Maharashtra Board Class 9 Maths Part 1 Algebra Chapter 4 Ratio and Proportion Set 4.2 Solutions

Question 1. Using the property \( \frac{a}{b} = \frac{ak}{bk} \), fill in the blanks by substituting proper numbers in the following.
(i) \( \frac{5}{7} = \frac{....}{28} = \frac{35}{....} = \frac{....}{3.5} \)
(ii) \( \frac{9}{14} = \frac{4.5}{....} = \frac{....}{42} = \frac{....}{3.5} \)
Answer:
(i) \( \frac{5}{7} = \frac{5 \times 4}{7 \times 4} = \frac{20}{28} \)
\( \frac{5}{7} = \frac{5 \times 7}{7 \times 7} = \frac{35}{49} \)
\( \frac{5}{7} = \frac{5 \times 0.5}{7 \times 0.5} = \frac{2.5}{3.5} \)
\( \therefore \frac{5}{7} = \frac{20}{28} = \frac{35}{49} = \frac{2.5}{3.5} \)
(ii) \( \frac{9}{14} = \frac{9 \times 0.5}{14 \times 0.5} = \frac{4.5}{7} \)
\( \frac{9}{14} = \frac{9 \times 3}{14 \times 3} = \frac{27}{42} \)
\( \frac{9}{14} = \frac{9 \times 0.25}{14 \times 0.25} = \frac{2.25}{3.5} \)
\( \therefore \frac{9}{14} = \frac{4.5}{7} = \frac{27}{42} = \frac{2.25}{3.5} \)
In simple words: This question demonstrates the fundamental property of ratios where multiplying both the numerator and denominator by the same non-zero number does not change the ratio's value, allowing for filling in equivalent fractions.

🎯 Exam Tip: Always show the multiplication factor (k) clearly when filling the blanks to demonstrate understanding of equivalent ratios.

Question 2. Find the following ratios.
(i) The ratio of radius to circumference of the circle.
(ii) The ratio of circumference of circle with radius r to its area.
(iii) The ratio of diagonal of a square to its side, if the length of side is 7 cm.
(iv) The lengths of sides of a rectangle are 5 cm and 3.5 cm. Find the ratio of numbers denoting its perimeter to area.
Answer:
(i) Let the radius of circle be r.
then, its circumference = \( 2\pi r \)
Ratio of radius to circumference of the circle \( = \frac{\text{radius}}{\text{circumference}} = \frac{r}{2\pi r} = \frac{1}{2\pi} \)
\( \implies \) The ratio of radius to circumference of the circle is 1 : \( 2\pi \).
(ii) Let the radius of the circle is r.
\( \therefore \) circumference = \( 2\pi r \) and area = \( \pi r^2 \)
Ratio of circumference to the area of circle \( = \frac{\text{circumference}}{\text{area}} = \frac{2\pi r}{\pi r^2} = \frac{2}{r} \)
\( \implies \) The ratio of circumference of circle with radius r to its area is 2 : r.
(iii) Length of side of square = 7 cm
\( \therefore \) Diagonal of square = \( \sqrt{2} \) x side = \( \sqrt{2} \) x 7 = \( 7\sqrt{2} \) cm
Ratio of diagonal of a square to its side \( = \frac{\text{diagonal}}{\text{side}} = \frac{7\sqrt{2}}{7} = \frac{\sqrt{2}}{1} \)
\( \implies \) The ratio of diagonal of a square to its side is \( \sqrt{2} : 1 \).
(iv) Length of rectangle = (l) = 5 cm,
Breadth of rectangle = (b) = 3.5 cm
Perimeter of the rectangle = \( 2(l + b) = 2(5 + 3.5) = 2 \times 8.5 = 17 \) cm
Area of the rectangle = \( l \times b = 5 \times 3.5 = 17.5 \) cm\( ^2 \)
Ratio of numbers denoting perimeter to the area of rectangle \( = \frac{\text{perimeter}}{\text{area}} = \frac{17}{17.5} \)
\( = \frac{17 \times 10}{17.5 \times 10} = \frac{170}{175} = \frac{5 \times 34}{5 \times 35} = \frac{34}{35} \)
\( \implies \) Ratio of numbers denoting perimeter to the area of rectangle is 34 : 35.
In simple words: This question requires applying basic geometric formulas for circles, squares, and rectangles to calculate and express various ratios involving their dimensions, circumference, area, perimeter, and diagonal.

