PE 4 - Average | Arithmetic - Average

Answer: Option B

Explanation :

Let ‘n’ be the number of numbers Anjali wanted to write on the board (including the missed number).

Let ‘m’ be the missed number.

Sum of all number, excluding the missed number = n(n + 1)/2 – m.

Average of these numbers (excluding the missed number) = $38\frac{1}{11}$ = $\frac{419}{11}$ = $\frac{{\displaystyle \frac{n(n+1)}{2}}-m}{n}$

The total no. of natural numbers she wanted to write on the black board (i.e., n) should be a multiple of 11. 

∴ n = 11k   …(1)

Now we know that the average of first n natural numbers is $\frac{(n+1)}{2}$. [Since the sum of first n natural nos. is $\frac{n(n+1)}{2}$].

Hence, no. of natural number = 2 × average – 1.

Even if she misses out one of the numbers the final average would be slightly less than $\frac{(n+1)}{2}$.

∴ The number of natural number written on the board would be around 2 × $\frac{419}{11}$ – 1 ≈ 75   …(2)

From (1) and (2)

The nearest possible number of 11k form around 75 is 77.

Hence, she wanted to write 77 numbers on the board.

∴ Sum of first 77 natural number = $\frac{77\times 78}{2}$ = 3003.

Sum of the number which she added up = 77 × $\frac{419}{11}$ = 2933.

Hence the number erased = 3003 - 2933 = 70.

Hence, option (b).

Answer: Option B

Explanation :

The average percentage of votes for the top three candidates is 30/3 = 10%. One of them must, therefore, have a percentage of votes less than or equal to 10%.

The average percentage of votes for the bottom six candidates is 45/6 = 7.5%. At least one of them must, therefore, have a percentage of votes greater than or equal to 7.5%.

Thus, the average percentage of the other candidates must lie between 10% and 7.5%, otherwise one of the other candidates would be either in the top three or the bottom six.

The total percentage of votes got by the other candidates is 100% - 30% - 45% = 25%. If there are N other candidates, their average percentage of votes is 25/N.

We have to solve the inequality 7.5 < 25/N < 10, for integral N.

This is possible only if N = 3.

Hence, option (b).

Answer: Option C

Explanation :

Let the average marks of the three sections A, B and C be a, b and c respectively.
Let the number of students in the three sections A, B and C be n_a, n_b and n_c respectively.

Average marks of sections A and B combined is 56.
⇒ 56 = $\frac{\mathrm{a}\times {\mathrm{n}}_{\mathrm{a}}+\mathrm{b}\times {\mathrm{n}}_{\mathrm{b}}}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}}$
⇒ an_a + bn_b = 56n_a + 56n_b   ...(1)

Similarly, average marks of sections B and C combined is 60.
⇒ bn_b + cn_c = 60n_b + 60n_c  ...(2)

Similarly, average marks of sections C and A combined is 64.
⇒ cn_c + an_a = 64n_c + 64n_a  ...(3)

Adding (1), (2) and (3), we get
⇒ 2an_a + 2bn_b + 2cn_c = 120n_a + 116n_b + 124n_c
⇒ an_a + bn_b + cn_c = 60n_a + 58n_b + 62n_c

Now, average of all three sections combined = $\frac{\mathrm{a}\times {\mathrm{n}}_{\mathrm{a}}+\mathrm{b}\times {\mathrm{n}}_{\mathrm{b}}+\mathrm{c}\times {\mathrm{n}}_{\mathrm{c}}}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}}}$

⇒ $\frac{\mathrm{a}\times {\mathrm{n}}_{\mathrm{a}}+\mathrm{b}\times {\mathrm{n}}_{\mathrm{b}}+\mathrm{c}\times {\mathrm{n}}_{\mathrm{c}}}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}}}$ = $\frac{60{\mathrm{n}}_{\mathrm{a}}+58{\mathrm{n}}_{\mathrm{b}}+62{\mathrm{n}}_{\mathrm{c}}}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}}}$

