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17 Jan 2010, 11:28
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Translate the Following Statements into equaions and /or inequalities
1.There are twice as many Computers as there are printers.
my answer was
assumimg p = printer and c= computer
p =2c,
but after rethinking found out that it is Wrong the Answer should be
c =2p
2.There are 10 more grapes than apples,and one fourth as many appples as pears.
assuming g= grape,a = apple ,p = pears
my answer was,
g = 10+a,
p = 1/4 a Which was wrong again I dont know how to decode this..
Drill 2: Translate and Solve the Following probems
Three Friends sit down to eat 14 slices of Pizza.If two of the Friends eat the same number of slices ,and the third eats two more slices than each of the other two,how many slices are eaten thrid friend.
Solution:
assume
Variables
f1 = friend1
f2 = friend2
f3 = friend3
Equations
f1+f2+f3 = 14 slices ---- eqn1
hint:
"two of the Friends eat the same number of slices"
=> f1 = f2 ----- eqn2
"the third eats two more slices than each of the other two"
=> f3 eats (f1+2) and (f2+2)
=> f3 = f1+2+f2+2 = f1+f2+4 Now apply eqn 2
=> f3 = 2f2 +4 ---->eqn 3
Apply f3 in eqn 1 also apply eqn 2 in eqn 1
f1+f2+f3 = 14
=> f2 + f2 +2f2+4 = 14
=> 4f2 + 4 = 14 => 4f2 = 14 - 4 => f2 = 10/4 => f2 = 2.5
f3 = 2f2 + 4 => 5 + 4 = 9
f3 = 9 and f1 = 2.5,f2 = 2.5 do a addition i get 2.5+2.5+9 = 14
But the Solution was Different..
f3 = 6 slices of pizza..
Please Help...
1.There are twice as many Computers as there are printers.
my answer was
assumimg p = printer and c= computer
p =2c,
but after rethinking found out that it is Wrong the Answer should be
c =2p
2.There are 10 more grapes than apples,and one fourth as many appples as pears.
assuming g= grape,a = apple ,p = pears
my answer was,
g = 10+a,
p = 1/4 a Which was wrong again I dont know how to decode this..
Drill 2: Translate and Solve the Following probems
Three Friends sit down to eat 14 slices of Pizza.If two of the Friends eat the same number of slices ,and the third eats two more slices than each of the other two,how many slices are eaten thrid friend.
Solution:
assume
Variables
f1 = friend1
f2 = friend2
f3 = friend3
Equations
f1+f2+f3 = 14 slices ---- eqn1
hint:
"two of the Friends eat the same number of slices"
=> f1 = f2 ----- eqn2
"the third eats two more slices than each of the other two"
=> f3 eats (f1+2) and (f2+2)
=> f3 = f1+2+f2+2 = f1+f2+4 Now apply eqn 2
=> f3 = 2f2 +4 ---->eqn 3
Apply f3 in eqn 1 also apply eqn 2 in eqn 1
f1+f2+f3 = 14
=> f2 + f2 +2f2+4 = 14
=> 4f2 + 4 = 14 => 4f2 = 14 - 4 => f2 = 10/4 => f2 = 2.5
f3 = 2f2 + 4 => 5 + 4 = 9
f3 = 9 and f1 = 2.5,f2 = 2.5 do a addition i get 2.5+2.5+9 = 14
But the Solution was Different..
f3 = 6 slices of pizza..
Please Help...
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18 Jan 2010, 09:38
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These are good exercises. Here's how to think about them:
computers = 2 * (printers)
c = 2p
the sentence: "there are twice as many computers as there are printers" implies there are more computers than there are printers. Check your equation --does it make sense? which side is larger? c? or 2p? It takes 2 p's to equal a 'c.' So 'c' is bigger. This makes sense!
grapes = 10 more than apples
g = 10 + a
apples = 1/4 pears.
"one fourth as many apples as pears" implies there are fewer apples than pears. How much fewer? one fourth! Make sure your final equation shows that there are fewer apples than pears.
a = (1/4)p
apples = a fraction of number of pears--so this makes sense!
"Three Friends sit down to eat 14 slices of Pizza."
f1 + f2 + f3 = 14
"If two of the Friends eat the same number of slices"
f1 = f2
"third eats two more slices than each of the other two"
f3 = f2 + 2
f3 = f1 + f2
"how many slices are eaten by the third friend?"
f3 = ?
