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