Problem 24: Australian Mathematics Competition 2023 – Upper Primary
Join Trial or Access Free ResourcesJoin Trial or Access Free ResourcesLet's discuss a problem from the AMC 2023 Upper Primary: Problem 24 which revolves around basic algebra.
I have 4 whole numbers that add up to 98. If I were to add 6 to the first number, subtract 6 from the second number, multiply the third number by 6 and divide the fourth number by 6, the four answers would all be the same. What is the sum of the largest two of my original four numbers?
(A) 72 (B) 86 (C) 88 (D) 90 (E) 94
Let the four whole numbers be \(x, y, z, a\).
According to the given data, \(x + y + z + a = 98\) ----------------------------(1)
Again, it's mentioned if we add \(6\) to the first number : \(x + 6\),
If we subtract \(6\) from the second number : \(y - 6\),
If we multiply \(6\) with the third number : \( 6 \times z\),
If we divide the fourth number by\(6\): \(\frac {a}{6}\), then all these expressions are equal in value.
Thus, \(x + 6 = y - 6 = 6 \times z = \frac {a}{6}\) ----------- (2)
From the equation (2) we can write, \(6 \times z = \frac {a}{6}\).
\(\therefore\) \(a = 36 \times z\).
The value of \(a\) can be either \(36\) or \(72\) as the sum of all the numbers equal to 98. So any one number can't exceed the value \(98\).
If \(a\) is 36 then \(z\) has to be 1. If \(z \) is 1 then y is \( 6 \times 1 + 6 = 12\).
If \(y\) is 12 then \(x\) is = \(12 - 6- 6 = 0\).
Let's implement all the values in equation (1) -> \(36 + 12 + 0 + 1 = 59\).
Now if \(a\) is 72 then \(z\) has to be 2. If \(z \) is 2 then y is \( 6 \times 2 + 6 = 18\).
If \(y\) is 18 then \(x\) is = \(18 - 6 - 6 = 6\).
Let's implement all the values in equation (1) -> \(72 + 18 + 6 + 2 = 98\).
Thus, the two largest numbers are - \( 72 + 18 = 90\).
So the answer is option \(D ) 90\).
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.
