CBSE Class 9 Mathematics Ratio and Proportion MCQs Set 14

Answer: (a) \( 24\text{ years} \)

Explanation:
1. Let the ages be \( 3x \) and \( 4x \): \( \dfrac{3x+12}{4x+12}=\dfrac56 \).
2. \( 18x+72=20x+60 \), so \( x=6 \) and the elder brother is \( 4\times6=24 \) years old.

Teacher's Note:
1. Add 12 years to both ages before forming the new ratio.
2. Check: 30 : 36 = 5 : 6.

Answer: (c) \( \dfrac{10}7 \)

Explanation:
1. Numbers \( 7x,3x \): \( \dfrac{7x+10}{3x+10}=\dfrac12\cdot\dfrac73=\dfrac76 \).
2. \( 42x+60=21x+70 \) gives \( x=\dfrac{10}{21} \); numbers are \( \dfrac{10}3 \) and \( \dfrac{10}7 \), the smaller being \( \dfrac{10}7 \).

Teacher's Note:
1. Half of the ratio 7/3 is 7/6.
2. Check: (10/3 + 10) : (10/7 + 10) = 40/3 : 80/7 = 7 : 6.

Answer: (d) \( 10 \)

Explanation:
1. \( \dfrac{38-x}{31-x}=\dfrac43 \) gives \( 114-3x=124-4x \).
2. So \( x=10 \).

Teacher's Note:
1. Cross-multiply and collect x on one side.
2. 28 : 21 = 4 : 3.

Answer: (b) \( 7\dfrac12\% \)

Explanation:
1. \( \dfrac{3}{40}\times100=\dfrac{15}2=7\dfrac12 \).
2. So the ratio is \( 7\dfrac12\% \).

Teacher's Note:
1. A ratio a : b as a percentage is (a / b) x 100.
2. 3/40 = 0.075.

Answer: (d) the second, by 2.5 percentage points

Explanation:
1. First mixture: \( \dfrac38\times100=37.5\% \); second mixture: \( \dfrac25\times100=40\% \).
2. The second is stronger by \( 40-37.5=2.5 \) percentage points.

Teacher's Note:
1. Convert both fractions to percentages.
2. 3/8 = 0.375 and 2/5 = 0.4.

Answer: (a) 200

Explanation:
1. Chinese boys are \( 100-(45+20+13)=22\% \) of the total.
2. \( \dfrac{22}{100}x=44 \), so \( x=200 \).

Teacher's Note:
1. Find the percentage left for the Chinese boys first.
2. 22% of 200 = 44.

Answer: (d) \( 1:2\text{ or }1:1 \)

Explanation:
1. Divide by \( y^2 \) and put \( w=\dfrac xy \): \( 2w^2-3w+1=0 \).
2. \( (2w-1)(w-1)=0 \), so \( w=\dfrac12 \) or \( w=1 \).

Teacher's Note:
1. A homogeneous equation fixes only the ratio x : y.
2. Factor directly: (2x - y)(x - y) = 0.

Answer: (a) \( \dfrac47 \)

Explanation:
1. Cross-multiplying: \( 4(3x-2y)=5x-4y \), i.e. \( 12x-8y=5x-4y \).
2. \( 7x=4y \), so \( \dfrac xy=\dfrac47 \).

Teacher's Note:
1. Treat the ratio as a fraction and cross-multiply.
2. Check with x = 4k, y = 7k: (20k - 28k) : (12k - 14k) = -8k : -2k = 4 : 1.

Answer: (b) \( -1:1:1 \)

Explanation:
1. Multiply the first equation by 3 and the second by 2 and add: \( 7x+7z=0 \), so \( \dfrac xz=-1 \).
2. Multiply the first by 2 and subtract from the second: \( 7y-7z=0 \), so \( \dfrac yz=1 \). Hence \( x:y:z=-1:1:1 \).

Teacher's Note:
1. Eliminate one variable at a time to find x : z and y : z.
2. Check: -1 - 2 + 3 = 0 and -2 + 3 - 1 = 0.

