# Arithmetic - Percentage - Previous Year CAT/MBA Questions

The best way to prepare for Arithmetic - Percentage is by going through the previous year **Arithmetic - Percentage CAT questions**.
Here we bring you all previous year Arithmetic - Percentage CAT questions along with detailed solutions.

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**CAT 2023 QA Slot 1 | Arithmetic - Percentage CAT Question**

The salaries of three friends Sita, Gita and Mita are initially in the ratio 5 : 6 : 7, respectively. In the first year, they get salary hikes of 20%, 25% and 20%, respectively. In the second year, Sita and Mita get salary hikes of 40% and 25%, respectively, and the salary of Gita becomes equal to the mean salary of the three friends. The salary hike of Gita in the second year is

- (a)
26%

- (b)
25%

- (c)
30%

- (d)
28%

Answer: Option A

**Text Explanation** :

Let the initial salaries of A, B and C be 5x, 6x and 7x respectively.

Salary of A increases by 20% and then by 40%.

∴ Salary of A after 2 years = 5x × 1.2 × 1.4 = 8.4x

Salary of C increases by 20% and then by 25%.

∴ Salary of C after 2 years = 7x × 1.2 × 1.25 = 10.5x

At the end of 2 years, B's salary is average of all three, hence B's salary will also be average of salaries of A and C.

⇒ Salary of B after 2 years = (8.4x + 10.5x)/2 = 9.45

Let the % increase in B's salary be P% in 2^{nd} year. Increase in first year is 25%.

⇒ 9.45x = 6x × 1.25 × (1 + P/100)

⇒ (1 + P/100) = 9.45/7.5 = 1.26

⇒ P = 26%

Hence, option (a).

**Concept: **Percentage Change

**Concept**:

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**CAT 2023 QA Slot 3 | Arithmetic - Percentage CAT Question**

The population of a town in 2020 was 100000. The population decreased by y% from the year 2020 to 2021, and increased by x% from the year 2021 to 2022, where x and y are two natural numbers. If population in 2022 was greater than the population in 2020 and the difference between x and y is 10, then the lowest possible population of the town in 2021 was

- (a)
73000

- (b)
72000

- (c)
75000

- (d)
74000

Answer: Option A

**Text Explanation** :

Population at the end of 2021 = 1,00,000 × (1 - x/100)

Population at the end of 2022 = 1,00,000 × (1 - y/100) × (1 + x/100) ...(1)

∴ Overall % change = -y + x + (-y × x)/100 [using formula = a + b + ab/100]

Since population in 2022 is greater than that in 2020

⇒ x > y.

∴ x = y + 10

Hence, the population decreases by y% and then increases by (y + 10)%

∴ Overall % change = -y + (y + 10) + (-y × (y + 10))/100 = 10 - y(y + 10)/100 [using formula = a + b + ab/100]

Overall change should be positive

∴ Overall % change = 10 - y(y + 10)/100 > 0

⇒ y(y + 10) < 1000

Higheset value of y satisfying the above inequality is 27.

∴ The population will decrease by a maximum of 27%.

⇒ Least population in 2021 = 1,00,000 × (1 - 27%) = 73,000.

Hence, option (a).

**Concept**: Successive Percentage Change Formula

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**CAT 2023 QA Slot 3 | Arithmetic - Percentage CAT Question**

A fruit seller has a stock of mangoes, bananas and apples with at least one fruit of each type. At the beginning of a day, the number of mangoes make up 40% of his stock. That day, he sells half of the mangoes, 96 bananas and 40% of the apples. At the end of the day, he ends up selling 50% of the fruits. The smallest possible total number of fruits in the stock at the beginning of the day is

Answer: 340

**Text Explanation** :

Let total number of fruits at the beginning of the day = 5x

∴ Number of mangoes = 40% of 5x = 2x

Let the number of apples be '5a', hence the number of bananas = 5x - 5a.

⇒ He sells 50% of 2x mangoes, 96 bananas and 40% of 5a apples i.e., x + 96 + 2a fruits.

