JEE Mathematics Ratio and Proportion MCQs Set 01

Answer: (c) \( 1:5 \)

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

Teacher's Note:
1. Replace the given ratio by multiples of a common constant.
2. The constant cancels in the final ratio.

Answer: (b) \( 1:8 \)

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

Teacher's Note:
1. Convert a ratio equation into a linear equation by cross-multiplication.
2. Then express one variable in terms of the other.

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

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

Teacher's Note:
1. The constant terms cancel, which leaves a homogeneous equation in x and y.
2. A negative ratio simply means the two quantities have opposite signs.

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

Explanation:
1. Factorise: \( 3x^2-9xy-xy+3y^2=3x(x-3y)-y(x-3y)=(3x-y)(x-3y) \).
2. So \( x=3y \) or \( 3x=y \).
3. Hence \( x:y=3:1 \) or \( 1:3 \).

Teacher's Note:
1. A homogeneous quadratic gives two possible ratios.
2. Do not forget the second root.

Answer: (c) 93

Explanation:
1. Let passes be \( 3k \) and failures \( 2k \).
2. With the line 12 places lower, \( \dfrac{3k+12}{2k-12}=\dfrac{21}{10} \), so \( 30k+120=42k-252 \).
3. Hence \( k=31 \) and passes \( =3k=93 \).

Teacher's Note:
1. Total candidates stay the same, so 12 moves from the failures to the passes.
2. Define the unknowns with a common multiplier.

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

Explanation:
1. Cross-multiply: \( 3x^2-33y^2=(x+7y)(x-4y)=x^2+3xy-28y^2 \).
2. So \( 2x^2-3xy-5y^2=0 \), i.e. \( (2x-5y)(x+y)=0 \).
3. Hence \( x:y=5:2 \) or \( -1 \) (that is \( x=-y \)).

Teacher's Note:
1. Form a homogeneous quadratic and factorise it.
2. Each factor gives one possible ratio.

Answer: (d) Rs. 1,64,358

Explanation:
1. \( 30\% \) of Rs. 547860 is \( 0.30\times547860=164358 \).
2. The other shares are Rs. 136965 (25%) and Rs. 109572 (20%).
3. So the first partner receives Rs. 1,64,358.

Teacher's Note:
1. Convert each percentage to a decimal before multiplying.
2. Check that the three shares and the funds add up to the total profit.

Answer: (c) 72500

Explanation:
1. Five years ago the population was \( \dfrac{84564}{1.08}=78300 \).
2. Ten years ago it was \( \dfrac{78300}{1.08}=72500 \).
3. So the population was 72500.

Teacher's Note:
1. A rise of 8% means the new value is 1.08 times the old value.
2. Work backwards by dividing, not by subtracting 8%.

Answer: (a) the square of the base (direct variation)

Explanation:
1. Area \( =\dfrac{\sqrt3}{4}b^2 \).
2. The area is a constant multiple of \( b^2 \).
3. So the area varies directly as the square of the base.

Teacher's Note:
1. Identify the formula first, then compare the variables.
2. Inverse variation would mean the product is constant, which is not so here.

Answer: (b) 7

Explanation:
1. \( x=k_1y+\dfrac{k_2}{y^2} \). From the data \( k_1+k_2=37 \) and \( 2k_1+\dfrac{k_2}{4}=11 \).
2. Solving gives \( k_1=1,\ k_2=36 \).
3. At \( y=3 \), \( x=3+\dfrac{36}{9}=7 \).

Teacher's Note:
1. Each data pair gives one linear equation in the constants.
2. Solve the pair of equations and then evaluate.

Answer: (b) \( 3\sqrt2 \)

Explanation:
1. The mean proportional of \( p \) and \( q \) is \( \sqrt{pq} \).
2. \( \sqrt{12}\times\sqrt{27}=\sqrt{324}=18 \).
3. So the mean proportional is \( \sqrt{18}=3\sqrt2 \).

Teacher's Note:
1. Mean proportional means the geometric mean of the two numbers.
2. Simplify the surd by taking out the perfect square 9.

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

Explanation:
1. The product is \( (a-b)^3(a+b)^3=(a^2-b^2)^3 \).
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 the factors into powers of \( (a^2-b^2) \) before taking the root.
2. Use exponent rules carefully.

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

Explanation:
1. Terms in continued proportion form a geometric series with ratio \( r \). Then \( ar^2=b \), so \( r^2=\dfrac ba \).
2. The first term is \( \dfrac{a}{r^2}=\dfrac{a^2}{b} \).
3. The fourth term is \( ar=a\sqrt{\dfrac ba}=\sqrt{ab} \).

Teacher's Note:
1. Continued proportion is a geometric progression.
2. Express the unknown terms through the common ratio.

Answer: (d) 25

Explanation:
1. The duplicate ratio of \( 7:12 \) is \( 49:144 \).
2. \( 144(x+24)=49\times6(x-1) \), so \( 144x+3456=294x-294 \).
3. Hence \( 150x=3750 \) and \( x=25 \).

