Maths Question For RRB NTPC & Group D Exams 2019 | Set- 5

 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

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.

D. 2625

1.6/100 × x = 42

x = 4200/1.6 = 2625

 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

 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

 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.

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

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

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

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.