TCS NQT Foundation Numerical Ability Questions and Answers

Let the length of the track be x. When they meet at the first meeting point, Brinda has run 100 meters, while Shanti has run x/2 100 meters. By the second meeting point, Shanti has run 150 meters, while Brinda has run x 150 meters.

Since they run at a constant speed, we can set up a proportion:

Distance covered by Brinda during first Meet/Distance Covered by Shanti during first meet = Distance covered by Brinda during Second Meet/Distance Covered by Shanti during the second meet

100/ (x 150) = (x/2 100) / 150

We get, X = 350.

The only combination when Ria has twice as many marbles as Ruhi as they pick three and two bags respectively is when Ria picks the bags with 34, 25 and 19 bags, while Ruhi picks the bags with 21 and 18 marbles.

The bag left out has to be the bag with chipped marbles.

Thus, there are 23 chipped marbles.

After meeting together they meet again in 14-5 = 9 sec

so with in 9 sec they completed on a lap together

one lap = 135 m

a + b = 135/9

=> a + b = 15 m/s - eq 1

first time they meet after 5 sec

so ditance covered by both = (15) * 5 = 75m

Distance between them was 75m in that direction

in other direction distance was 135 - 75 = 60 m

Now if b had started in opposite direction so to meet a

he has to cover the distance covered by a + 60m

b*60 = (a)60 + 60

=> (b-a)60 = 60

=> b-a = 1 - eq 2

Adding eq 1 + eq 2

2b = 16 m/s

=> b = 8 m/sec

Let the length, breadth, and height of the room be 3, 2, 1 unit respectively.
Area of walls = 2(l + b) × h = 2(3 + 2) × 1 = 10 sq. unit.
Now, the length, breadth, and height of the room will become 6, 1, and (1/2) respectively.
Now, area of walls = 2(6+1)×(1/2) = 7 sq. unit.
% decrease in the area of walls = (10−7)×100/10 = 30%

Given 1 ,7 ,8 ,49 ,50 ,56 ,57 ,343 ,344 ,350 ,351 ,392 ,393 ,399 ,400 ,…….. The above sequence contains sums of distinct powers of 7 in the increasing order (7^0, 7^1, 7^1 + 7^0, 7^2, etc). What is the value of term number 36?

Now seven power is present in 1st, 2nd,4th, 8th, 16th, 32nd, 64th. Power of 7 will be 0,1,2,3,4,5,6

Now term 1,2,4,8,16,32 = power to 7-----------(1)

We need to find 38 th term

38 th term = 32 th term + 6 th term

By eqn 1 , 32 th term = 7^5 and 6 th term is 56

So 36 th term = 7^5 + 49 = 16856

= 16856

So we can do it as

32nd term = 7^5

33 rd term = 7^5 + 1

34 th term = 7^5 + 7

35 th term = 7^5 + 8

36 th term = 7^5 + 49 = 16,856

(I) 6p^{2} + p - 1 = 0

By middle term splitting (factorization)

(3p - 1)(2p + 1)

p = \frac{1}{3},-\frac{1}{2}

(II) 14q^{2}+9q+1 = 0

By middle term splitting (factorization)

(7q + 1)(2q + 1)

q = -\frac{1}{7},-\frac{1}{2}

therefore p≥q

Let the age of Anubhav be x years.

Then the age of Akshat is (56 - x) years.

After the time span of 4 years,

3(x + 4) = 56 - x + 4 = 48

4x = 56 + 4 - 12

x = 12 years

Let the price be Rs 100 per 10 kg packet and the sales be 100 packets.

So the money obtained after selling all the 100 packets

100 x 100 = Rs 10,000

Now, the increased price is Rs 110 and the sales are 95 packets

110 x 95 = Rs 10,450

profit = 10,450 - 10,000 = Rs 450

∴ % profit = \frac{450\times 100}{10000} = 4.5%

This problem can be solved using a theorem which states that

"If two Quantities X and Y are in the ratio x:y. Then

X + Y : X - Y :: x + y : x - y"

Therefore following the above theorem we have,

9 - 5 = Rs 4200 ∴ 5 = \frac{4200}{9 - 5}\times 5 = Rs 5250

For each selection of two persons, there will one "Wakanda Forever". So, no. of "Wakanda Forever" in the part ^{n}C_{2}, where n = no. of persons

Now, ^{n}C_{2} = 210 (given)

or, \frac{n\times (n - 1)}{2} = 210

or, n x (n - 1) = 2 x 2 x 3 x 5 x 7 = 21 x 20

n = 21