# 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 2021 QA Slot 1 | Arithmetic - Ratio, Proportion & Variation**

The amount Neeta and Geeta together earn in a day equals what Sita alone earns in 6 days. The amount Sita and Neeta together earn in a day equals what Geeta alone earns in 2 days. The ratio of the daily earnings of the one who earns the most to that of the one who earns the least is

- A.
7 : 3

- B.
3 : 2

- C.
11 : 3

- D.
11 : 7

Answer: Option C

**Explanation** :

Let the earnings of Neeta, Geeta and Sita be ‘n’, ‘g’ and ‘s’ respectively.

The amount Neeta and Geeta together earn in a day equals what Sita alone earns in 6 days

⇒ 6s = n + g …(1)

The amount Sita and Neeta together earn in a day equals what Geeta alone earns in 2 days

⇒ 2g = s + n

⇒ 2(6s – n) = s + n [From (1)]

⇒ 11s = 3n

⇒ n = 11s/3 …(2)

Substituting n = 11s/3 in (1)

⇒ 6s = 11s/3 + g

⇒ 7s/3 = g …(3)

From (2) and (3)

11s/3 > 7s/3 > s

∴ n > g > s

∴ The ratio of the daily earnings of the one who earns the most to that of the one who earns the least is = 11s/3 : s = 11 : 3

Hence, option (c).

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**CAT 2021 QA Slot 2 | Arithmetic - Ratio, Proportion & Variation**

Anil, Bobby and Chintu jointly invest in a business and agree to share the overall profit in proportion to their investments. Anil’s share of investment is 70%. His share of profit decreases by ₹ 420 if the overall profit goes down from 18% to 15%. Chintu’s share of profit increases by ₹ 80 if the overall profit goes up from 15% to 17%. The amount, in INR, invested by Bobby is

- A.
2000

- B.
2400

- C.
2200

- D.
1800

Answer: Option A

**Explanation** :

Let the total investment done by all three of them is I.

When total profit decreases by 3% (from 18% to 15%) Anil’s share decreases by 420.

Decrease in total profit = 3% of I.

Decrease in Anil’s profit will be 70% of decrease in total profit.

⇒ 420 = 70% of 3% of I

⇒ 420 = 70/100 × 3/100 × I

⇒ I = 20,000

Now, when total profit increases by 2% (from 15% to 17%) Charu’s share increases by 80.

Increase in total profit = 2% of I.

Increase in Charu’s profit = C% of I [C% = percentage on investment contribution by Charu]

⇒ 80 = C% of 2% of I

⇒ 80 = C/100 × 2/100 × 20000

⇒ C = 20%

∴ Bobby’s percentage share of investment = 100 – 70 – 20 = 10%

⇒ Investment done by Bobby = 10% of 20,000 = Rs. 2,000

Hence, option (a).

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**CAT 2021 QA Slot 3 | Arithmetic - Ratio, Proportion & Variation**

One part of a hostel’s monthly expenses is fixed, and the other part is proportional to the number of its boarders. The hostel collects ₹ 1600 per month from each boarder. When the number of boarders is 50, the profit of the hostel is ₹ 200 per boarder, and when the number of boarders is 75, the profit of the hostel is ₹ 250 per boarder. When the number of boarders is 80, the total profit of the hostel, in INR, will be

- A.
20000

- B.
20200

- C.
20800

- D.
20500

Answer: Option D

**Explanation** :

Total cost = Fixed cost + Variable cost

Variable cost = k × b

where, b = number of boarders and k = variable cost per boarder.

Total cost = FC + kb

Total collection = 1600b

**Case 1**: Total profit = 200 × 50 = 1600 × 50 – (FC + k × 50) …(1)

**Case 2**: Total profit = 250 × 75 = 1600 × 75 – (FC + k × 75) …(2)

(2) - (1), we get

18750 – 10000 = 1600 × 25 - 25k

⇒ k = 1250 and FC = 7,500

Now, total profit when there are 80 boarders = 1600 × 80 – (7500 + 1250 × 80) = 20,500

Alternately,

Total cost when there are 50 students = (1600 – 200) × 50 = 70,000

Total cost when there are 75 students = (1600 – 250) × 75 = 1,01,250

Due to 25 students cost increases by (101250 - 70000) = 31250

∴ Due to 5 students cost increases by = 31250/5 = 6250

∴ Total cost when there are 80 students = 1,01,250 + 6250 = 1,07,500

Total revenue for 80 students = 1600 × 80 = 1,28,000

∴ Profit = 1,28,000 – 1,07,500 = 20,500.

