Best Number System Questions for CAT
Download important CAT Best Number System Questions with Solutions PDF based on previously asked questions in CAT exam. Practice Best Number System Questions with Solutions for CAT exam.
Download Best Number System Questions for CAT
Question 1: 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 will be the number left on the board at the end?
a) 820
b) 821
c) 781
d) 819
e) 780
Question 2: How many three digit positive integers, with digits x, y and z in the hundred’s, ten’s and unit’s place respectively, exist such that x < y, z < y and x $\neq$ 0 ?
a) 245
b) 285
c) 240
d) 320
Question 3: In Nuts And Bolts factory, one machine produces only nuts at the rate of 100 nuts per minute and needs to be cleaned for 5 minutes after production of every 1000 nuts.
Another machine produces only bolts at the rate of 75 bolts per minute and needs to be cleaned for 10 minutes after production of every 1500 bolts. If both the machines start production at the same time, what is the minimum duration required for producing 9000 pairs of nuts and bolts?
a) 130 minutes
b) 135 minutes
c) 170 minutes
d) 180 minutes
Question 4: If a/b = 1/3, b/c = 2, c/d = 1/2 , d/e = 3 and e/f = 1/4, then what is the value of abc/def ?
a) 3/8
b) 27/8
c) 3/4
d) 27/4
e) 1/4
Question 5: The digits of a three-digit number A are written in the reverse order to form another three-digit number B. If B > A and B-A is perfectly divisible by 7, then which of the following is necessarily true?
a) 100<A<299
b) 106<A<305
c) 112<A<311
d) 118<A<317
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Question 6: A red light flashes three times per minute and a green light flashes five times in 2 min at regular intervals. If both lights start flashing at the same time, how many times do they flash together in each hour?
a) 30
b) 24
c) 20
d) 60
Question 7: Let $b$ be a positive integer and $a = b^2 – b$. If $b \geq 4$ , then $a^2 – 2a$ is divisible by
a) 15
b) 20
c) 24
d) All of these
Question 8: Let N = 1421 * 1423 * 1425. What is the remainder when N is divided by 12?
a) 0
b) 9
c) 3
d) 6
Question 9: When $2^{256}$ is divided by 17, the remainder would be
a) 1
b) 16
c) 14
d) None of these
Question 10: To decide whether a number of n digits is divisible by 7, we can define a process by which its magnitude is reduced as follows: $(i_{1}, i_{2}, i_{3}$,….. are the digits of the number, starting from the most significant digit). $i_{1} i_{2} … i_{n} => i_{1}.3^{n-1} + i_{2}.3^{n-2} + … + i_{n}.3^0$.
e.g. $259 => 2.3^2 + 5.3^1 + 9.3^0 = 18 + 15 + 9 = 42$
Ultimately the resulting number will be seven after repeating the above process a certain number of times. After how many such stages, does the number 203 reduce to 7?
a) 2
b) 3
c) 4
d) 1
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Answers & Solutions:
1) Answer (C)
Let the first operation be (1+40-1) = 40, the second operation be (2+39-1) = 40 and so on
So, after 20 operations, all the numbers are 40. After 10 more operations, all the numbers are 79
Proceeding this way, the last remaining number will be 781
2) Answer (C)
x, y and z in the hundred’s, ten’s and unit’s place. So y should start from 2
If y=2 , possible values of x=1 and z = 0,1 .So 2 cases 120,121.
Also if y=3 , possible values of x=1,2 and z=0,1,2.
Here 6 three digit nos. possible .
Similarly for next cases would be 3*4=12,4*5=20,5*6=30,…..,8*9=72 . Adding all we get 240 cases.
3) Answer (C)
Machine A takes 15 min to produce 1000 nuts with clean time. machine b takes 30 min to make 1500 nuts with clean time . So B is slower. So with B 900 nuts will be made in 180 mins but at last round cleaning time of 10 min no need to count hence 170 mins
4) Answer (A)
a/d = a/b * b/c * c/d = 1/3 * 2 * 1/2 = 1/3
Similarly, b/e and c/f are 3 and 3/8 respectively.
b/e = b/c*c/d*d/e = 3
c/f = c/d*d/e*e/f = 3/8
=> Value of abc/def = 1/3 * 3 * 3/8 = 3/8
5) Answer (B)
Let A = 100x + 10y + z and B = 100z + 10y + x .According to given condition B – A = 99(z – x) As (B – A) is divisible by 7 . So clearly (z – x) should be divisible by 7. z and x can have values 8,1 or 9,2 , such that 8-2=9-2=7 and y can have value from 0 to 9.
So Lowest possible value of A lowest x,y and z which is is 108 and the highest possible value of A is 299.
6) Answer (A)
A red light flashes three times per minute and a green light flashes five times in 2 min at regular intervals. So red light fashes after every 1/3 min and green light flashes every 2/5 min. LCM of both the fractions is 2 min .
Hence they flash together after every 2 min. So in an hour they flash together 30 times .
7) Answer (C)
We know that a=$b^2-b$.
So$a^2-a$ = b($b^3-2b^2-b+2$) . = (b – 2)(b – 1)( b)(b + 1)
The above given is a product of 4 consecutive numbers with the lowest number of the product being 2(given b >= 4)
In any set of four consecutive numbers, one of the numbers would be divisible by 3 and there would be two even numbers with the minimum value of the pair being (2,4).
Thus, for any value of b >=4, $a^2-4$ would be divisible by 3 x 2 x 4 = 24.
Thus, option C is the right choice. Options A and B are definitely wrong as a set of four consecutive numbers need not always include a multiple of 5 eg:(6,7,8,9)
8) Answer (C)
The numbers 1421, 1423 and 1425 when divided by 12 give remainder 5, 7 and 9 respectively.
5*7*9 mod 12 = 11 * 9 mod 12 = 99 mod 12 = 3
9) Answer (A)
$2^4 = 16 = -1$ (mod $17$)
So, $2^{256} = (-1)^{64} $(mod $17$)
$= 1$ (mod $17$)
Hence, the answer is 1. Option a).
10) Answer (A)
For 203 :
first step = $2\times 3^2 + 0 \times 3^1 + 3 \times 3^0$ = 21
second step = $2 \times 3^1 + 1 \times 3^0$ = 7
So two steps needed to reduce it to 7
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