RRB Junior Engineer Stage 1 Model Papers 8

Question : 29 of 100

Marks: +1, -0

What is the smallest value that must be added to 709, so that the resultant is a perfect square?

Solution:

According to question

26<√709<2

Now,

(27)^2=729

∴ 729 – 709 = 20

∴ 20 must be added to 709 to make it a perfect square.

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