TCS Smart Hiring Quantitative Aptitude Questions and Answers

In the mixture of 45 litre,

Milk = \frac{45}{5}\times 4 = 36ltr, Water = \frac{45}{5}\times 1 = 9ltr

New ratio,

= 9 + 3 : 36 + 4

= 12 : 40 = 3 : 10

Hence, option B is correct.

Product of numbers = 11 x 385 = 4235

Let the numbers be 11a and 11b . Then , 11a x 11b = 4235  =>  ab = 35

Now, co-primes with product  35 are (1,35) and (5,7)

So, the numbers are ( 11 x 1, 11 x 35)  and (11 x 5, 11 x 7)

Since one number lies 75 and 125, the suitable pair is  (55,77)

Hence , required number = 77

LCM of 2, 3, 7 is 42.

=> (777 – 333)/42 =10.57 => 10 Numbers.

Let the numbers be x-4, x-2, x, x+2 and x+4

So, difference between highest and lowest number is (x+4)-(x-4)=8

Explanation :

\frac{m \times T}{W} = constant

where, M= Men (no. of men)

T= Time taken

W= Work load

So, here we apply

\frac{M_{1} \times T_{1}}{W_{1}} = \frac{M_{2} \times T_{2}}{W_{2}}

Given that, M_{1} = 7 men, T_{1} = 12 hours ; T_{2} = 3 hours, we have to find M_{2} =?

Note that here, W_{1} = W_{2} = 1 road, ie. equal workload.

Clearly, substituting in the above equation we get, M_{2} = 28 men.

Total number of possible outcomes = 15+30 = 45

Total number of prizes = 15

p(E) = n(s) / n(E) = 15 / 45 = ⅓

Speed of the bus excluding stoppages = 54 kmph

Speed of the bus including stoppages = 45 kmph

Loss in speed when including stoppages = 54 - 45 = 9kmph

In 1 hour, the bus covers 9 km less due to stoppages.

Hence, time in which the bus stops per hour

= Time taken to cover 9 km

= \frac{distance}{speed} = \frac{9}{54} hour = \frac{1}{6} hour = \frac{60}{6} min = 10 min

3A = 4B = 7C = k,Then A = k/3, B = k/4 and C= k/7.

A : B : C = k/3 : k/4 : k/7 = 28:21 :12.

Cs share = Rs. [427 x (12/61)] = Rs. 84

Let the speed of the motorboat in still water be x kmph. Then,

Speed downstream = (x + 2) kmph; Speed upstream = (x – 2) kmph.

\therefore \frac{6}{x+2} + \frac{6}{x-2} = \frac{33}{60} \Leftrightarrow 11x^2 – 240x – 44 = 0

\Rightarrow 11x^2 – 242x + 2x – 44 = 0  \Leftrightarrow (x – 22)(11x + 2) = 0 \Leftrightarrow x = 22

Hence, the speed of a motorboat in still water = 22 kmph.

Hence, option D is correct.

Let the mother's present age be x years.

Then, the person's present age = \frac{2}{5} x years.

  \frac{2x}{5}+8=\frac{1}{2}(x+8)

2(2x + 40) = 5(x + 8)

=> x = 40.

28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)

Odd days in 1600 years = 0

Odd days in 400 years = 0

5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) = 6 odd days

(31[jan] + 28 [Feb]+ 31[Mar] + 30[April] + 28[May] ) = 148 days = (21 weeks + 1 day) = 1 odd day.

Total number of odd days = (0 + 0 + 6 + 1) = 7 = 0 odd days.

Given day is Sunday

Let the original number be x.

 Final number obtained = 120% of (80% of x) =(120/100 * 80/100 * x) = (96/100)x. 

x-(96/100)x=20

=> x =500

Let C.P. be Rs. x and S.P. be Rs. y.

Then, 3(y - x) = (2y - x)    y = 2x.

Profit = Rs. (y - x) = Rs. (2x - x) = Rs. x.

Profit % = \frac{x}{x} * 100% = 100%

There are 100 people in the room ,so our n value is 100

No. of handshakes   = ^{100}C_{2}

          = 4950

LCM or least common multiple, is the least number which is a multiple of both the original numbers. Hence, the LCM would factor out the HCF of the numbers.

LCM x HCF =  m x n

i.e. n = LCM x HCF / m

∴ The value of n = 3

To solve this question, we can apply the net% effect formula

Net % effect =

Here, x = y = 25%

Applying the net% effect, we get

= 50 + 6.25 = 56.25%

Hence, the total percentage change in area will be 56.25%

Ans: Let the total investment in 2nd scheme be Rs.x, then

(49000 - x) \times \frac{5}{100} + x \times \frac{12}{100} = 4900(49000−x)×1005​+x×10012​=4900

x= 35000

1 Bike + 1 Walk = 5 hours 40 minutes

2 Bike = 5 hours 40 minutes – 3 hours = 2 hours 40 minutes

Hence, 1 way Bike journey takes time = 2 hours 40 minutes ÷ 2 = 1 hour 20 minutes

So, 1 way walk should take (5 hours 40 minutes) – (1 hour 20 minutes) = 4 hours 20 minutes

2 way walk would take time = 4 hours 20 minutes × 2 = 8 hours 40 minutes

2S+F = 79

2F+S = 104

S =18, F = 43

18+43+M/3 = 32

M =35

Virat Kohli's Salary  = 50% of Pujara = 50% of (40% of Raina) 

= 50/100[(40/100)*100] % of Raina = 

 = 20% of Raina