Answer: (a) \( \dfrac{bc}a \)

Explanation:
1. \( \dfrac{a+c}{b+x}=\dfrac ab \) gives \( ab+ax=ab+bc \).
2. \( ax=bc \), so \( x=\dfrac{bc}a \).

Teacher's Note:
1. Equal ratios allow cross-multiplication.
2. Check with a = 2, b = 3, c = 4: x = 6 and 6 : 9 = 2 : 3.

Answer: (b) \( 9:16 \)

Explanation:
1. The L.C.M. of 16 and 24 is 48: \( \dfrac9{16}=\dfrac{27}{48} \) and \( \dfrac{13}{24}=\dfrac{26}{48} \).
2. Since \( 27\gt 26 \), \( 9:16 \) is the larger ratio.

Teacher's Note:
1. Reduce both fractions to a common denominator.
2. Decimals: 0.5625 and 0.5417.

Answer: (c) \( 3x\gt y \)

Explanation:
1. Cross-multiplying (all terms positive): \( (x-y)(x+3y)\gt (x+y)(x-2y) \).
2. \( x^2+2xy-3y^2\gt x^2-xy-2y^2 \), so \( 3xy\gt y^2 \), i.e. \( 3x\gt y \).

Teacher's Note:
1. Cross-multiplication preserves the inequality when the denominators are positive.
2. The ratio is smaller when 3x < y.

Answer: (b) \( x=\dfrac23,\ y=1 \)

Explanation:
1. Put \( x=2k,y=3k \): \( 12k+21k=11 \), so \( k=\dfrac13 \).
2. \( x=\dfrac23 \) and \( y=1 \).

Teacher's Note:
1. Use a common multiplier k.
2. Check: 6(2/3) + 7(1) = 4 + 7 = 11.

Answer: (c) \( \dfrac{30}{11} \)

Explanation:
1. Let \( x=3k,y=k,z=2k \): \( 6k-3k+8k=11k=10 \), so \( k=\dfrac{10}{11} \).
2. \( x=3k=\dfrac{30}{11} \).

Teacher's Note:
1. Introduce a common multiplier k.
2. Then y = 10/11 and z = 20/11.

Answer: (d) \( 3:1:5 \)

Explanation:
1. Put each ratio equal to \( k \): \( x+y=2k,\ y+z=3k,\ z+x=4k \); adding gives \( x+y+z=\dfrac{9k}2 \).
2. Then \( z=\dfrac52k,\ x=\dfrac32k,\ y=\dfrac12k \), so \( x:y:z=3:1:5 \).

Teacher's Note:
1. Add the three equations to find x + y + z.
2. Check: (3 + 1) : (1 + 5) : (5 + 3) = 4 : 6 : 8 = 2 : 3 : 4.

Answer: (c) \( \dfrac85 \)

Explanation:
1. By componendo and dividendo, \( \dfrac{\sqrt{2+x}}{\sqrt{2-x}}=\dfrac{2+1}{2-1}=3 \).
2. Squaring: \( 2+x=9(2-x) \), so \( 10x=16 \) and \( x=\dfrac85 \).

Teacher's Note:
1. Apply componendo and dividendo to remove the sum and difference.
2. Check: sqrt(18/5) = 3 sqrt(2/5).

Answer: (d) \( 1 \)

Explanation:
1. By componendo and dividendo: \( \dfrac{(x+3)^3}{(x-3)^3}=\dfrac{16}{-2}=-8 \).
2. Taking cube roots, \( \dfrac{x+3}{x-3}=-2 \), so \( x+3=-2x+6 \) and \( x=1 \).

Teacher's Note:
1. The numerator plus denominator forms the cube (x + 3)^3 and the difference forms (x - 3)^3.
2. Check: (1 + 27) / (9 + 27) = 28/36 = 7/9.

Answer: (a) 32

Explanation:
1. For \( 7:8=28:x \) we need \( 7x=8\times28 \).
2. \( x=\dfrac{8\times28}7=32 \).

Teacher's Note:
1. Fourth proportional d satisfies a : b = c : d.
2. Product of means 8 x 28 = 224 = product of extremes 7 x 32.

