# PE 4 - Ratio | Arithmetic - Ratio, Proportion & Variation

**PE 4 - Ratio | Arithmetic - Ratio, Proportion & Variation**

Let a, b, c, d, and e be integers such that a : b : c = 6 : 2 : 1, and b : d : e = 8 : 4 : 3. Then which of the following pairs contains a number that is not an integer?

- (a)
a/8, b/d

- (b)
a/6, c/d

- (c)
a/24, ad/36

- (d)
a/12, bd/32

- (e)
b/4, c/e

Answer: Option E

**Explanation** :

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).

Workspace:

**PE 4 - Ratio | Arithmetic - Ratio, Proportion & Variation**

A dingo pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 4 leaps of the dingo are equal to 5 of the hare. Compare the rates of dingo and the hare:

- (a)
25 : 24

- (b)
24 : 25

- (c)
6 : 7

- (d)
None

Answer: Option A

**Explanation** :

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).

Workspace:

**PE 4 - Ratio | Arithmetic - Ratio, Proportion & Variation**

1/3rd of population of country X live in rural areas rest live in urban cities. If 4/5 of the rural population has not filed a IT returns and the fraction of urban population students who have filed IT return is 3 times the fraction of the rural population who have filed the return, then what fraction of the population is urban population who have not filed the return?

- (a)
4/15

- (b)
1/5

- (c)
4/15

- (d)
3/5

Answer: Option A

**Explanation** :

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).

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**PE 4 - Ratio | Arithmetic - Ratio, Proportion & Variation**

A and B invests Rs. 3000 and Rs. 4000 respectively in a business. If A doubles his capital after 6 months, in what ratio should A and B divide that year’s profit.

- (a)
3 : 10

- (b)
9 : 8

- (c)
3 : 4

- (d)
4 : 3

Answer: Option B

**Explanation** :

(3000 × 6 + 6000 × 6) : 4000 × 12

⇒ 18000 + 36000 : 48000

⇒ 54000/48000 = 9/8

Hence, option (b).

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**PE 4 - Ratio | Arithmetic - Ratio, Proportion & Variation**

The distance travelled by a freely falling body is directly proportional to the square of the time taken. If a body falls 250 m in 5 seconds, then find the distance that the body fell in the 6th second.

Answer: 110

**Explanation** :

d ∝ t^{2}

⇒ d_{1}/(t_{1}^{2} ) = d_{2}/(t_{2}^{2})

⇒ 250/((5)^{2} )=d_{2}/((6)^{2})

⇒ d_{2 }= 10 × 36 = 360 m

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

Hence, 110.

Workspace:

**PE 4 - Ratio | Arithmetic - Ratio, Proportion & Variation**

A’s share in a business is 3/4th of B’s share. B got a profit of Rs. 1200 by investing Rs. 40,000 in the business. What will be the ratio of A’s profit of his investment in the business?

- (a)
100 : 3

- (b)
3 : 100

- (c)
3 : 10

- (d)
None

Answer: Option B

**Explanation** :

∵ 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).

Workspace:

**PE 4 - Ratio | Arithmetic - Ratio, Proportion & Variation**

If (a + b) varies directly as (a – b), then (a² + b²) will vary as

- (a)
a

^{2}- b^{2} - (b)
ab

- (c)
Both (A) and (B)

- (d)
None of these

Answer: Option C

**Explanation** :

(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

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

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

⇒ (a^{2 }+ b^{2}) = (a^{2} - b^{2}) × (a constant, let's say t)

∴ (a^{2 }+ b^{2}) = t × (a^{2} - b^{2})

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

∴ (a^{2 }+ b^{2}) α ab

∴ Both (A) and (B) are true.

Hence, option (c).

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**PE 4 - Ratio | Arithmetic - Ratio, Proportion & Variation**

A starts business with a capital of Rs. 1200. B and C join with some investments after 3 and 6 months respectively. If at the end of a year, the profit is divided in the ratio 2 : 3 : 5 respectively. What is B’s investment in the business?

- (a)
Rs. 2400

- (b)
Rs. 1800

- (c)
Rs. 3600

- (d)
Rs. 6000

Answer: Option A

**Explanation** :

1200 × 12 : x × 9 : y × 6 = 2 : 3 : 5

⇒ (1200 × 12) / (x × 9) = 2/3

⇒ x = 2400

Hence, option (a).

Workspace:

**PE 4 - Ratio | Arithmetic - Ratio, Proportion & Variation**

Jet Airline has a free luggage allowance for its passengers. If any passenger carries excess luggage, it is charged at a constant rate per kg. The total luggage charge paid by Ravi and Pranav is Rs. 1100. If both Ravi and Pranav had carried luggage twice the weight than they actually did, their luggage charges would have been Rs. 2000 and Rs. 1000 respectively. What was the charge levied on Ravi’s luggage?

- (a)
Rs. 800

- (b)
Rs. 700

- (c)
Rs. 600

- (d)
Rs. 900

Answer: Option A

**Explanation** :

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).

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