Also, he sold 50% of his fruits

⇒ x + 96 + 2a = 50% of 5x = 2.5x

⇒ 1.5x = 96 + 2a

⇒ x = 64 + 4a/3

⇒ 5x = 320 + 20a/3

To minimise 5x, we need to minimise 2aa/3.

Minimum integral value of 20a/3 will be 20 when a = 3.

∴ 5x = 320 + 20 = 340

Hence, 340.

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**CAT 2022 QA Slot 2 | Arithmetic - Percentage CAT Question**

In an election, there were four candidates and 80% of the registered voters casted their votes. One of the candidates received 30% of the casted votes while the other three candidates received the remaining casted votes in the proportion 1 : 2 : 3. If the winner of the election received 2512 votes more than the candidate with the second highest votes, then the number of registered voters was:

- (a)
60288

- (b)
50240

- (c)
40192

- (d)
62800

Answer: Option D

**Text Explanation** :

Let the total number of registered voters be 100x.

Number of votes casted = 80x

Votes for Candidate 1 = 30% of 80x = 24x

∴ Remaining three candidates will recieve = 80x - 24x = 56x votes.

Remaining 3 candidates get votes in the ratio of 1 : 2 : 3 of the remaining 56x votes.

⇒ Votes of Candidate 2 = $\frac{1}{6}$ × 56x = $\frac{28x}{3}$

⇒ Votes of Candidate 3 = $\frac{2}{6}$ × 56x = $\frac{56x}{3}$

⇒ Votes of Candidate 4 = $\frac{3}{6}$ × 56x = 28x

Highest number of votes is received by Candidate 4 while second highest is by Candidate 1.

⇒ 28x – 24x = 2512

⇒ x = 2512/4 = 628

∴ total number of registered votes = 100x = 62800

Hence, option (d).

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**CAT 2021 QA Slot 1 | Arithmetic - Percentage CAT Question**

Identical chocolate pieces are sold in boxes of two sizes, small and large. The large box is sold for twice the price of the small box. If the selling price per gram of chocolate in the large box is 12% less than that in the small box, then the percentage by which the weight of chocolate in the large box exceeds that in the small box is nearest to

- (a)
127

- (b)
135

- (c)
144

- (d)
124

Answer: Option A

**Text Explanation** :

Let the weight and price of small box is ‘w’ and ‘p’ respectively.

∴ Selling price per gram for small box = p/w

Now, selling price per gram for large box = 0.88p/w

Also, selling price for the whole box = 2p.

⇒ Weight of the large box = $\frac{2p}{0.88p/w}$ = $\frac{2w}{0.88}$ = $\frac{25w}{11}$

∴ Required percentage = $\frac{{\displaystyle \frac{25}{11}}w-w}{w}$ × 100 = 14/11 × 100 ≈ 127%

Hence, option (a).

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**CAT 2021 QA Slot 2 | Arithmetic - Percentage CAT Question**

Raj invested ₹ 10000 in a fund. At the end of first year, he incurred a loss but his balance was more than ₹ 5000. This balance, when invested for another year, grew and the percentage of growth in the second year was five times the percentage of loss in the first year. If the gain of Raj from the initial investment over the two year period is 35%, then the percentage of loss in the first year is

- (a)
10

- (b)
15

- (c)
5

- (d)
70

Answer: Option A

**Text Explanation** :

Let the loss incurred by Raj in first year = P%

∴ Amount remaining after 1st year = 10,000$\left(1-\frac{P}{100}\right)$ > 5,000 …(1)

Now the percentage growth next year = 5P%

∴ Amount after 2 years = 10,000$\left(1-\frac{P}{100}\right)$$\left(1+\frac{5P}{100}\right)$

Overall growth after 2 years is 35%, hence amount after 2 years should be 10,000 × 1.35

⇒ 10,000$\left(1-\frac{P}{100}\right)$$\left(1+\frac{5P}{100}\right)$ = 10,000 × 1.35

⇒ 10000 – 5P^{2} + 400P = 13500

⇒ 5P^{2} - 400P + 3500 = 0

⇒ P^{2} – 80P + 700 = 0

⇒ P = 10% or 70%.

