# Quantitative Aptitude Questions Daily Quiz Day – 73

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168 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.

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1) A certain sum becomes 7 times in 8 years, at simple interest, then in how many years it will become 19 times?

(a) 15

(b) 18

(c) 28

(d) 24

Direction (2-5): The table given below shows the runs scored by 5 players in four matches. 2) What is the difference between total runs scored by A and B in four matches?

(a) 111

(b) 123

(c) 98

(d) 135

3) Runs scored by C in match II is how much percent more than the runs scored by C in match I?

(a) 11.76

(b) 16.66

(c) 17.12

(d) 13.33

4) Runs scored by A in match II is what percent of the total runs scored by these five players in match II?

(a) 71.07

(b) 35.14

(c) 28.92

(d) 55.13

5) What is the correct order of averages of the given players in the 4 matches?

(a) E > A > C > D > B

(b) E > A > D > C > B

(c) A > E > C > D > B

(d) A > E > B > C > D

6) A solid metallic sphere of radius 21 cm is melted and recast into a cone with diameter of the base as 21 cm. What is the height (in cm) of the cone?

(a) 336

(b) 112

(c) 224

(d) 66

7) The radius of a wheel is 3.5 cm. What is the distance (in cm) travelled by the wheel in 20 revolutions?

(a) 220

(b) 440

(c) 880

(d) 1320

8) If the perimeter of a square is 44 cm, then what is the diagonal (in cm) of the square?

(a) 11√2

(b) 2√11

(c) 11

(d) 44√2

9) What is the curved surface area (in cm2) of a cylinder having radius of base as 14 cm and height as 10 cm?

(a) 440

(b) 880

(c) 220

(d) 1320

10) Three circles of radius 63 cm are placed in such a way that each circle touches the other two. What is the area of the portion enclosed by the three circles?

(a) 7938√3 – 4158

(b) 3969√3 – 4158

(c) 7938√3 – 6237

(d) 3969√3 – 6237 Total runs scored by A = 70 + 105 + 55 + 135

= 365

Total runs scored by B = 40 + 35 + 95 + 72

= 242

∴ Difference = 365 – 242

= 123

Required% = (68 – 60)/60 × 100

= 8/60 × 100

= 13.33   H = 336 cm

Distance travelled by the wheel in 20 revolutions

= 2 × 22/7 × 3.5 × 20

= 440

Perimeter = 44

∴ Side = 44/4 = 11

Diagonal = √2 × Side

=  11√2

C.S.A = 2πrh

= 2 × 22/7 × 14 × 10

=880 cm2  