Question

23____4____1____53____ ____ 3____41 (a)514322 (b) 513242 (c)254312 (d)514225​

Answer 1

(d)514225

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Answer 2

Given :- 23____4____1____53____ ____ 3____41

(a) 514322

(b) 513242

(c) 254312

(d) 514225

Solution :-

given series is ,

→ 23____4____1____53____ ____ 3____41

as we can see that, total terms in the series are 15 . so, breaking them in pair of 5 we get,

→ 23_4_ |||| 1_53_ ||| _3_41

as we can see that, in all three pairs number are from 1 to 5 and some are missing . we have to fill that .

then filling rest numbers we get,

→ First pair = 23541 or 23145

→ Second pair = 12534 or 14532

→ Third pair = 53241 or 23541

now, checking options we get, none of the option has 1 at first . Then, we can conclude that, 5 is fixed at 5 and first pair will be 23541 .

now, next term missing terms in second pair are 2 and 4 .

therefore, first three options are eliminated and only left option is (D) 514225 .

Checking now,

→ 23541 |||| 14532 ||| 23541

As we can see that, all 3 pairs have all 5 numbers from 1 to 5 .

Hence, we can conclude that, Option (D) 514225 is correct answer .

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