Home Railway RRB Group D 2018 Railway RRB Group D 2018 Mathematics Practice Questions | Set-2

Railway RRB Group D 2018 Mathematics Practice Questions | Set-2

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Railway RRB Group D 2018 Mathematics Practice Questions

Railway RRB Group D Mathematics Questions 2018. Welcome to the www.gsspecial.com online Railway RRB Group D section. If you are preparing Railway, RRB ALP and RRB Group D Exams 2018, you will come across a section on “General Mathematics”. Here we are providing you with “Important RRB Group D 2018 Mathematics Practice Questions” based on the latest pattern of your daily practice.

This “Railway RRB Group D 2018 Mathematics Practice Questions“is also important for other Railway exams such as RRB ALP & Technicians Exam 2018.

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RRB Group D 2018 Mathematics Practice Questions | Set-2

1.Find the least number which when divided by 12, 18, 36 and 45 leaves the remainder 8, 14, 32 and 41 respectively.

A. 186

B. 176

C. 180

D. 178

Show Correct Answers

B. 176

We observe that 12 – 8 = 4, 18 – 14 = 4, 36 – 32 = 4 and 45 – 41 = 4

Hence, the required number should be LCM (12, 18, 36, 45) – 4

The LCM of 12, 18, 36 is 36.

The LCM of 36 and 45 is 180.

Hence, required number = 180 – 4

= 176

2.The H.C.F and L.C.M of two numbers are 8 and 48 respectively. If one of the numbers is 24, then other number is:

A. 48

B. 36

C. 24

D. 16

Show Correct Answers

D. 16

H.C.F of two number = 8

L.C.M of two number = 48

One number = 24

Let, the other number is ‘x’

We know that,

Product of two numbers = product of L.C.M and H.C.F of two numbers

⇒ 24 × x = 48 × 8

⇒ x = 2 × 8 = 16

3. Find the wrong number in the sequence :

196, 169, 144, 121, 101

A. 101

B. 121

C. 169

D. 196

Show Correct Answers

A. 101

All are squares of a number except 101.

196 = 14²

169 = 13²

144 = 12²

121 = 11²

4.The difference between the greatest and least prime numbers which are less than 100 is

A. 97

B. 94

C. 96

D. 95

Show Correct Answers

D.95

The greatest prime number less than 100 is 97. The least prime number is 2.

Difference = 97 – 2 = 95

5.Two pipes can fill a tank in 10 hours and 12 hours respectively, while a third pipe empties the full tank in 20 hours. If all the three pipes operate simultaneously, in how much time the tank will be filled?

A. 7 hrs

B. 8 hrs

C. 7 hrs. 30 min.

D. 8 hrs. 30 min.

Show Correct Answers

C. 7 hrs. 30 min.

Time taken by A , B and C to fill/empty the tank completely = 10 , 12 and 20 hours respectively

Hence part of tank filled by A, B and emptied by C in 1 hour = 1/10, 1/12 and 1/20 respectively

Part of tank filled by all 3 pipes in 1 min $= \frac{1}{10} + \frac{1}{12} - \frac{1}{20}$

$= \frac{6}{60} + \frac{5}{60} - \frac{3}{20}$

$= \frac{8}{60}$

= 2/15

Hence it will take 15/2 hours ie 7 hours and 30 mins to completely fill the tank

6.In a group of 50 students, 25 play Hockey, 30 play Football, and 8 play neither game. What is the number of students who play both Hockey and Football?

A. 10

B. 11

C. 12

D. 13

Show Correct Answers

D. 13

Total students = 50

Student who play hockey = 25

Student who play football = 30

Students who play neither football nor hockey = 8

Total students = (Student who play hockey) + (Students who play football) – (Students who play both hockey and football) + (Students who neither play hockey nor football)

50 = 25 + 30 – (Students who play both hockey and football) + 8

Students who play both hockey and football = 13

So, total 13 students play both hockey and football.

7.Simplify: 150% of (440% of 15 – 300% of 10)

A. 54

B. 64

C. 24

D. 47

Show Correct Answers

A. 54

Given that, 150 % of (440% of 15 – 300% of 10)

= 150% × (440/100 × 15 – 300/100 × 10)

= 150/100 × (66 – 30)

= 1.5 × 36

= 54

8.Instead of dividing Rs. 234 among A, B, C in the ratio $\frac{1}{3} : \frac{1}{4} : \frac{1}{2}$ , by mistake it was divided in the ratio 3 : 4 : 2. Who gained in the transaction?

A. A and B

B. B and C

C. A and C

D. A, B and C

Show Correct Answers

A. A and B

Let, $\frac{1}{3} : \frac{1}{4} : \frac{1}{2}$ = 4 : 3 : 6

Actual amount they have to got –

$⟹ A = (234 \times \frac{4}{13}) = R s .72$

$⟹ B = (234 \times \frac{3}{13}) = R s .54$

$⟹ C = (234 \times \frac{6}{13}) = R s .108$

But mistakenly they got the amount in the ratio = 3 : 4 : 2

$⟹ A = (234 \times \frac{3}{9}) = R s .78$

$⟹ B = (234 \times \frac{4}{9}) = R s .104$

$⟹ C = (234 \times \frac{2}{9}) = R s .52$

Hence, A and B gained in the transaction.

9.A series is given, with one term missing. Choose amongst the given responses choose the meaningful one.

208, 225, 259, 310, ?, 463

A. 356

B. 378

C. 313

D. 349

Show Correct Answers

B. 378

206, 221, 251, 296, ?, 431

225 – 208= 17

259 – 225= 34

310 – 259= 51

The value increases in multiples of 17.

The next number will be increased by 68, 310 + 68 = 37

Hence the correct option is 378

10. Two numbers A and B are 22% and 34% less then a third number C. Find by what percentage is the number B less than the number A?

A. 15%

B. 18%

C. 9%

D. 11%

Show Correct Answers

A. 15%

A = C – 0.22C = 0.78C

B = C – 0.34C = 0.66C

∴ Required % = ((0.78C – 0.66C)/0.78C) × 100 = 15.3 ≈ 15%

∴ B is less than A by 15%

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