ML Aggarwal ICSE Solutions Class 10 Math 7th Chapter Ratio and Proportion
Class 10 Chapter 7 Ratio and ProportionExercise 7.1
(1) An alloy consists of 27 ½ Kg of copper and 2 ¾ Kg of tin find the ratio by weight of Tin to the alloy.
Solution:
An alloy consists of copper = 27 1/2 Kg = 55/2 kg and tin = 2 3/4 Kg = 11/4 kg
Total Wight of alloy = 55/2 + 11/4 = 110+11/4 = 121/4 kg
Thus, weight of tin/weight of alloy = (11/4)/(121/4) = 11/121 = 1/11
Thus, the required ratio of weight of tin to alloy is 1:11
(2) Find the compounded ratio of:
(i) 2:3 and 4:9
Solution:
Here, compound ratio = (2/3) × (4/9)
= 8/27
Compound ratio = 8:27
(ii) 4:5, 5:7 and 9:11
Solution:
Here, compound ratio = (4/5) × (5/7) (9/11)
= 36/77
Compound ratio = 36:77
(iii) (a – b) : (a + b), (a + b)2 : (a2 + b2) and (a4 – b4) : (a2 – b2)2
Solution:
Here, compound ratio = (a-b)/(a+b) × (a+b)2/(a2+b2) × (a4-b4)/(a2-b2)2
= (a-b) (a+b)(a4-b4)/(a2+b2)(a-b)2(a+b)2 = (a4-b4)/(a2+b2)(a-b)(a+b)
∵ a2-b2 = (a-b) (a+b)
Compound ratio = a4-b4/(a2+b2)(a-b)(a+b)
∵ a2 – b2 = (a-b) (a + b) => a4 – b4 = (a2 – b2) (a2 + b2)
=> Compound ratio = a4-b4/(a2+b2)(a-b)(a+b)
∵ a2 – b2 = (a – b)(a+b) => a4-b4 = (a2 – b2) (a2 + b2)
Compound Ratio = (a2-b2)(a2+b2)/(a2+b2)(a-b)(a+b)
= (a-b)(a+b)/(a-b)(a+b)
Compound ratio = 1/1 = 1:1
(3) Find the duplicate ratio of–
(i) 2:3
=> Duplicate ratio of 2:3 → (2)2: (3)2 = 4:9
(ii) √5:7
=> Duplicate ratio of √5:7 => (√5)2 : 72 = 5:49
iii) 5a: 6b
=> Duplicate ratio of 5a:6b => (5a)2 : (6b)2 = 25a2 : 36b2
(4) Find the triplicate ratio of
(i) 3:4
=> Triplicate ratio of 3:4 => (3)3:(4)3 = 27:64
(ii) 1/2 : 1/3
=> Triplicate ratio of 1/2 : 1/3 => (1/2)3:(1/3)3 = 1/8 : 1/27 = 27:8
(iii) 13 : 23
=> Triplicate ratio of 13 : 23 → (13) :(23) = 13: 83= 1:512
(5) Find the sub-duplicate ratio of
(i) 9:16
=> Sub duplicate ratio of 9:16 => √9 : √16 = √9 : √16 = 3:4
(ii) 1/4 : 1/9
=> Sub-duplicate ratio of 1/4:1/9 => 1/√4 : 1/√9 = 1/2 : 1/3 = 3:2
(iii) 27a3 : 9a2 : 49b2
=> Sub-duplicate ratio of 9a2 : 49b2 => √9a2 : √49b2 = 3a:7b
(6) Find the sub-triplicate ratio of
(i) 1:216
=> Sub-triplicate ratio of 1:216 => ∛1 : ∛216 = 1:6
(ii) 1/8 : 1/125
=> Sub-triplicate ratio of 1/8 : 1/125 => (1/8)1/3 : (1/125)1/3 = 1/2 : 1/ = 5:2
(iii) 27a3 : 64b3
=> Sub-triplicate ratio of 27a3 : 64b3 => (27a3)1/3 : (64b)3 = 3a : 4b
(7) Find the reciprocal ratio of
(i) 4 : 7
=> Reciprocal ratio of 4:7 => 7:4
(ii) 32 : 42
=> Reciprocal ratio of 32 : 42 => 42 : 32 = 16:9
(iii) 1/9 : 2
=> Reciprocal ratio of 1/9 : 2 => 2 : 1/9 = 18:1
(8) Arrange the following ratios in ascending order of magnitude 2:3, 17:21, 11:14 and 5:7
Solution:
Given ratios are => 2:3, 17:21, 11:14 and 5:7
Ratios in fractions => 2/3, 17/21, 11/14 & 5/7
Here, the LCM of 3, 21, 14 & 7 is 42
Thus, we cannot ratio as equivalent.
