1985 AJHSME Problems/Problem 17: Difference between revisions

Latest revision as of 07:14, 13 January 2023

Problem

If your average score on your first six mathematics tests was $84$ and your average score on your first seven mathematics tests was $85$ , then your score on the seventh test was

$\text{(A)}\ 86 \qquad \text{(B)}\ 88 \qquad \text{(C)}\ 90 \qquad \text{(D)}\ 91 \qquad \text{(E)}\ 92$

Solution 1

If the average score of the first six is $84$ , then the sum of those six scores is $6\times 84=504$ .

The average score of the first seven is $85$ , so the sum of the seven is $7\times 85=595$

Taking the difference leaves us with just the seventh score, which is $595-504=91$ , so the answer is $\boxed{\text{D}}$

Solution 2

Let's remove the condition that the average of the first seven tests is $85$ , and say the 7th test score was a $0$ . Then, the average of the first seven tests would be $\frac{6\times 84}{7}=72$

If we increase the seventh test score by $7$ , the average will increase by $\frac{7}{7}=1$ . We need the average to increase by $85-72=13$ , so the seventh test score is $7\times 13=91$ more than $0$ , which is clearly $91$ . This is choice $\boxed{\text{D}}$

Video Solution

https://youtu.be/PMlL7M0HzTY

~savannahsolver

1985 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

@@ Line 18: / Line 18: @@
 If we increase the seventh test score by <math>7</math>, the average will increase by <math>\frac{7}{7}=1</math>. We need the average to increase by <math>85-72=13</math>, so the seventh test score is <math>7\times 13=91</math> more than <math>0</math>, which is clearly <math>91</math>.  This is choice <math>\boxed{\text{D}}</math>
+==Video Solution==
+https://youtu.be/PMlL7M0HzTY
+~savannahsolver
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