RRB NTPC Mathematics Practice Question | Set- 1

 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

A. 75

Finding 300% of 150

= (300/100) × 150

= 450

Now 450 = a% of 600

450 = (a/100) × 600

A = 75

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

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

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

S.P of the 2^nd 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 1^st cow = 22,000

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

 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

 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

 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

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 1/15^th of the cistern

Tap 2 fills 1/12^th of the cistern

Tap 3 fills 1/10^th 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 2^nd tap is closed

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%

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

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.