Practice Questions for Arithmetic - Ratio, Proportion & Variation

1. PE 2 - Ratio | Arithmetic - Ratio, Proportion & Variation

Divide Rs. 6270 among A, B and C so that A shall receive 3/7 of as much as B and C together and B shall receive 2/9 of as much as A and C together:

(a)
1881, 1240, 3149
(b)
1881, 1040, 3349
(c)
1881, 1250, 3139
(d)
1881, 1140, 3249

2. PE 2 - Ratio | Arithmetic - Ratio, Proportion & Variation

If $\frac{x}{a}$ = $\frac{y}{b}$ = $\frac{z}{c}$ , then $\frac{a x - b y}{(a + b) (x -y)}$ + $\frac{b y - c z}{(b + c) (y - z)}$ + $\frac{c z - a x}{(c + a) (z - x)}$ is equal to

(a)
2
(b)
3
(c)
1
(d)
0

Answer: Option B

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Explanation :

Given: $\frac{x}{a}$ = $\frac{y}{b}$ = $\frac{z}{c}$ = k (say)

x = ak, y = bk, z = ck

$\frac{a x - b y}{(a + b) (x - y)}$ + $\frac{b y - c z}{(b + c) (y - z)}$ + $\frac{c z - a x}{(c + a) (z - x)}$

= $\frac{a \times a k - b \times b k}{(a + b) (a k - b k)}$ + $\frac{b \times b k - c \times c k}{(b + c) (b k - c k)}$ + $\frac{c \times c k - a \times a k}{(c + a) (c k - a k)}$

= $\frac{k (a^{2} - b^{2})}{k (a^{2} - b^{2})}$ + $\frac{k (b^{2} - c^{2})}{k (b^{2} - c^{2})}$ + $\frac{k (c^{2} - a^{2})}{k (c^{2} - a^{2})}$

= 1 + 1 + 1 = 3

Hence, option (b).

3. PE 2 - Ratio | Arithmetic - Ratio, Proportion & Variation

If a/b = c/d = e/f, then the value of which of the following is equal to the value of $\frac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}}$

(a)
1/2
(b)
$\frac{a + c + e}{b + d + f}$
(c)
$\frac{a c e}{b d f}$
(d)
None of these

Answer: Option C

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Explanation :

If a/b = c/d = e/f, then

$\frac{a^{3}}{b^{3}} = \frac{c^{3}}{d^{3}} = \frac{e^{3}}{f^{3}}$ which is also equal to $\frac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}}$

(When 2 or more ratios are equal, they are also equal to sum of their Numerators divided by sum of their Denominators)

Now, $\frac{a^{3} + c^{3} + e^{3}}{b^{3} + d^{3} + f^{3}}$ = $\frac{a^{3}}{b^{3}}$ = $\frac{a}{b} \times \frac{a}{b} \times \frac{a}{b}$ = $\frac{a}{b} \times \frac{c}{d} \times \frac{e}{f}$ = $\frac{a c e}{b d f}$

Hence, option (c).

4. PE 2 - Ratio | Arithmetic - Ratio, Proportion & Variation

Between two stations the first, second and third class fares were fixed in the ratio 8 : 6 : 3, but afterwards the first class fares were reduced by 1/6 and the second class by 1/12. In a year, number of first, second and third class passengers were respectively 9 : 12 : 26 and the money at the booking offices was Rs. 1088. How much was paid by the first class passengers?