🎯 Exam Tip: Remember to write down the formulas for each geometric shape and its properties before substituting values to avoid errors and ensure clarity in your solution.

Question 3. Compare the following
(i) \( \frac{\sqrt{5}}{3}, \frac{3}{\sqrt{7}} \)
(ii) \( \frac{3\sqrt{5}}{5\sqrt{7}}, \frac{\sqrt{63}}{\sqrt{125}} \)
(iii) \( \frac{5}{18}, \frac{17}{121} \)
(iv) \( \frac{\sqrt{80}}{\sqrt{48}}, \frac{\sqrt{45}}{\sqrt{27}} \)
(v) \( \frac{9.2}{5.1}, \frac{3.4}{7.1} \)
Answer:
(i) Compare \( \frac{\sqrt{5}}{3} \) and \( \frac{3}{\sqrt{7}} \)
\( \frac{\sqrt{5}}{3} = \frac{\sqrt{5} \times \sqrt{7}}{3 \times \sqrt{7}} = \frac{\sqrt{35}}{3\sqrt{7}} \)
\( \frac{3}{\sqrt{7}} = \frac{3 \times 3}{\sqrt{7} \times 3} = \frac{9}{3\sqrt{7}} = \frac{\sqrt{81}}{3\sqrt{7}} \)
Here, \( 35 < 81 \)
\( \implies \sqrt{35} < \sqrt{81} \)
\( \implies \frac{\sqrt{35}}{3\sqrt{7}} < \frac{\sqrt{81}}{3\sqrt{7}} \)
\( \therefore \frac{\sqrt{5}}{3} < \frac{3}{\sqrt{7}} \)
(ii) Compare \( \frac{3\sqrt{5}}{5\sqrt{7}} \) and \( \frac{\sqrt{63}}{\sqrt{125}} \)
\( \frac{3\sqrt{5}}{5\sqrt{7}} = \frac{3\sqrt{5} \times \sqrt{125}}{5\sqrt{7} \times \sqrt{125}} = \frac{3\sqrt{5 \times 125}}{5\sqrt{7 \times 125}} = \frac{3\sqrt{625}}{5\sqrt{875}} = \frac{3 \times 25}{5\sqrt{875}} = \frac{75}{5\sqrt{875}} \)
\( \frac{\sqrt{63}}{\sqrt{125}} = \frac{\sqrt{63} \times 5\sqrt{7}}{\sqrt{125} \times 5\sqrt{7}} = \frac{5\sqrt{63 \times 7}}{5\sqrt{125 \times 7}} = \frac{5\sqrt{441}}{5\sqrt{875}} = \frac{5 \times 21}{5\sqrt{875}} = \frac{105}{5\sqrt{875}} \)
Alternatively for comparison:
Consider \( 3\sqrt{5} \times \sqrt{125} = 3\sqrt{5 \times 125} = 3\sqrt{625} = 3 \times 25 = 75 \)
Consider \( 5\sqrt{7} \times \sqrt{63} = 5\sqrt{7 \times 63} = 5\sqrt{441} = 5 \times 21 = 105 \)
Here, \( 75 < 105 \)
\( \implies \frac{3\sqrt{5}}{5\sqrt{7}} < \frac{\sqrt{63}}{\sqrt{125}} \)
(iii) Compare \( \frac{5}{18} \) and \( \frac{17}{121} \)
\( 5 \times 121 = 605 \)
\( 18 \times 17 = 306 \)
Here, \( 605 > 306 \)
\( \implies \frac{5}{18} > \frac{17}{121} \)
(iv) Compare \( \frac{\sqrt{80}}{\sqrt{48}} \) and \( \frac{\sqrt{45}}{\sqrt{27}} \)
Consider cross-multiplication:
\( \sqrt{80} \times \sqrt{27} = \sqrt{80 \times 27} = \sqrt{2160} \)
\( \sqrt{48} \times \sqrt{45} = \sqrt{48 \times 45} = \sqrt{2160} \)
Here, \( 2160 = 2160 \)
\( \implies \sqrt{2160} = \sqrt{2160} \)
\( \therefore \frac{\sqrt{80}}{\sqrt{48}} = \frac{\sqrt{45}}{\sqrt{27}} \)
Alternate method:
\( \frac{\sqrt{80}}{\sqrt{48}} = \frac{\sqrt{16 \times 5}}{\sqrt{16 \times 3}} = \frac{4\sqrt{5}}{4\sqrt{3}} = \frac{\sqrt{5}}{\sqrt{3}} \)
\( \frac{\sqrt{45}}{\sqrt{27}} = \frac{\sqrt{9 \times 5}}{\sqrt{9 \times 3}} = \frac{3\sqrt{5}}{3\sqrt{3}} = \frac{\sqrt{5}}{\sqrt{3}} \)
Here, \( \frac{\sqrt{5}}{\sqrt{3}} = \frac{\sqrt{5}}{\sqrt{3}} \)
\( \therefore \frac{\sqrt{80}}{\sqrt{48}} = \frac{\sqrt{45}}{\sqrt{27}} \)
(v) Compare \( \frac{9.2}{5.1} \) and \( \frac{3.4}{7.1} \)
Consider cross-multiplication:
\( 9.2 \times 7.1 = 65.32 \)
\( 5.1 \times 3.4 = 17.34 \)
Here, \( 65.32 > 17.34 \)
\( \implies \frac{9.2}{5.1} > \frac{3.4}{7.1} \)
In simple words: This question involves comparing different types of ratios, including those with square roots and decimals, by converting them to a common denominator, cross-multiplication, or simplifying square roots to determine which ratio is greater, smaller, or if they are equal.

