OP Malhotra Class 10 Maths Solutions Chapter 6 Ratio and Proportion Exercise 6 (A)

S Chand Class 10 ICSE Maths Solutions Chapter 6 Ratio And Proportion Ex 6(A)

Question 1. Find the value of x in each case :
(i) 8 : 14 :: x : 28
(ii) x : 9 :: 5 : 3
(iii) 12 : x :: 4 : 15
Answer:
(i) For a proportion a : b :: c : d, the product of the extremes equals the product of the means, so `ad = bc`.
8 : 14 :: x : 28
\( \implies \) \( 8 \times 28 = 14 \times x \)
\( \implies \) \( x = \frac{8 \times 28}{14} \)
\( \implies \) \( x = 8 \times 2 \)
\( \implies \) \( x = 16 \)
(ii) x : 9 :: 5 : 3
\( \implies \) \( x \times 3 = 9 \times 5 \)
\( \implies \) \( x = \frac{9 \times 5}{3} \)
\( \implies \) \( x = 3 \times 5 \)
\( \implies \) \( x = 15 \)
(iii) 12 : x :: 4 : 15
\( \implies \) \( 12 \times 15 = x \times 4 \)
\( \implies \) \( x = \frac{12 \times 15}{4} \)
\( \implies \) \( x = 3 \times 15 \)
\( \implies \) \( x = 45 \)
In simple words: In any set of four numbers that are in proportion, if you multiply the first and last numbers, you will get the same answer as when you multiply the two middle numbers. This rule helps us find a missing number.

🎯 Exam Tip: Remember the "product of extremes equals product of means" rule (\( ad = bc \)) for proportions. This is key for solving these types of problems quickly.

Question 2. Find the fourth proportional to :
(i) 25, 15, 40
(ii) \( 3a^2 b^2, a^3, b^3 \)
(iii) \( a^2 – 5a + 6, a^2 + a – 6, a^2 – 9 \)
Answer:
We know that if \( a : b :: c : d \), then the fourth proportional \( d = \frac{b \times c}{a} \). This is a useful formula to remember.
(i) 25, 15, 40
Here, \( a = 25, b = 15, c = 40 \).
Fourth proportional \( = \frac{15 \times 40}{25} = \frac{600}{25} = 24 \)
(ii) \( 3a^2 b^2, a^3, b^3 \)
Here, \( a = 3a^2 b^2, b = a^3, c = b^3 \).
Fourth proportional \( = \frac{a^3 \times b^3}{3a^2 b^2} = \frac{a^{3-2} b^{3-2}}{3} = \frac{ab}{3} \)
(iii) \( a^2 – 5a + 6, a^2 + a – 6, a^2 – 9 \)
First, we factorize the given expressions:
\( a^2 – 5a + 6 = a^2 – 3a – 2a + 6 = a(a – 3) – 2(a – 3) = (a – 3)(a – 2) \)
\( a^2 + a – 6 = a^2 + 3a – 2a – 6 = a(a + 3) – 2(a + 3) = (a + 3)(a – 2) \)
\( a^2 – 9 = (a + 3)(a – 3) \) (using \( x^2 - y^2 = (x-y)(x+y) \))
Now, let \( A = (a – 3)(a – 2), B = (a + 3)(a – 2), C = (a + 3)(a – 3) \).
Fourth proportional \( = \frac{B \times C}{A} \)
\( = \frac{(a + 3)(a – 2) \times (a + 3)(a – 3)}{(a – 3)(a – 2)} \)
\( = (a + 3)(a + 3) \)
\( = (a + 3)^2 \)
In simple words: The fourth proportional is the last number in a set of four numbers that are in proportion. To find it, you multiply the second and third numbers, and then divide the result by the first number. This works even when the numbers are algebraic expressions; you just need to simplify them first.

🎯 Exam Tip: When dealing with algebraic expressions, always factorize them completely before applying the proportional formulas. This simplifies calculations and helps avoid errors.

