OFFICIAL GUIDE EXPLANATIONAlgebra Simultaneous equations
Let a, b, c, and d be the number, respectively, of 5-cent, 10-cent, 25-cent, and 50-cent coins. We are given that \(a + b + c + d = 50\) and \(5a + 10b + 25c + 50d = 1,000\), or \(a + 2b + 5c + 10d = 200\). Determine the value of a.
\(a
+
b
+
c
+
d
=
50
a
+
2
b
+
5
c
+
10
d
=
200\)
Tip: Note that each of a, b, c, and d must be a nonnegative integer, and so care must be taken in deducing non-sufficiency. For example, there are many real number pairs (x,y) that satisfy the equation 2x + y = 1, but if each of x and y must be a nonnegative integer, then x = 0 and y = 1 is the only solution.
We are given that c = 10 and d = 10. Substituting c = 10 and d = 10 into the two equations displayed above and combining terms gives\( a + b = 30\) and \(a + 2b = 50\). Subtracting these last two equations gives b = 20, and hence it follows that a = 10; SUFFICIENT.
We are given that b = 2a. Substituting b = 2a into the two equations displayed above and combining terms gives \(a + 2a + c + d = 50\) and \(a + 4a + 5c + 10d = 200\), which are equivalent to the following two equations.
\(3
a
+
c
+
d
=
50
a
+
c
+
2
d
=
40\)
Subtracting these two equations gives 2a – d = 10, or 2a = d + 10. Since 2a is an even integer, d must be an even integer. At this point it is probably simplest to choose various nonnegative even integers for d to determine whether solutions for a, b, c, and d exist that have different values for a. Note that it is not enough to find different nonnegative integer solutions to \(2a = d + 10\), since we must also ensure that c and d are nonnegative integers. If d = 8, then 2a = 8 + 10 = 18, and we have a = 9, b = 18, c = 15, and d = 8. However, if d = 10, then 2a = 10 + 10 = 20, and we have a = 10, b = 20, c = 10, and d = 10; NOT sufficient.
Statement 1 alone is sufficient.