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the five consecutive even

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by sanju09 » Thu Apr 08, 2010 3:54 am
What is the average (arithmetic mean) of the five consecutive even integers?

(1) The product of the smallest and the greatest of the five consecutive even integers is 48.

(2) The product of the greatest two of the five consecutive even integers is 24.
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by neoreaves » Thu Apr 08, 2010 4:49 am
we can write the integers as


a-4,a-2,a,a+2,a+4

we need to find the average = a = ?

1) (a-2)(a+2) = 48

so a = -8 or a = 8
Insufficient

2) (a+4)(a+2) = 24
so a = -8 or a = -2


C) a = -8

Sufficient

Thus answer should be C
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by sanju09 » Thu Apr 08, 2010 5:00 am
neoreaves wrote:we can write the integers as


a-4,a-2,a,a+2,a+4

we need to find the average = a = ?

1) (a-2)(a+2) = 48

so a = -8 or a = 8
Insufficient

2) (a+4)(a+2) = 24
so a = -8 or a = -2


C) a = -8

Sufficient

Thus answer should be C
great, only the bold part is devious to me.
Last edited by sanju09 on Thu Apr 08, 2010 5:12 am, edited 1 time in total.
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by sanju09 » Thu Apr 08, 2010 5:08 am
repeat post regretted
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by neoreaves » Thu Apr 08, 2010 5:14 am
Sorry my bad ..

i meant to say

1) (a-4)(a+4) = 48

--> a^2 - 16 = 48
--> a^2 = 64
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by asfa » Thu Apr 08, 2010 3:37 pm
IMO, The way this question is written the answer should be E.
the answer could be 8 or -8

However if states in the question stem that consecutive integers are positive or negative then the answer is D
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by pops » Thu Apr 08, 2010 9:31 pm
sanju09 wrote:What is the average (arithmetic mean) of the five consecutive even integers?

(1) The product of the smallest and the greatest of the five consecutive even integers is 48.

(2) The product of the greatest two of the five consecutive even integers is 24.
statement1:
agree with the first reply that a can be +8 or -8


statement2:
(a+4)*(a+2)=24
=a^2 + 6a + 8 = 24
=a^2 + 6a -16 = 0
=a=8 or -2


combining both: a = 8 :)
hence C !
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by sanju09 » Fri Apr 09, 2010 12:24 am
pops wrote:
sanju09 wrote:What is the average (arithmetic mean) of the five consecutive even integers?

(1) The product of the smallest and the greatest of the five consecutive even integers is 48.

(2) The product of the greatest two of the five consecutive even integers is 24.
statement1:
agree with the first reply that a can be +8 or -8


statement2:
(a+4)*(a+2)=24
=a^2 + 6a + 8 = 24
=a^2 + 6a -16 = 0
=a=8 or -2


combining both: a = 8 :)
hence C !
How can a^2 + 6 a - 16 = 0 give us a = 8 or -2?
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by yatinbassi » Sat Apr 10, 2010 6:29 am
Why have we taken the nos as a-4,a-2,a,a+2..etc? Can we not simply take it as a,a+2,a+4,a+6,a+8 ?

1) a(a+8)=48
=> a=4, or a=-12
Mean=+8 or -8
Insuff

2) (a+6)(a+8)=24
=>a=-12 or a=-2
Mean=2. or -8
Insuff

Combining these 2 the common solution is mean = -8
Hence, C

Is that a way to solve this?
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by this_time_i_will » Sat Apr 10, 2010 7:26 am
sanju09 wrote:What is the average (arithmetic mean) of the five consecutive even integers?

(1) The product of the smallest and the greatest of the five consecutive even integers is 48.

(2) The product of the greatest two of the five consecutive even integers is 24.
Five consecutive even integers should be taken as 2x, 2x+2,2x+4,2x+6 and 2x+8.

I: 2x(2x+8) = 48. solving x=-4,3. INSUFF.

II: (2x+6)(2x+8)=24. solving x=-6,-1 INSUFF.

since no value is common in I and II, C ruled out.

So, E
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by liferocks » Sat Apr 10, 2010 9:00 am
IMO ans should be C
let the numbers be 2n-4,2n-2,2n,2n+2 and 2n+4
statement 1 will give
(2n-4)*(2n+4)=48
or n=-4 or +4

statement 2 will give
(2n+2)(2n+4)=24
or n=1 or -4

both statements gives two solution for the value of n
but combining we get n=-4
hence mean=2n or 2*(-4)
hence ans C
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by liferocks » Sat Apr 10, 2010 9:03 am
this_time_i_will wrote:
sanju09 wrote:What is the average (arithmetic mean) of the five consecutive even integers?

(1) The product of the smallest and the greatest of the five consecutive even integers is 48.

(2) The product of the greatest two of the five consecutive even integers is 24.
Five consecutive even integers should be taken as 2x, 2x+2,2x+4,2x+6 and 2x+8.

I: 2x(2x+8) = 48. solving x=-4,3. INSUFF.

II: (2x+6)(2x+8)=24. solving x=-6,-1 INSUFF.

since no value is common in I and II, C ruled out.

So, E
in this calculation for the first point x will be -6 or 2 so combining 1 and 2 we will get the ans.
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by this_time_i_will » Sat Apr 10, 2010 5:41 pm
liferocks wrote:
this_time_i_will wrote:
sanju09 wrote:What is the average (arithmetic mean) of the five consecutive even integers?

(1) The product of the smallest and the greatest of the five consecutive even integers is 48.

(2) The product of the greatest two of the five consecutive even integers is 24.
Five consecutive even integers should be taken as 2x, 2x+2,2x+4,2x+6 and 2x+8.

I: 2x(2x+8) = 48. solving x=-4,3. INSUFF.

II: (2x+6)(2x+8)=24. solving x=-6,-1 INSUFF.

since no value is common in I and II, C ruled out.

So, E
in this calculation for the first point x will be -6 or 2 so combining 1 and 2 we will get the ans.
You are write...seems i was sleeping while solving this prob!!
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