Question 3. Find the third proportional to :
(i) 16 and 36
(ii) \( \frac { x }{ y } + \frac { y }{ x } \) and \( \frac { x }{ y } \)
(iii) \( a^2 – b^2 \) and \( a + b \)
Answer:
If \( a, b, c \) are in continued proportion, then \( a : b :: b : c \), which means \( b^2 = ac \). So, the third proportional \( c = \frac{b^2}{a} \). This formula defines a continued proportion.
(i) 16 and 36
Here, \( a = 16, b = 36 \).
Third proportional \( = \frac{36^2}{16} = \frac{36 \times 36}{16} = \frac{1296}{16} = 81 \)
(ii) \( \frac { x }{ y } + \frac { y }{ x } \) and \( \frac { x }{ y } \)
Here, \( a = \frac{x}{y} + \frac{y}{x} = \frac{x^2+y^2}{xy} \) and \( b = \frac{x}{y} \).
Third proportional \( = \frac{b^2}{a} \)
\( = \frac{\left(\frac{x}{y}\right)^2}{\frac{x^2+y^2}{xy}} \)
\( = \frac{\frac{x^2}{y^2}}{\frac{x^2+y^2}{xy}} \)
\( = \frac{x^2}{y^2} \times \frac{xy}{x^2+y^2} \)
\( = \frac{x^3 y}{xy^2(x^2+y^2)} \)
\( = \frac{x^3}{y(x^2+y^2)} \)
(iii) \( a^2 – b^2 \) and \( a + b \)
Here, \( A = a^2 – b^2 \) and \( B = a + b \).
Third proportional \( = \frac{B^2}{A} \)
\( = \frac{(a + b)^2}{a^2 – b^2} \)
\( = \frac{(a + b)(a + b)}{(a – b)(a + b)} \)
\( = \frac{a + b}{a – b} \)
In simple words: When three numbers are in continued proportion, the middle number squared is equal to the product of the first and third numbers. To find the third number, you simply square the second number and then divide it by the first number. This concept helps extend a two-number sequence into a proportional series.

🎯 Exam Tip: Pay close attention to simplifying complex fractions and algebraic expressions carefully. A common error is incorrect cancellation or factorization, especially with fractions.

Question 4. Find the mean proportional between the following:
(i) 5 and 80
(ii) \( 360a^4 \) and \( 250a^2 b^2 \)
(iii) \( (x – y) \) and \( (x^3 – x^2y) \)
Answer:
If \( b \) is the mean proportional between \( a \) and \( c \), then \( a : b :: b : c \), which means \( b^2 = ac \). So, \( b = \sqrt{ac} \). The mean proportional is like the "average" in terms of multiplication.
(i) 5 and 80
Mean proportional \( = \sqrt{5 \times 80} = \sqrt{400} = 20 \)
(ii) \( 360a^4 \) and \( 250a^2 b^2 \)
Mean proportional \( = \sqrt{360a^4 \times 250a^2 b^2} \)
\( = \sqrt{360 \times 250 \times a^{4+2} \times b^2} \)
\( = \sqrt{90000 \times a^6 \times b^2} \)
\( = \sqrt{90000} \times \sqrt{a^6} \times \sqrt{b^2} \)
\( = 300 \times a^3 \times b \)
\( = 300a^3b \)
(iii) \( (x – y) \) and \( (x^3 – x^2y) \)
Mean proportional \( = \sqrt{(x – y)(x^3 – x^2y)} \)
First, factorize \( x^3 – x^2y = x^2(x – y) \). This makes the expression simpler.
Mean proportional \( = \sqrt{(x – y) \times x^2(x – y)} \)
\( = \sqrt{x^2 (x – y)^2} \)
\( = x(x – y) \)
In simple words: The mean proportional between two numbers is found by multiplying them together and then taking the square root of that product. This is a special type of average that relates to proportions. When dealing with variables, always simplify expressions inside the square root first.

🎯 Exam Tip: Remember that \( \sqrt{A \times B} = \sqrt{A} \times \sqrt{B} \). This property is very useful when simplifying expressions under a square root, especially in algebraic problems.

