CBSE Class 10 Mathematics Ratio and Proportion MCQs Set 05

Answer: (c) 9, 18, 36 or 36, 18, 9

Explanation:
1. Let the extremes be \(a\) and \(c\). \(ac=18^2=324\) and \(a+c=45\).
2. They are the roots of \(t^2-45t+324=0\), \((t-9)(t-36)=0\).
3. The numbers are 9, 18, 36 (or in reverse order).

Teacher's Note:
1. In continued proportion the middle term squared equals the product of the extremes.
2. Check: \(9:18=18:36\).

Answer: (a) 90 litres

Explanation:
1. Wine left after both operations \(=V\left(1-\dfrac{18}V\right)^2\), so wine fraction \(=\left(1-\dfrac{18}V\right)^2=\dfrac{16}{25}\).
2. \(1-\dfrac{18}V=\dfrac45\), so \(\dfrac{18}{V}=\dfrac15\).
3. \(V=90\) litres.

Teacher's Note:
1. The wine fraction is multiplied by \(\left(1-\dfrac{18}{V}\right)\) at each replacement.
2. Wine : water \(=16:9\) means wine fraction \(\dfrac{16}{25}\).

Answer: (b) 60

Explanation:
1. \(P=\dfrac{kQ}{R}\). \(k=\dfrac{PR}{Q}=\dfrac{(4/3)(9/14)}{3/7}=2\).
2. \(Q=\dfrac{PR}{k}=\dfrac{\sqrt{96}\sqrt{150}}{2}=\dfrac{\sqrt{14400}}{2}\).
3. \(Q=\dfrac{120}{2}=60\).

Teacher's Note:
1. \(\sqrt{96\times150}=\sqrt{14400}=120\).
2. Find \(k\) first, then substitute.

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

Explanation:
1. \(y=ax+\dfrac bx\). \(6=4a+\dfrac b4\) gives \(16a+b=24\).
2. \(\dfrac{10}3=3a+\dfrac b3\) gives \(9a+b=10\).
3. Subtract: \(7a=14\), \(a=2\), \(b=-8\), so \(y=2x-\dfrac8x\).

Teacher's Note:
1. A constant of variation can be negative.
2. Check: \(x=4\): \(8-2=6\).

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

Explanation:
1. \(y=ax+bx^2+cx^3\). \(a+b+c=4\), \(2a+4b+8c=8\), \(3a+9b+27c=18\).
2. Dividing the last two by 2 and 3: \(a+2b+4c=4\) and \(a+3b+9c=6\), so \(b+3c=0\) and \(b+5c=2\). Hence \(c=1,\ b=-3\) and \(a=4-b-c=6\).
3. \(y=6x-3x^2+x^3\).

Teacher's Note:
1. Check: \(x=3\): \(18-27+27=18\).
2. Three data points fix three constants.

Answer: (c) 452.39 cm\(^2\)

Explanation:
1. Area \(\propto r^2\), so \(\dfrac{A_2}{A_1}=\dfrac{144}{100}\).
2. \(A_2=314.159\times1.44=452.39\) cm\(^2\).

Teacher's Note:
1. This agrees with \(\pi r^2\).
2. Scale factor squared for area.

Answer: (c) Rs. 264

Explanation:
1. \(0.85\times\text{cost}=204\), so cost \(=\) Rs. 240.
2. For a 10% gain sell at \(1.10\times240=\) Rs. 264.

Teacher's Note:
1. Loss and gain percentages are based on the cost price.
2. The difference in selling prices is Rs. 60.

Answer: (b) neither direct nor inverse variation

Explanation:
1. \(CI=P\left[(1+r)^n-1\right]\).
2. This is not a constant multiple of \(n\) and not a constant divided by \(n\).
3. So it is not a direct or inverse variation.

Teacher's Note:
1. Direct variation needs \(CI/n\) to be constant.
2. Simple interest, not compound interest, varies directly with time.

Answer: (d) inversely as the square root of \(y\)

Explanation:
1. \(y=\dfrac k{x^2}\), so \(x^2=\dfrac ky\).
2. \(x=\sqrt k\cdot\dfrac1{\sqrt y}\).
3. So \(x\propto\dfrac{1}{\sqrt y}\).

Teacher's Note:
1. Take the square root on both sides.
2. The inverse square relation is not symmetric.

Answer: (c) \(2:5\)

Explanation:
1. \(\dfrac1{3x-5y}=-\dfrac1{2x+3y}\), so \(2x+3y=-(3x-5y)\).
2. \(5x=2y\), so \(x:y=2:5\).

Teacher's Note:
1. Two reciprocals sum to zero when the denominators are negatives.
2. Check: \(x=2,y=5\): \(\dfrac1{-19}+\dfrac1{19}=0\).

Answer: (b) \(\dfrac{c^2d}{a}\)

Explanation:
1. \(ab:bc=cd:z\), so \(z=\dfrac{bc\cdot cd}{ab}\).
2. \(z=\dfrac{c^2d}{a}\).

