Railway RRB Group D 2018 Mathematics Practice Questions | Set-7

 B. 216 sq. m

Now the volume of cube = (side)³

As three cubes whose edges are 3 cm, 4 cm and 5 cm respectively are melted to from a single cube

Volume of first cube = 3³ = 27 cm³

Volume of second cube = 4³ = 64 cm³

Volume of third cube = 5³ = 125 cm³

∴ Volume of the single resultant cube = 27 + 64 + 125 = 216 cm³

∴ Side of resultant cube = (216)^1/3 = 6 cm

∴ Surface area of the new cube = 6 × 6² = 216 cm³

B. 1

⇒3/7÷9/14×3/2

⇒2/3×3/2

⇒ 1

∴ 3/7÷9/14×3/2=1

C. Rs. 4000

If Akbar and Birbal get a share of 1/3 and 1/4 in profits, then after taking their LCM, their combined profits will be 7/12 and hence Chanakya’s share will be 5/12.

As given in the question, Chanakya earns rupees 5000 as profit, then the total profit will be

5000 × 12/5 = 12000.

Akbar’s share = 1/3 of the profits = 12000 × (1/3) = Rs. 4000.

 C. 3 : 2

Let the amount of cheese and oil taken be ‘a’ and ‘b’ kg respectively.

Given, pure cheese costs Rs.100 per kg and the oil costs Rs.50 per kg.

Total cost price = 100 × a + 50 × b = Rs. (100a + 50b)

He sells them at Rs. 96 per kg, thereby making a profit of 20%.

∴ 96(a + b) – (100a + 50b) = 20% of (100a + 50b)

⇒ 46b – 4a = 20a + 10b

⇒ 36b = 24a

⇒ a : b = 3 : 2

 A. 2/7

We know that if father will be selected, the son will be rejected and if the son will be selected, the father will be rejected.

Then, required probability = (1/5)×(6/7)+(4/5)×(1/7)

=6/35+4/35=10/35=2/7

 C. Rs. 14,720

Given that,

Discount = 8% of marked price

= Rs. {20000 × 8/100} = Rs. 1600

∴ Selling price = Marked price – discount = Rs. (20,000 – 1600) = Rs. 18,400

Now, according to the question, a profit of 25% is gained on selling the radio.

We know that, formula:

Profit percentage = (Profit/CP) × 100%

∴ Required CP = Rs. {18400 × 100/125} = Rs. 14,720.

Hence, the cost price of the radio was Rs. 14,720

A. 30

Let the original length of side of a square be ‘a’

⇒ Area of original square = a²

⇒ Area of the new square = a²(1 + 69/100)

⇒ Area of the new square = 1.69a²

⇒ Length of side of new square = √(1.69a²)

⇒ Length of side of new square = 1.3a

Since, % increase in the length of side is k

⇒ k = [(1.3a – a)/a] × 100

⇒ k = 30%

∴ the correct option is 1)

D. 4 min

Let the time taken by exhaust tap be x min

Total time taken by all to fill = 1/(1/10 + 1/5 – 1/x)

(1/10 + 1/5 – 1/x) = 20

(1/10 + 1/5 – 1/x) = 1/20

1/x = 1/10 + 1/5 – 1/20

x = 4 min

B. 133

From the given data,

We can write it as LCM × HCF = first number × other number

⇒ 19 × 665 = 95 × other number

⇒ Other number = 12635/95 = 133

D. 96

Number of soldiers must be multiple of 6, 8, 12 and 16 or should be completely divisible by LCM of 6, 8, 12 and 16.

LCM of 6, 8, 12, 16 = 48

Checking the options, we found 96 is completely divisible by 48.

∴ number of soldiers = 96.