# ML Aggarwal Solutions Class 10 Maths Chapter 7 Ratio and Proportion

1. An alloy consists of 27 ½ kg of copper and 2 ¾ kg of tin. Find out the ratio by weight of tin to the alloy.
Solution:
As we have given that
Copper = 27 ½ kg = 55/2 kg
Tin = 2 ¾ kg = 11/4 kg

As We know that
Total alloy = 55/2 + 11/4
Taking LCM
= (110 + 11)/ 4
= 121/4 kg

Here
Ratio between tin and alloy = 11/4 kg: 121/4 kg
So we get
= 11: 121
Hence = 1: 11

2 Find out the compounded ratio of:
(i) 2: 3 and 4: 9
(ii) 4: 5, 5: 7 and 9: 11
(iii) (a – b): (a + b),(a + b)2 ∶ (a2  + b2  ) and (a4   – b4  ): (a2  – b2  )2
Solution:

(i) 2: 3 and 4: 9
As We know that
Compound ratio = 2/3 × 4/9
= 8/27
= 8: 27

(ii) 4: 5,5: 7 and 9: 11
As We know that
Compound ratio = 4/5 × 5/7 × 9/11
=36/77= 36: 77

3. Find out the duplicate ratio of
(i) 2: 3
(ii) √5: 7
(iii) 5a: 6b
Solution:
(i) 2: 3
As We know that
Duplicate ratio of 2: 3 = 22: 32  = 4: 9

(ii) √5: 7
As We know that
Duplicate ratio of √5: 7 = √52 ∶ 72  = 5: 49

(iii) 5a: 6b
As We know that
Duplicate ratio of 5a: 6b = (5a)2 ∶ (6b)2  = 25a2 ∶ 36b2

4. Find out the triplicate ratio of
(i) 3: 4
(ii) ½: 1/3
(iii) 13 ∶ 23
Solution:

(i) 3: 4
As We know that
Triplicate ratio of
3: 4 = 33∶ 43 = 27: 64

(ii) ½: 1/3
As We know that

5. Find out the sub-duplicate ratio of
(i) 9: 16
(ii) ¼: 1/9
(iii)9a2 ∶ 49b2
Solution:

(i) 9: 16
As We know that
Sub-duplicate ratio of 9: 16 = √9: √16 = 3: 4

(ii) ¼: 1/9
As We know that
Sub-duplicate ratio of ¼: 1/9 = √1/4: √1/9
So we  will get = ½:¹/³
= 3: 2

(iii) 9a2: 49b
As We know that
Sub-duplicate ratio of 9a2 ∶ 49b2  = √9a2 ∶ √49b2  = 3a: 7b

6. Find out the sub-triplicate ratio of
(i) 1: 216
(ii) 1/8: 1/125
(iii)27a3 ∶ 64b3
Solution:

(i) 1: 216
As We know that
Sub-triplicate ratio of 1: 216 = ∛1: ∛216
By further calculation
= (13  )(1/3)  ∶ (63  )(1/3)
= 1: 6

(ii) 1/8: 1/125
As We know that

7. Find out the reciprocal ratio of
(i) 4: 7
(ii) 32 ∶ 42
(iii) 1/9: 2
Solution:

(i) 4: 7
As We know that
Reciprocal ratio of 4: 7 = 7: 4

(ii) 32 ∶ 4^2
As We know that
Reciprocal ratio of 32 ∶ 42    = 42   ∶ 32  = 16: 9

(iii) 1/9: 2
As We know that
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:
As It is given that
2: 3,17: 21,11: 14 and 5: 7

thus We can write it in fractions as
2/3,17/21,11/14,5/7
Here the LCM of 3,21,14 and 7 is 42

By converting the 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 writing it in ascending order
28/42,30/42,33/42,34/42

By further simplification
2/3,5/7,11/14,17/21

So we will get
2: 3,5: 7,11: 14 and 17: 21

9. (i) If A: B = 2: 3,B: C = 4: 5 and C: D = 6: 7,find A: D.
(ii) If x: y = 2: 3 and y: z = 4: 7,find x: y: z.
Solution:
(i) As we have given that
A: B = 2: 3,B: C = 4: 5 and C: D = 6: 7
We can write it as
A/ B = 2/3,B/C = 4/5,C/D = 6/7
By multiplication
A/B × B/C × C/D = 2/3 × 4/5 × 6/7
So we  will get
A/D = 16/35
A: D = 16: 35

