Wednesday, August 1, 2018

Notes to 2018 Mathcounts chapter more interesting problems

2018 Mathcounts Chapter Spring problems : solutions down below 
Thanks to a boy mathlete who tried these problems and e-mail me for feedback. 

Please try these problems first before reading the explanations. :D

#25 : Three employees split a bonus valued at some number of dollars. Arman first receives $10 more than one third of the total amount. Bernardo then receives $3 more than one half of what was left. Carson receives the remaining $25. What is the total dollar value of the bonus?

#27: For a particular list of four distinct integers the mean, median and range have the same value. If the least integer in the list is 10, what is the greatest value for an integer in the list?

#29: There are two values of x such that \( |\dfrac {x-2018} {x-2019}|=\dfrac {1} {6}\). . What is the absolute difference between these two values of x? Express your answer as a common fraction.

Target #8 : Four congruent circles of radius 2 cm intersect with their centers at intersection points as shown. What is the area of the shaded region? Express your answer in terms of π.



























#25 : It's easier if you go backward and use inverse operations to solve this question. 
(25 + 3) *2 = 56 and 56 + 10 = 66, which is \( \dfrac {2} {3}\) of the original bonus value, or 
what is left after \( \dfrac {1} {3}\) was given out. 

\( \dfrac {2} {3}\) of bonus is 66 dollars, so the answer is 99. 

#27: Let the average be x and the three other numbers be a, b, c and \( a < b < c \).
The least number is 10 (given), so 
\( 10+a+b+c \) = \( 4x \)---> equation 1 
\( \dfrac {a+b} {2}\) = \( x \) (how to find the median), so \( a+b\) = \( 2x \) --- (2)
\(c-10 = x\) (given because it's the range), \(c= x + 10\) ---(3)
Substitute (2) and (3) to equation one and you have \(10 + 2x + x + 10 = 4x\), so \(x = 20 \)
\(C = 20 + 10 = 30\), the answer 

#28 : Let \(x - 2018 = y\) , then \(x - 2019 = y -1\)
We then have either 
\( |\dfrac {y} {y -1}|=\dfrac {1} {6}\)  \(\rightarrow\) \(y\) = \( \dfrac {-1} {5}\)

or \( |\dfrac {y} {y-1}|=\dfrac {-1} {6}\) \(\rightarrow\) \(y\) = \( \dfrac {1} {7}\)

Their positive difference is \( \dfrac {12} {35}\) , the answer. 


Target #8 : Thanks to a 5th grader girl mathlete's solution: 




If you move parts around, you'll see the answer is exactly a semi-circle with a radius 2, so the answer is \(2\pi\), the answer. 






or check out another solution from me: 

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