# PE 5 - Ratio | Arithmetic - Ratio, Proportion & Variation

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**PE 5 - Ratio | Arithmetic - Ratio, Proportion & Variation**

Volume of a sphere varies directly with cube of its radius. Three spheres with radii in the ratio 3 : 4 : 5 are melted to form a large sphere. What is the ratio of radius of smallest sphere to the new sphere formed.

- (a)
1 : 2

- (b)
1 : 3

- (c)
3 : 7

- (d)
None of these

Answer: Option A

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**Explanation** :

Given, volume is directly proportional to cube of radius.

⇒ V = kr^{3}

Let the radii of the original cubes be 3r, 4r and 5r and that of the larger cube be R.

Since the three original cubes are melted to form the larger cube, total volume of original three cubes will be same as volume of the larger cube.

∴ k(3r)^{3} + k(4r)^{3} + k(5r)^{3} = k(R)^{3}

⇒ 27r^{3} + 64r^{3} + 125r^{3} = R^{3}

⇒ 216r^{3} = R^{3}

⇒ R = 6r

∴ Radius of smallest sphere : Radius of the new sphere = 3r : 6r = 1 : 2

Hence, option (a).

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**PE 5 - Ratio | Arithmetic - Ratio, Proportion & Variation**

A, B, C and D enter into partnership. A subscribes 1/3th of the capital, B, 1/4th and C ,1/5th and D the rest. What is the share of D out of a profit of Rs. 6,000?

- (a)
Rs. 2000

- (b)
Rs. 1600

- (c)
Rs. 1200

- (d)
Rs. 1300

Answer: Option D

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**Explanation** :

D’s capital = $1-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}$ = $\frac{60-20-15-12}{60}$ = $\frac{60-47}{60}$ = $\frac{13}{60}$

∴ Ratio of A, B, C and D is 1/3 : 1/4 : 1/5 : 13/60 = 20 ∶ 15 ∶ 12 ∶ 13

∴ D’s profit = 6000 × 13/60 = 1300

Hence, option (d).

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**PE 5 - Ratio | Arithmetic - Ratio, Proportion & Variation**

Ankur has five times as much money with him as Varun does. Each day, Ankur spends a constant amount of money while Varun earns a fifth of the amount that Ankur spends. After 10 days, the ratio of the amounts with Ankur and Varun is 15 : 7. After-how many more days will the ratio of the amounts with them be 5 : 9?

- (a)
20

- (b)
15

- (c)
5

- (d)
10

Answer: Option D

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**Explanation** :

Let Ankur and Varun have Rs. 5A and Rs. A respectively.

If Ankur spends Rs. 5x every day, Varun earns Rs. x every day.

Given, (5S - 50x) / (S + 10x) = 15/7

⇒ S = 25x.

Let their amounts be in the ratio 5 : 9, n days from the start

$\frac{5S-n\times 5x}{S+nx}=\frac{125x-5nx}{25x+nx}=\frac{5}{9}$

⇒ n = 20

∴ After 10 more days, the ratio will be 5 : 9

Hence, option (d).

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**PE 5 - Ratio | Arithmetic - Ratio, Proportion & Variation**

A is a working partner and B is a sleeping partner in a business. A puts in Rs. 5000 and B puts in Rs. 6000. A receives 12 1/2% of the profit for managing, the rest being divided in proportion of their capital. What is the share of A out of a profit of Rs. 4400?

- (a)
Rs. 2100

- (b)
Rs. 2300

- (c)
Rs. 2200

- (d)
Rs. 2150

Answer: Option B

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**Explanation** :

A receives 12.5% for managing the business i.e. 4400 × 25/200 = 550

Ration of investment of A : B = 5000 : 6000 = 5 : 6

∴ A’s share in remaining profit = 5/11 × (4400 - 550) = 5 × 350 = 1750

Thus A’s total profit = 1750 + 550 = 2300

Hence, option (b).

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**PE 5 - Ratio | Arithmetic - Ratio, Proportion & Variation**

A, B, C and D entered into a partnership by investing the capital in the ratio of 2 : 3 : 5 : 4. If A invested the money for 9 months, B for 8 months, C for 4 months and D for 6 months. Find the share of C in the annual profit of Rs. 64500.

- (a)
Rs. 18000

- (b)
Rs. 15000

- (c)
Rs. 21000

- (d)
Rs. 16500

Answer: Option B

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**Explanation** :

Ratio of capitals = (2 × 9) : (3 × 8) : (5 × 4) : (4 × 6) = 9 : 12 : 10 : 12

∴ C’s profit = 10/43 × 64500 = 15000.

Hence, option (b).

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**PE 5 - Ratio | Arithmetic - Ratio, Proportion & Variation**

There are five identical cups containing milk in the ratio 4 : 5 : 6 : 7 : 8. How many cups are at least half full of milk if the total volume of milk in the cups is three-fifth of the total volume of the five cups?