(ii) as We know that the LCM of y terms 3 and 4 is 12
Now making equals of y as 12
x/y = 2/3 = (2 × 4)/ (3 × 4) = 8/12 = 8: 12
y/z = 4/7 × 3/3 = 12/21 = 12: 21
So 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.
(ii) If 3A = 4B = 6C,find A: B: C
Solution:
(i) As We know that
A: B = 1/4 × 5/1 = 5/4
B: C = 1/7 × 6/1 = 6/7
Here the LCM of B terms 4 and 6 is 12
Now making terms of B as 12

A/B = (5 × 3)/ (4 × 3) = 15/12 = 15: 12
B/C = (6 × 2)/ (7 × 2) = 12/14 = 12: 14
So A: B: C = 15: 12: 14

(ii) As we have given that
3A = 4B
Thus  We can write it as
A/B = 4/3
A: B = 4: 3
Similarly 4B = 6C
We can write it as
B/C = 6/4 = 3/2
B:C = 3: 2
So we will get
A: B: C = 4: 3: 2

11. (i) If 3x + 5y/ 3x – 5y = 7/3,find x: y.
(ii) If a: b = 3: 11,find (15a – 3b): (9a + 5b).
Solution:
(i) 3x + 5y/ 3x – 5y = 7/3
By cross multiplication
9x + 15y = 21x – 35y
By further simplification
21x – 9x = 15y + 35y
12x = 50y
So we will get
x/y = 50/12
= 25/6
Therefore, x: y = 25: 6

(ii) As we have given that
a: b = 3: 11
a/b = 3/11
It is given that
(15a – 3b)/ (9a + 5b)
Now dividing both numerator and denominator by b
= [15a/b – 3b/b]/ [9a/b + 5b/b]
By further calculation
= [15a/b – 3]/ [9a/b + 5]
Substituting the value of a/ b
= [15 × 3/11 – 3]/ [9 × 3/11 + 5]
So we will get
= [45/11 – 3]/ [27/11 + 5]
Taking LCM
= [(45 – 33)/ 11]/ [(27 + 55)/ 11]
= 12/11/ 82/11
thus We can write it as
= 12/11 × 11/82
= 12/82
= 6/41
Hence, (15a – 3b): (9a + 5b) = 6: 41.

We will get
(x + 2y)/ (2x + y) = 11/10 or 14/13
(x + 2y): (2x + y) = 11: 10 or 14: 13

(ii) y (3x – y): x (4x + y) = 5: 12
Thus It can be written as

Taking common terms
5 (x/y) [4 (x/y) – 3] – 4 [4 (x/y) – 3] = 0
[4 (x/y) – 3] [5 (x/y) – 4] = 0

Here
4 (x/y) – 3 = 0
So we get
4 (x/y) = 3
x/y = 3/4

Similarly
5 (x/y) – 4 = 0
So we will get
5 (x/y) = 4
x/y = 4/5

(a) x/y = 3/4
We know that

13. (i) If (x – 9): (3x + 6) is the duplicate ratio of 4: 9, Find out the value of x.
(ii) If (3x + 1): (5x + 3) is the triplicate ratio of 3: 4, Find out the value of x.
(iii) If (x + 2y): (2x – y) is equal to the duplicate ratio of 3: 2, Find out x: y.
Solution:

So we will get
(x + 2y)/ (2x – y) = 9/4
By cross multiplication
9 (2x – y) = 4 (x + 2y)
18x – 9y = 4x + 8y
18x = 4x = 8y + 9y
So we get
14x = 17y
x/y = 17/14
x: y = 17: 14

14. (i) Find out two numbers in the ratio of 8: 7 such that when each is decreased by 12 ½, they are in the ratio 11: 9.
(ii) The income of a man is increased in the ratio of 10: 11. If the increase in his income is Rs 600 per month, Find out
Solution :

(i) Ratio = 8: 7
Lets assumed  the numbers as 8x and 7x
Using the condition
[8x – 25/2]/ [7x – 25/2] = 11/9
Taking LCM
[(16x – 25)/ 2]/ [(14x – 25)/ 2] = 11/9
By further calculation
[(16x – 25) × 2]/ [2 (14x – 25)] = 11/9
(16x – 25)/ (14x – 25) = 11/9
By cross multiplication
154x – 275 = 144x – 225
154x – 144x = 275 – 225
10x = 50
x = 50/10
= 5

