Deloitte NLA Quantitative Ability Questions and Answers 2025

To make things easy, lets set some notations:
Alipur - A ; Balipur - B; Chandipur - C

Let distance between A and C be z
∴ Distance between B and C = 72 – z
Distance travelled by Arun = AC = z km
Distance travelled by Varun = AB + BC = (72+(72-z)) = (144-z) km

We know, time for both is same
∴ T = T

∴ z = 68 km
∴ Distance between B and C = 72 - z = 72 - 68 = 4 km

Now, if it is asked to calculate total distance traveled by Varun before meeting Arun, do as follows -
Distance travelled by Varun before meeting Arun = AB + BC = 72 + 4 = 76 km

There are 4 stops of 15 min each. So total time spent in stops = 15 x 4 = 60 minutes = 1hour.
Let time taken without stopping be \'T\' hours
So time with stopping = (T+1) hours
Distance is same
∴ D = D
∴ S1 x T1 = S2 x T2
∴ 50 x T = 40 x (T+1)
∴ T = 4 hours
Distance = 50 x 4 = 200 km.

Let Tap A fill the cistern completely in A hours.

Together they fill the cistern in 2.4 hours

∴ Pipe A can fill 1/6^th tank in 1 hour
∴ Pipe a fills the tank completely in 6 hours.
It has a rate of 100-liter water per hour,
So, in 6 hours it gives out 6 x 100 = 600 liters
In 6 hours cistern is full, so capacity = 600 liters

Let the third proportional be y.

A completes the work in 12 days at 8 hour per day = 12 x 8 = 96 hours

B completes the work in 8 days (2 days less than A) working 10 hours every day (i.e. 2 hours more than A per day) = 8 x 10 = 80 hours

50% more than x is 1.5x. Simple, but very useful idea that might help you in solving these kinds of problems.

Let number of girls in class A = x
Let number of boys in class A = y
Total number of students = x + y
Proportion of girls = x/x+y

Number of boys in class B = x
Total number of students in class B = 1.5(x + y)
Proportion of girls = 1 - x/1.5(x+y)

Percentage of boys in the overall student community = x+y/2.5* (x + y) * 100 = 40%

Hence, the answer is 40%

Choice B is the correct answer.

First express the rate of interest as a fraction.

∴		5	x	11		x Installment = Sum
		6		6

Here Sum is the borrowed amount
Installment is the value of each installment (including interest on the installment)

∴		5	x	11		x Installment = 22000
		6		6

∴ Installment = Rs. 14400

Let Oil B quantity be B liters.
Final mixture has '60+B' litres
There is no profit no loss
So, (Oil 1 quantity * rate) + (Oil 2 quantity * rate) = Mix quantity * rate
Oil B is 9 Rs/- cheap. So, the rate of Oil B per litre = 23 Rs/litre
∴ 60 x 32 + B x 23 = (60+B) x 28
∴ B = 48 litres = Oil B quantity

Given :

Let the side of a square inscribed in the circle be ‘a’.

Circumference of a circle ,C = 100 cm

Circumference of a circle = πd

100 = πd

Diameter of a circle ,d = 100/π m

Diagonal of a square = diameter of a circle

√2a = 100/π

[Diagonal of a square = √2a]

√2a = 100/π

a = 100/π × 1/√2

a = 100/√2π

a = (100 ×√2)/ (√2π× √2)

[By rationalising the denominator]

a = 100√2/(2π)

a = (50√2)/π cm

Side of a square = (50√2)/π cm
Hence, the side of a square inscribed in the circle is (50√2)/π cm.

(x - 4) is HCF i.e. a factor of both equations
∴ The equations must get satisfied for x = 4
Also, when x = 4, both equations are equal in value.
Putting x = 4 in both equations
4² - 8(4) + 15 = 4² - p(4) +1
∴ 31 - 32 = 17 - 4p
∴ p = 4