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Re: Unitary method problem:easy problem but confusing answer [#permalink]
15 Apr 2012, 04:07
The question again is very badly worded. What they want you to work out is:
that 9/20 of the class are boys = 40 x 9/20 = 18
4/5 of the class are right handed = 40 x 4/5 = 32
All 18 boys could be right handed or a minimum of 10 could be right handed as 8 people in the class are left handed. If all the left handed pupils are boys then 18 - 8 = 10 right handers.
Answer is C between 18 and 10 The way they right it leads you to believe that 4/5 of the boys are right handed but you can't have 14.4 boys. The starting question needs to be alterred so it states 4/5 of the class are right handed.
There are 40 students in a classroom, 9/20 of them are boys and 4/5 of them are right-handed. How many right-handed boys are there in the classroom?
A. Between 10 and 32. B. Between 14 and 32. C. Between 10 and 18. D. Between 14 and 18. E. Between 18 and 36.
this is a very easy problem but i am confused............ between two answers C and D
I agree that the wording of the question could have been better. For example the question should ask what is the minimum and the maximum # of right-handed boys possible, in this case the answer would be: 10 and 18.
Given that there are 18 boys (9/20*40=18) and 22 girls in the class. Also we know that out of 40 students 32 are right-handed (4/5*40=32).
Maximum number of right-handed boys possible is if ALL 18 boys are right-handed; Minimum number of right-handed boys possible is if ALL 22 girls are right-handed, so in this case 32-22=10 boys would be right-handed.
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