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SSC CHSL 11 Aug 2021 Shift 2 Paper

Section: Quantitative Aptitude

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Question : 67 of 100

Marks: +1, -0

If

a-b=7

a^{2}+b^{2}=169

a, b > 0

, then the value of

3 a+b

is

44
41
38
46

Solution:

Given,

a-b=7

. . . (i)

We know,

(a-b)^{2}=a^{2}+b^{2}+2 a b

(7)^{2}=169-2 a b

\left[ \because a^{2}+b^{2}=169 \; \text{ (given) } \; \right]

2 a b=169-49=120

\therefore \; \text{ We know, } \; (a+b)^{2}=a^{2}+b^{2}+2 a b

a+b=\sqrt{169+120}=\sqrt{289}=17

a+b=17

On adding Eqs. (i) and (ii);

2 a=24

\Rightarrow \; \; a = \; \frac{24}{2} = 12

and

b=17-12=5 \; \;

[from Eq. (ii)]

Now, the value of

3 a+b \; \; =3 \times 12+5

\; \; =36+5=41

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