RRB Group D Model Paper 8 with solutions

Question : 69 of 100

Marks: +1, -0

The maximum value of F in the 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 + 2F8 + 307 = 1114, 2F8 = 1114 – 816 = 298
Therefore,

\F_{\max}

= 9.

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