(a)
10 × 2²
(b)
10 × 2³
(c)
10 × 2⁴
(d)
10 × 2⁵

Answer: Option D

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Explanation :

Fare ratio before reduction 8 : 6 : 3

After reduction in first class fare by 1/6 (i.e. actual fare is now 5/6 of original fare), we have (8 × 5)/6 ∶ 6 ∶ 3

⇒ 40/6 ∶ 6 ∶ 3

After reduction of second class fare by 1/12 (i.e. actual fare is now 11/12 of original fare), we have 40/6 ∶ (6 × 11)/12 ∶ 3

⇒ 40/6 ∶ 11/2 ∶ 3 = 80/6 ∶ 11 ∶ 6 = 80 ∶ 66 ∶ 36 = 40 ∶ 33 ∶ 18

Amount of fare collected depends upon fare/passenger×no.of passengers. Thus, combined fare collected ratio is

40 × 9 : 12 × 33 : 26 × 18 = 40 × 3 : 12 × 11 : 26 × 6

⇒ 40 : 4 × 11 : 26 × 2 = 40 : 44 : 52

Total first class fare collected then is $\frac{40}{40 + 44 + 52} \times 1088$ = 40/136 × 1088 = 40 × 8 = Rs. 320 or 10 × 2⁵

Hence, option (d).

5. PE 2 - Ratio | Arithmetic - Ratio, Proportion & Variation

There are two persons P and Q and P’s present age is less than Q’s present age. Which of the following will be the relation between cubes of the ratio of their ages after 4 years (M) and the ratio of their ages after 8 years (N)?

(a)
M < N
(b)
M = N
(c)
M > N
(d)
M ≤ N

Answer: Option A

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Explanation :

Let the present ages of A and B be a years and b years respectively. Given p < q

$\frac{p}{q} < \frac{p + 4}{q + 4} < \frac{p + 8}{q + 8}$

Hence, ${(\frac{p + 4}{q + 4})}^{3} < {(\frac{p + 8}{q + 8})}^{3}$

⇒ M < N

Hence, option (a).

6. PE 2 - Ratio | Arithmetic - Ratio, Proportion & Variation

The ratio of the present ages of a man and his wife is 6 : 5. Which of the following can’t be a possible ratio of their ages 10 years ago?

(a)
8 : 7
(b)
3 : 2
(c)
13 : 10
(d)
7 : 5

Answer: Option A

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Explanation :

Let their present ages be 6x years and 5x years respectively.

Ratio of their ages 10 years ago = $\frac{6 x - 10}{5 x - 10} = \frac{6 (x - 2) + 2}{5 (x - 2)}$ which is more than 6/5.

Only choice (a) violates this condition.

Hence, option (a).

7. PE 2 - Ratio | Arithmetic - Ratio, Proportion & Variation

The reduction in speed of a locomotive engine is directly proportional to square root of no. of wagons attached to it. Speed of engine was 240 km/hr when no wagons were attached to it. When 49 wagons were attached the speed became 100 km/hr. Maximum how many wagons can be attached to engine such that engine keeps on moving?

(a)
145
(b)
144
(c)
143
(d)
147

8. PE 2 - Ratio | Arithmetic - Ratio, Proportion & Variation

A sick man’s will stipulated that if his pregnant wife gave birth to a boy, then she should get two-fifth of his property and the child should get the remaining three-fifth. On the other hand, if she gave birth to a girl child, then one-fourth should go to his wife and the remaining three-fourth to the child. The sick man died after a few weeks. His wife delivered twins, a boy and a girl. How much should the wife get if his net-worth is Rs. 1.32 crores.

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9. PE 2 - Ratio | Arithmetic - Ratio, Proportion & Variation

A certain sum is divided among A, B and C in a manner that for every rupee that A gets, B gets 50 paise and for every rupee that B gets, C gets 25 paise. If C’s share in the total sum is Rs. 520, then find the share of A (in Rs.).

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10. PE 2 - Ratio | Arithmetic - Ratio, Proportion & Variation

Three friends count the number of coins they have. They found that the ratio of the total number of coins of 25 p, 50 p and Re 1 is 2 : 3 : 5. They have 25 p coins in the ratio 1 : 2 : 3, 50 p coins in the ratio 2 : 3 : 4 and Re 1 coins in the ratio 4 : 5 : 6. Find the amount with the second person (in Rs.), if they have money only in the denomination of 25 p, 50 p and Re 1 and the total number of coins with them is 540.

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PE 2 - Ratio | Arithmetic - Ratio, Proportion & Variation

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