# Quantitative Aptitude Questions Daily Quiz Day – 21

0
236 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) If (b – a) = b2/a, then what is the value of (a4 + a3b + ab3 + b4)?

a) 1

b) 1/2

c) 2

d) 0

2) Find the value of (718 × 718 × 718 + 134 × 134 × 134)/ (718 × 718 – 718 × 134 + 134 × 134)

a) 738

b) 956

c) 1024

d) 852

3)  7.474747.. is express in fractional form as a/99 and 9.070707.. is expressed in fractional form as 898/b, what is the value of a – b?

a) 641

b) 631

c) 671

d) 681

4) If 16A3 is completely divisible by 7, what is the value of A?

a) 0

b) 1

c) 2

d) 3

5) What is the value of

6/(1+√3) + 6/(√3+√5) + 6/(√5+√7) + .. +6/(√79 + √81)?

a) 18

b) 6

c) 12

d) 24

6) If (8 – √3)/(8 + √3) = a – b√3, then find the value of (a – b).

a) 41/51

b) 51/61

c) 61/71

d) 51/81

7) Area of a rectangular seminar hall is 0.5 acres. If we want to put some marbles having dimensions of 60 cm X 30 cm on its floor then how many approximate marbles will be needed? (1 acre = 4046.86 m2)

a) 11372

b) 10800

c) 12421

d) 11242

8) Find the curved surface area of the cone having radius 12 cm and height 9 cm. [Use π = 3]

a) 640 cm2

b) 560 cm2

c) 480 cm2

d) 540 cm2

9) The ratio of the length and breadth of the rectangle is 18:7 and the area of the rectangle is 126 sq. cm. What is the perimeter of the rectangle?

a) 35 cm

b) 50 cm

c) 65 cm

d) None of these

10) The area of the rectangular field is thrice the area of the square field and the perimeter of the square field is 80 meters. If the length of the rectangular field is 50% more than the breadth of the rectangular field, then find the breadth of the rectangular field.

a) 22 meters

b) √800 meters

c) √520 meters

d) 25 meters

(b – a) = b2/a

a(b – a) = b2

ab – a2 = b2

(a2 + b2 – ab) = 0 …….. (1)

= (a4 + a3b + ab3 + b4)

= a3(a + b) + b3(a + b)

= (a + b)(a3 + b3)

= (a + b)(a + b)(a2 – ab + b2)

= (a + b)2(a2 – ab + b2)

From equation (1)-

= (a + b)2 * 0

= 0

(x3 + y3)/(x2 – xy + y2) = x + y

So, (718 × 718 × 718 + 134 × 134 × 134)/ (718 × 718 – 718 × 134 + 134 × 134)

= (7183 + 1343)/ (7182 – 718 × 134 + 1342)

= 718 + 134 = 852

Let X = 7.474747.. (I)

=> 100X = 747.474747.. (II)

(II) – (I) gives,

99X = 740

=> X = 740/99 = a/99

∴ a = 740

Let Y = 9.070707.. (III)

=> 100Y = 907.070707.. (IV)

(IV) – (III) gives,

99Y = 898

=> Y = 898/99 = 898/b

∴ b = 99

∴ a – b = 740 – 99 = 641

16A3 is completely divisible by 7, among the given options A can only be equal to 0.

Then, 1603 / 7 = 229

6/(1+√3) + 6/(√3+√5) + 6/(√5+√7) + .. +6/(√79 + √81)

= 6(√3 – 1)/2 + 6(√5 − √3)/2 + 6/(√7 − √5)/2 + .. + 6/(√81 − √79)/2

= 3(√3 – 1 + √5 − √3 + √7 − √5 + .. + √81 − √79)

= 3(√81 – 1)

= 3(9 – 1)

= 24

(8 – √3)/(8 + √3) = a – b√3

[(8 – √3)/(8 + √3)] * [ [(8 – √3)/(8 – √3)] = a – b√3 (Rationalising the denominator)

[(8 – √3)2]/(64 – 3) = a – b√3

(64 + 3 – 16√3)/61 = a – b√3

(67/61) – (16√3/61) = a – b√3

By comparing the both side-

a = 67/61, b = 16/61

Now, the value of (a – b) = (67/61) – (16/61) = 51/61

We know that, 1 Acre = 4046.86 m2

Hence 0.5 Acre = 0.5 X 4046.86 = 2023.42 m2

Area of the one marble = 0.6 X 0.3 = 0.18 m2

Hence required number of marbles = 2023.42/0.18 = 11242 (approx)

Slant height of the cone = √(122 + 92) = √(144 + 81) = √225 = 15 cm

Curved surface area of the cone = πrl = 3 × 12 × 15 = 540 cm2

Given l/b = 18/7

l = 18b/7

Area of rectangle = l*b = (18b/7)*b = 126

b2 = 49

b = 7 cm

l = 18b/7 = 18 cm

Perimeter = 2(l + b) = 2(18 + 7) = 50 cm

Let the side of the square field be ‘a’ cm.

Given, perimeter of the square field = 80 meters

4a = 80, a = 20 meters

So, area of the square field = a2 = 202 = 400 m2

Let the breadth of the rectangular field be ‘x’ meters.

Then, the length of the rectangular field = ‘1.5x’ meters

So, area of the rectangular field = 3 × 400 = 1200 m2

x × 1.5x = 1200

1.5x2 = 1200

x2 = 800

x = √800 meters