Averages & Data

The total distance Anna wants to walk is (31 × 5) km = 155 km. Since she has walked 95 km up to the 16th of March, she has (155 − 95) km = 60 km to walk in (31 − 16) days = 15 days. Therefore the distance, in km, that she needs to average per day is 60 ÷ 15 = 4.

From the given information, the sum of Megan’s first three numbers is −9, the sum of her first four numbers is 16, and the sum of her first five numbers is −25.

Therefore the fourth number is 16 − (−9) = 25, and the fifth number is −25 − 16 = −41. The difference between them is 25 − (−41) = 66.

Let the number of jumps Viola has already made be \(n\) and let the total distance she has jumped in these \(n\) jumps be \(T\) m. Since the total distance jumped is equal to the average distance jumped multiplied by the number of jumps, the information in the question tells us that \(T = 3.8n\) and that \(T + 3.99 = 3.81(n + 1)\). Therefore \(3.8n + 3.99 = 3.81n + 3.81\) and hence \(0.18 = 0.01n\), which has solution \(n = 18\). In order to increase her average distance to 3.82 m with her final jump, she must jump \(x\) m, where \(T + 3.99 + x = 3.82(n + 2)\). Therefore \(3.8n + 3.99 + x = 3.82n + 7.64\) and hence \(x = 0.02n + 3.65\). Since we have already shown \(n = 18\), we have \(x = 0.36 + 3.65 = 4.01\). Hence the distance she needs to jump with her final jump is 4.01 m.

As the mode of the list is 5, at least two of the integers in the list are 5. The median of the list is 5, so when the list is written in ascending order the third integer is 5. So, since the range of the list is 5, the two possible lists, in ascending order, are \(x, y, 5, 5, x + 5\) or \(x, 5, 5, y, x + 5\), where \(x\) and \(y\) are positive integers and \(x \leqslant y \leqslant x + 5\). The mean of the integers is 5, so their sum is 5 × 5 = 25, for each possible list. Therefore \(x + y + 5 + 5 + x + 5 = 25\). So \(2x + y = 10\). The positive integer solutions of this Diophantine equation are (1, 8), (2, 6), (3, 4), (4, 2). Note that neither \(x\) nor \(y\) equals 5, so there are exactly two 5s in the list. Of the solutions, neither (1, 8) nor (4, 2) satisfy the condition \(x \leqslant y \leqslant x + 5\).
So the only two possible lists are 2, 5, 5, 6, 7 and 3, 4, 5, 5, 8.

Let the four numbers be \(a\), \(b\), \(c\), \(d\) with \(a \leqslant b \leqslant c \leqslant d\). Abdul chooses the numbers in turn, giving
\[\begin{array}{lll} a + \tfrac{1}{3}(b + c + d) = 17 \quad (1) & \qquad & b + \tfrac{1}{3}(a + c + d) = 21 \quad (2) \\ c + \tfrac{1}{3}(a + b + d) = 23 \quad (3) & \qquad & d + \tfrac{1}{3}(a + b + c) = 29 \quad (4) \end{array}\]
Adding these equations gives:
\[\begin{array}{lr} & a + b + c + d + \tfrac{1}{3}(3a + 3b + 3c + 3d) = 90 \\ \text{that is} & 2a + 2b + 2c + 2d = 90. \end{array}\]
Hence \(a + b + c + d = 45\). Subtracting a third of this from equation (4), we get
\[d + \tfrac{1}{3}(a + b + c) - \tfrac{1}{3}(a + b + c + d) = 29 - 15.\]
Which gives \(\tfrac{2}{3}d = 14\), so \(d = 21\). We can continue in a similar way to find \(a = 3\), \(b = 9\) and \(c = 12\).

Let \(k\) cm be the difference between the heights.
Then the heights in cm are: Tobias 184, Victor \(184 + k\), Peter \(184 - k\), and Oscar \(184 - 2k\). The mean is 178 so \(\tfrac{1}{4}\left(184 + 184 + k + 184 - k + 184 - 2k\right) = 178\). Hence \(4 \times 184 - 2k = 4 \times 178\), giving \(2k = 4 \times 184 - 4 \times 178 = 4 \times 6 = 24\). Hence \(k = 12\).
Therefore Oscar's height in cm is \(184 - 2 \times 12 = 160\) cm.

The total amount of the money the five girls have is (20 + 4 × 10) euros = 60 euros. Therefore, if all five girls are to have the same amount, they need to have (60 ÷ 5) euros = 12 euros each. Since each of Olivia's sisters currently has 10 euros, Olivia would need to give each of them (12 − 10) euros = 2 euros.

The smallest possible range of four positive integers is 3. Let these integers be \(n\), \(n + 1\), \(n + 2\), \(n + 3\). For the mean of these four integers to equal 2017, we have \(\frac{n + n + 1 + n + 2 + n + 3}{4} = 2017\), that is \(n = 2015\frac{1}{2}\). So the smallest possible range is not 3. However, note that the integers 2015, 2016, 2018 and 2019 have mean 2017 and range 4. So the smallest possible range of the integers is 4.

From the information given, there are at least two 9s in the list, since 9 is the mode, and at least one number greater than 10, since 10 is the mean. So there are at least three numbers greater than 8 in the list. Therefore the list must contain at least six numbers, as the median of the numbers is 8. Moreover, it is possible to find suitable lists of six numbers with sum 60 (as the mean is 10), for example 1, 2, 7, 9, 9, 32.

