Percentage - Aptitude MCQ Questions and Solutions with Explanations

Money spent originally = Rs. 140
Less Money to be spent now
= 10% of 140
= Rs. 14
Rs. 14 now yield 500 gm sugar
So, Present rate of sugar = Rs. 28 per kg.
If the present value is Rs. 90, the original value = Rs. 100
If the present value is Rs. 28 the original value
$$ = {\text{Rs}}{\text{. }}\frac{{100}}{{90}} \times 28 = {\text{Rs}}{\text{. }}31.11$$

$$\eqalign{ & {\text{Let third number is x}}. \cr & {\text{Then}}\,{\text{first}}\,{\text{no}}{\text{.}} \cr & 20\% \,{\text{of}}\,x = \frac{{20x}}{{100}} \cr & {\text{Second}}\,{\text{number}} \cr & = 50\% \,{\text{of}}\,x = \frac{{50x}}{{100}} \cr & {\text{Percent of first no of second no,}} \cr & = {\frac{{ {\frac{{20x}}{{100}}} }}{{ {\frac{{50x}}{{100}}} }}} \times 100 \cr & = \frac{{ {2 \times 100} }}{{20}} \cr & = 40\% \cr} $$

Let the capacity of the tank be 100 litres
Initially: A type petrol = 100 litres

After first operation:
A type petrol = $$\frac{{100}}{2}$$ = 50 litres
B type petrol = 50 litres

After second operation:
A type petrol = $$\frac{{50}}{2}$$ + 50 = 75 litres
B type petrol = $$\frac{{50}}{2}$$ = 25 litres

After third operation:
A type petrol = $$\frac{{75}}{2}$$ = 37.5 litres
B type petrol = $$\frac{{25}}{2}$$ + 50 = 62.5 litres
Required percentage = 37.5%

36% marks = 113 + 85
⇒ 36% marks = 198
So,
⇒ 1% marks = $$\frac{{198}}{{36}}$$ = 5.5
⇒ 100% marks = 5.5 × 100 = 550

$$\eqalign{ & 1\% \,{\text{of}}\,1\% \,{\text{of}}\,25\% \,1000 \cr & = 1\% \,{\text{of}}\,1\% \,{\text{of}}\,\, {\frac{{ {25 \times 1000} }}{{100}}} \cr & = 1\% \,{\text{of}}\,1\% \,{\text{of}}\,250 \cr & = 1\% \,{\text{of}}\,\, {\frac{{ {1 \times 200} }}{{100}}} \cr & = 1\% \,{\text{of}}\,\,2.5 \cr & = \frac{{2.5}}{{100}} \cr & = 0.025 \cr} $$

Using compounding formula
$${\text{a}} + {\text{b}} + \frac{{{\text{ab}}}}{{100}}$$
we have
$$\eqalign{ & 5 + 5 + \frac{{25}}{{100}} \cr & = 10.25\% \cr} $$
so if the population 2 yrs ago be x
then
$$\eqalign{ & {\text{x}} + \frac{{10.25x}}{{100}} = 4410 \cr & {\text{or}},\,110.25{\text{x}} = 441000 \cr & \therefore {\text{x}} = 4000 \cr} $$

$$\eqalign{ & {\text{Height}}\,{\text{of}}\,{\text{the}}\,{\text{Pole}} = 192\,m. \cr & {\text{Spider}}\,{\text{covered}}\,{\text{in}}\,{\text{first}}\,{\text{hour}}, \cr & = \frac{{125}}{2}\% \,of\,192 \cr & = \frac{{ {125 \times 192} }}{{ {2 \times 100} }} \cr & = 120\,m \cr & {\text{Remaining Pole}} = 192 - 120 = 72\,m \cr & {\text{Spider}}\,{\text{covered}}\,{\text{in}}\,{\text{second}}\,{\text{hour}}, \cr & = \frac{{25}}{2}\% \,{\text{of}}\,{\text{ remaining}}\,{\text{ height}} \cr & = \frac{{ {25 \times 72} }}{{ {2 \times 100} }} \cr & = 9\,m \cr} $$

$$\eqalign{ & a\,{\text{percent}}\,{\text{of}}\,b \cr & = {\frac{a}{{100}}} \times b = {\frac{{ab}}{{100}}} \cr & b\,{\text{percent}}\,{\text{of}}\,a \cr & = {\frac{b}{{100}}} \times a = {\frac{{ab}}{{100}}} \cr & {\text{on}}\,{\text{division}}\,{\text{we}}\,{\text{get}}\,1 \cr} $$

Winner gets 55% of votes
As 4% votes were declared invalid so 96% would be the valid votes
So,
Winner gets 55% of 96% valid votes
Winner gets % valid votes = $$\frac{{55 \times 96}}{{100}}$$   = 52.8% votes
Loser gets = 96 - 52.8 = 43.2% votes
Difference = 9.6%
Now, 9.6% = 4200
So, 1% = $$\frac{{4200}}{{9.66}}$$
Thus, 100% Votes = $$\frac{{4200 \times 100}}{{9.66}}$$   = 43750
Hence, Total Voters = 43,750

Alternatively,
Let total number of voters were X.
Invalid Votes = 4%
Valid Votes = 96%
Total Valid Votes = 96% of X = $$\frac{{96{\text{x}}}}{{100}}$$  = 0.96X
Winner gets 55% of Valid Votes,
= $$\frac{{0.96 \times 55}}{{100}}$$   = 0.528X votes
Loser gets = (0.96X - 0.528X) = 0.432X
Difference = 0. 528 - 0.432 = 0. 096X
0.096X = 4200
∴ X = 43750

Let Mixture was X initially
Initial Salt = 15% of X = 0.15X
Initial Water = 85% of X = 0.85X
30 liters of water evaporated
So, Final water = (0.85X - 30)
Now, salt rises to 20% of the mixture, it means as water dried up salt rises
Now, Total Mixture = (0.85X -30 + 0.15X)
Final salt = 20% of(0.85X - 30 + 0.15X)

As, Initial Amount of salt = Final amount of salt, we have
0.15X = 20% of(0.85X - 30 + 0.15X)
0.15X = 0.20X - 6
Or, 0.5X = 6
Thus, X = 120
Initial amount of the mixture was 120 liters