+33 There were 3 friends, A, B and C, sharing the cost of a toy.
The ratio of A's share to the total of B and C is 1:3.
The ratio of B's share to the total of A and C is 1:5
C spent $80 more than B. Find the cost of the toy.
+2668 Write as a system of 3 equations:
\(3a = b+c\) (Equation 1)
\(5b = a+c\) (Equation 2)
\(c=80+b\) (Equation 3)
Sub in (Equation 3) into (Equation 2): \(5b = a+b+80\)
Simplify: \(b = {a \over 4}+20\)
Now, sub what we found for b, and (Equation 3) into (Equation 1): \(3a= {a \over 4} +20+80+ {a \over 4} +20\)
Solving, we find \(a = 48\)
Subbing the value of \(a\) we just found and (Equation 3) into (Equation 2), we get: \(5b = 48+80+b\)
Solving, we find \(b = 32\)
Plugging in what we found for the value of \(b\) into (Equation 3), we find that \(c = 112\).
Now do \(a + b+c\) to find your answer.
3a + b + c -------------- (1)
5b = a + c -------------- (2)
c = 80 + b -------------- (3)
put c = 80 + b in equation (2)
5b = a + b + 80
5b = b = a + 80 => 4b = a + 80
b = \( {a \over 4}\) + \( {80 \over 4}\) = \( {a \over 4}\) + 20
put b = \( {a \over 4}\) + 20 and c = 80 + b in Eq (1)
3a = \( {a \over 4}\) + 20 + 80 + \( {a \over 4}\) + 20
3a = \( {a \over 4}\) + \( {a \over 4}\) + (80 + 20 + 20)
3a = 2 * \( {a \over 4} \) + 120 => 3a - \( {2a \over 4}\) = 120
\( {12a - 2a \over 4}\) = 120 => \( {10a \over 4}\) = 120 => a = \( {120 * 4 \over 10}\) = 48
put a = 48 and c = 80 + b in equation (2)
we get 5b = 48 + 80 + b => 5b - b = 128
4b = 128 => b = \( {128 \over 4}\) = 32
pluf the value of b in equation (3) we get
c = 80 + 32 = 112
Now do a + b + c to find your answer
a = 48, b = 32, c = 112
80 a + b + c = 48 + 32 + 112
= 192
80 a + b + c = 192 (Ans)
+37146 From BB:
a = (b+c)/3 and a = 5b - c <======equate
5b-c = (b+c)/3
re-arrange to 16b - 4c = 0 sub in c = 80 + b
16 b - 4(80+b) = 0 results in b = 32 then c = 112 then a = 48 summed = 192
