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

The best way to prepare for Arithmetic - Ratio, Proportion & Variation is by going through the previous year Arithmetic - Ratio, Proportion & Variation CAT questions. Here we bring you all previous year Arithmetic - Ratio, Proportion & Variation CAT questions along with detailed solutions.

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26. CAT 2001 QA | Arithmetic - Ratio, Proportion & Variation CAT Question

Three friends, returning from a movie, stopped to eat at a restaurant. After dinner, they paid their bill and noticed a bowl of mints at the front counter. Sita took 1/3 of the mints, but returned four because she had a momentary pang of guilt. Fatima then took 1/4 of what was left but returned three for similar reasons. Eswari then took half of the remainder but threw two back into the bowl. The bowl had only 17 mints left when the raid was over. How many mints were originally in the bowl?

(a)
38
(b)
31
(c)
41
(d)
None of these

Answer: Option D

Text Explanation :

Sita takes 1/3rd of the total mints which implies that the total number of mints in the bowl should be a multiple of 3. None of the options is a multiple of 3.

Hence, option (d).

Alternatively

This problem can be best solved by working backwards.

Number of mints before Eswari = (17 − 2) × 2 = 30

∴ Number of mints before a Fatima = $\frac{(30 - 3) \times 4}{3} = 36$

∴ Number of mints before Sita = $\frac{(36 - 4) \times 3}{2} = 48$

∴ Total number of mints in the bowl = 48

Hence, option (d).

27. CAT 2000 QA | Arithmetic - Ratio, Proportion & Variation CAT Question

A truck travelling at 70 kilometres per hour uses 30% more diesel to travel a certain distance than it does when it travels at the speed of 50 kilometres per hour. If the truck can travel 19.5 kilometres on a litre of diesel at 50 kilometres per hour, how far can the truck travel on 10 litres of diesel at a speed of 70 kilometres per hour?

(a)
130
(b)
140
(c)
150
(d)
175

28. CAT 1999 QA | Arithmetic - Ratio, Proportion & Variation CAT Question

The speed of a railway engine is 42 kmph when no compartment is attached, and the reduction in speed is directly proportional to the square root of the number of compartments attached. If the speed of the train carried by this engine is 24 kmph when 9 compartments are attached, the maximum number of compartments that can be carried by the engine is

(a)
49
(b)
48
(c)
46
(d)
47

Answer: Option B

Text Explanation :

18 ∝ $\sqrt{9}$

42 ∝ $\sqrt{x};$ Here x = number of compartments

$\frac{18}{42} = \frac{\sqrt{9}}{\sqrt{x}}$

Simplifying, x = 49, but this is with reference to maximum speed. Hence number of compartments would be one less in order to run i.e. 48.

Hence, option (b).

29. CAT 1999 QA | Arithmetic - Ratio, Proportion & Variation CAT Question

Total expenses of a boarding house are partly fixed and partly varying linearly with the number of boarders. The average expense per boarder is Rs. 700 when there are 25 boarders and Rs. 600 when there are 50 boarders. What is the average expense per boarder when there are 100 boarders?

(a)
550
(b)
580
(c)
540
(d)
570

30. CAT 1998 QA | Arithmetic - Ratio, Proportion & Variation CAT Question

I have one-rupee coins, 50-paisa coins and 25-paisa coins. The number of coins are in the ratio 2.5 : 3 : 4. If the total amount with me is Rs. 210, find the number of one-rupee coins.

(a)
90
(b)
85
(c)
100
(d)
105

Answer: Option D

Text Explanation :

Since the number of coins are in the ratio 2.5 : 3 : 4, the values of the coins will be in the ratio
(1 × 2.5) : (0.5 × 3) : (0.25 × 4) = 2.5 : 1.5 : 1 or 5 : 3 : 2
Since they totally amount to Rs. 210, if the value of each type of coins are assumed to be 5x, 3x and 2x, the average value per coin will be $\frac{210}{10 x}$ .

So the total value of one-rupee coins will be $5 \times (\frac{210}{10 x})$ = Rs. 105

So the total number of one-rupee coins will be 105.

31. CAT 1998 QA | Arithmetic - Ratio, Proportion & Variation CAT Question

Direction: Each question is followed by two statements, I and II. Answer the questions based on the statements and mark the answer as
1. if the question can be answered with the help of any one statement alone but not by the other statement.
2. if the question can be answered with the help of either of the statements taken individually.
3. if the question can be answered with the help of both statements together.
4. if the question cannot be answered even with the help of both statements together.

What is the value of ‘a’?
I. Ratio of a and b is 3 : 5, where b is positive.
II. Ratio of 2a and b is $\frac{12}{10},$ where a is positive.

Backspace

7 8 9 4 5 6 1 2 3 0 . -

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32. CAT 1997 QA | Arithmetic - Ratio, Proportion & Variation CAT Question

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

Text Explanation :

This equation is very straightforward. If the number is 'x', then $\frac{7 x}{8} - \frac{7 x}{18} = 770 .$ 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.

33. CAT 1997 QA | Arithmetic - Ratio, Proportion & Variation CAT Question

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

Text Explanation :

Let D₁, T₁ and D₂, T₂ denote the diameters and the thickness of the two coins respectively. If V₁ and V₂are the values of the two coins.

$\frac{V_{1}}{V_{2}} = (\frac{{D_{1}}^{2} T_{1}}{{D_{2}}^{2} T_{2}}) = {(\frac{D_{1}}{D_{2}})}^{2} (\frac{T_{1}}{T_{2}})$

Therefore, $\frac{4}{1} = {(\frac{4}{3})}^{2} (\frac{T_{1}}{T_{2}}) \Rightarrow (\frac{T_{1}}{T_{2}}) = \frac{9}{4}$

34. CAT 1996 QA | Arithmetic - Ratio, Proportion & Variation CAT Question

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

35. CAT 1996 QA | Arithmetic - Ratio, Proportion & Variation CAT Question

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

Text 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.

36. CAT 1993 QA | Arithmetic - Ratio, Proportion & Variation CAT Question

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

Text Explanation :

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

According to the question,

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

⇒ 2x - y = 6y - 3y

⇒ 2x = 4y

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

Hence, option (a).

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