Infosys aptitude questions

• Speed = 104 - 50 = 54 km/hr = \(54\times\frac{5}{18} m/s\)=15 m/s
• Distance = \(70+80 = 150\)m

• Time = Distance/Velocity

  Time= \(\frac{150 }{15} \) = 10 seconds.

• Apply VBodmas rule,

• 

• \(6.3\times50\div2.5+18-6\) = \(6.3\times20+18-6=126+18-6=138\)

• Answer is 138.

• \(4\frac{2}{4}+2\frac{1}{3}+2\frac{1}{2}\)=

  \(4+2+2 +(\frac{2}{4}+\frac{1}{3}+\frac{1}{2}\))

• \(8+1+\frac{1}{3}\)

  = \(9+\frac{1}{3}=9\frac{1}{3}\)

• Answer is \( 9\frac{1}{3}\) .

It can empty the tank in 4 hours 8 minutes.

Pipe A:PipeB:Pipe (A+B+C)

6 :8 : 12

Lcm of 6,8,12=24

Their efficiencies are in the ratio =4:3:2

The efficiency of C = 2-(4+3)=-5

therefore, it is an emptying pipe and it takes 4hrs 8minutes to empty the tank(\(\frac{24}{5}\)=4.8)

• \(\frac{5+1}{2}\) = 3

• 3rd number = average = 50

  ----, ----, 50 , ---- ,----

  46 , 48 , 50 , 52 , 54

• Smallest number = 46

• Largest number = 54

• Difference = 54 - 46 = 8.

The number of girls
= \(60 \times \frac{45}{100} =27\)

• \( a : b = 2 : 3 \)

• \( b : c = 4 : 7 \)

• \( a : b : c \) = \( 2×4 : 4×3 : 3×7 \) \( = 8 : 12 : 21 \)

• P\((1+\frac{r}{100})^{3}\)=2P

  \((1+\frac{r}{100})^{3}=2\) times

• After 12 years,

  \((1+\frac{r}{100})^{12}\)=\((1+\frac{r}{100})^{3\times4}=2^{4}\)=16 times

• Amount= \(100\times16=1600\)

• Let the principle = \( x \)

• Time = 20 years

  R =?

And amount A = 2\( x \)

We know that,

A = P + \(\frac{nPR}{100}\)

2\( x \)= \(x + \frac{20 x R}{100}\)

\(\frac{2x}{ x}\) = \(1 +\frac{ 20 R}{100}\)

\(2-1 = \frac{R}{5}\)

\(R =5\times1 = 5\)%

At 5% simple interest, a sum of money doubles itself in 20 years.

Easy method

N = Year

Rate of interest \(R = \frac{100}{N}\)

Rate of interest \(R = \frac{100}{20}= 5\)%.

• Take consecutive numbers such as:

  n(n+1) = 110

  Always remember that the answer is (n+1)

• let the infinite nested radical function = x

  then x is also also under the first root sign as: x =\(\sqrt{110+x}\)

  Then, squaring both sides:

• \(x^{2}\)= 110+ x

  x(x -1) = 110

  Now let x = n + 1

  Then , n(n+1) = 110 With the answer to the infinite nested radical function (x) being equal to n+1

  Solving this, we get n = 10 and n+1 = 11

  So,

• The answer is 11

• Difference=\(P (\frac{r}{100})^{2}\)

• 20=\(P (\frac{5}{100})^{2}\)

  p=\(\frac{20\times10000}{25}=8000\)

  Sum=8000

• Answer is 8000.

• \(\sqrt{1056+\sqrt{1056+\sqrt{1056+....}}}\) = x

• x² - 1056 -x = 0

  (x - 33)(x + 32)

• x = 33

S.I obtained on 1st half year=\(5\) percentage of \(1200=60\)(Half yearly so rate \((\frac{10}{2}=5\))<br>
S.I obtained on 2nd half year=5 percentage of \(1200=60\)(Half yearly so rate \(=5\))<br> Total S.I= \(120\)<br>
C.I obtained on 1st half year\(=5\) percentage of \(1200=60\) (Half yearly so rate \(5\))<br>
C.I obtained on 2nd half year\(=5\) percentage of \(1260=63\) (Half yearly so rate \(=5\))<br> Total C.I\(= 123\)<br>
C.I-S.I\(=123-120=3\)<br>

• Selling price = Cost price - 20 % loss

• 20 % loss means 80 % of cost price

• Here, we need to find out 80 % of 32000

• Selling price \( = \frac{80}{100}\times32000 \)

  = Rs. 25600 .

• Total work – 30\(\times 30\) = 900
• 1st 10 day work – 20\(\times 10\) = 200
• Remaining work = 700
• Now 700 work have to complete in 20 days.
• So Number of men = \(\frac{700}{20}\) = 35 .
• 30 men are already working so now only 5 men are needed.

• Selling price = Cost price - 18 % loss

  18 % loss means 82 % of cost price which is equalt to 2624

  \( \Rightarrow \) 82 % of selling price = 2624

  \( \therefore selling \ price = 2624 \times\frac{100}{18} = 32 \times100 \)

  = Rs. 3200

• \(\frac{4+1}{2}\) = 2.5

• ie,The value between 2nd and 3rd term = 29

• Hence 3rd term = 30

• Largest number = 4th term = 32.

• Let the ages of Shweta and Veena 6 years ago be 5x and 6x.

• Present ages is (5x + 6) and (6x +6) respectively, and their ages 4 years from now will be (5x + 10) and (6x + 10).

  \(\frac{5x+10}{6x+10}= \frac{10}{11}\)

  55x + 110 = 60x + 100

  5x = 10

  x = 2

• Present age of Veena = (6x + 6) = \(6 \times 2+6=12+6=18\).

• Number \( = x \)

  \(x \times \frac{32}{100}=448\) <br> \(x = \frac{44800}{32}=1400\)

  23 % of the number = \(1400 \times \frac{23}{100}=23 \times 14=322\)

• Simple trick

  Instead of taking 23% of a number a student took 32%

  i.e 32% \(\Rightarrow\)448<br> 23% \(\Rightarrow\) ?

• \(\frac{23\times448}{32}\) = 322

• Here man's rate in still water = (15 - 5.5) km/hr = 9.5 km/hr.

• Here minus sign is becasue its against the motion .

  Man's rate against the current = (9.5 - 5.5) km/hr = 4 km/hr.