# Arithmetic - Ratio, Proportion & Variation - Previous Year CAT/MBA Questions

You can practice all previous year OMET questions from the topic Arithmetic - Ratio, Proportion & Variation. This will help you understand the type of questions asked in OMET. It would be best if you clear your concepts before you practice previous year OMET questions.

**XAT 2020 QADI | Arithmetic - Ratio, Proportion & Variation**

A box contains 6 cricket balls, 5 tennis balls and 4 rubber balls. Of these, some balls are defective. The proportion of defective cricket balls is more than the proportion of defective tennis balls but less than the proportion of defective rubber balls. Moreover, the overall proportion of defective balls is twice the proportion of defective tennis balls. What BEST can be said about the number of defective rubber balls in the box?

- A.
It is exactly 3

- B.
It is either 3 or 4

- C.
It is exactly 2

- D.
It is either 2 or 3

- E.
It is either 0 or 1

Answer: Option A

**Explanation** :

Let the number of defective cricket, tennis and rubber balls = c, t and r respectively.

Given, r/4 > c/6 > t/5 ...(1)

and (c + t + r)/15 = 2 × t/5

⇒ c + r = 5t

Also, maximum value of c + r can be 5 + 4 = 9

⇒ The value of t can only be 1.

⇒ c + r = 5 …(2)

The only possible value to satisfy (1) and (2) is r = 3 and c = 2.

Hence, option (a).

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**XAT 2020 QADI | Arithmetic - Ratio, Proportion & Variation**

X, Y and Z start a web-based venture together. X invests Rs. 2.5 lakhs, Y invests Rs. 3.5 lakhs, and Z invests Rs. 4 lakhs. In the first year, the venture makes a profit of Rs. 2 lakhs. A part of the profit is shared between Y and Z in the ratio of 2 : 3, and the remaining profit is divided among X, Y and Z in the ratio of their initial investments.

The amount that Z receives is four times the amount that X receives. How much amount does Y receive?

- A.
Rs. 80,200

- B.
Rs. 75,000

- C.
Rs. 93,750

- D.
Rs. 74,250

- E.
Rs. 1,02,500

Answer: Option B

**Explanation** :

Let Rs. 5P is divided between Y and Z in the ratio of 2 : 3.

∴ Y received 2P and Z received 3P.

Remaining profit = 2,00,000 – 5P

This will be dived amongst X, Y and Z in the ratio 2.5 : 3.5 : 4 = 5 : 7 : 8.

∴ X receives 5/20 × (2,00,000 – 5P) and Z receives 8/20 × (2,00,000 – 5P)

Total amount received by X = 5/20 × (2,00,000 – 5P)

Total amount received by Z = 3P + 8/20 × (2,00,000 – 5P)

According to the question:

3P + 8/20 × (2,00,000 – 5P) = 4(5/20 × (2,00,000 – 5P))

⇒ 3P + 80,000 – 2P = 2,00,000 – 5P

⇒ 6P = 1,20,000

⇒ P = 20,000

∴ Y receives = 2P + 7/20 × (2,00,000 – 5P) = 40,000 + 7/20 × 1,00,000 = 75,000

Hence, option (b).

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**XAT 2019 QADI | Arithmetic - Ratio, Proportion & Variation**

Two numbers a and b are inversely proportional to each other. If a increases by 100%, then b decreases by

- A.
200%

- B.
50%

- C.
100%

- D.
80%

- E.
150%

Answer: Option B

**Explanation** :

Given that ‘a’ and ‘b’ are inversely proportional

∴ a ∝ 1/b

⇒ a × b = constant

Hence, if a is doubled (i.e., increased by 100%) b needs to become half.

∴ % change in b = (1/2 - 1) × 100% = 50%.

Hence, option (b).

Workspace:

**XAT 2018 QADI | Arithmetic - Ratio, Proportion & Variation**

The number of boys in a school was 30 more than the number of girls. Subsequently, a few more girls joined the same school. Consequently, the ratio of boys and girls became 3:5. Find the minimum number of girls, who joined subsequently.

- A.
31

- B.
51

- C.
52

- D.
55

- E.
Solution not possible

Answer: Option C

**Explanation** :

Let the number of girls in the school initially be ‘g’.

∴ Number of boys = g + 30.

If ‘x’ new girls who joined the school.

⇒ (g + 30)/(g + x) = 3/5

∴ 5g + 150 = 3g + 3x

∴ x = 2g/3 + 50

For x to be minimum g should be minimum. Also, g should be a multiple of 3 for x to be an integer.

⇒ Minimum value of g = 3

∴ The minimum value of x = 52.

Hence, option (c).

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**IIFT 2018 QA | Arithmetic - Ratio, Proportion & Variation**

Land Cruiser Prado, the latest SUV from Toyota Motors, consumes diesel at the rate of $\frac{1}{400}\left\{\frac{1000}{x}+x\right\}$ per km, when travelling at the speed of x km/hr. The diesel costs Rs. 65 per litre and the driver is paid Rs. 50 per hour. Find the steady speed that will minimize the total cost of a 1000 km trip?

