1.

(a) 4695(c) 4285(b) 4593(d) 438713. 4183, 4388, 4285, 4490, ? ,4592​

Answer»

ANSWER:

the alternate numbers are in series i.e

4183,4285,?......(1)

4388,4490,4592....(2)

4285-4183 =102

4490-4285 = 102

therefore in both series 102 is added.

we have to find ? so we need only 1 ST series.

4285+102 = 4387

therefore option (d) is the answer.

hope it helps you.



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