So the numbers are
8x = 8 × 5 = 40
7x = 7 × 5 = 35

(ii) Lets assumed the present income = 10x
Increased income = 11x
So the increase per month = 11x – 10x = x
Here x = Rs 600
New income = 11x = 11 × 600 = Rs 6600

15. (i) A woman reduces her weight in the ratio 7: 5. What does her weight become if originally it was 91 kg.
(ii) A school collected Rs 2100 for charity. It was decided to divide the money between an orphanage and a blind school in the ratio of 3: 4. How much money did each receive?
Solution:
(i) Ratio of original and reduced weight of woman = 7: 5
Lets assumed original weight = 7x
Reduced weight = 5x
Here original weight = 91 kg
So the reduced weight = (91 × 5x)/ 7x = 65 kg

(ii) Amount collected for charity = Rs 2100
Here the ratio between orphanage and a blind school = 3: 4
Sum of ratios = 3 + 4 = 7

As We know that
Orphanage schools share = 2100 × 3/7 = Rs 900
Blind schools share = 2100 × 4/7 = Rs 1200

16. (i) The sides of a triangle are in the ratio 7: 5: 3 and its perimeter is 30 cm. Find out the lengths of sides.
(ii) If the angles of a triangle are in the ratio 2: 3: 4, Find out the angles.
Solution:
(i) As we have given that
Perimeter of triangle = 30 cm
Ratio among sides = 7: 5: 3
Here the sum of ratios = 7 + 5 + 3 = 15

As We know that
Length of first side = 30 × 7/15 = 14 cm
Length of second side = 30 × 5/15 = 10 cm
Length of third side = 30 × 3/15 = 6 cm

Therefore, the sides are 14 cm, 10 cm and 6 cm.

(ii) as We know that
Sum of all the angles of a triangle = 180°
Here the ratio among angles = 2: 3: 4
Sum of ratios = 2 + 3 + 4 = 9

17. Three numbers are in the ratio 1/2: 1/3: ¼. If the sum of their squares is 244, Find out the numbers.
Solution:
As we have given that
Ratio of three numbers = 1/2: 1/3: 1/4
= (6: 4: 3)/ 12
= 6: 4: 3

Lets assumed first number = 6x
Second number = 4x
Third number = 3x
So based on the condition
(6x)2  + (4x)2  + (3x)2  = 244
36x + 16x2  + 9x2   = 244
So we will get
61x = 244
x2  = 244/61 = 4 = 22
x = 2

Here
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 Rs 500 more than C, Find out the total sum divided.
(ii) In a business, A invests Rs 50000 for 6 months, B Rs 60000 for 4 months and C Rs 80000 for 5 months. If they together earn Rs 18800 Find out the share of each.
Solution:
(i) As we have given that
Ratio between A,B and C = 7: 5: 4
Consider A share = 7x
B share = 5x
C share = 4x
So the total sum = 7x + 5x + 4x = 16x

Based on the condition
5x – 4x = 500
x = 500
So the total sum = 16x = 16 × 500 = Rs 8000

(ii) 6 months investment of A = Rs 50000
1 month investment of A = 50000 × 6 = Rs 300000

4 months investment of B = Rs 60000
1 month investment of B = 60000 × 4 = Rs 240000

5 months investment of C = Rs 80000
1 month investment of C = 80000 × 5 = Rs 400000

Here the ratio between their investments = 300000: 240000: 400000 = 30: 24: 40
Sum of ratio = 30 = 24 + 40 = 94
Total earnings = Rs 18800

So we will get
A share = 30/94 × 18800 = Rs 6000
B share = 24/94 × 18800 = Rs 4800
C share = 40/94 = 18800 = Rs 8000

19. (i) In a mixture of 45 litres, 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?
(ii) The ratio of the number of boys to the numbers of girls in a school of 560 pupils is 5: 3. If 10 new boys are admitted, Find out how many new girls may be admitted so that the ratio of the number of boys to the number of girls may change to 3: 2.
Solution:
(i) As we have given that
Mixture of milk to water = 45 litres
Ratio of milk to water = 13: 2
Sum of ratio = 13 + 2 = 15
Here the quantity of milk = (45 × 13)/ 15 = 39 litres
Quantity of water = 45 × 2/15 = 6 litres

Lets assumed  x litre of water to be added,then water = (6 + x) litres
Here the new ratio = 3: 1
39: (6 + x) = 3: 1
We can write it as
39/ (6 + x) = 3/1
By cross multiplication
39 = 18 + 3x
3x = 39 – 18 = 21
x = 21/3
= 7 litres
Hence, 7 litres of water is to be added to the mixture.

