Class 7 Ariyhmatic Chapter 2 Ratio
Question 1. Find the compound ratio of the following ratios and identify if these are ratios of greater or lesser inequality or ratio of equality.
Solution:
1. \(\frac{1}{2}: \frac{1}{3}, \frac{1}{3}: \frac{1}{4} \text { and } \frac{1}{4}: \frac{1}{5}\)
⇒ \(\frac{1}{2}: \frac{1}{3}\)
⇒ \(\frac{\frac{1}{2}}{\frac{1}{3}} \Rightarrow \frac{1}{2} \times \frac{3}{1} \Rightarrow \frac{3}{2}\)
⇒ \(\frac{1}{3}: \frac{1}{4} \Rightarrow \frac{\frac{1}{3}}{\frac{1}{4}} \Rightarrow \frac{1}{3} \times \frac{4}{1} \Rightarrow \frac{4}{3}\)
⇒ \(\frac{1}{4}: \frac{1}{5} \Rightarrow \frac{\frac{1}{4}}{\frac{1}{5}} \Rightarrow \frac{1}{4} \times \frac{5}{1} \Rightarrow \frac{5}{4}\)
Multiply these ratios to find a compound ratio
⇒ \(\frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \Rightarrow\left(\frac{3}{2} \times \frac{4}{3}\right) \times \frac{5}{4} \Rightarrow \frac{3 \times 4}{2 \times 3} \times \frac{5}{4} \Rightarrow \frac{4}{2} \times \frac{5}{4}=2 \times \frac{5}{4}=\frac{5}{2}\)
⇒ \(\left(\frac{3}{2} \times \frac{4}{3} \times \frac{5}{4}\right)=\frac{5}{2}\)
∴ The compound ratio of the given ratios is \(\frac{5}{2}\)
Read and Learn More Class 7 Maths Solutions
Since \(\frac{5}{2}>1\) (because 5>2) the compound ratio \(\frac{5}{2}\) is a ratio of greater inequality
2. \(\frac{2}{3}: 3, \frac{1}{5}: 5 \text { and } 15: \frac{1}{12}\)
⇒ \(\frac{2}{3}: 3 \Rightarrow \frac{2}{3}: \frac{3}{1} \Rightarrow \frac{2 / 3}{3 / 1} \Rightarrow \frac{2}{3} \times \frac{1}{3} \Rightarrow \frac{2}{9}\)
⇒ \(15: \frac{1}{12} \Rightarrow \frac{15}{1}: \frac{1}{12} \Rightarrow \frac{\frac{15}{1}}{\frac{1}{12}} \Rightarrow \frac{15}{1} \times \frac{12}{1} \Rightarrow 180\)
Multiply these ratios to find a compound ratio
⇒ \(\frac{2}{9} \times \frac{1}{25} \times 180 \Rightarrow\left(\frac{2}{9} \times \frac{1}{25}\right) \times 180 \Rightarrow \frac{2 \times 1}{9 \times 25} \times 180 \Rightarrow \frac{2}{225} \times 180 \Rightarrow \frac{360}{225}\)
The greatest common divisor (GCD) of 360 and 225 is 45.
⇒ \(\frac{360 \div 45}{225 \div 45}=\frac{8}{5}\)
The compound ratio is 8:5
The ratio of greater inequality is those where the compound ratio is either much greater than 1, or much less than 1
The compound ratio 8:5 is greater than 1, indicating it is a ratio of greater inequality.
Class 7 Maths Arithmetic Solutions
3. 4:6, 3:5, and 10:4
⇒ \(\frac{4}{6}, \frac{3}{5}, \frac{10}{4}\)
Multiply these ratios to find a compound ratio
⇒ \(\frac{4}{6} \times \frac{3}{5} \times \frac{10}{4} \Rightarrow\left(\frac{4}{6} \times \frac{3}{5}\right) \times \frac{10}{4} \Rightarrow \frac{4 \times 3}{6 \times 5} \times \frac{10}{4} \Rightarrow \frac{12}{30} \times \frac{10}{4}=\frac{120}{120}= \frac{1}{1}=1\)
So the compound ratio is 1:1
Ratios of equality occur when the compound ratio is exactly ‘1’.
