CBSE Class 10 Mathematics Ratio and Proportion MCQs Set 03

Download CBSE MCQs for Class 10 Mathematics: Ratio and Proportion

Review structured MCQ sets for Class 10 Mathematics Ratio and Proportion. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Chapter-wise Objective Questions: Ratio and Proportion

Access the complete set of multiple-choice questions for Ratio and Proportion below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official CBSE textbooks.

Multiple Choice Questions

Question 1: If \(a:b=3:4\), the value of \((8a-5b):(2b-4a)\) is
(a) \(1\)
(b) \(-\dfrac15\)
(c) \(-1\)
(d) \(\dfrac15\)
Show Answer & Explanation

Answer: (c) \(-1\)

Explanation:
1. Put \(a=3k,\ b=4k\).
2. \(8a-5b=24k-20k=4k\) and \(2b-4a=8k-12k=-4k\).
3. The ratio is \(4k:(-4k)=-1\).

Teacher's Note:
1. Replace the ratio by actual multiples \(3k\) and \(4k\).
2. The value is negative because the second term is negative.

Question 2: If \(0(a) \((1-x):1\), \((1+x):(1+2x)\), \(1:(1+x)\)
(b) \((1+x):(1+2x)\), \(1:(1+x)\), \((1-x):1\)
(c) \(1:(1+x)\), \((1+x):(1+2x)\), \((1-x):1\)
(d) \((1-x):1\), \(1:(1+x)\), \((1+x):(1+2x)\)
Show Answer & Explanation

Answer: (b) \((1+x):(1+2x)\), \(1:(1+x)\), \((1-x):1\)

Explanation:
1. Write each as a fraction: \(\dfrac{1}{1+x}\), \(\dfrac{1+x}{1+2x}\), \(1-x\).
2. Test \(x=\dfrac12\): \(0.667,\ 0.75,\ 0.5\).
3. Order: \(0.75>0.667>0.5\). Algebraically \(\dfrac{1+x}{1+2x}-\dfrac1{1+x}=\dfrac{x^2}{(1+x)(1+2x)}>0\) and \(\dfrac1{1+x}-(1-x)=\dfrac{x^2}{1+x}>0\).

Teacher's Note:
1. A test value gives the order quickly, and the algebra proves it for all \(x\).
2. All three ratios are less than 1.

Question 3: If \((3x-2y):(x-5y)=1:3\), then \(x:y\) is
(a) \(1:8\)
(b) \(1:3\)
(c) \(8:1\)
(d) \(2:5\)
Show Answer & Explanation

Answer: (a) \(1:8\)

Explanation:
1. \(3(3x-2y)=x-5y\), so \(9x-6y=x-5y\).
2. \(8x=y\), so \(x:y=1:8\).

Teacher's Note:
1. Cross-multiply a ratio equation.
2. Check: \(x=1,y=8\): \((3-16):(1-40)=-13:-39=1:3\).

Question 4: If \((3x+y+6):(2x-7y+11)=6:11\), then \(x:y\) is
(a) \(-53:21\)
(b) \(53:21\)
(c) \(-21:53\)
(d) \(5:3\)
Show Answer & Explanation

Answer: (a) \(-53:21\)

Explanation:
1. \(11(3x+y+6)=6(2x-7y+11)\).
2. \(33x+11y+66=12x-42y+66\).
3. \(21x=-53y\), so \(x:y=-53:21\).

Teacher's Note:
1. The constants \(66\) cancel on both sides.
2. A ratio can be negative when one variable is negative.

Question 5: If \(\dfrac{a+5b}{3a+b}=\dfrac43\), the value of \(\dfrac{a^2-ab+b^2}{a^2+ab+b^2}\) is
(a) \(\dfrac{13}{37}\)
(b) \(\dfrac{91}{281}\)
(c) \(\dfrac{101}{303}\)
(d) \(\dfrac{103}{301}\)
Show Answer & Explanation

Answer: (d) \(\dfrac{103}{301}\)

Explanation:
1. \(3a+15b=12a+4b\), so \(9a=11b\) and \(a:b=11:9\).
2. Put \(a=11,b=9\): numerator \(121-99+81=103\).
3. Denominator \(121+99+81=301\), so the value is \(\dfrac{103}{301}\).