(ii) As we have given that
Ratio between boys and girls = 5: 3
Number of pupils = 560
So the sum of ratios = 5 + 3 = 8

As We know that
Number of boys = 5/8 × 560 = 350
Number of girls = 3/8 × 560 = 210
Number of new boys admitted = 10
So the total number of boys = 350 + 10 = 360

Lets assumed x as the number of girls admitted
Total number of girls = 210 + x
Based on the condition
360: 210 + x = 3: 2
We can write it as
360/ 210 + x = 3/2
By cross multiplication
630 + 3x = 720
3x = 720 – 630 = 90
So we will get
x = 90/3 = 30
Hence, 30 new girls are to be admitted.

20. (i) The monthly pocket money of Ravi and Sanjeev are in the ratio 5: 7. Their expenditures are in the ratio 3: 5. If each saves Rs 80 per month, Find out their monthly pocket money.
(ii) In class X of a school, the ratio of the number of boys to that of the girls is 4: 3. If there were 20 more boys and 12 less girls, then the ratio would have been 2: 1. How many students were there in the class?
Solution:
(i) Lets assumed the monthly pocket money of Ravi and Sanjeev as 5x and 7x
Their expenditure is 3y and 5y respectively.
5x – 3y = 80 …… (1)
7x – 5y = 80 …… (2)
Now multiply equation (1) by 7 and (2) by 5
Subtracting both the equations
35x – 21y = 560
35x – 25y = 400
So we  will get
4y = 160
y = 40

In equation (1)
5x = 80 + 3 × 40
= 200
x = 40
Here the monthly pocket money of Ravi = 5  × 40 = 200

(ii) Lets assumed x as the number of students in class
Ratio of boys and girls = 4: 3
Number of boys = 4x/7
Number of girls = 3x/7

Based on the problem
(4x/7 + 20): (3x/7 – 12) = 2: 1
We can write it as
(4x + 140)/ 7: (3x – 84)/ 7 = 2: 1
So we get
(4x + 140)/ 7 × 7/ (3x – 84) = 2/1
(4x + 140)/ (3x – 84) = 2/1
6x – 168 = 4x + 140
6x – 4x = 140 + 168
2x = 308
x = 308/2
= 154
Therefore, 154 students were there in the class.

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:
Lets assumed the number of passes = 4x
Number of failures = x
Total number of students appeared = 4x + x = 5x

In case 2
Number of students appeared = 5x – 30
Number of passes = 4x – 20
So the number of failures = (5x – 30) – (4x – 20)
By further calculation
= 5x – 30 – 4x + 20
= x – 10

Based on the condition
c(4x – 20)/ (x – 10) = 5/1
By cross multiplication
5x – 50 = 4x – 20
5x – 4x = - 20 + 50
x = 30

Hence No. of students appeared = 5x = 5 × 30 = 150

Exercise 7.2

Find out the value of x in the following proportions:
(i) 10: 35 = x: 42
(ii) 3: x = 24: 2
(iii) 2.5: 1.5 = x: 3
(iv) x: 50∶: 3: 2
Solution:
(i) 10: 35 = x: 42
Thus We can write it as
35 × x = 10 × 42
So we will get
x = (10 × 42)/ 35
x = 2 × 6 x = 12

(ii) 3: x = 24: 2
Thus We can write it as
x × 24 = 3 × 2
So we will get
x = (3 × 2)/ 24
x = ¼

iii) 2.5: 1.5 = x: 3
Thus We can write it as
1.5 × x = 2.5 × 3
So we will get
x = (2.5 × 3)/ 1.5
x = 5.0

(iv) x: 50∶: 3: 2
Thus We can write it as
x × 2 = 50 × 3
So we will get
x = (50 × 3)/ 2
x = 75

