 Mathematics Question For RRB NTPC & RRC Group D Exams 2019

## RRB NTPC Mathematics Question

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

This “RRB NTPC Mathematics Quiz 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 Mathematics Quiz on New Pattern | Set – 1

1.The children in three societies A, B and C are in the ratio 2 : 3 : 5. If 10 children are increased in each society, the ratio changes to 4 : 5 : 7. The total number of children before the increase were

A. 5

B. 45

C. 50

D. 30

C. 50

Let the number of children in society A, B and C be 2x, 3x and 5x respectively.

Now, after increasing 10 children in each society, the number of children in A, B and C is 2x + 10, 3x + 10 and 5x + 10 respectively

According to the problem, 2x + 10 : 3x + 10 : 5x + 10 = 4 : 5 : 7

Equating the first of both sides, we get : (2x + 10)/(3x + 10) = 4/5

⇒ x = 5

∴ total number of children initially = 2x + 3x + 5x = 50 children

2.What should come in place of the question mark (?) in the following question?

300% of 150 = ?% of 600

A. 75

B. 45

C. 450

D. 133.5

A. 75

Finding 300% of 150

= (300/100) × 150

= 450

Now 450 = a% of 600

450 = (a/100) × 600

A = 75

3.A man buys two cows for Rs. 50,000. He then sold one cow at a loss of 10% and other cow at a profit of 10%, thereby gaining Rs. 600. Find the cost price of each cow.

A. Rs. 22,000, Rs. 28,000

B. Rs. 26,000, Rs. 24,000

C. Rs. 22,500, Rs. 27,500

D. Rs. 17,500, Rs. 32,500

A. Rs. 22,000, Rs. 28,000

Let the cost price (C.P) of the 1st cow be ‘x’ and that of the 2nd cow be (50,000 – x)

Selling Price (S.P) of the 1st cow = x – 10% of x = 0.9x

S.P of the 2nd cow = (50,000 – x) + 10% of (50,000-x) = 1.1 × (50,000 – x)

∴ According to the given condition,

0.9x + 1.1 × (50,000 – x) = 50,000 + 600

∴ -0.2x = -4,400

∴ x = 22,000

∴ C.P of the 1st cow = 22,000

And C.P of the 2nd cow = 50,000 – 22,000 = 28,000

4.In an examination a student attempted 20 questions correctly and secured 60 marks. If there were two types of questions (1 marks and 5 marks questions), how many questions of 1 marks did he attempt correctly?

A. 5

B. 10

C. 20

D. 40

B. 10

Let x be the number of questions of 1 mark

From the given data, we get

⇒ x + 5 (20 – x) = 60

⇒ x + 100 – 5x = 60

⇒ – 4x = – 40

⇒ x = 10

∴ He has attempted 10 questions of one mark correctly

5.Flipkart offers a discount of 20% on its product during a sale. It further offers a discount of 10% on payment through credit/debit card. It charges Rs. 60 as shipping cost. Find the final price of the mobile phone with price marked as Rs. 13,999 if payment is done through HDFC credit card

A. 11997

B. 10931

C. 11119

D. 10139

D. 10139

M.P = Rs. 13,999 and discount = 20% and 10%

⇒ S.P with 20% discount = 13999 – 20% of 13999

⇒ S.P = 13999 – 2799.8 = 11,199.2

Now, S.P with 10% discount = 11199.2 – 10% of 11199.2

⇒ S.P = 10079.28

With shipping charges included,

∴ S.P = 10079.28 + 60 = 10139.28 ≈ Rs. 10139

6.Fifteen identical robots, each working at the same constant rate, can assemble 50 toys in 8 minutes. How many minutes would it take to 8 such machines to assemble 72 toys?

A. 25

B. 21.6

C. 28.2

D. 30

B. 21.6

To solve Rates and Work problem with multiple workers, modify the standard formula, Work = Rate × Time to this:

Work = Rate × Number of Workers × Time

Solving for an individual robot’s rate, using the fact that 15 robots can assemble 50 toys in 8 minutes

Work = Rate × Number of Workers × Time

⇒ 50 = Rate × 15 × 8

⇒ Rate = 512512 toys per minute

Now,

∵ each robot can assemble 5/12 of a toy in 1 minute. Plug that rate back into the formula, but using the details from the latter part of question

⇒ 72 = 512512 × 8 × Time

∴ Time = 21.6 minutes

7.Three taps can fill a cistern in 15 minutes, 12 minutes and 10 minutes respectively. The cistern being empty, all the three taps are kept open. After 3 minutes, the second tap is closed. Counting time from that moment, the cistern will be full in

A. 5 minutes

B. 2 minutes

C. 3 minutes

D. 1.5 minutes

D. 1.5 minutes

Given, three taps can fill a cistern in 15 minutes, 12 minutes and 10 minutes respectively.

In 1 min,

Tap 1 fills 1/15th of the cistern

Tap 2 fills 1/12th of the cistern

Tap 3 fills 1/10th of the cistern

Also, they are opened at the same time and after 3 minutes, the second tap is closed.

Let the total time taken to fill the cistern after first three minutes be ‘d’ minutes.

d+315+312+d+310=1∴d+315+312+d+310=1

⇒ 4d + 12 + 15 + 6d + 18 = 60

⇒ 10d = 15

⇒ d = 1.5 mins
Thus, the cistern will be filled in 1.5 mins after the 2nd tap is closed

8.Sushi bought 1 sac of rice at Rs. 36/Kg and another sac at Rs. 42/kg. If she mixed both kind of rice and sold it at Rs. 40/kg then find her profit or loss percentage.

Note: 1 sac of rice contains 25kg of rice.

A. 2.5 % loss

B. 2.5% profit

C. No profit no loss

D. 3% profit

B. 2.5% profit

1 sac of rice contains 25 kg.

Hence, her total C.P = 36 × 25 + 42 × 25

⇒ C.P = 1050 + 900 = 1950

She mixes both kind of rice.

S.P = 40 × (25 + 25)

⇒ S.P = 2000

Thus, profit = 2000 – 1950 = 50

∴ Profit % = 50 × 100/1950 = 2.5%

9. A train x meter long runs at 120 km/h. If it cross a platform (x + 12) meter long in 51 seconds, find the length of the platform?

A. 835.5

B. 847.5

C. 912

D. 900.5

B. 847.5

Time taken = Distance/ Speed

Distance covered by train = Length of train + Length of platform

⇒ x + x + 12 = 2x + 12

Speed of train = 120 km/h = (120 × 1000)/3600 ≈ 33 m/s

Time taken by train to cross a platform = Distance covered by train/Speed of train

⇒ 51 = (2x + 12)/33

⇒ x = 835.5 m

∴ Length of the platform = x + 12 = 835.5 + 12 = 847.5 m

10.A mixture of wine and water contains 80% wine. In 50 liters of such a mixture, how many liters of water are required to increase the percentage of water to 50%?

A. 20

B. 30

C. 10

D. 40

B. 30

Total mixture = 50 liters, then,

⇒ Amount of wine in the mixture =

⇒ Amount of water in the mixture (50 – 40) = liters,

Initially we have 40 liters wine and 10 liters water.

Now we are required to have 50% wine and 50% water in new mixture. So we have to have 40 liters of water, equal to the amount of pure wine (which is constant) available in the mixture. Thus we have to add up 30 liters (40 – 10 = 30) water in the original mixture.

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