Maharashtra Board Class 9 Math Solution Chapter 4 – Ratio & Proportion
Balbharati Maharashtra Board Class 9 Math Solution Chapter 4: Ratio & Proportion. Marathi or English Medium Students of Class 9 get here Ratio & Proportion full Exercise Solution.
Std |
Maharashtra Class 9 |
Subject |
Math Solution |
Chapter |
Ratio & Proportion |
Practice Set 4.1
(1) From the following pairs of numbers, find the reduced form of ratio of first number to second number.
(i) 72, 60
Ans: 72/60 = 6×12/5×12 = 6/5
(ii) 38, 57
Ans: 38/57 = 2×19/3×19 = 2/3
(iii) 52, 78
Ans: 52/78 = 4×13/6×13 = 2×26/3×26 = 2/3
(2) Find the reduced form of the ratio of the first quantity to second quantity
(i) 700 RS, 308 RS
Ans: 700RS/308RS = 28×25/28×11 = 25/11
(ii) 14 RS, 12.40 paise
Ans: Converting rupees into paise
We know, 1 rupee 100 paise
14RS = 14×100 paise
12.40 paise = 12×100 + 40 paise
= 1240 paise
∴ 14RS/12RS.40 Paise = 14×100/1240 = 4×35/4×31 = 35/31
(iii) 5 litre, 2500 ml
Ans: We know, 1 litre = 1000 ml
∴ 5 litre = 5000 ml
∴ 5 litre/2500 ml = 5000/2500 = 2/1
(iv) 3 years 4 months, 5 years 8 months
Ans: We know, 1 year = 12 months
∴ 3 years 4 months = 3×12+4 months
= 40 months
5 years 8 months = 5×12+8 months
= 68 months
∴ 3 years 4 months/5 years 8 months = 40 months/68 months
= 4×10/4×17 = 10/17
(v) 3.8 kg, 1900 gm
Ans: We know,
1kg = 1000 gm
∴ 3.8 kg = 3800 gm
∴ 3.8kg/1900gm = 3800gm/1900gm = 2/1
(vi) 7 minutes 20 seconds, 5 minutes 6 seconds
Ans: We know,
1 minutes = 60 seconds
∴ 7 minutes 20 seconds = 7×60+20 seconds
= 440 seconds
5 minutes 6 seconds = 5×60+6 seconds
= 306 seconds
∴ 7 minutes 20 seconds/5 minutes 6 seconds = 440 seconds/306 seconds
= 220×2/153×2 = 220/153
(3) Express the following percentages as ratios in the reduced form.
(i) 75:100
Ans: 75/100 = 25×3/25×4 = 3/4
(ii) 44:100
Ans: 44/100 = 4×11/4×25 = 11/25
(iii) 6.25%
Ans: 6.25% = 6.25/100 = 625/10000 = 1/16
(iv) 52:100
Ans: 52/100 = 4×13/4×25 = 13/25
(v) 0.64%
Ans: 0.64% = 0.64/100 = 64/10000 = 16×4/16×625 = 4/625
(4) Three persons can build a small house in 8 days. To build the same house in 6 days, how many persons are required?
Ans: Let, x number persons be required to build the hence in 6 days.
A/Q,
8×3 = 6 × x
= 6x = 24
= x = 4
∴ 4 persons can build the house in 6 days
(5) Convert following ratios into percentage.
(i) 15:25
Ans: 15/25 = 15×4/25×4 = 60/100 = 60%
(ii) 47:50
Ans: 47/50 = 47×2/50×2 = 94/100 = 94%
(iii) 7/10
Ans: 7/10 = 7×10/10×10 = 70/100 = 70%
(iv) 546/600
Ans: 546/600 = (546×1/6)/(600×1/6) = 91/100 = 91%
(v) 7/16
Ans: 7/16 = 7/16 = (7×25/4)/(16×25/4) = 43.75/100 = 43.75%
(6) The ratio of ages of Abha and her mother is 2:5. At the time of Abha’s birth her mother’s age was 27 year. Find the present ages of Abha and her mother.
Ans: Let, the present age of Abha and her mother be x and x+27 years respectively
A/Q, 2/5 = x/x+27
= 2(x+27) = 5x
= 2x + 54 = 5x
= 3x = 54
x = 18
∴ Abha is 18 years old and her mother is 18+27 = 45 years old.