2. Find out the fourth proportional to
(i) 3,12,15
(ii) 1/3,1/4,1/5
(iii) 1.5,2.5,4.5
(iv) 9.6 kg,7.2 kg,28.8 kg
Solution:
(i) 3,12,15
Lets assumed x as the fourth proportional to 3,12 and 15
3: 12∶: 15: x
Thus We can write it as
3 × x = 12 × 15
So we will get
x =   (12 × 15)/ 3
x = 60

(ii) 1/3,1/4,1/5
Lets as sumed  x as the fourth proportional to 1/3,1/4 and 1/5
1/3: 1/4:: 1/5: x
thus We can write it as
1/3 × x = 1/4 × 1/5
So we will get
x = 1/4 × 1/5 × 3/1
x = 3/20

(iii) 1.5,2.5,4.5
Lets assumed  x as the fourth proportional to 1,5,2.5 and 4.5
1.5: 2.5∶: 4.5: x
Thus We can write it as
1.5 × x = 2.5 × 4.5
So we will get
x = ( 2.5 × 4.5)/ 1.5
x = 7.5

(iv) 9.6 kg,7.2 kg,28.8 kg
Lets asumed  x as the fourth proportional to 9.6,7.2 and 28.8
9.6: 7.2∶: 28.8: x
Thus We can write it as
9.6 × x = 7.2 × 28.8
So we willget
x = (7.2 × 28.8)/ 9.6
x = 21.6

3. Find out the third proportional to
(i) 5,10
(ii) 0.24,0.6
(iii) Rs.3,Rs.12
(iv) 5 ¼ and 7.
Solution:

(i) Lets assumed  x as the third proportional to 5, 10
5: 10∶: 10: x
Thus It can be written as
5 × x = 10 × 10
x = (10 × 10)/ 5 = 20

Hence, the third proportional to 5, 10 is 20.

(ii) Lets assumed  x as the third proportional to 0.24,0.6
0.24: 0.6∶: 0.6: x
Thus It can be written as
0.24 × x = 0.6 × 0.6
x = (0.6 × 0.6)/ 0.24 = 1.5

Hence, the third proportional to 0.24, 0.6 is 1.5.

(iii) Lets assumed  x as the third proportional to Rs. 3 and Rs. 12
3: 12∶: 12: x
thus It can be written as
3 × x = 12 × 12
x = (12 × 12)/ 3 = 48

Hence, the third proportional to Rs. 3 and Rs. 12 is Rs. 48

(iv) Lets assumed x as the third proportional to 5 ¼ and 7
5 ¼: 7∶: 7: x
thus It can be written as
21/4 × x = 7 × 7
x = (7 × 7 × 4)/ 21 = 28/3 = 9 1/3

Hence, the third proportional to 5 ¼ and 7 is 9 1/3.

4. Find out the mean proportion of:
(i) 5 and 80
(ii) 1/12 and 1/75
(iii) 8.1 and 2.5
(iv) (a – b) and (a3  – a2  b),a ˃ b
Solution:

(i) Lets assumed  x as the mean proportion of 5 and 80
5: x∶: x: 80
thus It can be written as
x2  = 5 × 80 = 400
x = √400 = 20

Therefore, mean proportion of 5 and 80 is 20.

(ii) Lets assumed x as the mean proportion of 1/12 and 1/75
1/12: x∶: x: 1/75
thus It can be written as
x2  = 1/12 × 1/75 = 1/900
x = √1/900 = 1/30

Therefore, mean proportion of 1/12 and 1/75 is 1/30.

(iii) Lets assumed  x as the mean proportion of 8.1 and 2.5
8.1: x∶: x: 2.5
Thus It can be written as
x2  = 8.1 × 2.5 = 20.25
x = √20.25 = 4.5
Therefore, mean proportion of 8.1 and 2.5 is 4.5.

(iv) lets assumed  x as the mean proportion of (a – b) and (a3  – a2  b),a ˃ b
(a – b): x∶: (a3  – a2 b)
thus It can be written as
x2  = (a – b) (a3   – a2  b)
So we will get
x2  = (a – b) a2  (a – b)

x2   = a2  (a – b)2
Here
x = a (a – b)
Therefore, mean proportion of (a – b) and (a3 – a2 b),a ˃ b is a (a – b).