=> 2/3 = 2×14/3×14 = 28/42
17/21 = 17×2/21×2 = 34/42
11/14 = 11×3/14×3 = 33/42
5/7 = 5×6/7×6 = 30/42
Now, the ratios in ascending order will be as follows:
28/42, 30/42, 33/42, 34/42 => 2/3, 5/7, 11/14, 17/21
(9) (i) If A:B = 2:3, B:C = 4:5 and C:D = 6:7 find A:D
Solution:
Given that, A:B = 2:3, B:C = 4:5 and C:D = 6:7
Then, A/B = 2/3, B/C = 4/5 and C/D = 6/7
We find A:D => A/B × B/C × C/D = A/D
(2/3) × (4/5) × (6/7) = A/D
16/35 = A/D
A:D = 16:35
(ii) If x:y = 2:3 and Y:Z = 4:7, find X:Y:Z
Solution:
Given that, x:y = 2:3, y:z = 4:7
Then, x/y = 2/3 and y/z = 4/7
LCM of 3 & 4 is 12 => By making equals of Y as 12.
=> x/y = 2×4/3×4 = 8/12 => x:y = 8:12
y/z = 4×3/7×3 = 12/21 => y:z = 12:21
Thus, x:y:z = 8:12:21
(10) (i) If A:B = 1/4 : 1/5 and B:C = 1/7 : 1/6, find A:B:C
Solution:
Given that, A:B = 1/4 : 1/5 = 5:4 and B:C = 1/7:1/6 = 6:7
Hence, A/B = 5/4 and B/C = 6/7
Making equals of B as 12, (LCM of 6&4 is 12)
=> A/B = 5×3/4×3 = 15/12 => A:B = 15:12
B/C = 6×2/7×2 = 12/14 => B:C = 12:14
Thus, A:B:C = 15:12:14
(ii) If 3A = 4B = 6C find A:B:C
Solution:
Given that, 3A = 4B = 6C
Now, 3A = 4B and 4B = 6C
A/B = 4/3 and B/C = 6/4 = 3/2
Thus, A:B:C = 4:8:2
(11) (i) If 3x+5y/3x-5y = 7/3, find x:y
Solution:
Given that, (3x+5y)/(3x-5y) = 7/3
9x + 15y = 21x – 35y
12x = 50y
x/y = 50/12 = 25/6 => x:y = 25:6
(ii) If a:b = 3:11, find (15a – 3b) : (9a + 5b)
Solution:
Given that, a:b = 3:11
= a/b = 3/11
Now, (15a-3b)/(9a+5b) = (15a/b-3)/(9a/b+5) Dividing N or & D or by b.
= (15×3/11-3)/(9×3/11+5) ∵ a/b = 3/11
(15a-3b)/(9a+5b) = (45-33)/(27+55) = (12)/(82) = 6/41
Thus, (15a-3b) : (9a+5b) = 6:41
(12) (i) If (4x2 + xy) : (3xy – y2) = 12:5, find (x +2y) : (2x + y)
Solution:
Given that, (4x2 + xy) : (3xy – y2) = 12:5
(4x2+xy)/(3xy-y2) = 12/5
20x2+ 5xy = 36xy – 12y2
20x2-31xy+12y2 = 0
20 x2/y2– 31xy/y2+ 12y2/y2 = 0
20 (x/y)2– 31(x/y) + 12 = 0
20 (x/y)2– 15(x/y) – 16 (x/y) +12 = 0
5 (x/y) [4 (x/y) – 3] – 4 [4(x/y)-3] = 0
[4(x/y) – 3] [5 (x/y) – 4] = 0
4 (x/y) – 3 = 0 or 5 (x/y) – 4 = 0
4 (x/y) = 3 or 5 (x/y) = 4
x/y = 3/4 or x/y = 4/5
Now, (x+2y)/(2x+y) = (x/y+2)/(2x/y+1)
(a) If x/y = 3/9 then (x/y+2)/(2x/y+1) = (3/4+2)/(6/4+1) =(3+8)/(6+4) = 11/20
Thus, (x + 2y) : (2x + y) = 11:20
(b) If x/y = 4/5 then (x/y+2)(2x/y+1) = (4/5+2)/(8/5+1) = (4+10)/(8+5) = 14/13
Thus, (x + 2y) : (2x + y) = 14 : 13
(13) (i) if (x-9) : (3x+6) is the duplicate ratio of 4:9, find the value of x.
