CRE 3 - Variation

Let the required cost be Rs. x. Then,

Cost ∝ No. of toys

(When 2 quantities are directly proportional, their ratio is constant)

⇒ $\frac{25}{35}=\frac{390}{x}$ 

⇒ (25 × x) = (35 × 390) 

⇒ x = (35 × 390)/25 = 546

∴ the cost of 35 toys is Rs. 546.

Hence, 546.

p α q²

$\frac{p_1}{{q_1}^2}=\frac{p_2}{{q_2}^2}$

$\frac{48}{4^2}=\frac{p_2}{5^2}$

p₂ = 75

Hence, option (c).

a ∝ 1/b

a₁b₁ = a₂b₂

If a₁ = 4, b₁ = 480 and a₂ = 3,

b₂ = $\frac{a_1{b}_2}{a_2}$ = $\frac{4\times 480}{3}$ = 640.

Hence, 640.

A ∝ (B + C)

Let A = K(B + C) where K is the proportionality constant

If B₂ = B₁ + 3 and C₂ = C₁+ 6,

A₁ = K(B₁ + C₁) and A₂ = K(B₁ + 3 + C₁ + 6)

A₂ - A₁= 9K

As K is unknown, A₂- A₁ cannot be found.

Hence, option (d).

A ∝ 1/BC

$\frac{A_1}{A_2}=\frac{B_2{C}_2}{B_1{C}_1}$

If A₁ = 150, B₁ = 3, C₁ = 24, B₂ = 10 and C₂ = 5,

A₂ = $\frac{A_1{B}_1{C}_1}{B_2{C}_2}$ = $\frac{150\times 3\times 24}{10\times 5}$ = 216.

Hence, 216.

Given that A α B and A α 1/C.

A ∝ B/C ⇒ A = k × B/C

Given that A = 32 when B is 14 and C is 14.

So, 32 = k × 14/14 ⇒ k = 32.

If B is 18 and C is 3, then

A = k × B/C = 32 × 18/3 = 192

Hence, 192.

Given, x ∝ y³

⇒ $\frac{x_1}{{y_1}^3}=\frac{x_2}{{y_2}^3}$

⇒ $\frac{4}{2^3}=\frac{32}{{y_2}^3}$ 

⇒ y₂³ = 64

⇒ y₂ = 4

Hence, 4

Let 3 engines of former type consume 1 unit in 1 hour.

Then, 4 engines of latter type consume 1 unit in 1 hour.

Therefore 1 engine of former type consumes (1/3) unit in 1 hour.

1 engine of latter type consumes (1/4) unit in 1 hour.

Let the required consumption of coal be x units.

Less engines, Less coal consumed (direct proportion)

More working hours, More coal consumed (direct proportion)

Less rate of consumption, Less coal consumed (direct proportion)

⇒ Coal consumption ∝ engines × working hours/day × rate of consumption

⇒ $\frac{(Coal consumption)}{(engines \times working hours per day\times rate of consumption)}$ = constant

⇒ $\frac{12}{9\times 8\times {\displaystyle \frac{1}{3}}}=\frac{x}{4\times 13\times {\displaystyle \frac{1}{4}}}$

⇒ x = 26

Hence, the required consumption of coal = 26/4 = 6.5 metric ton.

Hence, 6.5.

Radii of the two pipes are 3/2 cm and 6 cm.

Square of the radii of the pipes are 9/4 and 36/4.

Given, flow of water in pipes ∝ 1/(radius)²

∴ Required ratio of rates of flow in the two pipes = $\frac{1}{9/4}$ : $\frac{1}{36/4}$ = $\frac{1}{9}$ : $\frac{1}{36}$ = 4 : 1

Hence, option (d).

Initial values are 10, 30 and 50.

Thus, we have 10 × 30 = k × 50.

Hence, k = 6

Thus, the equation is AB = 6C.

