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

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

**CAT 1997 QA | Arithmetic - Ratio, Proportion & Variation**

A student instead of finding the value of $\frac{7}{8}$ of a number, found the value of $\frac{7}{18}$ of the number. If his answer differed from the actual one by 770, find the number.

- A.
1584

- B.
2520

- C.
1728

- D.
1656

Answer: Option A

**Explanation** :

This equation is very straightforward. If the number is 'x', then $\frac{7x}{8}-\frac{7x}{18}=770.$ On solving this equation, we get x = 1584. Hint: Students please note that if the difference in $\frac{7}{8}$ and $\frac{7}{18}$ of a number is 770, then the difference in $\frac{1}{8}$ and $\frac{1}{18}$ of the number should be 110. If we express this as an equation, we get $\frac{x}{8}-\frac{x}{18}=110$

or 10x = 110 × 18 × 8

or x = 11 × 18 × 8

You can further proceed from here in two ways: (i) the last digit of the required answer should be (1 × 8 × 8) = 4, (ii) number should be divisible by 11. In both cases, the answer that is obtained from the given choices is 1584.

Workspace:

**CAT 1997 QA | Arithmetic - Ratio, Proportion & Variation**

The value of each of a set of coins varies as the square of its diameter, if its thickness remains constant, and it varies as the thickness, if the diameter remains constant. If the diameter of two coins are in the ratio 4 : 3, what should be the ratio of their thickness if the value of the first is four times that of the second?

- A.
16 : 9

- B.
9 : 4

- C.
9 : 16

- D.
4 : 9

Answer: Option B

**Explanation** :

Let D_{1}, T_{1} and D_{2}, T_{2} denote the diameters and the thickness of the two coins respectively. If V_{1} and V_{2 }are the values of the two coins.

$\frac{{V}_{1}}{{V}_{2}}=\left(\frac{{{D}_{1}}^{2}{T}_{1}}{{{D}_{2}}^{2}{T}_{2}}\right)={\left(\frac{{D}_{1}}{{D}_{2}}\right)}^{2}\left(\frac{{T}_{1}}{{T}_{2}}\right)$

Therefore, $\frac{4}{1}={\left(\frac{4}{3}\right)}^{2}\left(\frac{{T}_{1}}{{T}_{2}}\right)\Rightarrow \left(\frac{{T}_{1}}{{T}_{2}}\right)=\frac{9}{4}$

Workspace:

**CAT 1996 QA | Arithmetic - Ratio, Proportion & Variation**

The cost of diamond varies directly as the square of its weight. Once, this diamond broke into four pieces with weights in the ratio 1 : 2 : 3 : 4. When the pieces were sold, the merchant got Rs. 70,000 less. Find the original price of the diamond.

- A.
Rs. 1.4 lakh

- B.
Rs. 2 lakh

- C.
Rs. 1 lakh

- D.
Rs. 2.1 lakh

Answer: Option C

**Explanation** :

Let the original weight of the diamond be 10x. Hence, its original price will be k(100x2) . . . where k is a constant.

The weights of the pieces after breaking are x, 2x, 3x and 4x. Therefore, their prices will be kx^{2}, 4kx^{2}, 9kx^{2} and 16kx^{2}. So the total price of the pieces = (1 + 4 + 9 + 16)kx^{2 }= 30kx^{2}. Hence, the difference in the price of the original diamond and its pieces = 100kx^{2} – 30kx^{2} = 70kx^{2} = 70000.

Hence, kx^{2} = 1000 and the original price = 100kx^{2 }= 100 × 1000 = 100000 = Rs. 1 lakh.

Workspace:

**CAT 1996 QA | Arithmetic - Ratio, Proportion & Variation**

Out of two-thirds of the total number of basketball matches, a team has won 17 matches and lost 3 of them. What is the maximum number of matches that the team can lose and still win more than threefourths of the total number of matches, if it is true that no match can end in a tie?

- A.
4

- B.
6

- C.
5

- D.
3

Answer: Option A

**Explanation** :

The team has played a total of (17 + 3) = 20 matches. This constitutes $\frac{2}{3}$ of the matches. Hence, total number of matches played = 30. To win $\frac{3}{4}$ of them, a team has to win 22.5, i.e. at least win 23 of them. In other words, the team has to win a minimum of 6 matches (since it has already won 17) out of remaining 10. So it can lose a maximum of 4 of them.

Workspace:

**CAT 1993 QA | Arithmetic - Ratio, Proportion & Variation**

From each of the two given numbers, half the smaller number is subtracted. Of the resulting numbers the larger one is three times as large as the smaller. What is the ratio of the two numbers?

- A.
2 : 1

- B.
3 : 1

- C.
3 : 2

- D.
None

Answer: Option A

**Explanation** :

Let the two given numbers be x and y such that x > y.

According to the question,

x - $\frac{y}{2}=3\left(y-\frac{y}{2}\right)$

$\Rightarrow \frac{x}{y}=\frac{2}{1}.$

Workspace:

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