**CAT Questions On Base System PDF:**

Download CAT Questions On Base System PDF. Practice important problems with detailed explanations for CAT on Base system related to conversions and sums.

Download CAT Questions On Base System PDF

Download All Quantitative Aptitude important Questions PDF

Take Free Mock Test for CAT 2018

**Question 1:** What will be the number of zeroes in$ (2000!)_{34}$. Here 34 is the base in which the number is written.

a) 122

b) 123

c) 124

d) 125

**Question 2:**Â $(245)_{x}+(162)_{x}-(427)_{x}=0$ in some base x. What is the value of x?

a) 8

b) 9

c) 10

d) Cannot be determined

**Question 3:Â **A number N has 4 digits when expressed in base 8, 3 digits when expressed in base 11 and 3 digits when expressed in base 26. How many values can N take?

a) 164

b) 298

c) 474

d) 655

**Question 4:Â **How many 3 digit numbers in decimal system are 3 digit numbers in both base 9 and base 11?

a) 625

b) 507

c) 528

d) 608

**Question 5:Â **A two digit number A in base 11 is one-third of the number formed by reversing its digits when considered in base 19. How many such numbers are possible?

a) 2

b) 1

c) 5

d) 4

CAT Mocks for just Rs. 1180 – Last Day

**Answers & Solutions:**

**1) Answer (B)**

34 = 17*2

So we have to find the highest power of 17 in 2000!. We need not find the power of 2 because of 2 will be greater than the power of 17. Thus the power of 17 will act as the limiting value.

Thus the highest power of 17 in 2000! is

[2000/17] + [ 2000/289] + [2000/4913], [] is greatest integer function

= 117 + 6 + 0 = 123

Thus the required number of zeroes is 123.

**2) Answer (A)**

$(245)_{x}+(162)_{x}-(427)_{x}=0$

$=>(2x^2+4x+5)+(x^2+6x+2)-(4x^2+2x+7)=0$

$=>x^2-8x=0$

$=>x=8$

**3) Answer (D)**

$8^3 = 512$ and $8^4 = 4096$ => The minimum value and maximum value of a number in base 10 to have 4 digits in base 8 are 512 and 4095 respectively => range = (512,4095).

$11^2 = 121$ and $11^3 = 1331$ => The minimum value and maximum value of a number in base 10 to have 3 digits in base 11 are 121 and 1330 respectively => range = (121,1330).

$26^2 = 676$ and $26^3 = 17576$ => The minimum value and maximum value of a number in base 10 to have 3 digits in base 26 are 676 and 17575 respectively => range = (676,17575).

Common range = (676, 1330)

=> Number of values that N can take = 1330 – 676 + 1 = 655

**4) Answer (D)**

Smallest 3 digit numbers in base 9 is $1*9^{2} + 0* 9 + 0* 9^{0} = 81$, Largest 3 digit number in base 9 is $8*9^{2} + 8*9 + 8*9 = 728$

Smallest 3 digit number in base 11 is is $1*11^{2} + 0* 11 + 0* 11^{0} = 121$

Largest 3 digit number in base 11 is $10*11^{2} + 10*11 + 10*11^{0}= 1330$

So 3 digit numbers in base 10 which are also 3 digit numbers in base 9 and base 11 are

from 121 to 728 = 608

**5) Answer (C)**

Let the number be xy.

=> $(xy)_{11}=\frac{1}{3}(yx)_{19}$

=> $y+11x=\frac{1}{3}(x+19y)$

=> $3y+33x=x+19y$

=> $32x=16y$ or $2x=y$

So, (x,y) can be (2,1)(4,2)(6,3)(8,4)(10,5)

Numbers such as 10, 11, 12.. etc. are represented as alphabets a,b,c etc in higher bases. Thus, the case of (10,5) also satisfies our requirements.

The next pair (12,6) canâ€™t be considered since 12 is not a single digit number in base 11 and thus, all other cases greater than (10,5) can be excluded.

Thus, the answer is 5.