HCL aptitude questions

• Even numbers between 50 and 150 is 52,54,.....,148.

• Average = \(\frac{52+148}{2}\)

• Average = \(\frac{200}{2}\) = 100

  = \(\frac{200}{2}\) = 100.

• Average of first 140 odd numbers = 140,

• ie, the same number itself.

• Answer is 140.

• Ratio of speed=51:34=3:2

• Since distance is same,speed ant time is inversely related.

  Ratio of time=2:3

  Time taken to travel A to B=2 unit=\(\frac{10}{5}\times2=4 hr\)

  Distance=\(Time\times speed=4\times 51=204km\)

• Answer is 204 km.

• Let the alcohol added be x liter

• ATQ

  \(\frac{9+x}{30+x}=\frac{2}{5}\)

  \(\Rightarrow\) \(45+5 x=60+2 x\)

  \(\Rightarrow\) 3 x = 15

  \(\Rightarrow\) x = 5 liter.

• \(A \rightarrow \)125, \(B\rightarrow 100\)

  B's salary is 25 rupees less than A's salary.

  \(\frac{25}{125}\times100=20\)% less.

• Easy Method :

  Take B's salary as 100

  Then the salary of A is 125

• In such questions, first write as follows

  

  B's salary is 25 rupees less than A's salary.

• Divide the difference 25 by the non-100 number 125. Then multiply by 100

  \(\frac{25}{125}\times100=20\)% less.

• \((a+b)(a-b)=a^2-b^2\)

• \((\sqrt{6}+\sqrt{3})\times(\sqrt{6}-\sqrt{3}) =(\sqrt{6})^2-(\sqrt{3})^2\)

  \( 6-3=3 \)

• Answer is 3.

• SI for 2 year=200

• SI for 1 year=\(\frac{200}{2}=100\)

  CI for first year-SI for first year=100

  CI for second year=\(100+(100\times4\%)=104\)

• Total CI=100+104=204

• Ratio of Suman's and Sumati's 1 day's work \(=\)

  \(\frac{1}{3}: \frac{1}{2}=2: 3\)

• Sum of the ratios \(=2+3=5\)

• Suman's share \(=\frac{2}{5} \times 150=Rs. 60\)

• Here downstream speed will be =(12+6)=18kmph

  Now here 12 minutes=\( \frac{12}{60} \)hours

  Now distance travelled=\( 18\times \)\( \frac{12}{60} \)

  =3.6km

• Distance travelled downstream is 3.6km

• Answer is 3.6 km.

• Here speed along the stream=2.5kmph

  Here speed against the stream=1.5kmph

• Speed of the boat in still water=\( \frac{2.5+1.5}{2} \)

  = 2 kmph

• Answer is 2 kmph

• Total age of father and mother =\( 38×2= 76 \)

• Total age of father, mother and daughter =\( 28×3= 84 \)

• Therefore age of daughter =\( 84−76= 8 \)

• \(a^{2}-b^{2} = (a+b)(a-b)\)

• Similarly ,\((0.85)^{2}-(0.31)^{2}=(0.85+0.31)(0.85-0.31)=1.16\times0.54=0.6264\)

  \(\frac{6.264}{(0.85)^{2}-(0.31)^{2}} =\frac{6.264}{0.6264}=10\)

• Answer is 10.

Calculation :

• Time taken by A to finish the work \(=40\) days

  Time taken by B to finish the work \(=60\) days

• Total time taken \(=\frac{x \times y}{x+y}\) days

  \(=\frac{40 \times 60}{40+60}\)

  \(=\frac{40 \times 60}{100}=24\) days

Hence the answer is option C

• \(a^3+b^3=(a+b)a^2-ab+b^2\),Put this in numerator of question.

• \(\frac{(0.3)^3+(0.7)^3}{(0.3)^2-0.21+(0.7)^2}\)=\(\frac{(0.3+0.7)((0.3)^2-0.21+(0.7)^2}{(0.3)^2-0.21+(0.7)^2}\)

  = 0.3+0.7=1

• Answer is 1.

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

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

  =\(\sqrt{9((25+4})\)

  =\(\sqrt{9×29}\)

  = 3\(\sqrt{29}\)

• Answer is 3\(\sqrt{29}\)

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

• Here,

  (\(\sqrt{7}\)-\(\sqrt{3}\))(\(\sqrt{7}\)+\(\sqrt{3}\))

  ∴ (\(\sqrt{7})^{2}\)-(\(\sqrt{3})^{2}\)

  ⇒ \(7 - 3 = 4\)

• Answer is 4.

• If we take the number as \(x \), then half of the number is \(\frac{x}{2}\) <br>

  \( x \times \frac{39}{100} + 88\) = \(\frac{x}{2} \)

  \( \frac{39x}{100} + 88\) = \(\frac{x}{2} \)

  \(\frac{x}{2} \) - \( \frac{39x}{100}\) = 88

  \(\frac{50x-39x}{100} \) = 88

  \(\frac{x \times11}{100} \) = 88

  That is, 11% of x is 88 <br>

• 11% \( \rightarrow\) 88 <br> 20 % \(\rightarrow\)?

• 20% of the number = <br>

  \(\frac{88\times 20}{11} \)= 160

Easy Method :

• Suppose the number is \( x \).

  Adding 39% of \( x \) to 88 gives half of that number.

  That is, half of \( x \) gets \( \Rightarrow\) 50% of \( x \)

  39% + 88 = 50% of \( x \)

  \( \Rightarrow\) 88 = 11%

  11% \( \rightarrow\) 88 <br> 20 % \(\rightarrow\)?

• To find 20% of the number, just multiply them crosswise and divide them as shown below.

  

• 20% of the number \( \Rightarrow\) \(\frac{88\times 20}{11} \)= 160

• Interest for 3 years=8850 - 8175 = 675

  Amount of 3 years=8175

  Principal=8175-675 = 7500

  Interest for 1 year=\(\frac{675}{3}=225\)

• rate of interest=\(\frac{225}{7500}\times 100=3\%\)

• Answer is 3%.

• A : B = 2 : 3 ---(1)

• B : C = 4 : 5 ---(2)

• C : D = 6 : 7 ---(3)

• Multiply by 8 in equation (1), multiply by 6 in equation (2) and multiply by 5 in equation (3).

  A : B : C : D = 16 : 24 : 30 : 35

• Let A = 16, B = 24, C = 30 and D = 35

• Now,

  (A + B + C)/D

  ⇒ (16 + 24 + 30)/35

  ⇒ 70/35

  ⇒ 2.

• 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.