# PE 1 - Average | Arithmetic - Average

**Answer the next 4 questions based on the information given below.**

There are 2 classes having 20 and 25 students respectively having the average marks in an examination as 40 and 50 respectively. If the two classes are represented by A and B. It is given that

A ⟶ Highest score 44, Lowest score 36

B ⟶ Highest score 62, Lowest score 46

If 5 people are transferred from B to A and another independent set of 5 people are transferred back from A to B, then after this operation (Assume that the set transferred from B to A contains none from the set of students that came to B from A)

**PE 1 - Average | Arithmetic - Average**

What will happen to A’s average?

- (a)
Increase if B’s average decreases

- (b)
Increases

- (c)
Decrease if B’s average increases

- (d)
Cannot be determined

Answer: Option B

**Explanation** :

Given, the highest score in A is less than the lowest score in B.

When 5 people are transferred from B to A, all these 5 people will have score more than or equal to 46.

When 5 people are transferred back from A to B, all these 5 people will have score less than or equal to 44.

Hence, for A, people leaving have score ≤ 44 and people coming in have score ≥ 46.

∴ Average for A will definitely increase.

Hence, option (b).

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**PE 1 - Average | Arithmetic - Average**

What can be said about B’s average?

- (a)
Decreases

- (b)
Increases

- (c)
May increase or decrease

- (d)
Will increase only if B's average decreases

Answer: Option A

**Explanation** :

Consider the solution to first question of this set.

For B, people coming in have score ≤ 44 and people leaving have score ≥ 46.

∴ B’s average will definitely decrease.

Hence, option (a).

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**PE 1 - Average | Arithmetic - Average**

At the end of the 2 steps mentioned above (in the directions) what could be the maximum value of the average of class B?

- (a)
50.8

- (b)
50

- (c)
49.6

- (d)
49.2

Answer: Option C

**Explanation** :

For B’s average to be maximum, 5 people leaving should weigh 46 each and 5 people coming in should weight 44 each.

∴ New average of B = (50 × 25 - 46 × 5 + 44 × 5)/25 = 49.6

**Alternately**,

The net decrease in total for B will be 5 × 2 = 10.

∴ Decrease in average = 10/25 = 0.4

∴ New average = 50 - 0.4 = 49.6

Hence, option (c).

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**PE 1 - Average | Arithmetic - Average**

At the end of the 2 steps mentioned above (in the directions) what could be the minimum value of the average of class B?

- (a)
44.8

- (b)
48.4

- (c)
50

- (d)
46

Answer: Option A

**Explanation** :

For B’s average to be minimum, 5 people leaving should weigh 62 each and 5 people coming in should weight 36 each.

∴ New average of B = (50 × 25 - 62 × 5 + 36 × 5)/25 = 44.8

**Alternately**,

The net decrease in total for B will be 5 × (62 - 36) = 130.

∴ Decrease in average = 130/25 = 5.2

∴ New average = 50 - 5.2 = 44.8

Hence, option (a).

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**PE 1 - Average | Arithmetic - Average**

A certain company employed 600 men and 400 women and the average wage was Rs. 2.55 per day. If a woman got 50 paise less than what a man got, then what were their daily wages?

- (a)
Man - Rs. 3.00 and woman - Rs. 2.50

- (b)
Man - Rs. 3.50 and woman - Rs. 3.00

- (c)
Man - Rs. 2.75 and woman - Rs. 2.25

- (d)
Man - Rs. 3.25 and woman - Rs. 2.75

- (e)
Man - Rs. 2.25 and woman - Rs. 2.75

Answer: Option C

**Explanation** :

Total employees = 600 + 400 = 1000

Total income at an average Rs. 2.55/day = 2.55 × 1000 = Rs. 2550

Let the daily wage of a man be Rs. x.

Then, their total wages = Rs. 600x

A woman got 50 paise less than a man.

∴ Their total wages = 400(x - 0.5)

⇒ 600x + 400(x - 0.5) = 2550

⇒ 600x + 400x - 200 = 2550

⇒ 1000x = 2750

⇒ x = Rs. 2.75

∴ Man’s daily wage = Rs. 2.75

Woman’s daily wage = 2.75 - 0.50 = Rs. 2.25

Hence, option (c).

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**PE 1 - Average | Arithmetic - Average**

A shipping clerk has five boxes of different but unknown weights, each weighing less than 100 kg. The clerk weighs the boxes in pairs. The weights obtained are 110, 112, 113, 114, 115, 116, 117, 118, 120 and 121 kg. What is the weight of the second heaviest box?

- (a)
62 kgs

- (b)
61 kgs

- (c)
60 kgs

- (d)
59 kgs

Answer: Option D

**Explanation** :

Let the weights in increasing order be a, b, c, d and e. (a is the lightest and e the heaviest)

Since all possible pairs are formed, each box will form a pair with each of the other 4 boxes. Hence, each box will have 4 pairs.

∴ When we add all such pairs, each box will get added 4 times.

∴ 4(a + b + c + d + e) = 110 + 112 + 113 + 114 + 115 + 116 + 117 + 118 + 120 + 121.

