Practice Percentage - quantitative aptitude Online Quiz (set-2) For All Competitive Exams
Q-1) If 60% of A's income is equal to 75% of B's income, then B's income is equal to x% of A's income. The value of x is :
(a)
(b)
(c)
(d)
Let A's income = Rs.a
and B's income = Rs.b
a × 60% = b × 75%
a × 4 = 5 × b
$b/a = 4/5$
Now, b = a × x%
$b/a = x/100 ⇒ x/100 = 4/5$
$x = 4/5×$ 100 = 80
Q-2) In the annual examination Mahuya got 10% less marks than Supriyo in Mathematics. Mahuya got 81 marks. The marks of Supriyo are
(a)
(b)
(c)
(d)
Let marks obtained by Supriyo = x
${9x}/10 = 81 ⇒ x = {81 × 10}/9 = 90$
Q-3) In a city, 40% of the people are illiterate and 60% are poor. Among the rich, 10% are illiterate. The percentage of the illiterate poor population is
(a)
(b)
(c)
(d)
Let the population of the city be 100.
Total illiterate people = 40
Poor people = 60; Rich people = 40
Illiterate rich people = ${40 × 10}/100$ = 4
Illiterate poor people = 40 – 4 = 36
Required percent = $36/60 × 100$ = 60%
Q-4) If 50 % of P = 25% of Q, then P = x% of Q. Find x.
(a)
(b)
(c)
(d)
P × $50/100 = Q × 25/100$
P × 50 = Q × 25
P = ${Q × 25}/50 ⇒ P = Q/2$
P = Q x %
Q × $x/100 =Q/2$
x = $100/2$ = 50
Q-5) What is 20% of 25% of 300?
(a)
(b)
(c)
(d)
20% of 25% of 300 =$20/100 × 25/100 × 300$
= $1/5 × 1/4 × 300$= 15
Q-6) In 2001, the price of a building was 80% of its original price. In 2002, the price was 60% of its original price. By what percent did the price decrease ?
(a)
(b)
(c)
(d)
Original price of building = Rs. 100 (let)
Its price in 2001 = Rs. 80
Its price in 2002 = Rs. 60
Required percentage decrease = $({80 - 60}/80) × 100$
= $200/8 = 25%$
Q-7) x is 5 times longer than y. The percentage by which y is less than x is :
(a)
(b)
(c)
(d)
Length of Y = 1 foot
Length of X = 5 feet
Required percent = $({5 – 1}/5) × 100 = 80%$
Q-8) If the monthly salary of an employee is increased by 2$2/3$% , he gets 72 rupees more. His monthly salary (in rupees) is
(a)
(b)
(c)
(d)
Suppose monthly income = Rs.x
Then, $8/3$% of x = 72
x × $8/300$ = 72
${72× 300}/8$ = Rs.2700
Q-9) The allowances of an empolyee constitutes 165% of his basic pay. If he receives Rs.11925 as gross salary, then his basic pay is (in Rs.) :
(a)
(b)
(c)
(d)
Basic pay of the employee
= $11925 × 100/265$ = Rs.4500
Q-10) In an examination, 60% of the candidates passed in English and 70% of the candidates passed in Mathematics, but 20% failed in both of these subjects. If 2500 candidates passed in both the subjects, the number of candidates who appeared at the examination was
(a)
(b)
(c)
(d)
Let the total number of candidates = x
Number of candidates passed in English = 0.6x
Number of candidates passed in Maths = 0.7x
Number of candidates failed in both subjects = 0.2x
Number of candidates passed in atleast one subject
= x – 0.2x = 0.8x
0.6 x + 0.7x – 2500 = 0.8 x
1.3x – 0.8x = 2500
0.5x = 2500 ⇒ $x = 2500/0.5 = 5000$
Q-11) If A exceeds B by 40%, B is less than C by 20%, then A : C is :
(a)
(b)
(c)
(d)
Let B = 100
According to question,
A is 40% greater than B. ∴ A = 140
B is 20% less than C
⇒ 0.8C = 100∴ C = 125
∴ A : C = 140 : 125 = 28 : 25
Q-12) If 50% of x = 30% y, then x : y is
(a)
(b)
(c)
(d)
50% of x = 30% of y
${x × 50}/100 = {y × 30}/100$
$x/y = 30/50 = 3/5$ = 3 : 5
Q-13) Two numbers are respectively 20% and 50% more than a third number. The ratio of the two numbers is
(a)
(b)
(c)
(d)
Let the third number be 100.
First number = 120; Second number = 150
Required ratio = $120/150$
= $4/5$ or 4 : 5
Q-14) 0.001 is equivalent to
(a)
(b)
(c)
(d)
0.1% = $0.1/100$ = 0.001
Q-15) The bus fare and train fare of a place from Kolkata were Rs.20 and Rs.30 respectively. Train fare has been increased by 20% and the bus fare has been increased by 10%. The ratio of new train fare to new bus fare is
(a)
(b)
(c)
(d)
Increased train fare = Rs.$(120/100 × 30)$= Rs.36
Increased bus fare = Rs.$(110/100 × 20)$ = Rs.22
Required ratio = 36 : 22 = 18 : 11
Q-16) The population of a town 2 years ago was 62,500. Due to migration to big cities, it decreases every year at the rate of 4%. The present population of the town is:
(a)
(b)
(c)
(d)
Using Rule 18,
Let the present population be P.
P = $62500(1 - 4/100)^2$
= $62500 × 24/25 × 24/25$ = 57600
Q-17) 40 litres of a mixture of milk and water contains 10% of water, the water to be added, to make the water content 20% in the new mixture is :
(a)
(b)
(c)
(d)
Water content in 40 litres of mixture = $40 × 10/100$ = 4 litres
Milk content = 40 – 4 = 36 litres
Let x litres of water is mixed.
Then, ${4 + x}/{40 + x} = 20/100$
${4 + x}/{40 + x} = 1/5$
20 + 5x = 40 + x
4x = 20 ⇒ x = 5 litres
Q-18) 15 litres of a mixture contains alcohol and water in the ratio 1 : 4. If 3 litres of Water is mixed in it, the percentage of alcohol in the new mixture will be
(a)
(b)
(c)
(d)
Alcohol =15 × $1/5$ = 3 litres
Water =15 × $4/5$ = 12 litres
Required percentage = $3/{15 + 3} × 100$
= $50/3 = 16{2}/3%$
Q-19) One type of liquid contains 20% water and the second type of liquid contains 35% of water. A glass is filled with 10 parts of first liquid and 4 parts of second liquid. The water in the new mixture in the glass is
(a)
(b)
(c)
(d)
In 10 litres of first type of liquid,
Water = $1/5×10$ = 2 litres
In 4 litres of second type of liquid,
Water = $4 × 35/100 = 7/5$ litres
Total amount of water = 2 + $7/5 = 17/5$ litres
Required percentage = ${17/5}/14 × 100$
= $170/7 = 24{2}/7%$
Q-20) 1 litre of water is added to 5 litres of alcohol-water solution containing 40% alcohol strength. The strength of alcohol in the new solution will be
(a)
(b)
(c)
(d)
Alcohol in original solution = $40/100 × 5$ = 2 litres
Water in original solution = 3 litres
On adding 1 litre water, water becomes 4 litres.
Now, 6 litres of solution contains 2 litres of alcohol.
100 litres of solution contains = $2/6 × 100$
= $100/3 = 33{1}/3%$ alcohol.