Q21Written answer
________ is a reduction on the marked price of the article.
Show solutionHide solution
Worked solution
Solution: Discount
NCERT Exemplar · Class 8 · Maths
Chapter 9: Comparing Quantities
Practice each question, then open the answer or worked solution.
Questions
50 questionsQ22Written answer
Increase of a number from 150 to 162 is equal to increase of ____ per cent.
Show solutionHide solution
Worked solution
Solution: 8%
Explanation: Increase of a number from 150 to 162 = 162-150 = 12
Percentage of increased number = 12/150 × 100 = 120/15 = 8%
Q23Written answer
15% increase in price of an article, which is Rs.1,620, is the increase of ____.
Show solutionHide solution
Worked solution
Solution: Rs.212
Explanation: Let x is the price of the article.
Thus, as per given question;
1620 = x + x × (15/100)
1620 = 115x/100
115x = 1620 × 100
x = (1620×100)/115
x = 1408
Hence, increase in price = 1620 – 1408 = 212.
Q24Written answer
Discount = _______-_______.
Show solutionHide solution
Worked solution
Solution: Discount = Marked Price – Selling Price.
Q25Written answer
Discount = Discount % of ________
Show solutionHide solution
Worked solution
Solution: Discount = Discount % of Marked Price.
Q26Written answer
_______ is charged on the sale of an item by the government and is added to the bill amount.
Show solutionHide solution
Worked solution
Solution: Sales tax
Q27Written answer
Amount, when interest is compounded annually, is given by the formula_____
Show solutionHide solution
Worked solution
A = P(1+R/100)T [P = Principal, R = Rate, T = time]
Q28Written answer
Sales tax = tax % of _______
Show solutionHide solution
Worked solution
Solution: Bill amount
Q29Written answer
The time period after which the interest is added each time to form a new principal is called the __________
Show solutionHide solution
Worked solution
Solution: Conversion period
Q30Written answer
________ expenses are the additional expenses incurred by a buyer for an item over and above its cost of purchase.
Show solutionHide solution
Worked solution
Solution: Overhead