Since the compound ratio is 1:1, it is a ratio of equality.
4. 2x:3x, 5a:6a, and 4p:9p
2x:3x ⇒ \(\frac{2 x}{3 x} \Rightarrow \frac{2}{3}\)
5a:6a ⇒ \(\frac{5 a}{6 a} \Rightarrow \frac{5}{6}\)
4p:9p ⇒ \(\frac{4 p}{9 p} \Rightarrow \frac{4}{q}\)
Multiply these ratios to find the compound ratio
⇒ \(\frac{2}{3} \times \frac{5}{6} \times \frac{4}{9} \Rightarrow\left(\frac{2}{3} \times \frac{5}{6}\right) \times \frac{4}{9} \Rightarrow \frac{2 \times 5}{3 \times 6} \times \frac{4}{9} \Rightarrow \frac{10}{18} \times \frac{4}{9} \Rightarrow \frac{20}{81}\)
∴ The compound ratio is 20:81
Ratios of greater inequality are those where the compound ratio is either much greater than 1 or much less than 1.
The compound ratio 20:81 is less than 1, indicating that it is a ratio of greater inequality.
Class 7 Arithmetic Chapter Solutions
Question 2. Reduce the following into the ratio of the whole numbers and Find their inverse ratios.
Solution:
1. 18Pq: 24pq
18pq: 24Pq [Dividing each term by 6Pq]
⇒ \(\frac{18 p q}{6 p q}: \frac{24}{6 p q}\)
⇒ 3:4
The inverse ratio of 3:4 is 4:3
2. 4.5:6
4·5:6
⇒ \(\frac{4.5}{6}=\frac{4.5 \times 10}{6 \times 10}=\frac{45}{60}\)
⇒ 3:4
The inverse ratio of 3:4 is 4:3
3. 25a: 40a
25a: 40a [Dividing each term by 5a]
⇒ \(\frac{25 a}{5 a}: \frac{40 a}{5 a}\)
⇒ 5:8
The Inverse ratio of 5:8 is 8:5
Question 3. “Reduce the Following into a ratio of the whole numbers and find their inverse ratio:
1. 0·3:0.5
⇒ \(\frac{0.3}{0.5}=\frac{0.3 \times 10}{0.5 \times 10}=\frac{3}{5}\)
⇒ 3:5
The inverse ratio of 0·3:0.5 is 5:3
2. \(3 \frac{1}{4}: 4 \frac{1}{5}\)
⇒ \(3 \frac{1}{4}: 4 \frac{1}{5}\)
⇒ \(\frac{13}{4}: \frac{21}{5}\) [LCM of 4 and 5 is 20 Multiply by L.C.M=20]
⇒ \(\frac{13}{4} \times 20: \frac{21}{5} \times 20\)
⇒ 13×5: 21×4
⇒ 65:84
The inverse ratio of 65:84 is 84:65
3. \(2 \frac{1}{3}: 6 \frac{2}{3}\)
⇒ \(2 \frac{1}{3}: 6 \frac{2}{3}\)
⇒ \(\frac{7}{3}: \frac{20}{3}\) (equal denominators so Numarators are ratios)
⇒ 7:20
The inverse ratio of 7:20 is 20:7
Class 7 Maths Arithmetic Problems
Question 4. In two types of ‘Sherbat,’ the ratios of Syrup and water are 7:5 and 5:3 Find which one is sweeter.
Solution:
The total parts in the mixture are syrup
7+5 = 12.
syrup = \(\frac{7}{12}\)
The total parts in Syrup to water mixtures.