🎯 Exam Tip: When comparing ratios with square roots, try to rationalize the denominator or simplify the roots first. For decimal comparisons, cross-multiplication is often the most straightforward method.

Question 4. Solve.
ABCD is a parallelogram. The ratio of \( \angle A \) and \( \angle B \) of this parallelogram is 5 : 4. Find the measure of \( \angle B \).
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समांतर चतुर्भुज ABCD को दर्शाता है। इसमें कोण A को 5x डिग्री और कोण B को 4x डिग्री के रूप में चिह्नित किया गया है। चित्र में A और B शीर्षों पर कोणों को दिखाया गया है, जो एक समांतर चतुर्भुज के आसन्न कोणों का प्रतिनिधित्व करते हैं।
Ratio of \( \angle A \) and \( \angle B \) for given parallelogram is 5 : 4
Let the common multiple be x.
\( \implies m\angle A = 5x^\circ \) and \( m\angle B = 4x^\circ \)
Now, \( m\angle A + m\angle B = 180^\circ \) ...[Adjacent angles of a parallelogram are supplementary]
\( \therefore 5x^\circ + 4x^\circ = 180^\circ \)
\( \therefore 9x^\circ = 180^\circ \)
\( \therefore x^\circ = \frac{180^\circ}{9} \)
\( \therefore x^\circ = 20^\circ \)
\( \therefore m\angle B = 4x^\circ = 4 \times 20^\circ = 80^\circ \)
\( \therefore \) The measure of \( \angle B \) is \( 80^\circ \).
In simple words: Given a parallelogram where the ratio of two adjacent angles is 5:4, we use the property that adjacent angles in a parallelogram are supplementary (sum to 180 degrees) to find the individual angle measures.

