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

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**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

Answer: Option D

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**Explanation** :

Here the simple fact to recognize is to interpret the problem in terms of ratios. When A receives 3/7 of B + C, means, if B + C receive 1 Rs., then, A receives 3/7th of the rupee.

Thus, A/(B + C) = 3/7

⇒ A/(6270 - A) = 3/7

(∴ A + B + C = 6270)

⇒ A = Rs. 1881

Similarly. B/(A + C) = 2/9

⇒ B/(6270 - B) = 2/9

⇒ B = Rs. 1140

∴ C = Rs. 3249

Hence, option (d).

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

If $\frac{x}{a}$ = $\frac{y}{b}$ = $\frac{z}{c}$, then $\frac{ax-by}{(a+b)(x-y)}$ + $\frac{by-cz}{(b+c)(y-z)}$ + $\frac{cz-ax}{(c+a)(z-x)}$ 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

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

= $\frac{a\times ak-b\times bk}{(a+b)(ak-bk)}$ + $\frac{b\times bk-c\times ck}{(b+c)(bk-ck)}$ + $\frac{c\times ck-a\times ak}{(c+a)(ck-ak)}$

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

= 1 + 1 + 1 = 3

Hence, option (b).

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**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)
$\frac{ace}{bdf}$

- (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, $\frac{{a}^{3}+{c}^{3}+{e}^{3}}{{b}^{3}+{d}^{3}+{f}^{3}}$ = $\frac{{a}^{3}}{{b}^{3}}$ = $\frac{a}{b}\times \frac{a}{b}\times \frac{a}{b}$ = $\frac{a}{b}\times \frac{c}{d}\times \frac{e}{f}$ = $\frac{ace}{bdf}$

Hence, option (c).

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**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 × 2

^{2} - (b)
10 × 2

^{3} - (c)
10 × 2

^{4} - (d)
10 × 2

^{5}

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 × 2^{5}

Hence, option (d).

Workspace:

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

⇒ M < N

Hence, option (a).

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**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 = $\frac{6x-10}{5x-10}=\frac{6(x-2)+2}{5(x-2)}$ which is more than 6/5.

Only choice (a) violates this condition.

Hence, option (a).

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**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

Answer: Option C

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**Explanation** :

Without any wagons, the speed is 240 km/hr. The speed diminishes by a quantity that varies (directly) as the square root of the number of wagons attached (∵ By default, variation is direct).

Let the quantity (by which the speed diminishes) be Q, the number of wagons attached be N.

Q ∝ √N

∴ Q = k√N

When Q = 240 – 100 = 140, N = 49. ∴ k = 20

When the train can just move, its speed = 240 – 20√N, would be slightly more than 0.

∴ 240 – 20√N > 0

∴ √N < 12.

∴ N < 144

∴ The greatest value of N is 143.

Hence, option (c).

Workspace:

**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.

Answer: 2400000

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**Explanation** :

If a son if born wife and the son will receive amounts in the ratio 2 : 3.

If a girl if born wife and the girl will receive amounts in the ratio 1 : 3.

∴ Mother : Son : Daughter = 2 : 3 : 2 × 3 = 2 : 3 : 6.

Wife's share = 2/11 × 1,32,00,000 = Rs. 24,00,000

Hence, 2400000.

Workspace:

**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.).

Answer: 4160

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**Explanation** :

A : B = 1 : ½ = 2 : 1

B : C = 1 : ¼ = 4 : 1

∴ A : B : C = 8 : 4 : 1

Given C’s share = Rs. 520

A’s share = 8/1 × 520 = Rs.4160

Hence, 4160.

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**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.

Answer: 126

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**Explanation** :

Total number of coins = 540

The ratio of 25 p, 50 p and Re 1 coins is 2 : 3 : 5.

Number of 25 p coins = 2/10 × 540 = 108

Number of 25 p coins = 3/10 × 540 = 162

Number of 25 p coins = 5/10 × 540 = 270

Number of 25 p coins with 2^{nd} person = 2/6 × 108 = 36

Number of 25 p coins with 2^{nd} person = 3/9 × 162 = 54

Number of 25 p coins with 2^{nd} person = 5/15 × 270 = 90

Total amount with 2nd person = 36 × ¼ + 54 × ½ + 90 = 9 + 27 + 90 = 126.

Hence, 126.

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