JEE Mathematics Ratio and Proportion MCQs Set 02

Answer: (a) 4400

Explanation:
1. After the deaths \( 95\% \) remain, and after the departures \( 85\% \) of those remain.
2. So \( 3553=x\times0.95\times0.85=0.8075x \).
3. Hence \( x=\dfrac{3553}{0.8075}=4400 \).

Teacher's Note:
1. Successive percentage changes multiply, they do not add.
2. Divide by the combined factor to go back to the original.

Answer: (c) Rs. 264

Explanation:
1. Loss of 15% means selling price \( =0.85\times \) cost, so cost \( =\dfrac{204}{0.85}=240 \).
2. For a gain of 10% the price is \( 1.10\times240 \).
3. So the goods must be sold for Rs. 264.

Teacher's Note:
1. Percentages of outlay are always taken on the cost price.
2. Find the cost price first and then the new selling price.

Answer: (d) Simple interest and the number of years

Explanation:
1. Simple interest \( =\dfrac{PRT}{100} \) is a constant multiple of \( T \), so it varies directly as \( T \).
2. Compound interest grows by the factor \( (1+r)^T \), so it is not directly proportional to \( T \).
3. Hence simple interest and time vary directly.

Teacher's Note:
1. Direct variation means the ratio of the two quantities is constant.
2. Compound interest grows faster than a straight-line relation.

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

Explanation:
1. \( y=\dfrac{k}{x^2} \), so \( x^2=\dfrac ky \).
2. Taking the square root, \( x=\sqrt k\cdot y^{-1/2} \).
3. So \( x \) varies inversely as the square root of \( y \).

Teacher's Note:
1. Make the other variable the subject and read off the power.
2. The power of y is \( -\tfrac12 \), not \( -2 \).

Answer: (a) \( 8:5 \)

Explanation:
1. The two fractions are negatives of each other, so \( 2x-3y=-(3x-5y) \).
2. This gives \( 5x=8y \).
3. Hence \( x:y=8:5 \).

Teacher's Note:
1. A sum of two reciprocals is zero only if the denominators are equal and opposite.
2. Check with \( x=8,y=5 \): the denominators are \( -1 \) and \( 1 \).

Answer: (b) \( \dfrac{9}{80} \)

Explanation:
1. Let \( p=\sqrt{5x+1} \) and \( q=\sqrt{5x} \). By componendo and dividendo, \( \dfrac{p-q}{p+q}=\dfrac14 \) gives \( \dfrac pq=\dfrac{4+1}{4-1}=\dfrac53 \).
2. So \( \dfrac{5x+1}{5x}=\dfrac{25}{9} \), i.e. \( 9(5x+1)=125x \).
3. Hence \( 80x=9 \) and \( x=\dfrac9{80} \).

Teacher's Note:
1. Componendo and dividendo turns \( \dfrac{p-q}{p+q}=\dfrac mn \) into \( \dfrac pq=\dfrac{n+m}{n-m} \).
2. Verify by substituting \( 5x=\dfrac9{16} \) back into the surds.

Answer: (c) 12, 9, 21

Explanation:
1. Let the numbers be \( 4k,3k,7k \).
2. \( 16k^2+9k^2+49k^2=74k^2=666 \), so \( k^2=9 \) and \( k=3 \).
3. The numbers are 12, 9, 21.

Teacher's Note:
1. Use a common multiplier for numbers given in a ratio.
2. Check: 144 + 81 + 441 = 666.

Answer: (c) 56 : 55

Explanation:
1. The new number of employees is \( \tfrac{15}{16} \) of the old, and the new wage is \( \tfrac{22}{21} \) of the old.
2. New total \( =\tfrac{15}{16}\times\tfrac{22}{21}=\tfrac{330}{336}=\tfrac{55}{56} \) of the old total.
3. So the total wages decrease in the ratio 56 : 55.

Teacher's Note:
1. Ratios compound by multiplication.
2. Name the decrease as old : new.

Answer: (a) 7

Explanation:
1. Let the number be \( x \). Then \( (37-x)(79-x)=(52-x)(55-x) \).
2. \( 2923-116x=2860-107x \), so \( 9x=63 \).
3. Hence \( x=7 \); the remainders 30, 45, 48, 72 are in proportion.

Teacher's Note:
1. Four numbers a, b, c, d are in proportion when \( ad=bc \).
2. The \( x^2 \) terms cancel, leaving a linear equation.

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

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

Teacher's Note:
1. If a fraction of one quantity equals a fraction of another, the quantities are in proportion to the denominators.
2. Do not invert the ratio.

Answer: (d) 75

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

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

Answer: (c) \( \dfrac12 \)

Explanation:
1. Joint variation means \( a=kbc \), so \( \dfrac{a}{bc}=k \).
2. \( k=\dfrac{6}{4\times3}=\dfrac12 \).

Teacher's Note:
1. The ratio \( \dfrac{a}{bc} \) is the constant of variation.
2. Joint variation means direct variation with each variable.