Hence, option (d).

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**CAT 2020 QA Slot 2 | Arithmetic - Ratio, Proportion & Variation**

A sum of money is split among Amal, Sunil and Mita so that the ratio of the shares of Amal and Sunil is 3 : 2, while the ratio of the shares of Sunil and Mita is 4 : 5. If the difference between the largest and the smallest of these three shares is Rs 400, then Sunil’s share, in rupees, is

Answer: 800

**Explanation** :

Let the amount received by them be a, s and m respectively.

Given, a : s = 3 : 2 and s : m = 4 : 5

∴ a : s : m = 6 : 4 : 5

⇒ Amal’s share = 6x and Sunil’s share = 4x.

Given, difference between the largest and the smallest of these three shares is Rs 400.

⇒ 6x – 4x = 400

⇒ x = 200.

∴ Sunil’s share = 4x = Rs. 800.

Hence, 800.

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**CAT 2019 QA Slot 2 | Arithmetic - Ratio, Proportion & Variation**

In an examination, Rama's score was one-twelfth of the sum of the scores of Mohan and Anjali. After a review, the score of each of them increased by 6. The revised scores of Anjali, Mohan, and Rama were in the ratio 11 : 10 : 3. Then Anjali's score exceeded Rama's score by

- A.
26

- B.
32

- C.
24

- D.
35

Answer: Option B

**Explanation** :

It is given that the scores of Anjali, Mohan and Rama after review were in the ratio 11 : 10 : 3

So, let their values be 11x, 10x and 3x respectively.

It is known that their score increased by 6 after review.

So, scores before review = 11x - 6, 10x - 6 and 3x - 6 respectively

Now, from the data given : (11x - 6 + 10x - 6) × 1/12 = 3x - 6

⇒ 21x - 12 = 36x - 72

⇒ 60 = 15x

⇒ x = 4

So, marks after revision are 44, 40 and 12 respectively.

Therefore, Anjali's score exceeded Rama's by 44 - 12 = 32 marks

Hence, option (b).

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**CAT 2018 QA Slot 1 | Arithmetic - Ratio, Proportion & Variation**

Raju and Lalitha originally had marbles in the ratio 4 : 9. Then Lalitha gave some of her marbles to Raju. As a result, the ratio of the number of marbles with Raju to that with Lalitha became 5 : 6. What fraction of her original number of marbles was given by Lalitha to Raju?

- A.
6/19

- B.
1/5

- C.
7/33

- D.
1/4

Answer: Option C

**Explanation** :

Let Raju and Lalitha originally had (4x) and (9x) marbles. Later Lalitha gave (y) marbles to Raju.

Therefore, $\frac{4x+y}{9x-y}$ = $\frac{5}{6}$

Solving this, we get

y = 21x/11

The required fraction = $\frac{{\displaystyle \frac{21}{11}}x}{9x}$ = $\frac{7}{33}$

Hence, option 3.

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**CAT 2018 QA Slot 2 | Arithmetic - Ratio, Proportion & Variation**

The scores of Amal and Bimal in an examination are in the ratio 11 : 14. After an appeal, their scores increase by the same amount and their new scores are in the ratio 47 : 56. The ratio of Bimal’s new score to that of his original score is

- A.
4 : 3

- B.
3 : 2

- C.
8 : 5

- D.
5 : 4

Answer: Option A

**Explanation** :

Suppose the original scores of Amal and Bimal are 11x and 14x respectively.

Suppose ‘y’ is the increase in their marks.

Therefore the new marks of Amal and Bimal are ‘11x + y’ and ‘14x + y’ respectively.

Therefore, we have $\frac{11x+y}{14x+y}=\frac{47}{56}$

∴ 616x + 56y = 658x + 47y

Solving for x and y, we get 14x = 3y

Therefore Bimal’s old marks = 14x = 3y and his new marks = 14x + y = 4y. Therefore the required ratio = 4:3.