Teacher's Note:
1. Fourth proportional \(=\) (second \(\times\) third) \(\div\) first.
2. The simpler answer \(ad\) is wrong.

Answer: (c) 0

Explanation:
1. Let each ratio be \(k\): \(x=k(b+c-a)\), etc.
2. Sum \(=k\left[(b-c)(b+c-a)+(c-a)(c+a-b)+(a-b)(a+b-c)\right]\).
3. Expanding, \((b^2-c^2)-a(b-c)+(c^2-a^2)-b(c-a)+(a^2-b^2)-c(a-b)=0\).

Teacher's Note:
1. Use the constant \(k\) for equal ratios.
2. The terms cancel in pairs.

Answer: (d) \(56:55\)

Explanation:
1. New total \(\propto15\times22=330\), old total \(\propto16\times21=336\).
2. The ratio of old to new is \(336:330=56:55\).
3. So total wages decrease in the ratio \(56:55\).

Teacher's Note:
1. A decrease in the ratio \(a:b\) means new \(=\dfrac ba\times\) old.
2. Multiply the ratios of employees and wages.

Answer: (d) 7

Explanation:
1. \((37-x):(52-x)=(55-x):(79-x)\).
2. \((37-x)(79-x)=(52-x)(55-x)\), so \(2923-116x=2860-107x\).
3. \(63=9x\), so \(x=7\).

Teacher's Note:
1. Check: \(30:45=48:72\), both equal \(2:3\).
2. Use product of extremes \(=\) product of means.

Answer: (d) \(4:5:7\)

Explanation:
1. Let each equal \(k\): \(A=4k,\ B=5k,\ C=7k\).
2. So \(A:B:C=4:5:7\).

Teacher's Note:
1. If a fraction of \(A\) equals a fraction of \(B\), the larger divisor goes with the larger quantity.
2. Check: \(\dfrac{4}{4}=\dfrac55=\dfrac77\).

Answer: (c) 75

Explanation:
1. Let the numbers be \(3k,4k,5k,6k\). Value \(=3k+2k+0.5k+0.3k=5.8k\) rupees.
2. \(5.8k=87\), so \(k=15\).
3. The 10-paise coins number \(5k=75\).

Teacher's Note:
1. Convert each coin to rupees.
2. Check: \(45+30+7.5+4.5=87\).

Answer: (c) \(-\dfrac{71}{13}\)

Explanation:
1. Put \(a=2,b=9\).
2. \(5a-9b=10-81=-71\) and \(2a+b=4+9=13\).
3. The value is \(-\dfrac{71}{13}\).

Teacher's Note:
1. Divide numerator and denominator by \(b\) to use \(a/b\) directly.
2. The result is negative because \(9b>5a\).

Answer: (a) 24 and 28

Explanation:
1. Let the numbers be \(6k\) and \(7k\).
2. \(7k-6k=4\), so \(k=4\).
3. The numbers are 24 and 28.

Teacher's Note:
1. The difference equals one unit of the ratio.
2. Check: \(24:28=6:7\).

Answer: (d) \(\dfrac12\) or \(-1\)

Explanation:
1. Each ratio \(=k\) gives \(a=k(b+c)\), \(b=k(c+a)\), \(c=k(a+b)\).
2. Adding: \(a+b+c=2k(a+b+c)\).
3. If \(a+b+c\neq0\), \(k=\dfrac12\); if \(a+b+c=0\), \(k=\dfrac{a}{-a}=-1\).

Teacher's Note:
1. Both cases must be considered.
2. The equal-ratio property \(k=\dfrac{a+b+c}{2(a+b+c)}\) fails when the sum is zero.

Answer: (a) Rs. 264

Explanation:
1. Let \(S=s\). \(R=1.5s\), \(Q=4.5s\) and \(R+S=2.5s\).
2. \(P+Q=10s\), so \(P=5.5s\).
3. Total \(=P+Q+R+S=12.5s=600\), so \(s=48\) and \(P=264\).

Teacher's Note:
1. Express everything in terms of the smallest share.
2. Check: \(264+216+72+48=600\).

Answer: (d) Rs. 153

Explanation:
1. Expenses \(=a+bn\). \(a+7b=65\) and \(a+11b=97\).
2. Subtract: \(4b=32\), \(b=8\) and \(a=9\).
3. For 18 guests: \(9+8\times18=153\).

Teacher's Note:
1. A fixed part plus a per-guest part is a linear relation.
2. Check: \(9+56=65\).

Answer: (d) Rs. 8000

Explanation:
1. After insurance he has \(0.95I\); after tax \(0.95\times0.95I=0.9025I\).
2. \(0.9025I=0.9I+20\), so \(0.0025I=20\).
3. \(I=8000\).

Teacher's Note:
1. Check: \(0.9025\times8000=7220=7200+20\).
2. The tax is on the remainder, not on the whole income.