CBSE Class 9 Mathematics Ratio and Proportion MCQs Set 10

Multiple Choice Questions (MCQs) for Class 9 Mathematics: Ratio and Proportion

Review structured MCQ sets for Class 9 Mathematics Ratio and Proportion. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Practice Ratio and Proportion MCQs for Class 9 Mathematics

Navigate directly to the 50 objective questions for Ratio and Proportion using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: If \( a:b=3:4 \), then \( (8a-5b):(2b+4a) \) equals
(a) \( 1:5 \)
(b) \( 5:1 \)
(c) \( 4:5 \)
(d) \( 1:4 \)
Show Answer & Explanation

Answer: (a) \( 1:5 \)

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

Teacher's Note:
1. Express both quantities with the same multiplier k.
2. k cancels in the ratio.

Question 2: For \( 0\lt x\lt 1 \), the ratios \( 1:(1+x) \), \( (1+x):(1+2x) \) and \( (1-x):1 \) in ascending order are
(a) \( 1:(1+x)\lt (1-x):1\lt (1+x):(1+2x) \)
(b) \( (1-x):1\lt 1:(1+x)\lt (1+x):(1+2x) \)
(c) \( (1+x):(1+2x)\lt 1:(1+x)\lt (1-x):1 \)
(d) \( (1-x):1\lt (1+x):(1+2x)\lt 1:(1+x) \)
Show Answer & Explanation

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

Explanation:
1. \( 1-x\lt \dfrac1{1+x} \) because \( 1-x^2\lt 1 \).
2. \( \dfrac1{1+x}\lt \dfrac{1+x}{1+2x} \) because \( 1+2x\lt (1+x)^2=1+2x+x^2 \).

Teacher's Note:
1. Compare the ratios as fractions by cross-multiplying.
2. Test with x = 0.5: 0.5, 0.667, 0.75.

Question 3: If \( \dfrac{a+5b}{3a+b}=\dfrac43 \), then \( \dfrac{a^2-ab+b^2}{a^2+ab+b^2} \) equals
(a) \( \dfrac{91}{301} \)
(b) \( \dfrac{103}{99} \)
(c) \( \dfrac{103}{301} \)
(d) \( \dfrac{81}{121} \)
Show Answer & Explanation

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

Explanation:
1. Cross-multiplying: \( 3a+15b=12a+4b \), so \( 9a=11b \), i.e. \( a:b=11:9 \).
2. With \( a=11,\ b=9 \): \( \dfrac{121-99+81}{121+99+81}=\dfrac{103}{301} \).

Teacher's Note:
1. First find the ratio a : b.
2. Substitute convenient numbers for a and b.

Question 4: For the ratio \( (a-c):(b-c) \) to be the duplicate ratio of \( a:b \), the relation between \( a,b,c \) is
(a) \( c^2=ab \)
(b) \( a+b=2c \)
(c) \( c=a+b \)
(d) \( c(a+b)=ab \)
Show Answer & Explanation

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

Explanation:
1. \( \dfrac{a-c}{b-c}=\dfrac{a^2}{b^2} \) gives \( b^2(a-c)=a^2(b-c) \).
2. \( ab(b-a)=c(b^2-a^2) \), so (for \( a\ne b \)) \( ab=c(a+b) \).

Teacher's Note:
1. Cross-multiply and factor out (b - a).
2. Equivalent form: 1/c = 1/a + 1/b.

Question 5: If \( 3x^2-10xy+3y^2=0 \), then \( x:y \) is
(a) \( 3:1\text{ or }1:3 \)
(b) \( 3:1\text{ only} \)
(c) \( 1:3\text{ only} \)
(d) \( 2:3\text{ or }3:2 \)
Show Answer & Explanation

Answer: (a) \( 3:1\text{ or }1:3 \)

Explanation:
1. Divide by \( y^2 \) with \( t=x/y \): \( 3t^2-10t+3=0 \).
2. \( (3t-1)(t-3)=0 \), so \( t=3 \) or \( t=\dfrac13 \).

Teacher's Note:
1. The equation is homogeneous, so it fixes only the ratio.
2. Both roots give valid ratios.

Question 6: In an examination the passes and failures are in the ratio 3:2. If the pass mark had been 12 marks lower, the ratio of passes to failures would have been 21:10. The number of students who passed is
(a) 62
(b) 93
(c) 105
(d) 31
Show Answer & Explanation

Answer: (b) 93

Explanation:
1. Let passes \( =3k \) and failures \( =2k \); lowering the line moves 12 students: \( \dfrac{3k+12}{2k-12}=\dfrac{21}{10} \).
2. \( 30k+120=42k-252 \), so \( k=31 \) and passes \( =3\times31=93 \).

