7 RATIO AND PROPORTION

Sylabus: (a) Proportion, continued proportion, mean proporion.

invertendo, properties and their
(0) Componendo, dividendo, alternendo,
combinations.
(c) Direct simple applications on proportions only.
Ratio
and in the same units, then
1 dO are two non-zero quantities of the same kind
the fraction Ls called the ratio between a andb. written as a:b (read as 'a is to b).
b
Terms of a Ratio
In the ratio a : b, we call a. the first term or antecedent and b, the second term
or consequent.
Example 1. Amit is 16 years old and his sister Pamela is 23 years old. We say that the
ages of Amit and Pamela are in the ratio 16 : 23.
Example 2. The cost of a pencil is 1 and that of an eraser is 75 paise. We say that the cost
of a pencil and that of an eraser are in the ratio 100 : 75. [:1= 100P]
Example 3. There is no ratio between 8 km and 12 kg, as these quantities are not of the
same kind and have different units, which are not inter-convertible.
An Important Property
The value of aratio remains unchanged, if both of its terms are multiplied or divided
by the Same non-zero number.
Proof : Let a : b be the given ratio and let m be a non-zero number.
a
Then, we may write, a : b as
a ma
(i) Clearly, we have mb
This shows that a :b is the same as mna : mb.
(a/m)
(iü) We may write,
b (b/m)
This shows that a:b is the same as b
m m
ARatio in Its Lowest Terms
When a and bhave no common factor, other than 1, we say that a:b is in
its lowest
terms.
1. To convert an ordinary ratio into its lowest terms, we
divide both the terms by
their H.C.F.
32 40
Thus, 32 : 40 = = 4:5. (in lowest terms)
8 8

82 Foundation Mathematics for Class X

 2. If the terms of a given ratio are fractions, then we convert them into whole
numbers by multiplying each term by the L.C.M. of their denominators.
1 1
Thus,
23
=

x66-3:2
Comparison of Ratios
()(a : b) > c: d) ad > bc.
d
(ü) (a :b) = (c : d) * = ad = bc.

(ii)(a : b) <(c : d) b < ad < bc.

When more than two fractions are given, we convert them into equivalent fractions with
the same denominator, and then compare them.

SOLVED EXAMPLES

Example 1. Find the ratio between

()) 45 minutes and 1 hour 12 minutes
(ü) 10months and 1year 3 months.
Solution: Changing the given quantities in same units, we have
(i) 1hour 12minutes = (60 min. + 12 min.) = 72 min.
:. Required ratio =45 : 72 = 5:8.
(ü) 1 year 3 months = (12 months + 3 months) = 15 months.
.:. Required ratio =10: 15 =2:3.
Example 2. If A:B = 4:9 and B: C = 6:5, find A: C.
Solution: A:B= 4:9 and B:C 6:5 (Given)

nd B_6

A A
15
A:C= 8: 15.
Hence, A: C = 8: 15.
Example 3. IfA:B= 3: 4 and B:C =6:5, find A: B:C.
Solution: A:B = 3:4 and B: C = 6:5
’ A: B= 3:4 and B: C = 1: 6

’A:B =3: 4 and B:C = 4:

10
A: B:C =3 :4: ’A: B:C = 9: 12 : 10
3

Hence, A:B:C=9:12 : 10.

83

Ratio and Proportion

 w et 2A = 4C -k (nay). Then,
A And C *
4
1 1 1
A: B: C= 3 4
4
12)
|:: LC.M. of 2, 3, 4 is

= 6: 4 : 3.
Hence, A: B: C = 6:4: 3.
A B C = k. Then,
(ü) Let 3
=

4
and C = 5k.
A = 3k, B= 4k
..A: B:C= 3k: 4k :
5k = 3:4:5.
Hence, A: B: C = 3:4:5.
A:B:C.
and 4B - 5C, find (i) A: Cand (ü)
Example 5. If 2A = 3B
A
Solution: 2A = 3B ’
B 2
B 5
4B = 5C ’ =

C 4

Hence, A:C= 15:8.

4 8
=5:4=1:= = 2: 5
3 ’ A: B = 3:2 and B:C 5
(ü) and C
’ A:B:C=3:2: = 15:10:8.
5

:8.
Hence, A: B:C= 15: 10
(2002)
76).
Example 6. Ifa :b =
5:3, find (5a + 86) : (6a
Then
Solution: Let a 5k andb = 3k.
5a + 8b 25k + 24k 49k 49
=
30k - 21k 9 9
6a -76
76) = 49: 9,
.. (5a + 86) : (6a -
+ 5b) = 11: 17, find
a:b.
If (4a + 36): (6a
Example 7.
56) = 11: 17.
Solution: (4a + 36) : (6a +
4a + 3b 11
6a + 56 17
66a + 555
68a + 516 =
4 2 ’ a:b= 2:1
2a= 4b ’ * =
’ 62 1
Foundation Mathematics for Class X
 Example 8. If (4x t xy): (3xy -
y') = 12 : 5, find x : y.
Solution:
(4x2 + xy): (3xy - y') = 12 : 5
2
4
4x + xy 12
+

12
3xy - y 5 (On dividing Nr & Dr by y)
3 1

4a +a 12
-, where
3a -1 5
20a² + 5a = 36a - 12
’ 20a' - 31a + 12 = 0
’ 20a2 16a
l5a + 12 = 0 ’ 4a (5a - 4) - 3(5a - 4) =0
’ (5a - 4) (4a - 3) = 0

5a 4= 0 or 4a - 3 = 0 4 3
a= Or a =
5 4
4 3
5 ’ x : y= 4:5 or x:y = 3: 4.
y 4
Hence, (x: y) = (4 : 5) or (x: y) = (3: 4).

Example 9. If 5x2 - 13xy + 6y = 0, find x : y.