Solution:
Given that, (x – 9) : (3x+6) is the duplicate ratio of 4:9
But, duplicate ratio of 4:9 = 42:92 = 16:18
Thus, (x-9)/(3x+6) = 16/81
81x – 9 (81) = 48x + 96
81x – 48 x = 96 + 9 (81)
33x = 96 + 729
33x = 825
x = 825/33
x = 25
(ii) If (3x+1) : (5x+3) is the triplicate ratio of 3:4, find the value of x.
Solution:
Given that, (3x+1) : (5x+3) is the triplicate ratio of 3:4.
But, the triplicate ratio of 3: 4 = 33 : 43 = 27 : 64
Thus, (3x+1)/(5x+3) = 27/64
192x + 64 = 135x + 81
192x – 135x = 81 – 64
51x = 17
x = 17/57 is the required value of x.
(iii) If (x +2y) : (2x-y) is equal to the duplicate ratio of 3:2, find x:y.
Solution:
Given that, (x+2y) : (2x-y) is equal to the duplicate ratio of 3:2
But, duplicate ratio of 3:2 = 32:22 = 9:4
Thus, (x+2y)/(2x-y) = 9/4
4x + 8y = 18x – 9y
18x – 4x = 8y + 9y
14x = 17y
x/y = 17/14
Thus, x:y = 17:14
(14) (i) Find two numbers in the ratio of 8:7 such that when each is decreased by 12 1/2, they are in the ratio 11:9.
Solution:
Given that, two numbers in the ratio of 8:7 such that when each is decreased by 12 1/2, they are in the ratio 11:9.
Let us consider the numbers 8x & 7x.
From given condition,
(8x-25/2/(7x-25/2) = 11/9
After talking LCM, ((16x-25)/2)/((14x-25)/2) = 11/9
(16x-25)/(14x-25) = 11/9
154x – 144x = 275 – 225
10x = 50
x = 50/10 = 5
x = 5
Thus, 8x = 8×5 = 40
7x = 7×5 = 35
Thus, the required numbers are 40 & 35 respectively.
(ii) The income of a man is increased in the ratio of 10:11, if the increases in his income is Rs.600 per month, find his new income.
Solution:
Let us consider the present income be ‘10x’ and increased income be ‘11x’
Increase in has income per month = 600
Thus, 11x – 10x = x =600 Rs.
Hence, new income = 11x = 11×600 = 6600.
15) (i) A women reduces her weight in the ratio 7: 5, What does her weight become, if originally it was 91 kg.
Solution:
(i) Given that, ratio between the original weight & reduced weight = 7: 5.
Let us consider, the original weight = 7x and reduced weight = 5x
If original weight is 91 kg, then, 91kg = 7x
x =13
Hence, the reduced weight = 5x = 13×5 = 65 kg
(ii) A school collected Rs. 2100 for charity. It was decided to divide money between an orphanage and a blind school in the ratio of 3:4. How much money did each receive?
Solution:
Amount collected for charity = Rs. 2100/-
Given that, the ratio between orphanage & a blind school is 3:4
Then, sum of ratio = 3 + 4 = 7
Thus, Orphanage school share = 2100 × 3/7 = Rs. 900
And blind school share = 2100 × 4/7 = Rs. 1200
(16) The sides of a triangle are in the ratio 7:5:3 & its perimeter is 30 cm. Find the lengths of sides.
Solution:
Given that, Perimeter of triangle = 30cm
Ratio among the sides = 7:5:3
Sum of ratios = 7+5+3 = 15
Thus, Length of first side = 90×7/15 = 14 cm
Length of second side = 30×5/15 = 10 cm
Length of third side = 30×3/15 = 6 cm
Thus, the sides of triangle are 14 cm, 10 cm, 6 cm.
(17) Three numbers are in the ratio 1/2 : 1/3 : 1/4 if the sum of their squares is 244, find the numbers.
Solution:
Given that, ratio of three numbers = 1/2 : 1/3 : 1/4
= (6:4:3)/12
= 6:4:3
Then, First number = 6x
Second number = 4x
Third number = 3x
From given condition,
(6x)2 + (4x)2 + (3x)2 = 244
36x2 + 16x2 + 9x2 = 244
61x2 = 244
x2 = 244/61 => 4 = 22
x = 2
Thus, first number = 6x = 6×2 = 12
Second number = 4x = 4×2 = 8
Third number = 3x = 3×2 = 6
(18) (i) A certain sum was divided among A, B and C in the ratio 7:5:4, If B got 500 more than C, find the total sum divided.
Solution:
Given that, Ratio between A, B and C = 7:5:4
Let us consider, A’s share = 7x
B’s share = 5x and C’s share = 4x
Then, total sum = 7x + 5x + 4x = 16x
= 5x – 4x = 500
x = 500
Hence, the total sum = 16x = 16×500 = Rs. 8000.