Teacher's Note:
1. Find the ratio \(a:b\) first, then substitute actual numbers.
2. The expression is homogeneous, so only the ratio matters.

Question 6: The relation among \(a,b,c\) so that the ratio \((a-c):(b-c)\) is the duplicate of \(a:b\) is
(a) \(c(a+b)=ab\)
(b) \(c=a+b\)
(c) \(c(a-b)=ab\)
(d) \(2c=a+b\)
Show Answer & Explanation

Answer: (a) \(c(a+b)=ab\)

Explanation:
1. Duplicate ratio of \(a:b\) is \(a^2:b^2\), so \(\dfrac{a-c}{b-c}=\dfrac{a^2}{b^2}\).
2. \(b^2(a-c)=a^2(b-c)\), so \(c(a^2-b^2)=ab(a-b)\).
3. Divide by \(a-b\) (\(a\neq b\)): \(c(a+b)=ab\).

Teacher's Note:
1. Duplicate ratio means the squares of the terms.
2. Sub-duplicate would mean the square roots.

Question 7: If \(3x^2-10xy+3y^2=0\), the ratio \(x:y\) is
(a) \(3:1\) only
(b) \(-3:1\) or \(-1:3\)
(c) \(3:1\) or \(1:3\)
(d) \(1:3\) only
Show Answer & Explanation

Answer: (c) \(3:1\) or \(1:3\)

Explanation:
1. Divide by \(y^2\): \(3\left(\dfrac xy\right)^2-10\dfrac xy+3=0\).
2. \(\left(3\dfrac xy-1\right)\left(\dfrac xy-3\right)=0\).
3. So \(\dfrac xy=3\) or \(\dfrac13\).

Teacher's Note:
1. A homogeneous equation gives a quadratic in the ratio.
2. Both ratios are valid.

Question 8: If \(\dfrac{a+b}{a-b}=\dfrac54\), the value of \(\dfrac{a^2+b^2}{a^2-b^2}\) is
(a) \(\dfrac{25}{16}\)
(b) \(\dfrac{41}{40}\)
(c) \(\dfrac{40}{41}\)
(d) \(\dfrac{9}{4}\)
Show Answer & Explanation

Answer: (b) \(\dfrac{41}{40}\)

Explanation:
1. \(4(a+b)=5(a-b)\), so \(a=9b\).
2. \(\dfrac{a^2+b^2}{a^2-b^2}=\dfrac{81+1}{81-1}=\dfrac{82}{80}=\dfrac{41}{40}\).

Teacher's Note:
1. Componendo and dividendo also gives \(a:b=9:1\).
2. Check: \(\dfrac{10}{8}=\dfrac54\).

Question 9: In an examination the numbers of passes and failures are in the ratio \(3:2\). If the pass line had been drawn 12 places lower, the ratio would have been \(21:10\). The number who passed is
(a) 90
(b) 84
(c) 93
(d) 63
Show Answer & Explanation

Answer: (c) 93

Explanation:
1. Let passes \(=3k\) and failures \(=2k\). Lowering the line by 12 places: \(\dfrac{3k+12}{2k-12}=\dfrac{21}{10}\).
2. \(30k+120=42k-252\), so \(12k=372\) and \(k=31\).
3. Passes \(=3\times31=93\).

Teacher's Note:
1. Lowering the pass line moves 12 students from failure to pass.
2. Check: \(105:50=21:10\).

Question 10: If \(\dfrac{3x^2-33y^2}{x-4y}=x+7y\), the possible values of \(x:y\) are
(a) \(5:2\) or \(-1\)
(b) \(-5:2\) or \(1\)
(c) \(2:5\) or \(1\)
(d) \(5:2\) only
Show Answer & Explanation

Answer: (a) \(5:2\) or \(-1\)

Explanation:
1. \(3x^2-33y^2=(x+7y)(x-4y)=x^2+3xy-28y^2\).
2. \(2x^2-3xy-5y^2=0\), so \((2x-5y)(x+y)=0\).
3. \(x:y=5:2\) or \(-1:1\).

Teacher's Note:
1. Rewrite as a homogeneous quadratic and factorise.
2. Check \(x=5,y=2\): \(\dfrac{75-132}{5-8}=19=5+14\).

