PE 3 - Average | Arithmetic - Average

Answer: Option C

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Explanation :

Let the first number be 'a' and the second number be 'b'.

When first number is doubled.
⇒ $\frac{2\mathrm{a}+\mathrm{b}}{2}$ = $\frac{\mathrm{a}+\mathrm{b}}{2}$ + 15
⇒ a = 30

When second number is doubled.
⇒ $\frac{\mathrm{a}+2\mathrm{b}}{2}$ = $\frac{\mathrm{a}+\mathrm{b}}{2}$ + 20
⇒ a = 40

Alternately,
When one of the numbers is doubled, the average of 2 numbers increases by 15, hence the sum of these two numbers increases by 30.

⇒ The first number must be 30.

Similarly, when the other number is doubled, the average increases by 20, which implies that the second number must be 40. (∵ 20 × 2 = 40)

Thus, the required difference = 40 - 30 = 10

Hence, option (c).

Answer: Option A

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Explanation :

Let the initial average mess expenditure per student be x.

Then, total expenditure = 42x

After 26 students are left, new expenditure = 42x - 110

According to the question, 42x – 110 = 26(x + 5)

⇒ 42x - 110 = 26x + 130

⇒ 16x = 240 

⇒ x = 15

So, sum of original expenditure and the new expenditure = 42 × 15 + 26 × 20 = Rs. 1150

Hence, option (a).

Answer: Option A

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Explanation :

Sum of all 10 numbers  (S) = 10 × 27 = 350

Let the smallest number be s and the largest number be L.

So, S - s = 9x   ... (1)

Also, S - L = 9y ... (2)

Adding (1) and (2), we get 

2S - (s + L) = 9(x + y) = 9(20) = 180

⇒ 2S - 180 = (s + L)

⇒ S - 90 = (s + L)/2

⇒ 350 - 90 = 260 = (s + L)/2 = Average of the smallest and the largest numbers 

Hence, option (a).

Answer: Option B

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Explanation :

n consecutive numbers will be in an AP and we know the average of numbers in AP is same as the average of two extreme numbers, i.e., the average of n consecutive numbers lies exactly at the middle. 

So, even if the highest number is deleted, the average will be the middle number of the remaining (n – 1) numbers, which is just 0.5 less than the average of n numbers. 

∴ The answer is (a – ½)

Hence, option (b).

Answer: Option A

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Explanation :

A₁ = $\frac{a+b+c}{3}$

A₂ = avg $\left(\frac{a+b}{2},\frac{b+c}{2},\frac{c+a}{2}\right)$ = $\frac{a+b+c}{3}$

⇒ A₁ = A₂

Hence, option (a).

Answer: Option C

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Explanation :

Average of 2 numbers is greater than 50 means that the sum should be greater than 100. 

Different pairs are as follows:
(1, 100) … (1 pair)
(2, 99), (2, 100) … (2 pairs)
(3, 98), (3, 99), (3, 100) … (3 pairs)

... so on

(49, 52), (49, 53) ... (49, 100) ... (49 pairs)
(50, 51), (50, 52 … (50, 100) ... (50 pairs)

So, 1 + 2 + 3 + 4 + ... + 50 = 1275

Again
(51, 51), (51, 52), (51, 53), ... (51, 100) ... (50 pairs)
(52, 52), (52, 53), (52, 54), ... (52, 100) ... (49 pairs)
(53, 53), (53, 54), (53, 55), ... (53, 100) ... (48 pairs)

... so on

(99, 99), (99, 100) ... (2 pairs)
(100, 100) ... (1 pair)

So, 1 + 2 + 3 + 4 + ... + 50 = 1275

Hence, total = 1275 + 1275 = 2550 pairs

Hence, option (c).

Answer: Option D

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Explanation :

Assume a, b and c are the numbers. a > b > c 

y = (a + c)/2 ⇒ a + c = 2y

x = (a + b + c)/3 ⇒ a + b + c = 3x

⇒ a + c = 3x – b 

⇒ 2y = 3x – b 

⇒ 2(x – 2) = 3x – b 

2x – 4 = 3x – b  

⇒ x – b = - 4

So, the difference between x and b is 4.

Hence, option (d).

Answer: Option C

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Explanation :

As the mean is 'a', the weight of all the boxes should be equal to 7a.

∴ Sum of the 4 heaviest boxes will be less than 7a.

∴ Options (1) and (2) cannot be the answer. 

Also, sum of the 4 heaviest boxes has to be greater than the mean of the given 7 numbers, i.e. 3.5a. 

∴ Options (4) and (5) cannot be the answer. 

Hence, option (c).

Answer: Option D

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Explanation :

A₁ = $\frac{(x_1+x_2+x_3...+x_n)}{n}$

A₂ = avg$\left(\frac{x_1+x_2}{2},\frac{x_2+x_3}{2},...,\frac{x_n+x_1}{2}\right)$ = $\frac{\left(\frac{x_1+x_2}{2},\frac{x_2+x_3}{2},...,\frac{x_n+x_1}{2}\right)}{n}$ = $\frac{\left(\frac{2(x_1+x_2+x_3...+x_n)}{2}\right)}{n}$ = $\frac{(x_1+x_2+x_3...+x_n)}{n}$

A₃ = avg$\left(\frac{x_1+x_2+x_3}{3},\frac{x_2+x_3+x_4}{3},...,\frac{x_n+x_1+x_2}{3}\right)$ = $\frac{\left(\frac{x_1+x_2+x3}{3},\frac{x_2+x_3+x4}{3},...,\frac{x_n+x_1+x_2}{3}\right)}{n}$ = $\frac{\left(\frac{3(x_1+x_2+x_3...+x_n)}{3}\right)}{n}$ = $\frac{(x_1+x_2+x_3...+x_n)}{n}$

A₁ = A₂ = A₃

Hence, option (d).

Answer: Option A

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Explanation :

A(n) is the average of all natrual numbers less than n.

∴ A(x) is average of all natrula numbers from 1 till x - 1 = $\frac{1+x-1}{2}$ = $\frac{x}{2}$

Similarly,  A(y) is average of all natrula numbers from 1 till y - 1 = $\frac{y}{2}$

and A(x + y) is average of all natrula numbers from 1 till x + y - 1 = $\frac{x+y}{2}$

∴ A(x) + A(y) = A(x + y)

Hence, option (a).