Teacher's Note:
1. Duplicate means the ratio of squares.
2. Keep track of the factor 6 in the second term.

Answer: (b) 9, 18, 36

Explanation:
1. If the extremes are \( p,q \) then \( pq=18^2=324 \) and \( p+q=45 \).
2. So \( p,q \) are roots of \( t^2-45t+324=0 \), i.e. 9 and 36.
3. The numbers are 9, 18, 36.

Teacher's Note:
1. In continued proportion the middle term squared equals the product of the extremes.
2. Form a quadratic from the sum and product.

Answer: (d) \( (ab+cd)^2 \)

Explanation:
1. Let \( a=bk,\ c=dk \).
2. Then \( (a^2+c^2)(b^2+d^2)=k^2(b^2+d^2)^2 \).
3. Also \( ab+cd=k(b^2+d^2) \), so the product equals \( (ab+cd)^2 \).

Teacher's Note:
1. Put a common ratio k in a proportion to prove identities.
2. Square the expression \( ab+cd \) to compare.

Answer: (a) \( \dfrac{(a+c)^3}{(b+d)^3} \)

Explanation:
1. Let \( a=bk,\ c=dk \).
2. Then \( \dfrac{a^3+c^3}{b^3+d^3}=\dfrac{k^3(b^3+d^3)}{b^3+d^3}=k^3 \).
3. Also \( \dfrac{(a+c)^3}{(b+d)^3}=k^3 \), so the answer is \( \dfrac{(a+c)^3}{(b+d)^3} \).

Teacher's Note:
1. Both sides reduce to the cube of the common ratio.
2. Compare the powers carefully to avoid choosing the square.

Answer: (a) \( 10\tfrac12 \) m

Explanation:
1. The volume varies jointly as \( r^2 \) and \( h \): \( V=kr^2h \).
2. From \( 1078=k\cdot49\cdot21 \), \( k=\dfrac{22}{21} \).
3. Then \( 275=\dfrac{22}{21}\cdot25\cdot h \), so \( h=\dfrac{21}{2}=10\tfrac12 \) m.

Teacher's Note:
1. With height varying as well, the volume varies jointly with \( r^2 \) and \( h \).
2. The constant is \( \tfrac{\pi}{3} \) with \( \pi=\tfrac{22}{7} \).

Answer: (d) 44.1 m

Explanation:
1. \( s=kt^2 \) with \( 490=100k \), so \( k=4.9 \).
2. Distance in 5 seconds is \( 4.9\times25=122.5 \) m, and in 4 seconds it is \( 4.9\times16=78.4 \) m.
3. The distance in the 5th second is \( 122.5-78.4=44.1 \) m.

Teacher's Note:
1. The distance in the nth second is the difference of the distances in n and (n - 1) seconds.
2. Do not confuse it with the total distance in 5 seconds.

Answer: (b) 180

Explanation:
1. \( P=\dfrac{kQ}{R} \), so \( k=\dfrac{PR}{Q}=\dfrac{(4/9)(9/14)}{3/7}=\dfrac23 \).
2. Then \( Q=\dfrac{PR}{k}=\dfrac32\sqrt{96\times150}=\dfrac32\times120 \).
3. So \( Q=180 \).

Teacher's Note:
1. Find the constant of variation first.
2. \( \sqrt{96\times150}=\sqrt{14400}=120 \).

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

Explanation:
1. Let \( y=k_1x+\dfrac{k_2}{x} \). Then \( 16k_1+k_2=24 \) and \( 9k_1+k_2=10 \).
2. Subtracting gives \( 7k_1=14 \), so \( k_1=2 \) and \( k_2=-8 \).
3. The relation is \( y=2x-\dfrac8x \).

Teacher's Note:
1. Multiply each equation by x (or 3x) to clear the denominators.
2. The constant k can be negative.

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

Explanation:
1. Write \( y=k_1x+k_2x^2+k_3x^3 \), giving \( k_1+k_2+k_3=4 \), \( 2k_1+4k_2+8k_3=8 \), \( 3k_1+9k_2+27k_3=18 \).
2. Solving, \( k_1=6,\ k_2=-3,\ k_3=1 \).
3. So \( y=6x-3x^2+x^3 \); it fits all three data pairs.

Teacher's Note:
1. Three unknown constants need three data pairs.
2. Substitute back to verify each value of y.

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

Explanation:
1. Let each ratio be \( k \). By the property of equal ratios, if \( a+b+c\ne0 \), \( k=\dfrac{a+b+c}{2(a+b+c)}=\dfrac12 \).
2. If \( a+b+c=0 \) then \( b+c=-a \), so \( k=\dfrac{a}{-a}=-1 \).
3. So each ratio is \( \dfrac12 \) or \( -1 \).

Teacher's Note:
1. Use the rule that equal ratios equal the ratio of the sum of numerators to the sum of denominators.
2. Always examine the special case when the sum of numerators is zero.