3 pupils have the same birthday as the head teacher. The product of the ages of the people is 2652, the sum of the ages of the pupils adds up 2 the head teachers age. How old is the head teacher?
2006-11-06 07:26:58 · 12 answers · asked by gavin 1 in Education & Reference ➔ Homework Help
I think something may be missing here.
The factors of 2652 are: 2^2 * 3 * 13 * 17.
Since you have four people, four of those factors would have to be taken up (assuming we can rule out 1, as that is too young for this word problem). Logically, the teacher would be 2*13, 2*17, 3*13, or 3*17. Let's examine each of those:
2*13 = 26. This implies that the students are 2, 3, and 17. They do not add up to 26.
2*17 = 34. This implies that the students are 2, 3, and 13. They do not add up to 34.
3*13 = 39. This implies that the students are 2, 2, and 17. They do not add up to 39.
3*17 = 51. This implies that the students are 2, 2, and 13. They do not add up to 51.
Now, happytraveler's answer of 12, 13, and 17 are some very nice numbers. In fact, I'd be inclined to go with him, except for two things:
The problem states that the product of all four is 2652. This sequence of numbers implies that the four people would be aged 1, 12, 13, and 17.
We have no reason to assume 12, 13, and 17, even if we were looking at just three students. There's no reason we couldn't go with 6, 17, and 26 instead.
I think you don't have the full information, so you cannot solve this question without making some wild assumptions.
Edit: Judging from the use of the phrase, "product of the ages of the people is 2652," I'm willing to bet you meant, "product of the ages of the pupils is 2652." This makes a lot of sense if that's the case. The numbers 12, 13, and 17 would fit that criterion well, giving an age of 42 for the teacher.
But it's not the only right answer. The question needs to give you more. As I said, 6, 17, and 26 fulfill this criterion. Does the question have any other qualifiers? Are all three students minors? If so, then 12, 13, and 17 are the only values that fulfill this criterion. Without more information, you have an open-ended question where the teacher can be 68, 42, or even 2654.
2006-11-06 07:37:36 · answer #1 · answered by Rev Kev 5 · 1⤊ 0⤋
Factor 2652 to get possible ages.
2652/2=1326
1326/2= 663
663/3=221
221/17 =13
therefore the only ages that make sense are 13, 17 and 12
The teacher's age is thus 62
2006-11-06 07:47:02 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
33
2006-11-06 07:39:57 · answer #3 · answered by alec c 4 · 0⤊ 0⤋
Since everyone bar few has beaten me to the answer to this question I'll respond to SundaeG1r's instead. I think they'd probably have Jacobs 'cream crackered' biscuits.
2006-11-06 11:08:56 · answer #4 · answered by saljegi 3 · 0⤊ 0⤋
3x = 2652
x = 884
That means that each pupil is 884 years old, which certainly isn't right. Maybe you typed the problem wrong?
2006-11-06 07:34:09 · answer #5 · answered by Anonymous · 0⤊ 1⤋
God, I hate these questions. They're designed to mess with your head.
"If it takes two men 4 hours to carry 25 bags of cement up 6 flights of stairs, what sort of biscuits did they have on their teabreak?"
2006-11-06 07:39:25 · answer #6 · answered by Anonymous · 0⤊ 1⤋
Gavin - your figures have to be wrong - otherwise the four people are all aged (on average) 663 years each.
2006-11-06 07:33:21 · answer #7 · answered by thomasrobinsonantonio 7 · 2⤊ 0⤋
34 and 3days
2006-11-06 07:35:52 · answer #8 · answered by scallywag 3 · 0⤊ 0⤋
42. students are 12,13 and 17.
2006-11-06 07:33:55 · answer #9 · answered by happytraveler 4 · 0⤊ 0⤋
Too hard for me I'm afraid.
2006-11-06 07:31:00 · answer #10 · answered by Ally 5 · 0⤊ 2⤋