SSC CHSL 11 Aug 2021 Shift 2 Paper

Section: Quantitative Aptitude

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Question : 61 of 100

Marks: +1, -0

A, B

and

C

are three positive numbers such that

A

is

70 \%

of

B

, and

B

is

40 \%

of

C

. If the sum of all three numbers is 336 , then

15 \%

of the sum of

B

and

C

is

42
44
48
32

Solution:

Given,

A=70 \%

of

B=0.7 B

,

B=40 \%

of

C=0.4 C

and

A+B+C=336

⇒ 0.7 × 0.4 C+0.4 C+C=336

⇒ \;\; C=\;{336}/{1.68}=200

and

B=0.4 C=0.4 × 200=80

Hence,

15 \%

of

(B+C) \;\; =\;{15}/{100} ×(200+80)

\;\; =\;{3}/{2} × 28=42

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