# CRE 3 - Variation | Arithmetic - Ratio, Proportion & Variation

**CRE 3 - Variation | Arithmetic - Ratio, Proportion & Variation**

If 25 toys cost Rs. 390, what do 35 toys cost?

Answer: 546

**Explanation** :

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.

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**CRE 3 - Variation | Arithmetic - Ratio, Proportion & Variation**

p varies directly as the square of q. When q = 4, p = 48. Find p when q = 5.

- A.
25

- B.
15

- C.
75

- D.
50

Answer: Option C

**Explanation** :

p α q²

$\frac{{p}_{1}}{{{q}_{1}}^{2}}=\frac{{p}_{2}}{{{q}_{2}}^{2}}$

$\frac{48}{{4}^{2}}=\frac{{p}_{2}}{{5}^{2}}$

p_{2} = 75

Hence, option (c).

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**CRE 3 - Variation | Arithmetic - Ratio, Proportion & Variation**

Quantities a and b are inversely proportional to each other. When a = 4, b = 480. Find b when a = 3.

Answer: 640

**Explanation** :

a ∝ 1/b

a_{1}b_{1} = a_{2}b_{2}

If a_{1} = 4, b_{1} = 480 and a_{2} = 3,

b_{2 }= $\frac{{a}_{1}{b}_{2}}{{a}_{2}}$ = $\frac{4\times 480}{3}$ = 640.

Hence, 640.

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**CRE 3 - Variation | Arithmetic - Ratio, Proportion & Variation**

Quantity A varies directly with the sum of the quantities B and C. If B increases by 3 and C increases by 6, by how much does A increase?

- A.
2

- B.
4

- C.
6

- D.
Cannot be determined

Answer: Option D

**Explanation** :

A ∝ (B + C)

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

If B_{2} = B_{1} + 3 and C_{2} = C_{1}+ 6,

A_{1} = K(B_{1} + C_{1}) and A_{2} = K(B_{1} + 3 + C_{1} + 6)

A_{2} - A_{1}= 9K

As K is unknown, A_{2}- A_{1} cannot be found.

Hence, option (d).

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**CRE 3 - Variation | Arithmetic - Ratio, Proportion & Variation**

Quantity A varies inversely with product of B and C. When B = 3 and C = 24, A = 150. Find A when B = 10 and C = 5.

Answer: 216

**Explanation** :

A ∝ 1/BC

$\frac{{A}_{1}}{{A}_{2}}=\frac{{B}_{2}{C}_{2}}{{B}_{1}{C}_{1}}$

If A_{1} = 150, B_{1} = 3, C_{1} = 24, B_{2} = 10 and C_{2} = 5,

A_{2} = $\frac{{A}_{1}{B}_{1}{C}_{1}}{{B}_{2}{C}_{2}}$ = $\frac{150\times 3\times 24}{10\times 5}$ = 216.

Hence, 216.

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**CRE 3 - Variation | Arithmetic - Ratio, Proportion & Variation**

A varies directly with B when C is constant and inversely with C when B is constant. A is 32, when B is 14 and C is 14. Find the value of A, when B is 18 and C is 3.

Answer: 192

**Explanation** :

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.

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**CRE 3 - Variation | Arithmetic - Ratio, Proportion & Variation**

A variable x varies directly with the cube of another variable y. When x = 4, y = 2. Find y when x = 32.

Answer: 4

**Explanation** :

Given, x ∝ y^{3}

⇒ $\frac{{x}_{1}}{{{y}_{1}}^{3}}=\frac{{x}_{2}}{{{y}_{2}}^{3}}$

⇒ $\frac{4}{{2}^{3}}=\frac{32}{{{y}_{2}}^{3}}$

⇒ y_{2}^{3} = 64

⇒ y_{2} = 4

Hence, 4

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**CRE 3 - Variation | Arithmetic - Ratio, Proportion & Variation**

If 9 engines consume 12 metric ton of coal, when each is working 8 hours day, how much coal will be required for 4 engines, each running 13 hours a day, it being given that 3 engines of former type consume as much as 4 engines of latter type? (in metric tons)

Answer: 6.5

**Explanation** :

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

⇒ $\frac{(Coalconsumption)}{(engines\times workinghoursperday\times rateofconsumption)}$ = 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.

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**CRE 3 - Variation | Arithmetic - Ratio, Proportion & Variation**

The ratio of the rate of flow of water in pipes varies inversely as the square of the radii of the pipes. What is the ratio of the rates of flow in two pipes of diameters 3 cm and 6 cm, respectively?

- A.
1 : 2

- B.
2 : 1

- C.
1 : 8

- D.
4 : 1

Answer: Option D

**Explanation** :

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).

