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In a set of three variables, the average of the first two variables is

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In a set of three variables, the average of the first two variables is  [#permalink]

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New post 19 Apr 2017, 05:19
1
5
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

57% (02:01) correct 43% (02:45) wrong based on 72 sessions

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In a set of three variables, the average of the first two variables is greater than 2, the average of the last two variables is great than or equal to 3 , the average of the first and third variable is 4.Which of the following could be the average of the three variables?

A)1
B)1.5
C)2
D)3
E)3.5

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Re: In a set of three variables, the average of the first two variables is  [#permalink]

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New post 19 Apr 2017, 08:55
stonecold wrote:
In a set of three variables, the average of the first two variables is greater than 2, the average of the last two variables is great than or equal to 3 , the average of the first and third variable is 4.Which of the following could be the average of the three variables?

A)1
B)1.5
C)2
D)3
E)3.5



quiet difficult to get avg below 4
i can get a set {0.7 , 3.4 , 7.3} having avg = 3.8
i will go with E

+1 for alternative approach :)
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In a set of three variables, the average of the first two variables is  [#permalink]

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New post 19 Apr 2017, 12:00
3
Ahh.
Thats not a precise solution rohit8865
Did you try adding the in-equations/equations ?
Can we ever get a mean below 3?

Never.
To put it simply => For mean to be 3.5 => Sum must be 10.5
As x+z=8=> y must be 2.5






Here is what i did in this question =>

Let the 3 variables be x,y,z

As per the question =>
x+y>4
y+z≥6
x+z=8


Lets just add them up.

Adding them =>
2(x+y+z)>18 ((notice the sign))

x+y+z>9

Hence x+y+z/3 >3

So mean must be greater than 3.

Only option that obeys that is E.

SMASH that E.



Alternatively we could just use the equation solving technique

using x+z=8 in x+y>4
y-z>-4
y+z>6
Adding them => 2y>2
y>1

Hence y>1

So the least average would be when y>1
At y=1 => Mean = x+y+z/3 =>9/3=3 which is not possible as y>1

Hence mean >3


Again => Smash that E.

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Re: In a set of three variables, the average of the first two variables is  [#permalink]

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New post 19 Apr 2017, 12:34
1
stonecold wrote:
Ahh.
Thats not a precise solution rohit8865
Did you try adding the in-equations/equations ?
Can we ever get a mean below 3?

Never.
To put it simply => For mean to be 3.5 => Sum must be 10.5
As x+z=8=> y must be 1.5






Here is what i did in this question =>

Let the 3 variables be x,y,z

As per the question =>
x+y>4
y+z≥6
x+z>8


Lets just add them up.

Adding them =>
2(x+y+z)>18 ((notice the sign))

x+y+z>9

Hence x+y+z/3 >3

So mean must be greater than 3.

Only option that obeys that is E.

SMASH that E.



Alternatively we could just use the equation solving technique

using x+z=8 in x+y>4
y-z>-4
y+z>6
Adding them => 2y>2
y>1

Hence y>1

So the least average would be when y>1
At y=1 => Mean = x+y+z/3 =>9/3=3 which is not possible as y>1

Hence mean >3


Again => Smash that E.



thanks

however my guess was also > 3 because getting a avg < 3 is impossible

just correct the above 2 typo... :)
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Re: In a set of three variables, the average of the first two variables is  [#permalink]

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New post 12 Mar 2018, 10:55
stonecold wrote:
In a set of three variables, the average of the first two variables is greater than 2, the average of the last two variables is great than or equal to 3 , the average of the first and third variable is 4.Which of the following could be the average of the three variables?

A)1
B)1.5
C)2
D)3
E)3.5


We can let the three variables be a, b and c. So we have:

(a + b)/2 > 2

a + b > 4

and

(b + c)/2 ≥ 3

b + c ≥ 6

and

(a + c)/2 = 4

a + c = 8.

If we add up these three inequalities/equations, we have:

2a + 2b + 2c > 18

a + b + c > 9

(a + b + c)/3 > 3

The only number is greater than 3 is 3.5, thus it’s the answer.

Answer: E
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Re: In a set of three variables, the average of the first two variables is  [#permalink]

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New post 12 Mar 2018, 22:06
Let first variable a, second - b, third = c

given: a + b > 4 , b+c >= 6, a + c = 8

average: (a + b + c)/3 = (8 + b)/3

a + b > 4 ----(1)
b + (8-a) >= 6
a - b <=2 ---(2)

Let a - b = 2 from (2)
then subsituting a = 2 + b in ---(1), 2 + b + b > 4 => 2b > 2 => b > 1
Let a - b < 2 from (2)
Adding (1) + (2)
a + b - (a - b) > 4 - 2 => 2b > 2 => b > 1

so either case b > 1,
so average , (8 + (>1))/3 => should be greater than 3, only one option stands (E)
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Re: In a set of three variables, the average of the first two variables is  [#permalink]

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Re: In a set of three variables, the average of the first two variables is   [#permalink] 10 Apr 2019, 01:01
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