# Railway RRB Group D 2018 Mathematics Practice Questions | Set-6

## 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-6

1.Pick the odd number from the sequence below:

2, 3, 6, 7, 11, 15, 30

A. 7

B. 11

C. 6

D. 30

B. 11

First: The number is first added to same number.

Second: The answer should added by 1.

Like wise,

⇒ 1 + 1 = 2

⇒ 2 + 1 = 3

⇒ 3 + 3 = 6

⇒ 6 + 1 = 7

⇒ 7 + 7 = 14

⇒ 14 + 1 = 15

⇒ 15 + 15 = 30

So, here 11 is not coming. Odd number is 11

2. A number when divided by 49 leaves 32 as remainder. This number when divided by 7 will have the reminder as

A. 4

B. 3

C. 2

D. 5

A. 4

Let X is the number and when it is divided by 49 it gives remainder 32 and quotient Y.

We know that,

Dividend = Divisor × Quotient + Remainder    ……………. Eq (1)

According to the question:

⇒ X = 49 × Y +32

This equation can be written as:

⇒ X = 7 × (7 × Y) + 28 + 4

⇒ X = 7 × Y’ + 7 × 4 + 4

Taking out 7 common, we get

⇒ X = 7(Y’ + 4) + 4

⇒ X = 7 × Y’’ + 4          ……………….eq (2)

Hence on comparing eq (1) and eq (2) we find that when the same number X is divided by 7 we get remainder as 4.

3. 42 is 1.6% of?

A. 1050

B. 5250

C. 7875

D. 2625

D. 2625

1.6/100 × x = 42

x = 4200/1.6 = 2625

4.Ravi and Ajay start simultaneously from a place A towards B, 60 km apart. Ravi’s speed is 4 kmph less than that of Ajay’s. Ajay after reaching B turns back and meet Ravi at a place 12 km away from B. Ravi’s speed is :

A. 12 kmph

B. 10 kmph

C. 8 kmph

D. 6 kmph

C. 8 kmph

We know that

Speed = Distance/Time

Let speed of Ravi be b kmph.

Then speed of Ajay will be b + 4 kmph.

We have total distance covered by B = 60 – 12 = 48 km

Distance covered by A in same time = 60 + 12 = 72 km.

Hence we have

⇒ 48/b = 72/(b + 4)

⇒ 2/b = 3/(b + 4)

⇒ 2(b + 4) = 3b

⇒ 2b + 8 = 3b

⇒ b = 8 km

5.Golu can do a work in 5 days and Rohit in 15 days. How many days they work together to complete the Two – fifth work?

A. 20

B. 3/2

C. 75/4

D. 15/4

B. 3/2

Golu complete work in = 5 days

Golu’s 1 – day work = 1/5

Rohit complete work in = 15 days

Rohit’s 1 – day work = 1/15

Both working together for 1 hour = (1/5) + (1/15)

Both working together for 1 hour = (3 + 1)/15

Time to complete the work together = 15/4 days

∴ Time to complete (2/5) of total work together = 15/4 × (2/5) = 3/2 days

6.Two numbers are in the ratio 4 : 3 and their LCM is 96. The larger number is

A. 24

B. 30

C. 32

D. 40

C. 32

Let the numbers be 4 k and 3 k for some parameter k.

As, 4 and 3 are co-prime the HCF of the numbers is k.

Also, HCF × LCM = Product of the numbers

⇒ k × 96 = 4 k × 3 k

⇒ 96/12 = k

⇒ k = 8

∴ the numbers are 32 and 24.

7.A boat can reach the destination after traveling 50 hours downstream. The speed of the boat in still water is 100 km/hr. One day after traveling half of the distance the speed of the boat reduced by 40% due to some mechanical problem and reaches 8 hours late than the usual time. Find the speed of the river?

A. 120 km/h

B. 150 km/h

C. 65 km/h

D. 60 km/h

C. 65 km/h

Let speed of the river be R

Downstream Speed of the boat = (100 + R)

Distance travelled = 50 × (100 + R)

Speed of the boat after 40% reduction in speed = 60 km/h

Time taken to complete half of the distance after 40% reduction in speed of the boat = (25 + 8)

50×(100+R)2=33×(60+R)⇒50×(100+R)2=33×(60+R)

∴ R = 65 km/h

8.If x = (0.08)2 , y = 1/(0.08)2 and z = (1 – 0.08)2 – 1, then out of the following, the true relation is

A. y < x and x = z

B. x < y and x = z

C. y < x < z

D. z < x < y

D. z < x < y

x = 0.082

y = 1/(0.08)2 = 10000/64 = 156.25

z = (1 – 0.08)2 – 1

= 1 + 0.082 – 2 × 0.08 – 1

= 0.082 – 2 × 0.08 < x

So, clearly z < x < y

9. A sum of money at simple interest amounts to Rs. 5000 in 3 years and Rs. 6000 in the 5 years. The sum is –

A. Rs. 3500

B. Rs. 3600

C. Rs. 3200

D. Rs. 3000

A. 3500

Simple interest = (P × R × t)/100

Where, P is principal, R is rate and t is time.

Given, sum of money at simple interest amounts to Rs. 5000 in 3 years and Rs. 6000 in the 5 years.

P×R×3100+P=5000∴P×R×3100+P=5000 and P×R×5100+P=6000P×R×5100+P=6000

Subtracting the two equations

⇒ 2PR/100 = 1000

⇒ PR = 50000

Substituting in any one of the equation, we get

P+50000×3100=5000P+50000×3100=5000

⇒ P = Rs. 3500

10. X and Y are two stations which are 320 miles apart. A train starts at a certain time from X and travels towards Y at 70 mph. After 2 hours, another train starts from Y and travels towards X at 20 mph. At what time do they meet?

A. 2 hours

B. 3 hours

C. 4 hours

D. 5 hours

C. 4 hours

⇒ Let the time after which they meet be‘t’ hours.

⇒ Then the time travelled by second train becomes ‘t – 2’.

⇒ Now, Distance covered by first train + Distance covered by second train = 320 miles

⇒ 70t + 20(t – 2) = 320

⇒ 70t + 20t – 40 = 320

⇒ 90t = 360

∴ t = 4

⇒ So, the two trains meet after 4 hours.

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