Solution: We have
5x2- 13xy + 6y² = 0 5x - 10xy - 3ey + 6y = 0
’ 5x (x - 2y) - 3y (x- 2y) = 0
’ (« - 2y) (5x 3y) = 0
’ X - 2y = 0 or 5x - 3y = 0
’x= 2y or 5x = 3y
2
=
1
Or
5 ’ (*:y) =(2:1) or (x:y) =(3:5).
Example 10. A ratio is equal to 5:7. If its antecedent is 35, what is the consequent?
Solution: Let the consequent be x. Then,
5 35
5:7= 35 : x
7
7x 35
’ 5x = (7×35) ’ x= = 49
5
Hence, the consequent is 49.
Example 11. Two numbers are in the ratio 3 : 5. If 8 is added to each number, the ratio
becomes 2: 3. Find the numbers. (2001)
Solution: Let the required numbers be 3x and 5x.
3x +8 2
Then, 3
5x +8
[By cross multiplication]
’ 3x (3x + 8) = 2 x (5x + 8)
’ 9x + 24 = 10x + 16 ’ x = 8.
and second number = (5 x 8) =40.
.. First number = (3 x 8) = 24
Hence, the required numbers are 24 and 40.
85
Ratio and Proportion
 number must be
added each term
to of the ratio 3 : 5to
Example 12.
What Then make it
be added be x. 11:12
Solution:
Let the number to
x)= 11: 12
(3 + r) : (5 +
3+x 11
12 x (3 + x) = 11 x (5 + x)
5+x 12 IBy cross multiplication
36 + 12x = 55 + 1lx = 19.
Hence, the required number is 19.
1
Example 13. Find the number which bears the same ratioto that does to 5
27 7
Solution: Let the required number be x. Then,. 9
1 3 5
27 79
27x: 1= 27:35
27x 27 27 1
1 35 35 27 35
1
Hence, the required number is
35
Fyample 14. The ratio between two numbers is 5:6 and
numbers.
their LCM is 150. Find the
Solution: Let the required numbers be 5x and 6:x.
Then, LCM of 5x and 6x is 30x.
Now, 30x = 150x= 150 =5.
30
.:. One number = (5 x 5) = 25, other number = (6 x 5)
= 30.
Hence, the required numbers are 25 and 30.
Example 15. The ages of Aand B are in the ratio 7:8. Six years ago, their
the ratio 5 : 6. Find their present ages. ages were in
Solution: Let the present ages of A and B be (7x) years and (8x) years respectively.
Then,
A's age 6 years ago = (Tx - 6) years.
B's age 6 years ago = (&r - 6) years.
7x-6
:. (7x - 6) : (&¢ -6) = 5: 6
8x-6 6
’6 x (7% - 6) = 5 x (8x - 6) By cross multiplication)
’ 42x 36 = 40x 30 2x = 6 ’x=3.
.. A's present age = (7 x 3) years = 21 years.
B's present age = (8 x 3) years = 24 years.
Example 16. Divide 1740 between Aand Bin the ratio 5: 7.
Solution: Given amount = 1740.
Sum of their ratio terms = 5 + 7= 12.
5
: A's share=1740 x =7 725.

B's share= 1 7 4 012x = 1015.

1.1 1

Example 17. Divide 3645 among A, B, C in the ratio 34 6

1 1
=4:3: 2.
Given ratio = 6
Solution:

Foundation Mathematics for Class

X
 Sum of ratio
terms = (4 + 3 + 2)
= 9.
A's share =
3645 x ={ 1620.
9

B's share =?3645x3

9 =? 1215.
C's share =3645 x
g = 810
Example 18. Divide R 3740 into
third of the second three parts in such a
part and one-sixth uway that
Solution : Let the required of the third half of the first part, one-
parts be A, B and c. part are equal.
A B C Then,
2 3 *(say)
:. A = 2x, B = 3x
and C = 6x.
Now, A +B + C= 3740
2x + 3x + 6x = 3740
1lr = 3740 ’x=
:.A = (2 x 340) = 680, B 340.
=(3 x 340) = 1020
:. First part = 680, and C = (6 x 340) =
second part = 1020 and 2040.
Example 19. The monthly pocket third part = 2040.
money of Ravi and Sanjeev are in
expenditures are in the ratio 3: 5. If each the ratio 5 :7.
monthly pocket money of each. saves 80 every month, findTheir
the
Solution : Let the monthly pocket (2012)
money of Ravi and Sanjeev be
respectively. 5x and 7x
Let their monthly
expenditures be 3y
Then, their monthly savings are (5x and 5y respectively.
- 3y) and (7x - 5y)
5x-3y = 80 ...i) respectively.
|7x-5y =80 ...(ü)
On multiplying (i) by 5 and (ii)
by 3 and subtracting, we get :
(25x - 21x) = (400 - 240) ’
4x= 160 =i= 40.
Monthly pocket money of Ravi = (5 x 40) = 200.
Monthly pocket money of Sanjeev = R(7 x 40)
=280.
Example 20. The work done by ( - 3) men in (2x
+ 1) days
men in (* + 4) days are in the ratio of 3 : 10. and the work done by (2x+ 1)
Solution : Work done by (x 3) Find the value of x. (2003)
men in (2x + 1) days
= (x - 3) (2% + 1) times the work done by 1 man in 1
day.
Work done by (2 + 1) men in ( + 4) days
=(2x + 1) (x + 4) times the work done by 1 man in 1 day.
(x-3)(2x+1) 3 x-3
(2x + 1)(x+4) 10 +4 10
’ 10 (x- 3) = 3 (x + 4)
10x 30 = 3x + 12
’ 7x 42 ’ x= 6.
Hence, x = 6.
87
Ratio and Proportion
 is 5:2. How much wate,
Examplo 21. In a mixture of 63 litres, the ratio of milk and water
must be added to this mixture to make the ratio 3 : 2
Solution: Ratio of milk and water in the given mixture = 5 : Z.
5 litres = 45 litres.
Quantity of milk in it = 63 x
Quantity of water in it = (63- 45) litres = 18 litres.
Let the quantity of water to be added be x litres.
Then quantity of milk in new mixture = 45 litres.
Quantity of water in new mixture = (18 + x) litres.
45
’ 54 + 3* = 90 ’ 3« = 36 ’ x = 12.
18 + x 2
* Required quantity of water to be added = 12 litres.
Example 22. The ratio of males to females in a committee of48 members is 3:1. How
more ladies be added to the committee so that the ratio of males to females moymanyh
9:5?
Solution: Total number of members of the committee = 48.
Ratio of males to females = 3: 1.
Sum of ratio terms = (3 + 1) = 4.

