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Class 8th CBSE , Percentage, Questions and answers :-
Q1. Divide 456 into two parts such that one part is 14% of the other
part.
Ans-
Let’s
denote the two parts as x and y, where x is the larger part and y is the
smaller part. According to the given information, we have two equations:
- The sum of
the two parts is 456: x + y = 456.
- One part
is 14% of the other part: y = 0.14x.
By
substituting the value of y from the second equation into the first equation,
we get x + 0.14x = 456, which simplifies to 1.14x = 456. Solving for x, we get
x = 456 / 1.14 ≈ 400. Therefore, the larger part is
approximately 400.
The
smaller part is y = 0.14x ≈ 0.14 * 400 = 56. Therefore, the
smaller part is approximately 56.
Q2. If 15% of a number is 25% of 900, find the number.
Ans
Let’s denote the number as x.
According to the given information, we have the equation 0.15x = 0.25 * 900.
Solving for x, we get x = (0.25 * 900) / 0.15 = 1500. Therefore, the number is 1500.
Q3. 35% marks is the
pass marks in an examination. A student secured 280 marks and failed by 35
marks. Find the full marks of the examinations.
Ans
Let’s denote the full marks of
the examination as x. According to the given information, we have the equation
0.35x = 280 + 35. Solving for x, we get x = (280 + 35) / 0.35 ≈ 900. Therefore, the full
marks of the examination are approximately 900.
Q4. Chalk contains 10% calcium, 3% carbon and 12% oxygen, Find the
amount in grams of each of these in 10 kg. chalk.
Ans
Let’s
calculate the amount of each element in 10 kg of chalk. Since 10 kg is equal to
10,000 grams, we can calculate the amount of each element as follows:
Calcium:
10% of 10,000 grams = (10/100) * 10,000 grams = 1,000 grams
Carbon:
3% of 10,000 grams = (3/100) * 10,000 grams = 300 grams
Oxygen:
12% of 10,000 grams = (12/100) * 10,000 grams = 1,200 grams
So,
in 10 kg of chalk, there are 1,000 grams of calcium, 300 grams of carbon and
1,200 grams of oxygen.
Q5. The population of a town increases 5% each year. If the present
population is 3,44,100, what will be its population next year?
Ans
The
population of the town next year will be calculated as follows:
Let’s
assume that the present population of the town is P and the rate of increase in
population is R. The population of the town next year will be P * (1 + R).
Since
the present population is 3,44,100 and the rate of increase in population is 5%
or 0.05, the population of the town next year will be 3,44,100 * (1 + 0.05)
which is equal to 3,61,305.
So,
the population of the town next year will be 3,61,305.
Q6. There was direct contest
between two students for the post of president of student council of a school.
If the winner got 53% vote and won by 48 votes then (i) how many valid votes
were cast and (ii) how many votes the runner received?
Ans
Let’s
assume that the total number of valid votes cast is x. Since the winner
got 53% of the votes, the winner received 0.53x votes. The runner
received the remaining 0.47x votes. The difference between the winner
and the runner is 0.53x - 0.47x = 0.06x, which is equal to 48 votes
according to the given information. Solving for x, we get x = 48 /
0.06 = 800. So, there were a total of 800 valid votes cast.
The
runner received 0.47 * 800 = 376 votes. So, the runner
received 376 votes.
Q7. Shivendu secured 696 marks in the annual examination. If full
marks of the examination was 800, what percentage of marks did Shivendu secure?
Ans
Shivendu
secured 696 marks out of a total of 800 marks. The
percentage of marks that Shivendu secured can be calculated
as (696/800)*100 = 87%. So, Shivendu secured 87% in the annual
examination.
Q8. An examinee secured 27%
marks in an examination and failed by 15 marks. A second examinee secured 35%
marks in the same examination and got 25 marks more than the minimum pass
marks. Find (i) the maximum marks and (ii) pass marks of the examination.
Ans
Let’s
denote the maximum marks of the examination as x.
(i)
The first examinee secured 27% of the maximum marks and failed by 15 marks. So,
the pass mark is 0.27x + 15.
(ii)
The second examinee secured 35% of the maximum marks and got 25 marks more than
the pass mark. So, 0.35x = 0.27x + 15 + 25.
By
solving this equation, we can find the value of x (the maximum marks)
and the pass mark.