Question 5.
(i) If x, 16, 48, y are in continued proportion, find the value of x and y.
(ii) If x, 9, y, 16 are in continued proportion, then find the value of x and y.
Answer:
(i) If x, 16, 48, y are in continued proportion, it means the ratio between consecutive terms is constant.
So, \( \frac{x}{16} = \frac{16}{48} = \frac{48}{y} \). We can split this into two parts to solve.
First, consider the ratio \( \frac{x}{16} = \frac{16}{48} \)
\( \implies \) \( x = \frac{16 \times 16}{48} \)
\( \implies \) \( x = \frac{256}{48} \)
\( \implies \) \( x = \frac{16}{3} \)
Next, consider the ratio \( \frac{16}{48} = \frac{48}{y} \)
\( \implies \) \( 16y = 48 \times 48 \)
\( \implies \) \( y = \frac{48 \times 48}{16} \)
\( \implies \) \( y = 3 \times 48 \)
\( \implies \) \( y = 144 \)
Therefore, \( x = \frac{16}{3} \) and \( y = 144 \).
(ii) If x, 9, y, 16 are in continued proportion, then:
\( \frac{x}{9} = \frac{9}{y} = \frac{y}{16} \)
First, consider \( \frac{9}{y} = \frac{y}{16} \)
\( \implies \) \( y^2 = 9 \times 16 \)
\( \implies \) \( y^2 = 144 \)
\( \implies \) \( y = \sqrt{144} \)
\( \implies \) \( y = 12 \)
Next, consider \( \frac{x}{9} = \frac{9}{y} \). Since we found \( y = 12 \), substitute this value.
\( \implies \) \( \frac{x}{9} = \frac{9}{12} \)
\( \implies \) \( x = \frac{9 \times 9}{12} \)
\( \implies \) \( x = \frac{81}{12} \)
\( \implies \) \( x = \frac{27}{4} \)
Therefore, \( x = \frac{27}{4} \) and \( y = 12 \).
In simple words: When numbers are in continued proportion, it means the ratio between the first and second number is the same as the ratio between the second and third, and so on. To find missing numbers, you can set up these equal ratios and solve them one by one.

🎯 Exam Tip: For continued proportion problems, clearly write out all the ratios as equal fractions. Then, break it down into pairs of equations to solve for each unknown variable separately.

Question 6. What number must be added to 3, 5, 7, 10 each in order to get four numbers in proportion?
Answer:
Let the number to be added be \( x \). This means each original number will increase by the same amount.
So, the new numbers are \( (3+x), (5+x), (7+x), (10+x) \).
For these numbers to be in proportion, the ratio of the first two must be equal to the ratio of the last two:
\( \frac{3+x}{5+x} = \frac{7+x}{10+x} \)
Now, we cross-multiply:
\( \implies \) \( (3+x)(10+x) = (7+x)(5+x) \)
\( \implies \) \( 30 + 3x + 10x + x^2 = 35 + 7x + 5x + x^2 \)
\( \implies \) \( 30 + 13x + x^2 = 35 + 12x + x^2 \)
Subtract \( x^2 \) from both sides:
\( \implies \) \( 30 + 13x = 35 + 12x \)
Rearrange the terms to solve for \( x \):
\( \implies \) \( 13x – 12x = 35 – 30 \)
\( \implies \) \( x = 5 \)
Therefore, 5 must be added to each number.
In simple words: We are looking for a single number that, when added to each of the four given numbers, makes them into a proportional set. This means the first two new numbers will have the same division result as the last two new numbers. We use algebra to find this special number.

🎯 Exam Tip: Carefully expand the binomials during cross-multiplication and combine like terms. This is where most students make calculation errors in such problems.

Question 7. What number must be subtracted from each of the numbers 28, 53, 19, 35 so that they are in proportion?
Answer:
Let the number to be subtracted be \( x \). Each original number will decrease by this amount.
So, the new numbers are \( (28-x), (53-x), (19-x), (35-x) \).
For these numbers to be in proportion, their ratios must be equal:
\( \frac{28-x}{53-x} = \frac{19-x}{35-x} \)
Cross-multiply to solve for \( x \):
\( \implies \) \( (28-x)(35-x) = (19-x)(53-x) \)
\( \implies \) \( 980 – 28x – 35x + x^2 = 1007 – 19x – 53x + x^2 \)
\( \implies \) \( 980 – 63x + x^2 = 1007 – 72x + x^2 \)
Subtract \( x^2 \) from both sides:
\( \implies \) \( 980 – 63x = 1007 – 72x \)
Rearrange the terms:
\( \implies \) \( 72x – 63x = 1007 – 980 \)
\( \implies \) \( 9x = 27 \)
\( \implies \) \( x = \frac{27}{9} \)
\( \implies \) \( x = 3 \)
Therefore, 3 must be subtracted from each number.
In simple words: We need to find a number that, when taken away from each of the four given numbers, makes them a proportional set. This means the ratio of the first two new numbers will match the ratio of the last two new numbers. We use algebra to find this specific number.

