34. If 6a + b2 =97,where a and b are the digits, then the values of a and b respectively are :
1) 5 and 3
2) 3 and 5
3) 2 and 3
4) 5 and 4

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Solution

$(10 \times 6) + a + (10 \times b) + 2 = 97$
$\Rightarrow 60 + a + 10 b + 2 = 97$
$\Rightarrow 62 + 10 b + a = 97$
$\Rightarrow 10 b + a = 35$
Now, check from options,
$a = 5 a n d b = 3$

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34. If 6a + b2 =97,where a and b are the digits, then the values of a and b respectively are : 1) 5 and 3 2) 3 and 5 3) 2 and 3 4) 5 and 4

(10×6)+a+(10×b)+2=97⇒60+a+10b+2=97⇒62+10b+a=97⇒10b+a=35Now, check from options,a=5 and b=3

34. If 6a + b2 =97,where a and b are the digits, then the values of a and b respectively are :
1) 5 and 3
2) 3 and 5
3) 2 and 3
4) 5 and 4

$(10 \times 6) + a + (10 \times b) + 2 = 97$
$\Rightarrow 60 + a + 10 b + 2 = 97$
$\Rightarrow 62 + 10 b + a = 97$
$\Rightarrow 10 b + a = 35$
Now, check from options,
$a = 5 a n d b = 3$