5. If a, 12, 16 and b are in continued proportion Find out a and b.
Solution:
As we have given that
a, 12, 16 and b are in continued proportion
a/12 = 12/16 = 16/b

as We know that
a/12 = 12/16
By cross multiplication
16a = 144
a = 144/16
= 9

Similarly
12/16 = 16/b
By cross multiplication
12b = 16 × 16 = 256
b = 256/12
= 64/3
= 21 1/3
Therefore,a = 9 and b = 64/3 or 21 1/3.

6. What number must be added to each of the numbers 5, 11, 19 and 37 so that they are in proportion?
Solution:
Lets assumed  x to be added to 5,11,19 and 37 to make them in proportion
5 + x: 11 + x∶: 19 + x: 37 + x
thus It can be written as
(5 + x) (37 + x) = (11 + x) (19 + x)
By further calculation
185 + 5x + 37x + x2  = 209 + 11x + 19x + x2
185 + 42x + x2   = 209 + 30x + x2
So we will get
42x – 30x +  x2   – x2   = 209 – 185
12x = 24
x = 2
Hence, the least number to be added is 2.

7. What number should be subtracted from each of the numbers 23, 30, 57 and 78 so that the remainders are in proportion?
Solution:
Lets assumed  x be subtracted from each term
23 – x,30 – x,57 – x and 78 – x are proportional
thus It can be written as
23 – x: 30 – x∶: 57 – x: 78 – x
(23 – x)/ (30 – x) = (57 – x)/ (78 – x)
By cross multiplication
(23 – x) (78 – x) = (30 – x) (57 – x)
By further calculation
1794 – 23x – 78x + x2   = 1710 – 30x – 57x + x2
x2   – 101x + 1794 – x2   + 87x – 1710 = 0
So we will get
- 14x + 84 = 0
14x = 84
x = 84/14
= 6
Therefore, 6 is the number to be subtracted from each of the numbers.

8. If 2x – 1,5x – 6,6x + 2 and 15x – 9 are in proportion, Find out the value of x.
Solution:
As we have given that
2x – 1,5x – 6,6x + 2 and 15x – 9 are in proportion
Thus We can write it as
(2x – 1) (15x – 9) = (5x – 6) (6x + 2)
By further calculation
30x2   – 18x – 15x + 9 = 30x2   + 10x – 36x – 12
30x2  – 33x + 9 = 30x2   – 26x – 12
30x2   – 33x – 30x2   + 26x = - 12 – 9
So we will get
-7x = - 21
x = -21/-7
= 3
Therefore, the value of x is 3.

9. If x + 5 is the mean proportion between x + 2 and x + 9, Find out the value of x.
Solution:
As we have given that
x + 5 is the mean proportion between x + 2 and x + 9
thus We can write it as
(x + 5)2  = (x + 2) (x + 9)
By further calculation
x2   + 10x + 25 = x2   + 11x + 18
x2    + 10x – x2   – 11x = 18 – 25
So we will get
- x = - 7
x = 7
Hence, the value of x is 7.

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:
Lets assumed  x be added to each number
16 + x ,26 + x and 40 + x are in continued proportion
Thus It can be written as
(16 + x)/ (26 + x) = (26 + x)/ (40 + x)
By cross multiplication
(16 + x) (40 + x) = (26 + x) (26 + x)
On further calculation
640 + 16x + 40x + x2   = 676 + 26x + 26x + x2
640 + 56x + x2   = 676 + 52x +x2
56x + x2   – 52x – x2   = 676 – 640
So we will get
4x = 36
x = 36/4
= 9
Hence, 9 is the number to be added to each of the numbers.

11. Find out two numbers such that the mean proportional between them is 28 and the third proportional to them is 224.
Solution:
Lets assumed  a and b as the two numbers
As we have given that 28 is the mean proportional
a: 28∶: 28: b
We get
ab = 282  = 784
Here a = 784/b …… (1)

As We know that 224 is the third proportional

a: b∶: b: 224
So we will get
b2  = 224a …..(2)

Now by substituting the value of a in equation (2)
b2  = 224 × 784/b
So we will get
b3  = 224 × 784
b3= 175616 = 563
b = 56

By substituting the value of b in equation (1)
a = 784/56 = 14

Therefore, 14 and 56 are the two numbers.