🎯 Exam Tip: Clearly state the geometric property being used (e.g., adjacent angles of a parallelogram are supplementary) to justify your steps and ensure full marks.

Question 4. ii. The ratio of present ages of Albert and Salim is 5 : 9. Five years hence ratio of their ages will be 3 : 5. Find their present ages.
Answer:
The ratio of present ages of Albert and Salim is 5 : 9
Let the common multiple be x.
\( \therefore \) Present age of Albert = 5x years and
Present age of Salim = 9x years
After 5 years,
Albert's age = (5x + 5) years and
Salim's age = (9x + 5) years
According to the given condition,
Five years hence ratio of their ages will be 3 : 5
\( \therefore \frac{5x+5}{9x+5} = \frac{3}{5} \)
\( \therefore 5(5x + 5) = 3(9x + 5) \)
\( \therefore 25x + 25 = 27x + 15 \)
\( \therefore 25 - 15 = 27x - 25x \)
\( \therefore 10 = 2x \)
\( \therefore x = \frac{10}{2} \)
\( \therefore x = 5 \)
\( \therefore \) Present age of Albert = \( 5x = 5 \times 5 = 25 \) years
Present age of Salim = \( 9x = 9 \times 5 = 45 \) years
\( \therefore \) The present ages of Albert and Salim are 25 years and 45 years respectively.
In simple words: This problem involves setting up and solving a linear equation based on age ratios at two different points in time to determine the current ages of two individuals.

🎯 Exam Tip: Define variables clearly for the common multiple and ages at both time points. Pay close attention to the "after X years" part to correctly set up the equations.

Question 4. iii. The ratio of length and breadth of a rectangle is 3 : 1, and its perimeter is 36 cm. Find the length and breadth of the rectangle.
Answer:
The ratio of length and breadth of a rectangle is 3 : 1
Let the common multiple be x.
Length of the rectangle (l) = 3x cm
and Breadth of the rectangle (b) = x cm
Given, perimeter of the rectangle = 36 cm
Since, Perimeter of the rectangle = \( 2(l + b) \)
\( \therefore 36 = 2(3x + x) \)
\( \therefore 36 = 2(4x) \)
\( \therefore 36 = 8x \)
\( \therefore x = \frac{36}{8} = \frac{9}{2} = 4.5 \)
Length of the rectangle = \( 3x = 3 \times 4.5 = 13.5 \) cm
\( \therefore \) The length of the rectangle is 13.5 cm and its breadth is 4.5 cm.
In simple words: We use the given ratio of length to breadth and the perimeter of a rectangle to set up an equation, solve for the common multiple, and then find the actual dimensions of the rectangle.

🎯 Exam Tip: Remember the perimeter formula for a rectangle: P = 2(L+B). Correctly substituting the ratio-based expressions for length and breadth is crucial.

Question 4. iv. The ratio of two numbers is 31: 23 and their sum is 216. Find these numbers.
Answer:
The ratio of two numbers is 31: 23
Let the common multiple be x.
\( \therefore \) First number = 31x and
Second number = 23x
According to the given condition,
Sum of the numbers is 216
\( \therefore 31x + 23x = 216 \)
\( \therefore 54x = 216 \)
\( \therefore x = \frac{216}{54} \)
\( \therefore x = 4 \)
\( \therefore \) First number = \( 31x = 31 \times 4 = 124 \)
Second number = \( 23x = 23 \times 4 = 92 \)
\( \therefore \) The two numbers are 124 and 92.
In simple words: This problem involves finding two numbers whose ratio and sum are given by representing them with a common multiple, forming a linear equation, and solving for the numbers.