⇒ a + b + c + d + e = 289 …(1)

Since a and b are the two lightest boxes

∴ a + b = 110 …(2)

Also, since d and e are the two heaviest boxes

∴ d + e = 121 …(3)

From (1), (2) and (3) we get,

c = 289 – 110 – 121 = 58

The second lightest pair will be a and c.

∴ a + c = 112

⇒ a = 112 – 58 = 54 …(4)

From (2) and (4) we get,

b = 110 – 54 = 56

The second heaviest pair will be c and e.

∴ e + c = 120

⇒ e = 120 – 58 = 62 …(5)

From (3) and (5) we get,

d = 121 – 62 = 59

∴ The second heaviest box is 59 kgs.

Hence, option (d).

Workspace:

**PE 1 - Average | Arithmetic - Average**

An enterprise got a bonus and decided to share it in equal parts between the workers. Two more workers joined the enterprise. Hence, each of them got 2 rupees less. The administration had found the possibility to increase the total sum of the bonus by 80 rupees and as a result, each worker including the two new workers got 22 rupees. If the numbers of workers at any point of time were more than 10, how many total workers are there now?

- (a)
15

- (b)
18

- (c)
20

- (d)
16

Answer: Option C

**Explanation** :

Let initial bonus = x and initial number of workers = n

Each worker gets x/n.

Condition (1): $\frac{x}{n+2}=\frac{x}{n}-2$ ... (1)

Condition (2): $\frac{x+80}{n+2}=22$ …(2)

Solving (2), we get

x = 22n - 36

From (1), we get

$\frac{22n-36}{n+2}=\frac{22n-36}{n}-2$

⇒ 22n^{2} - 36n = 22n^{2} - 36n + 44n - 72 - 2n^{2} - 4n

⇒ n^{2} - 20n + 36 = 0

⇒ n = 2 or 18

Since number of workers is more than 10, we will reject n = 2.

∴ Number of workers now = 18 + 2 = 20.

Hence, option (c).

Workspace:

**PE 1 - Average | Arithmetic - Average**

A publisher incurs a fixed expense of Rs. 1,00,000 towards composing and other activities in printing books. A variable cost of Rs. 12.5 per book is incurred towards paper, binding charges etc. What is the average cost per book if 80,000 copies are printed?

- (a)
Rs. 13.25

- (b)
Rs. 13.50

- (c)
Rs. 13.75

- (d)
Rs. 12.40

- (e)
None of these

Answer: Option C

**Explanation** :

Fixed expense = Rs. 1,00,000

Variable cost per book = Rs. 12.5

∴ Total variable cost for 80,000 books = Rs. 12.5 × 80,000 = Rs. 10,00,000

Total cost of 80,000 books = Rs. 10,00,000 + Rs. 1,00,000 = Rs. 11,00,000

Average cost per book = 11,00,000/80,000 = Rs. 13.75.

**Alternately**,

Average fixed cost per book = 1,00,000/80,000 = 1.25

∴ Average total cost per book = 1.25 + 12.5 = 13.75.

Hence, option (c).

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**PE 1 - Average | Arithmetic - Average**

If the average of 20 consecutive numbers, starting from x, is equal to the average of 30 consecutive numbers, starting from y, then x - y is equal to

Answer: 5

**Explanation** :

20 consecutive number starting from x are: x, x + 1, x + 2, …, x + 19

∴ The average of 20 consecutive numbers, starting from x is (x + x + 19)/2 = x + 9.5 (∵ average of all the terms in an A.P. is same as the average of first and the last terms)

Similarly, the average of 30 consecutive numbers, starting from y is y + 14.5.

So, x + 9.5 = y + 14.5

⇒ x - y = 5

Hence, 5.

Workspace:

**PE 1 - Average | Arithmetic - Average**

If A(a_{1}, a_{2}, …a_{n}) = Average of (a_{1}, a_{2}, …, a_{n}) and A’_{n} = A(n, n^{2}), then what is the value of A(A’_{1}, A’_{2}, …, A’_{n}), where n = 100?

Answer: 1717

**Explanation** :

A’_{n} = A(n, n^{2}) = (n + n^{2})/2,

A(A’_{1}, A’_{2}, …, A’_{100}) = A$\left(\frac{1+{1}^{2}}{2},\frac{2+{2}^{2}}{2},\frac{3+{3}^{2}}{2},...,\frac{100+{100}^{2}}{2}\right)$.

= $\frac{\left(\frac{1+{1}^{2}}{2},\frac{2+{2}^{2}}{2},\frac{3+{3}^{2}}{2},...,\frac{100+{100}^{2}}{2}\right)}{100}$

= $\frac{(1+2+...+100)+({1}^{2}+{2}^{2}+...+{100}^{2})}{200}$

= $\frac{{\displaystyle \frac{100\times 101}{2}}+{\displaystyle \frac{100\times 101\times 201}{6}}}{200}$

= $\frac{101}{4}\left(1+\frac{201}{3}\right)$

= $\frac{101}{4}\times 68$

= 1717

Hence, 1717.

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