Mathematics Corner

Here is a list of questions and corresponding solutions. The solutions have not been verified and therefore may not be correct. If you think they are not well presented, a mistake is made or improvement is required, please advise me so that I may update accordingly. Thanks.

Question

In year N, the 300th day of the year is a Tuesday. In year N+1, the 200th day is also a Tuesday. On what day of the the week does the 100th day of the year N-1 occur?

Solution

For year N, assume the number of days for the period from the 300th day of the year to the end of the year = 365 - 300 = 65 . (Note the word assume as it could be a leap year). (Equation 1)

For year N+1, the number of days from the beginning of the year to the 200th day = 200 . (Equation 2)

From Equations 1 and 2, the sum of the two periods = 265 days.

Since the 300th day of year N is a Tuesday and the 200th day of year N + 1 is also a Tuesday, the number of days for the period in between must be divisible by 7. But 265 cannot be divided by 7 to give an integer number. However 266 can.

Therefore year N is a leap year (366 days).

Therefore year N - 1 is not a leap year.

Therefore the number of days for the period between the 100th day of the year N -1 and the 300th day of year N = (365 - 100) + 300 = 565 .

The modulus of dividing 565 by 7 is 5. In other words, when 560 is divided by 7 it gives an integer.

Therefore the 105th day of year N - 1 is a Tuesday.

Therefore the 100th day of year N - 1 is a Thursday.

Question

When the mean, median, and mode of the list: 10, 2, 5, 2, 4, 2, x are arranged in increasing order, they form an arithmetic progression. What is the sum of all possible values of x?

Solution

Consider the following different values of x:

When x is less than 0, the list is: x, 2, 2, 2, 4, 5, 10 .
Therefore mean = (x + 2 + 2 + 2 + 4 + 5 + 10) / 7 = (x + 25) / 7 ,
median = 2 ,
mode = 2 (in fact, mode is always 2 for any value of x).
For (x + 25) / 7, 2, 2 to be an arithmetic progression,
2 - (25 + x) / 7 = 2 - 2 .
Therefore 14 - 25 - x = 0 .
Therefore x = -11 is the first possible value of x .

When x is equal to 0, the list is: 0, 2, 2, 2, 4, 5, 10 .
Since (2 + 2 + 2 + 4 + 5 + 10) / 7, 2, 2 is not an arithmetic progression, 0 is not a possible value of x .

When x is greater than 0 but less than 2, the list is: x, 2, 2, 2, 4, 5, 10 .
For (x + 2 + 2 + 2 + 4 + 5 + 10) / 7, 2, 2 to be an arithmetic progression,
2 - (x + 25) / 7 = 2 - 2 .
Therefore 14 - x - 25 = 0 .
Therefore x = -11 .
But this value of x contradicts the assumed range of values of x and therefore is not a possible value (in this case only).

When x is equal to 2, the list is: 2, 2, 2, 2, 4, 5, 10 .
Since (2 + 2 + 2 + 2 + 4 + 5 + 10) / 7, 2, 2 is not an arithmetic progression, 2 is not a possible value of x .

When x is greater than 2 but less than 4, the list is: 2, 2, 2, x, 4, 5, 10 .
For (2 + 2 + 2 + x + 4 + 5 + 10) / 7, x, 2 to be an arithmetic progression,
x - (25 + x) / 7 = 2 - x .
Therefore 7x - 25 - x = 14 - 7x .
Therefore 13x = 39 .
Therefore x = 3 is the second possible value of x .

When x is equal to 4, the list is: 2, 2, 2, 4, 4, 5, 10 .
Since (2 + 2 + 2 + 4 + 4 + 5 + 10) / 7, 4, 2 is not an arithmetic progression, 4 is not a possible value of x .

When x is greater than 4 but less than 5, the list is: 2, 2, 2, 4, x, 5, 10 .
For (2 + 2 + 2 + 4 + x + 5 + 10) / 7, 4, 2 to be an arithmetic progression,
4 - (25 + x) / 7 = 2 - 4 .
Therefore 28 - 25 - x = - 14 .
Therefore x = 17 .
But this value of x contradicts the assumed range of values of x and therefore is not a possible value (in this case only).

When x is equal to 5, the list is: 2, 2, 2, 4, 5, 5, 10 .
Since (2 + 2 + 2 + 4 + 5 + 5 + 10) / 7, 4, 2 is not an arithmetic progression, 5 is not a possible value of x .

When x is greater than 5 but less than 10, the list is: 2, 2, 2, 4, 5, x, 10 .
For (2 + 2 + 2 + 4 + 5 + x + 10) / 7, 4, 2 to be an arithmetic progression,
4 - (25 + x) / 7 = 2 - 4 .
Therefore 28 - 25 - x = - 14 .
Therefore x = 17 .
But this value of x contradicts the assumed range of values of x and therefore is not a possible value (in this case only).

When x is equal to 10, the list is: 2, 2, 2, 4, 5, 10, 10 .
Since (2 + 2 + 2 + 4 + 5 + 10 + 10) / 7, 4, 2 is not an arithmetic progression, 10 is not a possible value of x .

When x is greater than 10, the list is: 2, 2, 2, 4, 5, 10, x .
For (2 + 2 + 2 + 4 + 5 + 10 + x) / 7, 4, 2 to be an arithmetic progression,
4 - ( 25 + x ) / 7 = 2 - 4 .
Therefore 28 - 25 - x = - 14 .
Therefore x = 17 is the third possible value of x .

Therefore there are three possible values of x, i.e. -11, 3 and 17 .

Therefore the sum of all possible values of x = -11 + 3 + 17 = 9 .

Question

Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangle. The area of one of the the two small right triangles is m times the area of the square. What is the ratio of the area of the other small right triangle to the area of the square?

Solution

It is good to draw a diagram with a scale in accordance with the dimensions stated in the question.

Draw a horizontal line CF from left to right.

Draw a vertical perpendicular line from above C and let it intersect the horizontal line at C. Label the top point of the vertical line as E.

Join E and F so that it is the hypotenuse of triangle ECF.

Through a point on EF a horizontal line AB is drawn in such a way that it intersects the vetical line EC at A and intersects the hypotenuse EF at B and AB = AC.

Note that the sequence of points on the vertical line, from top to bottom, is E, A and C.

Draw a vertical line from B and let it intersect the horizontal line CF at D.

Note that ABDC is a square.

There are three triangles: a big triangle ECF, a small triangle EAB on top of square ABDC and another small triangle BDF on the left of square ABDC.

Let the length of AB, BD, CD and AC be s and EA = h and DF = k .

Let triangle EAB be the triangle the area of which is m times the area of the square.
Therefore, 1/2 * s * h = m (s * s) .
Therefore, h = 2 * m * s (Equation 1).

The area of triangle ECF is equal to the area of the two smaller triangles and the square, therefore
1/2 * (s + k) * (s + h) = 1/2 * s * h + 1/2 * s * k + s * s .
Therefore (s + k) * (s + h) = s * h + s * k + 2 * s * s .
Therefore s * s + s * h + s * k + k * h = s * h + s * k + 2 * s * s .
Therefore k * h = s * s (Equation 2).

Let the ratio of the area of the other small right triangle to the area of the square be R.
Therefore R = (1/2 * s * k) / s * s .

From Equation 2, R = (1/2 * s * k) / (k * h) = s / (2 * h) .

From Equation 1, R = s / (2 * 2 * m * s) = 1 / (4 * m) .

Therefore the ratio of the area of the other small right triangle to the area of the square is 1 / (4 * m) .

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