Look at the information you have. Do you have enough information? Do you the number of equations equal the number of variables?
1) f1 + f2 + f3 = 14
2) f1 = f2
3) f3 = f2 + 2
4) f3 = f1 + f2
Well, you have 3 unknowns and 4 equations. More equations than unknowns means you can answer this question! So now let's do it!
since f1 = f2, you can rewrite #1:
f2 + f2 + f3 = 14
Since we want to solve for f3, we want to leave all the f3's in there, but express f2 with f3 inside of it. Look for an equation that has both f2 and f3 in it
Aha! f3 = f2+2
express f2 as a function of f3 so that you can plug it in later
f2 = f3 - 2
Now plug it in to: f2 + f2 + f3 = 14, or 2f2 + f3 = 14
2 (f3-2) + f3 = 14
2f3 - 4 + f3 = 14
3f3 = 18
f3 = 6
Voila! The third guy ate six slices!!
And of course the other two guys must have eaten 2 less, so that would be 6-2 = 4 slices
Quote:
Number of computers is twice that of printers.
computers = 2 * (printers)
c = 2p
the sentence: "there are twice as many computers as there are printers" implies there are more computers than there are printers. Check your equation --does it make sense? which side is larger? c? or 2p? It takes 2 p's to equal a 'c.' So 'c' is bigger. This makes sense!
Quote:
grapes = 10 more than apples
g = 10 + a
apples = 1/4 pears.
"one fourth as many apples as pears" implies there are fewer apples than pears. How much fewer? one fourth! Make sure your final equation shows that there are fewer apples than pears.
a = (1/4)p
apples = a fraction of number of pears--so this makes sense!
Quote:
"Three Friends sit down to eat 14 slices of Pizza."
f1 + f2 + f3 = 14
"If two of the Friends eat the same number of slices"
f1 = f2
"third eats two more slices than each of the other two"
f3 = f2 + 2
f3 = f1 + f2
"how many slices are eaten by the third friend?"
f3 = ?
Look at the information you have. Do you have enough information? Do you the number of equations equal the number of variables?
1) f1 + f2 + f3 = 14
2) f1 = f2
3) f3 = f2 + 2
4) f3 = f1 + f2
Well, you have 3 unknowns and 4 equations. More equations than unknowns means you can answer this question! So now let's do it!
since f1 = f2, you can rewrite #1:
f2 + f2 + f3 = 14
Since we want to solve for f3, we want to leave all the f3's in there, but express f2 with f3 inside of it. Look for an equation that has both f2 and f3 in it
Aha! f3 = f2+2
express f2 as a function of f3 so that you can plug it in later
f2 = f3 - 2
Now plug it in to: f2 + f2 + f3 = 14, or 2f2 + f3 = 14
2 (f3-2) + f3 = 14
2f3 - 4 + f3 = 14
3f3 = 18
f3 = 6
Voila! The third guy ate six slices!!
And of course the other two guys must have eaten 2 less, so that would be 6-2 = 4 slices
19 Jan 2010, 10:49
Kudos
Bookmarks
Hi GMATpill,
thanks for the Reply i do understand the Last before with bit more clearer now.
but in the Last one Question
The values we got was 6 for the two guys and
Does this 4 slices is for both (f1+f2 +2) or it for f1 = f1+2 and f2 =f1+2..
But I see The question says as
"the third eats two more slices than each of the other two" So i cannot think other than that
f3 consumes f1 +2 and f2 +2 say f3 = f1+2+f2+2
but your Solution some how makes to think that (f1+f2)+2 = f3...what i mean by that is f3 eats 2 more than his friends.I know being a dull head here,its not with the Concept it is surely decoding the question wrongly.
do you think the solution can be Explained in any other Way..will appreciate the help.
thanks for the Reply i do understand the Last before with bit more clearer now.
but in the Last one Question
The values we got was 6 for the two guys and
Does this 4 slices is for both (f1+f2 +2) or it for f1 = f1+2 and f2 =f1+2..
But I see The question says as
"the third eats two more slices than each of the other two" So i cannot think other than that
f3 consumes f1 +2 and f2 +2 say f3 = f1+2+f2+2
but your Solution some how makes to think that (f1+f2)+2 = f3...what i mean by that is f3 eats 2 more than his friends.I know being a dull head here,its not with the Concept it is surely decoding the question wrongly.
do you think the solution can be Explained in any other Way..will appreciate the help.