Teacher's Note:
1. The total number of students stays the same.
2. New figures 105 and 50 are in the ratio 21 : 10.

Question 7: Profits of Rs. 547860 are shared: three partners get 30%, 25% and 20%, and the rest is divided between two funds in the ratio 2:1. The amount received by the larger fund is
(a) Rs. 45655
(b) Rs. 136965
(c) Rs. 91310
(d) Rs. 109572
Show Answer & Explanation

Answer: (c) Rs. 91310

Explanation:
1. Partners take \( 30+25+20=75\% \), so the funds get \( 25\% \) of 547860 \( =136965 \).
2. The larger fund gets \( \dfrac23\times136965=91310 \).

Teacher's Note:
1. First find what is left for the funds.
2. The other fund gets Rs. 45655, and the partners get Rs. 164358, Rs. 136965 and Rs. 109572.

Question 8: The population of a town increases by 8% in 5 years. If it is 84564 now, the population 10 years ago was
(a) 73000
(b) 76000
(c) 70000
(d) 72500
Show Answer & Explanation

Answer: (d) 72500

Explanation:
1. 10 years is two 5-year periods, so \( 84564=P\,(1.08)^2=1.1664P \).
2. \( P=\dfrac{84564}{1.1664}=72500 \).

Teacher's Note:
1. Apply the 8% increase twice, once for each 5-year period.
2. 72500 x 1.08 = 78300 and 78300 x 1.08 = 84564.

Question 9: If \( x \) varies directly as \( y \) and inversely as \( z^2 \), and \( x=1 \) when \( y=2,\ z=3 \), then \( x \) when \( y=3,\ z=2 \) is
(a) \( \dfrac{27}8 \)
(b) \( 3 \)
(c) \( \dfrac98 \)
(d) \( \dfrac{8}{27} \)
Show Answer & Explanation

Answer: (a) \( \dfrac{27}8 \)

Explanation:
1. \( x=\dfrac{ky}{z^2} \); \( 1=\dfrac{2k}{9} \) gives \( k=\dfrac92 \).
2. \( x=\dfrac92\cdot\dfrac34=\dfrac{27}8 \).

Teacher's Note:
1. Write the joint variation as x = k y / z^2.
2. Find k from the first set of values.

Question 10: (i) The area of an equilateral triangle and its base; (ii) the speed of a train and the time taken over a fixed distance. The types of variation are
(a) (i) direct variation with the base; (ii) direct variation
(b) (i) direct variation with the square of the base; (ii) inverse variation
(c) (i) inverse variation; (ii) direct variation
(d) (i) direct variation with the square of the base; (ii) direct variation
Show Answer & Explanation

Answer: (b) (i) direct variation with the square of the base; (ii) inverse variation

Explanation:
1. Area \( =\dfrac{\sqrt3}4\,\text{base}^2 \), so area varies directly as the square of the base.
2. Speed \( =\dfrac{\text{distance}}{\text{time}} \), so for a fixed distance speed varies inversely as time.

Teacher's Note:
1. Write each as a formula to see the type.
2. Doubling the base makes the area four times as large.

Question 11: \( x \) 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 \( y=2 \), then \( x \) when \( y=3 \) is
(a) 12
(b) 10
(c) 7
(d) 9
Show Answer & Explanation

Answer: (c) 7

Explanation:
1. \( x=ay+\dfrac b{y^2} \): \( a+b=37 \) and \( 2a+\dfrac b4=11 \).
2. \( a=1,\ b=36 \); at \( y=3 \): \( x=3+\dfrac{36}9=7 \).

Teacher's Note:
1. Set up two equations from the two given pairs.
2. Substitute b = 37 - a in the second equation.

Question 12: The mean proportional between \( \sqrt{12} \) and \( \sqrt{27} \) is
(a) \( \sqrt{18}\cdot2 \)
(b) \( 6 \)
(c) \( \sqrt{39} \)
(d) \( 3\sqrt2 \)
Show Answer & Explanation

Answer: (d) \( 3\sqrt2 \)

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

Teacher's Note:
1. The mean proportional of p and q is the square root of pq.
2. sqrt12 x sqrt27 = sqrt324 = 18.

Question 13: If \( a \) is the third and \( b \) the fifth term of a series in continued proportion, the first and fourth terms are
(a) \( \dfrac{a^2}b\text{ and }\sqrt{ab} \)
(b) \( \dfrac{b^2}a\text{ and }\sqrt{ab} \)
(c) \( \dfrac{a^2}b\text{ and }\dfrac ab \)
(d) \( \dfrac ab\text{ and }\sqrt{ab} \)
Show Answer & Explanation

Answer: (a) \( \dfrac{a^2}b\text{ and }\sqrt{ab} \)

Explanation:
1. Terms are \( t,\ tr,\ tr^2,\ tr^3,\ tr^4 \) with \( tr^2=a \), \( tr^4=b \), so \( r^2=\dfrac ba \).
2. First term \( t=\dfrac a{r^2}=\dfrac{a^2}b \); fourth \( tr^3=a\,r=a\sqrt{b/a}=\sqrt{ab} \).