P cannot be 70% since amount remaining after 1st year has to be greater than 5000 [from (1)]

∴ P = 10%

Hence, option (a).

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**CAT 2021 QA Slot 3 | Arithmetic - Percentage CAT Question**

In a tournament, a team has played 40 matches so far and won 30% of them. If they win 60% of the remaining matches, their overall win percentage will be 50%. Suppose they win 90% of the remaining matches, then the total number of matches won by the team in the tournament will be

- (a)
78

- (b)
84

- (c)
80

- (d)
86

Answer: Option B

**Text Explanation** :

Matches so far:

Total played = 40

Won = 30% of 40 = 12

Let the number of remaining matches be ‘n’.

The team wins 60% i.e., 0.6n matches.

There overall win percentage is 50% i.e., half

⇒ 12 + 0.6n = ½ (40 + n)

⇒ 24 + 1.2n = 40 + n

⇒ n = 80

Now, if the team wins 90% of these 80 remaining matches, they will win 90% of 80 = 72 more matches.

∴ Total matches won = 12 + 72 = 84

Hence, option (b).

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**CAT 2021 QA Slot 3 | Arithmetic - Percentage CAT Question**

The total of male and female populations in a city increased by 25% from 1970 to 1980. During the same period, the male population increased by 40% while the female population increased by 20%. From 1980 to 1990, the female population increased by 25%. In 1990, if the female population is twice the male population, then the percentage increase in the total of male and female populations in the city from 1970 to 1990 is

- (a)
68.50

- (b)
69.25

- (c)
68.75

- (d)
68.25

Answer: Option C

**Text Explanation** :

Let the male and female population be ‘m’ and ‘f’ in 1970.

Using alligation

$\frac{m}{f}$ = $\frac{25-20}{40-25}$ = $\frac{1}{3}$

∴ f = 3m

Total population in 1970 = f + m = 4m

⇒ Number of females in 1980 = 1.2f = 3.6m and number of males in 1980 = 1.4m

Population in 1990

Number of females = 1.25 × 3.6m = 4.5m

Number of males = ½ × 4.5m = 2.25m

Total population = 4.5m + 2.25m = 6.75m

⇒ Total population increases from 4m in 1970 to 6.75m in 1990.

∴ Percentage increase = 2.75m/4m×100 = 68.75%.

**Alternately**,

Let the number of males and females in 1970 be 'm' and 'f' respectively.

∴ In 1980, number of males = 1.4m, females = 1.2f and total population = 1.25(m + f)

⇒ 1.4m + 1.2f = 1.25(m + f)

⇒ 0.15m = 0.05f

⇒ m : f = 1 : 3

∴ In 1990, number of females = 1.25 × 1.2f = 1.5f, number of males = 1/2 × 1.5f = 0.75f

∴ Total population in 1990 = 1.5f + 0.75f = 2.25f

Also, total population in 1970 = m + f = f + f/3 = 4f/3

⇒ Required percentage = $\left(\frac{2.25f-4f/3}{4f/3}\right)$ × 100% = $\frac{11}{16}$ × 100% = 68.75%

Hence, option (c).

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**CAT 2020 QA Slot 1 | Arithmetic - Percentage CAT Question**

In a group of people, 28% of the members are young while the rest are old. If 65% of the members are literates, and 25% of the literates are young, then the percentage of old people among the illiterates is nearest to

- (a)
55

- (b)
66

- (c)
59

- (d)
62

Answer: Option B

**Text Explanation** :

According to the information given in the question, we can make the following table.

∴ % old illiterates among illiterates = 23.25/35 × 100 = 66.4%

Hence, option (b).

