# PE 4 - Percentage | Arithmetic - Percentage

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**PE 4 - Percentage | Arithmetic - Percentage**

The number of votes not cast for the PJB increased by 30% in the 2023 MP Assembly elections over those not cast for it in the 2018 Assembly elections, and the PJB lost by a majority thrice as large as that by which it had won the 2018 Assembly Polls. If a total of 4,30,000 people voted each time, how many voted for the PJB in the 2018 Assembly Elections?

Answer: 230000

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**Explanation** :

Let the number of votes not cast in the 2018 Assembly Polls be denoted by x,

∴ The number of votes not cast in the 2023 Assembly Election = 1.3x.

The total number of voters in both cases is given to be 430000.

Thus, the number of votes cast for the party in 2018 Assembly Polls = 4,30,000 – x.

Also, the number of votes cast for the party in the 2023 Assembly Election = 4,30,000 –1.3v.

Majority of votes in 2018 Assembly Polls = (4,30,000 – x) – x = 4,30,000 – 2x.

Majority of votes in 2023 Assembly Election = 1.3x – (4,30,000 – 1.3x) = 2.6x - 4,30,000

⇒ 2.6x - 4,30,000 = 3 × (4,30,000 - 2x)

⇒ 2.6x - 4,30,000 = 12,90,000 - 6x

⇒ 8.6x = 17,20,000

⇒ x = 17,20,000/8.6 = 2,00,000

Thus, the number of votes cast for the party in 2018 Assembly Polls = 4,30,000 – x = 4,30,000 – 2,00,000 = 2,30,000.

Hence, 230000.

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**PE 4 - Percentage | Arithmetic - Percentage**

There are four sections A, B, C and D in a school. If the percentage of students passing in sections A, B and C is 40%, that of sections A, B, D is 45%, that of A, C, D is 50% and that of B, C, D is 55% then which of the following cannot be the pass percentage of sections A, B, C and D combined together.

- (a)
45.66%

- (b)
49.85%

- (c)
51.3%

- (d)
47.28%

Answer: Option C

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**Explanation** :

Let the number of students in sections A, B, C and D be a, b, c and d respectively.

Let the number of students passing in sections A, B, C and D be p, q, r and w respectively.

Total number of students passing from sections A, B and C = p + q + r = (a + b + c) × 0.4

Total number of students passing from sections A, B and D = p + q + s = (a + b + d) × 0.45

Total number of students passing from sections A, C and D = p + r + s = (a + c + d) × 0.50

Total number of students passing from sections B, C and D = q + r + s = (b + c + d) × 0.55

Adding the 4 equation, we get

3 × (p + q + r + s) = 1.35a + 1.4b + 1.45c + 1.5d

This can be written as:

⇒ 3 × (p + q + r + s) = 1.35(a + b + c + d) + 0.05b + 0.1c + 0.15d

⇒ (p + q + r + s) = 0.45(a + b + c + d) + $\frac{(0.05\mathrm{b}+0.1\mathrm{c}+0.15\mathrm{d})}{3}$

⇒ $\frac{(\mathrm{p}+\mathrm{q}+\mathrm{r}+\mathrm{s})}{(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d})}$ = 0.45 + $\frac{(0.05\mathrm{b}+0.1\mathrm{c}+0.15\mathrm{d})}{3(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d})}$

⇒ $\frac{(\mathrm{p}+\mathrm{q}+\mathrm{r}+\mathrm{s})}{(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d})}\times 100\%$ = 45% + $\frac{(0.05\mathrm{b}+0.1\mathrm{c}+0.15\mathrm{d})}{3(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d})}\times 100$%

⇒ Overall pass percentage is definitely greater than 45%.

This can be also be written as:

⇒ 3 × (p + q + r + s) = 1.5(a + b + c + d) - (0.15a + 0.1b + 0.05c)

⇒ (p + q + r + s) = 0.5(a + b + c + d) - $\frac{(0.15\mathrm{a}+0.1\mathrm{b}+0.05\mathrm{c})}{3}$

⇒ $\frac{(\mathrm{p}+\mathrm{q}+\mathrm{r}+\mathrm{s})}{(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d})}$ = 0.5 - $\frac{(0.15\mathrm{a}+0.1\mathrm{b}+0.05\mathrm{c})}{3(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d})}$

⇒ $\frac{(\mathrm{p}+\mathrm{q}+\mathrm{r}+\mathrm{s})}{(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d})}\times 100\%$ = 50% - $\frac{(0.15\mathrm{a}+0.1\mathrm{b}+0.05\mathrm{c})}{3(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d})}$%

⇒ Overall pass percentage is definitely less than 50%.

∴ Overall pass percentage off all 4 sections combined should be between 45% and 50% (excluding both) .

∴ Overall pass percentage cannot be option (c) 51.3%

Hence, option (c).

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**PE 4 - Percentage | Arithmetic - Percentage**

After the elections, a firm conducted a survey. Exactly 3% of the male voters said they had voted for candidate A, while exactly 47% of the female voters said they had voted for candidate A. If in total, exactly 19% of the people said they had voted for the Candidate. What is the minimum number of people who were surveyed?

Answer: 1100

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**Explanation** :

Let the number of men be m and women be w.

