Marion and Hilda have some pens. If Marion gives 27 pens to Hilda, Hilda will have 5 times the number of pens as Marion. If Hilda gives 15 pens to Marion, Marion will have
23 as many pens as Hilda. How many pens does Marion have at first?
| |
Case 1 |
Case 2 |
| |
Marion |
Hilda |
Marion |
Hilda |
| Before |
5 u + 27 |
25 u - 27 |
12 u - 15 |
18 u + 15 |
| Change |
- 27 |
+ 27 |
+ 15 |
- 15 |
| After |
1x5 =
5 u |
5x5 =
25 u |
2x6 =
12 u |
3x6 =
18 u |
The total number of pens in both cases remains unchanged. Make the total number of pens in both cases the same. LCM of 6 and 5 is 30.
Number of pens that Marion had at first is the same in both cases.
12 u - 15 = 5 u + 27
12 u - 5 u = 27 + 15
7 u = 42
1 u = 6
Number of pens that Marion had at first
= 5 u + 27
= 5 x 6 + 27
= 30 + 27
= 57
Answer(s): 57