🎯 Exam Tip: Ensure that after finding the common multiple (x), you substitute it back into the expressions for the numbers (31x and 23x) to give the final answer, not just x.

Question 4. v. If the product of two numbers is 360 and their ratio is 10 : 9, then find the numbers.
Answer:
Ratio of two numbers is 10:9
Let the common multiple be x.
\( \therefore \) First number = 10x and
Second number = 9x
According to the given condition,
Product of two numbers is 360
\( \therefore (10x)(9x) = 360 \)
\( \therefore 90x^2 = 360 \)
\( \therefore x^2 = \frac{360}{90} \)
\( \therefore x^2 = 4 \)
\( \therefore x = 2 \) .... [Taking positive square root on both sides]
\( \therefore \) First number = \( 10x = 10 \times 2 = 20 \)
Second number = \( 9x = 9 \times 2 = 18 \)
\( \therefore \) The two numbers are 20 and 18.
In simple words: This question involves finding two numbers whose ratio and product are given, by using a common multiple to form a quadratic equation and solving for the numbers.

🎯 Exam Tip: When dealing with products, the common multiple will be squared (e.g., \( (10x)(9x) = 90x^2 \)). Remember to take only the positive square root for 'x' in this context, as numbers usually refer to positive values unless specified.

Question 5. If a : b = 3 : 1 and b : c = 5 : 1, then find the value of
(i) \( \frac{a^3}{15b^2c} \)
(ii) \( \frac{a^2}{7bc} \)
Answer:
Given, a : b = 3 : 1
\( \therefore \frac{a}{b} = \frac{3}{1} \)
\( \therefore a = 3b \) ....(i)
and b : c = 5 : 1
\( \therefore \frac{b}{c} = \frac{5}{1} \)
\( \therefore b = 5c \) .....(ii)
Substituting (ii) in (i),
we get \( a = 3(5c) \)
\( \therefore a = 15c \) ...(iii)
(i) \( \frac{a^3}{15b^2c} \)
Substituting (ii) and (iii):
\( \frac{a^3}{15b^2c} = \frac{(15c)^3}{15(5c)^2c} \)
\( = \frac{15c \times 15c \times 15c}{15 \times 5c \times 5c \times c} \)
\( = \frac{15 \times 15 \times 15 \times c^3}{15 \times 25 \times c^3} \)
\( = \frac{15 \times 15}{25} \)
\( = \frac{3 \times 5 \times 3 \times 5}{5 \times 5} \)
\( = 3 \times 3 = 9 \)
(Alternatively: \( = (3 \times 3)^3 / (15 \times 25) \) which is wrong.)
\( = (3 \times 3) = 9 \) -- Recheck calculation in OCR. The OCR has \( (3 \times 3)^3 = (9)^3 = 729 \). Let's re-calculate. \( \frac{15c \times 15c \times 15c}{15 \times 5c \times 5c \times c} = \frac{15^3 \times c^3}{15 \times 5^2 \times c^3} = \frac{15^3}{15 \times 25} = \frac{15 \times 15 \times 15}{15 \times 25} = \frac{15 \times 15}{25} = \frac{225}{25} = 9 \) The OCR output `(3 x 3)³ = (9)³ = 729` is an error in calculation or presentation. The intermediate simplification `(3 x 3)` seems to be \( \frac{15}{5} \times \frac{15}{5} \). Let's stick to the correct calculation. \( \frac{a^3}{15b^2c} = \frac{(15c)^3}{15(5c)^2c} = \frac{15^3 c^3}{15 \times 5^2 c^2 \times c} = \frac{15^3 c^3}{15 \times 25 c^3} = \frac{15 \times 15 \times 15}{15 \times 25} = \frac{15 \times 15}{25} = \frac{225}{25} = 9 \) So the correct answer is 9. The OCR content showing `(3 x 3)³ = (9)³ = 729` is incorrect. I will write the final value as 9. (ii) \( \frac{a^2}{7bc} \)
Substituting (ii) and (iii):
\( \frac{a^2}{7bc} = \frac{(15c)^2}{7(5c)c} \)
\( = \frac{15c \times 15c}{7 \times 5c \times c} \)
\( = \frac{15 \times 15 \times c^2}{7 \times 5 \times c^2} \)
\( = \frac{15 \times 15}{7 \times 5} \)
\( = \frac{3 \times 5 \times 15}{7 \times 5} \)
\( = \frac{3 \times 15}{7} = \frac{45}{7} \)
In simple words: This problem involves finding the value of algebraic expressions by first expressing all variables in terms of a single common variable using the given ratios.

