MATHS EX 7B
Chapter 7 - Ratio and Proportion (Including Properties and Uses) Exercise Ex. 7(B)
Question 1
Find the fourth proportional to:
(i) 1.5, 4.5 and 3.5 (ii) 3a, 6a2 and 2ab2
Solution 1
(i) Let the fourth proportional to 1.5, 4.5 and 3.5 be x.
 1.5 : 4.5 = 3.5 : x
 1.5 : 4.5 = 3.5 : x 1.5
 1.5  x = 3.5
 x = 3.5  4.5
 4.5 x = 10.5
 x = 10.5
(i) Let the fourth proportional to 3a, 6a2 and 2ab2 be x.
 3a : 6a2 = 2ab2 : x
 3a : 6a2 = 2ab2 : x 3a
 3a  x = 2ab2
 x = 2ab2 6a2
 6a2 3a
 3a  x = 12a3b2
 x = 12a3b2 x = 4a2b2
 x = 4a2b2
Question 2
Find the third proportional to:
(i) 2 and 4 (ii) a - b and a2 - b2
 and 4 (ii) a - b and a2 - b2
 and 4 (ii) a - b and a2 - b2
 and 4 (ii) a - b and a2 - b2
Solution 2
(i) Let the third proportional to 2 and 4 be x.
 and 4 be x.
 and 4 be x.
 and 4 be x. 2
 2 , 4, x are in continued proportion.
, 4, x are in continued proportion. 2
 2 : 4 = 4 : x
 : 4 = 4 : x
(ii) Let the third proportional to a - b and a2 - b2 be x.
 a - b, a2 - b2, x are in continued proportion.
 a - b, a2 - b2, x are in continued proportion. a - b : a2 - b2 = a2 - b2 : x
 a - b : a2 - b2 = a2 - b2 : x
Question 3
Find the mean proportional between:
(i) 6 + 3 and 8 - 4
 and 8 - 4
 and 8 - 4
 and 8 - 4
(ii) a - b and a3 - a2b
Solution 3
(i) Let the mean proportional between 6 + 3 and 8 - 4
 and 8 - 4 be x.
 be x.
 and 8 - 4
 and 8 - 4 be x.
 be x. 6 + 3
 6 + 3 , x and 8 - 4
, x and 8 - 4 are in continued proportion.
 are in continued proportion. 6 + 3
 6 + 3 : x = x : 8 - 4
 : x = x : 8 - 4
 x
 x  x = (6 + 3
x = (6 + 3 ) (8 - 4
) (8 - 4 )
) x2 = 48 + 24
 x2 = 48 + 24 - 24
- 24 - 36
 - 36 x2 = 12
 x2 = 12 x= 2
 x= 2
(ii) Let the mean proportional between a - b and a3 - a2b be x.
 a - b, x, a3 - a2b are in continued proportion.
 a - b, x, a3 - a2b are in continued proportion. a - b : x = x : a3 - a2b
 a - b : x = x : a3 - a2b x
 x  x = (a - b) (a3 - a2b)
x = (a - b) (a3 - a2b) x2 = (a - b) a2(a - b) = [a(a - b)]2
 x2 = (a - b) a2(a - b) = [a(a - b)]2 x = a(a - b)
 x = a(a - b)
Question 4
If x + 5 is the mean proportional between x + 2 and x + 9; find the value of x.
Solution 4
Given, x + 5 is the mean proportional between x + 2 and x + 9.
 (x + 2), (x + 5) and (x + 9) are in continued proportion.
 (x + 2), (x + 5) and (x + 9) are in continued proportion. (x + 2) : (x + 5) = (x + 5) : (x + 9)
 (x + 2) : (x + 5) = (x + 5) : (x + 9) (x + 5)2 = (x + 2)(x + 9)
 (x + 5)2 = (x + 2)(x + 9) x2 + 25 + 10x = x2 + 2x + 9x + 18
 x2 + 25 + 10x = x2 + 2x + 9x + 18 25 - 18 = 11x - 10x
 25 - 18 = 11x - 10x x = 7
 x = 7
Question 5
If x2, 4 and 9 are in continued proportion, find x.
Solution 5
Question 6
What least number must be added to each of the numbers 6, 15, 20 and 43 to make them proportional?
Solution 6
Let the number added be x.
 (6 + x) : (15 + x) :: (20 + x) (43 + x)
 (6 + x) : (15 + x) :: (20 + x) (43 + x)
Thus, the required number which should be added is 3.
Question 7(i)
Solution 7(i)
Question 7(ii)
Solution 7(ii)
Question 7(iii)
Solution 7(iii)
Question 8
What least number must be subtracted from each of the numbers 7, 17 and 47 so that the remainders are in continued proportion?
Solution 8
Let the number subtracted be x.
 (7 - x) : (17 - x) :: (17 - x) (47 - x)
 (7 - x) : (17 - x) :: (17 - x) (47 - x)
Thus, the required number which should be subtracted is 2.
Question 9
If y is the mean proportional between x and z; show that xy + yz is the mean proportional between x2+y2 and y2+z2.
Solution 9
Since y is the mean proportion between x and z
Therefore, y2 = xz
Now, we have to prove that xy+yz is the mean proportional between x2+y2 and y2+z2, i.e.,

LHS = RHS
Hence, proved.
Question 10
If q is the mean proportional between p and r, show that:
pqr (p + q + r)3 = (pq + qr + rp)3.
Solution 10
Given, q is the mean proportional between p and r.
 q2 = pr
q2 = pr
Question 11
If three quantities are in continued proportion; show that the ratio of the first to the third is the duplicate ratio of the first to the second.
Solution 11
Let x, y and z be the three quantities which are in continued proportion.
Then, x : y :: y : z  y2 = xz ....(1)
 y2 = xz ....(1)
 y2 = xz ....(1)
 y2 = xz ....(1)
Now, we have to prove that
x : z = x2 : y2
That is we need to prove that
xy2 = x2z
LHS = xy2 = x(xz) = x2z = RHS [Using (1)]
Hence, proved.
Question 12
If y is the mean proportional between x and z, prove that:

Solution 12
Given, y is the mean proportional between x and z.
 y2 = xz
y2 = xz
Question 13
Given four quantities a, b, c and d are in proportion. Show that:

Solution 13



 LHS = RHS
LHS = RHS
Hence proved.
Question 14
Find two numbers such that the mean mean proportional between them is 12 and the third proportional to them is 96.
Solution 14
Let a and b be the two numbers, whose mean proportional is 12.

Now, third proportional is 96

Therefore, the numbers are 6 and 24.
Question 15
Find the third proportional to 

Solution 15
Let the required third proportional be p.

 , p are in continued proportion.
, p are in continued proportion.
Question 16
If p: q = r: s; then show that:
mp + nq : q = mr + ns : s.
Solution 16

Hence, mp + nq : q = mr + ns : s.
Question 17
If p + r = mq and  ; then prove that p : q = r : s.
; then prove that p : q = r : s.
 ; then prove that p : q = r : s.
; then prove that p : q = r : s.
Solution 17

Hence, proved.
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