## 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-7

1. 3 cubes whose edges are 3 cm, 4 cm and 5 cm respectively are melted to from a single cube. The surface area of the new cube will be.

A. 196 sq. m

B. 216 sq. m

C. 169 sq. m

D. 343 sq. m

B. 216 sq. m

Now the volume of cube = (side)3

As three cubes whose edges are 3 cm, 4 cm and 5 cm respectively are melted to from a single cube

Volume of first cube = 33 = 27 cm3

Volume of second cube = 43 = 64 cm3

Volume of third cube = 53 = 125 cm3

∴ Volume of the single resultant cube = 27 + 64 + 125 = 216 cm3

∴ Side of resultant cube = (216)1/3 = 6 cm

∴ Surface area of the new cube = 6 × 62 = 216 cm3

2. Simplify

3/7÷9/14×3/2

A. 5/2

B. 1

C. 3

D. 4/3

B. 1

3/7÷9/14×3/2

2/3×3/2

⇒ 1

∴ 3/7÷9/14×3/2=1

3. Akbar, Birbal, Chanakya are partners in a business. During a particular year, Akbar received one third of the profit, Birbal received one fourth of the profit and Chanakya received the remaining Rs. 5000. How much amount of money did Akbar receive?

A. Rs. 1000

B. Rs. 3000

C. Rs. 4000

D. Rs. 5000

C. Rs. 4000

If Akbar and Birbal get a share of 1/3 and 1/4 in profits, then after taking their LCM, their combined profits will be 7/12 and hence Chanakya’s share will be 5/12.

As given in the question, Chanakya earns rupees 5000 as profit, then the total profit will be

5000 × 12/5 = 12000.

Akbar’s share = 1/3 of the profits = 12000 × (1/3) = Rs. 4000.

4.Pure cheese costs Rs.100 per kg. After adulterating it with oil costing Rs.50 per kg, a shopkeeper sells the mixture at the rate of Rs.96 per kg, thereby making a profit of 20%. In what ratio does he mix the two?

A. 4 : 3

B. 23 : 2

C. 3 : 2

D. 3 : 4

C. 3 : 2

Let the amount of cheese and oil taken be ‘a’ and ‘b’ kg respectively.

Given, pure cheese costs Rs.100 per kg and the oil costs Rs.50 per kg.

Total cost price = 100 × a + 50 × b = Rs. (100a + 50b)

He sells them at Rs. 96 per kg, thereby making a profit of 20%.

∴ 96(a + b) – (100a + 50b) = 20% of (100a + 50b)

⇒ 46b – 4a = 20a + 10b

⇒ 36b = 24a

⇒ a : b = 3 : 2

5.A father and a son appear in an interview for two vacancies for the same post. The probability of father’s selection is (1/7) and that of the son’s selection is (1/5). What is the probability that only one will be selected?

A. 2/7

B. 1/8

C. 3/4

D. 9/11

A. 2/7

We know that if father will be selected, the son will be rejected and if the son will be selected, the father will be rejected.

Then, required probability = (1/5)×(6/7)+(4/5)×(1/7)

=6/35+4/35=10/35=2/7

6.A shopkeeper sells a radio on a discount of 8% on marked price and gains a profit of Rs. 25%. If marked price was Rs. 20,000 then, what was the cost price?

A. Rs. 14,750

B. Rs. 14,552

C. Rs. 14,720

D. Rs. 14,850

C. Rs. 14,720

Given that,

Discount = 8% of marked price

= Rs. {20000 × 8/100} = Rs. 1600

∴ Selling price = Marked price – discount = Rs. (20,000 – 1600) = Rs. 18,400

Now, according to the question, a profit of 25% is gained on selling the radio.

We know that, formula:

Profit percentage = (Profit/CP) × 100%

∴ Required CP = Rs. {18400 × 100/125} = Rs. 14,720.

Hence, the cost price of the radio was Rs. 14,720

7.Due to increase of k% in the side, the area of a square increases by 69%. What is the value of k?

A. 30

B. 33

C. 34.5

D. 35

A. 30

Let the original length of side of a square be ‘a’

⇒ Area of original square = a2

⇒ Area of the new square = a2(1 + 69/100)

⇒ Area of the new square = 1.69a2

⇒ Length of side of new square = √(1.69a2)

⇒ Length of side of new square = 1.3a

Since, % increase in the length of side is k

⇒ k = [(1.3a – a)/a] × 100

⇒ k = 30%

∴ the correct option is 1)

8.A cistern has two taps (which fill it in 10 min and 5 min, respectively) and an exhaust tap. When all three taps are opened together, it takes 20 min to fill the empty cistern. How long will the exhaust tap take to empty it?

A. 20 min

B. 10 min

C. 5 min

D. 4 min

D. 4 min

Let the time taken by exhaust tap be x min

Total time taken by all to fill = 1/(1/10 + 1/5 – 1/x)

(1/10 + 1/5 – 1/x) = 20

(1/10 + 1/5 – 1/x) = 1/20

1/x = 1/10 + 1/5 – 1/20

x = 4 min

9. HCF of two numbers is 19 and their LCM is 665. If one of the number is 95, find the other.

A. 19

B. 133

C. 190

D. 77

B. 133

From the given data,

We can write it as LCM × HCF = first number × other number

⇒ 19 × 665 = 95 × other number

⇒ Other number = 12635/95 = 133

10.If n soldiers are arranged in rows of 6, 8, 12 or 16, no soldier is left behind. Then find the possible values of n.

A. 60

B. 72

C. 80

D. 96

D. 96

Number of soldiers must be multiple of 6, 8, 12 and 16 or should be completely divisible by LCM of 6, 8, 12 and 16.

LCM of 6, 8, 12, 16 = 48

Checking the options, we found 96 is completely divisible by 48.

∴ number of soldiers = 96.