Cathy, Jane and Diana have equal number of buttons. Cathy packs all her buttons equally into 10 packets. Jane packs all her buttons equally into 5 packets. Diana packs all her buttons equally into 6 packets. 2 packets of Cathy's buttons, 3 packets of Jane's buttons and 5 packets of Diana's buttons add up to 147 buttons. How many buttons do they have altogether?
| |
Cathy |
Jane |
Diana |
| Number of packets |
10 |
5 |
6 |
| Number of buttons |
30 u |
30 u |
30 u |
| Number of buttons in each packet |
3 u |
6 u |
5 u |
All the buttons can be put into the packets without remainder.
All the children have equal numbers of buttons.
Make the number of buttons that each child has the same. LCM of 10, 5 and 6 = 30
Number of buttons that each child has = 30 u
Number of buttons in 1 packet of Cathy's buttons = 30 u ÷ 10 = 3 u
Number of buttons in 1 packet of Jane's buttons = 30 u ÷ 5 = 6 u
Number of buttons in 1 packet of Diana's buttons = 30 u ÷ 6 = 5 u
Number of buttons in 2 packets of Cathy's buttons, 3 packets of Jane's buttons and 5 packets of Diana's buttons
= (2 x 3 u) + (3 x 6 u) + (5 x 5 u)
= 6 u + 18 u + 25 u
= 49 u
49 u = 147
1 u = 147 ÷ 49 = 3
Total number of buttons that they have
= 3 x 30 u
= 90 u
= 90 x 3
= 270
Answer(s): 270
