Arithmetic Mean - GMAT Quantitative

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Question

Salaries for employees at ABC Company: 1 employee makes $25,000 per year, 4 employees make $40,000 per year, 2 employees make $50,000 per year and 5 employees make $75,000 per year.

What is the average (arithmetic mean) salary for the employees at ABC Company?

Answer

The average is found by calculating the total payroll and then dividing by the total number of employees. $\frac{(1\cdot 25,000)+(4\cdot 40,000)+(2\cdot 50,000)+(5\cdot 75,000)}{1+4+2+5}$

$\frac{25,000+160,000+100,000+375,000}{12} = \frac{660,000}{12}= $55,000$

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Question

A bowler had an average (arithmetic mean) score of 215 on the first 5 games she bowled. What must she bowl on the 6th game to average 220 overall?

Answer

For the first 5 games the bowler has averaged 215. The equation to calculate the answer is

$\frac{(215\cdot 5)+x}{6}=220$

where $\dpi{100} \small x$ is the score for the sixth game. Next, to solve for the score for the 6th game $\dpi{100} \small (x)$ multiply both sides by 6:

$(215\cdot 5)+x =1,320$

which simplifies to:

$1,075+x =1,320$

After subtracting 1,075 from each side we reach the answer:

$x =1,320 - 1, 075 = 245$

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Question

Ashley averaged a score of 87 on her first 5 tests. She scored a 93 on her 6th test. What is her average test score, assuming all 6 tests are weighted equally?

Answer

We can't just average 87 and 93! This will give the wrong answer! The average formula is $\dpi{100} \small average = \frac{sum}{number\ of\ terms}$ .

For the first 5 tests, $\dpi{100} \small 87=\frac{sum}{5}$ . Then $\dpi{100} \small sum=87\times 5=435$ .

Now combine that with the 6th test to find the overall average.

$\dpi{100} \small average = \frac{435+93}{6}=88$

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Question

Sabrina made $3,000 a month for three months, $4,000 the next month, and $5,200 a month for the following two months. What was her average monthly income for the 6 month period?

Answer

$\dpi{100} \small average = \frac{3\times 3000 + 4000 + 2\times 5200}{6} = $ 3900$

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Question

Luke counts the number of gummy bears he eats every day for 1 week: {39, 18, 24, 51, 40, 15, 23}. On average, how many gummy bears does Luke eat each day?

Answer

$average = \frac{39+18+24+51+40+15+23}{7} = 30$

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Question

The average of 10, 25, and 70 is 10 more than the average of 15, 30, and x. What is the missing number?

Answer

The average of 10, 25, and 70 is 35: $\frac{10+25+70}{3}=35$

So the average of 15, 30, and the unknown number is 25 or, 10 less than the average of 10, 25, and 70 (= 35)

so $\frac{15+30+x}{3}=25$

$15+30+x=75$

$45+x=75$

$x=30$

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Question

What is the average of 2_x_, 3_x_ + 2, and 7_x_ +4?

Answer

$average = \frac{sum}{terms} = \frac{2x + 3x + 2 + 7x + 4}{3} = \frac{12x + 6}{3} = 4x +2$

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Question

The average high temperature for the week is 85. The first six days of the week have high temperatures of 89, 76, 92, 90, 80, and 84, respectively. What is the high temperature on the seventh day of the week?

Answer

$average = \frac{sum}{days}$

$85=\frac{89+76+92+90+80+84+x}{7}=\frac{511+x}{7}$

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Question

Jimmy's grade in his finance class is based on six equally-weighted tests. If Jimmy scored 98, 64, 82, 90, 70, and 88 on the six tests, what was his grade in the class?

Answer

$avg=\frac{98+64+82+90+70+88}{6}=\frac{492}{6}=82$

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Question

Find the mean of the sample data set.

Answer

The mean of a sample data set is the sum of all of the values divided by the total number of values. In this case:

$\frac{28+30+31+35+41}{5}=\frac{165}{5}=33$

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Question

Sandra's grade in economics depends on seven tests - five hourly tests, a midterm, and a final exam. The midterm counts twice as much as an hourly test; the final, three times as much.

Sandra's grades on the five hourly tests are 84, 86, 76, 89, and 93; her grade on the midterm was 72. What score out of 100 must she achieve on the final exam so that her average score at the end of the term is at least 80?

Answer

This is a weighted mean, with the hourly tests assigned a weight of 1, the midterm assigned a weight of 2, and the final assigned a weight of 3. The total of the weights will be

.

