# Infosys aptitude questions

Infosys recruiting tests often include a variety of aptitude questions intended to assess a candidate's eligibility for a position. The tests vary depending upon the position applied for and consist of questions related to verbal ability, numerical ability, logical reasoning, data interpretation, attitude, and decision-making. To give yourself the best chance of success, we will look at some of the aptitude questions that are commonly asked in Infosys recruitment tests. Practice aptitude tests to become familiar with the type and level of questions posed. Explore Now!

1. Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:

Part filled in 2 hours = $\frac {2} {6} = \frac {1} {3}$
Remaining part = $( 1 - \frac {1} {3} ) = \frac {2} {3}$
$\therefore$ (A + B)'s 7 hour's work = $\frac {2} {3}$
(A + B)'s 1 hour's work = $\frac {2} {21}$
$\therefore$ C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }
= $(\frac {1} {6} - \frac {2} {21}) = \frac {1} {14}$
$\therefore$ C alone can fill the tank in 14 hours.

1. One runs from one end of a gound to the other at a speed of 6 kmph. At a certain time he reached the other end, and when he ran back at 10 kmph, he arrived 3 minutes earlier than he had run earlier. What is the total distance he covered during his run ?

• Distance =$\frac{s_{1}\times s_{2}}{s_{1}-s_{2}}$

$s_{1}$ - speed 1 = 6kmph

$s_{2}$ - speed 2 = 10 kmph

${s_{1}-s_{2}}$ - speed difference = 10 - 6 = 4

• One is early and the other is late, Therefore t = 3 min

To convert this to hours =$\frac{3}{60}$

Distance =$\frac{10\times6}{4}$$\times\frac{3}{60}$

• Distance = 0.75 km = 750 m

1. The respective ratio of the present ages of Sita and Rita is 4 : 5. Six years hence the respective ratio of their ages will be 6 : 7. What is the difference between their ages ?

Let the age of Sita be x and Rita be y.
Ratio of the present ages of Sita and Rita = $\frac{x}{y}$ = $\frac{4}{5}$
$\therefore$ 5x = 4y
$\Rightarrow x = \frac{4}{5} y$................ (i)
Six years hence the respective ratio of their ages will be $\frac{x + 6}{y + 6}$ = $\frac{6}{7}$

$\therefore$ 7(x+6) = 6(y+6)................... (ii)
7x+42=6y+36
7x+6=6y
$7\times \frac{4}{5}y+6=6y$
$28y+30=30y$
$2y=30$
Solving equation (i) and (ii), we get x= 12, y=15
So, After 6 years there ages will be x+6 = 12 +6 =18 year, y+6 = 15 +6 = 21 year
$\therefore$ Difference between their ages = 21 - 18 = 3 yr

1. If the difference between 72% and 54% of a number is 432 then what is 55% of that number?

• Number $= x$

• $x \times \frac{72}{100}- x \times \frac{54}{100}=432$ <br> $x \times \frac{18}{100}=432$ <br> $x= \frac{43200}{18}=2400$ <br>

• 55 % of the number = $2400 \times \frac{55}{100}= 24 \times 55=1320$

• Easy trick

• 72% - 54% = 18%

18% $\Rightarrow$432<br> 55% $\Rightarrow$?

• = $\frac{55 \times432}{18}$ = 1320

1. Anu can do a job in 30 days, Binu in 45 days and Ramu in 60 days. If Anu is helped by Binu and Ramu every 3rd day, how long will it take for them to complete the job?

• Total work = 180
• Efficiency of Anu = 6, Efficiency Binu = 4, Efficiency Ramu = 3,
• 1st two days =6$\times 2$ =12unit work completed.

3rd day (6+4+3)= 13unit

• For every 3 days

12+13=25unit completed

3 days ==>25 unit

21 days ==>175 units

Remaining 5 unit done by Anu

Then $\frac {5}{6}$ days

• Then total 21$\frac {5}{6}$ days.