5+3=8
second type = \(\frac{5}{8}\)
To compare \(\frac{7}{12}\), and \(\frac{5}{8}\)
The least common multiple (L.CM) OF 12 and 8 is 24
⇒ \(\frac{7}{12}=\frac{7 \times 2}{12 \times 2}=\frac{14}{24}\)
convert \(\frac{5}{8}\) to a fraction with denominator 24.
⇒ \(\frac{5}{8}=\frac{5 \times 3}{8 \times 3}=\frac{15}{24}\)
Question 5. Comparing \(\frac{14}{24} \text { and } \frac{15}{24}\) we see that
Solution:
⇒ \(\frac{14}{24}<\frac{15}{24}\)
Convert each fraction to a decimal.
⇒ \(\frac{7}{12} \approx 0.5833\)
⇒ \(\frac{\bar{s}}{8}=0.625 .\)
Comparing 0.5833 and 0.625 we see that
0.5833 < 0.625
Since \(\frac{5}{8}\)(0.625) is greater than \(\frac{7}{12}\)(0.5833), the
The second type of sherbet is sweeter.
Question 6. The ratio of measurement of the three angles of a triangle is 4:5:6 Find the measurement of the angles.
Solution:
Given
The three angles of a triangle are 4:5:6.
The three angles are represented as 4x, 5x6x.
Some of the angles in a triangle are 180°
∴ 4x+5x+6x=180°
15x = 180°
x = \(\frac{180^{\circ}}{15}\)
x = 12°
Substitute the ‘x’ value in the above equation.
42 = 4×12 = 48°
5x = 5 ×12 = 60°
6x = 6 ×12 = 72°
∴ The measurement of three angles of a triangle is 48°, 60°, 72°
Class 7 Maths Chapter 2 Questions
Question 7. If the ratio of two numbers is 4:7 and their difference is 45 then Find the numbers.
Solution:
Given:
The ratio of the two numbers is 4:7.
Let the number be represented as 4x, 7x
where ‘X’ is a common multiplier.
the difference between the two numbers is given by:-
7x-4x = 45
3x = 45
x = \(\frac{45}{3}\)
x = 15
Now substitute ‘X’ into the expression. For the two numbers
7x = 7×15 = 105
4x = 4 ×15 = 60
The two numbers are 60,105.
Question 8. The ratio of the two numbers is 5:6 and their sum is 44. Find their H.C.F
Solution:
Given:-
The ratio of the two numbers is 5:6.
Let’s represent the two numbers as 5x and 6x
The sum of the two numbers is 44
⇒ 5x+6x = 44
⇒ 11x = 44
x = \(\frac{44}{11}\)
x = 4
∴ The numbers are
5x ⇒ 5×4 = 20
6x ⇒ 6×4 = 24
The Prime Factors of 20 are 22 × 5
The prime factors of 24 are 23 × 3
The minimum power of the factor 2 is 22
Hence the H.C.F is 22 = 4.
∴ The H.C.F of 20 and 24 is 4
Class 7 Maths Textbook Solutions
Question 9. The ratio of the circumference and diameter of a circle is 22:7. If the length of the circumference is 154 cm. then Find the length of its diameter.
Solution:
Given:
The ratio of the circumference and diameter of a circle is 22:7
This can be expressed as
⇒ \(\frac{C}{D}-\frac{22}{7}\)
Given that the circumference is 154 cm.
⇒ \(C-\frac{22}{7} D\)
⇒ \(154-\frac{22}{7} D\)
⇒ 154×7 – 22D
⇒ \(\frac{1078}{22}-0\)
D-49
∴ The length of the diameter is 49 cm
Question 10. The ratio of two numbers is 8:9 and their H.C.F is 15 Find the numbers and also. Find their LCM
Solution:
Given
The ratio of the two numbers is 8:9 & their H.C.F is 15.
Let the numbers be 8% and 9x, where x’ is a common factor.
