Bosch aptitude questions

• Let \(5x\) and \(7x\) be the present ages of \(P\) and \(Q\)

  \(7x-(5x+6)=2\)

  \(2x=8\)

  \(x=4\)

• Present ages \(=20,28\)

  the sum \(=48\)

• Answer is 48.

• Rate=time=R

  Interest=\(\frac{360\times R\times R}{100}=291.6\)

  \(R^{2}=\frac{29160}{360}=\frac{486}{6}=81\)

  R=\(\sqrt{81}=9\)

• Rate=9%

• Answer is 9%.

• \((a+b)^2-(a-b)^2=4ab\)

• \(\frac{(18+0.5)^2-(18-0.5)^2}{0.5}=\frac{4\times18\times0.5}{0.5}\)

  \(4\times18=72\)

• Answer is 72.

• \(4+66\div3 of 2=4+66\div(3\times2)\)

• \(4+66\div6 = 4+11=15\)

• Answer is 15.

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

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

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

• 100% of 2400 =2400

  10% of 2400=240

  5% of 2400=120

  115% of 2400=2400+240+120=2760

  100% of 4800=4800

  10% of 4800 =480

  20% of 4800 =960

  5% of 4800 =240

• 75% of 4800=4800-960-240=3600

• 115 % of 2400 - 75% of 4800 =2760-3600=-840

• Profit / Loss Percentage = <br> \(\left [x-y -\frac{xy }{100 } \right ]\)%

  \(\left [10-20-\frac{200 }{100 } \right ]\)%

  \(\left [-10-2 \right ]\)%

  = -12 %

  - 12% means 12% loss

Easy method :

• Suppose the price of the item is 100.

  When the price increased by 10%, the price went up to 110.

  After that 20% discount was given.

  That is, a 20% discount of 110 was given.

  20% of 110 \( \Rightarrow \)\( 110 \times \frac{20}{100} = 22 \)

  That is, the selling price is \( \Rightarrow \) 110 - 22 = 88

• The goods were sold at a loss of Rs 12 over the original price of Rs 100.

• This is 12% of the original price of 100. So there was a 12% loss in sales.

• \( \dfrac{AB}{ A - B} \) = \( \dfrac{50\times30}{50 - 30 } \) = 75
• 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.

• The average marks of 5 boys in the class = 85
• The total marks of 5 boys in the class \(= 5\times85=425 \)

• The total marks of remaining 4 boys in the class \(= 425-100=325 \)

• The new average when a boy who got 100 marks is replaced by another boy = 83

• The new total marks of 5 boys in the class \(= 5\times83 =415\)

• Marks obtained by the new boy \(=415-325=90 \)

• ALTERNATIVE METHOD

• The given question is the replacement type.

• As the average mark is decreased, the marks for the new boy will be less than 100. <br>

• \(A-B=\) number of students \(\times \)change in average <br>

  Where, \(A\) is the boy who got replaced by the boy \(B\)<br> \(100-B=5\times 2\)<br> \(B=100-10=90 \)<br>

  Marks obtained by the new boy = 90

• SI for 2 year=100

• SI for 1 year=50

  CI for first year=SI for first year=50

  CI for second year=50+\((50\times5\%)=52.5\)

  Total CI=50+52.5=102.5

• Answer is 102.5

• Let the number of coins be in the ratio

  8x:9x:11x

• Then

  (₹1)8x+(₹0.5)9x+(₹0.25)11x=366

  8x+4.5x+ 2.75x=366

  15.25x=366

  x=366/15.25

  x=24

• That is,

  1rupee coin : 8x = 8x24= 192 coins

  50 paise coin : 9x = 9 x 24 = 216 coins

  25 paise coin : 11x = 11 x 24 = 264 coins

• Let pipe M fills the cistern in x minutes.

  Therefore, pipe N will fill the cistern in (x+5) minutes.

  Now,

  \( \frac{1}{x} \)+\( \frac{1}{x+5} \)=\( \frac{1}{16} \)

  → x = 29.69 minutes

  Thus, the pipe M can fill in 29.6 minutes,

• so N can fill in 29.6+5 =34.6 minutes.

• Answer 34.6 min.

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

• Second convert time into hour

  =\(\frac{54\times45}{9}\times \frac{5-(-7)}{60}\)

  =\(\frac{54\times45}{9}\times \frac{12}{60}\)

  =54 km

• Answer is 54 km

• as given two pipes can fill a tank in X min now \( \frac{1}{x-40} \)-\( \frac{1}{x-30} \)=\( \frac{1}{x} \)

• After solving we will get

  x= 60 min

• Answer is 60 min.

• Amount = average \(\times \) number

  Average = 80

  Number = 50

• \(\therefore \) Total marks = \(80 \times 50 \) = 4000

• Answer is 4000.

• Let third pipe can fill or empty in x hours.

• \( \frac{(12)(15)(x)}{12(15)+12x+15x} \) = 10

  90x = -1800

  x = -20

• -ve sign, so third pipe can empty alone the tank in alone 20 hours.

• Answer is 20 hour.

• These are consecutive odd numbers.
• average =\(\frac{23+29}{2} = \frac{52}{2}= 26\)
• Answer is 26.

• Given \(\sqrt{4^{2}× 2^{3}+2^{3}× 3^{2}}\)

• \(\sqrt{4^{2}× 2^{3}+2^{3}× 3^{2}}\) =\(\sqrt{2^{3}(4^{2}+3^{2}})\)

  = \(\sqrt{8(16+9})\)

  = \(\sqrt{8(25})\)

  = \(\sqrt{200}\)

  = \(\sqrt{100×2}\)

  = 10\(\sqrt{2}\)

• Answer is 10\(\sqrt{2}\).

• Apply bodmas 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.

• \((891\div297)^2=3^2=9\)

  \(9=\sqrt81\)

• Answer is 81.

• CP = 100% , Profit = 15% \(\Rightarrow\) SP = 115%

  115% = 230

  100% = \(x\)

  \(x\times 115 = 230\times 100\)

  \(x\) = 200

• CP = 200

  200 - 195 = 5 rupees loss

  Loss % = \(\frac{loss}{CP}\times 100\)

  = \(\frac{5}{200}\times 100\)

  = 2.5%.

• Answer is 2.5%.