Let's discuss a problem from the AMC 2023 Upper Primary: Problem 29 which revolves around basic algebra.
Yifan has a construction set consisting of red, blue and yellow rods. All rods of the same colour are the same length, but differently coloured rods are different lengths. She wants to make quadrilaterals using these rods.
What number do you get when you multiply the lengths of one red rod, one blue rod and one yellow rod?
Let's consider the Red rods to be \(R\), the Blue rods to be \(B\) and the Yellow rods to be \(Y\).
According to the 1st point :\(2R + B + Y = 36\)............... (1)
According to the 2nd point: \(2B + R + Y = 35\)..............(2)
According to the 3rd point: \(2Y + B + R = 33\)..............(3)
We have to find the individual values of \(R, Y, B\).
Let's add all the three equation \((1), (2)\) & \((3)\), we get :
\(4R + 4B + 4Y = 36 + 35 + 33\)
\(\therefore\) \(4R + 4B + 4Y = 104\)
\(\therefore\) \(R + B + Y = \frac {104}{4}\)
\(\therefore\) \(R + B + Y = 26\)......................(4)
Now, subtracting (4) from (1):
\(2R + B + Y - R - B - Y= 36 - 26\)
\(R = 10\)
Now, subtracting (4) from (2):
\(2B + R + Y - R - B - Y= 35 - 26 \)
\(B = 9\).
Now implementing the values of \(R\) and \(B\) in the (1) equation we will get:
\(2R + B + Y = 36\)
\(\therefore\) \( 20 + 9 + Y = 36\)
\(\therefore\) \(Y = 36 - 29 = 7\).
So, when we multiply the length of \(R, Y , B\) we get : \( 7 \times 10 \times 9 = 630\).
So the answer is \(630\).
What is AMC (Australian Mathematics Competition)?
The Australian Mathematics Competition (AMC) is one of Australia's largest and oldest annual mathematics competitions, aimed at fostering interest and excellence in mathematics among students.
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