Henry spent 3/8 of his money on a badminton racket. He spent 1/8 of his money on a watch. The badminton racket cost $14 more than the watch. How much money did Henry spend on the two items altogether?
3/8 ths x on racket - 1/8x on watch = 14
3/8 x - 1/8 x = 14
2/8 x = 14
x = 14 * 8/2 = 14 * 4 = 56 dollars initially
(3/8 + 1/4) 56 = 35 dollars total
How much money did Henry spend on the two items altogether?
Hello Guest!
Henry owns x.
\(\color{blue}\frac{x}{8}+$14=\frac{3x}{8}\\ \frac{3x-x}{8}=$14\\ 2x=$112\)
\(x=$56\)
\((\frac{1}{8}+\frac{3}{8})\cdot $56=\color{blue}$28\)
$28 did Henry spend on the two items altogether.
!
Straightforward. First, assign variables to the unknown values:
Let H = Henry’s money
Let B = Badminton Racket
Let W = Watch
We are given:
B = 3/8 H
W = 1/8 H
B = W + $14
So:
Substitute B and W:
3/8 H = 1/8 H + $14 [Cost of Badminton racket = $14 more than cost of watch]
Solving:
3/8 H - 1/8 H = $14
(3/8–1/8) H = $14
2/8 H = $14
1/4 H= $14
H = 4 * $14 = $56 [Henry’s total money]
B = 3/8 * $56 = $21 [Cost of badminton racket]
W = 1/8 *$56 = $7 [Cost of watch]
Check:
B = W + $14
$21 =? $7 + $14
Yes!
Henry spent B+W on his items altogether, which is $21+$7 = $28.
Another way of expressing it is:
2/8 H = $14
1/8 H = $7
B+W = 3/8 H + 1/8 H = (3+1)/8 H = 4/8 H = 1/2 H.
H = $56
B+W = 1/2 H = $28