(19) (i) In the mixture of 45 liters, the ratio of milk to water is 13:2. How, much water must be added to this mixture to make the Ratio of milk to water as 3:1.
Solution:
Given that, mixture of milk & water = 45 liters
And ratio of milk & water = 13:2
Sum of ratio = 13 + 2 = 15
Then, Quantity of milk = (45×13)/15 = 39 liters.
Quantity of water = (45×2)/15 = 6 liters.
Let us consider, x be the water to be added
Thus, water = (6+x) liters
Thus, the new ratio = 3:1
39 : (6+x) = 3:1
39/(6+x) = 3/1
39 = 18 + 3x
3x = 39 –18 = 21
x = 21/3 = 7 liters
Thus, 7 liter water is added to make the ratio of milk to water as 3:1.
(20) The monthly pocket money of Ravi & Sanjeev is in the ratio 5:7. There expenditures are in the ratio 3:5, If each save Rs. 80 every month, find their monthly pocket money.
Solution:
Let us consider the monthly pocket money of Ravi & Sanjeev be 5x and 7x respectively.
Then, their expenditure be 3y and 5y respectively.
Thus, 5x – 3y = 80
7x – 5y = 80
(1) × 7 => 35x – 21y = 560
(2) × 5 => 35x – 25y = 400
(2) – (1) => 4y = 160
y = 40
Thus, 5x = 80+3×40 = 200
x = 40
Thus, monthly pocket money of Ravi is = 5×40 = 200.
(21) In an examination, the ratio of passes to failures was 4:1. If 30 less had appeared and 20 less passed, the ratio of passes To failures would have been 5:1, How many students appeared For the examination.
Solution:
Let us consider, the no. of passes = 4x and no. of failures = x
Thus, the total no. of students appeared = 4x + x = 5x
From second condition,
The no. of students appeared = 5x – 30 and no. of passes = 4x – 20
Thus, the no. of failures found to be = (5x – 30) – (4x – 20)
= 5x – 30 – 4x + 20
= x – 10
Thus, (4x-20)/(x-10) = 5/1
5x-50 = 4x-20
5x-4x = -20+50
x = 30
Thus, the no. of students appeared are = 5x = 5×30 = 150.
Exercise 7.2
(1) Find the value of x in the following proportions:
(i) 10:35 = x:42
Solution:
Given that, 10:35 = x:42
10/35 = x/42
35x = 420
x = 420/35
x = 12
This is the required value.
(ii) 3:x = 24:2
Solution:
Given that, 3:x = 24:2
3/x = 24/2
6 = 24x
x = 6/24
x = 1/4
This is the required value.
(iii) 2.5 : 1.5 = x:3
Solution:
2.5/1.5 = x/3
7.5 = 1.5x
x = 7.5/15 = 75/15
x = 5
This is the required value.
(iv) x:50 = 3:2
Solution:
Given that x:50 = 3:2
x/50 = 3/2
2x = 150
x = 150/2
x = 75
This is the required value.
(2) Find the fourth proportional to
(i) 3, 12, 15
Solution:
Let us consider fourth proportional to 3, 12, 15 be x.
Then, 3:12 :: 15:x
3x = 12 ×15
x = 60
(ii) 1/3, 1/4, 1/5
Solution:
Let us consider fourth proportional to 1/3, 1/4, 1/5 be x
Then, 1/3 : 1/4 :: 1/5 : x
1/3 : 1/4 :: 1/5 : x
1/3 (x) = (1/4) (1/5)
x/3 = 1/20
x = 3/20
This is the required value.
(iii) 1.5, 2.5, 4.5
Solution:
Let us consider fourth proportional to 1.5, 2.5, 4.5 be x
Then, 1.5 : 2.5 :: 4:5 : x
1.5 (x) = 2.5 (4.5)
x = (2.5 ×4.5)/1.5
x = 7.5
This is the required value.
(3) Find the third proportion to,
(i) 5, 10
Solution:
Let us consider third proportional to 5, 10 b x.
Then, 5:10::10:x
=> (5x) = 10×10
x = 100/5
x = 20
Thus, the third proportional to 5, 10 is 20.
(ii) Rs. 3, Rs. 12
Solution:
Let us consider third proportional to 3, 12 be x.
=> 3x = 12 × 12
x = 144/9
x = 48
Thus, the third proportional to 3, 12 is 48.
(4) Find the mean proportion of
(i) 5 and 80
Solution:
Let us consider mean proportion of 5 & 80 be x.