Question 11: If \(x:y=5:3\), the value of \(\dfrac{(5x-y)^2}{(3x+2y)(x-3y)}\) is
(a) \(-\dfrac{121}{84}\)
(b) \(-\dfrac{11}{21}\)
(c) \(\dfrac{121}{21}\)
(d) \(-\dfrac{121}{21}\)
Show Answer & Explanation

Answer: (d) \(-\dfrac{121}{21}\)

Explanation:
1. Take \(x=5,y=3\). \(5x-y=22\), so the numerator is \(484\).
2. \(3x+2y=21\) and \(x-3y=-4\), so the denominator is \(-84\).
3. \(\dfrac{484}{-84}=-\dfrac{121}{21}\).

Teacher's Note:
1. Only the ratio matters because the expression is homogeneous of degree 0.
2. The negative sign comes from \(x-3y<0\).

Question 12: Three partners receive 30%, 25% and 20% of the annual profits and the rest is divided between a Reserve Fund and an Employees Pension Fund in the ratio \(2:1\). If the total profit is Rs. 547 860, the Pension Fund gets
(a) Rs. 91 310
(b) Rs. 54 786
(c) Rs. 45 655
(d) Rs. 136 965
Show Answer & Explanation

Answer: (c) Rs. 45 655

Explanation:
1. The partners take \(30+25+20=75\%\), so the funds get \(25\%=\) Rs. 136 965.
2. Divide in the ratio \(2:1\): the Reserve Fund gets Rs. 91 310 and the Pension Fund gets Rs. 45 655.
3. The partners get Rs. 164 358, Rs. 136 965 and Rs. 109 572.

Teacher's Note:
1. Find the remaining percentage first.
2. Check: \(91\,310+45\,655=136\,965\).

Question 13: A town's population increases by 8% in five years. The present population is 84 564. Its population 10 years ago was
(a) 78 300
(b) 72 500
(c) 70 000
(d) 77 600
Show Answer & Explanation

Answer: (b) 72 500

Explanation:
1. Two five-year periods, so present \(=P\times1.08^2\).
2. \(P=\dfrac{84564}{1.1664}=72\,500\).

Teacher's Note:
1. Repeated percentage change is multiplicative, not additive.
2. Check: \(72\,500\times1.08=78\,300\) and \(78\,300\times1.08=84\,564\).

Question 14: Water weighs 1000 g per litre and ether 716 g per litre. Five litres of a mixture weigh 4.3 kg. The quantity of water in it is
(a) \(1\dfrac{33}{71}\) litres
(b) 3 litres
(c) \(2\dfrac{1}{2}\) litres
(d) \(2\dfrac{38}{71}\) litres
Show Answer & Explanation

Answer: (d) \(2\dfrac{38}{71}\) litres

Explanation:
1. Let water be \(w\) litres, ether \(5-w\). \(1000w+716(5-w)=4300\).
2. \(284w=720\), so \(w=\dfrac{180}{71}\).
3. \(w=2\dfrac{38}{71}\) litres.

Teacher's Note:
1. Convert 4.3 kg to 4300 g.
2. The mixture is a little more than half water by volume.

Question 15: \(x\) varies directly as \(y\) and inversely as \(z^2\). If \(x=1\) when \(y=2\) and \(z=3\), the value of \(x\) when \(y=3\) and \(z=2\) is
(a) \(2\dfrac14\)
(b) \(3\dfrac38\)
(c) \(\dfrac92\)
(d) \(\dfrac{8}{27}\)
Show Answer & Explanation

Answer: (b) \(3\dfrac38\)

Explanation:
1. \(x=\dfrac{ky}{z^2}\). From the first case \(1=\dfrac{2k}{9}\), so \(k=\dfrac92\).
2. Then \(x=\dfrac{9}{2}\cdot\dfrac{3}{4}=\dfrac{27}{8}\).
3. This is \(3\dfrac38\).

Teacher's Note:
1. Write the constant of variation first.
2. The ratio method gives \(\dfrac{3/4}{2/9}=\dfrac{27}{8}\).

