#IDETICEDU #MATH #RRBNTPCQUIZ

Mathematics Quiz For RRB NTPC

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1.I know many numbers, which when divided by 12 and 16, leave the same remainder in each case, and are exactly divisible by 11. Find the least of such numbers.
1. 99 

2. 55                     

 3. 132                       

4. 176

 

2. The average age of a family of 5 persons, 5 years back was 25 years. Due to the birth of two children, the average age of the family now is 22 years. If the age difference between the two children is 2 years, their ages in years are:
1. 15.5 and 13.5

2. 3 and 1

3. 5 and 3

4. 8.5 and 6.5

 

3. rectangular piece of wood has its length and breadth cut by 10% and 30% respectively. The percentage of area cut off is –
1. 20

2. 25

3. 37

4. None of these

 

4. If 20! is divided by 6, which of the following will be the remainder?
1. 0

2. 1

3. 2

4. 4

 

5. In a parking lot, every third car is red and every fourth car is white. What could be the maximum number of cars in that parking lot?
1. 12

2. 16

3. 13

4. 11

 

6. What is the total number of positive integer solutions of the form of (p, q) that satisfy the equation 2p + 1.5q = 60?
1. 9

2. 11

3. 10

4. 8

 

7. The ratio between a two-digit number and the sum of the digits of that number is 4 : 1. If the digits in the unit’s place is 3 more than the digit in the ten’s place, find the number.
1. 36

2. 63

3. 48

4. 84

 

8. A certain even number K is given, which is not divisible by 3. What will be the remainder if this number will be divided by 6?
1. 2

2. 4
3. either 1st or 2nd option

4. Any natural number < 6

 

9. How many ordered integer solutions of the form of (P, Q) are there, which satisfy the equation
|P| +|Q| =  7

1. 26

2. 28

3. 22

4. 30

 

10. There exist a number N. N > 1545. If N – 6 is a multiple of 13, then the largest number that will always divide (N + 7) * (N + 20) is
1. 26

2. 169

3. 338

4. 13

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Answer

1.Try with options. 1st option and 2nd option give same remainders when divided by 12 and 16. But 2nd
option is smaller than 1st. So 2nd is the answer.

 

2. Sum of their ages 5 years back = 125.
Sum of present ages of 5 members = 125 + 25 = 150. Total age of 7 members = 22  7=154.
So sum of ages of 2 children = 154 – 150 = 4. Diff. of ages of 2 children = 2  their ages are 3, 1.

 

3.Net area left = 0.9 * 0.7 = 0.63, ; area cut off = 1 – 0.63 = 0.37 = 37 %.

 

4.The factorial of all the natural numbers ≥ 3 is divisible by 6. Therefore 20! will be exactly divisible by
6, hence no remainder shall be there.

 

5.Every 3rd car is red and every fourth car is white.
On the face of it seems that the data is inadequate to answer this question. But take the LCM of 3 and 4
i.e. 12. ⇒ 12th car is red as well as white, which can’t be true. The maximum number of cars in parking
lot is 11.

 

6.The equation can be rewritten as 4p + 3q = 120. The smallest value of p and the greatest value of q that
satisfies this equation is (0, 40). The greatest value of p and the smallest value of q possible is (30, 0).
Now after taking p as 0, the next p which will make it possible is p = 3, then p = 6 and so on the last
will be p = 30 i.e. 11 values of p can make this equation right.
But the questions states positive integers only thus we have to exclude two sets of solutions, which
include a zero i.e. (0, 40) and (30, 0). Thus remaining there are 9 solutions.
Thus 1st option is the answer.

 

7. Let the unit’s digit be U and the ten’s digit is T. The equation will be 10T + U = 4 (T + U) ⇒ 6T = 3U
⇒ 2T = U. Their difference is given to be 3. Solving you get U = 6 and T = 3. Thus first option.

 

8. As it is an even number, it must be either a multiple of 6, or (a multiple of 6) + 2, or (a multiple of 6) +
4. It cannot be a multiple of 6, as the question states that it is not divisible by 3. Thus the only possible
remainders are now 2 or 4. Thus 3rd option is the answer.

 

9. The following are the cases for (a, b) which make this equation right. (0, 7) (0, -7) (1, 6) (1, -6) (-1, 6)
(-1, -6) (2, 5) (-2, 5) (2, -5) (-2, -5) (3, 4) (-3, 4) (3, -4) (-3, -4). These 14 cases and their reverse 14
cases.
Thus 28 solutions are there.

 

10. As N – 6 is a multiple of 13, thus (N + 7) and N + 20 should also be divisible by 13. Because there are
respectively 13 and 26 more than N – 6. Now (N + 7) and (N + 20) are two consecutive multiples of 13,
one of them must be even. Thus their product would always be divisible by 13 × 13 × 2 = 338. Hence
3rd option is the answer

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