Mindtree aptitude questions

• Percentage difference <br> 20% - 12% ⟹ 24

  8% ⟹ 24

  The number is 100%.

  8% ⟹ 24 <br> 100% ⟹ ?

• To find the number, simply cross-multiply them and then divide them as shown below.

• 

• The number = $\frac{24 \times 100}{8}$= 300

• Average = $\frac{n+1}{2}$

• $\frac{111+1}{2}$ = $\frac{112}{2}$ = 56

• Answer is 56.

• Given $\sqrt{3} ×\sqrt{12} ×\sqrt{27} ×\sqrt{48}$

• $\sqrt{3} ×\sqrt{12} ×\sqrt{27} ×\sqrt{48}$ =$\sqrt{3} ×\sqrt{4 × 3} ×\sqrt{9 × 3} ×\sqrt{16 ×3}$

  =$\sqrt{3} ×2\sqrt{3} ×3\sqrt{3} ×4\sqrt{3}$

  = $24 × 3 × 3$

  = $216$

• Answer is 216

• Percentage of girls = 100 - 60 = 40%

  40% = 500

  60% = $x$

  $x\times40 = 500\times60$

  $x$ = $\frac{500\times60}{40 }$ = 750

  Number of boys = 60% = 750.

• Easy method :

  Boys are 60%.

  Percentage of girls = 100 - 60 = 40%

  40% = 500

  60% =?

• To find the number of boys, just multiply them crosswise and divide them as shown below.

  

• Number of boys = $\frac{500\times 60}{40}=750$

• Let X is the sum.

• SI=$\frac{x \times6\times5}{100}=1635$

  x = $\frac{163500}{30}=5450$

• Answer is 5450.

• b is in mean proportion to a and c, so $\frac{a}{b} = \frac{b}{c}$ = k

• a = bk and b = ck

  $\frac{ (a-b)^{3} }{ (b-c)^{3} }$ = $\frac{ (bk-b)^{3} }{ (ck-c)^{3} }$ = $\frac{ b^{3} }{ c^{3} }$

  But, $\frac{a}{b} = \frac{b}{c}$

• $\frac{ (a-b)^{3} }{ (b-c)^{3} }$ = $a^{3}$ : $b^{3}$

• Let weigth be 10 gm

• Price = 10$^{3}$ = 1000

• Weigth of their pices = 3 : 4 : 3

• Their values = 3gm, 4gm, 3gm

• Their prices = 3$^{3}$ + 4$^{3}$ + 3$^{3}$

  =27+64+27=118

  Difference =1000-118=882

  % of difference = 882/1000 x 100 = 88.2%

• First year interest=$3500\times\frac{6}{100}=210$

• Second year interest=$3500+210\times\frac{6}{100}=222.6$

• Total interest=210+222.6=432.6

• We know that

  average speed =$\frac{2xy}{x+y}$

• now ATQ,

  average speed =$\frac{2 \times 36 \times 24}{36+24}$

  =$\frac{72 \times 24}{60}$=28.8 km/hr

• Answer is 28.8 km/hr.

• Total age of father and mother =$38×2= 76$

• Total age of father, mother and daughter =$28×3= 84$

• Therefore age of daughter =$84−76= 8$

• Distance =$\frac{s_{1}\times s_{2}}{s_{1}−s_{2}}\ $ $\times$ differenceintime

• Convert 15 minutes into hour.

  =$\frac{12\times 8}{12−8}\times\frac{15}{60}$

  =$\frac{12\times 8}{4}\times\frac{1}{4}=6km$

• Answer is 6 km

• Speed of Emin =120m/12 sec=10m/sec

• Distance completed by Emin in 8 seconds=$speed\times time =10\times8=80$ meter

• Answer is 40 meter.

Remaining distance to finishing point=120-80= 40 meter

• Sum of n consecutive odd numbers =$n^{2}$

  sum of first 45 odd numbers =$45^{2}$=2025

• Sum of n consecutive even numbers =n(n+1)

  sum of first 45 even numbers =45(45+1)= $45\times46$=2070

• difference =2070-2025=45

• $(a+b)^{2}=a^{2}+b^{2}+2ab$,

• Question show the format,$a^{2}+2ab+b^{2}$

$6.4\times6.4+2\times55.04+8.6\times8.6 =(6.4+8.6)^{2}$

=$15^{2}=225$

• Answer is 15.

• Work done by tap A in one hour = $\frac{1}{50}$

• Work done by tap B in one hour = $\frac{1}{60}$

• Work done by tap C in one hour =$\frac{1}{75}$

• Work done if all the three taps are opened together = $\frac{1}{50}$ + $\frac{1}{60}$+ $\frac{1}{75}$ $= \frac{6+5+4}{300} = \frac{15}{300} = \frac{1}{20}$

• Time taken to fill the tank if all of them are opened together =$\frac{1}{\frac{1}{20}} = 20$ hours

• A can fill the tank in 10 hours

  A's one hour work = $\frac{1}{10}$

• B can fill the tank in 12 hours

  B's one hour work = $\frac{1}{12}$

• C can fill the tank in 15 hours

  C's one hour hour work = $\frac{1}{15}$

  Then (A + B + C)'s one hour work = $\frac{1}{10}$+$\frac{1}{12}$ + $\frac{1}{15}$= $\left ( \frac{6+5+4}{60}\right )$ = $\frac{1}{4}$

• Hence, (A+B+C) can fill the tank in 4 hours.

• Speed downstream = (15 + 6) km/hr = 21 km/hr.

• Time taken to travel 84 km downstream hrs=($\frac{84}{21}$)=4hrs

• Answer is 4 hours.

• A’s 2 days work = B’s 3 days work

• $\therefore$ Time taken by A = 8 days

• $\therefore$ Time taken by B =$\dfrac{8}{2} \times 3 = 12$days

• Ratio = 11x : 6x

  Five years ago (11x - 5) = 2(6x -5)

  11x - 5 = 12x - 10

  x = 5

  So, the age of mom = 55 and age of son = 30

• From 20 years, mom = 75 and son = 50

  So, Son's age will the Mother's age be twenty years from now will be 1.5 times

• Answer is 1.5 times.

• $\sqrt{56+\sqrt{56+\sqrt{56+....}}}$ = x

• x² - 56 -x = 0

  (x - 8)(x + 7)

• x = 8 (Take the largest number in the case of addition )