Hence, option 1.

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**CAT 2017 QA Slot 1 | Arithmetic - Ratio, Proportion & Variation**

Suppose, C_{1}, C_{2}, C_{3}, C_{4}, and C_{5} are five companies. The profits made by C_{1}, C_{2}, and C_{3} are in the ratio 9 : 10 : 8 while the profits made by C_{2}, C_{4}, and C_{5} are in the ratio 18 : 19 : 20. If C_{5} has made a profit of Rs 19 crore more than C1, then the total profit (in Rs) made by all five companies is:

- A.
438 crore

- B.
435 crore

- C.
348 crore

- D.
345 crore

Answer: Option A

**Explanation** :

C_{5} – C_{1} = 19.

The numbers above are the actual profits (and not just the ratio).

The total profit = 438 crore.

Hence, option (a).

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**CAT 2017 QA Slot 1 | Arithmetic - Ratio, Proportion & Variation**

A stall sells popcorn and chips in packets of three sizes: large, super, and jumbo. The numbers of large, super, and jumbo packets in its stock are in the ratio 7 : 17 : 16 for popcorn and 6 : 15 : 14 for chips. If the total number of popcorn packets in its stock is the same as that of chips packets, then the numbers of jumbo popcorn packets and jumbo chips packets are in the ratio:

- A.
1 : 1

- B.
8 : 7

- C.
4 : 3

- D.
6 : 5

Answer: Option A

**Explanation** :

Let the total number of popcorn packets in stock be T.

Total number of chips packets in stock = T

Required ratio = $\frac{16}{40}T:\frac{14}{35}T$ = 1 : 1.

Hence, option 1.

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**CAT 2017 QA Slot 2 | Arithmetic - Ratio, Proportion & Variation**

If a, b, c are three positive integers such that a and b are in the ratio 3 : 4 while b and c are in the ratio 2 : 1, then which one of the following is a possible value of (a + b + c)?

- A.
201

- B.
205

- C.
207

- D.
210

Answer: Option C

**Explanation** :

To get a common ratio we take L.C. M(2, 4) = 4

So, we multiply each of b and c by 2

Hence, ratio of b : c = 4 : 2

∴ Ratio of a : b : c = 4 : 2

Now let a = 3x, b = c = 3 : 4 : 2

So a + b + c =3x + 4x + 2x = 9x

As, a, b and c are all positively integers, it implies that the sum of a, b and c will be a positive integer and a multiple of 9.

Out of the given options, only 207 i.e, option [3] is a multiple of 9.

Hence, option 3.

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**CAT 2006 QA | Arithmetic - Ratio, Proportion & Variation**

If $\frac{a}{b}=\frac{1}{3},\frac{b}{c}=2,\frac{c}{d}=\frac{1}{2},\frac{d}{e}=3$ and $\frac{e}{f}=\frac{1}{4}$ then what is the value of $\frac{\mathit{a}\mathit{b}\mathit{c}}{\mathit{d}\mathit{e}\mathit{f}}$?

- A.
$\frac{3}{8}$

- B.
$\frac{27}{8}$

- C.
$\frac{3}{4}$

- D.
$\frac{27}{4}$

- E.
$\frac{1}{4}$

Answer: Option A

**Explanation** :

$\frac{\mathrm{a}}{\mathrm{d}}=\frac{\mathrm{a}}{\mathrm{b}}\times \frac{\mathrm{b}}{\mathrm{c}}\times \frac{\mathrm{c}}{\mathrm{d}}=\frac{1}{3}\times 2\times \frac{1}{2}=\frac{1}{3}$

$\frac{\mathrm{b}}{\mathrm{e}}=\frac{\mathrm{b}}{\mathrm{c}}\times \frac{\mathrm{c}}{\mathrm{d}}\times \frac{\mathrm{d}}{\mathrm{e}}=2\times \frac{1}{2}\times 3=3$

$\frac{\mathrm{c}}{\mathrm{f}}=\frac{\mathrm{c}}{\mathrm{d}}\times \frac{\mathrm{d}}{\mathrm{e}}\times \frac{\mathrm{e}}{\mathrm{f}}=\frac{1}{2}\times 3\times \frac{1}{4}=\frac{3}{8}$

$\therefore \frac{\mathit{a}\mathit{b}\mathit{c}}{\mathit{d}\mathit{e}\mathit{f}}=\frac{1}{3}\times 3\times \frac{3}{8}=\frac{3}{8}$

Hence, option (c).

