PE 4 - Ratio

We are given a : b : c = 6 : 2 : 1  …..(1), and 

b : d : e = 8 : 4 : 3.                           …..(2)

Combined ratio, a : b : c : d : e = 24 : 8 : 4 : 4 : 3

Go with the options.

Let`s take option (e)

By multiplying both c/e = 4/3 ……not an integer.

Hence, option (e).

4 leaps of dingo = 5 leaps of hare

5 leaps of dingo = 5 × 5/4 leaps of hare

Move rate of dingo : move rate of hare = 5 × 5/4 : 6

⇒ 25 : 24

Hence, option (a).

Let the total population be 300.

Rural population = 100

Urban population = 200

Fraction of rural population who have filed IT return = (1 – 4/5) = 1/5

Fraction of urban population who have filed IT return = 3 × 1/5 = 3/5

∴ Urban population who have filed IT return = 3/5 × 200 = 120

∴ Urban population who have not filed IT return = 200 - 120 = 80.

∴ Required fraction = 80/300 = 4/15

Hence, option (a).

(3000 × 6 + 6000 × 6) : 4000 × 12

⇒ 18000 + 36000 : 48000

⇒ 54000/48000 = 9/8

Hence, option (b).

d ∝ t²

⇒ d₁/(t₁² ) = d₂/(t₂²)

⇒ 250/((5)² )=d₂/((6)²)

⇒ d₂ = 10 × 36 = 360 m

∴ Distance travelled in the 6^th second = 360 – 250 = 110 m

Hence, 110.

∵ 4/7 x = 1200 ⇒ x = 2100

A’s profit = 900

A’s investment = 3/4 × 40,000 = 30,000

∴ Ratio = 900/30000 = 3 ∶ 100

Hence, option (b).

(a + b) α (a - b)

⇒ (a + b) = k(a - b), when k is a constant

⇒ (a + b)/(a - b) = k

Applying componendo and dividendo,

a/b = (k + 1)/(k - 1) = x (say)

Squaring both sides; a²/b² = x²

Applying componendo and dividendo

$\frac{a^2+b^2}{a^2-b^2}=\frac{x^2+1}{x^2-1}$

⇒ $a^2+b^2=(a^2-b^2)\times \frac{x^2+1}{x^2-1}$

⇒ (a² + b²) = (a² - b²) × (a constant, let's say t)

∴ (a² + b²) = t × (a² - b²)

Also, $\frac{a^2+b^2}{ab}=\frac{x^2{b}^2+b^2}{xb\times b}=\frac{b^2(x^2+1)}{b^{2}x}$ = $\frac{1+x^2}{x}$ = constant

∴ (a² + b²) α ab

∴ Both (A) and (B) are true.

Hence, option (c).

1200 × 12 : x × 9 : y × 6 = 2 : 3 : 5

⇒ (1200 × 12) / (x × 9) = 2/3

⇒ x = 2400

Hence, option (a).

Let the free luggage allowance be ‘f’ kg. Let the weight of the luggage carried by Ravi be ‘r’ kg and the weight of the luggage carried by Pranav be ‘p’ kg. Thus, the excess luggage weights carried by Ravi and Pranav respectively are (r – f)kg and (p – f)kg. 

Thus, the total luggage charge for both would be (r – f)k + (p – f)k if k is the charge per kg.

Thus, (r – f)k + (p – f)k = 1100.

(r + p – 2f)k = 1100   ...(1)

If Ravi carried twice the luggage weight he actually did, i.e., if he carried 2r kg, then the excess luggage weight he carried would have been 2r – f and the corresponding charge would have been (2r – f) k.

Therefore, (2r – f)k = 2000   ...(2)

Likewise, If Pranav carried twice the luggage he actually did i.e., if he carried 2p kg, then the excess luggage he carried would have been 2p –f and the corresponding charge would have been (2p – f) k.

Therefore, (2p – f)k = 1000   ...(3)

Adding (2) and (3) and simplifying, we get,

(r + p – f)k = 1500   ...(4)

Dividing (4) by (1) and simplifying, we get,

19f = 4r + 4p   ...(5)

Dividing (2) by (3) and simplifying, we get,

–f = 2r – 4p   ...(6)

Solving (5) and (6) for r, we get,

r = 3f   ...(7)

Subtracting (1) from (4) and simplifying, we get,

fk = 400.

Ravi’s luggage charge = (r – f)k.

But, according to equation (7), r = 3f. Therefore, Ravi’s luggage charge = 2fk

But, fk = 400. Therefore, Ravi’s luggage charge = Rs. 800.

Hence, option (a).