TCS aptitude questions

• New speed = \(\frac{6}{7}\) of usual speed

• Distance = speed \(\times\)Time ------(1)

• Here distance = \(\frac{6}{7}\)S x (T+ 12 ) -------(2)

Comparing above two equations ,

S\(\times\)T =\(\frac{6}{7}\)ST + \(\frac{72S}{7}\)

ST - \(\frac{6}{7}\)ST = \(\frac{72S}{7}\)

\(\frac{ST}{7} \) = \(\frac{72S}{7}\)

= 72 minutes

• Usual time = 1 hour 12 minutes.

• If \(x+\frac{1}{x} = a\) then,\(x^{2}+\frac{1}{x^{2}} = a^{2}-2\)

• In this way,\(x+\frac{1}{x} = 7\),then \(x^{2}+\frac{1}{x^{2}} = 7^{2}-2\)

  =47

• We can find the same by using the equation,\((a+b)^{2}=a^{2}+b^{2}+2ab\)

• Simple interest = \(I= \frac{PNR}{100}\)

• \(I= \frac{12000 \times 3 \times 12}{100}=120 \times 3 \times 12=4320\)

• Answer is \( 4320 \).

• 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=\(130\times-1=-130\)

• Let the number be \(x\)

  \(x \times 66\frac{2}{3}\) % = 96

• \(x \times \frac{1}{100} \times \frac{200}{3}\) = 96

  \( x = \frac{96 \times 3}{2}\) = 48 \(\times\) 3 = 144

  Number = 144

• 25% of the number =

  144 \(\times \frac{25}{100} = \frac{144\times25}{100} = \frac{3600}{100}\)= 36

  Easy trick

• \(\frac{100}{3}\) = \(33\frac{1}{3}\)

  \(33\frac{1}{3}\)+\(33\frac{1}{3}\)= \(66\frac{2}{3}\)

• \(33\frac{1}{3}\)+\(33\frac{1}{3}\)+\(33\frac{1}{3}\) =100%

  \(66\frac{2}{3}\) \(\Rightarrow\) 96

  \(33\frac{1}{3}\)\(\Rightarrow\) \(\frac{96}{2}\) \(\Rightarrow\)48

• 100% \(\Rightarrow\)48 + 48 + 48 = 144

  25% \(\Rightarrow\)\(\frac{144}{4}\) \(\Rightarrow\)36

Suppose pipe A alone takes x hours to fill the tank. <br> Then, pipes B and C will take \( \frac {x} {2} \) and \( \frac {x} {4} \) hours respectively to fill the tank. <br> \( \therefore \frac {1} {x} + \frac {2} {x} + \frac {4} {x} = \frac {1} {5} \) <br> \(\Rightarrow \frac{7}{x} = \frac{1}{x} \) <br> \(\Rightarrow x = 35\) hrs.

Productivity is inversely proportional to the time taken.Thus Productivity of A:B=8;6=4:3 Thus 140 is divided in the ratio 4:3 Thus A get 140x4/7=80 Rs

• 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

• Suppose, first pipe alone takes x hours to fill the tank.

• Then, second and third pipes will take (x -5) and (x - 9) hours respectively to fill the tank.

  \(\frac{1}{x} + \frac{1}{(x-5)} = \frac{1}{(x-9)} \)

  \( \Rightarrow \frac{x-5+x}{x(x-5)} = \frac{1}{(x-9) } \)

  \( \Rightarrow (2x-5) (x-9) = x(x-5) \)

  \( \Rightarrow x^{2} - 18x + 45 = 0\)

  \( (x-15) (x-3) = 0 \)

  \( \Rightarrow x = 15 \) [neglecting \(x\) = 3].

• Take the sum of mony as 100

• At the end of \(5^{th}\) years it becomes 300

• Here interest= 300- 100 = 200

  1 year interest=\(\frac{200}{5}=40\)

• Rate of interest=\(\frac{40}{100}\times100\) =

  40 %.

• Net effect (overall change) when selling at 30% discount after 60% price increase,

  Let, Added price percentage = \( x \)

  Discount price percentage = \( y \)

  Profit / Loss Percentage =\((x-y) - \frac{xy}{100}\)

  \(60-30-\frac{1800}{100}=30-18=12 \%\)

  That is, a formula can be used to these type of questions,

  If \(b \%\) value increases after \(a \%\) increase, net effect = \(a+b+\frac{ab}{100}\)

  If \(b \%\) value decreases after \(a \%\) increase, net effect = \(a-b-\frac{ab}{100}\)

  Profit will be '+' and loss will be '-'

• Here it is \(+ 12 \%\). Therefore it's profit.

  Easy method :

• Suppose the price of a shoe is 100.

  When the price increased by 60%, the price went up to 160.

  After that a 30% discount was given.

  That is, a 30% discount of 160 was given.

  30% of 160 \( \Rightarrow \)\( 160 \times \frac{30}{100} = 48 \)

  i.e. selling price \( \Rightarrow \) 160 - 48 = 112

  This is 12% of the original price of 100. So there was a 12% profit on sales.

• 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

• Simple Interest \(I=\frac {PNR}{100}\)

• \(1800= \frac{P \times 3 \times 8}{100}\)

  \(180000=24P\)

• \(P= \frac{180000}{24}=7500\)

• First year interest=\(2500\times\frac{4}{100}=100\)

• Second year interest=\((2500+100)\times\frac{4}{100}=104\)

• Total interest=100+104=204.

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

• \(\frac{5+10+15+20+x}{5} = 18\)

• \(50+x=90\)

  \(x = 90 - 50 \)

• \(x\) = 40.

A take 6 days to complete the work.B take 8 days to complete the work.Let total work be 24
Then A do 4(24/6) part of work per day and B do 3(24/8) parts of work per day. Now in 3 days they complete 3x(4+3)=21 part of work out of 24. Remaining work=24-21=3 parts In 1 day B do 3 parts of work Then time required for B to complete remaining work = 3/3=1 day

\(A:B=2:3=8:12\)

\(B:C=4:5=12:15\)

Here the values of B are changed to the LCM of 3 and 4. \(A:B:C=8:12:15\)

or Equalize the value of B that is common in both ratios

A : B = 2 : 3 = 8 : 12 (Multiply by 4)

B : C = 4 : 5 = 12 : 15 (Multiply by 3)

A : B :C = 8 : 12 : 15

Sum of present ages of Prabha and Raju = \(8+10 =18\)

Sum of ages needs to be \(30\) .

Difference = \(30-18 =12\)

Since there are two persons, Sum of their ages will be \(30\) after \(\frac{12}{2} = 6\) years.

• While emptying the pipe we use this formula

  \( \frac{1}{x} \)-\( \frac{1}{y} \)=\( \frac{1}{z} \)

  \( \frac{1}{30} \)-\( \frac{1}{20} \)= -\( \frac{1}{60} \)

• Full tank can be emptied 60 Minutes

• In 24 minutes 40% of tank can be emptied.