Q1: Meeta saves ₹ 4000 from her salary. Suppose this is 10% of her salary. Then What is her total salary?
View AnswerLet us assume that
Meeta’s salary be ₹ x,
Now,
10% of ₹ x = ₹ 4000
(10/100) × (x) = 4000
X = 4000 × (100/10)
X = 4000 × 10
X = ₹ 40000
Hence Meeta’s salary is ₹ 40000.
Q2: Tell the profit or loss in the following transactions. Also, find the profit per cent or the loss per cent in each of the below cases.
(i) Gardening shears bought for ₹ 250 and sold for ₹ 325.
From the above question, it is given that
The cost price of the gardening shears = ₹ 250
The selling price of the gardening shears = ₹ 325
Since (SP) > (CP), hence there is a profit.
Profit is = (SP) – (CP)
= ₹ (325 – 250)
= ₹ 75
Profit % = {(Profit/CP) × 100}
= {(75/250) × 100}
= {7500/250}
= 750/25
= 30%
(ii) A refrigerator bought for ₹ 12,000 and sold at ₹ 13,500.
From the above question, it is given that the
The cost price of the refrigerator is = ₹ 12000
The selling price of the refrigerator is = ₹ 13500
Since (SP) > (CP), hence there is a profit
Profit = (SP) – (CP)
= ₹ (13500 – 12000)
= ₹ 1500
Profit % = {(Profit/CP) × 100}
= {(1500/12000) × 100}
= {150000/12000}
= 150/12
= 12.5%
(iii) A cupboard bought for ₹ 2,500 and sold at ₹ 3,000.
From the above question, it is given that
The cost price of the cupboard is = ₹ 2500
The selling price of the cupboard is = ₹ 3000
Since (SP) > (CP), hence there is a profit.
Profit = (SP) – (CP)
= ₹ (3000 – 2500)
= ₹ 500
Profit % = {(Profit/CP) × 100}
= {(500/2500) × 100}
= {50000/2500}
= 500/25
= 20%
(iv) A skirt bought for ₹ 250 and sold at ₹ 150.
Since (SP) < (CP), hence there is a loss.
Loss is = (CP) – (SP)
= ₹ (250 – 150)
= ₹ 100
Loss % = {(Loss/CP) × 100}
= {(100/250) × 100}
= {10000/250}
= 40%
Q3: The population of a city gets decreases from 25,000 to 24,500. Find the percentage decrease.
From the above question, it is given that.
The Initial population of the city is = 25000
The final population of the city is = 24500
Population decrease is equal to = Initial population – Final population
= 25000 – 24500
= 500
Now,
Percentage decrease in the population = (population decrease/Initial population) × 100
= (500/25000) × 100
= (50000/25000)
= 50/25
= 2%
Q4: Juhi sells a washing machine for ₹ 13,500. She loses 20% in the bargain. What was the initial price at which she bought it?
From the above question, it is given that.
The selling price of the washing machine is = ₹ 13500
Percentage of loss is = 20%
Now, we have to find the initial cost of price washing machine.
By using the given formula, we have:
CP = ₹ {(100/ (100 – loss %)) × SP}
= {(100/ (100 – 20)) × 13500}
= {(100/ 80) × 13500}
= {1350000/80}
= {135000/8}
= ₹ 16875
Q5: Amina buys a book for ₹ 275 and then sells it at a loss of 15%. How much does she initially sell it for?
From the above question, it is given that
The cost price of the book = ₹ 275
Percentage of loss of the book = 15%
Now, we have to find the selling price of a book,
By using the below formula, we have:
SP = {((100 – loss %) /100) × CP)}
= {((100 – 15) /100) × 275)}
= {(85 /100) × 275}
= 23375/100
= ₹ 233.75
Q6: What rate gives ₹ 280 as interest on a sum of ₹ 56,000 in two years?
Given:
P(principal) = ₹ 56000, SI = ₹ 280, t = 2 years.
We know the formula,
R (rate) = (100 × SI) / (P × T)
= (100 × 280)/ (56000 × 2)
= (1 × 28) / (56 × 2)
= (1 × 14) / (56 × 1)
= (1 × 1) / (4 × 1)
= (1/ 4)
= 0.25%
Q7: A local cricket team played around 20 matches in one season. They won 25% of the matches. Hence how many matches did they win?
From the above question, it is given that the
Total matches played by the local team = 20
Percentage of matches won by these local teams = 25%
Now,
The number of matches won by the team is = 25% of the 20
= (25/100) × 20
= 25/5
= 5 matches.
