## Railway RRB Group D 2018 Mathematics Practice Questions

Railway RRB Group D Mathematics Questions 2018. Welcome to the www.letsstudytogether.co 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.

## RRB Group D 2018 Mathematics Practice Questions | Set-3

1.Simplify up to three decimal places

√17.64 – (2.3)2 – 38.39 × 2.3 + (43.5) × (3.7)

A. 65.238

B. 71.563

C. 82.525

D. 98.643

B. 71.563

⇒ √17.64 – (2.3)2 – 38.39 × 2.3 + (43.5) × (3.7)

⇒ 4.2 – 5.29 – 88.297 + 160.95

⇒ 165.15 – 93.587

⇒ 71.563

∴ √17.64 – (2.3) – 38.39 × 2.3 + (43.5) × (3.7) = 71.563

2.A and B invest in a business in the ratio 3 : 2. If 5% of the total profit goes to charity and A’s share in profit is Rs. 8,550, then total profit is

A. Rs. 14,250

B. Rs. 15,760

C. Rs. 15,735

D. Rs. 15,000

D. Rs. 15,000

Let the total profit be P. Given, 5% of it goes to charity,

⇒ Profit left for sharing =P(15100)=1920P=P(1−5100)=1920P

Given, this profit is shared in the ratio 3:2

⇒ Share of A = 35×1920P=57100P35×1920P=57100P

57100P=8550⇒57100P=8550

⇒ P = 15000

3. What is the discount percentage offered on a book having marked price Rs. 2150 being sold at Rs. 1892?

A. 12

B. 13

C. 14

D. 16

A. 12

Discount can be calculated as (marked price – selling price) × 100/(marked price)

⇒ Discount = (2150 – 1892) × 100/ 2150

⇒ Discount = 12%

∴ the discount is 12%

4.Rakesh can do the 6 times the actual work in 36 days while Jitu can do the actual work in 3 days. In how many days will both working together complete the 3 times of the original work?

A. 6

B. 10

C. 12

D. 15

A. 6

Rakesh can thus complete the work in 6 days.

Jitu can complete the work in 3 days.

Let ‘x’ be the required number of days

∴ x × (1/6 + 1/3) = 3

∴ x = 6

∴ Together they can do 3 times the original work in 6 days.

5.Pipe A alone can fill a tank in 8 hours. Pipe B alone can fill it in 6 hours. If both the pipes are opened and after 2 hours Pipe A is closed, the other pipe will fill the tank is

A. 6 hours

B. 3 ½ hours

C. 4 hours

D. 2 ½ hours

D. 2 ½ hours

⇒ Pipe A filled in 8 hours = 1

⇒ Pipe A filled in 1 hour = 1/8

⇒ Pipe B filled in 6 hours = 1

⇒ Pipe B filled in 1 hour = 1/6

⇒ Both pipe fill in 1 hour = (1/6) + (1/8) = (14/48)

⇒ Both pipe fill in 2 hours = 2 × (14/48) = 7/12

⇒ Tank needs to be filled by B alone = 1 – (7/12) = 5/12

⇒ Time to fill the remaining part = (5/12)/(1/6)

∴ Time to fill the remaining part = 2 (1/2) hours

6.Find the average of 45 natural numbers starting from 454.

A. 475

B. 476

C. 477

D. 478

B. 476

The 45 natural numbers start from 454 and ends with 498 (= 454 – 1 + 45)

As we know, sum of n natural numbers is computed as,

Sum = n(n + 1)/2

Now,

Sum of numbers from 454 to 498 = sum of 498 numbers – sum of 453 numbers

⇒ [(498 × 499)/2] – [(453 × 454)/2]

⇒ 124251 – 102831

⇒ 21420

∴ Required average = 21420/45 = 476

7.Find the area of the largest square that can be made inside a right angle triangle having sides 6 cm, 8 cm & 10 cm when one of vertices of the square coincide with the vertex of right angle of the triangle?

A. 576/49 cm2

B. 24 cm2

D. None of these

A. 576/49 cm2

Length of side of the square with maximum area = (product of the sides forming the right angle)/(Sum of the sides forming the right angle)

Side of the square = (8 × 6)/(8 + 6) = 48/14 = 24/7

∴ The area of the square of maximum size is 576/49 sq.cm

8.The age of a man is 4 times that of his son. 10 yrs ago, the man was nine times as old as his son was at that time. What is the present age of the son?

A. 16 yrs

B. 32 yrs

C. 64 yrs

D. 24 yrs

A. 16 yrs

⇒ Let son’s present age be x, man’s age = 4x

⇒ 10 years ago,

⇒ Man’s age = 4x – 10, son’s age = x – 10

⇒ According to question

⇒ 4x – 10 = 9(x – 10)

⇒ 4x – 10 = 9x – 90

⇒ 5x = 80

⇒ x = 16

∴ Son’s present age is 16 years

9.A boat covers 9 km upstream and 38 km downstream in 3 hrs, while it covers 45 km upstream and 57 km downstream in 8 hrs. What is the speed of the current?

A. 2 km/hr

B. 3 km/hr

C.4 km/hr

D.5 km/hr

D. 5 km/hr

Let the speed of the current be c and the speed of the boat be s Km/hr.

While going upstream, the effective speed of the boat is s – c Km/hr.

While going downstream, the effective speed of the boat is s + c Km/hr.

Time = Distance/Speed

According to the question

9/(s – c) + 38/(s + c) = 3      —-(i)

45/(s – c) + 57/(s + c) = 8      —-(ii)

To solve these equations, we first multiply (i) by 5 and subtract (ii) from (i) i.e. (i) × 5 – (ii)

⇒ 45/(s – c) + 38 × 5/(s + c) – 45/(s – c) – 57/(s + c) = 15 – 8

⇒ (5 × 38 – 57)/(s + c) = 7

⇒ 133/(s + c) = 7

⇒ s + c = 19      —-(iii)

Put s + c = 19 in eqn(i) we get,

⇒ s – c = 9      —-(iv)

Solving eqn (iii) and eqn (iv) simultaneously we get,

⇒ c = 5 Km/hr

Therefore, the speed of the current is 5 Km/hr

10. In question, a series is given, with one term missing. Choose the correct alternative from the given ones that will complete the series.

1, 3, 4, 8, 15, 27, ?

A. 64

B. 50

C. 216

D. 224

B. 50

The sum of any three consecutive terms of the series gives the next term.

Thus, 1 + 3 + 4 = 8;

3 + 4 + 8 = 15;

4 + 8 + 15 = 27;

Thus missing number = 8 + 15 + 27 = 50

Clearly, the next number is 50.

## Railway RRB Group D Study Material  