Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 8 Comparing Quantities Ex 8.1 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.1
1. Find the ratio of :
Question (i).
₹ 5 to 50 paise
Solution:
To find ratio of ₹ 5 to 50 paise.
Firstly convert both the quantities into same units.
₹ 1 = 100 paise
₹ 5 = 5 × 100 paise = 500 paise.
So, ratio of 500 paise to 50 paise
= \(\frac{500}{50}=\frac{10}{1}\)
= 10 : 1
Hence, required ratio is 10 : 1
Question (ii).
15 kg to 210 g
Solution:
To find ratio of 15 kg to 210 g
Firstly, convert both the quantities into same unit
1 kg = 1000 g
15 kg = 15 × 1000 g = 15000 g.
So, ratio of 15000 to 210 g
= \(\frac{15000}{210}=\frac{500}{7}\)
= 500 : 7
Hence, required ratio is 500 : 7
Question (iii).
4 m to 400 cm
Solution:
To find ratio of 4 m to 400 cm
Firstly convert both the quantities into same unit.
1 m = 100 cm
4 m = 4 × 100 cm = 400 cm
So, ratio of 400 cm to 400 cm
= \(\frac{400}{400}=\frac{1}{1}\)
= 1 : 1
Hence, required ratio is 1 : 1
Question (iv).
30 days to 36 hours
Solution:
To find ratio of 30 days to 36 hours
1 day = 24 hours
30 days = 30 × 24 hours
= 720 hours
So, ratio of 720 hours to 36 hours
= \(\frac{720}{36}=\frac{20}{1}\)
= 20 : 1
Hence required ratio is 20 : 1
2. Are the ratios 1: 2 and 2 :3 equivalent ?
Solution:
To check this we need to know whether
1 : 2 and 2 : 3 are equal
First convert the ratios into fractions
1 : 2 is written as \(\frac {1}{2}\)
2 : 3 is written as \(\frac {2}{3}\)
Now to change these fraction into like fraction.
Make denominator of both the fraction same.
\(\frac{1}{2}=\frac{1}{2} \times \frac{3}{3}=\frac{3}{6}\) and \(\frac{2}{3}=\frac{2}{3} \times \frac{2}{2}=\frac{4}{6}\)
4 > 3
\(\frac{4}{6}>\frac{3}{6}\)
Hence 1 : 2 and 2 : 1 are not equivalent
3. If the cost of 6 toys is ₹ 240, find the cost of 21 toys.
Solution:
The more the number of toys one purchases, the more is the amount to be paid.
Therefore, there is a direct proportion between the number of toys and the amount to be paid.
Let x be number of toys to be purchased
∴ 6 : 240 : : 21 : x
\(\frac{6}{240}=\frac{21}{x}\)
x = \(\frac{21 \times 240}{6}\) = ₹ 840
Thus cost of 21 toys is ₹ 840.
4. The car that I own can go 150 km with 25 litres of petrol. How far can it go with 30 litres of petrol ?
Solution:
More km ↔ More petrol
So, the consumption of petrol and the distance travelled by a car is a case of direct proportion.
Let the distance travelled be x km.
∴ 150 : 25 : : x : 30
\(\frac{150}{25}=\frac{x}{30}\)
x = \(\frac{150 \times 30}{25}\)
x = 180
Thus, it can go 180 km is 30 litres of petrol.
5. In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students ?
Solution:
The more the number of students, the more is the number of computer needed.
More students ↔ More will be the computed needed.
There is a direct proportion between the number of students and the number of computers needed. Let computer needed will be x.
6 : 3 : : 24 : x
\(\frac{6}{3}=\frac{24}{x}\)
6 × x = 24 × 3
x = \(\frac{24 \times 3}{6}\) = 12
Thus, 12 computers will be needed.