Number of male members =48x 4,

= 36.
Number of female members = (48 -36) = 12.
Let x ladies be added to the
committee.
Then, number of females = 12 + x.
36 9
12 + x ’ 9(12+ x) = 180 ’ 9x = (180 - 108) =
72 x= 8.
Hence, the required number of ladies
to be added to the committee = 8.
Example 23. An employer reduces the
increases their wages in thenumber
ratio
of employees in the
ratio of 10:7 and
increased or decreased? of 14 : 15. In what ratio,
Solution: Initially, the wage bill is
let the number of
Initially, the total wage bill employees
=
be 10x and wages
per head be 14y.
At present, the (10x x 14y) = (140cy)
number of employees = 7x
At present, the total and
wage bill = ? (7x x 15y) wages per head = 15y.
Ratio in which the wage = (105xy).
bill is decreased =
Hence, the wage bill is 140xy : 105xy = 4: 3.
Example 24. Compare the ratios decreased in the ratio 4: 3.
Solution: The given (4: 7) and (5 :9).
ratios are (4:7) and (5 :
Let a = 4, b = 7, c= 5 9).
and d = 9.
.. (a x d) =(4 x
9) = 36 and bxc=
(7 x 5) = 35
Thus, ad > be ’ >(a:b)
d >(e:d) ’ (4:7)> (5:9)
Hence (4: 7) >(5: 9).

Foundation Mathematics for Class

 25. Arrange the following
Example
ratios
7: 10, 11: 15, 13 20 and in ascending order of magnitude :
17: 25
7 11 13
Solution : Given ratios are 17
15' 20 and
10 25
L.C.M. of 10, 15, 20, 25 = 300.
7 7x 30 210 11 11x 20 220
10 10 x 30 300 15 15 x 20 300
13 13 x 15 195 17 17 x 12 204
20 20x 15 300 25 25 x 12 300
195 204 210 220
Clearly, 300 300 300 300 [: 195<204 < 210 < 220)
13 177 11
<
20 25 10 15

’ (13 : 20) < (17 : 25) << (7: 10) < (11:
15).
Hence, the given ratios in ascending order are :
(13 : 20), (17 : 25), (7 : 10) and (11: 15).
Alternative Method :
7 11
7: 10 = -| = 0.7: 11: 15 = = 0.733;
10 15
13 17
13: 20 = = 0.65; 17: 25 = =0.68.
20 25
.i. 0,65 < 0.68 < 0.7 < 0.733,
Hence, (13 : 20) <(17:25) <(7: 10) <(11: 15).

EXERCISE 7A
1. Find the ratio between :
() 60 paise and 1.35 (iü) 1.8 m and 75 cm
3
(iüü) 1.5 kg and 600 g (iv) 35 min and 1 hrs
4
2. If A: B = 5:6 and B :C=9: 11, find (i) A: C (ü) A: B: C.
3. If P:Q=7:4 and Q: R= 5: 14, find (i) R: P (ü) P:Q:R.
4. If 3A = 5B = 6C, find A: B: C.
5. Ir A B C find A: B:C.
2 3 6
6. If a:b=8: 5. find (7a + 55): (8a 96).
I. f*:y = 3:2, find (5x - 3y) : (7x + 2y).
b. If *:y = 10 : 3, find (3¢2 + 2y2) : (3¢- 2y).
O. lf a:b= 2: 5. fnd (3a2 - 2ah + 562) : (a2 + 7ab - 26).
10, If (5x + 2y): (7x + 4y) = 13 : 20, find x:y.
d lf (3x + 5y) : (3* - 5y) =7:3, find x : y.
89

Ratio and Proportion

 dA 9, lnd y.

V.

rtI nri muat tn 4 rits consequent is 144, what is its antecedent?

