Difference between revisions of "1995 AJHSME Problems/Problem 16"
Soccerpro101 (talk | contribs) (→Problem) |
(Undo revision 80697 by Soccerpro101 (talk)) |
||
Line 7: | Line 7: | ||
*Five students from Carver school worked for <math>9</math> days. | *Five students from Carver school worked for <math>9</math> days. | ||
− | The total amount paid for the students' work was <dollar>< | + | The total amount paid for the students' work was <dollar><math>744. Assuming each student received the same amount for a day's work, how much did the students from Balboa school earn altogether? |
− | <math>\text{(A)}\ 9.00\text{ dollars} \qquad \text{(B)}\ 48.38\text{ dollars} \qquad \text{(C)}\ 180.00\text{ dollars} \qquad \text{(D)}\ 193.50\text{ dollars} \qquad \text{(E)}\ 258.00\text{ dollars}< | + | </math>\text{(A)}\ 9.00\text{ dollars} \qquad \text{(B)}\ 48.38\text{ dollars} \qquad \text{(C)}\ 180.00\text{ dollars} \qquad \text{(D)}\ 193.50\text{ dollars} \qquad \text{(E)}\ 258.00\text{ dollars}<math> |
==Solution== | ==Solution== | ||
− | Altogether, the summer project totaled <math>(7)(3)+(4)(5)+(5)(9)=21+20+45=86< | + | Altogether, the summer project totaled </math>(7)(3)+(4)(5)+(5)(9)=21+20+45=86<math> days of work for a single student. This equals </math>744/86=9<math> dollars per day per student. The students from Balboa school earned </math>9(4)(5)=\boxed{\text{(C)}\ 180.00\ \text{dollars}}$. |
==See Also== | ==See Also== | ||
{{AJHSME box|year=1995|num-b=15|num-a=17}} | {{AJHSME box|year=1995|num-b=15|num-a=17}} |
Revision as of 12:03, 20 October 2016
Problem
Students from three middle schools worked on a summer project.
- Seven students from Allen school worked for days.
- Four students from Balboa school worked for days.
- Five students from Carver school worked for days.
The total amount paid for the students' work was <dollar>\text{(A)}\ 9.00\text{ dollars} \qquad \text{(B)}\ 48.38\text{ dollars} \qquad \text{(C)}\ 180.00\text{ dollars} \qquad \text{(D)}\ 193.50\text{ dollars} \qquad \text{(E)}\ 258.00\text{ dollars}(7)(3)+(4)(5)+(5)(9)=21+20+45=86744/86=99(4)(5)=\boxed{\text{(C)}\ 180.00\ \text{dollars}}$.
See Also
1995 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |