Arithmetic - Percentage - Previous Year CAT/MBA Questions

Answer: Option D

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).

Answer: Option A

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).

Answer: Option A

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 – 5P² + 400P = 13500

⇒ 5P² - 400P + 3500 = 0

⇒ P² – 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).

Answer: Option B

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).

Answer: Option C

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).

Answer: Option B

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).

Answer: Option B

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).

Answer: Option A

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).

Answer: Option C

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).

Answer: Option A

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).

Answer: Option D

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).

Answer: 20

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.

Answer: Option A

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).

Answer: 80

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.

Answer: Option A

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).

Answer: Option C

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).

Answer: 20

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.

Answer: 70000

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.

Answer: Option D

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).

Answer: Option B

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).

Answer: Option C

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 p₁ is population after increase.

1.4f = p₁(1.27c) 

$\frac{1.4f}{1.27c}$ = p₁ 

⇒ p₁ = $\left(\frac{1.4}{1.27}\right)\left(\frac{f}{c}\right)$

or p₁ = $\left(\frac{1.4}{1.27}\right)$(p) [$\left(\because \mathrm{ p}=\frac{f}{c}\right)$ ]

$\Rightarrow$ p₁ ≃ 1.1p 

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

Hence, option (c).

Answer: Option C

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).

Answer: Option B

Explanation :

As shown by the table formulated in the first question, maximum returns are guaranteed by investing 36% in option B and 64% in option C.

Hence, option (b).

Answer: Option C

Explanation :

Let there be 100x employees. So, 30x are male and 70x are female.

∴ 7x female employees have an engineering background.

From statement I, 25x employees have an engineering background.

∴ 18x male employees have an engineering background.

Required percentage = 18x × 100/ 30x

Statement I is sufficient.

From Statement II, Number of male employees having an engineering background = 1.2 × 7x

Required percentage = 1.2 × 7x × 100/30x

Hence, option (c).

Answer: Option A

Explanation :

Let there be a, b and c questions in groups A, B and C respectively.

Then, a + b + c =100

Total marks = a + 2b + 3c

Also, $\frac{a}{a+2b+3c}\ge \frac{60}{100}$

b = 23

⇒ a + c = 77

⇒ a = 77 – c

⇒ $\frac{\mathrm{a}}{\mathrm{a}+46+3\mathrm{c}}\ge 0.6$

⇒ a ≥ 0.6a + 27.6 + 1.8c

⇒ 4a ≥ 276 + 18c

⇒ a ≥ 69 + 4.5c

⇒ a – 4.5c ≥ 69

⇒ 77 – c – 4.5c ≥ 69

⇒ 77 ≥ 69 + 5.5c

⇒ c ≤ 8/5.5

⇒ c = 1

Hence, option (a).

Alternatively,
Evaluating options,

If C has 1 question then B and A have 23 and 76 questions respectively.
The total number of marks = 125
⇒ Group A has 60.8% marks.
⇒ Option (a) is possible.

If C has 2 questions then B and A have 23 and 75 questions respectively.
The total number of marks = 127
⇒ Group A has 59.05% marks.
⇒ Option (b) is not possible.

If C has 3 questions then B and A have 23 and 74 questions respectively.
The total number of marks = 129
⇒ Group A has 57.36% marks.
⇒ Option (c) is not possible.

Hence, option (a).