Answer: (d) \( 2:3=4:6 \)

Explanation:
1. Means \( b+c=7 \), extremes \( a+d=8 \) and \( bc=ad \). From \( (b+c)^2+(a+d)^2=113 \) and the sum of squares 65, \( 2(bc+ad)=48 \), so \( bc=ad=12 \).
2. Then \( b,c=3,4 \) and \( a,d=2,6 \), giving \( 2:3=4:6 \) (or \( 6:4=3:2 \)).

Teacher's Note:
1. Use ad = bc together with the two sums.
2. Check: 4 + 9 + 16 + 36 = 65.

Answer: (a) 5

Explanation:
1. \( \dfrac{5+x}{9+x}\gt \dfrac7{10} \) gives \( 10(5+x)\gt 7(9+x) \), i.e. \( 3x\gt 13 \).
2. So \( x\gt 4\tfrac13 \) and the least integer is \( 5 \).

Teacher's Note:
1. Both denominators are positive, so cross-multiplication keeps the inequality.
2. For x = 5: 10/14 = 5/7 = 0.714 > 0.7; for x = 4: 9/13 = 0.692.

Answer: (d) \( 133\dfrac23 \)

Explanation:
1. \( y=ax^{3/2}+\dfrac b{x^{1/2}} \): \( a+b=1 \) and \( 8a+\dfrac b2=38 \), i.e. \( 16a+b=76 \).
2. So \( a=5,\ b=-4 \), and at \( x=9 \): \( y=5\cdot27-\dfrac43=133\dfrac23 \).

Teacher's Note:
1. Form two equations from the two given pairs of values.
2. The arithmetic: 135 - 4/3 = 133 2/3.

Answer: (a) 25:32

Explanation:
1. \( v=kt\,d^2 \), so \( \dfrac{v_1}{v_2}=\dfrac{t_1}{t_2}\left(\dfrac45\right)^2=\dfrac12 \).
2. \( \dfrac{t_1}{t_2}=\dfrac12\cdot\dfrac{25}{16}=\dfrac{25}{32} \).

Teacher's Note:
1. Divide the two variation equations.
2. The first coin is thinner than the second, so t1 : t2 is less than 1.

Answer: (b) \( \sqrt2:1 \)

Explanation:
1. \( S=kR^2 \) and \( 2S=kR_1^2 \), so \( \dfrac{R_1^2}{R^2}=2 \).
2. \( R_1:R=\sqrt2:1 \).

Teacher's Note:
1. Doubling an area multiplies lengths by the square root of 2.
2. Radius up by 41% roughly.

Answer: (a) \( \dfrac14\% \)

Explanation:
1. Inside: \( w=kx \); at the surface \( 6400k \) and at depth 16 km \( 6384k \), a loss of \( 16k \).
2. Loss \( =\dfrac{16}{6400}\times100=\dfrac14\% \).

Teacher's Note:
1. Compare weight at distance 6384 km with weight at 6400 km.
2. 16/6400 = 1/400.

Answer: (b) \( \dfrac12\% \)

Explanation:
1. Weight ratio \( =\left(\dfrac{6400}{6416}\right)^2=\left(\dfrac{400}{401}\right)^2 \).
2. Loss \( =\dfrac{401^2-400^2}{401^2}\times100=\dfrac{80100}{160801}\approx0.5\% \).

Teacher's Note:
1. Distance from the centre becomes 6416 km.
2. 401^2 - 400^2 = 801.

Answer: (c) 350 cubic metres

Explanation:
1. \( V=kr^2h \): \( 88=k\cdot4\cdot7 \), so \( k=\dfrac{22}7 \).
2. \( V=\dfrac{22}7\cdot49\cdot\dfrac{25}{11}=350 \).

Teacher's Note:
1. Find k using the first cylinder.
2. Here k equals pi (22/7).

Answer: (b) a ratio smaller than \( 4:3 \), namely \( 5:4 \)

Explanation:
1. \( \dfrac{4+1}{3+1}=\dfrac54=1.25 \), while \( \dfrac43\approx1.33 \).
2. A ratio of greater inequality is diminished when the same positive number is added to both terms.

Teacher's Note:
1. Compare 5/4 with 4/3 by cross-multiplication: 15 against 16.
2. If the first term is smaller the ratio is increased instead.