Number Set I(Exams > Cat > Quantitaitve Aptitude ) Questions and answers for exam Preparation

Exams > Cat > Quantitaitve Aptitude

NUMBER SET I MCQs

Total Questions : 90 | Page 7 of 9 pages

Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares? (CAT 2007)

Discuss Question

Answer: Option D. -> 1
:
D
Soln:

Using the technique of finding squares of numbers upto 100 by keeping the base of 50 and 100( refer to the e-booklet for further details). We see that only squares of 12 will end with 144 (or the last 2 digits being the same). Hence we will have to find out if the first 2 digits will be the same for squares of 38, 62 (absolute difference of 12 from 50) and also for 88 (absolute difference of 12 from 100), we see that only 88 satisfies the condition, hence option 4.

Question 62.

The integers 1,2,...... 40 are written on a blackboard. The following operation is then repeated 39 times; in each repetition, any two numbers, say a and b, currently on the blackboard, are erased and a new number a+b-1 is written. What is the number left on the board at the end? (CAT 2007)

Discuss Question

Answer: Option C. -> 781
:
C

option (c) 781

Here, in each step we are adding two number and reducing the sum by 1. So after 39 operations, we will have the sum of all the numbers from 1 to 40 reduced by 39. Hence the final number will be $S_{40} - 39 = 781.$

Question 63.

What are the last two digits of $7^{2008} ?$ (CAT 2008)

Discuss Question

Answer: Option C. -> 01
:
C
Ans: (c) 01

The last two digits of a number is nothing but the remainder obtained when the number is divided by 100.This number leaves a remainder 1 when divided by 4 as well as 25. Hence the remainder obtained when this number is divided by 100 is also 1. Hence the last two digits of this number are 01.

Alternatively

Using Chinese remainder theorem: as we have to divide by 100 and find the reminder by 4 and then by 25 ,

i.e., 4A+1=25B+1, finding integer solutions we see that A=25 and B=4 will hence give a remainder of 01.

Alternatively, :

Use the last 2 digit rule for 7. Even here u will get 01 as the last 2 digits.

Question 64.

$5^{6} - 1$ is divisible by : (CAT 1995)

13
31
5
None of these
81

Discuss Question

Answer: Option B. -> 31
:
B
Ans: (b)

$(5^{6} - 1) = (5^{3})^{2} - (1)^{2} = (125)^{2} - (1)^{2} = (125 + 1) (125 - 1) = 126 * 124 = 31 * 4 * 126.$ It is therefore clear that the expression is divisible by 31.

Question 65.

Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is a perfect square whose square root equals the total of the three original integers. Which of the following best describes the minimum value, say m, of these three integers?(CAT 2008)

$1 <= m <= 3$
$4 <= m <= 6$
7
10
$13 <= m <= 15$

Discuss Question

Answer: Option A. ->

1 <= m <= 3

:
A
Ans: a
It's an easy question. Trial with some numbers will give you the solution

1 <= m <= 3

3^{1} + 4^{2} + 5^{3} = 144 = 12^{2} = (3 + 4 + 5)^{2} .

Question 66.

When $2^{256}$ is divided by 17, the remainder would be : (CAT 2002)

1
16
14
None of these
81

Discuss Question

Answer: Option A. -> 1
:
A

Ans: (a)

$2^{256}$ can be written as $(2^{4})^{64} = (17 - 1)^{64} .$

In the expansion of $(17 - 1)^{64}$ every term is divisible by $17$ except $(- 1)^{64}$ . Hence remainder is 1.

Or directly:

Euler's number of 17 is 16 and 256 is a multiple of 16, hence the remainder is 1.

Question 67.

$7^{6 n} - 6^{6 n},$ where n is an integer $> 0$ , is divisible by (CAT 2002)

13
127
559
All of these
81

Discuss Question

Answer: Option D. -> All of these
:
D
Ans: (d)

For $n = 1, 7^{6} - 6^{6} = (7^{3})^{2} - (6^{3})^{2};$

$= (7^{3} - 6^{3}) (7^{3} + 6^{3}) = (343 - 216) (343 + 216); = 127 * 559 = 127 * 13 * 43.$ Hence it is divisible by $127, 13, 559.$

Question 68.

For two positive integers a and b, define the function h (a,b) as the greatest common factor (GCF) of a, b. Let A be a set of n positive integers G(A), the GCF of the elements of set A is computed by repeatedly using the function h. The minimum number of times h is required to be used to compute G is : (CAT 1999)

n
(n - 1)
n
None of these
81

Discuss Question

Answer: Option B. -> (n - 1)
:
B
Ans:

It is clear that for n positive integers function h (a,b) has to be used one time less than the number of integers, i.e., (n-1) times.

Question 69.

What is the remainder when $4^{96}$ is divided by 6? (CAT 2003)

Discuss Question

Answer: Option D. -> 4
:
D
Ans: (d)

$4^{2} = 4$ (mod 6)

$4^{4} = 4$ (mod 6)

$4^{6} = 4$ (mod 6)

and so on. The answer will remain the same.

Question 70.

If x and y are integers then the equation 5x + 19y = 64 has (CAT 2003)

no solution for $x < 300$ and $y < 0$
no solution for $x > 250$ and $y > - 100$
a solution for $250 < x < 300$
a solution for $- 59 < y < - 56$
97 : 84

Discuss Question

Answer: Option C. -> a solution for

250 < x < 300

:
C

Soln:

We see that if y = 1, we get an integer solution for x = 9, now if y changes (increases or decreases) by 5, x will change (decrease or increase) by 19.

Looking at options, if x = 256 we get y = 64.

Using these values we see option 1, 2 and 4 are eliminated and also that these exists a solution for $250 < x < 300.$