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**CAT 2020 QA Slot 2 | Arithmetic - Percentage CAT Question**

In May, John bought the same amount of rice and the same amount of wheat as he had bought in April, but spent ₹ 150 more due to price increase of rice and wheat by 20% and 12%, respectively. If John had spent ₹ 450 on rice in April, then how much did he spend on wheat in May?

- (a)
Rs. 580

- (b)
Rs. 560

- (c)
Rs. 590

- (d)
Rs. 570

Answer: Option B

**Text Explanation** :

John spent Rs. 450 on rice in April hence he will spend 450 × 1.2 = 540 on rice in May.

He spends Rs. 90 extra on rice.

He also spends a total of Rs. 150 more in May compared to April

⇒ He spends 150 – 90 = Rs. 60 more on wheat in May compared to April.

∴ His expenditure on wheat in April on wheat = 60/0.12 = Rs. 500.

⇒ His expenditure on wheat in May = 500 + 60 = Rs. 560.

Hence, option (b).

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**CAT 2020 QA Slot 3 | Arithmetic - Percentage CAT Question**

In the final examination, Bishnu scored 52% and Asha scored 64%. The marks obtained by Bishnu is 23 less, and that by Asha is 34 more than the marks obtained by Ramesh. The marks obtained by Geeta, who scored 84%, is

- (a)
399

- (b)
439

- (c)
417

- (d)
357

Answer: Option A

**Text Explanation** :

Let the maximum marks be 100x.

Marks scored by Bishnu = 52x and Asha = 64x

Bishnu scored 23 less, Asha scored 34 more than the marks obtained by Ramesh

∴ Asha – Bishnu = 34 + 23 = 57

⇒ 64x – 52x = 57

⇒ 12x = 57

Now, Geeta scored 84%, her score = 84x = 7 × 12x

= 7 × 57 = 399.

Hence, option (a).

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**CAT 2019 QA Slot 1 | Arithmetic - Percentage CAT Question**

Meena scores 40% in an examination and after review, even though her score is increased by 50%, she fails by 35 marks. If her post-review score is increased by 20%, she will have 7 marks more than the passing score. The percentage score needed for passing the examination is

- (a)
60

- (b)
80

- (c)
70

- (d)
75

Answer: Option C

**Text Explanation** :

Let the total marks be 100x.

Meena's score = 40x.

Meena's score after review = 40x + [(40x)/2] = 60x.

Passing marks = 60x + 35.

Post review score × (6/5) = 7 + Passing marks

∴ 60x × (6/5) = 60x + 42.

Solving this equation we get; x = 3.5.

So, passing marks = 60x + 35 = (60 × 3.5) + 35 = 245 and total marks = 100x = 100 × 3.5 = 350.

Percentage score needed to pass the examination = (Passing marks/Total marks) × 100 = (245/350) × 100 = 70%.

Hence, option (c).

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**CAT 2019 QA Slot 1 | Arithmetic - Percentage CAT Question**

The income of Amala is 20% more than that of Bimala and 20% less than that of Kamala. If Kamala's income goes down by 4% and Bimala's goes up by 10%, then the percentage by which Kamala's income would exceed Bimala's is nearest to

- (a)
31

- (b)
29

- (c)
28

- (d)
32

Answer: Option A

**Text Explanation** :

Let incomes of Amala, Bimala and Kamala be a, b and k respectively.

∴ a = $\frac{6}{5}$ × b = $\frac{4}{5}$ × k.

∴ $\frac{\mathrm{k}}{\mathrm{b}}$ = $\frac{3}{2}$

Let Kamala's income be 300 and Bimala's income be 200.

∴ Kamala's new income = 300 × 0.96 = 288 and Bimla's new income = 200 × 1.1 = 220.

∴ Required percentage = $\left(\frac{288-220}{220}\right)$ × 100 = $\left(\frac{68}{220}\right)$ × 100 = 30.9 ≈ 31%.

Hence, option (a).

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**CAT 2019 QA Slot 1 | Arithmetic - Percentage CAT Question**

On selling a pen at 5% loss and a book at 15% gain, Karim gains Rs. 7. If he sells the pen at 5% gain and the book at 10% gain, he gains Rs. 13. What is the cost price of the book in Rupees?