h humbe a in the ratio 8: 13. If 14 is added to each of the numbers, the ratio
m9 3 nd the numbers.
IN ?hn mbers are in the ratio 5:7. If 8 is subtracted from each, the ratio becomes
§ Nnd the numbers.
19 What least number must be added to each term of the ratio 5:7 to make it 8: 9?
20. Out of the monthly income of 45000, Rahul spends 31500 and the rest he saves,
Fnd the ratio of his
()income to expenditure () income to savings (üi) savings to expenditure.
21. The cost of making an umbrella is divided between material, labour and overheads in
the ratio 6:4:1.Ifthe material costs 132, find the cost of production of an umbrella.
22. Divide 6720 in the ratio 5:3.
23. Divide ? 11620 among A. B and C in the ratio 35 : 28 : 20.
24. Divide 782 among P, Q and R in the ratio 12 3
234
(Hint : Given ratio = 6: 8:9.J
25. If 5100 be divided among A, B, C in such a way that A gets 2
of what B gets and
1
B gets of what Cgets, find their respective shares.
4
26. Divide 8300 among A, Band C such that 4 times A's share, 5 times B's share and
7 times C's share may all be equal.
27. A sum of money is divided between A and B in the ratio 6 : 11. If B's share is? 7315.
find (i) A's share (iü) the total amount of money.
28. The ages of Tanvy and Divya are in the ratio 5:7. Five years
be in the ratio 3:4. Find their present ages. hence, their ages will
29. One year ago, the ratio of Amit's and Arun's ages
was 6: 7 respectively. Four years
hence, their ages will be in the ratio 7 : 8. How old is Amit?
30 Reena reduces her weight in the ratio 5 : 4.
70 kg? What is her weight now, if originally it was
31, 68 kg of amixture contains milk and
to be added to this mixture to get a new
water in the ratio 27:7. How much more
water
3: 1? mixture containing milk and water in the ratiois
32. A mixture contains milk and
ratio of milk to water becomeswater in the ratio 5: 1. On adding 5
5 :2. Find the quantity of milk in litres of water, the
33. In an examination, the ratio of the original mixture.
passes to failures was 4: 1. Had 30
to failures would have been 5: less
and 20 less passed, the ratio of passes appeared
students appeared for the examination? 1. How many
34. Find the angles of a triangle which are in the ratio
5 : 4:3.
35. The sides of a triangle are in the ratio ::*
lengths of the sides of the triangle. 23 4 and its perimeter is 91 cm, Find the
36. In aschool, the boys and girls are in the ratio 9 : 5. If there are 425 girls.
what is
the total number of students in the school?
Foundation Mathematics for Class X
37.
Compare the following ratios:
(i) (7: 9) and (11: 16) (ii) (19: 25) and (17: 20) (iüi) and (5: 2)
38. Arrange the following ratios in descending order of magnitudes:
(i) (5 : 6), (8 : 9), (13 : 18) and (19 : 24)
(i) (6: 7), (13 : 14), (19 : 21) and (23 : 28)
(iii) (7: 12), (9 : 16), (13 : 20) and (5 : 8)
39. Arrange the following ratios in ascending order of magnitudes :
() (4: 9), (6 : 11), (7 : 13) and (27 : 50)
ii) (2: 3), (8 : l5), (11 : 12) and (7: 16)
(iii) (3 : 5), (4 : 9), (5 : 11) and (10 : 17).
A0. If (3a + 26) : (5a + 3b) = 18: 29, find (a : ) (2016)

Proportion

An equality of two ratios is called a proportion. Four non-zero quantities a, b, c and d

aro soid to be in proportion if a:b = c:d and we write a:b:: c: d.
We read it as, 'a is to b as c is to d'.
If a:b ::c:d, then :
(i) a, b, c, d are known as first, second, third and fourth term respectively.
(ii)) d is called the fourth proportional to a, b, c.
(iii) a and d are called extremes, while b and c are called means.

An Important Property
Ifa :b : c: d, then the product of extremes is equal to the product of
means.
Proof : a:b:: c:d ad = bc.
.:. Product of extremes = product of means.

Continued Proportion
We say that a, b, C, d, e, f etc., are in continued proportion, if b =de

b c d e f
Mean Proportion (or Geometric Mean)
Let a, b, c be in continued proportion.

Then, b
or b = 1c or b = Vac

ere, b 1s called the mean proportion or geometric mean between a and e.

Third Proportional
Ta:6 =b:c, then cis called the third proportional to aand b.
Clearly, it is the fourth proportional to a, b, b.
91
Ratio and Proportion
 SOLVED EXAMPLES

Example 1. Ifx :8:3:4, fnd the value ofx.

Solution: I:8:: 3:4 (Given)
’ x 4 = 8x 3 |: Product of extremes =
Product of means]

Example 2. If x, 4 and 9 are n continued

proportion, fnd x
Solution: 2, 4 and 9 are in continued
4
proportion.
4 9r = 16 ’ = 16 16 4
4 9
Hence, x = V9 3
3
Example 3. What least number must
so that the resulting be added to each of the
Solution: numbers are proportional? numbers 6, l5, 20 and 43,
Let the required number
to be added be x. (2013)
(6 + x) : (15 + x) :: (20 Then,
+ x): (43
(6+*)\43 + x) = (15 + xX20 + x)+(:x)
’ + 49x + 258 = r2 + 35x Product of extremes = Product of means
+ 300
’ 49%- 35x = 300 -
258
’ 14x =42 ’ x =
3.
Hence, the required number is 3.
Example 4. Find the :
(i) fourth proportional to 7, 13
and 35;
(ii) third proportional to 9 and
(iii) mean proportion between 15;
and 80.
Solution: (i) Let the fourth
proportional to 7, 13 and 35 be x.
Then, 7: 13 : 35 :x
7x x= 13 x 35
[:: Product of extremes =
13 x 35 Product of means)
= 65.
7
Hence, the fourth proportional to
7, 13 and 35 is 65.
(ü) Let the third proportional to 9 and 15
Then, 9: 15 :: 15:x be x.
9 x x= 15 x 15
[:: Product of extremes
~ l5x15 = Product of means]
= 25
9
Hence, the third proportional to and
15 is 25.
(iii) Mean proportion between 5
and 80 = J5x 80 = 400 - 20.
Example 5. Find two numbers
is 224. whose mean proportion is 28
and the third proportional
Solution : Let the
required numbers be a and b. Then,
Mean proportion between them is yab
.:. Vab = 28 ’ ab = (28 x
28) ..i)
92
Foundation Mathematics for Class X
 Also, it is given that the third
proportional to a and b is 224.
a:d ::b: 224 ’ b = 224a

= 224 ...(i)
a
On multiplying the corresponding sides of (i)
and (ii), we get :
ab x -(28 x 28 x 224) b = (28 x 28 x 28 x 2 x 2 x 2)
’b = (28 x 2) =56.
Substituting b = 56 in (), we get :
ax 56 = (28 x 28) ’ a= 28 x 28 = 14.
56
Thus a = 14 and b = 56
Hence, the required numbers are 14 and 56.
Example 6. If b is the mean proportion between a and c, prove that :
a'-b +
=b*.
a-6+e2
Solution: Since b is the mean proportion between a and c, we have b = ac.
a'-6 +? (a'b.2)?-6 +)
1 1.1 6-a?,2 +ab)

