{"id":4405,"date":"2017-10-24T15:42:23","date_gmt":"2017-10-24T10:12:23","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=4405"},"modified":"2018-10-26T15:48:33","modified_gmt":"2018-10-26T10:18:33","slug":"number-system-questions-cat-solutions-test-2","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/number-system-questions-cat-solutions-test-2\/","title":{"rendered":"Number System Questions for CAT With Solutions Test-2"},"content":{"rendered":"

Number System is of the important topic for CAT. The number of questions asked from this topic is high in the exam.<\/p>\n

Number System Questions for CAT:<\/strong><\/p>\n

You can download Number System questions or you can go through the details below.<\/p>\n

Download Number System for CAT Set-2 PDF<\/a><\/p>\n

CAT 30 Days Crash Course<\/a><\/p>\n

Question 1:
\n<\/b>
\nA number is formed by writing first 54 natural numbers next to each other as 12345678910111213 … Find the remainder\u00a0when this number is divided by 8.<\/p>\n

A.\u00a0<\/b>1<\/p>\n

B.<\/b>\u00a07<\/p>\n

C.\u00a0<\/b>2<\/p>\n

D.\u00a0<\/b>0<\/p>\n

Question 2:<\/b><\/p>\n

How many five-digit numbers can be formed using the digits 2, 3, 8, 7, 5 exactly once such that the number is divisible by 125?
\n
\nA.\u00a0<\/b>0
\n
\nB.<\/b>\u00a01
\n
\nC.\u00a0<\/b>4
\n
\nD.\u00a0<\/b>3<\/p>\n

Question 3:<\/b><\/p>\n

How many five digit numbers can be formed from 1, 2, 3, 4,\u00a0 5, without repetition, when the digit at the unit\u2019s place must be greater than that in the ten\u2019s place?<\/p>\n

A.\u00a0<\/b>54<\/p>\n

B.\u00a0<\/b>60<\/p>\n

C.\u00a0<\/b>17<\/p>\n

D.\u00a0<\/b>2 x 4!<\/p>\n

Free Videos for CAT<\/a><\/p>\n

Question 4:
\n<\/b>
\nIf x is a positive integer such that 2x +12 is perfectly divisible by x, then the number of possible values of x is<\/p>\n

A.<\/b>\u00a02<\/p>\n

B.<\/b>\u00a05<\/p>\n

C.<\/b>\u00a06<\/p>\n

D.<\/b>\u00a012<\/p>\n

Question 5:<\/b><\/p>\n

Let k be a positive integer such that k+4 is divisible by 7.\u00a0 Then the smallest positive integer n, greater than 2, such that k+2n is divisible by 7 equals
\n
\nA.\u00a0<\/b>9<\/p>\n

B.\u00a0<\/b>7<\/p>\n

C.\u00a0<\/b>5<\/p>\n

D.\u00a0<\/b>3<\/p>\n

Take a Free Mock for CAT<\/a><\/p>\n

Solutions:<\/b><\/p>\n

1)<\/b>\u00a0Answer (C)<\/p>\n

For a number to be divisible by 8, last 3 digits must be
\ndivisible by 8.
\nLast 3 digits of this number are 354.
\n354 mod 8 = 2
\nHence, 2 is the remainder.<\/p>\n

2)<\/b>\u00a0Answer (C)<\/p>\n

As we know for a number to be divisible by 125, its last three digits should be divisible by 125 So for a five digit number, with digits 2,3,8,7,5 its last three\u00a0 digits should be 875 and 375 Hence only 4 numbers are possible with its three digits as\u00a0 875 and 375 I.e. 23875, 32875, 28375, 82375<\/p>\n

XAT Previous Papers with Solutions<\/a><\/p>\n

3) <\/strong>Answer (B)<\/p>\n

Possible numbers with unit’s place as 5 = $$4 \\times 3 \\times 2 \\times 1 = 24$$ Possible numbers with unit’s place as 4 and ten’s place 3,2,1 = $$3 \\times 3 \\times 2 \\times 1 = 18$$
\nPossible numbers with unit’s place as 3 and ten’s place 2,1 = $$2 \\times 3 \\times 2 \\times 1 = 12$$
\nPossible numbers with unit’s place as 3 and ten’s place 1 = $$1 \\times 3 \\times 2 \\times 1 = 6$$
\nTotal possible values = 24+18+12+6 = 60<\/p>\n

4)<\/b>\u00a0Answer (C)<\/p>\n

If 2x+12 is perfectly divisible by x, then 12 must be divisible by x.
\nHence, there are six possible values of x : (1,2,3,4,6,12)<\/p>\n

5)<\/b>\u00a0Answer (A)<\/p>\n

let’s say k+4 = 7m
\nk = 7m-4
\nNow for k+2n or 7m+(2n-4) is also \u00a0multiple of 7.
\nor 2n-4 should be a multiple of 7
\nSo 2n-4 = 7p
\nor 2n = 7p+4
\nFor p=2; n=9 (p cannot be 1 as n is an integer )<\/p>\n

 <\/p>\n

CAT Previous Solved Papers<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

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