠A = 7 / (7 + 6) x 1300

=> ∠A = 7/13 x 1300

=> ∠A = 700 Volume of the first container = π (12)² (26) = 3744π

Volume of the second container = π (6)² (58) = 2088π

Total volume = 3744 π + 2088 π = 5832 π

πr² (32) = 5832 π

=> r2 = 5832/32 = 182.25

Or, r = 13.5 cm By Pythagoras theorem

OD = √[(OE)2 + (DE)2]

=> OD = √[42 + (12)2]

=> OD = √(16 + 144)

=> OD = √160

=> OD = 4√10 cm

And

OP = √[(OQ)2 + (PQ)2]

=> OP = √(42 + 32)

=> OP = √(16 + 9)

=> OP = √25

=> OP = 5 cm

OD – OP = (4√10 – 5) cm

Let, l and b be the length and breadth of rectangle and ‘a’ be the side of cube.

So, length of rectangle = 1.4a

And, breadth of rectangle = 0.75a

According to question:

l × b + 6a2 = 2820

1.4a × 0.75a + 6a2 = 2820

1.05a2 + 6a2 = 2820

7.05a2 = 2820

a2 = 2820/7.05

a2 = 400

a = √400 = 20 cm

So, length of rectangle = 1.4 × 20 = 28 cm

And, breadth of rectangle = 0.75 × 20 = 15 cm

So, perimeter of rectangle = 2(28 + 15) = 86 cm

Let, r and h be the radius and height of cylinder, respectively.

So, according to question,

2πrh: 2πr (h + r) = 1760: 2992

h: (h + r) = 20: 34

h: (h+ r) = 10: 17

7h = 10r

r: h = 7: 10

100% = 360°

22.2% = 360/100 x 22.2= 79.92° ≈ 80°

80° is shown here for Science.