Practice Percentage - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) If a man receives on one-fourth of his capital 3% interest, on two third 5% and on the remainder 11%, the percentage he receives on the whole is
(a)
(b)
(c)
(d)
Required percent
= $1/4 × 3 + 2/3 × 5 + (1 - 1/4 - 2/3) × 11$
= $3/4 + 10/3 + 11/12 = {9 + 40 + 11}/12$ = 5%
Q-2) 2 is what percent of 50?
(a)
(b)
(c)
(d)
Let 2 be x% of 50
x% of 50 = 2
$x/100$× 50 = 2⇒ $x/2$ = 2
x = 4
Q-3) If 120% of a is equal to 80% of b, then $\text"b + a"/ \text"b - a"$ is equal to
(a)
(b)
(c)
(d)
a × $120/100 = b × 80/100$
$b/a = 120/80 = 3/2$
${b + a}/{b - a} = {b/a + 1}/{b/a - 1} = {3/2 + 1}/{3/2 - 1} = {5/2}/{1/2} = 5$
Q-4) The price of an edible oil is increased by 25%. To maintain the budget, Sushma reduces the consumption of this oil by 20%. The increase in expenditure due to this edible oil is:
(a)
(b)
(c)
(d)
Cost of edible oil = 100 per kg.
Consumption = 1 kg.
Again, New price = 125 per kg.
Consumption = 0.8 kg.
Expenditure = Rs.(125 × 0.8) = Rs.100
OR
Percentage effect = $(x + y + {xy}/100)%$
=$(25 - 20 - {25 × 20}/100)$% = 0%
Q-5) A's income is 25% more than B's income. B's income is what per cent of A's income ?
(a)
(b)
(c)
(d)
Let B's income be Rs. 100.
A's income = Rs. 125
Required percent = $(100/125 × 100)$ = 80%
Q-6) The difference between the value of the number increased by 20% and the value of the number decreased by 25% is 36. Find the number.
(a)
(b)
(c)
(d)
Let the number be x.
(20 + 25)% of x = 36
${45x}/100 = 36$
$x = {36 × 100}/45$ = 80
Q-7) A number reduced by 25% becomes 225. What per cent should it be increased so that it becomes 375?
(a)
(b)
(c)
(d)
Clearly, 75% of the number = 225
Number = ${225 × 100}/75$ = 300
Again, 125% of 300 = 375
Hence, the number should be increased by 25%
Q-8) The sum of two numbers is 520. If the bigger number is decreased by 4% and the smaller number is increased by 12%, then the numbers obtained are equal. The smaller number is
(a)
(b)
(c)
(d)
Larger number = x and smaller number = 520 – x
${96x}/100 = {520 - x}/100 × 112$
96x = 520 × 112 – 112x
112x + 96x = 520 × 112
208x = 520 × 112
$x = {520 × 112}/208 = 280$
Smaller number = 520 – 280 = 240
Q-9) The number of employees working in a farm is increased by 25% and the wages per head are decreased by 25%. If it results in x % decrease in total wages, then the value of x is
(a)
(b)
(c)
(d)
Let the original number of employees be 100 and wages per head be Rs.100.
