# PE 3 - Percentage | Arithmetic - Percentage

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**PE 3 - Percentage | Arithmetic - Percentage**

The value of a machine depreciates at a rate of 16% per annum. If the price of a new machine is Rs. 15,625, then its value after 3 years will be

Answer: 9261

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**Explanation** :

Value of machine after 3 years = 15,625 × ${\left(1-\frac{16}{100}\right)}^{3}$ = 15,625 × ${\left(\frac{21}{25}\right)}^{3}$ = Rs. 9,261.

Hence, option (c).

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**PE 3 - Percentage | Arithmetic - Percentage**

In an exhibition, 40% of the collection is from the sale of champaign bottles and rest is from the sale of wine bottles. Had 30 more wine bottles been sold, their contribution would have been 80%. If the number of wine bottles sold is thrice the number of champaign bottles sold, find the total number of champaign bottles sold.

- (a)
4

- (b)
7

- (c)
6

- (d)
8

- (e)
5

Answer: Option C

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**Explanation** :

Let the number of champaign bottles sold be x, and wine bottles sold is 3x.

Let the price of each bottle of champaign be 'c' and that of each wine bottle be 'w'.

Total collection from the sales of champaign and wine bottles = (cx + 3wx)

40% of the collection is from sale of champaigns (given).

∴ $\frac{40}{100}$ × (cx + 3wx) = cx.

⇒ 2cx + 6wx = 5cx

⇒ c = 2w

Now, had 30 more wine bottles been sold, their contribution would have been 80%.

∴ (3x + 30)w = $\frac{80}{100}$ × [(3x + 30)w + cx]

⇒ 15wx + 150w = 12wx + 120w + 4cx

⇒ 3wx + 30w = 4cx

⇒ 3wx + 30w = 8wx [∵ c = 2w]

⇒ 3x + 30 = 8x

⇒ x = 6.

Hence, option (c).

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**PE 3 - Percentage | Arithmetic - Percentage**

Sunita made 4 mistakes in an exam and obtained 75% marks. If she had attempted 6 more questions and made 2 mistakes from these 6 questions, then she would have obtained 85% marks. If all the questions carried equal marks and there was no negative marking for wrong answers, then how many questions were there in the exam?

- (a)
36

- (b)
35

- (c)
40

- (d)
50

- (e)
None of these

Answer: Option C

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**Explanation** :

Let the total number of questions be x.

Let Sunita attempt ‘a’ number of questions correct.

Then, $\frac{a}{x}$ = $\frac{75}{100}$ = $\frac{3}{4}$

⇒ 4a = 3x ...(1)

Also, according to the given condition,

$\frac{a+4}{x}$ = $\frac{85}{100}$ = $\frac{17}{20}$

[∵ Out of 6 additional questions attempted, she got 2 wrong i.e., she got 4 more correct answers]

⇒ 20a + 80 = 17x ...(2)

From (1) and (2),

⇒ 15x + 80 = 17x

⇒ x = 40

∴ There are a total 40 questions in the exam.

**Alternately**,

Because of 4 additional correct answers her marks increased by 10%.

⇒ 10% of total no. of questions = 4

⇒ total no. of questions = 4/10% = 40

Hence, option (c).

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**PE 3 - Percentage | Arithmetic - Percentage**

ABC ltd. is an IT company and places an order for ‘x’ computers. The manufacturer can produce a maximum of ‘y’ computers daily, out of which a% are rejected and cannot be delivered. How many days will the computer manufacturer take to complete the order of ABC ltd.?

- (a)
$\frac{100x}{{y}^{2}\left(100-a\right)}$

- (b)
$\frac{100x}{y\left(100-a\right)}$

- (c)
$\frac{75x}{y{\left(100-a\right)}^{3}}$

- (d)
$\frac{x}{y\left(100-a\right)}$

- (e)
$\frac{(100-a)x}{100y}$

Answer: Option B

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**Explanation** :

Since a% of daily production is rejected, hence

Daily supply = (100 - a)% of y = $\frac{\left(100-a\right)y}{100}$

Total requirement = x.

∴ Required days = $\frac{x}{\frac{\left(100-a\right)y}{100}}$ = $\frac{100x}{y\left(100-a\right)}$

Hence, option (b).

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**PE 3 - Percentage | Arithmetic - Percentage**

Dinesh, Prateek and Ankush go for a party in a restaurant. Dinesh has four times as many coins as Prateek and 9 more than Ankush. When Dinesh gives away 50% of his coins, Prateek and Ankush agree to split those coins up equally. Now, Ankush has twice as many coins as Prateek. How many coins did Prateek start with?

- (a)
9

- (b)
6

- (c)
10

- (d)
4

- (e)
8

Answer: Option A

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**Explanation** :

Let Prateek has x coins.

Dinesh has four times as many coins as Prateek i.e., number of coins with Dinesh = 4x.

Dinesh has 9 more coins than Ankush i.e., number of coins with Ankush = 4x – 9

Now, when Dinesh gave 50% i.e., 2x coins, Prateek and Ankush divided them equally among themselves.

∴ Each of them receives x coins and now Ankush has twice the number of coins as Prateek.

∴ (4x - 9 + x) = 2(x + x)

⇒ x = 9

∴ Prateek had 9 coins.

Hence, option (a).