⇒ $\frac{\mathrm{a}\times {\mathrm{n}}_{\mathrm{a}}+\mathrm{b}\times {\mathrm{n}}_{\mathrm{b}}+\mathrm{c}\times {\mathrm{n}}_{\mathrm{c}}}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}}}$ = $\frac{58({\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}})+2{\mathrm{n}}_{\mathrm{a}}+4{\mathrm{n}}_{\mathrm{c}}}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}}}$

⇒ $\frac{\mathrm{a}\times {\mathrm{n}}_{\mathrm{a}}+\mathrm{b}\times {\mathrm{n}}_{\mathrm{b}}+\mathrm{c}\times {\mathrm{n}}_{\mathrm{c}}}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}}}$ = $\frac{58({\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}})}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}}}$ + $\frac{2{\mathrm{n}}_{\mathrm{a}}+4{\mathrm{n}}_{\mathrm{c}}}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}}}$

⇒ $\frac{\mathrm{a}\times {\mathrm{n}}_{\mathrm{a}}+\mathrm{b}\times {\mathrm{n}}_{\mathrm{b}}+\mathrm{c}\times {\mathrm{n}}_{\mathrm{c}}}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}}}$ = 58 + (positive number)

⇒ $\frac{\mathrm{a}\times {\mathrm{n}}_{\mathrm{a}}+\mathrm{b}\times {\mathrm{n}}_{\mathrm{b}}+\mathrm{c}\times {\mathrm{n}}_{\mathrm{c}}}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}}}$ > 58

Also, 

⇒ $\frac{\mathrm{a}\times {\mathrm{n}}_{\mathrm{a}}+\mathrm{b}\times {\mathrm{n}}_{\mathrm{b}}+\mathrm{c}\times {\mathrm{n}}_{\mathrm{c}}}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}}}$ = $\frac{60{\mathrm{n}}_{\mathrm{a}}+58{\mathrm{n}}_{\mathrm{b}}+62{\mathrm{n}}_{\mathrm{c}}}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}}}$

⇒ $\frac{\mathrm{a}\times {\mathrm{n}}_{\mathrm{a}}+\mathrm{b}\times {\mathrm{n}}_{\mathrm{b}}+\mathrm{c}\times {\mathrm{n}}_{\mathrm{c}}}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}}}$ = $\frac{62({\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}})-2{\mathrm{n}}_{\mathrm{a}}-4{\mathrm{n}}_{\mathrm{b}}}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}}}$

⇒ $\frac{\mathrm{a}\times {\mathrm{n}}_{\mathrm{a}}+\mathrm{b}\times {\mathrm{n}}_{\mathrm{b}}+\mathrm{c}\times {\mathrm{n}}_{\mathrm{c}}}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}}}$ = $\frac{62({\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}})}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}}}$ - $\frac{2{\mathrm{n}}_{\mathrm{a}}+4{\mathrm{n}}_{\mathrm{b}}}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}}}$

⇒ $\frac{\mathrm{a}\times {\mathrm{n}}_{\mathrm{a}}+\mathrm{b}\times {\mathrm{n}}_{\mathrm{b}}+\mathrm{c}\times {\mathrm{n}}_{\mathrm{c}}}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}}}$ = 62 - (positive number)

⇒ $\frac{\mathrm{a}\times {\mathrm{n}}_{\mathrm{a}}+\mathrm{b}\times {\mathrm{n}}_{\mathrm{b}}+\mathrm{c}\times {\mathrm{n}}_{\mathrm{c}}}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}}}$ < 62

∴ 58 < $\frac{\mathrm{a}\times {\mathrm{n}}_{\mathrm{a}}+\mathrm{b}\times {\mathrm{n}}_{\mathrm{b}}+\mathrm{c}\times {\mathrm{n}}_{\mathrm{c}}}{{\mathrm{n}}_{\mathrm{a}}+{\mathrm{n}}_{\mathrm{b}}+{\mathrm{n}}_{\mathrm{c}}}$ < 62

Hence, option (c)