Question 1

Find the ratio of the following.
(a) Speed of a cycle 15 km per hour to the speed of a scooter 30 km per hour.
(b) 5 m to 10 km
(c) 50 paises to 5/5

Accepted Answer

Solution:
(a) Ratio = Speed of cycle : Speed of scooter = 15 km/h : 30 km/h = 15:30 = 1:2

(b) Convert 10 km to meters: 10 km = 10,000 m
Ratio = 5 m : 10,000 m = 5 : 10,000 = 1 : 2000

(c) 50 paises to 5/5 = 50 paises to 1 (since 5/5 = 1)
Ratio = 50 : 1 — Step-by-step:
(a) Simplify 15:30 by dividing both by 15.
(b) Convert km to m to have same units, then simplify.
(c) 5/5 = 1, so ratio is 50:1.

Question 2

Convert the following ratios to percentages.
(a) 3:4
(b) 2:3

Accepted Answer

Solution:
(a) 3:4 means 3/4 = 0.75
Percentage = 0.75 × 100% = 75%

(b) 2:3 means 2/3 ≈ 0.6667
Percentage ≈ 0.6667 × 100% ≈ 66.67% — Convert ratio a:b to fraction a/b, then multiply by 100 to get percentage.

Question 3

72% of 25 students are interested in mathematics. How many are not interested in mathematics?

Accepted Answer

Solution:
Number of students interested = 72% of 25 = (72/100) × 25 = 18
Number of students not interested = Total - Interested = 25 - 18 = 7 — Calculate 72% of 25 to find interested students, subtract from total to find not interested.

Question 4

A football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all?

Accepted Answer

Solution:
Let total matches played = x
Win percentage = (Number of wins / Total matches) × 100
40 = (10 / x) × 100
=> 40x = 1000
=> x = 1000 / 40 = 25
Total matches played = 25 — Use formula for percentage and solve for total matches.

Question 5

If Chameli had 600/50 left after spending 75% of her money, how much did she have in the beginning?

Accepted Answer

Solution:
Chameli had (600/50) = 12 left after spending 75% of her money.
Let total money = x
Money left = 25% of x = (25/100) × x = 12
=> (1/4) x = 12
=> x = 12 × 4 = 48
Chameli had ₹48 in the beginning. — Since 75% spent, 25% left equals 12. Use this to find total money.

Question 6

If 60% people in a city like cricket, 30% like football and the remaining like other games, then what per cent of the people like other games? If the total number of people is 50 lakh, find the exact number who like each type of game.

Accepted Answer

Solution:
Percentage liking other games = 100% - (60% + 30%) = 10%
Total people = 50 lakh = 5,000,000
Number liking cricket = 60% of 5,000,000 = 0.6 × 5,000,000 = 3,000,000
Number liking football = 30% of 5,000,000 = 0.3 × 5,000,000 = 1,500,000
Number liking other games = 10% of 5,000,000 = 0.1 × 5,000,000 = 500,000 — Subtract sum of given percentages from 100% to find remaining percentage, then calculate numbers accordingly.

Question 7

TRY THESE
1. A shop gives 20% discount. What would the sale price of each of these be?
(a) A dress marked at ₹ 120
(b) A pair of shoes marked at ₹ 750
(c) A bag marked at ₹ 250

Accepted Answer

Solution:
Discount = 20% of Marked Price (MP)
Sale Price = MP - Discount

(a) MP = ₹120
Discount = 20% of 120 = (20/100) × 120 = ₹24
Sale Price = 120 - 24 = ₹96

(b) MP = ₹750
Discount = 20% of 750 = (20/100) × 750 = ₹150
Sale Price = 750 - 150 = ₹600

(c) MP = ₹250
Discount = 20% of 250 = (20/100) × 250 = ₹50
Sale Price = 250 - 50 = ₹200 — Calculate 20% discount for each item and subtract from marked price to get sale price.

Question 8

A table marked at ₹ 15,000 is available for ₹ 14,400. Find the discount given and the discount per cent.

Accepted Answer

Solution:
Marked Price (MP) = ₹ 15,000
Sale Price (SP) = ₹ 14,400
Discount = MP - SP = 15,000 - 14,400 = ₹ 600
Discount % = (Discount / MP) × 100 = (600 / 15,000) × 100 = 4% — Subtract sale price from marked price to get discount, then calculate discount percentage.