- A.
33 km/hr

- B.
36 km/hr

- C.
39 km/hr

- D.
52 km/hr

Answer: Option B

**Explanation** :

Total cost incurred = total cost incurred due to petrol + total cost incurred due to driver charges

= (total litres consumed × cost per litre) + (total driver hours × cost per hour)

Total petrol consumed in 1000 km = 1000 × $\frac{1}{400}\left\{\frac{1000}{x}+x\right\}=2.5\left\{\frac{1000}{x}+x\right\}$

∴ Total petrol cost = $2.5\left\{\frac{1000}{x}+x\right\}\times 65$

$=162.5\times \left\{\frac{1000}{x}+x\right\}$

Time taken to travel 1000 km = 1000/x

∴ Total driver cost = (1000/x) × 50 = Rs. (50000/x)

∴ Total cost = 162.5 × $\left\{\frac{1000}{x}+x\right\}+\left(\frac{50000}{x}\right)$

$=\left(\frac{1000}{x}\right)(162.5+50)+162.5x$

$=\frac{212500}{x}+162.5x$

You can now substitute options and verify that the least cost is for x = 36.

Hence, option 2.

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**IIFT 2018 QA | Arithmetic - Ratio, Proportion & Variation**

A physical therapist of Russian football team knows that the team will play 40% of its matches on artificial turf, this season. Because of his vast experience, he knows that a football player’s chances of incurring a knee injury is 50% higher if he is playing on artificial turf instead of grass. If the player's chances of a knee injury is 0.42, what is the probability that a football player with knee injury, incurred the injury while playing on grass?

- A.
0.28

- B.
0.336

- C.
0.5

- D.
None of the above

Answer: Option C

**Explanation** :

P(playing on artificial turf) = 0.4

∴ P(playing on grass) = 1 – P(playing on artificial turf) = 1 – 0.4 = 0.6

P(injury/artificial turf) = 0.42

Also, P(injury/artificial turf) = 1.5 × P(injury/grass)

∴ P(injury/grass) = 0.42/1.5 = 0.28

Now, P(injury) = P(injury/grass) × P(playing on grass) + P(injury/artificial turf) × P(playing on artificial turf)

= (0.28 × 0.6) + (0.42 × 0.4)

= 0.168 + 0.168 = 0.336

Also, P(injury/grass) × P(playing on grass) = P(grass/injury) × P(injury)

∴ P(grass/injury) = 0.168/0.336 = 0.5

Hence, option 3.

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**IIFT 2017 QA | Arithmetic - Ratio, Proportion & Variation**

Suntex Company plans to manufacture a new product line of Razor next year and sell it at a price of Rs. 12 per unit. The variable costs per unit in each production run is estimated to be 50% of the selling price, and the fixed costs for each production run is estimated to be Rs. 50,400. Based om their estimated costs how many units of the new product will company Suntex need to manufacture and sell in order for their revenue to be equal to their total costs for each production run?

- A.
5400

- B.
4200

- C.
8400

- D.
2100

Answer: Option C

**Explanation** :

Let n units be sold.

Variable cost per unit = 0.5 × 12 = Rs. 6 per unit

∴ Total cost for one production run = Rs. (50400 + 6n)

Total revenue by selling n units = Rs. (12n)

When cost = revenue; 50400 + 6n = 12n

∴ 6n = 50400 i.e. n = 8400

Hence, option 3.

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**IIFT 2015 QA | Arithmetic - Ratio, Proportion & Variation**

Sailesh is working as a sales executive with a reputed FMCG Company in Hyderabad. As per the Company’s policy, Sailesh gets a commission of 6% on all sales upto Rs.1,00,000 and 5% on all sales in excess of this amount. If Sailesh remits Rs. 2,65,000 to the FMCG company after deducting his commission, his total slaes were worth:

- A.
Rs. 1,20,000

- B.
Rs. 2,90,526

- C.
Rs. 2,21,054

- D.
Rs. 2,80,000

Answer: Option D

**Explanation** :

Sailesh earns Rs. 6,000 as commission from first Rs. 1,00,000.

Let his total sales = X

∴ Total commission = 6000 + 0.05 (X – 100000) = 0.05X + 1000

∴ X – 0.05X + 1000 = 265000

Solving, we get X = 280000

Hence, option 4.

Workspace:

**XAT 2012 QA | Arithmetic - Ratio, Proportion & Variation**

Ram and Shyam form a partnership (with Shyam as working partner) and start a business by investing Rs. 4000 and Rs. 6000 respectively. The conditions of partnership were as follows:

- In case of profits till Rs. 200,000 per annum, profits would be shared in the radio of the invested capital.
- Profits from Rs. 200,001 till Rs. 400,000 Shyam would take 20% out of the profit, before the division of remaining profits, which will then be based on ratio of invested capital.
- Profits in excess of Rs. 400,000, Shyam would take 35% out of the profits beyond Rs. 400,000, before the division of remaining profits, which will then be based on ratio of invested capital.

If Shyam’s share in a particular year was Rs. 367000, which option indicates the total business profit (in Rs.) for that year?

- A.
520,000

- B.
530,000

- C.
540,000

- D.
550,000

- E.
None of the above

Answer: Option D

**Explanation** :

For first Rs. 200000, Shyam gets, 6000/(4000 + 6000) × 100 = 60% of the profit.

For next Rs. 200000, he gets 20% + plus 60% of the remaining profit.

i.e. 20% + 80 × 0.6% = 68%

Similarly, for a profit margin greater than Rs. 400000, he will get, 35% + 65 × 0.6% = 74% of the profits beyond Rs. 400000

Now, for a profit of first Rs. 400000, Shyam will receive 200000 × (68 + 60)/100 = 256000

But Shyam earns a total profit of 367000.

Let total profit earned by them be Rs. 400000 + x.

Hence, Shyam received 367000 – 256000 = Rs. 110000 from Rs. x profit.

i.e. Rs. 110000 is 74% of x.

Hence, x = 110000/0.74 = 150000

Hence, total profit earned by them = 400000 + 150000 = Rs. 550000

Hence, option 4.

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