Question 16: Over a given distance, the speed of a train varies with the time taken
(a) as the square of the time
(b) inversely
(c) it does not vary
(d) directly
Show Answer & Explanation

Answer: (b) inversely

Explanation:
1. Speed \(=\dfrac{\text{distance}}{\text{time}}\).
2. For a fixed distance speed \(\propto\dfrac1{\text{time}}\).

Teacher's Note:
1. Double the speed halves the time.
2. The product speed \(\times\) time is constant.

Question 17: \(x\) is the sum of two quantities, one varying directly as \(y\) and the other inversely as \(y^2\). If \(x=37\) when \(y=1\) and \(x=11\) when \(y=2\), the value of \(x\) when \(y=3\) is
(a) 5
(b) 7
(c) 13
(d) 9
Show Answer & Explanation

Answer: (b) 7

Explanation:
1. \(x=ky+\dfrac m{y^2}\). \(37=k+m\) and \(11=2k+\dfrac m4\).
2. From the first \(m=37-k\): \(11=2k+\dfrac{37-k}{4}\), so \(44=7k+37\), \(k=1\) and \(m=36\).
3. At \(y=3\), \(x=3+\dfrac{36}{9}=7\).

Teacher's Note:
1. Write the general form with constants \(k\) and \(m\) and use both data points.
2. Check: at \(y=2\), \(2+9=11\).

Question 18: The mean proportional between \(\sqrt{12}\) and \(\sqrt{27}\) is
(a) \(18\)
(b) \(\sqrt{39}\)
(c) \(3\sqrt2\)
(d) \(\sqrt{18}\cdot2\)
Show Answer & Explanation

Answer: (c) \(3\sqrt2\)

Explanation:
1. Mean proportional \(m=\sqrt{\sqrt{12}\cdot\sqrt{27}}\).
2. \(\sqrt{12}\cdot\sqrt{27}=\sqrt{324}=18\).
3. \(m=\sqrt{18}=3\sqrt2\).

Teacher's Note:
1. Mean proportional is the geometric mean of the two numbers.
2. The product of the extremes equals \(m^2\).

Question 19: The mean proportional between \((a-b)(a+b)^2\) and \((a+b)(a-b)^2\) is
(a) \((a-b)(a+b)\)
(b) \(a^2-b^2\)
(c) \((a^2-b^2)^{3/2}\)
(d) \((a^2-b^2)^3\)
Show Answer & Explanation

Answer: (c) \((a^2-b^2)^{3/2}\)

Explanation:
1. Product \(=(a-b)^3(a+b)^3=(a^2-b^2)^3\).
2. The mean proportional is the square root of the product.
3. So it is \((a^2-b^2)^{3/2}\).

Teacher's Note:
1. Combine like factors under the root.
2. The result can also be written \((a^2-b^2)\sqrt{a^2-b^2}\).

Question 20: Two numbers have their sum, difference and the sum of their squares in the ratio \(2:1:25\). The numbers are
(a) 12 and 8
(b) 20 and 5
(c) 15 and 5
(d) 10 and 5
Show Answer & Explanation

Answer: (c) 15 and 5

Explanation:
1. Let \(x+y=2k\) and \(x-y=k\), so \(x=\dfrac{3k}2\) and \(y=\dfrac k2\).
2. \(x^2+y^2=\dfrac{10k^2}{4}=25k\), so \(k=10\).
3. \(x=15\) and \(y=5\).

Teacher's Note:
1. Check: sum 20, difference 10, squares \(225+25=250\); \(20:10:250=2:1:25\).
2. Express the numbers through the common multiple \(k\).

Question 21: If \((x+24):6(x-1)\) is the duplicate ratio of \(7:12\), then \(x\) is
(a) 24
(b) 25
(c) 35
(d) 5
Show Answer & Explanation

Answer: (b) 25

Explanation:
1. Duplicate ratio of \(7:12\) is \(49:144\).
2. \(144(x+24)=49\cdot6(x-1)=294(x-1)\).
3. \(3456+144x=294x-294\), so \(150x=3750\) and \(x=25\).

Teacher's Note:
1. Check: \(49:144\) and \((49):(144)\) agree.
2. Duplicate means squared terms.

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Where can I access latest CBSE Class 10 Mathematics Ratio and Proportion MCQs Set 03?

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