PE 5 - Ratio

Given, volume is directly proportional to cube of radius.

⇒ V = kr³

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

⇒ 27r³ + 64r³ + 125r³ = R³

⇒ 216r³ = R³

⇒ R = 6r

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

Hence, option (a).

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).

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).

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).

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).

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.

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 = 200

∴ 216kx³ = 216 × 200 = Rs.43,200

Hence, option (b).

Let TSA = k₁rh + k₂r²

1440 = 5hk₁ + 25k₂   …(1)

5280 = 10hk₁ + 100k₂   …(2)

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

∴ TSA = 15hk₁ + 225k₂   ...(3)

(2) - (1) ⇒ 5hk₁ +  75k₂​​​​​​​ = 5280 - 1440 = 3840

Required area = 15k₁ + 225k₂ = 3 × 3840 = 11520

Hence, 11520.

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.

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).