MU Sigma aptitude questions

• Let equation a+b+\((\frac{a\times b}{100})\) %
  a=50
  b=50

• 50+50+\((\frac{50\times 50}{100})\) % = 125%

• Simple Trick

  length and breadth increased by 50% \(\Rightarrow\)150%

• Area = \(100 \times \frac{150}{100}\times\frac{150}{100}\) = 225

  Increased area = 225 - 100

  \(\Rightarrow\) 125%

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

• \(4+66\div6 = 4+11=15\)

• Answer is 15.

• 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

• easy method

• x - y -\( \frac{xy}{100}\)

  10 - 15 - \(\frac{10\times 15}{100}\) = -6.5 % , decrease of 6.5%

Simple trick

• Let the Orginal price = 100

  Increased by 10% = 100+10 = 110

  {110 \(\times \frac{15}{100 }=16.5\)}

  Decreased by \( 15% = 110 - 16.5 = 93.5 \)

  Total % decrease = 100% -93.5 = 6.5%

• Upstream speed = 1/1 =1 km/hr, Downstream speed = \( \frac{1}{20/60} \)=3 km/hr.

  Speed in still water =\( \frac{1}{2} \)(1+3)

  = 2 \(km/hr \)

• \( .˙. \) Time taken to cover 8 km in still water = \( \frac{8}{2} \)=4 hours

• Answer is 4 hour.

• Take the sum of money as 100

  At the end of third years it becomes \(100\times\frac{8}{5}\)=160

• here interest=160-100=60

  1 year interest=\(\frac{60}{3}=20\)

• rate of interest=\(\frac{20}{100}\times100=20\%\)

• Given, the numbers are 54, 71, 75, and 99.

• Assume that x be number subtracted from each of the numbers 54, 71, 75 and 99 so that their remainders will be a proportion.

  ∴ 54 - x , 71 - x , 75 - x , and 99 - x are in proportion.

  ∴ (54 − x)/(71 − x) = (75 − x)/(99 − x)

  ⇒ (54 − x)(99 − x) = (75 − x)/(71 − x)

  ⇒ x\(^{2}\)− 99x − 54x + 5346 = x\(^{2}\) − 75x − 71x + 5325

  ⇒ −153x + 146x = −5346 + 5325

  ⇒ −7x = −21

  ⇒ x = 3

• ∴ When x = 3 subtracted from each of the numbers so that the remainders will be a proportion.

• Let CP = 100% , MP = 140%

• Discount = 10% \(\Rightarrow\) SP = 140 - 14 = 126%

• Profit % = \(\frac{SP - CP}{CP}\times100\)

  = \( \frac{26}{100}\times100\) = 26%.

• x be Rajan's age and y be his son's

• Total age = \(x+y=2 \times 20\) \(x=3\times y\) ,\( y=10 \) , \( x=30 \) ,\( x+5=35. \)

• Age of Rajan = 35

• Initially alcohol = \(\frac{2}{5} \times 25\) ltr and water is 15 ltr.

• To make a solution of 60% alcohol = \(\frac{10+x}{25+x} = \frac{60}{100}\).

  \(100\times(10+x) = 60\times(25+x)\)

  \(1000+100x = 1500 + 60x\)

  \(100x - 60x = 1500 - 1000\)

  \(40x = 500\)

  \(x = 12.5\)

• 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

• Let CP = 100% , then MP =130%

• Discount = 20% \(\Rightarrow\) SP = 130 - 20% of 130

= 130 -26 = 104%

• Hence Profit = 4%.

• Age average of 20 people = 24

• Age average of 19 people = 25

  The average age of a person decreases by 1 year.

  ∴ (1 × 20 )+ (25 × -1) = -5

• Age can never be a negative number

• ∴ Age of lost child = 5

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

• x² - 1056 -x = 0

  (x - 33)(x + 32)

• x = 33

• In one hour, X can fill \( \frac{1}{6} \) of the tank

  In one hour, Y can fill \( \frac{1}{10} \)of the tank

  In one hour, Z can empty \( \frac{1}{4} \) of the tank

• After one hour when all X, Y and Z is open

  Level of water = \( \frac{1}{6} \) + \( \frac{1}{10} \) - \( \frac{1}{4} \) = \( \frac{1}{60} \)

  After 60 hours, Level of water = \( 60 \times \frac{1}{60} \)= 1

• It will take 60 hours to fill the tank when all three pipes are opened.

• Part of cistern filled in 6 minutes = \( \frac{2}{5} \)

  Part of cistern filled in 1 minute =\( \frac{2}{30} \) = \( \frac{1}{15} \)

• Total time required to fill the cistern = 15 minutes

  Remaining part to be filled = \( 1 -\frac{2}{5} = \frac{3}{5} \)

  Required time to fill the remaining part = \( 15 - 6 = 9 \) minutes

• Answer is 9 minutes.

• Let the speed of the stream x mph. Then ,

  Speed downstream =\( (10+x) \) mph,

• Speed upstream =(10-x) mph \( \frac{7.5}{10-x} \) -\( \frac{7.5}{10+x} \)=\( \frac{60}{60} \)

  \( \Rightarrow \)\( 15x= \)(100-x\( ^{2} \))

  \( \Rightarrow \) x\( ^{2} \) +15x-\( 100 \)

  \( \Rightarrow \) \( x \)\( (x+20)-5(x+20) \)

  \( \therefore \)\( (x+20)(x-5) \)

  so,\( x=5 \) mph

• Answer is 5mph.

• Distance=860m

  Relative speed(same direction)=\(S_{1}-S_{2}\)=101-11=90 km/h

• Convert into m/sec

  \(90\times\frac{5}{18}\)=25m/s

  Time=\(\frac{Distance}{Speed}\)

  =\(\frac{150}{25}\)=6 sec

• Answer is 6 sec.

• One day's work of Rama and Shyama =\( \frac{1}{12} \)

• two day's work of rama = \( \frac{2}{20} \) = \( \frac{1}{10} \)

• \( \therefore \) \( \frac{1}{10} \) work is done by by Rama and Shyama

• Required time taken = 12 + 12 x \( \frac{1}{10} \) = 13\( \frac{1}{5}\) days

• Given \(\sqrt{3^{2}×5^{2}+2^{6}×5^{2}}\)

• \(\sqrt{3^{2}×5^{2}+2^{6}×5^{2}}\) = \(\sqrt{5^{2}(3^{2}+2^{6})}\)

  = \(\sqrt{25(9+64)}\)

  = \(\sqrt{25 ×73}\)

  = \(\sqrt{1825}\)

• Answer is \(\sqrt{1825}\).