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Lisa gave four English tests last week. The score of her fourth test w
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10 Nov 2021, 01:30
10 Nov 2021, 01:30
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Lisa gave four English tests last week. The score of her fourth test was 8 more than average score of her first three tests. How much greater is her fourth test score than the average score of four tests?
A. 2
B. 4
C. 6
D. 8
E. 10
A. 2
B. 4
C. 6
D. 8
E. 10
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Re: Lisa gave four English tests last week. The score of her fourth test w
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10 Nov 2021, 06:02
10 Nov 2021, 06:02
1
Carcass wrote:
Lisa gave four English tests last week. The score of her fourth test was 8 more than average score of her first three tests. How much greater is her fourth test score than the average score of four tests?
A. 2
B. 4
C. 6
D. 8
E. 10
A. 2
B. 4
C. 6
D. 8
E. 10
Great question!
Lisa gave four English tests last week. The score of her fourth test was 8 more than average score of her first three tests.
Let A, B, C and D represent the scores of the 4 tests (D is the fourth test)
From the given information, we can write: (4th score) - (average of first three tests) = 8
Substitute to get: D - (A + B + C)/3 = 8
Multiply both sides of the equation by 3 to get: 3D - (A + B + C) = 24
How much greater is her fourth test score than the average score of four tests?
In other words, we want to find the value of: (4th score) - (average all four tests)
Let's let x be the desired value.
In other words: (4th score) - (average all four tests) = x (and we want to find the value of x)
Substitute to get: D - (A + B + C + D)/4 = x
Multiply both sides of the equation by 4 to get: 4D - (A + B + C + D) = 4x
Simplify to get: 3D - (A + B + C) = 4x
Since we already learned that 3D - (A + B + C) = 24, we can substitute 24 for 3D - (A + B + C) to get 24 = 4x
Solve: x =6
Answer: C
Re: Lisa gave four English tests last week. The score of her fourth test w
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10 Nov 2021, 22:32
10 Nov 2021, 22:32
1
\(Avg(a,b,c) = \frac{a + b + c}{3}\)
\(d = 8 + Avg(a,b,c)\)
Difference = \(d - Avg(a,b,c,d) = d - \frac{a + b + c + d}{4} = \frac{3d - (a + b + c)}{4} = \frac{3d - 3Avg(a,b,c)}{4}\)
\(\frac{3d - 3Avg(a,b,c)}{4} = \frac{3(8 + Avg(a,b,c)) - 3Avg(a,b,c)}{4} = \frac{3(8) + 3Avg(a,b,c) - 3Avg(a,b,c)}{4} = \frac{3(8)}{4}\) = 6
Hence, Answer is C
\(d = 8 + Avg(a,b,c)\)
Difference = \(d - Avg(a,b,c,d) = d - \frac{a + b + c + d}{4} = \frac{3d - (a + b + c)}{4} = \frac{3d - 3Avg(a,b,c)}{4}\)
\(\frac{3d - 3Avg(a,b,c)}{4} = \frac{3(8 + Avg(a,b,c)) - 3Avg(a,b,c)}{4} = \frac{3(8) + 3Avg(a,b,c) - 3Avg(a,b,c)}{4} = \frac{3(8)}{4}\) = 6
Hence, Answer is C