🎯 Exam Tip: To avoid errors, express all variables (a, b) in terms of the lowest common variable (c in this case) before substituting into the complex expression. Simplify carefully, cancelling common factors.

Question 6. If \( \sqrt{0.04 \times 0.4 \times a} = 0.4 \times 0.04 \times \sqrt{b} \), then find the ratio \( \frac{a}{b} \).
Answer:
Given, \( \sqrt{0.04 \times 0.4 \times a} = 0.4 \times 0.04 \times \sqrt{b} \)
Squaring both sides:
\( (\sqrt{0.04 \times 0.4 \times a})^2 = (0.4 \times 0.04 \times \sqrt{b})^2 \)
\( 0.04 \times 0.4 \times a = (0.4)^2 \times (0.04)^2 \times b \)
\( \implies \frac{a}{b} = \frac{(0.4)^2 \times (0.04)^2}{0.04 \times 0.4} \)
\( \implies \frac{a}{b} = \frac{(0.4 \times 0.4) \times (0.04 \times 0.04)}{0.04 \times 0.4} \)
\( \implies \frac{a}{b} = 0.4 \times 0.04 \)
\( \implies \frac{a}{b} = 0.016 \)
To express as a fraction:
\( \frac{a}{b} = \frac{16}{1000} \)
\( \implies \frac{a}{b} = \frac{8 \times 2}{8 \times 125} \)
\( \implies \frac{a}{b} = \frac{2}{125} \)
In simple words: This problem asks for the ratio of 'a' to 'b' given an equation involving square roots and decimal numbers. We solve it by squaring both sides of the equation and then rearranging the terms to isolate the ratio \( \frac{a}{b} \).

🎯 Exam Tip: When dealing with equations involving square roots, squaring both sides is often the first step. Remember to apply the square to all terms on both sides of the equation. Simplify decimals into fractions for easier manipulation if possible.

Question 7. \( (x + 3) : (x + 11) = (x - 2) : (x + 1) \), then find the value of x.
Answer:
Given, \( (x + 3) : (x + 11) = (x - 2) : (x + 1) \)
\( \implies \frac{x+3}{x+11} = \frac{x-2}{x+1} \)
Cross-multiplying:
\( \therefore (x + 3)(x + 1) = (x - 2)(x + 11) \)
Expand both sides:
\( \therefore x(x + 1) + 3(x + 1) = x(x + 11) - 2(x + 11) \)
\( \therefore x^2 + x + 3x + 3 = x^2 + 11x - 2x - 22 \)
Combine like terms:
\( \therefore x^2 + 4x + 3 = x^2 + 9x - 22 \)
Subtract \( x^2 \) from both sides:
\( \therefore 4x + 3 = 9x - 22 \)
Rearrange terms to solve for x:
\( \therefore 3 + 22 = 9x - 4x \)
\( \therefore 25 = 5x \)
\( \therefore x = \frac{25}{5} \)
\( \therefore x = 5 \)
In simple words: This problem requires solving a proportion involving algebraic expressions by cross-multiplication, expanding the terms, and then solving the resulting linear equation for x.

🎯 Exam Tip: Be careful with algebraic expansion, especially when dealing with negative signs. Group like terms correctly and perform operations systematically to avoid calculation errors.