RRB Mathematics Day – 01 | RRB JE Math Questions

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RRB JE Exam Mathematics Day – 01

maximum of 10 points
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1) Find the greatest 5 digit number which is divisible by 35?

a) 99995

b) 99990

c) 99985

d) 99980

2) What is the least number added to 5200 to get a number exactly divisible by 180?

a) 180

b) 160

c) 20

d) 60

3) If two-fifth of two-third of one-third of a number is 16, find the number?

a) 60

b) 180

c) 280

d) 90

4) A certain number consist of two digit number whose sum is 9. If the order of digit is    reversed, the new number is 9 less than the original number. What is the original number?

a) 45

b) 36

c) 54

d) 63

5) How many numbers up to 300 is divisible by 5 and 7 together?

a) 9

b) 10

c) 7

d) 8

6) The difference between the squares of two consecutive numbers is 39. What are the numbers?

a) 19,20

b) 20, 21

c) 18,19

d) 17,18

7) When a number is added to another number the total become 250% of the second number. What is the ratio between the first and the second number?

a) 3 : 2

b) 2 : 3

c) 4 : 3

d) 3 : 4

8) The sum of three consecutive odd numbers is equal to one-fourth of 156.what is the middle number?

a) 8

b) 10

c) 12

d) 13

9) A positive number which is when decreased by 4 is equal to 21 times the reciprocal of the same number. Find the number?

a) 3

b) 4

c) 7

d) 5

10) A number when divided by 1404, leave a remainder 93. What remainder would be obtained by dividing the same by 39?

a) 39

b) 29

c) 21

d) 15

Answers :

1) Answer: a)

Greatest 5 digit no = 99999

By dividing 999999 by 35 remainder = 4

So greatest 5 digit number divisible by 35 = 99999 – 4 = 99995

2) Answer: c)

After dividing 5200 by 180 Remainder =160

Required Number to be added = 180 – 160 = 20

3) Answer: b)

2/5×2/3×1/3 × Number = 16

X = 16×5×3×3/2×2 = 180

4) Answer: c)

Let the number be 10x + y

x + y =9

(10x + y) – (10y + x) = 9

x – y = 1

By solving these two we get x= 5 and y=4

5) Answer: d)

LCM of 5 and 7 = 35

300 = 8 × 35 +20

So number which are divided by 5 and 7 together is 8

6) Answer:  a)

The 1st number = x

Consecutive next number = x + 1

x2 – (x + 1)2 = 39

x2 – x2 + 1 – 2x = 39

1 – 2x = 39

x = 19

The numbers are 19, 20

Alternate method

(Number +1)/2 = 40/2 = 20

(Number – 1)/2 = 38/2 = 19

7) Answer: a)

8) Answer: d)

x + x + 2 + x + 4 = ¼ × 156

3x + 6 = 39

3x = 33

x = 11

x + 2 = 13

The middle value of the number = 13

9) Answer: c)

x  – 4 = 21(1/x)

x2 – 4x = 21

x2 – 4x – 21 = 0

Solving this equation

x2 + 3x – 7x – 21 = 0

x (x + 3) – 7 ( x + 3) = 0

(x – 7) (x + 3) = 0

So we get

x = 7 , – 4

Hence the positive number = 7

10) Answer:  d)

Number = 140 × quotient + 93

= 39 × 6 × quotient + 39 × 2 + 15 

Here 15 is remainder.

 

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