12. If b is the mean proportional between a and c, prove that a,c,a2  +  b2  and  b2   + c2 are proportional.
Solution:
As we have given that
b is the mean proportional between a and c
thus We can write it as
b2  = a × c
b= ac …..(1)

As We know that
a,c,a + b2  and  b2  + c2  are in proportion
thus It can be written as
a/c = (a2  + b2   )/ (b2   + c2  )
By cross multiplication
a (b2  + c2  ) = c (a2  + b2  )
Using equation (1)
a (ac + c2  ) = c (a2  + ac)
So we will  get
ac (a + c) = a2 c + ac2
Here ac (a + c) = ac (a + c) which is true.
Therefore, it is proved.

13. If b is the mean proportional between a and c, prove that (ab + bc) is the mean proportional between (a2  + b2  ) and (b2  + c).
Solution:
As we have given that
b is the mean proportional between a and c
b2  = ac ….(1)

14. If y is mean proportional between x and z, prove that xyz (x + y + z)3  = (xy + yz + zx)3  .
Solution:
As we have given that
y is mean proportional between x and z
thus We can write it as
y2  = xz …… (1)

Lets assumed
LHS = xyz (x + y + z)3
thus It can be written as
= xz.y (x + y + z)3
Using equation (1)
= y2  y (x + y + z)3
= y3   (x + y + z)
So we will get
= [y (x + y + z)]3

By further calculation
= (xy + y2  + yz)3
Using equation (1)
= (xy + yz + zx)3
= RHS Hence, it is proved.

15. If a + c = mb and 1/b + 1/d = m/c,prove that a,b,c and d are in proportion.
Solution:
As we have given that
a + c = mb and 1/b + 1/d = m/c
a + c = mb
Dividing the equation by b
a/b + c/d = m …….(1)

1/b + 1/d = m/c
Multiplying the equation by c
c/b + c/d = m …… (2)

Using equation (1) and (2)
a/b + c/b = c/b + c/d
So we get
a/b = c/d

Therefore, it is proved that a, b, c and d are in proportion.

RHS = 27 (a + b) (c + d) (e + f)
thus It can be written as
= 27 (bk + b) (dk + d) (fk + f)
Taking out the common terms
= 27 b (k + 1) d (k + 1) f (k + 1)
So we will get
= 27 bdf (k + 1)

Therefore, LHS = RHS.

Solution:

As we have given that
a,b,c,d are in proportion
Lets assumed a/b = c/d = k
a = b,c = dk

(i) LHS = (5a + 7b) (2c – 3d)
Substituting the values
= (5bk + 7b) (2dk – 3d)
Taking out the common terms
= k (5b + 7b) k (2d – 3d)
So we will get
= k2  (12b) (-d)
= - 12 bd k2

RHS = (5c + 7d) (2a – 3b)
Substituting the values
= (5dk + 7d) (2kb – 3b)
Taking out the common terms
= k (5d + 7d) k (2b – 3b)
So we will get
= k2  (12d) (-b)
= - 12 bd k2
Therefore, LHS = RHS.

(iii)(a4  + c4   ): (b4  + d4  )= a2  c2 ∶ b2  d
thus We can write it as

Solution:
As we have given that
a, b, c, d are in continued proportion
Here we will get
a/b = b/c = c/d = k
c = dk,b = ck = dk .k = dk
a = bk = dk  .k = dk

5. If (ma + nb): b∶: (mc + nd): d, prove that a, b, c, d are in proportion.
Solution:
As we have given that
(ma + nb): b∶: (mc + nd): d
thus We can write it as
(ma + nb)/ b = (mc + nd)/ d
By cross multiplication
mad + nbd = mbc + nbd
By further calculation
a/b = c/d
Therefore, it is proved that a, b, c, d are in proportion.

7. If (a + 3b + 2c + 6d) (a – 3b – 2c + 6d) = (a + 3b – 2c – 6d) (a – 3b + 2c – 6d), prove that a: b:: c: d.
Solution:
As we have given that
(a + 3b + 2c + 6d) (a – 3b – 2c + 6d) = (a + 3b – 2c – 6d) (a – 3b + 2c – 6d)
Thus, We can write it as

By cross multiplication
50x – 75 = 12x + 1
50x – 12x = 1 + 75
So we will get
38x = 76
x = 76/38
= 2

2. If (7p + 3q): (3p – 2q) = 43: 2, Find out p: q.
Solution:
As we have given that
(7p + 3q): (3p – 2q) = 43: 2
thus We can write it as
(7p + 3q)/ (3p – 2q) = 43/2
By cross multiplication
129p – 86q = 14p + 6q
129p – 14p = 6q + 86q
So we will get
115p = 92q
By division

