# Question

Benjamin spent 5/9 of his money on 5 toys and 5 erasers, and 25% of the remainder on 2 cards.
Each eraser cost ¼ as much as a toy.
Each card cost \$1.20 more than an eraser.
How much money did Benjamin spend on each toy?
How much money did Benjamin have in the end?

5u on 5(erasers + toys) and

1/4 of remainder (4u) on 2 cards, i.e. he spend 1u on them.

Eraser = 1/4 of toy,

Toy= 4 erasers

5u = 5 erasers + 5 toy = 5 erasers + 20 erasers

Therefore 5u = 25 erasers

1u = 5 erasers

1u= 2 cards= 2 erasers + \$2.40

So 5 erasers = 2 erasers + \$2.40

3 erasers = \$2.40

1 eraser = \$0.80.

A) 5 toys= 20 erasers = \$16 ##

Looking SAHmom’s solution, I realized I had a oversight. Question looking for one toy, not five.

One toy = 4 erasers = 4 X \$0.80 = \$3.20 ###

B) in the end he left 3u

3u = 15 erasers = \$12 ##

0 Replies 2 Likes ✔Accepted Answer

Just thought a diagram may help, especially those visual learners.

Step 1:  Branch out a diagram and write down information:

All money (           )

1st spend 5/9 (     )                                   Remainder 4/9 (       )

5 toys (      )               5 erasers (     )      2 cards 25%(=1/9) (     )   Balance (=3/9)(     )

Step 2:  Process “Each eraser cost ¼ as much as a toy“ – Let 1 eraser costs \$e

Process “Each card cost \$1.20 more than an eraser”

All money (           )

1st spend   5/9(#3. 25e )                                    Remainder 4/9 (       )

5 toys (#2. 20e )   5 erasers (#1. 5e  )    2 cards 1/9 (#4. 2e+2.40)  Balance 3/9 (    )

Step 3:  (Step back and) analyse the branch diagram… and Solve

All money (           )

1st spend  5/9 (25e )                                    Remainder 4/9 (     )

5 toys (20e )            5 erasers (5e)               2 cards 1/9 (2e+2.40)    Balance 3/9 (     )

Taking only 1/9 from each side:   25e/5 = 2e +2.40 –> e=0.80

Each toy = 4e = \$3.20#

Balance is 3/9 –> 5e x 3 = \$4 x3 = \$12#