Hence, The local team won 5 matches out of 20 matches.
Q8: Convert each part of the ratio given below to a percentage:
(i) 3: 1
We have to find the total parts by adding the given ratio = 3 + 1 = 4
1st part = ¾ = (¾) × 100 %
= 3 × 25%
= 75%
2nd part = ¼ = (¼) × 100%
= 1 × 25
= 25%
(ii) 2: 3: 5
We have to find the total parts by adding the given ratio as = 2 + 3 + 5 = 10
1st part = 2/10 = (2/10) × 100 %
= 2 × 10%
= 20%
2nd part = 3/10 = (3/10) × 100%
= 3 × 10
= 30%
3rd part = 5/10 = (5/10) × 100%
= 5 × 10
= 50%
(iii) 1:4
We have to find the total parts by adding the given ratio = 1 + 4 = 5
1st part = (1/5) = (1/5) × 100 %
= 1 × 20%
= 20%
2nd part = (4/5) = (4/5) × 100%
= 4 × 20
= 80%
(iv) 1: 2: 5
We have to find the total parts by adding the given ratio as = 1 + 2 + 5 = 8
1st part = 1/8 = (1/8) × 100 %
= (100/8) %
= 12.5%
2nd part = 2/8 = (2/8) × 100%
= (200/8)
= 25%
3rd part = 5/8 = (5/8) × 100%
= (500/8)
= 62.5%
Q9: Ishan buys a T.V. for ₹ 10,000 and then sells it at a profit of 20%. How much money did Ishan get for it?
From the above question, it is given that.
The cost price of the T.V. is = ₹ 10000
Percentage of profit is = 20%
Profit = (20/100) × 10000
= ₹ 2000
Then,
The selling price of the T.V. is = cost price + profit
= 10000 + 2000
= ₹ 12000
Hence Ishan will get it for ₹ 12000.
Q10: (a) Chalk contains calcium, carbon, and oxygen in the ratio of 10:3:12. Find the total percentage of carbon in chalk.
View AnswerFrom the above question, it is given that,
The ratio of calcium, carbon and oxygen in chalk is = 10: 3: 12
So, total part = 10 + 3 + 12 = 25
In that, the total part amount of carbon = 3/25
Then,
Percentage of carbon is = (3/25) × 100
= 3 × 4
= 12 %
(b) If in a stick of chalk, the amount of carbon is 3g, what is the weight of the chalk stick?
From the above question, it is given that,
The Weight of carbon in the chalk is = 3g
Let us assume that the weight of the stick is x
Then,
12% of x = 3
(12/100) × (x) = 3
X = 3 × (100/12)
X = 1 × (100/4)
X = 25g
∴The weight of the stick is 25g.
Q11: Find the final amount to be paid at the end of three years in each case:
(i) Principal = ₹ 1,200 at 12% p.a.
Given in the above question.
P (Principal) = ₹ 1200, Rate (R) = 12% p.a. and T (time) = 3years.
If the interest is calculated uniformly on an original principal throughout the loan period, then it is called Simple interest (SI).
SI = (P × R × T)/100
= (1200 × 12 × 3)/ 100
= (12 × 12 × 3)/ 1
= ₹432
Amount = (principal + SI)
= (1200 + 432)
= ₹ 1632
(ii) Principal = ₹ 7,500 at 5% p.a.
View AnswerGiven in the question: –
Principal (P) = ₹ 7500, Rate (R) = 5% p.a. and Time (T) = 3years.
If the interest is calculated uniformly on an original principal throughout the loan period, then it is called Simple interest (SI).
SI = (P × R × T)/100
= (7500 × 5 × 3)/ 100
= (75 × 5 × 3)/ 1
= ₹ 1125
Amount = (principal + SI)
= (7500 + 1125)
= ₹ 8625
Q12: If Meena gives an interest of ₹ 45 for one year at a 9% rate per annum. Then What is the sum she has borrowed?
Ans:
From the above question, it is given that the SI = ₹ 45, R = 9%, T = 1 year, P =?
SI = (P × R × T)/100
45 = (P × 9 × 1)/ 100
P = (45 ×100)/ 9
= 5 × 100
= ₹ 500
Hence, she borrowed ₹ 500.
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1. How do you compare quantities using the method of ratios? |
2. What is the difference between the method of fractions and the method of percentages in comparing quantities? |
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4. How do you compare quantities when dealing with different units of measurement? |
5. Is it possible to compare quantities without using mathematical methods? |
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