Number System Questions for RRB NTPC Set-2 PDF

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Number System Questions for RRB NTPC Set-2 PDF
Number System Questions for RRB NTPC Set-2 PDF

Number System Questions for RRB NTPC Set-2 PDF

Download RRB NTPC Number Syatem Set-2 Questions and Answers PDF. Top 15 RRB NTPC Number System Set-2 questions based on asked questions in previous exam papers very important for the Railway NTPC exam.

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Question 1: During the school assembly, if the teachers make the students stand in a row of 14 each, 10 students are left. Instead, if they decide to make the students stand in a row of 24 each, 20 students will be left. If they decide to make them stand in rows of 25 each then 21 students will be left. What is the minimum number of students present in the school?

a) 3742

b) 4196

c) 3192

d) 4852

Question 2: What is the largest possible number which will leave the same remainder on dividing 4048, 3096 and 2024?

a) 8

b) 16

c) 32

d) 24

Question 3: Three numbers A, B and C are such that B is 40 percent less than A and C is 60 percent more than B. Another number D is twice as big as the difference between A and C. Find the value of D as a percentage of C.

a) 12.5 %

b) 20 %

c) 16.67 %

d) 8.33 %

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Question 4: Ravi gives equal number of chocolates to some children. If he gives 4 chocolates to each children then he is left with 9 chocolates in the end. Instead if he decides to give 5 chocolates to each children then the number of chocolates fall short by 8. Find the sum of the number of chocolates with Ravi and the total number of children to whom he distributed the chocolates.

a) 69

b) 84

c) 94

d) 77

Question 5: When the ten’s and units digit of a two digit number are interchanged, the resulting number is 9 more than the original number. How many such two digit numbers exist?

a) 7

b) 8

c) 9

d) 10

Question 6: There are three bells which are named as A, B and C. Bell A rings after every 7 minutes. Bell B rings after every 9 minutes and bell C rings after every 10 minutes. The bells rang together at 12 pm on Friday. How many times will the three bells have rung together(exclude the one at 12 pm) by 6 pm on the next day (Saturday)?

a) 9

b) 7

c) 2

d) 5

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Question 7: Find the smallest number which leaves remainders 3, 5 and 7 when divided by 11, 13 and 15?

a) 135

b) 187

c) 2137

d) 2797

Question 8: The sum of two numbers x and y is multiplied with x and y separately. The product obtained in each of these cases will be 1080 and 520. Find the difference between the given numbers.

a) 17

b) 14

c) 11

d) 13

Question 9: The ratio of the sum of two numbers and the difference of the two numbers is 17:4. Which of the following cannot be the two numbers?

a) 63, 39

b) 210, 130

c) 189, 104

d) 147, 91

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Question 10: A number when successively divided by 7, 8, 9 leave remainders 3, 2 and 1 respectively. What will be the smallest such number?

a) 219

b) 52

c) 480

d) 73

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Question 11: A church has 3 bells. All of them start ringing together.The first bell rings every 5 seconds and the second bell rings every 7 seconds. The third bell rings every 15 seconds. Ram noticed that all the 3 bells rang together 210 seconds after he came home. When will the 3 bells ring together again?

a) 420 seconds after Ram came home

b) 305 seconds after Ram came home

c) 415 seconds after Ram came home

d) 315 seconds after Ram came home

Question 12: Geetha thinks of a number X. She notices that 2X has an odd number of factors. Which of the following statements about X is definitely true?

a) X is a perfect cube.

b) X is an odd number.

c) X is not a perfect square.

d) X can be expressed as the square of a prime number.

Question 13: Sheela has a certain number of marbles. She notices that she can divide the marbles in groups of 4 but not in groups of 8. She also notices that she can divide the marbles in groups of 5 but not in groups of 15. Which of the following cannot be the number of marbles with Sheela?

a) 100

b) 820

c) 340

d) 720

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Question 14: The ratio of the product of 2 numbers to their sum is 8:3. The difference between the 2 numbers is 4. What is the difference between the squares of the 2 numbers if it is known that both the numbers are whole numbers?

a) 56

b) 48

c) 64

d) 36

Question 15: The remainder when a number is divided by 289 is same as the remainder when the number is divided by 17. How many 3 digit numbers satisfy this condition?

a) 42

b) 48

c) 51

d) 68

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Answers & Solutions:

1) Answer (B)

We can observe that 14 – 10 = 4
24 – 20 = 4
and
25 – 21 = 4
Hence, the required number of students will be nothing but LCM ( 14, 24, 25) – 4
We have
14 = 2*7
24 = 8*3
25 = 5*5
Hence, the required LCM will be 8*25*3*7 = 600*7 = 4200
Hence, the required number of students will be 4200 – 4 = 4196

2) Answer (A)

We know that the largest number which leaves same remainder on dividing A, B and C is given by HCF ( B-A, C-B)
Thus, here we have A = 4048, B = 3096 and C =2024
So, the required number will be
HCF (952, 1072)
952 = 8*7*17
1072 = 16*67
Thus, we can see that the HCF is 8
Hence, option A is the correct answer.