(7) Present ages of Vatsala and Sara are 14 years and 10 years respectively. After how many years the ratio of their ages will become 5:4?
Ans: Let, the age of Vatsala and Sara be 5:4 after x years
∴ A/Q, 5/4 = 14+x/10+x = 5(10+x) = 4(14+x)
= 50+5x = 56+4x
= x = 6
∴ their ages will be 5:4 after 6 years
(8) The ratio of present ages of Rehana and her mother is 2:7. After 2 years, the ratio of their ages will be 1:3. What is Rehana’s present age?
Ans: Let, Rehana’s age be 2x, then mothers age be 7x
After 2 years,
Rehana’s age will be 2x, then mothers age will be (7x + 2)
∴ A/Q, 2x+2/7x+2 = 1/3
= 6x + 6 = 7x + 2
= x + 4
∴ Rehana’s present age is 2×4 = 8 years
Practice Set 4.2
(1) Using the property a/b = ak/bk, fill in the blanks substituting proper numbers in the following.
(i) 5/7 = /28 = 35/ = /3.5
Ans: 5/7 = 5×4/7×4 = 20/28
5/7 = 5×7/7×7 = 35/49
5/7 = (5× 5/10)/(7× 5/10) = 2.5/3.5
∴ 5/7 = 20/28 = 35/49 = 2.5/3.5
(ii) 9/14 = 4.5/ = /42 = /3.5
Ans: 9/14 = (9× 5/10)/(14× 5/10) = 4.5/7
9/14 = 9×3/14×3 = 27/42
9/14 = (9× 5/20)/(14× 5/20) = 2.25/3.5
∴ 9/14 = 4.5/7 = 27/42 = 2.25/3.5
(2) Find the following ratios.
(i) The ratio of radius to circumference of the circle.
Ans: Let, radius = r
We know, circumference = 2πr
∴ radius/circumference = r/2πr = 1/2π
(ii) The ratio of circumference of circle with radius r to its area.
Ans: Area of circle = πr²
∴ Circumference/area = 2πr/πr² = 2/r
(iii) The ratio of diagonal of a square to its side, if the length of side is 7 cm
Ans: We know,
Diagonal of a square = √2 × side of square
= √2 × 7
= 7√2
∴ Diagonal of a square/sides of a square = 7√2/7 = √2/1
(iv) The lengths of sides of a rectangle are 5cm and 3.5cm. Find ratio of its perimeter to area.
Ans: Perimeter/area = 2(5+3.5)/5×3.5 cm² = 17/17.5 = 170/175
= 34/35
(3) Compare the following pairs of ratios.
(i) √5/3, 3/√7
Ans: √5×√7 = √35
3×3 = 9 = √81
∵ 81 > 35
= √81 > √35
= √5/3 < 3/√7
(ii) 3√5/5√7, √63/√125
Ans: 3√5 × √125 = 75
5√7 × √63 = 105
∵ 75 < 105
= 3√5/5√7 < √63/√125
(iii) 5/18, 17/121
Ans: 5 × 121 = 605
17 × 18 = 306
∵ 605 > 306
= 5/18 > 17/121
(iv) √80/√48 , √45/√27
Ans: √80 × √27 = √2160
√48 × √45 = √2160
∴ √2160 = √2160
∴ √80/√48 = √45/√27
(v) 9.2/5.1, 3.4/7.1
Ans: 9.2 × 7.1 = 65.32
5.1 × 3.4 = 17.34
∴ 65.32 > 17.34
= 9.2/5.1 > 3.4/7.1
(4) (i) ▭ ABCD is a parallelogram. The ratio of ∠A and ∠B of this parallelogram is 5:4. Find the measure of ∠B.
Ans: Let, ∠A be 5x and ∠B be 4x
Now, Sum of angles of a parallelogram = 360°
∴ ∠A + ∠B + ∠C + ∠D = 360°
= 2 (5x + 4x) = 360° (∵ Opposite ∠S of 11gm are equal)
= 18x = 360°
= x = 20°
∴ ∠B = 4 × 20° = 80°
iii) The ratio of length and breadth of a rectangle is 3:1 and perimeter is 36 cm. Find length & breadth of the rectangle.
