Arithmetic - Ratio, Proportion & Variation - Previous Year CAT/MBA Questions
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A king has distributed all his rare jewels in three boxes. The first box contains 1/3 of the rare jewels, while the second box contains k/5 of the rare jewels, for some positive integer value of k. The third box contains 66 rare jewels. How many rare jewels does the king have?
- (a)
990
- (b)
660
- (c)
240
- (d)
1080
- (e)
Cannot be determined uniquely from the given information.
Answer: Option A
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Text Explanation :
Workspace:
Mr. Jose buys some eggs. After bringing the eggs home, he finds two to be rotten and throws them away. Of the remaining eggs, he puts five-ninth in his fridge, and brings the rest to his mother’s house. She cooks two eggs and puts the rest in her fridge. If her fridge cannot hold more than five eggs, what is the maximum possible number of eggs bought by Mr. Jose?
- (a)
9
- (b)
17
- (c)
11
- (d)
20
- (e)
29
Answer: Option C
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Text Explanation :
Let the number of eggs bought = 9x + 2
number of eggs left after throwing away 2 = 9x
number of eggs kept in fridge = 5x
number of eggs brought to his mothers' house = 4x
number of eggs left after cooking 2 which are kept in fridge = 4x - 2
Given, 4x - 2 <= 5
⇒ x ≤
Hence the max value of x is 1
Max number of eggs bought = 11
Workspace:
One third of the buses from City A to City B stop at City C, while the rest go non-stop to City B. One third of the passengers, in the buses stopping at City C, continue to City B, while the rest alight at City C. All the buses have equal capacity and always start full from City A. What proportion of the passengers going to City B from City A travel by a bus stopping at City C?
- (a)
- (b)
- (c)
- (d)
- (e)
Answer: Option A
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Text Explanation :
Let us assume there are 9 buses.
3 of them stop at C and 6 go non-stop
Given, One-third of the passengers, in the buses stopping at City C, continue to City B, while the rest alight at City C
⇒ Since all buses have equal capacity. we can say 2 will elite at C and 1 will proceed to B.
Hence required proportion = 1/7
Workspace:
The topmost point of a perfectly vertical pole is marked A. The pole stands on a flat ground at point D. The points B and C are somewhere between A and D on the pole. From a point E, located on the ground at a certain distance from D, the points A, B and C are at angles of 60, 45 and 30 degrees respectively. What is AB : BC : CD?
- (a)
(3 + √3) : (1 + √3) : 1
- (b)
(3 - √3) : 1 : (√3 - 1)
- (c)
1 : 1 : 1
- (d)
(3 - √3) : (√3 - 1) : 1
- (e)
(√3 - 1) : 1 : (3 - √3)
Answer: Option D
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Text Explanation :
Let ED = √3x
In triangle CDE, tan 30 = ⇒ CD = x
In triangle BDE, tan 45 = ⇒ BD = x ⇒ BC = x - x
In triangle ÖDE, tan 60 = ⇒ AD = 3x ⇒ AB = 3x - x
AB : BC : CD = (3 - √3) : (√3 - 1) : 1
Workspace:
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
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Text 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).
Workspace:
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
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Text 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).
Workspace:
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
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Text 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:
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
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Text 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).
Workspace:
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
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Text 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 (d).
Workspace:
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