Zoho aptitude questions

• \((a+b)^{2}=a^{2}+b^{2}+2ab\),

• Question show \(a^{2}+2ab+b^{2}\) format.

• \(8.2\times8.2+2\times8.2\times8.2+1.8\times1.8 =(8.2+1.8)^{2}\)

  =\(10^{2}=100\)

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

• 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.

• 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.

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

\(\frac{7+1}{2}\) = 4

ie, 4th term = 43

5th term = 45

6th term = 47

So, Largest number = 7 th term = 49

\(\frac{4+1}{2}\) = 2.5

ie,73.5 will be between the 2nd and 3rd term.

3rd term = 74

Hence, Largest number = 4th term = 75

• 50Kmph =50Km/hr = \(\frac{50\times1000}{3600}\) =\(\frac{500}{36}\) m/s.

  40Kmph =40Km/hr =\(\frac{40\times1000}{3600}\) =\(\frac{400}{36}\) m/s.

  (1Km =1000m, 1hr =3600s)

• When one object is moving with respect to the other moving obect,we have to consider the relative speed s1 and s2.

• If both are moving in same direction ,

  Relative speed =s1-s2

• If both are moving in opposite direction ,

  Relative speed =s1+s2

  time =\(\frac{distance}{relative speed}\)

  distance = length of train t1+ length of train t2

  = 60+90= 150m.

• Relative speed=\(\frac{500}{36}\) m/s +\(\frac{400}{36}\) m/s. =\(\frac{900}{36}\) m/s =25m/s

• Crossing time = \(\frac{150}{25}\)m/s =6m/s.

Let the present age of the man be x years and that of son be y years.

Then, from the question average of their ages be \(\frac{(x+y)}{2}= 44 \)

\(x+y = 88\) \( \to (i)\)

Now \(11\) years before, \(\frac{(x-11)}{(y-11)} = \frac{9}{2}\)

\(2(x-11) = 9(y-11)\)

\(2x-22 = 9y-99\)

\(2x -9y = -77\) \( \to (ii)\)

Solving (i) & (ii), We get

y = \( 23\) years

\(A \times \frac{40}{100} = \frac{2}{3}B\)

\(\frac{2A}{5} = \frac{2B}{3}\)

\(6A = 10B\)

\(A : B = 10 : 6 = 5 : 3 \)

• 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 \( 2:3 \).

(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

• First year CI=\(12000\times\frac{4}{100}=480\)

• Second year CI=\((12000+480)\times\frac{4}{100}=499.2\)

• Total CI=480+499.2

  = 979.2.

• \(\sqrt{72-\sqrt{72-\sqrt{72-....}}}\) = x

• x² - 72 -x = 0

  (x - 9)(x + 8)

• x = 8 ( We should take the smallest number in the case of subtraction ).

• Cost price = 100 %

• Selling price = Cost price + 38 % loss

• 38 % loss means 62 % of cost price which is equalt to Rs. 930

  \( \Rightarrow \) 930 = 62 % of Cost price

  \( \therefore\) cost price = 930 \(\times\frac{100}{62} \)

  \( = 15\times100 =\) Rs. 1500 .

Let total money be Rs. X per day wage of A =X/14 Rs per day wage of B = X/21 Rs (A+B)’s 1 day’s wage = Rs( X/14+X/21)=5X/42 Money is sufficient to pay the wages of both for 42/5=8.4 days

• \(\sqrt{56+\sqrt{56+\sqrt{56+....}}}\) = x

• x² - 56 -x = 0

  (x - 8)(x + 7)

• x = 8 (Take the largest number in the case of addition )

• Votes polled by winning candidate = 62%

• Votes polled by other candidate = 100 - 62 = 38%

• Majority = 62% - 38% = 24% = 144

• Total number of votes polled = 100% = \(x\)

• To find x, cross multiply in the below mentioned format.

  

  \(x = 144 \times \frac{100}{24} = 600 \).

Selling price = Cost price + profit

= 320 +15%

= \( 320 \times\frac{115}{100} \)

= 368 rupees

first year compound interest=\(10000\times\frac{6}{100}=600\)

second year compund interest=\(10000+600\times\frac{6}{100}=636\)

total CI=600+636=1236

SI for two year at 12%=\(\frac{1236}{2}=618\)

\(618=sum\times\frac{12}{100}\times2\)

\(618 \times\frac{100}{12} \)\( \times \ \) \(\frac{1}{2}\)

= 2575

Sum = 2575

• So here , as its a downstream motion

• Now the speed =(13+6)=19km/h

• Downstream speed is 19kmph

• Now time taken to cover 68 km with the speed of 19kmph

• Time=\( \frac{68}{19} \)=3.57 hours.