- (a)
95

- (b)
85

- (c)
100

- (d)
80

Answer: Option D

**Text Explanation** :

Let the cost price of one pen and one book be 100p and 100b respectively.

On selling a pen at 5% loss and a book at 15% gain, Karim gains Rs. 7.

∴ 95p + 115b = 7 + (100p + 100b) ⇒ 15b − 5p = 7 ...(1)

On selling the pen at 5% gain and the book at 10% gain, he gains Rs. 13.

∴ 105p + 110b = 13 + (100p + 100b) ⇒ 10b + 5p = 13 ...(2)

Solving (1) and (2), we get; b = 4/5.

So, cost price of one book = 100b = 100 × (4/5) = Rs. 80.

Hence, option (d).

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**CAT 2019 QA Slot 1 | Arithmetic - Percentage CAT Question**

In a class, 60% of the students are girls and the rest are boys. There are 30 more girls than boys. If 68% of the students, including 30 boys, pass an examination, the percentage of the girls who do not pass is

Answer: 20

**Text Explanation** :

Let the total number of students be 100x.

So, number of girls and boys are 60x and 40x respectively.

There are 30 more girls than boys ∴ 60x = 40x + 30.

∴ x = 3/2.

Number of students who passed = 68x = 68 × (3/2) = 102 out of which 30 boys passed.

So, number of girls who passed = 102 − 30 = 72

Number of girls who did not pass = 60x − 72 = 90 − 72 = 18.

Required percentage = [18/90] × 100 = 20%.

Hence, 20.

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**CAT 2019 QA Slot 2 | Arithmetic - Percentage CAT Question**

In 2010, a library contained a total of 11500 books in two categories - fiction and non-fiction. In 2015, the library contained a total of 12760 books in these two categories. During this period, there was 10% increase in the fiction category while there was 12% increase in the non-fiction category. How many fiction books were in the library in 2015?

- (a)
6600

- (b)
6160

- (c)
6000

- (d)
5500

Answer: Option A

**Text Explanation** :

In 2010 : Total books = 11500

Let the number of fiction books be f, non-fiction books will be (11500 - f)

In 2015 : Total books = 12760

Increase in total number of books = 12760 - 11500 = 1260

Fiction books increase by 10%, non-fiction books increase by 12%.

Hence, f × $\frac{10}{100}$ + (11500 - f) × $\frac{12}{100}$ = 1260

10f + 12 × 11500 – 12f = 126000

x = 6000 books

So, Fiction books in 2015, 6000 + 600 = 6600 books

Hence, option (a).

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**CAT 2019 QA Slot 2 | Arithmetic - Percentage CAT Question**

In an examination, the score of A was 10% less than that of B, the score of B was 25% more than that of C, and the score of C was 20% less than that of D. If A scored 72, then the score of D was

Answer: 80

**Text Explanation** :

Given: A scored 72

A's score was 10% less than B

So, Score of B = 72/0.9 = 80

We know that B was 25% more than C

So, C × 5/4 = 80

⇒ C = 64

Now, we know that C scored 20% less than D.

So, C = 4/5 × D

⇒ 64 = 4/5 × D

⇒ D = 80 marks

Hence, 80.

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**CAT 2019 QA Slot 2 | Arithmetic - Percentage CAT Question**

The salaries of Ramesh, Ganesh and Rajesh were in the ratio 6 : 5 : 7 in 2010, and in the ratio 3 : 4 : 3 in 2015. If Ramesh's salary increased by 25% during 2010-2015, then the percentage increase in Rajesh's salary during this period is closest to

- (a)
7

- (b)
8

- (c)
9

- (d)
10

Answer: Option A

**Text Explanation** :

The salaries of Ramesh, Ganesh and Rajesh were in the ratio 6 : 5 : 7 in 2010

In 2010, let salary of Ramesh = 6x, Ganesh = 5x & Rajesh = 7x

In 2015, their salaries are in the ratio 3 : 4 : 3 respectively.

