Maths Quiz - Solutions and New Task
In Hugi 23, I assigned you three mathematical tasks. Here are the correct solutions.
Task "Easy":
On her seventh birthday Sue receives 50 coins which she is allowed to spend freely. In the supermarket she notices delicious sweets. Each pack costs 15 coins. Sue buys as many of them as possible. For the remaining money she purchases lollipops - each piece is 2 coins. With how many packs of sweets, lollipops and coins will Sue return home?
Solution:
This task can probably be solved in less than 10 seconds. The number of sweets is int(50 / 15) = 3; from the remaining money, which is 50 mod 15 = 5 coins, she can buy int(5 / 2) = 2 lollipops; finally, 5 mod 2 = 1 coin remains.
Task "Medium":
Several years ago I met a woman at a party that took place in the evening before New Year's Day. She told me about her life: "At the age of 24 I married John, who was born in the year 1952. A year later our family was enriched by our first child, and after another two years our second child saw the light of the day. Oh, how many things have changed since then! Now the sum of the kids' ages is 10 years." After a short pause I made this remark: "Funny! Had you computed the sum of the ages of all your family members (your husband, your two children and yourself) three years ago, you'd have got a total of 70 years." I did not meet this woman again until recently, on the evening before January 1st, 2001. How old was she then? (All ages are positive integer numbers.)
Solution:
Transforming this into a system of linear equations and solving it is all you need.
Let's define some variables first:
Kid 1 = b
Kid 2 = g
Mother = m
Father = f
b is two years older than g, so b = g + 2. The current sum of the kids' ages is 10, so b + g = 10. From this we get g + 2 + g = 10 => 2 * g = 8 => g = 4 => b = 6.
The woman married when she was 24, her first child was born a year later - in other words, she is 25 years older than b. Thus, m = 25 + b = 25 + 6 = 31.
Three years before the narrator ("I") was told this story, the sum of all ages was 70. This means:
b + g + m + f - 3 * 4 = 70
10 + 31 + f = 82
f = 41
The father was born in the year 1952. As he was 41 years old when this meeting happened, the year of the meeting must have been 1993. Note that the day of the meeting was December 31st, so the man had definitely already celebrated his 41st birthday before.
The narrator met the woman again on December 31st, 2001 - that is 7 years later. Thus the woman was 38 years old at this point in time.
Task "Hard":
The side a of a triangular fence has a length of 3 metres, the side b a length of 2 metres. Regarding the angles of the triangle we know these facts: sin(alpha) = cos(beta) and sin(beta) > cos(alpha). How long is the fence (i.e., the perimeter of the triangle)? (a is the opposite side of alpha, b is the opposite side of beta; the third angle can be called gamma and its opposite side c.)"
Solution:
The first thing we can do is to derive the value of angle beta. In any triangle, this sine theorem applies:
a / sin(alpha) = b / sin(beta)
Since we know that in this triangle sin(alpha) = cos(beta), we can write:
a / cos(beta) = b / sin(beta)
By means of simple algebraic transformation, we get:
sin(beta) / cos(beta) = b / a
As the quotient of the sine and the cosine of an angle is the tangent of the angle, we can write:
tan(beta) = b / a
From here follows:
beta = atan(b / a) = atan(2 / 3) = 33.69006753°
To derive alpha, let's ponder over the properties of the angles. When does sin(alpha) = cos(beta) apply? A possible case is alpha + beta = 90°; it is a consequence of the rule sin(alpha) = cos(90° - alpha), whose truth can be proven using the unit circle or (for positive alpha and beta, and the condition alpha + beta < 180°) a rightangled triangle. However, in this case also sin(beta) = cos(alpha) would apply; this does not apply to the triangle we are trying to solve now.
So let's think about other cases in which sin(alpha) = cos(beta). As we know (or may derive from the unit circle), the following rule applies:
sin(angle) = sin(180° - angle)
Let's combine it with the formula sin(alpha) = cos(90° - alpha). In this way we derive:
sin(180° - alpha) = cos(90° - alpha)
sin(90° + 90° - alpha) = cos(90° - alpha)
If we introduce a new variable beta1 = 90° - alpha, we thus get:
sin(90° + beta1) = cos(beta1)
So alpha = 90° + beta is also a case in which sin(alpha) = cos(beta) applies. How about the second requirement, sin(beta) > cos(alpha)?
In a triangle, the sine of any angle is positive, as all angles are greater than 0° and lower than 180°. If it can be proven that cos(alpha) = cos(90° + beta) is negative, the second requirement is fulfilled.
In fact cos(90° + angle) always has the opposite sign to cos(angle), except if cos(angle) = 0, as zero has no sign. We can differentiate between three cases:
a) If 0° < beta < 90°, cos(beta) is positive; thus cos(alpha) = cos(90° + beta) is negative.
b) If 90° = beta, cos(beta) is zero; thus cos(alpha) = cos(90° + beta) is zero.
c) If 90° < beta < 180°, cos(beta) is negative; thus cos(alpha) = cos(90° + beta) is positive.
Cases b) and c) cannot apply, as alpha = 90° + beta would be equal to or greater than 180°, i.e. out of bounds. Thus, case a) must apply. In this case cos(alpha) becomes negative, and thus the requirement sin(beta) > cos(alpha) is fulfilled.
Using this information, we can easily solve the assignment.
As the sum of all angles in a triangle is 180°, we can calculate the value of gamma:
gamma = 180° - (alpha + beta) = 180° - (90° + beta + beta) = 90° - 2 * beta = 90° - 2 * 33.69006753° = 22.61986495°
Using the sine theorem we get c:
c / sin(gamma) = b / sin(beta) => c = b * sin(gamma) / sin(beta) = 1.386750491 m
And getting the perimeter is a simple addition:
Perimeter = a + b + c = 3 m + 2 m + 1.386750491 m = 6.386750491 m
Answer: The length of the fence is 6.386750491 metres.
The clever ones
The following people submitted correct solutions for all tasks:
1. Puthoon: August 13, 2001
2. The Update (Malte Clasen): August 14, 2001
3. Tor Myklebust: August 15, 2001
4. Nevidimka (Dmitry Kozhenikov): August 16, 2001
5. mados (Thiemo Mättig): August 17, 2001
6. Bonz (Paolo Bonzini): August 27, 2001
7. SunmaN: August 28, 2001
Congratulations!
New task
Here's one new task of approximately medium, or perhaps medium to hard, level.
If an aeroplane (e.g. the Concorde) flies with a velocity faster than the speed of sound, it flies away from the spherical sound waves it generates. When the sound waves reach the ground, the viewers on the ground get to hear an ultrasonic bang that makes windows burst. Let's say the velocity of the aeroplane, which is constantly flying at the same height, is twice as fast as the speed of sound. When the people at a certain location get to hear the ultrasonic sound, the distance from the aeroplane to the air above that location is already one kilometer. At what height is it flying?
Have fun!