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Chapter 4 Ratio and Proportion Practice Set 4.5 Solutions

Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 4 Ratio and Proportion Practice Set 4.5. Step-by-step solved exercises, numerical problems, and digest answers.

5 Solved Questions• 860 words

Practice Set 4.5 Algebra 9th Std Maths Part 1 Answers Chapter 4 Ratio and Proportion

Question 1 Maharashtra Board Solution
Which number should be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion?
Solution & Step-by-Step Answer:
Let the number to be subtracted be x. ∴ (12 – x), (16 – x) and (21 – x) are in continued proportion. ∴ 84 – 4x = 80 – 5x ∴ 5x – 4x = 80 – 84 ∴ x = -4 ∴ -4 should be subtracted from 12,16 and 21 so that the resultant numbers in continued proportion.
Question 2 Maharashtra Board Solution
If (28 – x) is the mean proportional of (23 – x) and (19 – x), then find the value ofx.
Solution & Step-by-Step Answer:
(28 – x) is the mean proportional of (23 – x) and (19-x). …[Given] ∴ -5(19 – x) = 9(28 – x) ∴ -95 + 5x = 252 – 9x ∴ 5x + 9x = 252 + 95 ∴ 14x = 347 ∴ x =
Question 3 Maharashtra Board Solution
Three numbers are in continued proportion, whose mean proportional is 12 and the sum of the remaining two numbers is 26, then find these numbers.
Solution & Step-by-Step Answer:
Let the first number be x. ∴ Third number = 26 – x 12 is the mean proportional of x and (26 – x). ∴ = ∴ x(26 – x) = 12 x 12 ∴ 26x – x2 = 144 ∴ x2 – 26x + 144 = 0 ∴ x2 – 18x – 8x + 144 = 0 ∴ x(x – 18) – 8(x – 18) = 0 ∴ (x – 18) (x – 8) = 0 ∴ x = 18 or x = 8 ∴ Third number = 26 – x = 26 – 18 = 8 or 26 – x = 26 – 8 = 18 ∴ The numbers are 18, 12, 8 or 8, 12, 18.
Question 4 Maharashtra Board Solution
If (a + b + c)(a – b + c) = a2 + b2 + c2, show that a, b, c are in continued proportion.
Solution & Step-by-Step Answer:
(a + b + c)(a – b + c) = a2 + b2 + c2 …[Given] ∴ a(a – b + c) + b(a – b + c) + c(a – b + c) = a2 + b2 + c2 ∴ a2 – ab + ac + ab – b2 + be + ac – be + c2 = a2 + b2 + c2 ∴ a2 + 2ac – b2 + c2 = a2 + b2 + c2 ∴ 2ac – b2 = b2 ∴ 2ac = 2b2 ∴ ac = b2 ∴ b2 = ac ∴ a, b, c are in continued proportion.
Question 5 Maharashtra Board Solution
If = and a, b, c > 0, then show that, i. (a + b + c)(b – c) = ab – c2 ii. (a2 + b2)(b2 + c2) = (ab + be)2 iii.
Solution & Step-by-Step Answer:
Let = = k ∴ b = ck ∴ a = bk =(ck)k ∴ a = ck2 …(ii)

i. (a + b + c)(b – c) = ab – c2
L.H.S = (a + b + c) (b – c)
= [ck2+ ck + c] [ck – c] … [From (i) and (ii)]
= c(k2+ k + 1) c (k – 1)
= c2(k2+ k + 1) (k – 1)
R.H.S = ab – c2
= (ck2) (ck) – c2… [From (i) and (ii)]
= c2k3– c2
= c2(k3– 1)
= c2(k – 1) (k2+ k + 1) … [a3– b3= (a – b) (a2+ ab + b2]
∴ L.H.S = R.H.S
∴ (a + b + c) (b – c) = ab – c2

ii. (a2+ b2)(b2+ c2) = (ab + bc)2
b = ck; a = ck2
L.H.S = (a2+ b2) (b2+ c2)
= [(ck2) + (ck)2] [(ck)2+ c2] … [From (i) and (ii)]
= [c2k4+ c2k2] [c2k2+ c2]
= c2k2(k2+ 1) c2(k2+ 1)
= c4k2(k2+ 1)2
R.H.S = (ab + bc)2
= [(ck2) (ck) + (ck)c]2…[From (i) and (ii)]
= [c2k3+ c2k]2
= [c2k (k2+ 1)]2 = c4(k2+ 1)2
∴ L.H.S = R.H.S
∴ (a2+ b2) (b2+ c2) = (ab + bc)2

iii.

9th Standard Algebra Practice Set 4.5 Question 6. Find mean proportional of .
Solution:
Let a be the mean proportional of and

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Chapter 4 Ratio and Proportion Problem Set 4