In 2015, let salary of Ramesh = 3y, Ganesh = 4y & Rajesh = 3y

Now, Ramesh’s salary increased by 25%

Hence, 3y = 5/4 × 6x

⇒ y = 2.5x

∴ Rajesh’s salary in 2015 = 3y = 3 × 2.5x = 7.5x

Percentage increase in Rajesh's salary during 2010 - 2015 = $\frac{7.5-7}{7}$ × 100% ≈ 7.

Hence, option (a).

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**CAT 2018 QA Slot 1 | Arithmetic - Percentage CAT Question**

In an examination, the maximum possible score is N while the pass mark is 45% of N. A candidate obtains 36 marks, but falls short of the pass mark by 68%. Which one of the following is then correct?

- (a)
N ≤ 200

- (b)
201 ≤ N ≤ 242

- (c)
243 ≤ N ≤ 252

- (d)
N ≥ 253

Answer: Option C

**Text Explanation** :

Passing marks = 45% of N

As 36 marks are short of the pass marks by 68%, it means the cadidate got (100 - 68 = ) 32% of the passing marks.

⇒ 36 = 32% of 45% of N

∴ 36 = 0.32 × 0.45 × N

∴ N = $\frac{36}{0.32\times 0.45}$ = 250

Hence, option (c).

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**CAT 2017 QA Slot 1 | Arithmetic - Percentage CAT Question**

Arun's present age in years is 40% of Barun's. In another few years, Arun's age will be half of Barun's. By what percentage will Barun's age increase during this period?

Answer: 20

**Text Explanation** :

Let Barun’s age be 10x. Arun’s age is 4x.

The difference of these ages in 6x, a constant.

When Arun’s age is 50% of Barun’s age, this difference also would be 50% i.e., Barun’s age, at that stage would be 12x.

It would be increase by 20%.

**Alternately**,

Let Barun’s age be 10x. Arun’s age is 4x.

In t years, Arun's age will be 50% or half of Barun's age

⇒ (4x + t) = 1/2 × (10x + t)

⇒ t = 2x

∴ Barun's age will become 12x.

⇒ Required % = (12x - 10x)/10x × 100% = 20%

Hence, 20.

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**CAT 2017 QA Slot 1 | Arithmetic - Percentage CAT Question**

Ravi invests 50% of his monthly savings in fixed deposits. Thirty percent of the rest of his savings is invested in stocks and the rest goes into Ravi's savings bank account. If the total amount deposited by him in the bank (for savings account and fixed deposits) is Rs 59,500, then Ravi's total monthly savings (in Rs) is:

Answer: 70000

**Text Explanation** :

Let the total monthly savings be S.

Investment in FD = 50% of S = 0.5S

Remaining savings = S - 0.5S = 0.5S

Investment in stocks = 30% of 0.5S = 0.15S

Total invested amount (FD + stocks) = 0.5S + 0.15S = 0.65S

Remaing amount that is invested in savings account = 0.35S.

⇒ 0.35S + 0.5S = 59,500

⇒ 0.85S = 59500.

⇒ S = 70,000.

Hence, 70,000.

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**CAT 2017 QA Slot 1 | Arithmetic - Percentage CAT Question**

The number of girls appearing for an admission test is twice the number of boys. If 30% of the girls and 45% of the boys get admission, the percentage of candidates who do not get admission is:

- (a)
35

- (b)
50

- (c)
60

- (d)
65

Answer: Option D

**Text Explanation** :

Let the number of boys appearing for the admission test be b.

Number of girs appearing for the admission test = 2b.

Total boys who got admission = 45% of b = 0.45b

Total girls who got admission = 30% of 2b = 0.6b

Total students who got admission = 0.45b + 0.6b = 1.05b

∴ Percentage of candidates who get admission = $\frac{1.05b}{2b+b}\times 100\%$ = 35%

∴ 65% of the candidates do not get admission.