The total number of cupcakes eaten by the girls is \(12 \times 1\tfrac{1}{2} = 18\). Two girls ate no cakes so the 18 cupcakes were eaten by 10 girls. Since no-one ate more than two cupcakes, the maximum number of cupcakes the 10 girls could have eaten is 10 × 2 = 20. For every girl who eats only one cupcake, the total of 20 is reduced by 1. Hence the number of girls who ate only one cupcake is 20 − 18 = 2 and hence the number of girls who ate two cupcakes is 8.

The sum of 1, 2, … \(n\) is \(\tfrac{1}{2}n(n + 1)\). So the smallest the sum could be if one number were deleted would be \(\tfrac{1}{2}n(n + 1) - n = \tfrac{1}{2}(n^2 + n - 2n) = \tfrac{1}{2}n(n - 1)\). Thus the mean of any \(n - 1\) of these numbers is at least \(\tfrac{1}{2}n\). So for the mean to be 4.75, we must have \(n \leqslant 9\).
Similarly, the largest the sum could be with one number deleted would be \(\tfrac{1}{2}n(n + 1) - 1 = \tfrac{1}{2}(n^2 + n - 2) = \tfrac{1}{2}(n - 1)(n + 2)\). Thus the mean of any \(n - 1\) of the numbers is at most \(\tfrac{1}{2}(n + 2)\). So we must have \(\tfrac{1}{2}(n + 2) \geqslant 4.75\) and so \(n \geqslant 8\).
Since \(4.75 = 19/4\) and the mean is obtained by dividing by \(n - 1\), we require \(n - 1\) to be a multiple of 4. So \(n = 9\). The sum of 1, … , 9 is 45 and we need the 8 numbers used to have a mean of 4.75. So their total is 38 and hence 7 is the number to be eliminated.

Suppose the first three terms of the sequence are \(a\), \(b\), \(c\). Then \(c = \tfrac{1}{2}(a + b)\) and so \(a + b = 2c\). The mean of the first three terms is then \(\tfrac{1}{3}(a + b + c) = \tfrac{1}{3}(2c + c) = c\), so the fourth term is \(c\). Similarly, the following terms are all equal to \(c\). Since one of these terms is 26 and \(a = 8\) then \(b = 2c - a = 52 - 8 = 44\).

Let the mean score of the 40% of the students who failed the test be \(x\). The information in the question tells us that \(0.6 \times 8 + 0.4 \times x = 6\). Hence \(0.4x = 1.2\) and so \(x = 3\).

As the 3 numbers have mean 7, their sum equals 3 × 7 = 21. For one of the numbers to be as large as possible the other two numbers must be as small as possible. They must also be different and so must be 1 and 2. Hence the largest possible of the three numbers equals 21 − (1 + 2) = 18.

The options given imply that the median cannot exceed 3.5, so we first look to see if the median can have this value. The median of 190 numbers is \(\tfrac{1}{2}(a + b)\) where, when the numbers are arranged in increasing order, \(a\) is the 95th number and \(b\) is the 96th number in the list. In this problem, \(a\) and \(b\) are positive integers with \(a \leqslant b\), so if \(\tfrac{1}{2}(a + b) = 3.5\) then \(a + b = 7\). The smallest total of all 190 numbers with median 3.5 occurs when \(a = 3\), \(b = 4\) and the numbers, in increasing order, are

In this case their total is 94 × 1 + 3 + 95 × 4 = 94 + 3 + 380 = 477, which is the total number of cups of coffee given in the question. So 3.5 is a possible value for the median. Note also, that, if the median were greater than 3.5, then at least one of the numbers in the above list would need to be larger and so the total would be larger than 477. So 3.5 really is the maximum possible value of the median.

The mean of the first three numbers is \(\tfrac{1}{3}(20 + x)\); the mean of the last four numbers is \(\tfrac{1}{4}(33 + x)\). Therefore \(4(20 + x) = 3(33 + x)\), that is \(80 + 4x = 99 + 3x\), so \(x = 99 - 80 = 19\).

The sum of all 64 numbers is 64 × 64 = 64². The sum of the first 36 numbers is 36 × 36 = 36². So the sum of the remaining 28 numbers is 64² − 36² = (64 + 36)(64 − 36) = 2800. Therefore the mean of these 28 numbers is 100.

The sum of the three numbers is \(3x\), so \(y + z = 2x\). Hence the mean of \(y\) and \(z\) is \(x\).

The difference between the largest and smallest of 9 consecutive whole numbers is 8, irrespective of the sum of the numbers. However in this case, since the average of the 9 numbers is 223, we can say they are 219, 220, …, 226, 227.

The mean of \(1.\dot{2}\) and \(2.\dot{1}\) is \((1.\dot{2} + 2.\dot{1}) \div 2 = 3.\dot{3} \div 2 = 1.\dot{6}\).

Whatever number is placed in the fifth box, the median and mode of the numbers will both be 7. For the mean to equal 7, the total of the 5 numbers must be 35. This means that the missing number is 9.

The four fractions total \(\dfrac{5}{4}\), so their mean = \(\dfrac{5}{4} \div 4 = \dfrac{5}{16}\).