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**Answer the following question based on the information given below.**

An airline has a certain free luggage allowance and charges for excess luggage at a fixed rate per kg. Two passengers, Raja and Praja have 60 kg of luggage between them, and are charged Rs. 1200 and Rs. 2400 respectively for excess luggage. Had the entire luggage belonged to one of them, the excess luggage charge would have been Rs. 5400.

**CAT 2006 QA | Arithmetic - Ratio, Proportion & Variation**

What is the weight of Praja‘s luggage?

- A.
20 kg

- B.
25 kg

- C.
30 kg

- D.
35 kg

- E.
40 kg

Answer: Option D

**Explanation** :

Let f kg be the free luggage allowance and let Raja and Praja have r kg and p kg excess luggage respectively.

Let x be the fixed rate per kg for excess luggage.

∴ 2f + r + p = 60 ... (i)

rx = 1200 ... (ii)

px = 2400 ... (iii)

(60 – f)x = 5400 ... (iv)

From (ii) and (iii),

p = 2r ... (v)

Substituting in (i),

2f + 3r = 60

∴ f = 30 – 3r/2 ... (vi)

Substituting in (iv),

(60 – 30 + 3r/2)x = 5400

∴ 30x + 3rx/2 = 5400

From (ii),

rx = 1200

∴30x = 3600

∴ x = 120

∴ r = 10, p = 20 and f = 15

∴ Weight of Praja's luggage = p + f = 35 kg

Hence, option (d).

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**CAT 2006 QA | Arithmetic - Ratio, Proportion & Variation**

What is the free luggage allowance?

- A.
10 kg

- B.
15 kg

- C.
20 kg

- D.
25 kg

- E.
30 kg

Answer: Option B

**Explanation** :

As calculated in the solution to the previous question, f = 15 kg

Hence, option (b).

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**CAT 2004 QA | Arithmetic - Ratio, Proportion & Variation**

If $\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=r,$ then r cannot take any value except _________.

- A.
$\frac{1}{2}$

- B.
-1

- C.
$\frac{1}{2}$ or -1

- D.
$-\frac{1}{2}$ or -1

Answer: Option C

**Explanation** :

$\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{c+a}=r$

By property of equal ratios,

$\frac{a+b+c}{b+c+c+a+a+b}=r$

$\therefore \frac{a+b+c}{2(a+b+c)}=r$

Assuming (*a *+ *b *+ *c *≠ 0),

∴ r = $\frac{1}{2}$

If a + b + c = 0, a = –(b + c)

Then, $r=\frac{a}{b+c}=\frac{-(b+c)}{(b+c)}=-1$

Hence, option 3.

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**CAT 2003 QA - Retake | Arithmetic - Ratio, Proportion & Variation**

In a coastal village, every year floods destroy exactly half of the huts. After the flood water recedes, twice the number of huts destroyed are rebuilt. The floods occurred consecutively in the last three years namely 2001, 2002 and 2003. If floods are again expected in 2004, the number of huts expected to be destroyed is:

- A.
Less than the number of huts existing at the beginning of 2001.

- B.
Less than the total number of huts destroyed by floods in 2001 and 2003.

- C.
Less than the total number of huts destroyed by floods in 2002 and 2003.

- D.
More than the total number of huts built in 2001 and 2002.

Answer: Option C

**Explanation** :

Let the number of huts at the beginning of 2001 be n.

From the table, it is clear that only option 3 is satisfied as per the above table.

Hence, option 3

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**CAT 2003 QA - Retake | Arithmetic - Ratio, Proportion & Variation**

Let a, b, c, d and e be integers such that a = 6b = 12c, and 2b = 9d = 12e. Then which of the following pairs contains a number that is not an integer?