Ans: Let, length & breadth of the rectangle be 3x and x respectively
2(3x + x) = 36
8x = 36
x = 4:5
Therefore, length = 3 x 4.5 cm = 13.5
Breadth = 4.5
- iv) The ratio of two numbers is 31 : 23 and their sum is 216. Find these numbers.
Ans: Let, the numbers be 31x and 23x = 216
54x = 216
x = 4
Therefore, The numbers are 31 x 4 = 124 and 23 x 4 = 92
v) If the product of two numbers is 360 and their ratio is 10 : 9, then find the numbers.
Ans: Let, the numbers be 10x and 9x
Therefore, 10X x 9X = 360
90x2 = 360
x2 = 360/90 = 4
x = ± 2
Therefore, x = 2
The numbers are 10 x 2 = 20 and 9 x 2 = 18
(5) If a : b = 3 : 1 and b : c = 5 : 1 then find the value of (i) (a3/15b2c) (ii) a2/7bc
Ans: a : b = 3 : 1 and b: c = 5 : 1
Let, a = 3k , b = k and b = 5l, C = l
Therefore, a = 3b, b = 5c
a = 3 x 5c = 15c and b = 5c
(a3/15b2c)3 = { (15c)3/15(5c)2c )
= (3375c3 /375c3)
= 93
= 729
(ii) a2/7bc
= (15c2)/7(5c)c
= 225c2/35c2
= 45/7
(6) If √0.04 x 0.4 x a = 0.4 x 0.04 x √b then find the ratio a/b
= If √0.04 x 0.4 x a = 0.4 x 0.04 x √b
= > √0.04 x √0.4 x √a = 0.4 x 0.04 x √b
= > √a/√b = 0.4 x 0.04 /√0.4 x √0.04
= > √a/√b = √0.4 x √0.04 = √0.016
Squaring both sides, We get =
A/b = 0.016 = 16/1000 = 2/125
(7) (x + 3) : (x + 11) = (x – 2) : (x + 1) then find the value of x.
Ans: x + 3 / x + 11 = x – 2/x + 1
= > (x + 3) (x + 1) = (x + 11) (x – 2)
= > x2 + 4x + 3 = x2 + 9x – 22
= > 5x = 19
= > x = 19/5
Practice set 4.3
1> If a/b = 7/3 then find the values of the following ratios.
i> 5a + 3b/5a-3b
Answer- a/b = 7/3
= > a/b = 7/3
= > 5a/3b = 35/9
therefore, By compounds & dividends
5a + 3b / 5a – 3b
= 35 + 9 / 35-9
= 44/26
= 22/13
ii> 2a2 + 3b2 / 2a2 – 3b2
answer- a/b = 7/3
= > 2a2 / 3b2 = 2 x 72 / 3 x 32
= 98 / 27
Therefore, 2a2 + 3b2/ 2a2– 3b2
= 98 + 27/ 98 – 27
= 125
(iii) a3 – b3 / b3
ans- a/b = 7/3
= a3/b3 = 343/27
Therefore, a3– b3/b3
= 343 – 27 / 27
= 316 / 27
iv) 7a + 9b / 7a – 9b
Ans- a/b = 7/3
= > 7a / 9b = 49/27
Therefore, 7a + 9b / 7a – 9b
= 49 + 27 / 49 – 27
= 76 / 22
= 38 / 11
2> If 15a2 + 4b2/15a2 – 4b2 = 47/7 then find the values of the following ratios.