For the problem, keep C constant at 50. Then, A × 18 = 6 × 50.

i.e. A = 300/18 = 16.66

Hence, option (b).

P = K × W²

⇒ 24,500 = K × 70² ⇒ K = 5.

Thus, our price and weight relationship is : P = 5W².

When the two pieces are in the ratio 2 : 5 (weight wise) then we know that their weights must be 20 grams and 50 grams respectively. Their values would be:

10 gram piece : 5 × 20² = 2000;

25 gram piece : 5 × 50² = 12,500.

Total Price = 2,000 + 12,500 = 14,500. From an initial value of 24,500, this represents a loss of 10,000.

Hence, option (d).

If T is the resistance, l is the length and r is radius.

R ∝ l/r² 

∴ R = $\frac{k\mathrm{l}}{r^2}$ (where k is a constant)

∴ $\frac{R_1}{R_2}=\frac{{\displaystyle \raisebox{1ex}{$l_1$}\!\left/ \!\raisebox{-1ex}{${r_1}^2$}\right.}}{\raisebox{1ex}{$l_2$}\!\left/ \!\raisebox{-1ex}{${r_2}^2$}\right.}=\frac{l_1\times {r_2}^2}{l_2\times {r_1}^2}$

∴ $\frac{1}{1}=\frac{64\times 9}{l_2\times 16}$ (∵ R₁ = R₂)

⇒ l₂ = (64 × 9)/16 = 36 cms.

Hence, option (c).

Let the required number of hours be x. Then,

If less men are employed it will take more hours to finish the same task.

Hence, no. of men and time is inversely proportion.

⇒ m₁ × t₁ = m₂ × t₂

⇒ (15 × x) = (36 × 20)

⇒ (36 × 20)/15 = 48

∴ 15 men can do it in 48 hours.

Hence, 48.

Let the required wages be Rs. x.

More men, More wages ⇒ wages ∝ men    (Direct Proportion)

Less days, Less wages ⇒ wages ∝ days    (Direct Proportion)

⇒ wages ∝ men × days

⇒ $\frac{wages}{men\times days}$ = constant

⇒ $\frac{4200}{6\times 15}=\frac{x}{9\times 12}$

⇒ x = 5040

∴ the required wages are Rs. 5040.

Hence, 5040.

Let the required length be x meters.

More men, More length built    (Direct Proportion)                                           

Less days, Less length built     (Direct Proportion)

⇒ length ∝ men × days

⇒ length/(men×days) = constant

⇒ $\frac{33}{20\times 6}=\frac{x}{30\times 8}$

⇒ x = 66

∴ the required length is 66 m. 

Hence, 66.

Let the required number of days be x.

More men, Less days ⇒ days ∝ 1/men  (Indirect Proportion)

Less hours per day, More days ⇒ days ∝ 1/hours (Indirect Proportion)

⇒ days ∝ 1/(men × hours)

⇒ M₁H₁D₁ = M₂H₂D₂

⇒ 15 × 9 × 8 = 9 × 8 × D₂

⇒ D₂ = 15

Hence, required number of days = 15.

Hence, 15.

If a month has D working days and Ashok’s salary in that month is Rs. S, S ∝ D

(S₁)/(S₂) = (D₁)/(D₂)

If S₁ = 9000, D₁ = 20 and D₂ = 21.

S₂ = $\frac{S_1{D}_2}{D_1}$ = $\frac{9000\times 21}{20}$ = Rs. 9,450

Hence, option (b).

Rent payable ∝ no. of cows,

Rent payable ∝ amount of time

⇒ Rent payable ∝ (no. of cows)  × (amount of time)

$\frac{Rent A pays}{Rent B\hspace{0.17em}pays}=\frac{A\text{'}s cows\times months}{B\text{'}s cows\times months}$

$\frac{2}{3}=\frac{80\times 7}{n\times 3}$

⇒ n = 280

Hence, option (d).