Teacher's Note:
1. A continued proportion is a geometric progression.
2. The ratio of consecutive terms is constant.

Question 14: Two numbers are such that their sum, their difference and the sum of their squares are in the ratio 2:1:25. The numbers are
(a) 20 and 10
(b) 15 and 5
(c) 25 and 5
(d) 10 and 5
Show Answer & Explanation

Answer: (b) 15 and 5

Explanation:
1. Let sum \( =2k \), difference \( =k \): numbers \( \dfrac{3k}2 \) and \( \dfrac k2 \), sum of squares \( =\dfrac{5k^2}2=25k \).
2. \( k=10 \), so the numbers are 15 and 5.

Teacher's Note:
1. Write both numbers in terms of k.
2. Check: 20 : 10 : 250 = 2 : 1 : 25.

Question 15: If \( (x+24):6(x-1) \) is the duplicate ratio of \( 7:12 \), then \( x \) equals
(a) 24
(b) 26
(c) 25
(d) 30
Show Answer & Explanation

Answer: (c) 25

Explanation:
1. Duplicate ratio is \( 49:144 \): \( \dfrac{x+24}{6(x-1)}=\dfrac{49}{144} \).
2. \( 144x+3456=294x-294 \), so \( 150x=3750 \) and \( x=25 \).

Teacher's Note:
1. Square both terms of 7 : 12.
2. Check: 49 : 144 = 49 : (6 x 24).

Question 16: Three numbers are in continued proportion. The middle number is 18 and the sum of the other two is 45. The numbers are
(a) 10, 18, 32
(b) 6, 18, 54
(c) 12, 18, 27
(d) 9, 18, 36 (in some order)
Show Answer & Explanation

Answer: (d) 9, 18, 36 (in some order)

Explanation:
1. The extremes satisfy \( ac=18^2=324 \) and \( a+c=45 \).
2. Roots of \( t^2-45t+324=0 \) are 9 and 36.

Teacher's Note:
1. In continued proportion the middle term squared equals the product of the extremes.
2. 9 x 36 = 324 and 9 + 36 = 45.

Question 17: 18 litres are drawn from a cask full of wine and replaced by water; then 18 litres of the mixture are drawn and replaced by water again. If wine and water are now in the ratio 16:9, the capacity of the cask is
(a) 90 litres
(b) 72 litres
(c) 60 litres
(d) 100 litres
Show Answer & Explanation

Answer: (a) 90 litres

Explanation:
1. Wine left \( =x\left(1-\dfrac{18}x\right)^2 \), so \( \left(1-\dfrac{18}x\right)^2=\dfrac{16}{25} \).
2. \( 1-\dfrac{18}x=\dfrac45 \), so \( x=90 \).

Teacher's Note:
1. The wine fraction is 16/25 of the whole.
2. The root x = 10 is rejected because 18 litres cannot be drawn from 10.

Question 18: The volume of a cone varies as the square of its base radius when the height is constant, and as the height when the radius is constant. A cone with radius 7 m and height 21 m has volume 1078 cubic m. The height of a cone of volume 275 cubic m with base radius 5 m is
(a) 10 m
(b) 10.5 m
(c) 12 m
(d) 11 m
Show Answer & Explanation

Answer: (b) 10.5 m

Explanation:
1. \( V=kr^2h \): \( 1078=k\cdot49\cdot21 \), so \( k=\dfrac{22}{21} \).
2. \( 275=\dfrac{22}{21}\cdot25\cdot h \) gives \( h=10.5 \).

Teacher's Note:
1. Joint variation: V = k r^2 h.
2. k = pi/3 with pi = 22/7.

Question 19: A body falling from rest covers a distance proportional to the square of the time. If it falls 490 m in 10 s, the distance covered in the 5th second alone is
(a) 122.5 m
(b) 78.4 m
(c) 44.1 m
(d) 49 m
Show Answer & Explanation

Answer: (c) 44.1 m

Explanation:
1. \( s=kt^2 \) with \( k=\dfrac{490}{100}=4.9 \).
2. \( s(5)-s(4)=4.9(25-16)=44.1 \) m.

Teacher's Note:
1. The 5th second is the interval from t = 4 to t = 5.
2. In 5 seconds the body falls 122.5 m.