(a)- ac +e)
(ae-a2 +a)
(a) - ac + ¢)
acla - ac +c)
Example 7. Ifb is the mean proportion between a and c, show that bla + c) is
the mean
proportion between (a² + b) and (b? + ).
Solution : Since b is the mean proportion between a and c, so we have b = ac.
:. Mean proportion between (a² + b²) and (b + c2)
=Vla+?)l% +) =la +ac)lae+) [:8 =ac
=Vala+o).cla +e) =yacta +e)² =y8 la +e
= bla + c)
Hence, the mean proportion between (a² + 6) and (b² + c) is bla + c).
EXample 8. If a b anda:b is the duplicate ratio of (a + c): (6 + c), prove that cis
the mean proportion between a and b.
Solution : Since the duplicate ratio of (a + c): (6 + c) is a : b, we have
(a+c? a +c + 2ac
(6+c B + + 2bc
[By cross multiplication]
ab'++2bc) =b(a+e +2ac)
(ab- ab²) +
(b-e'a) =0 93
Ratio and Proportion
 »ab- ab?)-(ea -b)=0
abla - b) - c(a-b) = 0
’ (a- b) (ab-e)=0 a’a-b 0]

’=ab ’c=ab. b.
and
between a
Hence, c is the mean proportion

prove that
:
Example 9. If x, y, z are in continued
proportion,

(x+y (2010)
(y+z)
continued proportion. Then,
Solution: Let , y, z be in
K:y = y:z ’
y² =
y

.:. LHS. =
(*+y
(y+2)
x+ xz + 2xy
x+y+ 2xy
xz + 2 + 2yz
y++2yz
x(* +2+2y) R.H.S.
z(* +2+ 2y)

(*+y?
Hence,
(y+2)²
prove that the first is to third
Example 10. If three quantities are in continued proportion,
is the duplicate ratioof the first to the second.
Solution : Let a, b, c be in continued proportion. Then,
b
a:b = b:c= =
..i)
C

,6_ a a a?
Now,
C C

a:c = g': 2

Hence, (First :Third) = Duplicate ratio of first to the second.

Problems Based on k-Method

Example 11. If a:bi: c:d, show that :
(a+ b):(c+ d) =ya? +b : Ve+a2
Soiution: Since a:6 = c:d, we have
C

b==k
d (say). Then, a= bk and c= dk.
Putting a = bk and c = dk, we get :
(a+b) (bk+ b) b(k +1) b
LHS =
lc+ d) (dk +d) d(k+1) d
 RHS = byk² +1
dyk +1 d
.. LHS =RHS.
Hence, (a + b) : (c + d) = va² + b2 : Je + d²
Example 12. If a:b = c:d, prove that :
(abcd)\a +b2 +c2 +d) =(4 +b² +e+d²).
Solution: Since a:b = c: d, we have
C
= k (say). Then, a = bk and c = dk.
Putting a =bh and c = dk, we get:
. LHS = abcdla +62 +e-2 +d2)
= abcd| ,1 1

1 1 1
- (bk)b (dk) d. (Putting a= bk, c = dk]
1 1 1 1
- (6a²2).
62
=d² +d²²+6² +b² =(1+ *²)(a² +6²).
RHS =(a? +b +2 +d?),
-(62² +6² +d? +d²)
[Putting a= bh, c = dk)
= (1 + kX? + b2).
.. LHS = RHS.

Hence, (abcd) la?+b +c2 +d2)=(a' +6 + e + d²).

3/2
Example 13. If=*
a
= prove that xyz
abc
Solution: Let = k.
C

Then, x = ka, y = kb and z = kc.

.. LHS = [Putting x = ka, y =kb, z = kc]

ka + kb + ke
2a+ b +c*)
klat +6 +e)
ka.kb.ke
RHS
abc
.:. LHS= RHS.
3/2
+ c2 yz
Hence, Vabc
a'x+by+ e°z
95
Ratio and Proportion
 Example 14 I| (h-c) (e-a) (a-b) Proue that I + by + 2=0
2
Solution : Lot
(h-c) (a-b) = k. Then,
xh (b - c), y=k c - a) and 2 = h la-h)
Substituting these values, we get :
ax + by + cz = ah (b - c)
+ bk (c- a) + ck la - b)
=k (a (b - c) + b
lc-a) + c (a - b))
= kx 0 = 0.
Hence, ax + by + cz = 0.
Example 15. If a, b, care in continued
proportion, prove that
2a -5ab +762
26 - 5bc+ 7e2
Solution:
1t 1s being given that a. b. care in continued proportion.

.i. ;=-=k
C
(say)
. b ck and a= bk = (ck)k = ck.
Substituting these values, we get:
2a - 5ab+ 762
L.H.S. =
262 - 5bc + 7e
2le2)' -5x ck' xck + 7(ck)'
2(ck) -5x ck xc+ 7¢
202k - 5ck +7e°k? =
(2k' -5k +7)
20242-5e'k+7e2 (22-5k +7)
R.H.S. = = k2.
C C

.:. LH.S. = R.H.S.

2a - 5ab + 76
Hence,
26 -5be +7e2 c
Example 16. If a, b, c, d are in continued proportion, prove that:
la-b)³
(6-c)
Solution: Let a, b, c, d be in continued proportion. Then
b C

b
= k (say).
C d

Then, c = kd, b = ke = kkd) = kd and a = kb = k(kd) = k3d.

