AP-II

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maihuna: Legendary Member; Posts: 1578; Joined: Sun Dec 28, 2008 1:49 amFollowed by: 9 members; Thanked: 82 times; GMAT Score:720

by maihuna » Sun Aug 16, 2009 3:31 am

There are two sets of numbers each consisting of 3 terms and sum of each set is 15. The common difference of the first is greater by 1 than the common difference of the second, and the product of first set is to the product of the second set as 7/8, what are the numbers.

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mohitsharda: Senior | Next Rank: 100 Posts; Posts: 64; Joined: Sat Aug 01, 2009 3:13 amFollowed by: 1 member; Thanked: 5 times; GMAT Score:740

Re: AP-II

by mohitsharda » Sun Aug 16, 2009 7:17 am

maihuna wrote:There are two sets of numbers each consisting of 3 terms and sum of each set is 15. The common difference of the first is greater by 1 than the common difference of the second, and the product of first set is to the product of the second set as 7/8, what are the numbers.

Let the two series be {a-r1,a,a+r1} and {b-r2,b,b+r2}

sum of the series 3a = 3b = 15

=> a = b= 5

Now, r1=r2 + 1

And, a(a-r1)(a+r1)/{b(b-r2)(b+r2)} = 7/8

use a=b =5 to get

{25-(r1^2)}/{25-(r2^2)} = 7/8

simplify to get 25 = 8 (r1^2) - 7 (r2^2)
use r1 = r2 + 1 to get a quadratic equation in r2

=> we get,

(r2^2)+ 16 r2 - 17 = 0

=> r2 = 1
=> r1 = 2

so the sets are (3,5,7) and (4,5,6)

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dtweah: Master | Next Rank: 500 Posts; Posts: 487; Joined: Fri Mar 27, 2009 5:49 am; Thanked: 36 times

Re: AP-II

by dtweah » Sun Aug 16, 2009 8:07 am

maihuna wrote:There are two sets of numbers each consisting of 3 terms and sum of each set is 15. The common difference of the first is greater by 1 than the common difference of the second, and the product of first set is to the product of the second set as 7/8, what are the numbers.

Let the respective sequence be be

A B C
E F G

Since the both sums are 15 we can equate the two sums and find a relationship between the two first terms.

3/2 ( 2A + 2(d+1))= 3/2( (2E +2d))

2A + 2 = 2E

A+1 = E

So we know the respective first terms differ by 1. Lets write out the series

A , A + d+1, A + 2(d+1)

(A+1 ) (A+ 1) + d, (A + 1) + 2d

Each should sum to 15, so

3A + 3d +3 =15

A+d =4, so we have

A , 5, 6 +d

A+1, 5, 5+d

Since A+d= 4, candidates (A,d)

(3, 1) (2, 2 ) (1, 3), (0 4) is excluded b/c that would make the ration zero

One of these will meet the sum and ratio conditions. Trying (3, 1) gives

3, 5, 7
4, 5, 6

3 x 5 x7/ 4 x 5 x 6= 7/8

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nano124: Junior | Next Rank: 30 Posts; Posts: 17; Joined: Sat Jun 20, 2009 2:57 pm

by nano124 » Sun Aug 16, 2009 10:06 am

This is one of those questions that doesnt really neatly fall into a specific category so i used a plug and chug method to solve it.

First clue> both sets are equal to 15: so using the starting point (5,5,5) for the second question you can come up with an instance when both sets are equal to 15 and then adjust till you come up with 7/8

First Set (diff b/w no.s is 1) (4,5,6) product = 120

Second Set (diff b/w no.s is 0) (5,5,5) product = 125

since this doesnt work, you know you have to decrease the second set and see what happens to the first.

First Set (diff b/w no.s is 2) (3,5,7) product = 105

Second Set (diff b/w no.s is 1) (4,5,6) product = 120

this gives you ur 7/8 and holds true for all the remaining conditions (sum is 15 etc). hence these sets are the answer.

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tohellandback: Legendary Member; Posts: 752; Joined: Sun May 17, 2009 11:04 pm; Location: Tokyo; Thanked: 81 times; GMAT Score:680

by tohellandback » Sun Aug 16, 2009 7:55 pm

let a be the first term of 1st series
b be the first term of second series
sum is equal:
3/2(2a+2(d+1)=3/2(2b+2d)
2a+2d+2=2b+2d
a+1=b
from here you can go the big algebra way.
but the possibilities are only the following:

168
258
357
456
numbers are 3,5,7 and 4,5,6

The powers of two are bloody impolite!!

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