Mettl Quantitative Aptitude Questions

Let Shobha's age = x,

her brother's age = 2 - 4

Sobha's father's age x +38
Sobha's mother's age = (x – 4) + 36 = x + 32

%3D

Sobha's father's age - Sobha's mother's age

= (x +38) - (x + 32)

=6

We do not have the number of boys and girls and hence average age of the whole class cannot be found.

The committee should have 4 engineers. But the committee must contain exactly 1 civil engineer.

Hence, select 3 software engineers from 10 software engineers and select 1 civil engineer from 6 civil engineers.

Number of ways this can be done

=^{10}C_{3}\times^{6}C_{1}
= 720

The word 'MATHEMATICS' has 11 letters. It has the vowels 'A','E','A','I' in it and these 4 vowels must always come together. Hence these 4 vowels can be grouped and considered as a single letter. That is, MTHMTCS(AEAI).

Hence we can assume total letters as 8. But in these 8 letters, 'M' occurs 2 times, 'T' occurs 2 times but rest of the letters are different.

Hence,number of ways to arrange these letters
=\frac{8!}{2!2!}=10080

In the 4 vowels (AEAI), 'A' occurs 2 times and rest of the vowels are different.

Number of ways to arrange these vowels among themselves = \frac{4!}{2!}=12

Hence, required number of ways = 180080*12 = 120960

water filled by the inlet pipe in 24 hours = water emptied by the leak in 24−6=18 hours.

Therefore, water emptied by the leak in 6 hours= water filled by the inlet pipe in 8 hours

i.e., capacity of the tank
= water filled by the inlet pipe in 8 hours=8×60×4=1920 litre.

Given that, time taken to travel upstream = 2 × time taken to travel downstream

When distance is constant, speed is inversely proportional to the time
Hence, 2 × speed upstream = speed downstream

Let speed upstream =x

Then speed downstream =2x

we have, 1/2(x+2x) = speed in still water
⇒1/2(3x)=7.5
⇒3x=15
⇒x=5

i.e., speed upstream = 5 km/hr

Rate of stream =1/2(2x−x)=2.5 km/hr

Speed downstream = 15 km/hr
Rate of the current= 3/2 km/hr

Speed in still water = 15 - 3/2 = 27/2 km/hr

Rate against the current = 27/2 km/hr - 3/2 = 12 km/hr

Let the marks secured by them be x and (x+9)

sum of their marks = x + (x+9) = 2x +9

Given that (x+9) was 56% of the sum of their marks.

= (x + 9) = 56/100(2x + 9)
= (x + 9) = 14/25(2x + 9)
= 25x + 225 = 28x + 126
= 3x = 99
= x = 33
Then (x+9) = 33 + 9 = 42
Hence their marks are 33 and 42

Initial investment capital ratio of A and B = 3 : 5

Let initial capital of A and B be 3x and 5x respectively.

Amount that C invested after 6 months =5x (since it is equal to B's investment)

Ratio in which profit should be distributed after 1 year
=3x*12:5x*12:5x*6
=6:10:5

8444 ÷ 7 = 1206, remainder = 2

7 - 2 = 5
Hence, 5 should be added to 8444 such that the sum is completely divisible by 7.

43÷6 = 7, remainder = 1. Hence 5 should be added to 43 to get the lowest number divisible by 6 between 43 and 200.

Therefore, lowest number divisible by 6 between 43 and 200
= 43 + 5 = 48

200÷6= 33, remainder = 2. Hence 2 should be subtracted from 200 to get the highest number divisible by 6 between 43 and 200.

Therefore, highest number divisible by 6, between 43 and 200
= 200 - 2 = 198

This is AP with a = 48, d =6, l = 198

Number of terms = \frac{(l-a)}{d} + 1
= 26

By rule of alligation,

% Concentration
of water in pure water (100) % Concentration
of water in the given mixture (10)
Mean % concentration
(20)
20 - 10 = 10 100 - 20 = 80
=> Quantity of water : Quantity of the mixture = 10 : 80 = 1 : 8

Here quantity of the mixture = 16 litres
=> Quantity of water : 16 = 1 : 8
=> Quantity of water
=16×1/8=2 litre

LCM of 12, 14, 18 and 22 = 2772

Hence the least number which will be exactly divisible by 12, 14, 18 and 22 = 2772

2772 ÷ 2 = 1386

1386 is the number which when doubled, we get 2772

Hence, 1386 is the least number which when doubled will be exactly divisible by 12, 14, 18 and 22.

LCM of 252, 308 and 198 = 2772

Hence they all will be again at the starting point after 2772 seconds.
i.e., after 46 minutes 12 seconds.

Total number of outcomes possible when a coin is tossed = 2 (∵ Head or Tail)
Hence, total number of outcomes possible when 5 coins are tossed, n(S) = 2^5

E = Event of getting exactly 2 heads when 5 coins are tossed.
n(E) = Number of ways of getting exactly 2 heads when 5 coins are tossed = ^{5}C_{2}

P(E = \frac{n(E)}{n(S)}=\frac{^{5}C_{2}}{2^{5}}=\frac{5}{16}