Substituting these values, we get :
la -b)
L.HS. =
(b-c
Foundation Mathematics for Class X
96
 (Wd-kd d- hd hl (h-1)
R.H.S,
d
:. L.H.. =
R.H.S.
Hence, la-b
(b-c d'
Example 17. If a, b, c, d are in
(a' + b + eXb2 + c2 4
continued proportion, prove that :
d') = (ab + be + cd.
Solution : Let a, b, c, d be in
C
continued proportion.Then
C

i c= kd, b = ke = klkd) =
k'd and a = kb = klkd) = kd.
L.H.S. =(a' +6 +e)(6² +e+ d')
- (#9 +kd² +a)a +*d
+d)
=ka(k +k+1).d( ++1)=(a\++1.
R.HS. =(ab + be +cdy² =|(*dx k'd) +kdx kd) +(kd x 2
d)

.i. L.H.S. = R.H.S.

Hence, (a² + 62 + cXb2 + c2 + d²) = (ab + be +
cd.
Example 18. If a, b, c, d are in continued proportion,
prove that :
Vab +oc - Ved = la +b- c)(6+c- d)
Solution: Let a, b, c, d be in continued proportion. Then
C
= k (say).
d
.. c= dk, b = ck = (dk xk) = dk and a =bk = (dk2)k = dk.
Substituting these values, we get :
LH.S. =Jab + vbc -ycd
= ydk x dk + ydk x dk - Jdk xd

- (d +dk - d)Wk =dt +*-1)/.

R.H.S. = Vla +b- c)(b+c-d)
97
Ratio and Proportion
 = VlJdk +dk- dk)\dk +dk-d)
- ydk(k +k- 1)xd+*-1)

=
d+k-1)Jk.
.:. LH.S. = R.HS.
Hence, Vab + vbe - ycd = a+b-c)(b+c-d).

An Important Property
a+c+e
C
=k., prove that each ratio is equal to b+d+f
Sum of antecedents
Le., each ratio Sum of consequents
Proof : Let h Then, a = bk, c = dk and e = fk.
f
a+c+e bk + dk + fk k(b+ d+f)k.
b+d+f b+d+f (b+d+f)
a+c+e
Hence, each ratio =
b+d+f
*+y y+z
Example 19. If ax + by ay + bz az + 6x'
2
prove that each of these ratios is equal to (a+b)
unless x + y t 2 = 0.
Sum of antecedents
Solution: Clearly, each ratio = Sum of consequents
*+y ytz (x+y) +(y+z) +(2+x)
ax + by ay + bz az + bx (a+b)(*+ y+z)
2(x +y +z) 0.
(a+6)(*+y+z) o+h)' When x+y+z#

1
Example 20. If prove that each fraction is equal to or -1.
(6+c) (c+a) (a +b)
Sum of antecedents
Solution : Clearly, each ratio = Sum of consequents
C a+b+c 1
(6+c) (c+ a) (a +b) 2%a +b +c) 2 when a + b + c0.
Now, when a + 6 + c = 0, we have
b+c =- 4, c + a = -6 anda +b = - c.
b b

b+c
a-1, C+a -b
= -1and
a+b -C
=-1.

-1.
Hence, in this case, each ratio is

Foundation Mathematics for Class X

 EXERCISE 7B
when :
1,.Find x,
(i) 3:4:2.4 : x (iiü) x : 1.5 :: 3 : 5
(ü)1:3::x:7
aBnd the fourth proportional to :
(i) 3, 8 and 21 () 1.4, 3.2 and 7 iii) 1.5, 4.5 and 3.6
(iv) a², ab and b2 (v) (a² - ab + b2), (a3 + b3) and (a - b)
Find the third proportional to : 2
(i) 9 and 6 (ü) 2, and 4 (iii) 1.6 and 2.4

(kw) (2+ da and (5 +448) (u)and ya² +b2

b a
A. Find the mean proportion between :
(i) 28 and 63 (iü) 2.5 and 0.9 (iiü) 6.25 and 1.6

(ie) (V26-N17 and (26 + 17) (o) (6+3/3) and(8 -4N3)

5.6 is the mean proportion between two numbers x and y and 48 is the third proportional
of xandy. Find the numbers. (2011)

Hint: Jxy =6’ xy = 36 ...)

36
x:y:y:48 ’ y = 48x = 48 x y = (48 x 36)
Find x and y.]
6. What least number must be added to each of the numbers 5, 11, 19 and 37, so that
the resulting numbers are proportional. (2009)
7. What number must be added to each of the numbers 4, 6, 8, 11 in order to get the
four numbers in proportion? (2023)
8. What least number must be subtracted from each of the numbers 23, 30, 57 and 78,
so that the remainders are in proportion? (2004)
9. If (x - 2), (x + 2), (2x + 1) and (2* + 19) are in proportion, find the value of x.
10. The following numbers, K + 3, K + 2, 3K 7 and 2K - 3 are in proportion. Find K.
(2019)
11. If (« + 5) is the geometric mean between (x + 2) and (* + 9), find the value of x.
12. Find two numbers whose mean proportion is 36 and the third proportional is 288.
13. If a:b:c:d, prove that :
@(a² +ab):(+od)=(8? - 2ab): (² - 2od)
(ü) (a² +b?):(+ d)= (ab +ad - be):(cd - ad +bc)
(üi) (a +ac +c):(a' - ac + )=(8 +bd + d²):(8 - bd +d²)
Hint : Use k-method in each.]
14. If a:b ::c: d, show that :

(i)
a+b 2a'+762 ma' +ne2 ya+t
(ü)
c+d
J2e+7d2 mb² +nd² J& +dt
la+c ala -c?
(üi)
a'+ ab +b? e+ cd +² (iv)
a-ab+ b? - cd+d? (6+ d) b6- d
Hint : Use k-method.]
99
Ratio and Proportion
 15,. If = ,prove that :