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

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

Teacher's Note:
1. Replace the ratio by multiples of a constant.
2. Watch the sign when the numerator becomes negative.

Answer: (b) \( (4b-5c):(3c-b):(2c-3b) \)

Explanation:
1. Take each ratio to be \( 1 \) and test option (a): \( x=4b-5c,\ y=3c-b,\ z=2c-3b \).
2. Then \( x+2y=2b+c \), \( 2y+3z=12c-11b \) and \( 3z+4x=7b-14c \), so each fraction equals 1.
3. Hence \( x:y:z=(4b-5c):(3c-b):(2c-3b) \).

Teacher's Note:
1. In equal-ratio problems the common value can be absorbed into the unknowns.
2. Verify the proposed values in all three fractions.

Answer: (b) 24 and 28

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

Teacher's Note:
1. The difference of the terms equals the common multiplier here.
2. Check that 24 : 28 reduces to 6 : 7.

Answer: (d) \( 3:2 \) or \( -6:5 \)

Explanation:
1. Cross-multiply: \( (a-3b)(2a+5b)=(2a+b)(3b-4a) \).
2. \( 2a^2-ab-15b^2=-8a^2+2ab+3b^2 \), so \( 10a^2-3ab-18b^2=0 \).
3. This factorises as \( (2a-3b)(5a+6b)=0 \), so \( a:b=3:2 \) or \( -6:5 \).

Teacher's Note:
1. Form a homogeneous quadratic in a and b and factorise.
2. Both roots satisfy the original equation.

Answer: (a) \( p:q=(a+b)(d-c):(a-b)(c+d) \)

Explanation:
1. The first person gets \( \dfrac{pa}{a+b}+\dfrac{qc}{c+d} \) and the second gets \( \dfrac{pb}{a+b}+\dfrac{qd}{c+d} \).
2. Equating, \( \dfrac{p(a-b)}{a+b}=\dfrac{q(d-c)}{c+d} \).
3. Hence \( p:q=(a+b)(d-c):(a-b)(c+d) \).

Teacher's Note:
1. Write each person's total share as a sum of two fractions.
2. Bring p terms to one side and q terms to the other.

Answer: (a) 225 cm

Explanation:
1. \( T=k\sqrt l \) with \( 1=k\sqrt{100} \), so \( k=\tfrac1{10} \).
2. For \( T=1.5 \), \( \sqrt l=15 \).
3. So \( l=225 \) cm.

Teacher's Note:
1. Square both sides to remove the root when finding the length.
2. Do not scale the length linearly with time.

Answer: (c) Rs. 26.40

Explanation:
1. Let \( S=2u,\ R=3u \). Then \( Q=9u \).
2. \( P+Q=4(R+S)=20u \), so \( P=11u \).
3. The total is \( 11u+9u+3u+2u=25u=60 \), so \( u=2.4 \) and \( P=Rs. 26.40 \).

Teacher's Note:
1. Start from the smallest share and express all others through it.
2. Check: P + Q = 48 = 4 x (7.20 + 4.80).

Answer: (d) Rs. 150

Explanation:
1. The distances are \( 30,48,18 \) km.
2. The charges per km are Rs. 2, Rs. 1.50 and Rs. 1.
3. Total \( =60+72+18=Rs. 150 \).

Teacher's Note:
1. Find the actual value of each ratio term before multiplying.
2. The charge per km scales with the taxi rate of Rs. 2 for the ratio term 4.

Answer: (a) 13

Explanation:
1. Work is proportional to men times days.
2. \( \dfrac{(x-2)(x+2)}{(x+3)(x-2)}=\dfrac{15}{16} \), so \( \dfrac{x+2}{x+3}=\dfrac{15}{16} \).
3. Then \( 16x+32=15x+45 \) and \( x=13 \).

Teacher's Note:
1. Cancel the common factor \( x-2 \) only after noting that \( x\ne2 \).
2. Work equals men multiplied by days.

Answer: (b) 36

Explanation:
1. Let the number be \( n \) and \( s=\sqrt n \). Then \( \tfrac{11}{2}s+3=s^2 \), i.e. \( 2s^2-11s-6=0 \).
2. This factorises as \( (2s+1)(s-6)=0 \), so \( s=6 \).
3. Hence \( n=36 \).

Teacher's Note:
1. Put the square root equal to a new variable to get an ordinary quadratic.
2. Reject the negative root because a square root is not negative.

Answer: (a) Rs. 4,500

Explanation:
1. Tax \( =k(I-10000)^2 \). With \( 500=k\times5000^2 \), \( k=\dfrac{500}{25\times10^6} \).
2. For \( I=25000 \), the excess is 15000 and the tax is \( 500\times\left(\dfrac{15000}{5000}\right)^2=500\times9 \).
3. The tax is Rs. 4,500.

Teacher's Note:
1. Compare the excess amounts, not the incomes themselves.
2. Squaring the ratio 3 gives 9.

Answer: (d) Rs. 8,000

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

Teacher's Note:
1. Successive deductions multiply the remaining fractions.
2. Compare with 90% of the original income, not of the remainder.