Tech Mahindra Placement Papers

Let us assume he buys n goods.
Total CP = 20n
Total SP = 2 + 4 + 6 + 8 ….n terms
Total SP should be at least 40% more than total CP
2 + 4 + 6 + 8 ….n terms ≥ 1.4 * 20 n
2 (1 + 2 + 3 + ….n terms) ≥ 28n
n(n + 1) ≥ 28n
n² + n ≥ 28n
n² - 27n ≥ 0
n ≥ 27

The question is " If he wants to make an overall profit of at least 40%, what is the minimum number of goods he should sell?"
He should sell a minimum of 27 goods.
Hence, the answer is 27.

Choice C is the correct answer

Let track length be equal to T.
Time taken to meet for the first time = T/(relativespeed) = T/6−b or T/b−6
Time taken for a lap for Ajay = T/6
Time taken for a lap for Vijay = T/b
So, time taken to meet for the first time at the starting point = LCM (T/6,T/b) = T/HCF(6,b)
Number of meeting points on the track = Time taken to meet at starting point/Time taken for first meeting = Relative speed / HCF (6,b).
So, in essence we have to find values for b such that 6−b/HCF(6,b) = 2,   or

b−6/HCF(6,b)  =   2

The question is " If two people cross each other at exactly two points on the circular track and b is a natural number less than 30, how many values can b take?"
b = 2, 10, 18 satisfy this equation. So, there are three different values that b can take.
Hence, the answer is 3.

Choice A is the correct answer.

If the overall average weight has to increase after the new people are added, the average weight of the new entrants has to be higher than 40.
So, n > 40

Consequently, m has to be < 10 (as n + m = 50)
Working with the “differences” approach, we know that the total additional weight added by “m” students would be (n - 40) each, above the already existing average of 40. m(n - 40) is the total extra additional weight added, which is shared amongst 40 + m students.
So, m * (n−40)/(m+40)
has to be maximum for the overall average to be maximum.

At this point, use the trial and error approach (or else, go with the answer options) to arrive at the answer.

The maximum average occurs when m = 5, and n = 45

And the average is 40 + (45 – 40) * 545 = 40 + 59

= 40.56 kgs

Choice B is the correct answer.

Let the distance from Delhi to Gurgaon be ‘d’ km.
The first 30 km he travels at his usual speed. However, the remaining ‘d-30’ km he travels at a reduced speed.

To travel ‘d’ km he usually takes T hr. Therefore, to travel ‘d - 30’ km he should ideally take (d−30)T/d
hr. However, this is only if he travels at his usual speed. It is given that he traveled only at 4/5th of his usual speed. Because of this he would have taken 5/4th of the time to travel the remaining distance, i.e., he takes 1/4th of the time extra. This is given to be 45 minutes (or 34th hr)
(1/4)*(d−30)T/d = 34 ........(1)

On the other hand, had the same thing happened after he travelled 48 km, he would have reached only 36 minutes or 34 hrs late. Hence,
(1/4)*(d−48)T/d = 35

........(2)

Dividing (1) by (2) and solving for d, we get d = 120 km.

The question is What is the distance between Delhi and Gurgaon? we get d = 120 km.
Hence, the answer is d = 120 km.

Choice B is the correct answer.

sin 945∘= sin(720∘+225∘)

sin(720∘+225∘) = sin(225∘+2∗360∘)

sin(225∘+2∗360∘) = sin 225∘

sin 225∘=  sin(225∘−360∘)

sin(225∘−360∘) = sin(−135∘) = −sin135∘

−sin135∘= -2/√2