Then, 5 : x :: x : 80
x2 = 5×80 = 400
x = 20
Thus, the mean proportion of 5&80 is 20.
(ii) 1/12 & 1/75
Solution:
Let us consider mean proportion of 1/12 & 1/75 be x.
Then, 1/12 : x :: x : 1/75
x2 = 1/12 × 1/75 = 1/900
x = 1/30
Thus, the mean proportion of 1/12 & 1/75 is 1/30.
(5) If a, 12, 16, and b are in continued proportion find a and b.
Solution:
Given that, a, 12, 16, and b are in continued proportion
Then,
a/12 = 12/16 = 16/b
a/12 = 12/16 and 12/16 = 16/b
16a = 144 and 12b = 256
a = 144/16, b = 256/12
a = 9, b = 64/3
Thus, the required values of a and b found to be 9 & 64/3 respectively.
(6) What number must be added to each of the no. 5, 11, 19, & 37 so that they are in proportion.
Solution:
Let us consider, ‘x’ be the number added to 5, 11, 19, & 37 to make in Proportion.
Then, (5+x) : (11+x) :: (19+x) : (37+x)
(5 + x) (37 + x) = (11 + x) (19 + x)
= 185 + 5x + 37x + x2= 209 + 11x + 19x + x2
185 + 42x + x2 = 209 + 30x + x2
42x – 30x + x2 – x2 = 209 – 185
12x = 24
x = 2
This is the required number.
(7) What number should be subtracted from each of the numbers 23, 30, 57 and 78 so that the remainders are in proportion.
Solution:
Let us consider ‘x’ be the number subtracted from each term.
Then, (23 – x), (30 – x), (57 – x) and (78 – x) are proportional.=
Thus, (23-x) : (30-x) : (57-x) : (78-x)
(23-x)/(30-x) = (57-x)/(78-x)
1794 -23x-78x+x2 = 1710-57x+x2
X2-101x+1794-x2+87x-1710 = 0
-14x+84=0
14x = 84
x = 84/14 = 6
X = 6
This is the required number.
(8) The following numbers (k+3), (k+2), (3k-7) and (2k-3) are in proportion find ‘k’.
Solution:
Given that, (k+3), (k+2), (3k-7) and (2k-3) are In proportion
Then, (k+3)(k+2) = (3k+7)/(2k-3)
= (k+3) (2k-3) = (k+2) (3k-7)
2k2-3k+6k-9 = 3k2-7k+6k-14
K2-4k-5=0
(k-5) (k+1) = 0
K = 5 or k = 1 is the required value of ‘k’.
(9) If (x+5) is the mean proportion between (x+2) and (x+9), Find the value of ‘x’.
Solution:
Given that, (x+5) is the mean proportion between (x+2) and (x+9)
Then, (x+5)2 = (x+2) (x+9)
X2+10x+25 = x2+11x+18 and x2+10x-x2-11x = 18-25
Thus, -x = -7
X = 7 is the required value.
(10) What number must be added to each of the numbers 16, 26 and 40 So that the resulting numbers may be in continued proportion.
Solution:
Let us consider ‘x’ be the number added to each term.
Then, (16 + x), (26 + x), and (40 + x) are in continued proportion.
= (16 + x)/(26+x) = (26+x)/(40+x)
= (16+x) (40+x) = (26+x) (26+x)
640 + 16x + 40x + x2 = 676 + 26x + 26x + x2
640 + 56x + x2 = 676 + 52x + x2
56x + x2 – 52x – x2 = 676 – 640
4x = 36
x = 36/4 = 9
x = 9 is the required value.
(12) If ‘b’ is the mean proportional between a and c, prove that a, c, (a2+b2) and (b2+c2) are proportional.
Solution:
Given that, b is the mean proportional between a and c.
Then, b2 = a c
And, a, c, (a2+b2) and (b2 + c2) are proportional
Thus, a/c = (a2+b2)/(b2+c2)
a (a2+c2) = c (b2+b2)
=> a (ac + c2) = c (a2 + ac)
ac (a + c) = a2c + ac2
Thus, ac (a + c) = ac (a + c)
(14) If y is mean proportional between x and z, prove that xyz (x + y + z)3 = (xy + yz + zx)3
Solution:
Given that, y is mean proportional between x and z.
Then, y2 = xz
Now L.H.S. = xyz (x + y + z)3
= xz . y (x + y + z)3
= y2 . y (x + y + z)3
= y3 (x +y +z)3
= [y (x + y + z)]3
= (xy+y2+yz)3
= (xy + xz + yz)3
(x + y + z)2 = (xy + yz + zx)3 (Hence Proved)
(15) If (a+c) = mb and (1/b + 1/d) = m/c, prove that a, b, c and d are in proportion.