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**PE 3 - Percentage | Arithmetic - Percentage**

At the end of every year, the value of a gold coin is 'c' percent more than that at the start of the year. If the value was 'k' dollars on January 1, 1992 and 'm' dollars on January 1, 1994, then in terms of 'm' and 'k', what was the value (in dollars) on January 1, 1995?

- (a)
m + $\frac{1}{2}$(m - k)

- (b)
m + $\frac{1}{2}\left(\frac{m-k}{k}\right)m$

- (c)
$m\sqrt{\frac{m}{k}}$

- (d)
$\frac{{m}^{2}}{2k}$

Answer: Option C

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**Explanation** :

Value on January 1, 1992 = k

Value on January 1, 1993 = k(1 + c/100)

Value on January 1, 1994 = k(1 + c/100)(1 + c/100) = k(1 + c/100)^{2}

As per the question,

⇒ m = k(1 + c/100)^{2}

⇒ (1 + c/100) = (m/k)^{1/2}

Value on January 1, 1995 = k(1 + c/100)^{3} = k(m/k)^{3/2} = (m√m/√k)

Hence, option (c).

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**PE 3 - Percentage | Arithmetic - Percentage**

For a road to be built in Alabama, the committee decided that each and every member of the town will give a fixed amount of money for the purpose. Thus, 75% of the amount needed was collected from the citizens of Alabama, which equaled $600 on an average. If it is known that people of Alabama represent 60% of the total population of the area that the road will serve and if the road committee has to raise exactly the amount needed for building the road, what should be the average donation from people of rest of the service area?

- (a)
$200

- (b)
$300

- (c)
$400

- (d)
$500

- (e)
$600

Answer: Option B

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**Explanation** :

Let x be the total number of people from which the committee will ask for donations.

People of Alabama = 0.6x and the remaining people = 0.4x

Amount raised from the people of Alabama = 600 × 0.6x = $360x

Now, 360x constitutes 75% of the amount.

Hence, remaining 25% = $120x

Average donation from the remaining people = 120x/0.4x = $300.

Hence, option (b).

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**PE 3 - Percentage | Arithmetic - Percentage**

x% of y is z and (x + 20)% of y is equal to 3z. Value of y, when expressed in terms of z, is

- (a)
2z

- (b)
5z

- (c)
10z

- (d)
z + 10

- (e)
None of these

Answer: Option C

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**Explanation** :

x% of y = z ... (1)

Also, (x + 20)% of y = 3z ... (2)

(2) - (1)

⇒ 20% of y = 2z

⇒ y = 10z

Hence, option (c).

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**PE 3 - Percentage | Arithmetic - Percentage**

The cost of manufacturing an article is made up of components A, B, C and D, which have a ratio of 1 : 2 : 3 : 4, respectively. If there are respective changes in the costs of these four components as -10%, +20%, +30% and -40%, then what would be the percentage change in the cost of the article?

- (a)
-6.11%

- (b)
3.33%

- (c)
-4%

- (d)
+4%

- (e)
0%

Answer: Option C

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**Explanation** :

Let the costs of components A, B, C and D be 100, 200, 300 and 400, respectively.

∴ Total cost = 1000.

Change in cost will be as follows.

For A = -10% of 100 = -10

For B = +20% of 200 = +40

For C = +30% of 300 = +90

For D = -40% of 400 = -160

∴ Total change in cost = -10 + 40 + 90 – 160 = -40

∴ % change = (-40)/1000 × 100 = -4%

Hence, option (c).

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**PE 3 - Percentage | Arithmetic - Percentage**

A company regularly changes price of its laptops according to market demand. In January, they increased the price by X% and decrease it by X% in February. Due to it, the price of the laptop was decreased by Rs. 200 with respect to its price in January. Company did the same thing in March and April, first increasing the price by X% and then decreasing it by X%. The final price after whole cycle was Rs. 19,602. What is the original price of the laptop before January?

- (a)
Rs. 19,778

- (b)
Rs. 20,000

- (c)
Rs. 20,166

- (d)
Rs. 22,000

- (e)
Rs. 22,250

Answer: Option B

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**Explanation** :

For the first increase decrease cycle of +x% and -x%, overall change would be (-x^{2}/100)%. This results in a decrease of 200 and the price becomes P – 200.

The next increase decrease cycle of +x% and -x% would be on a smaller base of (P – 200). The overall change would be the same(-x^{2}/100)%. The absolute change would be (P - 200) × x^{2}/100 which would definitely be less than 200.

Hence, the total change after 2 such cycles would definitely be more than 200 but less than 400. Only option (b) satisfies this criterion.

**Alternately,**

Let P be the original price of the laptop.

According to the question,

After one cycle, P$\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right)$ = P - 200

$\left(1+\frac{x}{100}\right)\left(1-\frac{x}{100}\right)$ = $\frac{P-200}{P}$ ...(1)

After second cycle, P${\left(1+\frac{x}{100}\right)}^{2}{\left(1-\frac{x}{100}\right)}^{2}$ = 19,602

From (1)

$\frac{{(P-200)}^{2}}{P}$ = 19602 ...(2)

⇒ (P – 200)^{2} – 19602P = 0

⇒ P^{2} – 20002P + 40000 = 0

⇒ P = 20,000 and 2

P = 2 is not possible.

Hence, P = Rs. 20,000

Hence, option (b).

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