Capgemini aptitude questions

Calculation :

• When four number a, b, c and d are in proportion we can write

  ab =cd

• Now here we have to find the no. which when subtracted from numerator and denominator leaves the remainder in proportion.

• Thus let that no. Be “x”.

  So the equation will be :

  \(\frac{(36 - x)}{(48- x)} = \frac{(56 - x)}{(76 - x)}\)

  ⇒\(\frac{(36 - x)}{12}= \frac{(56 - x)}{20}\) (if a/b=c/d , then a/(b-a)=c/(d-c))

  Or

• \(720 - 20 x = 672 - 12 x\)

  \(8 x = 48\)

  \(x = 6\)

• Required number = \(6\)

• Alternate Solution:

  \(x = \frac{(ad-bc)}{(a+d)-(b+c)}\)

  = \(\frac{(36x\times 76 - 56 x\times 48 )} {(36+ 76) - (56 +48 )}\)

  = \(\frac{48}{8}= 6\) .

Hence, the answer is option B

• Capacity of the tank (10, 15) = 30

• Time taken by A alone to fill the cistern = \( \frac{30}{3-2} \) = 30 minutes.
• Work Done = Time Taken × Rate of Work
• Rate of Work = 1 / Time Taken
• Time Taken = 1 / Rate of Work
• If a piece of work is done in x number of days, then the work done in one day = 1/x
• Total Wok Done = Number of Days × Efficiency
• Efficiency and Time are inversely proportional to each other
• x:y is the ratio of the number of men which are required to complete a piece of work, then the ratio of the time taken by them to complete the work will be y:x

• 

• Second pipe is twice efficient than the first pipe. So it will take half of the time as 1st pipe take to empty the tank.

• Together they can empty the tank in \( \frac{40}{3} \) min

• Distance =\( \frac{s_{1}\times s_{2}}{s_{1}-s_{2}} \)\( \times t \)

  \( s_{1} \) - speed 1 = 4 kmph

  \( s_{2} \) - speed 1 = 6 kmph

  \( {s_{1}-s_{2}} \) - speed difference = 6 - 4 = 2 kmph

• t = 5+10 = 15 minute ,one reaching early and one reaching late

  Converting it into hour, we get\( \frac{15}{60} \)

  Hence, distance = \( \frac{4\times 6}{2} \) \( \times\frac{15}{60} \) = 3 kilometer

• , Distance between home and school is 3 km.

• Distance travelled = \(\frac{x\times y\times Time Difference}{y-x}\), where \(x , y\) are the given speeds.

• \(\frac{12\times18\times20}{(18-12)60} \)

  \(\frac{12\times18\times20}{6\times60} \)

  12 km.

• Answer is 12 km.

• Let X is the time.

• Interest=\(\frac{2980\times x \times3}{100}=447\)

  x=\(\frac{44700}{8940}=5\)

• Answer is 5 years.

• Loss = CP-SP

• CP of 6 ball = CP of 12 ball – SP of 12 ball

• SP of 12 ball = CP of 6 ball

• Loss % = \(\frac{6}{12}\times 100\)= 50%

• Let their present ages be 4x, 7x and 9x years respectively.

  Then, (4x - 8) + (7x - 8) + (9x - 8) = 56

• 20x = 80

  x = 4.

• Their present ages are 4x = 16 years, 7x = 28 years and 9x = 36 years respectively.

• We know that,

  \(\sqrt{x\sqrt{x\sqrt{x.......}}}\) = x

• \(\sqrt{20\sqrt{20\sqrt{20.......}}}\) =20

• Answer is 20

• Interest of first year=\(\frac{6000\times10}{100}=600\)

• Interest of second year=\(\frac{(6000+600)\times10}{100}=660\)

• Total interest=600+660=1260

• Total amount recievable=6000+1260=7260

• The time in which P Q R S can complete the job are 96 hrs,100 hrs, 90 hrs , and 88 hrs . Therefore S will take the least time to complete the job .
• Work Done = Time Taken × Rate of Work
• Rate of Work = 1 / Time Taken
• Time Taken = 1 / Rate of Work
• If a piece of work is done in x number of days, then the work done in one day = 1/x
• Total Wok Done = Number of Days × Efficiency
• Efficiency and Time are inversely proportional to each other
• X:y is the ratio of the number of men which are required to complete a piece of work, then the ratio of the time taken by them to complete the work will be y:x.