Solution:
Given that, (a + c) = mb and (1/b + 1/d) = m/c
Divide by – b => (a+c)/b = m, multiply by c
a/b + c/b = m —- (1)
c/b + c/d = m —– (2)
From (1) & (2) => a/b + c/b= c/b + c/d
a/b = c/d
Thus, a, b, c and d are in proportion.
(16) If x/a = y/b = z/c prove that x3/a2 + y3/b2 + z3/c2 = (x + y + z)3/(a+b+c)2
Solution:
Let us consider, x/a = y/b = z/c = k
Then, x = ak, y = bk and z = ck
LHS = x3/a2 + y3/b2 + z3/c
LHS = a3k3/a2 + b3k3/b2 + c3k3/c2 = (a + b + c) k3 —— (1)
Now, RHS = (x+y+z)3/(a+b+c)2
= (ak+bk+ck)3/(a+b+c)2 = (k3(a+b+c)3/(a+b+c)2
RHS = k3 (a + b + c) —– (2)
From (1) and (2) => x3/a2 + y3/b2 + z3/c2 = (x+y+z)3/(a+b+c)2 (Hence Proved)
(17) If a/b = c/d = c/f then prove that
(b2 + d2 + f2) (a2 + c2 + c2) = (ab + cd + ef)2
Solution:
Let us consider, a/b = c/d = e/f = k
Then, a = bk, c = dk, e = fk
Now, L.H.S. = (b2+d2+f2) (a2+c2+e2)
= (b2+d2+f2) (b2k2+d2k2+f2k2)
= k2 (b2+d2+f2) (b2+d2+f2)
L.H.S. = k2 (b2+d2+f2)2 —– (1)
Now, RHS = (ab+cd+ef)2
= [(bk)b + (dk)d + (fk)f ]2
= (b2k + d2k + f2k)2
R.H.S. = k2 (b2 + d2 + f2)2 —- (2)
From (1) & (2) = (b2 + d2 + f2) (a2 + c2 + e2) = (ab + cd + ef)2 (Hence proved)
(18) If ax = by = cz then prove that x2/yz + y2/3x + z2/x = bc/a2 + ca/b2 + ab/c2
Solution:
(19) If a, b, c and d are in proportion. Prove that,
(i) (5a + 7b) (2c – 3d) = (5c + 7d) (2a – 3b)
Solution:
Given that, a, b, c and d are in proportion
Let us consider, a/b = c/d = k
= a = bk, c = dk
Now, L.H.S = (5a + 7b) (2c – 3d)
= (5bk+7b) (2dk-3d)
= k2 (5a+7b) (2d-3d)
= k2(12b) (-d)
L.H.S= -12bdk2 —— (1)
And R.H.S = (5c + 7d) (2a – 3b)
= (5dk + 7d) (2bk-3b)
= k2 (5d+7d) (2bk-3b)
= k2 (12d) (-b)
R.H.S = -12bdk2 —— (2)
From (1) & (2) = (5a+7b) (2c-3d) = (5c+7d) (2a-3b) Hence proved.
(20) If x, y, z, are in continued proportion, prove that,
(x+y)2/(y+z)2 = x/z
Solution:
Given that, x, y, z, are in continued proportion
Let us consider, x/y = y/z = k
Then, y = kz
x = yk = (kz) k = k2 z
Now, L.H.S = (x + y)2/(y + z)2
= (k2 z + kz)2/(kz + z)2
L.H.S = [kz (k + 1)2] / [z2(k+1)2]
L.H.S. = k2 —— (1)
Now, RHS = x/y = k2z/z = k2 ——- (2)
From (1) & (2) => (x+y)2/(y+z)2 = x/z (Hence Proved)
(22) If a, b, c are in continued proportion then prove that, (a+b)/(b+c) = a2(b-c)/b2(a-b)
Solution:
Given that, a, b, c are in continued proportion.
