Wipro aptitude questions

• BODMAS rule, the brackets have to be solved first followed by powers or roots (i.e. of), then Division, Multiplication, Addition, and at the end Subtraction.

• \((6\frac{1}{2}\times3\frac{1}{3})+(8\frac{2}{3}\times 6\frac{1}{2})=6\frac{1}{2}(3\frac{1}{3}+8\frac{2}{3})\)

  =\(\frac{13}{2}\times12=78\)

• Answer is 78.

• \(\frac{1}{2}\) of tank capacity = 120 litres

• Tank capacity = 240 litres

• \(\frac{1}{3}\) of tank Capacity

  = \(\frac{1}{3}\) \( \times \) 240 = 80 litres

• CP = 6000 , SP = 4000 , Loss = 6000 - 4000 = 2000

• Loss % = \(\frac{loss}{CP}\times 100\)

  = \(\frac{2000}{6000}\times 100\)

  = \(\frac{100}{3}\) = 33.33%.

• Answer is 33.33%

• Total number of children = 100% <br>

  Number of boys = 70% <br>

  Therefore, the number of girls = \(100 -70 = 30\)%<br>

  Girls = 30 % \(\rightarrow\)510<br> Boys = 70 % \(\rightarrow\) ?<br>

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

• 

• Number of boys = \(\frac{510\times70}{30}\)= 1190<br>

• 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.

\(\sqrt[3]{0.000001728 }\) = ?

0.012\(\times\)0.012\(\times\) 0.012 = 0.000001728

• Let their present ages be 4x, 7x and 9x years respectively.

  Then, (4x - 8) + (7x - 8) + (9x - 8) = 56

• 20x = 80

  x = 4.

• Their present ages are 4x = 16 years, 7x = 28 years and 9x = 36 years respectively.

• Let P = Rs. 100. Then, S.I. Rs. 60 and T = 6 years.

  \(R= \frac{100 \times 60}{100 \times 6}=10 \%\)

• Now, P = Rs. 12000. T = 3 years and R = 10% p.a.

  \(C.I = 12000[(1+ \frac{10}{100})^3-1]=3972\).

• Total age of 10 students = 150 years

• Total age of 15 students = 240 years

• Total age of 5 new students = 240 - 150 = 90 years

• Average age of 5 new students = \(\frac{90}{5}\) =18 years

• 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.

• P\((1+\frac{r}{100})^{3}\)=2P

  \((1+\frac{r}{100})^{3}=2\) times

• After 12 years,

  \((1+\frac{r}{100})^{12}\)=\((1+\frac{r}{100})^{3\times4}=2^{4}\)=16 times

• Amount= \(100\times16=1600\)

• Take consecutive numbers such as:

  n(n+1) = 110

  Always remember that the answer is (n+1)

• let the infinite nested radical function = x

  then x is also also under the first root sign as: x =\(\sqrt{110+x}\)

  Then, squaring both sides:

• \(x^{2}\)= 110+ x

  x(x -1) = 110

  Now let x = n + 1

  Then , n(n+1) = 110 With the answer to the infinite nested radical function (x) being equal to n+1

  Solving this, we get n = 10 and n+1 = 11

  So,

• The answer is 11

• Let the principal be P and rate of interest be R%.
• Required ratio = \(\frac{\frac{p \times r \times 6}{100}}{\frac{ p \times r \times 9}{100}}= \frac{6pr}{9pr}= \frac{6}{9}=2:3\)
• Answer is \( 2:3 \).

• CP = 2700

  Loss = 15%

• 15% of 2700 = (10% + 5%) of 2700

  = 270 + 135 = 405

• SP = 2700 - 405 = 2295.

• \(\sqrt{1056+\sqrt{1056+\sqrt{1056+....}}}\) = x

• x² - 1056 -x = 0

  (x - 33)(x + 32)

• x = 33

(A+B)’s 2 day’s work = ( 1/8 +1/6)=7/24 Work done in 3 pairs of days = ( 3 x7/24)=21/24 Remaining work =3/24=1/8 Work done by A on 7th day = Remaining work =1/8 Thus work is finished in 7 days

• Total age of 40 people = \(40\times10 = 400\)

• Total age of 41 people = \(41\times 11 = 451\)

• Difference = \(451-400 = 51\)

• \(\therefore \) Teacher age = 51.

(Find the total amount .. take the difference)

• Total work – 30\(\times 30\) = 900
• 1st 10 day work – 20\(\times 10\) = 200
• Remaining work = 700
• Now 700 work have to complete in 20 days.
• So Number of men = \(\frac{700}{20}\) = 35 .
• 30 men are already working so now only 5 men are needed.

• \(speed= 54 \ km/hr =54 \times 5/18 \ m/sec\ = 15 \ m/sec\)
• \(==> time = \frac{180}{15}=12\ sec\)
• Time can be defined as duration in which an object covers a given distance.
• Formula of Time - Time = Distance ÷ Speed.
• Formula of Distance - Distance = Speed × Time.

• Yearly basis Compound interest=\(3000\times8\%=240\)

• Half yearly basis compound interest\((3000\times4\%=120)+(3120\times4\%=124.8)\)=244.8

• Difference=244.8-240=4.8