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**CRE 3 - Variation | Arithmetic - Ratio, Proportion & Variation**

Three quantities A, B, C are such that AB = kC, where k is a constant. When A is kept constant, B varies directly as C; When B is kept constant, A varies directly as C and when C is kept constant, A varies inversely as B. Initially, A was at 10 and A : B : C was 1 : 3 : 5. Find the value of A when B equals 18 at constant C.

- A.
16

- B.
16.66

- C.
18

- D.
19

Answer: Option B

**Explanation** :

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).

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**CRE 3 - Variation | Arithmetic - Ratio, Proportion & Variation**

A precious stone weighing 70 grams worth 24,500 is accidentally dropped and gets broken into two pieces having weights in the ratio of 2 : 5. If the price varies as the square of the weight then find the loss incurred.

- A.
11,500

- B.
12,000

- C.
11,000

- D.
10,000

Answer: Option D

**Explanation** :

P = K × W^{2}

⇒ 24,500 = K × 70^{2} ⇒ K = 5.

Thus, our price and weight relationship is : P = 5W^{2}.

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 × 20^{2} = 2000;

25 gram piece : 5 × 50^{2} = 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).

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**CRE 3 - Variation | Arithmetic - Ratio, Proportion & Variation**

The resistance of a wire is proportional to its length and inversely proportional to the square of its radius. Two wires of the same material have the same resistance and their radii are in the ratio 4 : 3. If the length of the first wire is 64 cms., find the length of the other.

- A.
32 cm

- B.
30 cm

- C.
36 cm

- D.
25 cm

Answer: Option C

**Explanation** :

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

R ∝ l/r^{2}

∴ 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}$}\right.}=\frac{{l}_{1}\times {{r}_{2}}^{2}}{{l}_{2}\times {{r}_{1}}^{2}}$

∴ $\frac{1}{1}=\frac{64\times 9}{{l}_{2}\times 16}$ (∵ R_{1} = R_{2})

⇒ l_{2 }= (64 × 9)/16 = 36 cms.

Hence, option (c).

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**CRE 3 - Variation | Arithmetic - Ratio, Proportion & Variation**

If 36 men can do a piece of work in 20 hours, in how many hours will 15 men do it ?

Answer: 48

**Explanation** :

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.

⇒ m_{1 }× t_{1} = m_{2 }× t_{2}

⇒ (15 × x) = (36 × 20)

⇒ (36 × 20)/15 = 48

∴ 15 men can do it in 48 hours.

Hence, 48.

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**CRE 3 - Variation | Arithmetic - Ratio, Proportion & Variation**

If the wages of 6 men for 15 days be Rs. 4200, then find the wages of 9 men for 12 days.

Answer: 5040

**Explanation** :

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.

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**CRE 3 - Variation | Arithmetic - Ratio, Proportion & Variation**

If 20 men can build a wall 33 metres long in 6 days, what length of a similar nature can be built by 30 men in 8 days?

Answer: 66

**Explanation** :

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.

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**CRE 3 - Variation | Arithmetic - Ratio, Proportion & Variation**

If 15 men, working 9 hours a day, can reap a field in 8 days, in how many days will 9 men reap the field, working 8 hours a day?

Answer: 15

**Explanation** :

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)

⇒ M_{1}H_{1}D_{1} = M_{2}H_{2}D_{2}

⇒ 15 × 9 × 8 = 9 × 8 × D_{2}

⇒ D_{2} = 15

Hence, required number of days = 15.

Hence, 15.

Workspace:

**CRE 3 - Variation | Arithmetic - Ratio, Proportion & Variation**

Ashok’s monthly salary varies directly with the number of working days in that month. His salary would be Rs. 9000 if there are 20 working days in a month. Find his monthly salary if there are 21 working days in that month.

- A.
Rs. 9,550

- B.
Rs. 9,450

- C.
Rs. 9,650

- D.
Rs. 9,850

Answer: Option B

**Explanation** :

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

(S_{1})/(S_{2}) = (D_{1})/(D_{2})

If S_{1} = 9000, D_{1} = 20 and D_{2} = 21.

S_{2} = $\frac{{S}_{1}{D}_{2}}{{D}_{1}}$ = $\frac{9000\times 21}{20}$ = Rs. 9,450

Hence, option (b).

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**CRE 3 - Variation | Arithmetic - Ratio, Proportion & Variation**

A and B rent a pasture for 10 months; A puts in 80 cows for 7 months. How many can B put in for the remaining 3 months, if he pays half as much more as A?

- A.
120

- B.
180

- C.
200

- D.
280

Answer: Option D

**Explanation** :

Rent payable ∝ no. of cows,

Rent payable ∝ amount of time

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

$\frac{RentApays}{RentB\hspace{0.17em}pays}=\frac{A\text{'}scows\times months}{B\text{'}scows\times months}$

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

⇒ n = 280

Hence, option (d).

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