Answer: 4

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**Explanation** :

Let the quantities of milk in the cups be 4x ml, 5x ml, 6x ml, 7x ml and 8x ml. Let the volume of each cup be 100 ml.

Total volume of the cups = 500 ml

4x + 5x + 6x + 7x + 8x = 0.6 × 500 = 300

⇒ 18x = 300

⇒ x = 100/9

The quantities of milk in the cups are 400/9 ml, 500/9 ml, 600/9 ml, 700/9 ml and 800/9 ml

i.e., except the first cup all other are filled with milk at least to 50% of their capacity.

Hence, 4.

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**PE 5 - Ratio | Arithmetic - Ratio, Proportion & Variation**

The cost of a precious stone varies as the cube of its weight. A certain precious stone broke into three pieces whose weights are in the ratio 1 : 2 : 3, as a result of which its cost reduces by Rs. 36,000. What was the cost of the stone before breaking?

- (a)
Rs. 416000

- (b)
Rs. 432000

- (c)
Rs. 204800

- (d)
Rs. 216000

Answer: Option B

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**Explanation** :

Let the weights be x, 2x and 3x grams.

Previously, the weight of the stone was x + 2x + 3x = 6x gms

Cost of the stone before breaking = k(6x)³ = 216kx³

Cost of the stone after breaking = k[x³ + (2x)³ + (3x)³] = 36kx³

Now, 216kx³ - 36kx² = 36000

⇒ kx³ = 36000/180 = 2000

∴ 216kx³ = 216 × 2000 = Rs.4,32,000

Hence, option (b).

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**PE 5 - Ratio | Arithmetic - Ratio, Proportion & Variation**

The total surface area of a cylinder having a certain height is the sum of 2 parts. One of the parts varies directly with its radius when height is constant & directly with its height when radius is constant. The other part varies directly with the square of its radius. The total surface areas of two cylinders having the same height whose radii are 5 cm and 10 cm are 1440 sq.cm and 5280 sq.cm respectively. Find the total surface area in sq.cm) of a cylinder of the same height whose radius is 15 cm.

Answer: 11520

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**Explanation** :

Let TSA = k_{1}rh + k_{2}r²

1440 = 5hk_{1} + 25k_{2 }…(1)

5280 = 10hk_{1} + 100k_{2 }…(2)

We have to calculate the surface area when height = h and radius = 15

∴ TSA = 15hk_{1} + 225k_{2} ...(3)

(2) - (1) ⇒ 5hk_{1} + 75k_{2} = 5280 - 1440 = 3840

Required area = 15k_{1} + 225k_{2} = 3 × 3840 = 11520

Hence, 11520.

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**PE 5 - Ratio | Arithmetic - Ratio, Proportion & Variation**

A quantity Q equals the sum of three other quantities, the first of which is a constant, the second varies directly as x and the third varies directly as x². When x = 1; p = 52. when x = 2, p = 144 and when x = 3, p = 316. Find the constant.

Answer: 40

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**Explanation** :

Let p = A + Bx + Cx²

∴ A + B + C = 52 …(1)

A + 2B + 4C = 144 …(2)

A + 3B + 9C = 316 …(3)

(2) – (1) ⇒ B + 3C = 92 …(4)

(3) – (2) ⇒ B + 5C = 172 …(5)

(5) – (4) ⇒ 2C = 80

∴ C = 40 ⇒ B = -28

∴ A – 28 + 40 = 52

⇒ A = 40

Hence, 40.

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**PE 5 - Ratio | Arithmetic - Ratio, Proportion & Variation**

The consumption of petrol per hour of my car varies directly as the square of its speed. When the car is travelling at 25 kmph its consumption is 4 litres. If each litre costs Rs. 15 and other expenses per hour are Rs. 100, then what should be the speed to minimize the expenditure to cover a distance of 200 km?

- (a)
25√5/√3

- (b)
25

- (c)
25√5

- (d)
None of these

Answer: Option A

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**Explanation** :

Let consumption of fuel per hour be C ltrs and speed be V kmph.

C α V²

Or, C = kV² ⇒ 4 = k(25)²

⇒ k = 4/((25)²) = 4/625

∴ C = 4V²/625

Let the required velocity be V kmph

Time taken to cover 200 km at V kmph = 200/V hours

∴ Consumption of petrol in 200/V hours = 200C/V ltrs

∴ Cost of petrol = 200C/V × 15 = Rs. 3000C/V = 3000/V × 4V²/625 = 96V/5

and other expenses = 200/V × 100

∴ Total expenditure (T) = 96V/5 + 20000/V

Product of 96V/5 and 20000/V is a constant

⇒ Sum is minimum when 96V/5 = 20000/V ⇒ V =25√5/√3

⇒ V = 25√5/√3

Hence, option (a).

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