If we let be Sandra's final exam score, Sandra's final weighted average will be

$\small \frac{84 + 86 + 76+ 89+ 93 + 72 \cdot 2 + N \cdot 3}{10}$

$\small = \frac{84 + 86 + 76+ 89+ 93 + 144 + 3N}{10}$

$\small = \frac{572 + 3N}{10}$

For Sandra to get a final average of 80, then we set the above equal to 80 and calculate :

$\small \frac{572 + 3N}{10} = 80$

$\small 572 + 3N = 800$

$\small 3N = 228$

$\small N = 76$

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Question

Consider the following set of numbers:

85, 87, 87, 82, 89

What is the mean?

Answer

$mean=\frac{85+87+87+82+89}{5}=86$

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Question

Consider the following set of numbers:

85, 87, 87, 82, 89

What is the median?

Answer

Reorder the values in numerical order:82, 85, 87, 87, 89

The median is the center number, 87.

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Question

The average of the following 6 digits is 75. What is a possible value of ?

80, 78, 78, 70, 71,

Answer

Therefore, the sum of all 6 digits must equal 450.

Subtract 377 from both sides.

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Question

When assigning a score for the term, a professor takes the mean of all of a student's test scores except for his or her lowest score.

On the first five exams, Donna has achieved the following scores: 76, 84, 80, 65, 91. There is one more exam in the course. Assuming that 100 is the maximum possible score, what is the range of possible final averages she can achieve (nearest tenth, if applicable)?

Answer

The worst-case scenario is that she will score 65 or less, in which case her score will be the mean of the scores she has already achieved.

$\frac{76+ 84+80+65+ 91}{5} = 79.2$

The best-case scenario is that she will score 100, in which case the 65 will be dropped and her score will be the mean of the other five scores.

$\frac{76+ 84+80+ 91+100}{5} = 86.2$

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Question

If and , then give the mean of , , , , and .

Answer

The mean of , , , , and is $\frac{1}{5} \left (a + b + c + d + e \right )$

If you add both sides of each equation:

$\underline{5a + 4b + 3c + 2d + e = 5,300}$

or

$6\left (a + b + c + d + e \right )= 8,700$

Equivalently,

$6\left (a + b + c + d + e \right ) \div 6= 8,700 \div 6$

$\frac{1}{5} \left (a + b + c + d + e \right )= \frac{1}{5} \cdot 1,450 = 290$ ,

making 290 the mean.

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Question

What is the mean of the following data set in terms of and ?

$\left {a -x, a, a + x, a + 2x, 2a-x, 2a, 2a+x, 2a+2x \right }$

Answer

Add the expressions and divide by the number of terms, 8.

The sum of the expressions is:

$\left (a -x \right ) + a + \left ( a + x \right ) +\left ( a +2x \right ) + \left ( 2a-x \right ) + 2a + \left ( 2a+x \right ) +\left ( 2a+2x \right )$

Divide this by 8:

$\left ( 12a + 4x \right )\div 8 =\frac{3}{2} a +\frac{1}{2} x$

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Question

What is the mean of this data set?

$\left { 1, 0.1, 0.01, 0.001, 0.0001, 0.00001 \right }$

Answer

Add the numbers and divide by 6:

$\left (1 + 0.1 + 0.01 + 0.001+ 0.0001 + 0.00001 \right )\div 6$

$= 1.11111\div 6 = 0.185185$

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Question

Julie's grade in a psychology class depends on three tests, each of which are equally weighted; one term paper, which counts half as much as a test; one midterm, which counts for one and one-half as much as a test; and one final, which counts for twice as much as the other tests.

Julie has scored 85%, 84%, and 74% on her three tests, 90% on her term paper, and 72% on her midterm. She is going for an 80% in the course; what is the minimum percent she must score on the final (assuming that 100% is the maximum possible) to achieve this average?

Answer

Let be her final grade. Julie's final score is calculated as a weighted mean, so we can set up the following inequality:

$\frac{85 + 84 + 74 + 0.5 \cdot90 + 1.5 \cdot 72 + 2 \cdot X}{1 + 1 + 1 + 0.5 + 1.5 + 2} \geq 80$

Simplify and solve for :

$\frac{85 + 84 + 74 + 45+ 108 + 2 \cdot X}{7} \geq 80$

$\frac{396 + 2 X}{7} \geq 80$

$\frac{396 + 2 X}{7} \cdot 7 \geq 80 \cdot 7$

$396 + 2 X \geq 560$

$396 + 2 X -396 \geq 560 -396$

$2 X \geq 164$

$2 X \div 2 \geq 164 \div 2$

$X \geq 82$

Julie must make a minimum of 82% on the final to meet her goal.

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Question

The mean of the following terms is 11. Solve for .

15, 12, 9, 7, 17, 8,

Answer

The mean times the number of terms is equal to 77.

$7\times11=77$

The sum of all of the terms must equal 77.

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