# Cognizant aptitude questions

Cognizant recruitment tests usually contain aptitude questions to assess a candidate's ability to think logically and solve problems. To obtain a position at Cognizant, applicants must pass a written aptitude test. The aptitude test consists of questions covering topics such as Mathematics, English, problem-solving, and logical reasoning. The questions are designed to assess the candidate's ability to solve complex problems and think logically. To prepare for the aptitude test, candidates should practice sample aptitude questions to familiarize themselves with the format and question types. Knowing the topics that are likely to be covered and being well-prepared can help increase the chances of success.

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. $8.2\times8.2+2\times8.2\times1.8+1.8\times1.8 = ?$

• $(a+b)^{2}=a^{2}+b^{2}+2ab$,

• Question show $a^{2}+2ab+b^{2}$ format.

• $8.2\times8.2+2\times8.2\times8.2+1.8\times1.8 =(8.2+1.8)^{2}$

=$10^{2}=100$

1. If $x+\frac{1}{x} = 7$, then $x^{2}+\frac{1}{x^{2}} = ?$

• If $x+\frac{1}{x} = a$ then,$x^{2}+\frac{1}{x^{2}} = a^{2}-2$

• In this way,$x+\frac{1}{x} = 7$,then $x^{2}+\frac{1}{x^{2}} = 7^{2}-2$

=47

• We can find the same by using the equation,$(a+b)^{2}=a^{2}+b^{2}+2ab$

1. Two pipes A and B can fill a tank in 10 min and 20 min respectively. If both pipes are opened together, the time taken to fill the tank is:

• Total capacity = LCM (10, 20) = 20
• Efficiency of A $= \frac{20}{10} = 2$ and Efficiency of B $= \frac{10}{10} = 1$
• Time taken by (A+B) to fill the tank =$\frac{Total Capacity}{Total Efficiency }$= $\frac{20}{2+1}$ = $\frac{20}{3}$ min
1. Pipe A can completely fill an empty tank in 11 hours. Pipe B can empty the same completely filled tank in 15 hours. If they are opened together, in how much time the empty tank will get filled?

• Pipe A (Fill) = 11 hours
• Pipe B (Empty) = 15 hours

⇒ L.C.M. of 11 and 15 is 165

⇒ Let the total work be 165 units

⇒ Efficiency of pipe A = 165/11 = 15 units

⇒ Efficiency of pipe B = 165/15 = – 11 units ((Empty))

⇒ Pipe A and B can fill the tank = 165/(15 – 11) = 165/4 = 41.25 hours

⇒ Pipe A and B can fill the tank = 41 hours 15 minutes

∴ The tank will be filled by pipe A and pipe B in 41 hours 15 minutes.