You are not logged in.
Pages: 1
I was also wondering about solving the problem that way. Would it also be better to assume that boys = girls = 136/2 = 68, so start testing pairs starting from the middle of the ordered pairs? (if this question happened to appear in a test where external aids like calculators and excel spreadsheets are not allowed)
Is it possible to solve this question without the use of algebra?
"A class has 136 pupils. 5/8 of the boys and 3/4 of the girls wear glasses. There are 93 pupils who wear glasses. How many boys are there in the class?"
Thanks alot! Actually, I was in that dinner
I just thought of this question:
Scenario:
There are 350 people attending a celebration dinner.
Seated at every table are 10 people (35 tables).
One of the event is a lucky draw with 35 prizes to be given out by drawing numbers electronically.
Assumptions (if necessary):
1) Nobody can get a prize twice.
2) Everybody stays until the end of the event.
3) The drawing of the numbers is fair.
Question:
1) What is the chance that a person gets a prize?
2) What is the chance that at least one person from a table gets a prize?
3) What is the chance that everybody from a table gets a prize?
Anybody can help me? Please show your working and answers clearly documented.
Pages: 1