SSC CPO SI and ASI 2015 Paper

Question : 110 of 199

Marks: +1, -0

The maximum value of F in the following equation 5E9 + 2F8 + 3G7 = 1114 is (E, F, and G stand for the tens digits)

Solution:

(2) For F to be maximum, E and G must be minimum. So, put E = G = 0. Then,

509+2 F 8+307=1114,2 \ {F} 8=

1114-816=298

Therefore,

F_{\max }=9

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