Number of women selecting candidate A = 47% of w = 0.47w

Number of men selecting candidate A = 3% of m = 0.03m

⇒ 0.03m + 0.47w = 0.19(m + w)

⇒ 0.28w = 0.16m

⇒ w : m = 4 : 7

Hence, w = 4x, m = 7x where x is an integer.

Number of women selecting candidate A = 47% of 4x = 4x × 47/100 = 47x/25

Number of men selecting candidate A = 3% of 7x = 7x × 3/100 = 21x/100

Now, both 47x/25 and 21x/100 must be integers, hence least possible value of x = 100.

∴ Least Total population is 4x + 7x = 11x = 1100

Hence, 1100.

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**PE 4 - Percentage | Arithmetic - Percentage**

The daily % increase in the population of bacteria in a colony increases by 10% points every day. On the first day, the population of bacteria increases by 20%. Find the increase in the number of bacteria on the 2^{nd} day, if there were 1872 more bacteria at the end of the 3^{rd} day compared to the end of the 2^{nd} day.

Answer: 3000

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**Explanation** :

Percentage increase in bacteria on

Day 1 = 20%

Day 2 = 30%

Day 3 = 40%

... and so on

Let the number of bacteria at the end of Day 1 be x.

∴ Number of bacteria at the end of Day 2 = x × 1.3 = 1.3x

∴ Number of bacteria at the end of Day 2 = x × 1.3 × 1.4 = 1.82x

⇒ 1.82x - 1.3x = 1872

⇒ 0.52x = 1872

⇒ x = 1872/0.52 = 3600

∴ There were 3600 bacteria at the end of day 1.

⇒ Number of bacteria at the beginning of Day 1 = 3600/1.2 = 3000

Hence, 3000.

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**PE 4 - Percentage | Arithmetic - Percentage**

Roomesh bought 5 pens, 7 pencils and 4 erasers. Priya bought 6 pens, 8 erasers and 14 pencils for an amount which was half more what Roomesh paid. What % of the total amount paid by Roomesh was for the pens?

- (a)
37.5%

- (b)
62.5%

- (c)
50%

- (d)
None of these

Answer: Option B

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**Explanation** :

Roomesh bought 5 pens, 7 pencils and 4 erasers. Priya bought 6 pens, 8 erasers and 14 pencils for an amount which was half more what Roomesh paid. What % of the total amount paid by Roomesh was for the pens?

Let the cost of each pen, pencil and eraser be p, l and e respectively.

Amount paid by Priya is 1.5 times that of Roomesh

Let the total amount paid be Roomesh be T and that by Priya will be 1.5T.

Total cost of Roomesh = T = 5p + 7l + 4e ...(1)

Total cost of Priya = 1.5T = 6p + 14l + 8e ...(2)

(1) × 2 - (2)

⇒ 2T - 1.5T = 2(5p + 7l + 4e) - 6p + 14l + 8e

⇒ 0.5T = 4p

⇒ p/T = 1/8

Now, % of the total amount paid by Roomesh was for the pens = 5p/T × 100%

= 5 × 1/8 × 100% = 62.5%

Hence, option (b).

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**PE 4 - Percentage | Arithmetic - Percentage**

Vishal went to market with some money. With this money, he could buy either 50 oranges or 40 mangoes. He retained exactly 10% of the money for taxi fare. If he bought 25 mangoes, then maximum how many oranges can he buy?

- (a)
25

- (b)
16

- (c)
20

- (d)
None of these

Answer: Option D

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**Explanation** :

Let the price of each orange and mango be o and m respectively.

∴ Total amount with Vishal = 50o or 40m

⇒ 50o = 40m

⇒ 5o = 4m

⇒ o : m = 4 : 5

Let o = 4x and m = 5x and total amount = 50o = 200x

Vishal kept 10% of 200x i.e., 20x for taxi

Amount left now = 200x - 20x = 180x.

Then, he bought 25 mangoes, hence cost of these 25 mangoes = 25m = 125x

Amount left now = 180x - 125x = 55x.

Now, maximum oranges that he can buy = 55x/4x = 13.75 i.e., 13 oranges.

Hence, option (d).

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**PE 4 - Percentage | Arithmetic - Percentage**

Population of sheep in a farm at the beginning of the year was 1,25,000. The population of sheep increased by x% on 1st of every month and decreased by y% on 20th of every month. At the end of the year, there were 1,25,000 sheep in the farm. Which of the following is true.

- (a)
x = y

- (b)
x > y

- (c)
x < y

- (d)
Cannot be determined

Answer: Option B

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**Explanation** :

Every month the population increased by x% and then decreased by y%.

∴ overall % change every month = x - y + (x × -y)/100 %

Let this be p%

Now, if p > 0, then the population every month should increase and the population at the end of the year should be more than the starting population. Hence, p cannot be greater than 0.

Now, if p < 0, then the population every month should decrease and the population at the end of the year should be less than the starting population. Hence, p cannot be less than 0.

∴ p = 0%

⇒ x - y + (x × -y)/100 = 0

⇒ x - xy/100 = y

⇒ x(100 - y)/100 = y

⇒ x = $\frac{100}{100-y}\times y$

Now, y < 100 (the population cannot decrease by 100%)

∴ $\frac{100}{100-\mathrm{y}}>1$

⇒ x = $\frac{100}{100-y}\times y$ > y

Hence, option (b).

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