 Mathematics Question For RRB NTPC & RRC Group D Exams 2019

## Maths Question For RRB NTPC 2019

RRB NTPC Maths Quiz on New Pattern. RRC Group D and NTPC Maths  Study Material. RRB NTPC Mathematics Practice Question Set. RRB NTPC Study Material in Hindi language. Welcome to the www.letsstudytogether.co online Railway NTPC & Group D section. If you are preparing Railway, RRB NTPC, RRC Group D, ALP and RRB Group D Exams 201-20, you will come across a section on “Mathematics”. Here we are providing you with “Maths Question For RRB NTPC”   based on the New Pattern pattern of your daily practice.

This “Maths Question For RRB NTPC on New Pattern” is also important for other Railway exams such as RRB ALP & Technicians Exam 2019-20. आरआरबी NTPC Mathematics प्रश्न हिंदी में Attempt करने के लिये यहाँ क्लिक करें

## RRB NTPC Maths Questions | Set – 5

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.

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.

## RRC Group D Previous Year Question Papers PDF  