# Previous Year CAT Questions on Number System PDF

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## Previous Year CAT Questions on Number System PDF

Download important Number System Questions for CAT PDF based on previously asked questions in CAT exam. Practice Number System Questions PDF for CAT exam.

Question 1: If $N = (11^{p + 7})(7^{q – 2})(5^{r + 1})(3^{s})$ is a perfect cube, where $p, q, r$ and $s$ are positive integers, then the smallest value of $p + q + r + s$ is :

a) 5

b) 6

c) 7

d) 8

e) 9

Question 2: If $10^{67}- 87$ is written as an integer in base 10 notation, what is the sum of digits in that integer? ·

a) 683

b) 489

c) 583

d) 589

Question 3: $2-\frac{\sqrt{6407522209}}{\sqrt{3600840049}}=$

a) 0.666039

b) 0.666029

c) 0.666009

d) None of the above

Question 4: The sum of $1 – \frac{1}{6} + (\frac{1}{6} \times \frac{1}{4})-(\frac{1}{6} \times \frac{1}{4} \times \frac{5}{18})+$ …. is

a) $\frac{2}{3}$

b) $\frac{2}{\sqrt{3}}$

c) $\sqrt{\frac{2}{3}}$

d) $\frac{\sqrt{3}}{2}$

Question 5: A rod is cut into 3 equal parts. The resulting portions are then cut into 12, 18 and 32 equal parts, respectively. If each of the resulting portions have integer length, the minimum length of the rod is

a) 6912 units

b) 864 units

c) 288 units

d) 240 units

Question 6: If the product of $n$ positive integers is $n^n$, then their sum is

a) a negative integer

b) equal to n

c) n + $\frac{1}{n}$

d) never less than $n^2$

Question 7: If the product of the integers a, b, c and d is 3094 and if 1 < a < b < c < d, what is the product of b and c?

a) 26

b) 91

c) 133

d) 221

Question 8: A sum of Rs.1400 is divided amongst A, B, C and D such that A’s share : B’s share = B’s share : C’s share = C’s share = D’s share = $\frac{3}{4}$ how much is C’s share?

a) Rs.72

b) Rs.288

c) Rs.216

d) Rs.384

Question 9: Z is the product of first 31 natural numbers. If X = Z + 1, then the numbers of primes among X + 1, X + 2, …, X + 29, X + 30 is

a) 30

b) 2

c) Cannot be determined

d) None of the above

Question 10: In the following series, what numbers should replace the question marks?
-1, 0, 1, 0, 2, 4, 1, 6, 9, 2, 12, 16, ? ? ?

a) 11, 18, 27

b) -1, 0, 3

c) 3, 20, 25

d) Cannot be ascertained

It has been given that $N = (11^{p + 7})(7^{q – 2})(5^{r + 1})(3^{s})$ is a perfect cube. All the factors given are prime. Therefore, the power of each number should be a multiple of 3 or 0.

$p,q,r$ and $s$ are positive integers. Therefore, only the power of the expressions in which some number is subtracted from these variables or these variables are subtracted from some number can be made $0$.

$11^{p + 7}$:

This expression must be made a perfect cube. The nearest perfect cube is $11^9$. Therefore, the least value that $p$ can take is $9-7=2$.

$7^{q – 2}$

The least value that $q$ can take is 2. If $q=2$, then the value of the expression $7^{q-2}$ will become $7^0=1$, without preventing the product from becoming a perfect cube.

$5^{r+1}$:

The least value that $r$ can take is $2$.

$3^{s})$:

The least value that $s$ can take is $3$.

Therefore, the least value of the expression $p + q + r + s$ is $2+2+2+3=9$.
Therefore, option E is the right answer.

$10^{67}- 87$ = $9999….99913$ (total 67 digits)

Sum of digits =$65*9 + 1 + 3$ = 589

$2-\frac{\sqrt{6407522209}}{\sqrt{3600840049}}=2-\frac{80047}{60007}$
=$2-1.3339610$
$=0.666039$
Therefore, option A is the right answer.

We can see that the magnitude in each succeeding term is less than that of preceding term.

Hence, we can say that for S = $1 – \frac{1}{6} + (\frac{1}{6} \times \frac{1}{4})-(\frac{1}{6} \times \frac{1}{4} \times \frac{5}{18})+$ …

The value will lie between (5/6, 1). We can check with option choices.

Option A: $\frac{2}{3}$ < $\frac{5}{6}$. Hence, this can’t be the answer.

Option B: $\frac{2}{\sqrt{3}}$ = 1.155 > $1$. Hence, this can’t be the answer.

Option C: $\sqrt{\frac{2}{3}}$ = 0.8164 < $\frac{5}{6}$. Hence, this can’t be the answer.

Option D: $\frac{\sqrt{3}}{2}$ = 0.866. This lies between (5/6, 1).Hence, this is the correct answer.

The rod is cut into 3 equal parts thus the length of the rod will be a multiple of 3.
Each part is then cut into $12 = 2^2*3$
$18 = 2*3^2$ and $32 = 2^5$ parts and thus, each part of rod has to be a multiple of $2^5*3^2 = 288$
Thus, the rod will be a multiple of $288*3 = 864$
Thus, the minimum length of the rod is 864 units.
Hence, option B is the correct answer.

We know that for a given product the sum of all the numbers will be the least when all the numbers are equal
Thus, the sum of all the number in $n^n$ will be least when
$n^n$ = $n*n*n*………..*(n \text{times})$
Thus, their sum $= n+n+……..+(n times) = n^2$
Hence, the sum can never be less than $n^2$
Hence, option D is the correct answer.

3094 = 2*7*13*17
Given, 1 < a < b < c < d
Thus, a = 2, b = 7, c = 13 and d = 17
Hence, b*c = 7*13 = 91
Hence, option B is the correct answer.

A:B = B:C = C:D = 3:4
Let C be 3x and D be 4x
B : C = 3 : 4 and C = 3x So, B = 9x/4
A : B = 3 : 4 and B = 9x/4 So, A = 27x/16
Therefore, A : B : C : D is 27x/16 : 9x/4 : 3x : 4x
Also, A + B + C + D = Rs.  1400
On solving, we get x = 128
Therefore, C’s share = 3x = Rs. 384
Hence, option D is the correct answer.

It is given that Z = 31!
X = 31! + 1
X+1 = 31!+2 this is divisible by 2
X+2 = 31!+3 this is divisible by 3
X+3 = 31!+4 this is divisible by 4
.
.
.
.
X+29 = 31!+30 this is divisible by 30
X+30 = 31!+31 this is divisible by 31
Hence, none of X + 1, X + 2, …, X + 29, X + 30 is a prime number.
Hence, option D is the correct answer.