Let us consider, a/b = b/c = k
= a = bk
b = ck
Then (a+b)/(b+c) = (bk+b)/(ck+c) = (k+1)b/(k+1)c = b/c —— (1)
Now, RHS = a²(b-c) /b2(a-b) = b2k2 (b-c) /b²(b2k2-b) = k2(b-c)/b(bk-1) = k —— (1)
From (1) & (2) => (a+b)/(b+c) = a2(b-c)/b2(a-b) (Hence proved)
(23) If a, b, c, d are in continued proportion, prove that (a3+b3+c3)/(b3+c3+d3) = a/b
Solution:
Given that, a, b, c, d are in continued proportion
Let us consider, a/b = b/c = c/d = k
∴ a = bk, c = dk, b = ck = dk²
a = bk = ck² = dk3
a = dk3, b = dk²
Exercise 7.3
(1) If a:b::c:d, prove that
(i) (2a+5b)/(2a-5b) = (2c+5d)/(2c-5d)
Solution:
Give that, a:b::c:d
Then, a/b = c/d
2a/5b = 2c/5d
By componendo & dividendo,
(2a+5b)/(2a-5b) = (2c+5d)/(2c-5d) (Hence Proved)
(ii) (5a+11b)/(5c+11d) = (5a-11b)/(5c-11d)
Solution:
Given that, a:b::c:d
Then, a/b = c/d
5a/11b = 5c/11d
By componendo & dividendo,
(5a+11b)/(5a-11b) = (5c+11d)/(5c-11d)
=> (5a+11b)/(5c+11d) = (5a-11b)/(5c-11d) (Hence Proved)
(2) If (5x+7y)/(5u+7v) = (5x-7y)/(5u-7v), show that x/y = u/v
Solution:
Given that, (5x+7y)/(5u+7v) = (5x-7y)/(5u-7v)
Then, by alternendo,
(5x+7y)/(5u+7y) = (5x-7y)/(5u-7v)
By componendo and dividendo,
(5x+7y+5x-7y)/(5x+7y -5x+7y) = (5u+7v+5u-7v)/(5u+7u-5u+7v)
10x/14y = 10u/14v
x/y = u/v (Hence Proved)
(3) If (4a+5b) (4c-5d) = (4a-5b) (4c+5d), prove that a, b, c, d, are in proportion.
Solution:
Given that, (4a+5b) (4c-5d) = (4a-5b) (4c+5d)
=> (4a+5b)/(4a-5b) = (4c+5d)/(4c-5d)
By Componendo & dividendo,
(4a+5b+4a-5b)/(4a+5b-4a+5b) = (4c+5d+4c-5d)/(4c+5d-4c+5d)
8a/10b = 8c/10d
a/b = c/d
a/b = c/d
Hence, a, b, c are in proportion proved.
(4) If (pa+qb) : (pc+qd) :: (pa-qb) : (pc-ad) prove that a:b::c:d
Solution:
Given that, (pa+qb) : (pc+qd) :: (pa-qb) : (pc-ad)
=> (pa+qd)/(pc+qd) = (pa-qb)/(pc-ad)
=> (pa+qb)/(pa-qd) = (pc+qd)/(pc-qd)
By componendo and dividendo,
(pa+qb+pa-qb)/(pa+qb-pa+qb) = (pc+qd+pc-qd)/(pc-qd-pc+qd)
(2pa)/(pqb) = (2pc)/(2qd)
=> a/b = c/d
Thus, a:b::c:d (Hence Proved)
(5) If (ma+nb) : b :: (mc+nd) : d, prove that a, b, c, d, are in proportion.
Solution:
Given that, (ma+nb) : b :: (mc+nd) : d
(ma+nb)/b = (mc+nd)/d
= mad+nbd = mbc+nbd
= mad+mbc
ad = bc
a/b = c/d
Thus, a:b::c:d
Hence, a, b, c, d, are in proportion.
(6) If (11a2 + 13b2) (11c2 – 13d2) = (11a2 – 13b2) (11c2 + 13d2), then prove that, a:b::c:d
Solution:
Given that, (11a2+13b2) (11c2-13d2) = (11a2-13b2) (11c2+13d2)
(7) If x = 2ab/(a+b) find the value of (x+a)/(x-a) + (x+b)/(x-b)
Solution:
Given that, x = 2ab/(a+b)
x/a = 2b/(a+b
(8) If x = 8ab/(a+b) find the value of (x+4a)/(x-4a) + (x+4b)/(x-4b)
Solution:
Given that, x = 8ab/(a+b)
x/4a = 2b/(a+b) By Componendo & dividendo,
(x+4a)/(x-4a) = (2b+a+b)/(2b-a-b) = (3b+a)/(b-a) —– (1)
Now, x/4b = 2a/(a+b) By Componendo & dividendo,
(x+4b)/(x-4b) = (2a+a+b)/(2a-a-b) = (3a+b)/(a-b) —- (2)
Now, (x+4a)/(x-4a) + (x+4b)/(x-4b) = (3b+a)/(b-a) + (3a+b)/(a-b)
= -3b-a+3a+b/(a-b)
= (-2b+2a)/(a-b) = 2(a-b)/(a-b) = 2
Thus, (x+4a)/(x-4a) + (x+4b)/(x-4b) = 2
(9) If, x = 4√6/√2+√3, then find the value of (x+2√2)/(x-2√2) + (x+2√3)/(x-2√3)
Solution:
(10) Using properties of proportion, find x from the following:
(i) √2-x + √2+x/√2-x – √2+x = 3
Solution:
(ii) √x+4 + √x-10/√x+4 – √x-10 = 5/2
Solution
(11) Using properties of proportion, solve for x.
