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At a certain school, the ratio of the number of second grad

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by mitzwillrockgmat » Sat Jun 05, 2010 10:06 am
At a certain school, the ratio of the number of second
graders to the number of fourth graders is 8 to 5, and
the ratio of the number of first graders to the number of
second graders is 3 to 4. If the ratio of the number of
third graders to the number of fourth graders is 3 to 2 ,
what is the ratio of the number of first graders to the
number of third graders?

a. 16 to 15
b. 9 to 5
c. 5 to 16
d. 5 to 4
e. 4 to 5

Hi, I want to know how i can do this question faster?
I can solve it but i take almost 2 minutes!!
any shortcuts for these types of questions?

2nd /4th = 8/5
1st /2nd = 3/4
3rd/4th =3/2

find 1st/3rd =?

3rd/4th = 3/2 * 5/5 = 15/10

2nd /4th = 8/5 * 2/2 = 16/10

1st /2nd = 3/4 *4/4 = 12/16

so 1st/3rd = 12/15 = 4/5

but how can i do this faster? including reading time im touching almost 2.5 minutes!
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by 4GMAT_Mumbai » Sat Jun 05, 2010 10:29 am
After 1st sentence, in one's mind,

1st:2nd:4th = 6:8:5

3rd:4th = 3:2

To match 1st and 2nd equations, multiply 1st eq by 2 and 2nd eq by 5

1st:2nd:3rd:4th = 12:16:15:10

From this, 1st:3rd = 12:15 = 4:5

Hope this helps. Thanks.
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by mitzwillrockgmat » Sat Jun 05, 2010 12:33 pm
4GMAT_Mumbai wrote:After 1st sentence, in one's mind,

1st:2nd:4th = 6:8:5 *******the 6 is a mistake right? should be 3:8:5 no?

3rd:4th = 3:2

To match 1st and 2nd equations, multiply 1st eq by 2 and 2nd eq by 5

1st:2nd:3rd:4th = 12:16:15:10

From this, 1st:3rd = 12:15 = 4:5

Hope this helps. Thanks.
Thanks! this is faster than what i was doing!
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by muralithe1 » Sun Jun 06, 2010 11:56 am
Hi,
This is my apporach..

2nd /4th = 8/5
1st /2nd = 3/4
3rd/4th =3/2


2nd/4th * 1st/2nd * 4th/3rd = 1st/3rd= 8/5*3/4*2/3= 4/5
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by mitzwillrockgmat » Sun Jun 06, 2010 12:56 pm
muralithe1 wrote:Hi,
This is my apporach..

2nd /4th = 8/5
1st /2nd = 3/4
3rd/4th =3/2


2nd/4th * 1st/2nd * 4th/3rd = 1st/3rd= 8/5*3/4*2/3= 4/5
Hi, this works!!! but im not sure if there's an actual mathematical concept here or that the answer is just a coincidence i.e. that all given ratio's product is the answer. can you explain why & how you knew that multiplying all the given ratios would give the ratio that is being asked?
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by muralithe1 » Sun Jun 06, 2010 1:25 pm
No its not coincidence...


Then the ratios as as A/B * B/C * C/D = A/D


here ofcouce we have given D/C ratio, so for our convenience while we multiply.. i inverted to C/D...so when i multiply...


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by nmerchant » Tue Jun 08, 2010 12:12 pm
muralithe1 wrote:Hi,
This is my apporach..

2nd /4th = 8/5
1st /2nd = 3/4
3rd/4th =3/2


2nd/4th * 1st/2nd * 4th/3rd = 1st/3rd= 8/5*3/4*2/3= 4/5
Yup.. I did the same thing.
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by eloka » Tue Jun 08, 2010 5:27 pm
Okay...this is my approach..eloka's way

x = 2nd grade
y= 4th grade
m= 1st grade
k= 3rd grade

x:y = 8:5 (*2) ===> x:y = 16:10======> x:y = 16:10
m:x=3:4 (*2)====> m : x = 6: 8 (*2) ==> m :x = 12: 16
k:y = 3:2(*5)====> k:y = 15:10======> k:y = 15:10

Now we know x and y are equal to all the ration. Therefore, we can find the ratio solution for m:k = 12: 15 = 4:5
m/k = 4/5
Warm wishes to all of you who are striving to conquer the GMAT

Eloka
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All I need is 450 GMAT score!! thats all, Geez
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by eloka » Tue Jun 08, 2010 5:31 pm
Okay...this is my approach..eloka's way

x = 2nd grade
y= 4th grade
m= 1st grade
k= 3rd grade

x:y = 8:5 multiply with (*2) ===> x:y = 16:10======> x:y = 16:10
m=3: 4 multiply with (*2)====> m : x = 6: 8 multiply again with (*2) ==> m = 12: 16
k:y = 3: 2 multiply with (*5)====> k:y = 15:10======> k:y = 15:10

keep multiplying until you find x and y in a ratio group shares the same number. As you can see

x/y = 16/10
m/x = 12/16
k/y=15/10

x is 16 and y is 10....so now we can find ratio of m to k


Now we know x and y are equal to all the ration. Therefore, we can find the ratio solution for m:k = 12: 15 = 4:5
m/k = 4/5
Warm wishes to all of you who are striving to conquer the GMAT

Eloka
Indonesia
https://lavachoco.blocked
All I need is 450 GMAT score!! thats all, Geez
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