H.C.F.(8x, 9x) = x
Given H.C.F is 15
∴ x = 15
using x = 15, the numbers are
8x = 8×15= 120
9x = 9×15=135
∴ The numbers are 120 and 135.
L.C.M.(a,b) = \(\frac{a \times b}{H \cdot c \cdot f(a, b)}\)
⇒ \(\frac{120 \times 135}{15}=\frac{16200}{15}=1080\)
L.C.M(120, 135) = 1080
∴ The L.C.M of 120 and 135 is 1080
Question 11. out of 150 Sums, Arka got 80 sums correct, and out of 120 sums, Soumya got 70 sums correct
Solution:
Express them in ratio to find who got more sums correct.
For Arka:
Total sums attempted: 150
Sums correct: 80
The ratio of correct sums to total sums: \(\frac{80}{150}\)
Dividing both the numerator and the denominator by their greatest common divisor (GCD). The GCD of 80 and 150 is 10.
⇒ \(\frac{80 \div 10}{150 \div 10}=\frac{8}{15}\)
For Soumya
Total sums attempted: 120
sums correct 70
The ratio of correct sums to total sums: \(\frac{70}{120}\)
The GED OF 70 and 120 is 10.
∴ \(\frac{70 \div 10}{120 \div 10}=\frac{7}{12}\)
Class 7 Maths Textbook Solutions
Question 12. Now compare the ratios \(\frac{8}{15} \text { and } \frac{7}{12}\)
Solution:
The LCM of 15 and 12 is 60
Convert \(\frac{8}{15}\) to a fraction with denominator of 60:
⇒ \(\frac{8}{15}=\frac{8 \times 4}{15 \times 4}=\frac{32}{60}\)
Convert \(\frac{7}{12}\) to a fraction with a denominator of 60;
⇒ \(\frac{7}{12}=\frac{7 \times 5}{12 \times 5}=\frac{35}{60}\)
Now compare \(\frac{32}{60} \text { and } \frac{35}{60}\) since 35 is greater than 32.
⇒ \(\frac{35}{60}>\frac{32}{60}\)
∴ Soumya has a higher ratio of sums correct to total sums attempted.
Soumya got more sums correct in terms of ratio.
Question 13. ₹260 is divided among Rupak, Anirban, and Apurba in the ratio \(1: \frac{2}{3}: 2\). Calculate how much each will get
Solution:
The given amount is ₹7260
Rupak, Anirban and Apurba in ratio = \(1: \frac{2}{3}: 2\)
⇒ \(1: \frac{2}{3}: 2 \Rightarrow(1 \times 3):\left(\frac{2}{3} \times 3\right):(2 \times 3)\)
⇒ 3:2:6
Rupak gets = \(₹\left(7260 \times \frac{3}{3+2+6}\right)=7260 \times \frac{3}{11}= ₹1980\)
Anirban gets = \(₹\left(7260 \times \frac{2}{3+2+6}\right)=7260 \times \frac{2}{11}= ₹1320\)
Apurba gets = \(₹\left(7260 \times \frac{6}{3+2+6}\right)=7260 \times \frac{6}{11}= ₹3960\)
∴ Rupak gets = ₹1980.
Anirban gets = ₹1320
Apurba gets = ₹3960
Question 14. The ratio between the speeds of two buses is 3:5 If the first bus runs 120 km in 4 hours. Find the Speed of the Second bus.
Solution:
Given:
The ratio between the speeds of the two buses is 3:5 first bus runs 120 km in 4 hours.
Speed of the first bus = \(\frac{\text { Distance }}{\text { Time }}\)
⇒ \(\frac{120 \mathrm{~km}}{4 \mathrm{~h}}\)
⇒ 30km/h
Let the speed of the second bus be ‘x’ km/h.