3) Answer (D)

Let A be 100x. So B will be 60x. C in this case will be 60x*1.6 = 96x
Hence, the difference between A and C is 4x. Thus, D must be 8x. Hence, D as a percentage of C will be 8x*100/96x = 8.33 %
Thus, option D is the correct answer.

4) Answer (C)

Let the number of children be x and number of chocolates be y. So we have
4x + 9 = y
and
5x – 8 = y
Solving both equations, we get
4x + 9 = 5x – 8
=> x = 17
Hence, there are 17 children. Hence, the number of chocolates must be 17*4 + 9 = 77
Thus, the required sum is 77 + 17 = 94

5) Answer (B)

Let the given two digit number be 10x + y
So after interchanging the digits, the new number will be 10y + x
We have been given that
10x + y + 9 = 10y + x
9y – 9x = 9
=> y – x = 1
Since both x and y are single digit natural numbers so y can take values from 2 to 9. Consequently, x will vary from 1 to 8. Hence, there are 8 possible numbers which satisfy the given conditions.

6) Answer (C)

Bell A rings after every 7 minutes, bell B rings after every 9 minutes and bell C rings after every 10 minutes. Hence, the three bells will ring together after an interval of LCM(7, 9, 10) minutes. Hence, the three bells will ring together after every 630 minutes. Thus they will ring together after every 10 hours and 30 minutes. Hence, they will ring together 2 times in the next 30 hours. Thus, option C is correct.

7) Answer (C)

We can observe that in each of the cases the difference between the divisor and quotient is constant.
11 – 3 = 8
13 – 5 = 8
15 – 7 = 8
Hence, the smallest number which satisfies the given conditions will be of the form
LCM(11, 13, 15) – 8
Thus, the required number will be
11*13*15 – 8 = 2145 – 8 = 2137

8) Answer (B)

We have been given that (x + y)*x = 1080 and (x + y)*y = 520. The HCF of 1080 and 520 will be the value of x + y. 1080 = 40*27 and 520 = 40*13
Hence, the HCF will be 40. Thus, x = 27 and y = 13
Hence, x – y = 14

9) Answer (C)

Let the two numbers be x and y.
We have been given that
$\frac{x+y}{x-y} = \frac{17}{4}$
=> $4x + 4y = 17x – 17y$
=> 13x = 21y
=> x/y = 21/13
Hence, the numbers must be in this ratio.
Numbers in option C are not in this ratio. Hence, it is our answer.

10) Answer (D)

The number on division with 9 leaves are remainder of 1. Since we are looking for smallest number so the number must be 1. This must have been the quotient on division by 8. Hence, the number which would have been divided by 8 would have been 8*1 + 2 = 10.
Thus, the number which must have been divided by 7 will be 10*7 + 3 = 73
Hence, option D is the correct answer.

11) Answer (D)

We know that the three bells ring together at intervals of their LCMs.
The LCM of 5, 7 and 15 seconds is 105 seconds.
If the 3 bells rang together at 210 seconds, they will ring together again at 210+105 = 315 seconds.
Hence, option D is the right answer.

12) Answer (C)

2X has an odd number of factors. Therefore, in 2X, the power of 2 must be even. This means that the power of 2 in X must have been an odd number. Therefore, X is definitely an even number. We can eliminate option B.

X must have been an even number. We know that 2X has an odd number of factors. Also, we know that only perfect squares have an odd number of factors. If 2X is a perfect square, X could not have been a perfect square for sure. Therefore, option C is the right answer.

13) Answer (D)

The number of marbles with Sheela is a multiple of 4 and 5, but are not multiples of 8 and 15. Therefore, the number of marbles must be of the form 20k, where k must neither be even nor must be a multiple of 3.

720 is a multiple of 3. Therefore, the number of marbles with Sheela cannot be 720. Hence, option D is the right answer.

14) Answer (B)

The difference between the 2 numbers is 4.
Let the numbers be $x$ and $(x+4)$.
We know that $\frac{x(x+4)}{x+x+4} = \frac{8}{3}$
$\frac{x^2+4x}{2x+4} = \frac{8}{3}$
$3x^2+12x = 16x+32$
$3x^2-4x-32 = 0$
Applying $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, we get,
Roots = $\frac{4\pm\sqrt{4^2+4*3*32}}{6}$
= $\frac{4\pm\sqrt{400}}{6}$
Since we know that $x$ is a whole number, $x$ must be $4$.
=> The other number is $x+4$ = $8$.
Difference between the squares = $8^2 – 4^2$ = $64-16$ = $48$.
Therefore, option B is the right answer.

15) Answer (C)

When a number is divided by 17, the remainder can range from 0 to 16 (17 values).
When a number is divided by 289, the remainder can range from 0 to 288 (289 values).
Numbers of the form 289x+k, where k is less than 17 will leave the same remainder when divided by both 17 and 289.
The question asks for 3 digit numbers.
Therefore, x can take the values 1,2 and 3. For every value of ‘x’, 17 values of k will satisfy this condition. Therefore, the number of 3 digit numbers that satisfy the given condition is 3*17 = 51.
Hence, option C is the right answer.

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