Students in three different classes wrote the same Calculus exam. The exam was marked out of 100.
One class had 22 students in it. After writing the exam, their class average on
the exam was reported as 87%. The second class had 27 students in it. After
wiſiting the exam, their class average on the exam was reported as 83%. The
third class had 31 students in it. After writing the exam, their class average on
the exam was reported as 81%.
Three students, Alf, Bet, and Tildi, discussed their results. Alf obtained a mark
one less than Bet and Tildi obtained a mark one more than Bet.
Upon reviewing their papers, Alf and Bet both discovered addition errors on
their papers. Both of their marks increased to 92. Tildi discovered that one of
her questions had not been marked. This review resulted in her mark increasing
to 92 as well.
These changes resulted in the exam average for all of the students in the three
classes combined changing to exactly 84%.
What marks did Alf, Bet and Tildi originally have on their papers before the
errors were corrected?

Alf’s original mark was 73

Bet’s original mark was 74

Tildi’s original mark was 75

Step-by-step explanation:

From the question, we have

Three students, Alf, Bet, and Tildi, discussed their results. Alf obtained a mark one less than Bet and Tildi obtained a mark one more than Bet.

Hence,

Let Bet’s original mark be x .

Alf’s original mark is x - 1

Tildi’s original mark is x + 1.

We are told in the question also that:

First class had 22 students in it. After writing the exam, their class average on the exam

was reported as 87%.

Total score for the First class

= 22 × 87 = 1914

The second class had 27 students in it. After writing the exam, their

class average on the exam was reported as 83%.

Total score for the second class

=27 × 83 = 2241

The third class had 31 students in it. After

writing the exam, their class average on the exam was reported as 81%.

Total score for the third class =

31 × 81 = 2511.

Hence, the total number of scores before the correction of errors

= 1914 + 2241 + 2511 = 6666.

The total number of students in the three classes =

22 + 27 + 31 = 80.

From the question, we are told that the the average after correcting the three errors = 84%

Hence, the total number of marks for the three classes

= Number of students × Average

= 80 × 84 = 6720.

The difference in the total marks before and after correction =

6720 - 6666 = 54.

But the difference in the total marks can also be calculated by subtracting the old marks of Bet, Alf, Tildi from 92 and then adding the three differences.

Where,

Bet’s original mark be x .

Alf’s original mark is x - 1

Tildi’s original mark is x + 1.

So:

[92 - (x - 1)] + [92 - x] + [92 - (x + 1)] = 54.

92 - x + 1 + 92 - x + 92 - x - 1 = 54

We would collect like terms

92 + 92 + 92 - x - x - x + 1 - 1 = 54

276 - 3x = 54

222 = 3x

222/3 = x

74 = x

Since x = 74,

Bet’s original mark be x .

Hence, Bet's mark = 74

Alf’s original mark is x - 1

Alf's mark = 74 - 1 = 73

Tildi’s original mark is x + 1.

Tildi's mark = 74 + 1 = 75

Therefore, Alf’s original mark was 73, Bet’s original mark was 74 and Tildi’s original mark was 75

56.6

step-by-step explanation:

the ratio of the perimeters is equal to the scale.

the ratio of the areas is equal to the square of the scale.

34 / 20 = k

a / 19.6 = k²

substituting:

a / 19.6 = (34 / 20)²

a = 19.6 (34 / 20)²

a ≈ 56.6