Question 9

An almirah is sold at ₹ 5,225 after allowing a discount of 5%. Find its marked price.

Accepted Answer

Solution:
Sale Price (SP) = ₹ 5,225
Discount = 5%
Let Marked Price (MP) = x
Since discount is 5%, SP = 95% of MP
=> 5225 = (95/100) × x
=> x = 5225 × (100/95) = ₹ 5,500 — Use the relation SP = (100 - discount%) × MP / 100 and solve for MP.

Question 10

Try estimating 20% of the same bill amount (₹ 577.80).

Accepted Answer

Solution:
(i) Round off ₹ 577.80 to nearest tens = ₹ 580
(ii) Find 10% of ₹ 580 = (10/100) × 580 = ₹ 58
(iii) 20% = 2 × 10%, so 20% of ₹ 580 = 2 × ₹ 58 = ₹ 116
Estimated amount to be paid after 20% discount = 580 - 116 = ₹ 464 — Use rounding and simple percentage calculations to estimate 20% discount.

Question 11

Try finding 15% of ₹ 375.

Accepted Answer

Solution:
15% of ₹ 375 = (15/100) × 375 = ₹ 56.25 — Multiply 375 by 15% to get the value.

Question 12

I purchased a hair-dryer for ₹ 5,400 including 8% VAT. Find the price before VAT was added.

Accepted Answer

Let the price before VAT be ₹ x.
VAT is 8% of the price before VAT, so VAT amount = 8% of x = 0.08x.
Total price including VAT = price before VAT + VAT = x + 0.08x = 1.08x.
Given total price including VAT = ₹ 5400.
So, 1.08x = 5400
=> x = 5400 / 1.08 = ₹ 5000.
Therefore, the price before VAT was ₹ 5000. — To find the price before VAT, divide the total price by (1 + VAT rate). Here, VAT rate is 8%, so divide by 1.08.

Question 13

An article was purchased for ₹ 1239 including GST of 18%. Find the price of the article before GST was added?

Accepted Answer

Let the price before GST be ₹ x.
GST is 18% of the price before GST, so GST amount = 18% of x = 0.18x.
Total price including GST = price before GST + GST = x + 0.18x = 1.18x.
Given total price including GST = ₹ 1239.
So, 1.18x = 1239
=> x = 1239 / 1.18 = ₹ 1050.
Therefore, the price before GST was ₹ 1050. — To find the price before GST, divide the total price by (1 + GST rate). Here, GST rate is 18%, so divide by 1.18.

Question 14

The population of a place increased to 54,000 in 2003 at a rate of 5% per annum
(i) find the population in 2001.
(ii) what would be its population in 2005?

Accepted Answer

Let the population in 2001 be P.
Given the population increases at 5% per annum.
From 2001 to 2003 is 2 years.
Population in 2003 = P × (1 + 5/100)^2 = 54000
=> P × (1.05)^2 = 54000
=> P × 1.1025 = 54000
=> P = 54000 / 1.1025 = 48986 (approx)

(ii) Population in 2005 (4 years from 2001) = P × (1.05)^4
= 48986 × (1.21550625) ≈ 59500 (approx)

Hence,
(i) Population in 2001 ≈ 48,986
(ii) Population in 2005 ≈ 59,500 — We use the compound interest formula for population growth:
Population after n years = Initial population × (1 + rate/100)^n
Here, rate = 5%, n = number of years.
(i) For 2003, n=2, so we find initial population by dividing 54000 by (1.05)^2.
(ii) For 2005, n=4, so multiply initial population by (1.05)^4.

Question 15

In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.

Accepted Answer

Initial count of bacteria, P = 506,000
Rate of increase, r = 2.5% per hour
Time, n = 2 hours
Using compound growth formula:
Bacteria after 2 hours = P × (1 + r/100)^n
= 506,000 × (1 + 2.5/100)^2
= 506,000 × (1.025)^2
= 506,000 × 1.050625
= 531,821 (approx)

Therefore, the bacteria count after 2 hours is approximately 531,821. — Since the bacteria count increases at a compound rate, we use the formula:
Final count = Initial count × (1 + rate/100)^time
Substitute the given values and calculate stepwise.

Chapter 7

Chapter 7 — Study Notes

7.1 Recalling Ratios and Percentages

Definition of Ratio

Definition of Percentage

Practice Questions — Chapter 7

All 13 Chapters in Mathematics