⇒ \(\frac{\text { speed of first bus }}{\text { speed of second bus }}=\frac{3}{5}\)
⇒ \(\frac{30}{x}=\frac{3}{5}\)
3×x = 30×5
3x = 150
x = \(\frac{150}{3}\)
x = 50km/h
∴ The Speed of the second bus is 2 = 50km/h
WBBSE Class 7 Maths Guide
Question 15. If A:B=4:5, B:C= 6:7 then Find A:B:C:
Solution:
Given
A:B = 4:5
B:C = 6:7
The ratio of B in both equations is the same.
L.C.M OF 5 and 6 is 30.
For A: B we multiply by 6
A : B = 4×6:5×6
A : B = 24:30
For B:c we multiply by 5
B:C = 6×5:7×5
B:C = 30:35
Now, we have;
A:B = 24:30
B:C = 30:35
∴ The ratios to Find A: B: C:
A: B: C = 24:30:35
Question 16. If AB = 3:4, B: C = 5:6 and C:D = 7:8 then find A:D and A:B: C: D.
Solution:
Givent
A:B = 3:4
B:C = 5:6
C:D = 7:8
A:B = 3:4
⇒ \(\frac{A}{B}=\frac{3}{4}\)
⇒ \(A=\frac{3}{4} B\)
B:C = 5:6
⇒ \(\frac{B}{C}=\frac{5}{6}\)
⇒ \(B=\frac{5}{6} c\)
C:D = 7:8
⇒ \(\frac{C}{D}=\frac{7}{8}\)
⇒ \(c=\frac{7}{8} D\)
Substitute \(C=\frac{7}{8} D \text { into } B=\frac{5}{6} C \text { : }\)
B = \(\frac{5}{6} c\)
⇒ \(\frac{5}{6}\left(\frac{7}{8} D\right)\)
⇒ \(\frac{5 \times 7}{6 \times 8} D\)
B = \(\frac{35}{48} D\)
Substitute B = \(B=\frac{35}{48} D \text { into } A=\frac{3}{4} B\)
WBBSE Class 7 Maths Guide
Question 17. \(A=\frac{3}{4} B \quad B=\frac{35}{48} D\)
A = \(\frac{3}{4}\left(\frac{35}{48}\right) D\)
A = \(\frac{3 \times 35}{4 \times 48} D\)
A= \(\frac{105}{192} D\)
⇒ \(\frac{A}{D}=\frac{105}{192} D\)
⇒ \(\frac{A}{D}=\frac{35}{64}\)
∴ \(A: D=\frac{35}{64}\)
We have
A = \(\frac{105}{192} D\)
B = \(\frac{35}{48} D\)
C = \(\frac{7}{8} D\)
and D = D
Convert these into a common base with D = 192
∴ A = \(\frac{105}{192} \times 192\) =105
B = \(\frac{35}{48} \times 192=35 \times 4\) = 140
C = \(\frac{7}{8} \times 192=7 \times 24\) = 168
D = 192
Hence the combined ratio A:B: C:D is 105: 140: 168: 192
Question 18. During retirement, Amal Babu gets ₹256000. He donated 7 16,000 to a club and the remaining amount he divided among his wife son and daughter in a ratio of 5:4:3 to find how much money he gave each of them.
Solution:
Retirement Amal baby get ₹256000
Amal babu donated = ₹16,000.
Remaining amount = 256000 – 16,000
= 240000
The ratio of distribution among his wife, son, and daughter is 5:4:3
Total parts = 5+4+3 = 12.
240000 Remaining amount is divided into 12 parts:
value of one part = \(\frac{2.40000}{12}\) = 20,000
Question 19. Calculate the amounts for each family member.
Solution:
Wife( 5 parts):
wife’s shone = 5x 20,000 = 1,00,000
Son (4 parts):
Son’s share = 4x 20,000 = 80,000
Daughter (3 parts):
Daughter’s share = 3x 20,000 60,000.
∴ The amounts given to each Family member are
wife: ₹1,00,000
Son: ₹80,000
Daughter: ₹760,000.