The sum of the ages of 22 boys and 24 girls is 160.What is the sum of ages of one boy plus one girl, if all the boys are of the same age and all the girls are of the same age, and only full years are counted?
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sum of the ages
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truplayer256
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This is more of a guess and check question.
X= Age of one boy
Y= Age of one girl
22X+24Y=160
We want X+Y=?
Just to make the numbers more simpler and easier to work with, let's factor out a 2.
2(11X+12Y)=160
11X+12Y=80
X=4--> 12Y=36--> Y=3
X+Y=4+3=7 B
X= Age of one boy
Y= Age of one girl
22X+24Y=160
We want X+Y=?
Just to make the numbers more simpler and easier to work with, let's factor out a 2.
2(11X+12Y)=160
11X+12Y=80
X=4--> 12Y=36--> Y=3
X+Y=4+3=7 B
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shanmugam.d
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to continue with trueplayer:
11X+12Y=80 to find the combination which makes up 80
just list multiples of 11: 11 22 33 44 55
multiple of 12: 12 24 36 48 56
the combination is: 4 & 3 hence: 4+3 = 7 (B)
11X+12Y=80 to find the combination which makes up 80
just list multiples of 11: 11 22 33 44 55
multiple of 12: 12 24 36 48 56
the combination is: 4 & 3 hence: 4+3 = 7 (B)
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ajlcalbanese1
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if we have:
22X+24Y=160
can we find the value for X and then substitute in the ecuation?
22x=160-24y
x=160-24y/22
22(160-24y/22)=160
and from here find the value of y and substitute in the ecuation?
would that work? any comments about my aproach?
22X+24Y=160
can we find the value for X and then substitute in the ecuation?
22x=160-24y
x=160-24y/22
22(160-24y/22)=160
and from here find the value of y and substitute in the ecuation?
would that work? any comments about my aproach?
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scoobydooby
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22B+24G=160
=>11B+12G=80
80 is even, 12G is even=> 11B must be even (E+E=E)
since B, G must be integers,
G=(80-11B)/12 must be an integer
let B be 2, (B must be even)=>G=(80-22)/12=58/12 not an integer
let B be 4, (G=80-44)/12=36/12=3 integer. bingo!
so B+G=4+3=7
hence, B
=>11B+12G=80
80 is even, 12G is even=> 11B must be even (E+E=E)
since B, G must be integers,
G=(80-11B)/12 must be an integer
let B be 2, (B must be even)=>G=(80-22)/12=58/12 not an integer
let B be 4, (G=80-44)/12=36/12=3 integer. bingo!
so B+G=4+3=7
hence, B
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ghacker
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Another way
Girls = G ; Boys = B
22B+24G = 160 , B, G integers
22(B+G) +2G = 160 -------------> 11(B+G)+G = 80
Since B, G are integers B+G must also be an integers and the only positive integers which satisfy the equations are 7 and 3
So G = 3 and B= 4
Girls = G ; Boys = B
22B+24G = 160 , B, G integers
22(B+G) +2G = 160 -------------> 11(B+G)+G = 80
Since B, G are integers B+G must also be an integers and the only positive integers which satisfy the equations are 7 and 3
So G = 3 and B= 4