i) a/b
ans: 15a2 + 4b2 / 15 a2 – 4b2 = 47/7
= > (15a2 + 4b2) + (15a2 – 4b2) / (15a2 + 4b2) – (15a2 – 4b2) = 47 + 7 / 47 – 7
= > 30a2 / 8b2 = 54 / 40
= > a2/b2 = 54 / 40 x 8 / 30
= > a2/b2 = 9/25
= > a/b = 3/5
(ii) 7a – 3b / 7a + 3b
Ans- a/b = 3/5
= > 7a / 3b = 21 / 15
Therefore, 7a + 3b / 7a -3b = 36 / 6
= > 7a – 3b/ 7a + 3b = 1/6
iii) b2 – 2a2/b2 + 2a2
Ans- a/b = 3/5
= > a2/b2 = 9/25
= > b2/a2 = 25/9
= > b2/2a2 = 25/18
= > b2 / 2a2 / b2 – 2a2 = 25 + 18 / 25 – 18 = 43/7
= > b2 – 2a2 / b2 + 2a2 = 7 / 43
iv) b3 – 2a3/b3 + 2a3
ans- a/b = 3/5
= > a3 / b3 = 27/125
= > 2a3/b3 = 54/125
= > b3 / 2a3 = 125 / 54
= > b3 + 2a3 / b3 – 2a3 = 125 + 54 / 125 – 54
= > b3 – 2a3 / b3 + 2a3 = 71/179
3) If 3a + 7b/3a – 7b = 4/3 then find the value of the ratio 3a2 – 7b2 / 3a2 + 7b2
ans- 3a + 7b/3a-7b = 4/3
= > (3a + 7b) + (3a – 7b) / (3a + 7b) – (3a – 7b) = 4 + 3 / 4 – 3
= > 6a / 14b = 7/1 x 14/6
= > a/b = 49/3
Therefore, a = 49k and b = 3k
Therefore, 3a2 – 7b2 / 3a2 + 7b2
= 3 (49k)2 – 7(3k2) / 3(49k2) + 7 (3k)2
= 7203K2 – 63k2 / 7203k2 + 63k2
= 7140k2 / 7266k2
= 170/173
4> Solve the following equations
i> x2 + 12x – 20/ 3x – 5 = x2 + 8x + 12 / 2x + 3
A: Multiplying both sides by ¼
x2/4 + 3x – 5 /3x – 5 = x2/4 + 2x + 3 / 2x + 3
= > (x2/4 + 3x – 5) – (3x – 5) / 3x – 5 = (x2/4 + 2x + 3) – (2x + 3)/2x + 3
= > x2/4 / 3x – 5 = x2/4/2x + 3
x = 0 is not a solution of the equation
If x ≠ 0 then, x2/4 (2x + 3)
= x2/4 (3x -5)
= > 2x + 8 , 3x – 5
= > x = 8
Therefore, x = 8 is solutions of the equation.
(ii) 10x2 + 15x + 63 / 5x2 – 25x + 12 = 2x + 3/x – 5
Ans- Multiplying denominator by 2
10x2 + 15x + 63 / 10x2 – 50x + 24 = 2x + 3 / 2x – 10
By Components & dividends,
(10 x2 + 15x + 63) + (10x2 – 50x + 24) / (10 x2 + 15x + 63) – (10x2 – 50x + 24) = (2x + 3) + (2x – 10) / (2x + 3) – (2x – 10)
= > 20x2 – 35x + 87 / 65x + 39 = 4x – 7 / 13
= > 13 (20x2 – 35x + 87) = (4x -7) (65x + 39)
= > 260x2 – 455x + 1131 = 260x2 – 299x – 273
= > 156 x = 1404
= > x = 9
Therefore, x = 9 is a solution of the equation
(iii) (2x + 1)2 + (2x – 1)2 / (2x + 1)2 – (2x – 1)2 = 17 / 8
Ans- By components & dividends,
(2x + 1)2 + (2x – 1)2 + (2x + 1)2 – (2x – 1)2 /(2x + 1)2 + (2x – 1)2 – (2x + 1)2 – (2x – 1)2 } = 17 + 8 / 17 – 8
= > 2 (2x + 1)2 / 2 (2x – 1)2 = 25/9
= > (2x + 1)2 / (2x – 1)2 = 52/33
= > 2x + 1 / 2x – 1
= > 6x + 3 = 10 x – 5
= > 4x = 8
= > x = 2
Therefore x = 2 is a solution of the equation.
