PE 1 - Average

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).

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).

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).

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).

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).

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.

a + b = 110
a + c = 112
...
c + e = 120
d + e = 121

Box 'a' will make a total of 4 pairs (one each with b, c, d and e)

∴ 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).

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 

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

⇒ 22n² - 36n = 22n² - 36n + 44n - 72 - 2n² - 4n 

⇒ n² - 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).

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).

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.

A’_n = A(n, n²) = (n + n²)/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.