(u) 3ryz
a+b* pa+ qb +r abc (2016)
3
(üi)
(+y+ *
(a+h+c
ar- by by - cz
(iv)
(a+b)(r - y) (6+c)(y- z) (c+ alz- *) =3
Hint : Use k-method in each.]
16.I f ì prove that :
(i) (62 + d + f) (a² + + e)= (ab + cd + ef)
(iü) ace a Ce ae
(ùi)

a+b c+d et f
(iv) (baf). b
= 27(a+b)lc+d)le+f)
[(Hint : Use k-method in each.]
7. If a, b, c are in continued proportion, prove that :
a+b a²b-c) a+b+c (a +b+c
() (iü)
b+c b² (a -b) a-b+c (a+b2 +)
a'+ ab + b? (iv) (a + b + c) (a -b + c) = (a² + b2 + c2)
(iüü)
6² +bc + c? c
(w) a622 (a + 6S+) = (a + b +c) (2015)
a
[Hint : ==kh= ck and a =ck']
18. If a, b, c, d are in continued proportion, prove that :
(i) (6 + c) (6 + d) = (c + a) (c + d)
(iü) (a + b) (6 + c) - (a + c) (6 + d) = (6 - c)2

Hint : ===k’c= dk, b= dk' and a=dk³.]

C

CO ab
19. If ax = by = cz, prove that + +
yz

Hint : ax = by= cz, ’ -=k’x = kbc, y = kca and z = kab.]

bo Ca ab

20. If *+y+
6+c-a C+a-b a+b-c
prove that each ratio is equal to
a+b+c
Also, show that (6 - c) x + (e - a)y+ (a - b)2 = 0.
*+y+z *+ y+z
[Hint : Each ratio = (6+c-a) +(c+ a-b) +(a+ b-c) a+b+e

Foundation Mathematics for Class X

100
 Let each ratio be equal to k. Then,
(b +c -
a)k, y = (c + a - b)k
and z = (a +b - c)k.]
b is the mean proportion
21.If between a and c, show that: (2017)

ruint:Substitute b = ac in L.H.S. to get

R.H.S.)

Some Useful Results on Ratio and Proportion

d
C Unvertendo)
b d c (Alternendo)
C a+6 C+d
(üi)
d (Componendo)
a-b C-d
(iv)
d (Dividendo)
a+b c+d
hd a-b c- d (Componendo & Dividendo)
SOLVED EXAMPLES

Example 1. If Iprove
e 3a- 5b 3c - 5d
d
that (2000)
3a + 56 3c + 5d

Solution :
C 3a 3c 3
bd 5b 5d [Multiplying both sides by
3a + 56 3c + 5d
3a -56 3c - 5d
By componendo & dividendo]
3a - 56 3c-5d
3a + 5b 3c + 5d
(By invertendo]
3a - 5b 3c-5d
Hence,
3a + 56 3c + 5d
Example 2. If a: b::c:d, prove that *0tb0 4a - 96
4c + 9d 4c-9d
Solution : a:b::c:d ’;=
C

4a 4c
96 9d Multiplying both sides byl
4a +9b 4c + 9d
4a -96 4c-9d By componendo & dividendo]
4a +96 4a-96
4c+9d 4c-9d (By alternendo]
4a + 96 4a-96
Hence,
4c + 9d 4c- 9d

101
Ratio and Proportion
 8a -55 8a + 56
Example 3. If 8c - 5d 8c + 5d
prove that =
d (2008)
Solution : We have
8a- 56 8a +56
8c -5d 8c +5d (Given]

8a- 5b 8c-5d
8a+56 (By alternendo]
8c+5d
8a+ 5b 8c +5d
(By invertendo]
8a-5b 8c-5d
(8a +5b)+(8a-55) (8c + 5d) +(8c-5d)
(8a +5b)-(8a-5b) (By componendo &dividendol
(8c + 5d)- (8c - 5d)
16a 16c C
On multiplying each side by 10
10b 10d 16
C
Hence, =
+3ab? 63
Example 4. Given that (2009)
b +3ab 62
Using Componendo and Dividendo, find a :b.
Solution: We have
a +3ab 63
(Given]
b + 3a b 62

(a+3ab?)+(6 +3a'b) 63 +62

(By Componendo &Dividendo|
(a+ 3ab?) -(6° +3a²b) 63 - 62
a +6 + 3abla + b) 125 (a+b)³
a-b - 3abla - b) 1 la-b)³
a+6)) a+b 5
l(a-b)) a-b 1

(a+b)+ (a-b) 5+1 2a 6

(a+b)-(a -b) 5-1 2 4 ’¡ a:b=3:2
Hence, a:b=3:2.
Example 5. If 7m + 2n 5
7m- 2n 3 use properties of proportion to find : (2017)

(i) m:n
Solution: m'-n2
(i) We have
7m + 2n 5
7m -2n 3 [Given]
(7m + 2n) +(7m -2n) 5+3
(7m + 2n)-(7m-2rn) 5-3 (By Componendo &Dividendol
14m 8 7m m 4 x2 8
=4’
4n 2 2n n 7 7
 64
72 49
m+n 64 +49 113
m? 64-49 15 By Componendo & Dwidendol
Example 6. Using componendo and
dividendo, find the value ofx when
/3x +4 +y3r-5
=9
J3x +4 - 3r -5
(2011)
Solution : We have

J3x +4 + 3x- 5 9
J3x + 4- J3x -5 1 IGiven)
/3x +4+ J3r -5 +(J3x+ 4 - V3x -5)
9+1
W3r +4+ 3r - 5)-(J3x +4-N3x -5) 9-1

(By Componendo & Dividendol

2/3x +4 10 V3x +4 5 3x+4 25
2/3x -5 V3x-5 4 3x-5 |On squaring both sides]
16
(3x +4) + (3x- 5) 25 + 16
(3* +4)-(3¢ -5) 25 -16 (By Componendo & Dividendo]
6x -1 41
9 9 ’6x-1=41’ 6x = 42 x=7.
Hence, x = 7.
Example 7. Using the properties of proportion, fnd the
value of x, when
x+1 17
2x? 8 (2013)
Solution : We have
x+1 17
212 (Given]
x +1+2x 17 +8
17 - 8 IBy Componendo & Dividendol