The
equation simplifies to 0.08x = 40, so x = 40 / 0.08 = 500. Therefore,
the maximum marks of the examination are 500.
The
pass mark is 0.27 * 500 + 15 = 150. Therefore, the pass marks of the
examination are 150.
Q9. The monthly income of a man is 52000. He spends 21840 per month.
Then what percentage of income does he save?
Ans
The man’s monthly income is 52000 and he
spends 21840 per month. So, he saves 52000 - 21840 = 30160 per month. The
percentage of income that he saves can be calculated as (30160/52000)*100 =
58%. So, the man saves 58% of his monthly income.
Q10. The price of a table and a chair is together 3600. If the price
of chair is 20% less than the price of table, find the price of the chair.
Ans
Let’s
denote the price of the table as x and the price of the chair as y. According
to the given information, we have two equations:
- The price
of a table and a chair is together 3600: x + y = 3600.
- The price
of chair is 20% less than the price of table: y = x - 0.2x = 0.8x.
By
substituting the value of y from the second equation into the first equation,
we get x + 0.8x = 3600, which simplifies to 1.8x = 3600. Solving for x, we get
x = 3600 / 1.8 = 2000. Therefore, the price of the table is 2000.
The
price of the chair is y = 0.8x = 0.8 * 2000 = 1600. Therefore, the price of the
chair is 1600.
Q11. If the price of sugar decreases by 15%, how much consumption is
increased so that the expenditure is constant?
Ans
Let’s
say the initial price of sugar is P and the initial consumption is C. The
initial expenditure on sugar is P * C.
If
the price of sugar decreases by 15%, the new price will be P - 0.15 * P = 0.85
* P.
Let’s
say the new consumption is C'. Since the expenditure is constant, we have 0.85
* P * C' = P * C, which means C' = (P * C) / (0.85 * P) = C / 0.85.
So,
the consumption must be increased by (C / 0.85 - C) / C = (1 / 0.85 - 1) * 100%
≈ 17.65% to keep the expenditure
constant.
In
other words, if the price of sugar decreases by 15%, the consumption must be
increased by approximately 17.65% to keep the expenditure constant.
Q12. If the price of ghee is increased by 25%, then to keep the
expenditure constant, how much consumption of ghee should be reduced?
Ans
Let’s
say the initial price of ghee is P and the initial consumption is C. The
initial expenditure on ghee is P * C. If the price of ghee is increased by 25%,
the new price will be P * 1.25. To keep the expenditure constant, the new
consumption should be (P * C) / (P * 1.25) = C / 1.25. So, the consumption of
ghee should be reduced by (1 - (1 / 1.25)) * 100% = 20%.
Q13. The age of Sarita is 8
Ans
Let’s
assume that the age of Babita is x years. Since the age of Sarita is
8.25% more than the age of Babita, we can calculate Sarita’s age as
follows: x + 0.0825x = 1.0825x. Therefore, Sarita’s age
is 1.0825x years.
Now,
to find out how much percent the age of Babita is less than the age of Sarita,
we can use the formula for percentage change: ((new value - original
value) / original value) * 100%. In this case, the new value is Babita’s age
(x) and the original value is Sarita’s age (1.0825x). Plugging these values
into the formula, we get: ((x - 1.0825x) / 1.0825x) * 100% = -7.6%. This
means that Babita’s age is approximately 7.6% less than Sarita’s age.
Q14. There are two candidates in an election, 10% voters did not
cast vote. The winner got 48% votes and won by 180 votes. Find how much votes
each candidate got?
Ans
Let’s
assume that the total number of voters is x. Since 10% of voters did not cast
their vote, the total number of votes cast is 0.9x. The winner got 48% of the
total votes cast, which is 0.48 * 0.9x = 0.432x. The loser got the remaining
votes, which is 0.9x - 0.432x = 0.468x. The difference between the winner and
loser’s votes is 180, so we can write the equation 0.432x - 0.468x = 180.
Solving for x, we get x = 3000.
Therefore,
the total number of voters is 3000. The winner got 0.432 * 3000 = 1296 votes and
the loser got 0.468 * 3000 = 1404 votes.
Q15. The value of a machine depreciates by 10% every year. If its
present value is 44,31,285, what was its value one year ago?
