PE 4 - Percentage | Arithmetic - Percentage
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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.
Workspace:
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) +
⇒ = 0.45 +
⇒ = 45% + %
⇒ 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) -
⇒ = 0.5 -
⇒ = 50% - %
⇒ 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).
Workspace:
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.
Workspace:
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 2nd day, if there were 1872 more bacteria at the end of the 3rd day compared to the end of the 2nd 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.
Workspace:
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).
Workspace:
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).
Workspace:
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 =
Now, y < 100 (the population cannot decrease by 100%)
∴
⇒ x = > y
Hence, option (b).
Workspace:
Answer the next 2 questions based on the information given below.
A shopkeeper sells only three types of items - pencil, sharper and eraser. On Monday morning, of the total items he had in the shop, 30% were pencils. On Monday he sold 40% pencils, 160 sharpners and 30% erasers. On Monday, he ends up selling 40% of items that he had in the morning.
What is the least number of items he could have had on Monday morning?
Answer: 650
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Explanation :
Number of pencils is 30% of total items = 3/10 × total items
Now, ‘3/10 × total items’ should be an integer, hence total items should be a multiple of 10.
Also, 40% of these are sold = 2/5 × 3/10 × total items = 3/25 × total items.
Now, ‘3/25 × total items’ should be an integer, hence total items should also be a multiple of 25.
⇒ Total items should be a multiple of LCM (10, 25) = 50.
Let the total number of items initially be ‘50i’. [i in an integer]
Now, 30% of erasers are sold = 3/10 × erasers
Now, ‘3/10 × erasers’ should be an integer, hence number of erasers should be a multiple of 10.
Let the number of erasers be ‘10e’. [e in an integer]
Now, items sold are
Pencils = 40% of 30% of 50i = 6i
Sharpeners = 160
30% of Erasers = 3e
Total 40% of items are sold
⇒ 40% of 50i = 6i + 160 + 3e
⇒ 20i = 6i + 160 + 3e
⇒ 14i = 160 + 3e
⇒ i =
⇒ i =
⇒ i = 11 + (4 + 3e)/14
Since, i & e are both integers, possible values of a should be such that (4 + 3e) should be divisible by 14.
∴ least possible value of ‘e’ = 8
⇒ Least possible value of i = 11 + 2 = 13
⇒ Least possible value of items = 50 × 13 = 650
Hence, 650
Workspace:
What is the lowest amount of sharpners he could have had on Monday morning?
Answer: 165
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Explanation :
Now, number of oranges = 50i – sharpeners – erasers
= 50i – 15i – 10e
= 35i – 10e
In the previous question we had,
⇒ i = 11 + (4 + 3e)/14
When e = 8, i = 13
⇒ Number of sharpeners = 35 × 13 – 10 × 8 = 375
When e = 22, i = 16
⇒ Number of sharpeners = 35 × 16 – 10 × 22 = 340
When e = 36, i = 19
⇒ Number of sharpeners = 35 × 19 – 10 × 36 = 305
i.e., Number of sharpeners decrease by 35 for subsequent values of e and i.
⇒ Number of sharpeners = 305 – 35x
Now, this has to be greater than or equal to 160 as 160 sharpeners are to be sold.
⇒ 305 – 35x ≥ 160
⇒ x ≤ 145/35 = 4.14
∴ Least number sharpeners on Monday morning = 305 – 35 × 4 = 165.
Hence, 165.
Workspace:
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