- A.
$\left[\frac{a}{27},\frac{b}{e}\right]$

- B.
$\left[\frac{a}{36},\frac{c}{e}\right]$

- C.
$\left[\frac{a}{12},\frac{bd}{18}\right]$

- D.
$\left[\frac{a}{6},\frac{c}{d}\right]$

Answer: Option D

**Explanation** :

Since a = 6b = 12c and 2b = 9d = 12e

∴ a : b : c = 12 : 2 : 1 and b : d : e = 18 : 4 : 3

∴ a : b : c : d : e = 108 : 18 : 9 : 4 : 3

∴ a = 108k; b = 18k; c = 9k; d = 4k and e = 3k where ‘k’ is an integer

Option 1 is: $\left(\frac{a}{27},\frac{b}{e}\right)$ = (4k, 6)

Option 2 is: $\left(\frac{a}{36},\frac{c}{e}\right)$ = (3k, 3)

Option 3 is: $\left(\frac{a}{12},\frac{bd}{18}\right)$ = (9k, 4k^{2})

Option 4 is: $\left(\frac{a}{6},\frac{c}{d}\right)=\left(18k,\frac{9}{4}\right)$

The 4th option contains $\frac{9}{4}$, which is not an integer.

Hence, option 4.

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**CAT 2002 QA | Arithmetic - Ratio, Proportion & Variation**

**Each question is followed by two statements A and B. Answer each question using the following instructions:**

**Answer (1) **if the question can be solved by any one of the statements, but not the other one.

**Answer (2) **if the question can be solved by using either of the two statements.

**Answer (3) **if the question can be solved by using both the statements together and not by any one of them.

**Answer (4) **if the question cannot be solved with the help of the given data and more data is required.

A sum of Rs. 38,500 was divided among Jagdish, Punit and Girish. Who received the minimum amount?

A. Jagdish received 2/9 of what Punit and Girish together received.

B. Punit received 3/11 of what Jagdish and Girish together received.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

**Explanation** :

**Consider statement A:**

Jagdish : (Punit + Girish) = 2 : 9

∴ Jagdish's share = $\frac{2}{11}$ × Total = 18.18% of the total

However, we don’t know the percentage distribution between Punit and Girish.

∴ Statement A alone is not sufficient.

**Consider statement B:**

Punit : (Jagdish + Girish) = 3 : 11

∴ Punit's share = $\frac{3}{14}$ × Total = 21.4% of the total

However, we do not know the percentage distribution between Jagdish and Girish.

∴ Statement B alone is not sufficient.

**Consider both statements together:**

Jagdish = 18.18% of the total amount

Punit = 21.4% of the total amount

Girish = 100 − 18.18 − 21.4 = 60.42% of the total amount

∴ Jagdish received the minimum amount.

Hence, option 3.

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**CAT 2001 QA | Arithmetic - Ratio, Proportion & Variation**

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

**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 4.

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 4.

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**CAT 2000 QA | Arithmetic - Ratio, Proportion & Variation**

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

Answer: Option C

**Explanation** :

At 50 km/hr, truck covers 19.5 km in 1 litre diesel.

∴ The truck will use 1.3 litre diesel to cover 19.5 km at 70 km/hr.

∴ At 70 km/hr, in 10 litres of diesel, the truck will cover (19.5 × 10)/1.3 = 150 km

Hence, option 3.

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**CAT 1999 QA | Arithmetic - Ratio, Proportion & Variation**

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

**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).

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**CAT 1999 QA | Arithmetic - Ratio, Proportion & Variation**

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

Answer: Option A

**Explanation** :

Let x be the fixed cost and y the variable cost 17500 = x + 25y … (i)

30000 = x + 50y … (ii)

Solving the equation (i) and (ii), we get

x = 5000, y = 500

Now if the average expense of 100 boarders be ‘A’.

Then

100 × A = 5000 + 500 × 100

∴ A = 550.

Hence, option (a).

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**CAT 1998 QA | Arithmetic - Ratio, Proportion & Variation**

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

**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 $\frac{210}{10x}$.

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

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

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**CAT 1998 QA | Arithmetic - Ratio, Proportion & Variation**

**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.

Answer: 4

**Explanation** :

Note that both the statements give the same piece of information that a : b = 3 : 5 and that a and b are both positive. But none of the statements either in itself or together can give the value of a.

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**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.

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

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