Ans
The
value of a machine depreciates by 10% every year, which means the present value
is 90% of the value one year ago. If we denote the value one year ago as V, we
have:
44,31,285=0.9×V
To
find V, we can rearrange the equation and solve for V:
V=44,31,285/0.9
Let’s
calculate this.
The
value of the machine one year ago was 49,23,650.
Q.16. The value of a car depreciates annually by 20%. If the present
value of the car is 6, 25,000, what will be its value after 2 years?
Ans
find
the value of the car after 2 years is to use the formula for calculating the
depreciation of an asset. The formula for calculating the depreciation of an
asset is:
V=P(1−r)n
Where V is the depreciated value of the
asset, P is the initial value of the asset, r is the rate
of depreciation, and n is the number of years.
In
this case, we know that the rate of depreciation r is 20%, or 0.2,
and the number of years n is 2. We also know that the present value
of the car P is 6,25,000. We can plug these values into the formula
to find the value of the car after 2 years V:
V=6,25,000(1−0.2)2
Let’s
calculate this.
The
value of the car after 2 years will be 4,00,000
Q.17. The population of a town increases 7% annually. If the present
population is 137602, what was it a year ago?
Ans
The
population of a town increases 7% annually, which means the present population
is 107% of the population one year ago. If we denote the population one year
ago as P, we have:
137602=1.07×P
To
find P, we can rearrange the equation and solve for P:
P=
Let’s
calculate this.
The
population of the town one year ago was 128600.
Q18. The population of a city increases 8% annually. If its present
population is 328500, what will be it next year?
Ans
The
population of a city increases 8% annually, which means the population next
year will be 108% of the present population. If we denote the present
population as P0 and the population
next year as P1, we have:
P1=P0×(1+0.08)
Substituting
the present population P0 as 328500, we get:
P1=328500×(1+0.08)
Let’s
calculate this.
The
population of the city next year will be 354840.
Hots
Q1. At 4.00 pm on a sunny day, a stick 2 feet tall casts a shadow 5
feet long. At the same time a tree nearby casts a shadow 55 feet long. What is
the height in feet of the tree and what percent is the shadow withrespect to
the tree?
Ans
Since
the stick and the tree are casting shadows at the same time, we can assume that
the angles of elevation of the sun for both the stick and the tree are the
same. This means that the ratio of the height of an object to the length of its
shadow is constant. Let’s denote the height of the tree as H:
2/5
= H/55
Solving
for H, we get:
H=2*55/5
=22
The
height of the tree is 22 feet.
The
shadow of the tree is 55 feet long, which is 250% of its height.
Q2. The average age of three girls is 14 years and their ages are in
the ratio 6 : 7: 8. Find the age of the youngest girl.
Ans
Let’s denote the ages of the three girls as A,
B, and C. Since their ages are in the ratio 6:7:8, we can write their ages as
6x, 7x, and 8x for some value of x. We also know that the average age of the
three girls is 14 years, so we can write:
(A+B+C)
/ 3 = 14
Substituting
the values of A, B, and C in terms of x, we get:
(6x+7x+8x)/3
=14
Solving
for x, we get:
x=14×3/(6+7+8
) = 2
The
youngest girl has an age of 6x which is equal to 6 \times 2 = 12.
The
age of the youngest girl is 12 years.
Q.3 A number exceeds 45% of itself by 110. Find the number.
Ans
Let’s
denote the number as N. Since the number exceeds 45% of itself by 110, we can
write:
N=0.45N+110
Solving
for N, we get:
N=110
/(1−0.45) = 200
The
number is 200.
Q4. The ratio of present age of A and B is 3: 5. After 10 years this
ratio becomes 5:7. Find their present ages.
Ans
Let’s
denote the present ages of A and B as 3x and 5x respectively, where x is a
common factor. After 10 years, the ages of A and B will be 3x + 10 and 5x + 10
respectively. We know that after 10 years, the ratio of A’s age to B’s age is
5:7. So we can write:
(3x
+ 10) / (5x + 10) = 5/7
Cross
multiplying and simplifying, we get:
21x+70=25x+50
Solving
for x, we get:
X = 5
Substituting
x = 5 into 3x and 5x, we get the present ages of A and B as 15 years and
25 years respectively.
So,
the present age of A is 15 years and the present age of B is 25 years.