🎯 Exam Tip: Be very careful with negative signs when expanding products like \( (-x)(-x) = x^2 \) and when combining terms like \( -28x - 35x = -63x \). Sign errors are common in these calculations.

Question 8.
(i) Find two numbers such that the mean proportional between them is 14 and the third proportional to them is 112.
(ii) Find two numbers such that the mean proportional between them is 18 and the third proportional to them is 144.
Answer:
(i) Let the two numbers be \( a \) and \( b \).
Mean proportional between \( a \) and \( b \) is 14:
\( \sqrt{ab} = 14 \)
\( \implies \) \( ab = 14^2 \)
\( \implies \) \( ab = 196 \) (Equation 1)
Third proportional to \( a \) and \( b \) is 112:
\( \frac{b^2}{a} = 112 \) (Equation 2)
From Equation 1, \( a = \frac{196}{b} \). Substitute this into Equation 2:
\( \implies \) \( \frac{b^2}{\frac{196}{b}} = 112 \)
\( \implies \) \( \frac{b^3}{196} = 112 \)
\( \implies \) \( b^3 = 112 \times 196 \)
\( \implies \) \( b^3 = 21952 \)
To find \( b \), we need the cube root of 21952. We can recognize that \( 20^3 = 8000 \) and \( 30^3 = 27000 \). The number ends in 2, so the cube root must end in 8 (since \( 8^3 = 512 \)). So, \( b = 28 \).
Now substitute \( b = 28 \) back into \( a = \frac{196}{b} \):
\( a = \frac{196}{28} = 7 \)
The two numbers are 7 and 28.
(ii) Let the two numbers be \( a \) and \( b \).
Mean proportional between \( a \) and \( b \) is 18:
\( \sqrt{ab} = 18 \)
\( \implies \) \( ab = 18^2 \)
\( \implies \) \( ab = 324 \) (Equation 3)
Third proportional to \( a \) and \( b \) is 144:
\( \frac{b^2}{a} = 144 \) (Equation 4)
From Equation 3, \( a = \frac{324}{b} \). Substitute this into Equation 4:
\( \implies \) \( \frac{b^2}{\frac{324}{b}} = 144 \)
\( \implies \) \( \frac{b^3}{324} = 144 \)
\( \implies \) \( b^3 = 144 \times 324 \)
\( \implies \) \( b^3 = 46656 \)
To find \( b \), we need the cube root of 46656. We can test numbers: \( 30^3 = 27000 \), \( 40^3 = 64000 \). The number ends in 6, so its cube root must end in 6 (since \( 6^3 = 216 \)). So, \( b = 36 \).
Now substitute \( b = 36 \) back into \( a = \frac{324}{b} \):
\( a = \frac{324}{36} = 9 \)
The two numbers are 9 and 36.
In simple words: This problem asks us to find two mystery numbers using two clues: their mean proportional (the square root of their product) and their third proportional (the square of the second number divided by the first). We set up two equations and solve them together to find the two numbers.

🎯 Exam Tip: These problems involve solving simultaneous equations. Always define your variables clearly and systematically substitute one equation into the other to reduce the number of unknowns.