As we have given that
5x + 12x + 13x = 360 cm
By further calculation
30x = 360
We will get
x = 360/30
= 12

Here the length of the longest side = 13x
Substituting the value of x
= 13 × 12
= 156 cm

5. The ratio of the pocket money saved by Lokesh and his sister is 5: 6. If the sister saves Rs 30 more, how much more the brother should save in order to keep the ratio of their savings unchanged?
Solution:
Lets assumed 5x and 6x as the savings of Lokesh and his sister.
Lokesh should save Rs y more
Based on the problem
(5x + y)/ (6x + 30) = 5/6
By cross multiplication
30x + 6y = 30x + 150
By further calculation
30x + 6y – 30x = 150
So we will get
6y = 150
y = 150/6
= 25
Therefore, Lokesh should save Rs 25 more than his sister.

6. In an examination, the number of those who passed and the number of those who failed were in the ratio of 3: 1. Had 8 more appeared, and 6 less passed, the ratio of passed to failures would have been 2: 1. Find out the number of candidates who appeared.
Solution:

7. What number must be added to each of the numbers 15, 17, 34 and 38 to make them proportional?
Solution:
Lets assumed  x be added to each number
So the numbers will be
15 + x,17 + x,34 + x and 38 + x
Based on the condition
(15 + x)/ (17 + x) = (34 + x)/ (38 + x)
By cross multiplication
(15 + x) (38 + x) = (34 + x) (17 + x)
By further calculation
570 + 53x +  x  = 578 + 51x + x
So we will get
x   + 53x –  x  – 51x = 578 – 570
2x = 8
x = 4
Hence, 4 must be added to each of the numbers.

8. If (a + 2b + c),(a – c) and (a – 2b + c) are in continued proportion, prove that b is the mean proportional between a and c.
Solution:
As we have given that
(a + 2b + c),(a – c) and (a – 2b + c)
are in continued proportion
thus We can write it as
(a + 2b + c)/ (a – c) = (a – c)/ (a – 2b + c)
By cross multiplication
(a + 2b + c)(a – 2b + c)= (a – c)
On further calculation
a2  – 2ab + ac + 2av – 4b2+ 2bc + ac – 2bc + c2  = a2  – 2ac + c2
So we will get
a2  – 2ab + ac + 2ab – 4b2  + 2bc + ac – 2bc + c2  – a2  + 2ac – c2  = 0
4ac – 4b2  = 0
Dividing by 4
ac – b2  = 0
b2  = ac
Therefore, it is proved that b is the mean proportional between a and c.

9. If 2,6,p,54 and q are in continued proportion, Find out the values of p and q.
Solution:
As we have given that
2,6,p,54 and q are in continued proportion
Thus We can write it as
2/6 = 6/p = p/54 = 54/q

(i) as We know that
2/6 = 6/p
By cross multiplication
2p = 36
p = 18

(ii) as We know that
p/54 = 54/q
By cross multiplication
pq = 54 × 54
Substituting the value of p
q = (54 × 54)/ 18 = 162
Therefore, the values of p and q are 18 and 162

10. If a, b, c, d, e are in continued proportion, prove that: a: e = a4 ∶ b4  .
Solution:
As we have given that
a, b, c, d, e are in continued proportion
thus We can write it as
a/b = b/c = c/d = d/e = k
d = ek,c = ek ,b = ek3  and a = ek4

Here
LHS = a/e
Substituting the values
= ek4  / e
= k

RHS = a4/b4
Substituting the values
= (ek4  )   /   (ek3 )4
So we get
= e4  k16/e4    k12
= k(16-12)
= k4
Hence, it is proved that a: e = a4 ∶ b .

11. Find out two numbers whose mean proportional is 16 and the third proportional is 128.
Solution:
Lets assumed  x and y as the two numbers
Mean proportion = 16
Third proportion = 128
√xy = 16
xy = 256
Here
x = 256/y …..(1)
y2  /x = 128
Here
x = y2  /128 ….(2)

Using both the equations
256/y =  y3  / 128
By cross multiplication
y3  = 256 × 128 = 32768
y3  = 323
y = 32

Substituting the value of y in equation (1)
x = 256/y
So we will get
x = 256/32 = 8
Hence, the two numbers are 8 and 32.