Hence, option (d).

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**CAT 2017 QA Slot 2 | Arithmetic - Percentage CAT Question**

Out of the shirts produced in a factory, 15% are defective, while 20% of the rest are sold in the domestic market. If the remaining 8840 shirts are left for export, then the number of shirts produced in the factory is

- (a)
13600

- (b)
13000

- (c)
13400

- (d)
14000

Answer: Option B

**Text Explanation** :

Let us assume that the number of shirts produced in the factory is ‘100x’. Now 15% of ‘100x’ or ‘15x’ shirts are defective.

So the number of remaining shirts = 100x – 15 = 85x

Now 20% of the remaining ‘85x’or ‘17x’ shirts are sold in the domestic market.

So number of shirts left for exports = 85x – 17x = 68x

As per given information, 8840 shirts are left for exports.

∴ 68x = 8840

⇒ x = 130 or 100x = 13000

So number of shirts produced in the factory is 13000.

Hence, option (b).

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**CAT 2017 QA Slot 2 | Arithmetic - Percentage CAT Question**

In a village, the production of food grains increased by 40% and the per capita production of food grains increased by 27% during a certain period. The percentage by which the population of the village increased during the same period is nearest to

- (a)
16

- (b)
13

- (c)
10

- (d)
7

Answer: Option C

**Text Explanation** :

Total Food Production = Total Population × Per Capita Food Production

Let the initial population be ‘p’ and the initial per capita food production be ‘c’

If f is the total food production, f = pc

This can also be rewritten as p = $\frac{f}{c}$

Now, as per given information, f increases to 1.4f and c increases to 1.27c.

So, if p_{1} is population after increase.

1.4f = p_{1}(1.27c)

$\frac{1.4f}{1.27c}$ = p_{1}

⇒ p_{1} = $\left(\frac{1.4}{1.27}\right)\left(\frac{f}{c}\right)$

or p_{1} = $\left(\frac{1.4}{1.27}\right)$(p) [$\left(\because p=\frac{f}{c}\right)$ ]

$\Rightarrow $ p_{1} ≃ 1.1p

This means that the population of the village has increased by approximately 10%.

Hence, option (c).

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**Answer the next 2 questions based on the information given below.**

Shabnam is considering three alternatives to invest her surplus cash for a week. She wishes to guarantee maximum returns on her investment. She has three options, each of which can be utilized fully or partially in conjunction with others.

Option A: Invest in a public sector bank. It promises a return of +0.10%

Option B: Invest in mutual funds of ABC Ltd. A rise in the stock market will result in a return of +5%, while a fall will entail a return of –3%

Option C: Invest in mutual funds of CBA Ltd. A rise in the stock market will result in a return of –2.5%, while a fall will entail a return of +2%

**CAT 2007 QA | Arithmetic - Percentage CAT Question**

The maximum guaranteed return to Shabnam is:

- (a)
0.25%

- (b)
0.10%

- (c)
0.20%

- (d)
0.15%

- (e)
0.30%

Answer: Option C

**Text Explanation** :

Let Shabnam have Rs. 100 to invest. Let Rs. x, Rs. y and Rs. z be invested in option A, B and C respectively.

∴ x + y + z = 100 ... (I)

If there is a rise in the stock market, returns = 0.001x + 0.05y – 0.025z

If there is a fall in the stock market, returns = 0.001x – 0.03y + 0.02z

Now, x, y and z should be such that regardless of whether the market rises or falls, they give the same return, which is the maximum guaranteed return.

∴ 0.001x + 0.05y – 0.025z = 0.001x – 0.03y + 0.02z

∴ y/z = 9/16

Now, consider different possible values of x, y and z. The returns are as follows:

We see that as the values of y and z increase, the returns increase.

∴ The returns are maximum when x = 0%, y = 36% and z = 64% (Note that the values of y and x are multiples of 9 and 16.)

The maximum returns are 0.2%.

Hence, option (c).

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