Ratio and Proportion Class 8 RS Aggarwal Exe-9B Goyal Brothers ICSE Maths Solutions Ch-9. We provide step by step Solutions of council prescribe textbook / publication to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-8 Mathematics.
Ratio and Proportion Class 8 RS Aggarwal Exe-9B Goyal Brothers ICSE Maths Solutions Ch-9
| Board | ICSE |
| Publications | Goyal Brothers Prakashan |
| Subject | Maths |
| Class | 8th |
| writer | RS Aggarwal |
| Book Name | Foundation |
| Ch-9 | Ratio and Proportion |
| Exe-9B | Proportion |
| Academic Session | 2024-2025 |
PROPORTION
when two ratios are equivalent, they are in proportion. the two given ratios are equivalent to each other. In other words, the proportion states the equality of the two fractions or the ratios. In proportion, if two sets of given numbers are increasing or decreasing in the same ratio, then the ratios are said to be directly proportional to each other
in proportion, the two ratios are a:b & c:d. The two terms ‘b’ and ‘c’ are called ‘means or mean term,’ whereas the terms ‘a’ and ‘d’ are known as ‘extremes or extreme terms.
Continued Proportion
Three quantities are said to be in continued proportion if the ratio between the first and the second is equal to the ratio between the second and the third. The middle term is called the mean proportional between the first the third terms.
Ex- Let a, b and c are in continued proportion, if a : b = b : c . Here The second quantity is called the mean proportional between the first and the third and The third quantity is called the third proportional to the first and the second. in above example c is the third proportional to a and b
Exercise- 9B
( Ratio and Proportion Class 8 RS Aggarwal Exe-9B Goyal Brothers ICSE Maths Solutions Ch-9 )
Que-1 : Which of the following statements are true ?
(i) 27:36 = 4.5:6 (ii) (3/4):(15/16) = (2/3):(5/6) (iii) Rs14 : Rs21 = 2pens : 3pens (iv) 6.5km : 2.6km = Rs60 : Rs24
Sol: (i) 27/36 = 3/4
4.5/6 = 45/60 = 3/4
Since both are equal, this statement is true.
(ii) (3/4)/(15/16) = (3×16)/(4×15)
= 4/5
(2/3)/(5/6) = (2×6)/(3×5)
= 4/5
Since both are equal, this statement is true.
(iii) 14/21 = 2/3
2/3
Since both are equal, this statement is true.
(iv) 6.5/2.6 = 2.5
60/24 = 2.5
Since both are equal, this statement is true.
Que-2: Check whether the following numbers are in proportion are not :
(i) 8,12,18,24 (ii) 6.4,3.6,4.8,2.7 (iii) 11*(1/3).9*(1/3),8*(1/2),7 (iv) 0.36,1.8,6.4,32 (v) (3/4),(5/6),(7/8),(9/10)
Sol: (i) Since product of extremes is equal to the product of means
12 × 18 = 24 × 8
216 = 192
since 216 is not equal to 192 they are not in proportion
(ii) Since product of extremes is equal to the product of means
3.6 × 4.8 = 6.4 × 2.7
17.28 = 17.28
since 17.28 is equal to 17.28 they are in proportion
(iii) 11*(1/3) = 34/3
9*(1/3) = 28/3
8*(1/2) = 17/2
7
Since product of extremes is equal to the product of means
(28/3)×(17/2) = (34/3)×7
238/3 = 238/3
since 238/3 is equal to 238/3 they are in proportion.
(iv) Since product of extremes is equal to the product of means
1.8 × 6.4 = 32 × 0.36
11.52 = 11.52
since 11.52 is equal to 11.52 they are in proportion
(v) (3/4),(5/6),(7/8),(9/10)
0.75, 0.83, 0.87, 0.9
Since product of extremes is equal to the product of means
0.83 × 0.87 = 0.9 × 0.75
0.7221 = 0.675
since 0.7221 is not equal to 0.675 they are not in proportion
Que-3: Find the value of x in each of the following :
(i) 8:x :: 6:27 (ii) 5.6:3.5 :: x:1.25 (iii) [1*(4/5)]:[2*(4/5)] :: x:[3*(1/2)] (iv) (2/3):(4/7) :: [1*(5/6)]:x
Sol: (i) 8/x = 6/27
x = (8×27)/6
x = 36
(ii) 5.6/3.5 = x/1.25
x = (5.6×1.25)/3.5
x = 2.00
(iii) [1*(4/5)]:[2*(4/5)] :: x:[3*(1/2)]
(9/5)/(14/5) = x/(7/2)
9/14 = 2x/7
x = (9×7)/(14×2)
x = 9/4 = 2*(1/4)
(iv) (iv) (2/3):(4/7) :: [1*(5/6)]:x
(2/3)/(4/7) = (11/6)/x
7/6 = 11/6x
x = (11×6)/(7×6)
x = 11/7 = 1*(4/7)
Que-4: Find the fourth proportional to :
(i) 2.8, 14 and 3.5 (ii) [3*(1/3)], [1*(2/3)] and [2*(1/2)] (iii) [1*(5/7)], [2*(3/14)] and [3*(3/5)] (iv) [1*(1/5)],[1*(3/5)] and 2.1
Sol: (i) Let the fourth number be x
Then,
2.8:14 :: 3.5:x
2.8/14 = 3.5/x
x = (3.5×14)/2.8
x = 35/2 = 17.5
(ii) Let the fourth number be x
Then,
10/3 : 5/3 :: 5/2 : x
(10/3)×x = (5/3)×(5/2)
x = (5/3)×(5/2)×(3/10)
x = 5/4 = 1*(1/4)
(iii) Let the fourth number be x
Then,
12/7 : 31/14 :: 18/5 : x
(12/7)×x = (31/14)×(18/5)
x = (31/14)×(18/5)×(7/21)
x = 558/120
x = 93/20 = 4*(13/20).