iv> √4x + 1 + √x + 3 / √4x + 1 – √x + 3 = 4/1
Ans- By Componends & dividends,
√4x + 1 + √x + 3 + √4x + 1 – √x + 3 / √4x + 1 + √x + 3 – (√4x + 1 – √x + 3) = 4 + 1 / 4 – 1
= > 2 √4x + 1 / 2 √x + 3 = 5 /3
= > √4x + 1 / √x + 3 = 5/3
= > 4x + 1 / x+3 = 25/9
= > 36x + 9 = 25x + 75
= > 11x = 66
= > x = 6
Therefore, x = 6 is a solution of the equation
—–
v> (4x + 1) + (2x + 3)2 / 4x2 + 12x + 9 = 61 / 36
ans- (4x + 1)2 + (2x + 3)2 / 4x 2 + 12x + 9 = 61/36
= > 16x2 + 1 + 8x + 4x2 + 9 + 12x / 4x2 + 12x + 9 = 61 /36
= > 20x2 + 20x + 10 / 4x2 + 12x + 9 = 61/36
= > 5(4x2 + 4x + 2) / 4x2 + 12x + 9 = 61/36
= > 4x2 + 4x + 2 / 4x2 + 12x + 9 = 61/180
= > 4x2 + 4x + 2 + 4x2 + 12x + 9 / 4x2 + 4x + 2 – (4x2 + 12x + 9) = 61 + 180 / 61 – 180
= > 8x2 + 16x + 11/ – (8x + 7) = 241/-119
= > 119 (8x2 + 16x + 11) = 241 (8x + 7)
= > 952x2 + 1904x + 1309 = 1928x + 1687
= > 952x2 – 24x – 378 = 0
= > 476x2 – 12x – 189 = 0
Therefore, x = – b ± √b2 – 4ac / 2a
= 12 ± √144 + 359856 / 952
= 12 ± √360000 / 952
= 12 ± 600 / 952
Taking positive sign,
therefore, x = 12 + 600 / 952
= 612 / 952
= 9 / 14
Vi) (3x – 4)3 – (x+1)3 / (3x – 4)3 + (x + 1)3 = 61 / 189
Ans: (3x – 4)3 – (x+1)3 / (3x – 4)3 + (x + 1)3 = 61 / 189
= (3x – 4)3 – (x+1)3 + (3x – 4)3 + (x+1)3 / (3x – 4)3 – (x+1)3 – (3x – 4)3 + (x+1)3 = 61 + 189 / 61 – 189 ( By components & dividends)
= > 2 (3x – 4)3 / -2 (x + 1)3 = 250/-128
= > (3x – 4)3 / (x + 1)3 = 250 / 128 = 125/64
Use your brain power
In a certain gymnasium, there are 35 girls and 42 boys in the kid’s section, 30 girls and 36 boys in the children’s section and 20 girls and 24 boys in the teens’ section. What is the ratio of the number of boys to the number of girls in every section ? For physical exercises, all three groups gathered on the ground. Now what is the ratio of number of boys to the number of girls ? From the answers of the above questions, did you verify the theorem of equal ratios ?
Ans- ratio of no. of boys to no. of girls is given by,
In kids section, ratio = 42/35 = 6/5
In Children section, ratio = 36 / 30 = 6/5
In teens section, ratio = 24 / 20 = 6/5
Clearly , all the ratios are equal.
Now, ratio of numbers of boys to the number of girls is given as,
Sum of number of boys in every section / Sum of number of girls in every section
= 42 + 36 + 24 / 35 + 30 + 20
= 7 x 6 + 6 x 6 + 4 x 6 / 7 x 5 + 6 x 5 + 4 x 5
= 102 / 85
= 6/5
Therefore, ratio in every section = ratio of all the section.
Practice set 4.4
(1) Fill in the blanks of the following
i> X / 7 = Y/3 = 3x + 5y/…. = 7x – 9y/…..
Therefore, 3x + 5y / 21 + 15 = 3x + 5y / 36
Therefore, x/7 = y/3 = 3x + 5y/36
X/7 = Y/3 = > 7x/49 = 9y/27
Therefore, 7x – 9y / 49 – 27 = 7x – 9y / 22
Therefore, x / 7 = y / 3 = 7x – 9y / 22
Therefore, x / 7 = y/3 = 3x + 5y / 36 = 7x – 9y/22
(ii) a/3 = b/4 = c/7 = a – 2b + 3c/….. = …./ 6 – 8 + 14
ans- a/3 = b/4 = c/7
= > a/3 = 2b/8 = 3c / 21
Therefore, a – 2b + 3c / 3 – 8 + 21 = a – 2b + 3c / 16
Therefore, a/3 = b/4 = c/7 = a-2b+3c/16
Also,
a/3 = b/4 = c/7 = > 2a/6 = 2b/8 = 2c/14
Therefore, 2a – 2b + 2c / 6 – 8 + 14 = 2 (a – b + c) / 6 – 8 + 14
Therefore, a/3 = b/4 = c/7 = a – 2b + 3c /16 = 2 (a-b+c) / 6 – 8 + 14
2> 5m -n = 3m +4n then find the values of the following expressions.