Question 9. If \( p + r = 2q \) and \( \frac { 1 }{ q } + \frac { 1 }{ s } = \frac { 2 }{ r } \), then prove that \( p : q = r : s \).
Answer:
We are given two conditions:
1. \( p + r = 2q \) (Equation i)
2. \( \frac { 1 }{ q } + \frac { 1 }{ s } = \frac { 2 }{ r } \) (Equation ii)
Our goal is to prove that \( p : q = r : s \), which means \( \frac{p}{q} = \frac{r}{s} \), or \( ps = qr \).
Let's simplify Equation (ii):
\( \frac{1}{q} + \frac{1}{s} = \frac{2}{r} \)
Combine the fractions on the left side:
\( \implies \) \( \frac{s+q}{qs} = \frac{2}{r} \)
Cross-multiply:
\( \implies \) \( r(s+q) = 2qs \)
Distribute \( r \) on the left side:
\( \implies \) \( rs + rq = 2qs \)
From Equation (i), we know that \( 2q = p+r \). Substitute this into the equation:
\( \implies \) \( rs + rq = (p+r)s \)
Distribute \( s \) on the right side:
\( \implies \) \( rs + rq = ps + rs \)
Subtract \( rs \) from both sides:
\( \implies \) \( rq = ps \)
Now, divide both sides by \( qs \) (assuming \( q \neq 0 \) and \( s \neq 0 \), which must be true for the original fractions to exist):
\( \implies \) \( \frac{rq}{qs} = \frac{ps}{qs} \)
\( \implies \) \( \frac{r}{s} = \frac{p}{q} \)
This means \( p : q = r : s \). Hence proved. The problem links arithmetic mean with harmonic progression properties.
In simple words: We are given two relationships between four numbers \( p, q, r, s \). The first one means \( q \) is the average of \( p \) and \( r \). The second one looks at their inverses. By cleverly using the first given fact to simplify the second, we can show that the ratio of \( p \) to \( q \) is the same as the ratio of \( r \) to \( s \).

🎯 Exam Tip: When proving algebraic identities or proportions, always start with the given conditions and systematically manipulate them using valid algebraic steps to reach the desired conclusion. Look for opportunities to substitute one condition into another.

Question 10. If b is the mean proportional between a and c, prove that \( a, c, a^2 + b^2 \) and \( b^2 + c^2 \) are proportional.
Answer:
Given that \( b \) is the mean proportional between \( a \) and \( c \).
This means \( b^2 = ac \) (Equation i). This is the definition of a mean proportional.
We need to prove that \( a, c, a^2 + b^2 \) and \( b^2 + c^2 \) are proportional.
This means we need to show that \( \frac{a}{c} = \frac{a^2 + b^2}{b^2 + c^2} \).
Let's start with the right-hand side (RHS) of the equation we want to prove:
RHS \( = \frac{a^2 + b^2}{b^2 + c^2} \)
Substitute \( b^2 = ac \) from Equation (i) into the RHS:
RHS \( = \frac{a^2 + ac}{ac + c^2} \)
Now, factor out common terms from the numerator and denominator:
Numerator: \( a^2 + ac = a(a + c) \)
Denominator: \( ac + c^2 = c(a + c) \)
So, RHS \( = \frac{a(a + c)}{c(a + c)} \)
Assuming \( a+c \neq 0 \) (otherwise \( a=0, c=0 \) or \( a=-c \), leading to trivial or undefined cases), we can cancel out \( (a+c) \):
RHS \( = \frac{a}{c} \)
This is equal to the left-hand side (LHS) of the proportion we wanted to prove.
Since LHS \( = \frac{a}{c} \) and RHS \( = \frac{a}{c} \), we have \( \frac{a}{c} = \frac{a^2 + b^2}{b^2 + c^2} \).
Thus, \( a, c, a^2 + b^2 \) and \( b^2 + c^2 \) are proportional. Hence proved. The property of mean proportion extends to other derived ratios.
In simple words: We are given that \( b \) is the middle number in a special kind of sequence with \( a \) and \( c \), meaning \( b \) squared is the same as \( a \) times \( c \). We use this fact to show that if we make two new numbers by adding \( b \) squared to \( a \) squared and \( c \) squared, these new numbers will keep the same ratio as \( a \) to \( c \).

🎯 Exam Tip: The most important step here is the substitution of \( b^2 = ac \). Always simplify expressions after substitution by factoring out common terms to reveal the required proportion.

