# Computations Involving the Mean, Sample Size, and Sum of a Data Set Online Quiz

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**Q 1 - The average of four numbers is 12. If three of those numbers are 7, 9 and 15, find the last number.**

### Answer : D

### Explanation

**Step 1:**

Let the last number be = x

Average = $ \frac{(7 + 9 + 15 + x)}{4}$ = 12.

**Step 2:**

Solving for x;

31 + x = 48; x = 48 – 31 = 17

**Q 2 - The mean of a, b and c is 26. The sum of a and twice the value of c is 28. What is a + 2b equal to?**

### Answer : C

### Explanation

**Step 1:**

Given a + b + c = 26 and a + 2c = 28

**Step 2:**

Subtracting the first from second we get c – b = 28 -26 = 2; c = 2 + b

**Step 3:**

Plugging value of c in first equation we get a + 2b = 24

**Q 3 - The average of x and 3 is equal to the average of x, 6 and 9. Find x.**

### Answer : B

### Explanation

**Step 1:**

Average of x and 3 = $\frac{(x + 3)}{2}$; Average of x, 6, and 9 = $\frac{(x + 6 + 9)}{3}$

**Step 2:**

Given $\frac{(x + 3)}{2} = \frac{(x + 15)}{3}$

Solving we get 3x + 9 = 2x + 30 or 3x – 2x = x = 30 – 9 = 21

**Step 3:**

So x = 21.

**Q 4 - 5 consecutive integers have an average of 123. What's the average of the first three of those integers?**

### Answer : A

### Explanation

**Step 1:**

Let the consecutive integers be a -2, a -1, a, a + 1 and a + 2

Average $\frac{(a -2 + a -1 + a + a + 1 + a + 2)}{5} = \frac{5a}{5}$ = 123

**Step 2:**

The numbers are 121, 122, 123, 124 and 125

Average of first three numbers = 122

**Q 5 - 5 consecutive odd integers have an average of 101. What's the average of the first three of those integers?**

### Answer : C

### Explanation

**Step 1:**

- Let the consecutive odd integers be a - 4, a -2, a, a + 2 and a + 4

Average $\frac{(a - 4 + a - 2 + a + a + 2 + a + 4)}{5} = \frac{5a}{5}$ = 101

**Step 2:**

The numbers are 97, 99, 101, 103 and 105

Average of first three numbers = 99

**Q 6 - P is a set of numbers with a mean of 10. Q is a set of numbers found by adding 2 to each of the numbers in set P. What is the average of the numbers in set Q?**

### Answer : D

### Explanation

**Step 1:**

Let there be n numbers x_{1}, x_{2}, ….x_{n} in set P. Their mean = 10.

Total in set P = (x_{1} + x_{2} + …x_{n}) = 10n

**Step 2:**

In set Q, n numbers x_{1}+2, x_{2}+2 ….x_{n}+2.

Their total (x_{1}+2, x_{2}+2 ….x_{n}+2) = (x_{1} + x_{2} + …x_{n}) + 2n = 10n + 2n = 12n

**Step 3**

Mean of set Q = $\frac{Total}{n}= \frac{12n}{n}$ = 12

**Q 7 - The average temperature over a 10-day period was 85 degrees. For the first six days, the average temperature was 87 degrees. Over the next 4 days, what was the average temperature?**

### Answer : B

### Explanation

**Step 1:**

Total degrees over 10 day period = 10 × 85 = 850

**Step 2:**

Let average temperature over 4 days = x

87 × 6 + 4 x = 850

Solving for x, 4x = 850 – 522 = 328

So x = $\frac{328}{4}$ = 82 degrees.

**Q 8 - The average of 14, 18 and 19 is 10 more than the average of 8 and what number?**

### Answer : A

### Explanation

**Step 1:**

Average of 14, 18 and 19 = $\frac{(14 + 18 + 19)}{3} = \frac{51}{3}$ = 17

**Step 2:**

Let required number by x

Average of x and 8 = $\frac{(8 + x)}{2}$

Given $\frac{(8 + x)}{2}$ + 10 = 17

So $\frac{(8 + x)}{2}$ = 17 – 10 = 7

8 + x = 14; x = 14 – 8 = 6.

**Step 3**

So required number = 6

**Q 9 - 7 consecutive even integers have an average of 48. Find the average of the greatest two of those integers.**

### Answer : C

### Explanation

**Step 1:**

Let the consecutive even integers be x – 6, x – 4, x – 2, x, x + 2, x + 4, x + 6

Their average = $\frac{(x – 6 + x – 4 + x – 2 + x + x + 2 + x + 4 + x + 6)}{7} = \frac{7x}{7}$ = 48. So x = 48.

**Step 2:**

So the numbers are 42, 44, 46, 48, 50, 52, 54

The average of the two greatest of these integers 52 and 54 is $\frac{(52 + 54)}{2}=$ = 53

**Q 10 - The average height of students in a school is 5 foot 6 inches. If the 200 girls at the school average 5 foot 4 inches, what is the average height of the 180 boys at the school?**

### Answer : D

### Explanation

**Step 1:**

Let the average height of the boys = x Then total height of 180 boys = 180x

Total height of 200 girls = 200 × $5\frac{4}{12} = \frac{3200}{3}$

**Step 2:**

Total height of boys and girls = 180x + $\frac{3200}{3}$

= Average x number of boys and girls = 5$\frac{6}{12}$ × 380

180x + $\frac{3200}{3}$ = 2090; 180x = $\frac{(6270 − 3200)}{3} = \frac{3070}{3}$; x = $\frac{3070}{540}$ = 5.69 feet.