PWC aptitude questions

• According to BODMAS 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.

• \(8+32\div(12-8)^2×32-20\)

  =\(8+32\div4^2\times32-20\)

  =\(8+2\times32-20\)

  =\( 52 \)

• Answer is 52.

• Ratio=\(\frac{\frac{P\times X \times 3}{100}}{\frac{P\times X \times 18}{100}}\)=\(\frac{3}{18}\)

• Ratio=1:6

• Answer is 1:6.

• Speed = 15m/s = \(15\times \frac{18}{5} = 54\)km/h

• 2 hours 45 minutes=\(2\frac{3}{4}=\frac{11}{4}\)h

• Distance=\(Time \times speed=\frac{11}{4}\times54=148.5 km\)

• Given \((\sqrt{5}-\sqrt{2})+(\sqrt{5}+\sqrt{2})\)

  =\((\sqrt{5})^{2}-(\sqrt{2})^{2}\)

  = 5 - 2 = 3

• Equation→ (a-b)(a+b)= \(a^{2}-b^{2}\)

• Correct answer is 3

• Interest for 5 year=849

• Interest for 1 year=2% of borrowing=\(\frac{849}{5}=169.8\)

• 100% of borrowing=\(\frac{169.8}{2}\times 100=8490\)

• Net effect (overall change) when selling at 30% discount after 60% price increase,

  Let, Added price percentage = \( x \)

  Discount price percentage = \( y \)

  Profit / Loss Percentage =\((x-y) - \frac{xy}{100}\)

  \(60-30-\frac{1800}{100}=30-18=12 \%\)

  That is, a formula can be used to these type of questions,

  If \(b \%\) value increases after \(a \%\) increase, net effect = \(a+b+\frac{ab}{100}\)

  If \(b \%\) value decreases after \(a \%\) increase, net effect = \(a-b-\frac{ab}{100}\)

  Profit will be '+' and loss will be '-'

• Here it is \(+ 12 \%\). Therefore it's profit.

  Easy method :

• Suppose the price of a shoe is 100.

  When the price increased by 60%, the price went up to 160.

  After that a 30% discount was given.

  That is, a 30% discount of 160 was given.

  30% of 160 \( \Rightarrow \)\( 160 \times \frac{30}{100} = 48 \)

  i.e. selling price \( \Rightarrow \) 160 - 48 = 112

  This is 12% of the original price of 100. So there was a 12% profit on sales.

• The average of first \(n\) odd numbers = \(n\)

• \(n\) = 101

\(\therefore\) The average of the first 101 odd numbers is 101.

• Speed (Rate) of stream =

  \( \frac{1}{2} \) (u-v) km/hr

  = \( \frac{1}{2} \)(27-11)=8km/hr

• Speed of the stream = 8kmph

• Answer is 8km/hr

• so ,here downstream speed = (8+1)=9kmph

• upstream speed=(8-1)=7kmph

  suppost the required distance is x km

  now \( \frac{x}{9} \)+\( \frac{x}{7} \)=1

  16X=63

  X=3.93KM

• Answer is 3.93 km.

• 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

• Average speed=\(\frac{2xy}{x+y}\)km/h

  =\(\frac{2\times 30\times20}{30+20}\)km/h

  =\(\frac{2\times 30\times20}{50}\)=24km/h

• Answer is 24 km/hr
• The average speed is the total distance traveled by the object in a particular time interval.

• No.of coins =\(1\times5x:2\times3x:4\times\;x \)

  \((5x+6x+4x)\rightarrow(465) \)

  15x = 465

  x=31.......(i)

• No.of coins continue =\(5\times31,6\times31,4\times31 \)

  =155,186,124

• Volume of mixture = 30 liters
• Volume of milk in the mixture = 7/10 x 30 = 21 litres
• Volume of water in the mixture = 9 liters
• Let x liters water be added in the mixture.

• Now, 21/ 9+ x = 1/2

  42 = 9 + x

  X = 33 liters.

• \((a+b)^2-(a-b)^2=4ab\)

• \(256-(a-b)^2=4\times48\)

  \(=256-192=(a-b)^2\)

  \((a-b)^2\)=64

• Answer is 64.

• Average=\(\frac{sum}{number}\)

• \(\frac{(35+38+42+25+30)}{5}\)=\(\frac{170}{5}\)=34

• Answer is 34.

• Mother =\( M \)
• Daughter =\( D \)
• Now \( M = 3D \) After twelve years ; \( M+12 = 2(D+12) \) \( M+12=2D+24 \) \( 3D+12=2D+24 \) \( 3D-2D=24-12 \) \( D=12 \) \( M=36 \)
• Answer is 12.

• According to BODMAS 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.

• \(648\div18\div18\)

  \(=36\div18=2\)

• Answer is 2.

• while emptying the pipe we use this formula

  \( \frac{1}{x} \)-\( \frac{1}{y} \)=\( \frac{1}{z} \)

  \( \frac{1}{30} \)-\( \frac{1}{20} \)= -\( \frac{1}{60} \)

• Full Tank can be emptied 60 Minutes

  In 24 minutes 40% of Tank can be emptied.

• Answer is 40%

• Present age \(= x - 10\)
• Ten years from now, he will be \(x - 10 + 10\) or \(x\) years old.
• Answer is \( x. \)

• Apply Vbodmas rule, According to BODMAS 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.

• \(0.5\times100\div\frac{160}{800}+6=0.5\times100\times\frac{800}{160}+6\)

  =\(0.5\times100\times5+6\)

  = \( 250+6=256 \) = \(4^4\)