In this CBSE Worksheet for chapter 3, we shall discuss more word problems based on percentage. Students who struggle to solve problems have arrived at the right platform. The solutions here are solved and prepared by our expert tutors at Physics Wallah in a step-by-step format to help students understand the concepts.
When preparing for the exam, students can refer to RS Aggarwal Solutions Class 8 , the best study material prepared by our experts by the latest CBSE patterns. For other CBSE Worksheet for Class 8 Mathematics check out main page of Physics Wallah.
Q1. Two fifth of one third of three seventh of a number is 15. What is 40 percent of that number?
Q2. If A is 20% more than B, by what percent is B less than A?
Q3. In order to compute 0.15% of a number, it must be multiplied by:
Q4. If 25% of a number is subtracted from a second number, the second number reduces to its five-sixth. What is the ratio of the first number to the second number?
Q5. A student has to obtain 33% of the total marks to pass. He got 125 marks and failed by 40 marks. The maximum marks are:
Q6. Peter earns 40% more money than Albert. Albert earned 20% less than Michael. Peter earned more than Michael by:
Q7. Milk contains 5% water. What quantity of pure milk should be added to 10 litres of milk to reduce this to 2%.
Q8. In an examination A gets 10% marks less than B and B gets 10% marks less than C. If A gets 810 marks, what marks does C get?
Q9. If the cost of the book worth Rs. 50 is increased by Rs. 25, the rate of increase is:
Q10. A man spends Rs, 3500 and saves 12 % of his income. His monthly salary is:
Q11. In an examination 52% candidates failed in English, 42% in Maths and 17% in both. The percentage of those passed in both the subjects is:
Q12. The numerator of a fraction is increased by 20% and its denominator be diminished by 10%. The value of new fraction is , then the fraction is:
Q13. If A is 25% of B then what % is B of A?
Q14. Let x and y be two numbers such that x = 6y. Then y is less than x by:
Q15. The price of sugar rises by 50%. By what factor a family reduces the consumption of sugar, so that the expenditure does not increase?
Word Problems
Q1. A coolie deposits Rs 150 monthly in his post office Savings Bank account. If this is 15% of his monthly income, find his monthly income.
Ans. Let the monthly income of a coolie be ‘x.’
Here, coolie deposits Rs 150 per month, which is 15% of his monthly income
From the above, we can derive that,
x × (15/100) =150
Which implies, x = (150 × 100) / 15
x = 15000/15
x = 1000
The monthly income is Rs 1000
Q2. Isha got 86.875% marks in the annual examination. If she got 695 marks, find the total number of student Isha marks for the examination.
Ans. Given, Marks scored by Isha are 695
And Percentage of marks Isha got was 86.875%
Now,
Let the total marks be ‘x.’
From the above, we can derive that,
x × (86.875/100) =695
Here, 86.875 can be expressed as 86875/1000
Which implies, x = (695 × 100 × 1000) / 86875
x = 6950000/86875 = 800
Total no: of Marks, x = 800
The total number of marks is 800marks
Q3. Aysha went to school for 216 days in a full year. If her attendance is 90%, find the days on which the school was opened.
Ans. Given the number of days Aysha went to school = 216 days
Aysha's Attendance percentage is = 90%
So, let the number of days when school remained open be x days
Hence,
(x × 90)/100 = 216
By using cross multiplication, we get,
x = (216×100)/90
= 240 days
Number of days the school remained open for 240 days
Q4. A garden has 2000 trees. 12% are mango trees, 18% are lemon, and the rest are orange trees. Find the number of orange trees.
Ans. Given details are,
Total number of trees = 2000
Number of mango trees = 12% of 2000
= (12/100) × 2000
= 240 trees
Number of lemon trees = 18% of 2000
= (18/100) × 2000
= 360 trees
Number of orange trees = 2000 – (Number of mango trees+ Number of lemon trees)
= 2000 – (240+360)
= 2000 – 600
= 1400 trees
The number of orange trees is 1400 trees
Q5. A balanced diet should contain 12% proteins, 25% fats, and 63% carbohydrates. If a child needs 2600 calories in this food daily, calculate the calories and the amount of each of these in his daily food intake.
Ans. The given details are,
Amount of calorie daily needed = 2600 calorie
Amount of protein needed = 12% of 2600
= (12/100) × 2600
= 312 calorie
Amount of fats needed = 25% of 2600
= (25/100) × 2600
= 650 calorie
Amount of carbohydrate needed = 63% of 2600
= (63/100) × 2600
= 1638 calorie
The amount of calories required in protein is 312 calories, fat is 650 calories, and carbohydrates is 1638 calories.