So S is the most efficient

• \(\sqrt{342-\sqrt{342-\sqrt{342-.........}}}\)

  We can also write it as \(\sqrt{342+x}\) = x

  As,the x is nested into it. Solve it

  \(\sqrt{342+x}\) = x

• Square both sides

  342 + x = \(x^{2}\)

  \(x^{2}\)- x - 342 = 0

• When we simplify this, we get

  ⇒ ( x+ 18)(x - 19) = 0

  From this, we get, x = 18 & 19

• We need the smallest number .i.e = 18

• the brackets have to be solved first followed by powers or roots (i.e. of), then Division, Multiplication, Addition, and at the end Subtraction.

• \(36\frac{1}{2}-24\frac{7}{14}-5\frac{1}{4}=(36-24-5)+(\frac{1}{2}-\frac{7}{14}-\frac{1}{4})\)

  =\(7-\frac{1}{4}=\frac{28-1}{4}\)

• Answer is \(\frac{27}{4}\)

• Apply Vbodmas rule,the brackets have to be solved first followed by powers or roots (i.e. of), then Division, Multiplication, Addition, and at the end Subtraction.

• \(720\div360\div120\div30\times18\)

  \(=\frac{720}{360\times120\times30}\times18\)

  =\(\frac{1}{5\times20}=\frac{1}{100}=0.01\)

• Answer is 0.01.

• If the third proportion of two numbers '\(a\)' and '\(b\)' is denoted to be that number 'c' such that,

  \(a:b = b:c\)

• Then, \(c\) = \(\frac{b^{2}}{a}\)

• Here \(a\) = 5 and \(b\) = 10

  \(\Rightarrow \frac{10^{2}}{5}\)

  = 20.

• The third proportion of two numbers '\(5\)' and '\(10\)' is 20.

• Let the person's present age be \( x \) years.
• \(\therefore x + 15 = 4(x - 15)\)

• \(x +15 = 4x - 60\) \(4x - x = 75\)

  \(3x = 75\)

  \(x = 25\)

• Value of rupee coins = 10 = 10 coins

• Value of 50 p coins = 4 = 8 coins

• Value of 25 p coins = Rs. 3 = 12 coins

• Ratio of coins = 10 : 8 : 12 ⇒ 5 : 4 : 6

• Number of rupee coins = \( \frac{5}{15} \times 300\) = 100

• Number of 50 P coins = \( \frac{4}{15} \times 300 \) = 80

• Number of 25 P coins = \( \frac{6}{15} \times 300 \) = 120.

42 × 18 = x × 21

(x = no. of women required)

\( \Rightarrow x\, = \,\frac{{42 \times 18}}{{21}}\)

x = 36 women

• Amount=2500+ 827.5=3327.5

  Then,\(2500[1+\frac{10}{100}]^{n}=3327.5\)

• \([1+\frac{10}{100}]^{n}=\frac{33275}{25000}\)

  \([\frac{110}{100}]^{n}=\frac{1331}{1000}=[\frac{11}{10}]^{3}\)

• n =3.

• Easy method :

  Trick 1

• Average of first \( n \) natural numbers = \(\frac{n+1}{2}\)

  = \(\frac{90+1}{2}\)

  = \(\frac{91}{2}\)= 45.5

Trick 2

• If the question is to find the average of the first \( n \) natural numbers, then half of the natural numbers + 0.5 is sufficient

• \(\frac{90}{2}\) = 45 + 0.5 = \( 45.5 \).