Total wages = Rs.(100 × 100) = Rs.10000
New number of employees = 125
New wages per head = Rs.75
Total new wages = Rs.(125 × 75) = Rs.9375
Decrease = Rs.(10000 – 9375) = Rs.625
Percentage decrease =$625/10000 × 100$
=$625/100 = 25/4%$
Q-10) The salary of a person is reduced by 20%. To restore the previous salary, his present salary is to be increased by
(a)
(b)
(c)
(d)
Required percentage increase = $x/{100 - x} × 100$
= $(20/{100 - 20}) × 100 = 20/80 × 100 = 25%$
Q-11) Two numbers A and B are such that the sum of 5% of A and 4% of B is $2/3$ rd of the sum of 6% of A and 8% of B. The ratio A : B is
(a)
(b)
(c)
(d)
Numbers ⇒ A and B
${A × 5}/100 + {B × 4}/100$
= $2/3({A × 6}/100 + {B × 8}/100)$
5A + 4B = ${12A + 16B}/3$
15A + 12B = 12A + 16B
15A – 12A = 16B – 12B
3A = 4B
$A/B = 4/3$ = 4 : 3
Q-12) A milkman mixed some water with milk to gain 25% by selling the mixture at the cost price. The ratio of water and milk is respectively
(a)
(b)
(c)
(d)
C.P. of 1 litre of milk= Rs. 100
Mixture sold for Rs. 125
= $125/100 = 5/4$ litre
Quantity of water = $5/4 - 1 = 1/4$ litre
Required ratio = $1/4$ : 1 = 1 : 4
Q-13) In what ratio must 25% hydrochloric acid be mixed with 60% hydrochloric acid to get a mixture of 40% hydrochloric acid ?
(a)
(b)
(c)
(d)
Required ratio = 15 : 20 = 3 : 4
Q-14) If 30% of (B – A) = 18% of (B + A), then the ratio A : B is equal to
(a)
(b)
(c)
(d)
$(B – A) × 30/100 = (B + A) × 18/100$
${B - A}/{B + A} = 18/30 = 3/5$
By componendo and dividendo,
${2B}/{- 2A} = {3 + 5}/{3 - 5} = 8/{-2} = 4/{-1}$
$B/A = 4/1$ = A : B = 1 : 4
Q-15) The ratio of two numbers is 4:5 when the first is increased by 20% and the second is decreased by 20%, the ratio of the resulting numbers is
(a)
(b)
(c)
(d)
Let the numbers be 4x and 5x.
After corresponding increase or decrease,
Required ratio = ${4x} × 120/100 : {5x} × 80/100$
= 12x : 10x = 6 : 5
Q-16) If A's income is 40% less than that of B, how much percent B's income is more than that of A?
(a)
(b)
(c)
(d)
Using Rule 9,
If ‘x' is A% less than ‘y', then y is more than ‘x' by
Required% = $(A/(100 - A)× 100)%$(increase)
Required percentage = $x/{100 - x} × 100$
= $40/60 × 100 = 200/3$= 66.66%
Q-17) In two successive years 100 and 75 students of a school appeared at the final examination. Respectively, 75% and 60% of them passed. The average rate of pass is
(a)
(b)
(c)
(d)
Number of students passed in first year = 75
Number of students passed in second year
= ${60 × 75}/100 = 45$
Total number of passed students = 75+45=120
Total number of appeared students =175
∴ Required percentage
= $120/175 × 100 = 68{4}/7%$
Q-18) The present population of a city is 180000. If it increases at the rate of 10% per annum, its population after 2 years will be :
(a)
(b)
(c)
(d)
Using Rule 17,
If the population/cost of a certain town/ article, is P and annual increament rate is r%, then
- After ‘t' years population/cost = $P(1 + r/100)^t$
- Before ‘t' years population/cost = $P/{(1 + r/100)^t}$
Required population after two years = $180000(1 + 10/100)^2$
= $180000 × 11/10 × 11/10 = 217800$
Q-19) The population of a town increases every year by 4%. If its present population is 50,000, then after 2 years it will be
(a)
(b)
(c)
(d)
Using Rule 17,
Required population = $50000(1 + 4/100)^2$
= $50000 × 26/25 × 26/25$= 54080
Q-20) The population of a town increases each year by 4% of its total at the beginning of the year. If the population on 1st January 2001 was 500000, what was it on 1st January, 2004 ?
(a)
(b)
(c)
(d)
Using Rule 17,
Required population = $P(1 + R/100)^T$
= $500000(1 + 4/100)^3$
= $500000 × (1 + 1/25)^3$
= $500000 × 26/25 × 26/25 × 26/25$ = 562432