In this article certain questions have been randomly taken from different topics of the quantitative section of CAT. Each question has been selected keeping in mind certain challenges you need to manage as a test taker. These challenges may range from concept basics to application orientation and test taking strategies.

1. 4

2. 6

3. 8

4. None of these

The two digits can be 4 and 7: Two possibilities 47 and 74.

The three digits can be 1, 4 and 7, so a total of 3! or 6 possibilities.

We cannot have three digits as (2, 2, 7) as the digits have to be distinct.

We cannot have numbers with 4 digits or more without repeating the digits.

So, there are 8 such numbers.

1. 07

2. 49

3. 43

4. 01

The last two digits of 7^{2} are 49.

The last two digits of 7^{3} are 43.

The last two digits of 7^{4} are 01.

The last two digits of powers of 7 go in a cycle - 07, 49, 43, 01.

So, to find the last two digits of 7^{47}, divide the index 47 by 4; check the remainder which is 3, at which level the actual remainder is 43

1. 33

2. 66

3. 55

4. None of these

Now (x + 2) and (x + 3) are two consecutive numbers. One of (x + 2) or (x + 3) has to be even.

We need x such that (x + 2) (x + 3) is a multiple of 3; which means one out of (x + 2) or (x + 3) should be a multiple of 3. Thus x leaves a remainder of 0 or 1 when divided by 3.

So x could be 1, 3, 4, 6, 7, 9, 10, 12, 13, 15, 16, 18 ....

So, from among the first 99 natural numbers, x cannot take 2, 5, 8, 11, 14…….98.

There are 33 numbers in this list – (3 × 0 + 2, 3 × 1 + 2, 3 × 3 + 2, ….3 × 32 + 2).

So, x can take 99 – 33 = 66 values.

1. 6

2. 8

3. 4

4. 7

Consider a number formed from the first 48 2s of the given number. This number would be a multiple of 13 (8 sets of 222222), so would a number comprising 36 2's.

Now we are left with a two digit number 2n, where n is the 50^{th} digit. 26 is a multiple of 13, so the 50^{th} digit should be 6.