Accenture aptitude questions

Easy Store sold a computer for 8% profit = 108 %

If it was sold for 23% profit = 123 %

Easy Store sold a computer for 8% profit. If it had been sold for 23% profit, they would have got Rs 12,600 more.

108% + 12600 = 123 %

123% - 108% = 12600

15 % = 12600

Purchase price = Actual price = 100%

15% = 12600

1% = \(\frac{12600}{15}\) = 840

100% = 84000

Actual price = 100% = 84000

• Apply vbodmas rule,

  V - Vinculum

  b - Bracket

  o - of

  d - Division

  m - Multiplication

  a - Addition

  s - Substraction

• \(\frac{3}{5}\times2\frac{2}{5}\div\frac{4}{25}+320=\frac{3}{5}\times\frac{12}{5}\div\frac{4}{25}+320\)

• \(\frac{3}{5}\times\frac{12}{5}\times\frac{25}{4}+320=9+320\)

• Answer is 329.

• Apply vbodmas rule,

  V - Vinculum

  b - Bracket

  o - of

  d - Division

  m - Multiplication

  a - Addition

  s - Substraction

• \(1832\div458\div8\times32=4\div8\times32\)

  =\(0.5\times32=16\)

• Answer is 16.

• Pipe A (Fill) = 11 hours

• Pipe B (Empty) = 15 hours

  ⇒ L.C.M. of 11 and 15 is 165

  ⇒ Let the total work be 165 units

  ⇒ Efficiency of pipe A = 165/11 = 15 units

  ⇒ Efficiency of pipe B = 165/15 = – 11 units ((Empty))

  ⇒ Pipe A and B can fill the tank = 165/(15 – 11) = 165/4 = 41.25 hours

  ⇒ Pipe A and B can fill the tank = 41 hours 15 minutes

  ∴ The tank will be filled by pipe A and pipe B in 41 hours 15 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.

• If compound interest is calculated every half year, \(P[1+\frac{R}{200}]^{2N}\)

  P=400, R=10, N =1

  \(400[1+\frac{10}{200}]^{2}=441\).

• If the number is = \(x \),

  \(\frac{80}{100}\times x+80=x\)

  \(\frac{80x}{100}+80=x\)

  \(\frac{80x + 8000}{100}=x\)

  \(80x + 8000 = 100 x\)

  \(8000 = 100x - 80x\)

  \(8000 = 20x\)

• \(x\) = \( \frac{8000}{20}\)= 400.

Easy Method :

• Suppose the number is \(x \).

  Adding 80% of x with 80 gives the same number.

  That is, 20% of x is 80.

  Number = 100%

  20% = 80

  100% =?

• To find the number, simply multiply them crosswise and divide them as shown below.

  

• Number = \(\frac{80\times 100}{20}=400\).

Let total work completed in x days. So A worked for all x days.
B for (x-6) days and C for (x-3) days.
So
x/12 + (x-6)/15 + (x-3)/25 = 1.
Solving for x, we get x = 8
So A completed x/12 = 8/12 = 2/3rd of work.

Let the principal be P,
Rate of interest be R% and Time = T
P = SI
N = R
SI = (P × T × R) / 100

\(P = P\times \frac{R^2}{100}\)
\(R^2 = 100\)
R = 10

• Here the efficiency of A = 6.
• Efficiency of B = 5.
• Efficiency of C = 4.
• Total efficiency is, 6 + 5 + 4 = 15.
• Then the total work is, 15 \(\times\) 24 = 360.
• B alone can do the work in, \(\frac{360}{5}\)= 72 days.

• We know that,

  \(\sqrt{x\sqrt{x\sqrt{x.......}}}\) = x

• \(\sqrt{20\sqrt{20\sqrt{20.......}}}\) =20

• Answer is 20

• \(\sqrt{342-\sqrt{342-\sqrt{342-.........}}}\)

  We can also write it as \(\sqrt{342+x}\) = x

  As,the x is nested into it. Solve it

  \(\sqrt{342+x}\) = x

• Square both sides

  342 + x = \(x^{2}\)

  \(x^{2}\)- x - 342 = 0

• When we simplify this, we get

  ⇒ ( x+ 18)(x - 19) = 0

  From this, we get, x = 18 & 19

• We need the smallest number .i.e = 18

• 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)

Purchase price = 100%

13% profit = 100% + 13% = 113%

113%=2034

1%=\(\frac{2034}{113}\)=18

100%=1800

Purchase price = 1800

Profit from selling them for Rs. 2448 = 2448-1800 = 648

Profit Percentage =\(\frac{profit}{purchase price}\times100\)

=\(\frac{648}{1800}\times100\)=36%.

• Number = \(x \) \(x \times \frac{40}{100} +42=x\) \(x \times \frac{2}{5} +42=x\) i.e, \(\frac{2x}{5}+42=x\) \( 2x + 210 = 5x \) \(210 = 3x \) \(\therefore x = 70 \)

Easy Method :

• Suppose the number is \( x \).

• Adding 40% of \( x \) to 42 gives the same number.

  That is, 60% of \( x \) is 42

  Number = 100%

  60% = 42

  100% =?

• To find the number, simply multiply them crosswise and divide them as shown below.

• 

• Number = \(\frac{42\times 100}{60}=70\)

Let the age of younger brother be x years. Hence, the age of the elder brother will be \((x + 8)\) years.

12 years ago, the younger brother was \((x – 12)\) years and the elder brother was\( (x-4)\) years of age.

\((x – 4) = 2 \times (x – 12) \)

\(x – 4 = 2x – 24\)

x = 20 years

• Let the son’s age 15 years ago be x years.

• Therefore, the father’s age 15 years ago was 3x years.

• The present age of father and son is (3x + 15) and (x + 15) respectively.

• 3 years hence their ages will be (3x + 18) and (x + 18) respectively. \((3x + 18) = 2 (x + 18) =3x + 18 = 2x + 36\)

• The sum of the present ages of father and son = (3x + 15) + (x + 15)

  = \(3 \times 18 + 15\)+ (18 + 15)

  = 102.

Distance =\( \frac{s_{1}\times s_{2}}{s_{1}-s_{2}} \)\( \times t \)

\( s_{1} \) - speed 1 = 70kmph

He increased the speed by 20% \( \Rightarrow \) \( 70\times\frac{120}{100} \) = 84 km/hr

\( {s_{1}-s_{2}} \) - speed difference = 84 - 70 = 14

Here ,both reached lately and hence t = 15-10 = 5 minute

Converting it into hour, we get\( \frac{5}{60} \)

Hence, distance = \( \frac{70\times 84}{14} \) \( \times\frac{5}{60} \) = 35 kilometer

Let us take 100 as deposited sum.in 4 years it becomes 100+40=140

here interest=140-100=40

interest in 1 year=\(\frac{40}{4}\)=10

rate of interest=\(\frac{10}{100}\times 100=10\%\)

copmound interest for \(1^{st} year=12000\times10\%=1200\)

copmound interest for \(2^{nd} year=(12000+1200)\times10\%=1320\)

Total compound interest=1200+1320=2520