Question 11. If \( x + 7 \) is the mean proportional between \( (x + 3) \) and \( (x + 12) \), find the value of x.
Answer:
If \( P \) is the mean proportional between \( A \) and \( B \), then \( P^2 = A \times B \). This is the definition we use.
Here, \( P = x + 7 \), \( A = x + 3 \), and \( B = x + 12 \).
So, we can write the equation:
\( (x + 7)^2 = (x + 3)(x + 12) \)
Expand both sides of the equation:
Left side: \( (x + 7)^2 = x^2 + 2(x)(7) + 7^2 = x^2 + 14x + 49 \)
Right side: \( (x + 3)(x + 12) = x(x + 12) + 3(x + 12) = x^2 + 12x + 3x + 36 = x^2 + 15x + 36 \)
Now, set the expanded sides equal to each other:
\( \implies \) \( x^2 + 14x + 49 = x^2 + 15x + 36 \)
Subtract \( x^2 \) from both sides:
\( \implies \) \( 14x + 49 = 15x + 36 \)
Now, gather the \( x \) terms on one side and the constant terms on the other:
\( \implies \) \( 49 – 36 = 15x – 14x \)
\( \implies \) \( 13 = x \)
Therefore, the value of \( x \) is 13. This type of problem often leads to a linear equation after initial quadratic terms cancel.
In simple words: We are told that \( (x+7) \) is the "middle" number in a special multiplication way between \( (x+3) \) and \( (x+12) \). This means \( (x+7) \) multiplied by itself is equal to \( (x+3) \) multiplied by \( (x+12) \). By solving this equation, we find the value of \( x \).

🎯 Exam Tip: Be careful when expanding \( (a+b)^2 \) and \( (a+b)(c+d) \). Make sure to apply the distributive property correctly on both sides of the equation to avoid algebraic errors.

Question 12. If \( \frac{a^2+c^2}{ab+cd} = \frac{ab+cd}{b^2+d^2} \), Prove that \( \frac { a }{ b } = \frac { c }{ d } \).
Answer:
Given the equation: \( \frac{a^2+c^2}{ab+cd} = \frac{ab+cd}{b^2+d^2} \)
This equation implies that \( (ab+cd) \) is the mean proportional between \( (a^2+c^2) \) and \( (b^2+d^2) \).
To prove \( \frac{a}{b} = \frac{c}{d} \), let's cross-multiply the given equation:
\( \implies \) \( (a^2+c^2)(b^2+d^2) = (ab+cd)^2 \)
Expand both sides:
Left side (LHS): \( a^2b^2 + a^2d^2 + c^2b^2 + c^2d^2 \)
Right side (RHS): \( (ab)^2 + 2(ab)(cd) + (cd)^2 = a^2b^2 + 2abcd + c^2d^2 \)
Set LHS equal to RHS:
\( \implies \) \( a^2b^2 + a^2d^2 + b^2c^2 + c^2d^2 = a^2b^2 + 2abcd + c^2d^2 \)
Now, subtract \( a^2b^2 \) and \( c^2d^2 \) from both sides:
\( \implies \) \( a^2d^2 + b^2c^2 = 2abcd \)
Rearrange the terms to one side to form a quadratic expression:
\( \implies \) \( a^2d^2 – 2abcd + b^2c^2 = 0 \)
This expression is a perfect square trinomial, in the form \( (x-y)^2 = x^2 - 2xy + y^2 \).
Here, \( x = ad \) and \( y = bc \).
So, \( \implies \) \( (ad – bc)^2 = 0 \)
Take the square root of both sides:
\( \implies \) \( ad – bc = 0 \)
\( \implies \) \( ad = bc \)
Divide both sides by \( bd \) (assuming \( b \neq 0 \) and \( d \neq 0 \), which must be true for the original fractions to be defined):
\( \implies \) \( \frac{ad}{bd} = \frac{bc}{bd} \)
\( \implies \) \( \frac{a}{b} = \frac{c}{d} \)
Hence proved. This demonstrates a key property of proportions involving mean proportionals.
In simple words: We start with a complex equation showing a special relationship between four numbers. By multiplying across and simplifying the equation, we find that a perfect square is formed. When we solve this, it shows us that the product of the first and last numbers is equal to the product of the two middle numbers, which is the definition of a simple proportion.

🎯 Exam Tip: Recognizing the perfect square trinomial \( (ad - bc)^2 = 0 \) is crucial for quickly solving this proof. Always be on the lookout for such patterns in algebraic manipulation.