If a-b=7 and a^{2}+b^{2}=169 where a, b > 0, then the value of 3 a+b is

SSC CHSL 11 Aug 2021 Shift 2 Paper

Section: Quantitative Aptitude

Question : 67 of 100

Marks: +1, -0

a-b=7

and

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

where

a, b > 0

, then the value of

3 a+b

Solution:

Given,

a-b=7

. . . (i)
We know,

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

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

{[ ∵ a^{2}+b^{2}=169 \;\text " (given) "\;]}

2 a b=169-49=120

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

a+b=√{169+120}=√{289}=17

a+b=17

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

2 a=24

⇒ \;\; {a}=\;{24}/{2}=12

and

b=17-12=5 \;\;

[from Eq. (ii)]
Now, the value of

3 a+b \;\; =3 × 12+5

\;\; =36+5=41

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