(i) 3x + √9x2-5/3x-√9x2-5 = 5
Solution:
12x = 6√9x2-5
2x = √9x2 – 5 squaring on both sides
4x2 = 9x2 – 5
9x2 – 4x2 – 5 =
5x2 = 5
x2 = 1
∴ x = ±1
But, when x = 1 => 3(1)+ √9-5/3(1)- √9-5 = 3+√4/3-√4 = 3+2/3-2 = 5/1
Hence, the required value of x is found to x = +1
(ii) 2x + √4x2-1/2x – √4x2-1 = 4
Solution:
Given that, (2x + √4x2-1)/(2x- √4x2-1) = 4
By Componendo & dividendo
= (2x+√4x2-1+2x-√4x2-1)/(2x+√4x2-1-2x+√4x2-1) = (4+1)/(4-1) = 5/3
=> (4x)/2√4x2-1 = 5/3
=> 12x/√4x2 – 1 = 10
=> 6x = 5√4x2-1
Squaring on both sides,
36x2 – 25 (x2 – 1)
=> 36x2 – 100x2 = – 25
-64x2 = – 25
x2 = +25/64
= x = 5/8
∵ x is positive
This is the required value of x
(12) Solve: (1+x+x2)/(1-x+x2) = 62(1+x)/63(1-x)
Solution:
Given that,
(1+x+x2)/(1-x+x2) = 62(1+x)/63(1-x)
(1-x)(1+x+x2)/(1+x)(1-x+x2) = 62/63
(1+x)3/(1-x3) = 63/62
∵ By simplifying
By componendo & dividendo,
(1+x3+1-x3)/(1+x3-1+x3) = (63+60)/(63+62)
2/2x3 = 125/1
1/x3 = 125
∴ x3 = 1/125 => x = 1/5
This is the required value of x.
(13) Solve for x: 16 (a-x3/a+x) = (a+x)/(a-x)
Solution:
Given that, 16 (a-x/a+x)3 = (a+x)/(a-x)
=> (a+x)3(a+x)/(a-x)(a-x)3 = 16
=> [(a+x)/(a-x)]4 = (±2)4
(a+x)/(a-x) = ±2
When (a+x)/(a-x) = 2/1
By componendo & dividendo
(a+x+a-x)/(a+x-a+x) = 3/1
= 2a/2x = 3
= a/x = 3
x = a/3
When (a+x)/(a-x) = -2/1
By componendo & dividendo
(a+x+a-x)/(a+x-a+x) = (-2+1)/(-2-1) = -1/-3 = 1/3
2a/2x = 1/3
=> a/x = 1/3
=> x = 3a
Hence, the required values of x are found to be a/3, 3a.
(14) If x = (√a+x + √a-1)/( √a+1-√a-1), Using properties of proportion, show that (x2 – 2ax + 1) = 0
Solution:
(15) Given, x = √a2+b2+√a2-b2/√a2+b2-√a2-b2 use componendo & dividendo to prove b2 = 2a2x/(x2+1)
Solution:
(16) Given that: (a3 + 3ab2)/(b3+3a2b) = 63/62, using componendo & dividendo find a:b.
Solution:
Given that, (a3 + 3ab2)/(b3+3a2b) = 63/62
By componendo & dividendo
(a3+3ab2+b3+3a2b)/(a3+3ab2-b3-3a2b) = (63+62)/(63-62) = 125/1
= (a+b)3/(a-b)3 = (5/1)3
= (a + b) = 5a-5b
5a – a – 5b – b = 0
4a – 6b = 0
4a = 6b
a/b = 6/4 = 3/2
Thus, a:b = 3:2
(17) Given (x3+12x)/(6x2+8) = (y3+27y)/(9y2+27), Using componendo & dividendo find x:y.
Solution:
(18) Using the properties of proportion, solve the following equation for x; Given => x3+3x/3x2+1 = 341/91
Solution:
∴ (x+1)/(x-1) = 6/5
6x – 6 = 5x + 5
=> 6x – 5x = 5+6
x = 11
This is the required value of x.
(19) If (x+y)/(ax+by) = (y+z)/(ay+bz) = (z+x)/(az+bx), prove that each of these ratio is equal to 2/(a+b) unless x+y+z = 0
Solution:
Also See: