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 $\sqrt{1825}$.