1
25 +1

*+1 5
x'-1 3* (On taking square root on each sidel

+1)+-1) 5+3 (By Componendo & Diwidendo)

+1)-(?-1) 5-3
242 8
= 4 ’*=4 ’x=t 2.
Hence, * = t 2.
103
Ratio and Proportion
 a +1 + a -1
Example 8. Ifx =
va + l - a - 1
that x*-2ax + 10
using properties of proportion, shoe (20121

Solution: We have ya+l t ya-1

1 ya+l - va -1 (Given)

.+1 (Va +1 t ya- 1) +(Va +1- Ja-1)

x-1 (ya +1 t ya-1)-(Wa +1- Ja -1)
[By Componendo & Dividendol
x+1_ 2va +1 X+1 Va+1
x-1 2Na -1 N-1 Va-1
2
I+1 (a +1)
|On squaring both sides)
la -1)
r+1+ 2x a+1
+1-2x a-1
(r+1+2x)+(+1- 2x) (a +1) +(a-1)
x+1+2x)-(² +1 - 2t) la+1) - (a -1)
[By Componendo & Dividendol
21? +1) 2a a+1
= a ’ x - 2ax + 1=0.
4x 2 2x
Hence, x? 2ax + 1 = 0.

Va+36 + va-36 0. (2007)

prove that 3br- 2ax + 36 =
Example 9. If x=
Va +36 - Va-36
Solution: We have
Va+36 + ya- 36 (Given]
1 ya+ 36 - va -36
+36 - a - 36)
x+1 (Va +36 + Ja - 36)+(Ja
36)
I-1 (Na +36 +Ja- 36) -(Va +36 - Ja - Componendo & Dividendo)
[By
2va+36 Va +36
x+1
x-1 Va-36
t-1 2Va -3b
sides]
(+1)² a+36
(On squaring both
’ a-36

(a+36) + (a-36) & Dividendol

(x+1) + (r-1) (By Componendo
(a+36) -(a-36)
(z+1 - (u-1'

2 +1) 2a +1)
6 2x
4x
3bx-2ax + 36 = 0.
2ax ’
3ox2 + 36 =
+ 36) =0. for Class X
Hence, (36r'
- 2ax Foundation
Mathematics
Example 10. Given x

Use Componendo and Divndenlo to prove that

b2 - 2'x +' 0 . (2010)
S o l u t i o n :
We have

Va' - |Givenl

-1
+6² + ya
\By Componendo& Dividendol
2ya +6
2ya -b
(x+1) a +1+ 2x
(x-1)² 2+1 2x |On squaring both sidesl
(x+1+2x) +(x +1 - 2x) (a² +62) +(a2 -62)
(a2 +1+ 2*) -(x +1- 2x) (a' +6²) -(a'-62)
[By Componendo & Dividendol
2( +1) 2a2
x +1
4x 262 2x 62
b2x2 + b' = 2a'x ’ bx2 - 2a'x + b2 = 0.
Hence, b2x2 - 2ax + b2 = 0.

Example 11. Solve for x when va+x + Va-x = b.

Va +x - ya -x
Solution: We have

Va+x + va-x
Va+x - Ja-x 1 (Given]

Na tx + va-*+Va+x - Na-x b+1

Wa +x + va-*-|Wa+*- va-x b-1
[By Componendo & Dividendol
2Va +x b+1 Va+x (b+ 1)
2ya-x b-1 Jn (b-1)
(6+1)
(6-1 (On squaring both sidesl
(a+x) + (a-x) (6+1) + (6-1)
(a + x) - (a- x)
(6+1) - (6-1)
[By Componendo & Dividendol
2a 2(6² +1) a(b + 1)
2x 4b 26
Ratio and Proportion 105
 2ab
x{b +1) =2ab x=
(6² +1)
2ab
Hence,
(6²+1)
4xy p+ 2x p+2y
Example 12. If p=
prove that + = 2,
(x+y)' p-2x p-2y

Solution : We have
4xy
p= |Given)
(x+ y)
P 2y and P 2x
2x (x+ y) 2y (x+y)
p+ 2x 2y + (x + y) p+2y 2x + (* + y)
and
p- 2x 2y - (* +y) p- 2y 2x - (* + y)
(By Componendo& Dividendol
p+ 2x 3y + x p+ 2y 3x + y
and
p-2* p- 2y
p+ 2x p+2y 3y +x, 3x +y
p- 2x p- 2y y- *
3x +y_ 3y + * (3x + y)- (3y +x) 2(*- y) - 2
(r- y) (x- y

Hence, p+2x, p+2y = 2.

p- 2x p-2y
+ 3ab? x+ 3*y prove that
Example 13. If
3a b+6° 32y + y
Soiution: We have
a'+3ab? +3xy? (Given]
3a'b+b³ 3*'y + y
(a+Sab?)+ (3a²b +b) +3ary2)+(3r²y +y)
(a+ 3ab?)-(3ab +b°) (+ 3xy²)-(3*y + y' )
’

Componendo & Dividendo) By

a +b +3abla + b) +y+ 3xy(x+ y)
a'-b -3abla -b) -y - 3xy(*- y)
3
(a+b)³ (*+y)° Jla + b)]
(a-b)³ (r-y l(a-b)J

(a+b) (*+y) (Taking cube root of each sidel

(a-b) (-y)
(a+b) +la-b) (x+ y)+(*-y) (By Componendo & Dividendol
(a+b) - (a-b) (x+y)-(r-y)
2a 2x
b
26 2y b y

Hence, for
Cass

Foundation Mathematics

106