(iv) Let the fourth number be x
Then,
6/5 : 8/5 :: 2.1 : x
(6/5)/(8/5) = 2.1/x
3/4 = 2.1/x
x = (2.1×4)/3
x = 2.8
Que-5: Find the third proportional to :
(i) 12,16 (ii) 4.5,6 (iii) [5*(1/2)],[16*(1/2)] (iv) [3*(1/2)],[8*(3/4)]
Sol: (i) Let the third proportional be x
12:16 :: 16:x
x = (16×16)/12
x = 64/3 = 21*(1/3)
(ii) Let the third proportional be x
4.5:6 :: 6:x
x = (6×6)/4.5
x = 8
(iii) Let the third proportional be x
11/2 : 33/2 :: 33/2 : x
x = [(33/2)×(33/2)]/(11/2)
x = 99/2 = 49*(1/2)
(iv) Let the third proportional be x
7/2 : 35/4 :: 35/4 : x
x = [(35/4)×(35/4)]/(7/2)
x = 175/8 = 21*(7/8)
Que-6: Find the mean proportion between :
(i) 8 and 18 (ii) 0.3 and 2.7 (iii) [66*(2/3)] and 6 (iv) 1.25 and 0.45 (v) 1/7 and 4/63
Sol: (i) Let mean proportional between 18 and 8 be x, then
18 : x : : x : 8
⇒ 18/x = x/8
⇒ x2 = 144
⇒ x = √144
⇒ x = 12
(ii) Mean proportion of two numbers means square root of product of two numbers.
Mean proportion of 0.3, 2.7 = √(0.3×2.7)
= √0.81
= √(0.9×0.9)
= 0.9
(iii) Let mean proportional between 200/3 and 6 be x, then
200/3 : x : : x : 6
⇒ (200/3)/x = x/6
⇒ x2 = 200/(3×6)
⇒ x = √400
⇒ x = 20
(iv) Mean proportion of two numbers means square root of product of two numbers.
Mean proportion of 1.25, 0.45 = √(1.25×0.45)
= √0.5625
= √(0.75×0.75)
= 0.75
(v) Mean proportion of two numbers means square root of product of two numbers.
Mean proportion of 1/7, 4/63 = √((1/7)×(4/63))
= √(4/441)
= √((2/21)×(2/21))
= 2/21
Que-7: If 28 is the third proportional to 7 and x, find the value of x,
Sol: Third proportional c = a : b : : b : c
Then, 7 : x : : x : 28
x² = 7×28
x² = 196
x = √196 = 14
Que-8: If 8,x,50 are in continued proportion, find the value of x.
Sol: As it is given that 18 , x and 50 are in continued proportion
by proportionality law if a ,b and c are in continued proportion then
a :b = b: c
18 : x :: x : 50
x² = 50×18
x = √900
x = 30
Que-9: A rod was cut into two pieces in the ratio 7:5. If the length of the smaller piece was 45.5 cm, then find the length of the longer piece.
Sol: Let the length of the longer piece be 7x and the length of the smaller piece be 5x.
As per the question-
5x = 45.5
x = 9.1 centimetres
So, the length of the longer piece = 7*9.1
Length of longer piece = 63.7 centimetres.
Que-10: The area of two rectangular fields are in the ratio 5:9. Find the area of the smaller field if that of the larger field is 2331 sq. metres.
Sol: Ratio of the area of rectangles = 5:9
Area of larger field Al = 2331 sq. metres
As/Al = 5/9
As/2331 = 5/9
As = (2331×5)/9
As = 259×5
As = 1295 sq. metres.
Que-11: What number must be subtracted from each of the numbers 41,55,36,48, so that the difference are proportional.
Sol: Let the required number be x
Then,
55-x, 48-x, 41-x, 36,x
The difference of the number are in proportional
(55-x)/(48-x) = (41-x)/(36-x)
(55-x)(36-x) = (41-x)(48-x)
1980-55x-36x+x² = 1968-41x-48x+x²
1980-91x = 1968-89x
1980-1968 = -89x+91x
12 = 2x
x = 6
Que-12: An alloy is to contain copper and zinc in the ratio 9:4. Find the quantity of zinc to be melted with 2*(2/5)kg of copper, to get the desired alloy.
Sol: Ratio of copper and zinc is 9:4
copper = 2*(2/5) = 12/5 kg
9:4 = (12/5):x
9/4 = 12/5x
x = (12×4)/(9×5)
x = 16/15 = 1*(1/15)kg
— : Ratio and Proportion Class 8 RS Aggarwal Exe-9B Goyal Brothers ICSE Maths Solutions Ch-9 :–
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