Question 20: P varies directly as Q and inversely as R. P = 4/9 when Q = 3/7 and R = 9/14. The value of Q when \( P=\sqrt{96} \) and \( R=\sqrt{150} \) is
(a) 120
(b) 90
(c) 360
(d) 180
Show Answer & Explanation

Answer: (d) 180

Explanation:
1. \( P=\dfrac{kQ}R \): \( \dfrac49=k\cdot\dfrac37\cdot\dfrac{14}9=\dfrac{2k}3 \), so \( k=\dfrac23 \).
2. \( Q=\dfrac{PR}k=\dfrac32\sqrt{96\cdot150}=\dfrac32\cdot120=180 \).

Teacher's Note:
1. Find the constant of variation first.
2. sqrt(96 x 150) = sqrt(14400) = 120.

Question 21: y is the sum of two quantities, one varying directly as x and the other inversely as x. If \( y=6 \) when \( x=4 \) and \( y=\dfrac{10}3 \) when \( x=3 \), the relation between x and y is
(a) \( y=2x-\dfrac8x \)
(b) \( y=2x+\dfrac8x \)
(c) \( y=3x-\dfrac{6}x \)
(d) \( y=x+\dfrac{8}{x} \)
Show Answer & Explanation

Answer: (a) \( y=2x-\dfrac8x \)

Explanation:
1. \( y=ax+\dfrac bx \): \( 16a+b=24 \) and \( 9a+b=10 \).
2. Subtracting: \( 7a=14 \), \( a=2 \), \( b=-8 \).

Teacher's Note:
1. Multiply each equation by x to clear the fraction.
2. Check at x = 4: 8 - 2 = 6.

Question 22: y is the sum of three quantities varying as \( x \), \( x^2 \) and \( x^3 \). If \( y=4,8,18 \) for \( x=1,2,3 \), then \( y \) equals
(a) \( x+3x^2+x^3 \)
(b) \( 6x-3x^2+x^3 \)
(c) \( 6x+3x^2-x^3 \)
(d) \( 4x-3x^2+3x^3 \)
Show Answer & Explanation

Answer: (b) \( 6x-3x^2+x^3 \)

Explanation:
1. \( y=ax+bx^2+cx^3 \): \( a+b+c=4 \), \( 2a+4b+8c=8 \), \( 3a+9b+27c=18 \).
2. Solving: \( a=6,\ b=-3,\ c=1 \).

Teacher's Note:
1. Solve the three linear equations.
2. Check at x = 2: 12 - 12 + 8 = 8.

Question 23: The area of a circle varies as the square of its radius. If the area of a circle of radius 10 cm is 314.159 sq cm, the area of a circle of radius 12 cm, in sq cm, is
(a) 376.991
(b) 628.318
(c) 452.389
(d) 435.4
Show Answer & Explanation

Answer: (c) 452.389

Explanation:
1. \( A=kr^2 \) with \( k=\dfrac{314.159}{100}=3.14159 \).
2. \( A=3.14159\times144=452.389 \).

Teacher's Note:
1. The area scales with the square of the radius ratio (12/10)^2 = 1.44.
2. 314.159 x 1.44 = 452.389.

Question 24: Five percent of the inhabitants of a town died of fever and 15 percent of the remaining left the town. If the population then fell to 3553, the original population was
(a) 4334
(b) 4000
(c) 4200
(d) 4400
Show Answer & Explanation

Answer: (d) 4400

Explanation:
1. After the deaths \( 0.95P \) remain; after the departures \( 0.95\times0.85P=0.8075P \).
2. \( 0.8075P=3553 \), so \( P=4400 \).

Teacher's Note:
1. Apply the two percentage changes one after the other.
2. 4400 x 0.8075 = 3553.

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Ratio and Proportion Objective Questions & Solutions for Class 9 Mathematics

Class 9 Mathematics Ratio and Proportion Objective Test Questions

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FAQs

Where can I access latest CBSE Class 9 Mathematics Ratio and Proportion MCQs Set 10?

You can get most exhaustive CBSE Class 9 Mathematics Ratio and Proportion MCQs Set 10 for free on StudiesToday.com. These MCQs for Class 9 Mathematics are updated for the 2026-27 academic session as per CBSE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Mathematics Class 9 material?

Yes, our CBSE Class 9 Mathematics Ratio and Proportion MCQs Set 10 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the CBSE paper is now competency-based.

How do practicing Mathematics MCQs help in scoring full marks in Class 9 exams?

By solving our CBSE Class 9 Mathematics Ratio and Proportion MCQs Set 10, Class 9 students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Mathematics.

Do you provide answers and explanations for CBSE Class 9 Mathematics Ratio and Proportion MCQs Set 10?

Yes, Mathematics MCQs for Class 9 have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused CBSE exams.

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