i> m2 + n2 / m2 – n2
Given, 5m – n = 3m + 4n
= > 2m = 5n
= > m/n = 5/2
Therefore, m2 / n2 = 25 / 4
By compounds & dividends
m2 + n2 / m2 – n2
= 25 + 4 / 25 – 4
= 29 / 21
ii> 3m + 4n / 3m – 4n
Ans – 5m – n = 3m + 4n
= > m/n = 5/2
= > 3m + 4n / 3m – 4n = 3 m/n + 4 / 3 m/n – 4
= 3 x 5/2 + 4 / 3 x 5/2 – 4
= 15 + 8 /2 / 15 – 8/2
= 23 / 7
(i) If a(y+z) = b(z+x) = c(x+y) and out of a, b, c no two of them are equal then Show that y – z/a(b-c) = z-x / b(c-a) = x + y / c( a – b)
given, a(y + 3) = b (Z + x) = c (x + y)
Dividing by abc,
y + z / bc = z + x /ac = x+y/ab = k
Therefore, K = (x + y) – (z+x) / ab – ac = y – z / a(b-c)
K = (y + z) – (x + y) / bc – ab = z –x/ b(c-a)
k = (z + x) – (y+z)/ac-bc = x-y/c(a-b)
therefore, y – z / a(b-c) = z-x/b(c-a) = x-y / c (a-b)
ii> If x / 3x – y – z = y /3y –z –x = z/3z – x – y and x+y +z ≠ 0 then show that the value of each ratio is equal to 1
Ans- Given, x / 3x – y – z = y / 3y – 3 – x = z / 3z – x – y = K (let)
Also, K = x + y + z / 3x – y – z + 3y – z – x + 3z – x – y
= x + y + z / 3 (x + y + z) -2 (x + y + z)
= x + y + z / x + y + z
Therefore, K = 1
Therefore, the value of each ratio is equal to 1
iii> If ax + by / x+y = bx + az/x+z = ay + bz / y + z and x + y + z ≠ 0
Then show that a+b/2
Ans: Given ax + by / x + y = bx + az / x +z = ay + bz / y + 3 = k (let)
Therefore, K = ax + by + bx + az + ay + bz / x + y + x + z + y + z
= (a + b)x + (a+b) y + (a+b) z / 2(x + y + z)
= (a + b) (x + y + z) / 2 (x + y + z)
= a + b / 2
Therefore, The value of each ratio is equal to a+b / 2
iv> If y + 3 / a = 3 + x/b = x + y / c Then show that x / b+c – a = y / c + a – b = z / a + b –c
Ans- given, y+3/a = 3 + x / b = x + y /c =K (let)
Therefore, K = 3 + x + x + y – (y + 3) / b + c – a
= 3 + x + x + y – y – z / b + c –a
= 2x / b + c – a
k = x + y + y + z – (z + x) /c + a – b = x + y + y + z- z – x / c + a –b = 2y / c + a – b
k = y + z + z + x – (x + y) /a + b – c = y + z + z + x – x – y / a + b + c = 2z / a + b – c
Therefore, 2x / b + c – a = 2y / c + a – b = 2y / a + b – c
v> If 3x – 5y / 5z + 3y = x + 5z / y – 5x = y –z / x –z then show that every ratio = x / y
= Given, 3x – 5y / 5z + 3y = x + 5z / y – 5x = y –z / x –z = K (let)
Therefore, (3x – 5y) + (x + 5z) + 5 (y + z) / 5z + 3y + y – 5x + 5 (x –z )
= 3x – 5y + x + 5z + 5y – 5z / 5z + 3y + y – 5x + 5x – 5z
= 4x / 4y
= x/y
Therefore every ratio = x/y
We will post more solutions very soon, thank you