Hexaware aptitude questions

Let the number be '\(a\)'

75% of a - 20% of \( a\) = 165

=> 55% of \(a\) = 165

=> \(\frac{55}{100}\times a\) = 165

=> \( a\) = \(165\times \frac{100}{55}\)

=> \(a \)= \(300\)

Now 40% of \( a\) =\(\frac{40}{100}\times 300\) = 120

Easy trick

75% - 20% = 55%

55% \(\Rightarrow\)165

40% \(\Rightarrow\)?

= \(\frac{40 \times165}{55}\) = 120

• We know that, \(a^{2}-b^{2} = (a+b)(a-b)\)

  In this way

• \(0.35^{2} -0.03^{2}\)

  \(=(0.35+0.03)(0.35-0.03)\)

  \(=0.38\times0.32=0.1216\)

• \(\frac{0.1216}{(0.35)^{2}-(0.03)^{2}} =\frac{0.1216}{0.1216}=1\)

• We know that, \(a^{2}-b^{2} =(a+b)(a-b)\)

• In this way, \(64^{2}-66^{2} = (64+66)(64-66)\)

  =\(130\times(-2)=2 x\)

  x=\(130\times-1=-130\)

• Part filled by A in 1 min = \( \frac {1} {20}\)

• Part filled by B in 1 min = \( \frac {1} {30}\)

• Part filled by (A + B) in 1 min = \( ( \frac {1} {20} +\frac {1} {30}) = \frac {1} {12} \)

• \( \therefore \) Both pipes can fill the tank in 12 minutes.

• Part filled by P in 1 hour = \( \frac {1} {7}\)

• Part filled by Q in 1 hour = \( \frac {1} {13}\)

• Part filled by (A + B) in 1 hour= \( ( \frac {1} {7} +\frac {1} {13}) = \frac {7+13} {7\times13} \)

  = \(\frac{20}{91}\)

• \( \therefore \) They together can fill the tank in 4 hours 33 min.

• 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

• Loss = 20%

• 20% of 1200 = 10% + 10% = 120 + 120 = 240

SP = 1200 - 240 = 960.

• Suppose the values of Rs.1 coins, 50 paise coins and 25 paise coins are 2x, 3x and 5x respectively.

  \(\therefore\) The number of coins respectively are

  2x, 3x \(\times\) 2, 5x \(\times\) 4 = 2x , 6x and 20x

• According to the question,

  2x + 6x + 20x = 420

  28x = 420 \(\Rightarrow\) x = 15

• Number of Rs.1 coins = 2x = 15 \(\times\) 2 = 30.

• 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.

• \(\dfrac{AB}{A + B } \)

  = \( \dfrac{20 \times 30}{20 + 30 } \)

  = 12 days

• Raman's age=15

• Krishnan's age= 18

• Total age = 15 +18 = 33

• Time to make a total age of 45 =

  \(\frac{45−33}{2}=\frac{12}{2}=6\) years

• Let the principle = \( x \)

• Time = 20 years

  R =?

And amount A = 2\( x \)

We know that,

A = P + \(\frac{nPR}{100}\)

2\( x \)= \(x + \frac{20 x R}{100}\)

\(\frac{2x}{ x}\) = \(1 +\frac{ 20 R}{100}\)

\(2-1 = \frac{R}{5}\)

\(R =5\times1 = 5\)%

At 5% simple interest, a sum of money doubles itself in 20 years.

Easy method

N = Year

Rate of interest \(R = \frac{100}{N}\)

Rate of interest \(R = \frac{100}{20}= 5\)%.

• Let three numbers be A ,B and C . So mean proportional is \( A : B = B : C \) , which implies \( B^2 = A\times C \)

• so \( A\times C = 100 \) . \( A = \frac{100}{C} .\)

• And consider three numbers A, C and D , D is the third proportional ,

  \( A : C = C : 80 \) , \( \frac{100}{C} = \frac{C^2}{80} \) , \( C ^3 = 8000 \) , \( C = 20 \) .

• Therefore \(A = 5\) .

• Ratio = \( \frac{1}{4}: \) \( \frac{1}{6}: \)\( \frac{1}{3} \)

  \( => \) \( \frac{1}{4} \times 12 : \) \( \frac{1}{6} \times 12 : \)\( \frac{1}{3} \times 12 \)

  \( => 3 : 2 : 4 \)

• Therefore, share of Ravi, Amit and Alok is \( 3x , 2x, 4x \).

• So, \( 3x + 2x + 4x = 630 \)

  \( x= 70 \).

• Alok Share = \( 4x \) = \( 4 \times 70 \) = \( 280 \) Rs.

Calculation :

• Number of digits =\( 6\)

  Average = \(25\)

  Sum =\( 6\times25 = 150\)

  \(\therefore\) \(40+10+25+20+35+x = 130+ x=150\)

  x = \(150-130 = 20\)

Hence the answer is option A

• Suppose the number is \(x \).

  \(x \times \frac{20 }{100 }= 160 \)

  \(x\) = \( 160 \times \frac{100 }{20 }\) = 800

Easy Method :

• Number = 100%

  20% = 160

  100% =?

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

  

• Number = \(\frac{160\times 100}{20}=800\)

• \(\frac{5+10+15+20+x}{5} = 18\)

• \(50+x=90\)

  \(x = 90 - 50 \)

• \(x\) = 40.

• Cost Price for article =70.
• \(20\)% of \(70 = (\frac{70}{100}) \times 20 = 7\times2=14\)
• selling price \(= 70-20\)% \(= 70-14 =56.\)

• Mohit's one hour's work = \( \dfrac{1}{3} \) - \( \dfrac{1}{6} \) .
• Raja 's one hour's work =
  \( \dfrac{1}{2} \) - \( \left( \dfrac{1}{3} - \dfrac{1}{6}\right) \) = \( \dfrac{1}{3} \)
• Required time taken by Raja = 3 hrs

• Let \(x\) must be subtracted
  Then \(\frac{14-x}{17-x}= \frac{34-x}{42-x}\)
• \((14-x)(42-x)=(34-x)(17-x)\)
• \(x^2-56x+588=x^2-51x+578\) <br> \(5x=10 \implies x=2\)

• Speed of boat in downstream = \( \frac{18}{2} \) =9kmph

• Speed of boat in upstream = \( \frac{18}{3} \)=6kmph

• Speed of boat in still water =\( \frac{1}{2} \) (9+6)=7.5kmph