Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 4.5 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 4 Ratio and Proportion.

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

Question 1.

Which number should be subtracted from 12, 16 and 21 so that resultant numbers are in continued proportion?

Solution:

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.

If (28 – x) is the mean proportional of (23 – x) and (19 – x), then find the value ofx.

Solution:

(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 = \(\frac { 347 }{ 14 }\)

Question 3.

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:

Let the first number be x.

∴ Third number = 26 – x

12 is the mean proportional of x and (26 – x).

∴ \(\frac { x }{ 12 }\) = \(\frac { 12 }{ 26 – x }\)

∴ x(26 – x) = 12 x 12

∴ 26x – x^{2} = 144

∴ x^{2} – 26x + 144 = 0

∴ x^{2} – 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.

If (a + b + c)(a – b + c) = a^{2} + b^{2} + c^{2}, show that a, b, c are in continued proportion.

Solution:

(a + b + c)(a – b + c) = a^{2} + b^{2} + c^{2} …[Given]

∴ a(a – b + c) + b(a – b + c) + c(a – b + c) = a2 + b2 + c2

∴ a^{2} – ab + ac + ab – b^{2} + be + ac – be + c^{2} = a^{2} + b^{2} + c^{2}

∴ a^{2} + 2ac – b^{2} + c^{2} = a^{2} + b^{2} + c^{2}

∴ 2ac – b^{2} = b^{2}

∴ 2ac = 2b^{2}

∴ ac = b^{2}

∴ b^{2} = ac

∴ a, b, c are in continued proportion.

Question 5.

If \(\frac { a }{ b }\) = \(\frac { b }{ c }\) and a, b, c > 0, then show that,

i. (a + b + c)(b – c) = ab – c^{2}

ii. (a^{2} + b^{2})(b^{2} + c^{2}) = (ab + be)^{2}

iii. \(\frac{a^{2}+b^{2}}{a b}=\frac{a+c}{b}\)

Solution:

Let \(\frac { a }{ b }\) = \(\frac { b }{ c }\) = k

∴ b = ck

∴ a = bk =(ck)k

∴ a = ck^{2} …(ii)

i. (a + b + c)(b – c) = ab – c^{2}

L.H.S = (a + b + c) (b – c)

= [ck^{2} + ck + c] [ck – c] … [From (i) and (ii)]

= c(k^{2} + k + 1) c (k – 1)

= c^{2} (k^{2} + k + 1) (k – 1)

R.H.S = ab – c^{2}

= (ck^{2}) (ck) – c^{2} … [From (i) and (ii)]

= c^{2}k^{3} – c^{2}

= c^{2}(k^{3} – 1)

= c^{2} (k – 1) (k^{2} + k + 1) … [a^{3} – b^{3} = (a – b) (a^{2} + ab + b^{2}]

∴ L.H.S = R.H.S

∴ (a + b + c) (b – c) = ab – c^{2}

ii. (a^{2} + b^{2})(b^{2} + c^{2}) = (ab + bc)^{2}

b = ck; a = ck^{2}

L.H.S = (a^{2} + b^{2}) (b^{2} + c^{2})

= [(ck^{2}) + (ck)^{2}] [(ck)^{2} + c^{2}] … [From (i) and (ii)]

= [c^{2}k^{4} + c^{2}k^{2}] [c^{2}k^{2} + c^{2}]

= c^{2}k^{2} (k^{2} + 1) c^{2} (k^{2} + 1)

= c4k^{2} (k^{2} + 1)^{2}

R.H.S = (ab + bc)^{2}

= [(ck^{2}) (ck) + (ck)c]^{2} …[From (i) and (ii)]

= [c^{2}k^{3} + c^{2}k]^{2}

= [c^{2}k (k^{2} + 1)]2 = c^{4}(k^{2} + 1)^{2}

∴ L.H.S = R.H.S

∴ (a^{2} + b^{2}) (b^{2} + c^{2}) = (ab + bc)^{2}

iii. \(\frac{a^{2}+b^{2}}{a b}=\frac{a+c}{b}\)

9th Standard Algebra Practice Set 4.5 Question 6. Find mean proportional of \(\frac{x+y}{x-y}, \frac{x^{2}-y^{2}}{x^{2} y^{2}}\).

Solution:

Let a be the mean proportional of \(\frac{x+y}{x-y}\) and \(\frac{x^{2}-y^{2}}{x^{2} y^{2}}\)