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Six numbers are randomly selected and placed within a set. If the set

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10 Jun 2018, 05:00
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Six numbers are randomly selected and placed within a set. If the set has a range of 16, a median of 6, a mean of 7 and a mode of 7, what is the greatest of the six numbers?

(1) The sum of the two smallest numbers is one-fifth of the sum of the two greatest numbers
(2) The middle two numbers are 5 and 7

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10 Jun 2018, 20:40
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Bunuel wrote:
Six numbers are randomly selected and placed within a set. If the set has a range of 16, a median of 6, a mean of 7 and a mode of 7, what is the greatest of the six numbers?

(1) The sum of the two smallest numbers is one-fifth of the sum of the two greatest numbers
(2) The middle two numbers are 5 and 7

Great question..
Let the numbers be a,b,c,d,e,f
Given info..
A) f-a=16
B) (c+d)/2=6
C) a,b,c are surely less than 7
D) mode means that 7 is atleast twice but the largest that is 'f' is >7 as it has to get the average to 7 and a,b,c are<7
E) so d and e are 7 each
F) now (c+d)/2=7 and d is 7 , so c is 5..
G) numbers become a,b,5,7,7,f

We have to find 'f'

Let's see what each statement tells us..

(1) The sum of the two smallest numbers is one-fifth of the sum of the two greatest numbers
This tells us that $$a+b=\frac{7+ f}{5}$$..(i)
Also $$\frac{a+b+5+7+7+f}{6}=7....a+b+f=42-(19)=23$$...(ii)
From (i) & (ii)..
$$\frac{(7+f)}{5}+f=23......7+f+5f=115.....6f=108...f=18$$
So largest number is 18
Rather all numbers can be found a=18-16=2 & b =3
2,3,5,7,7,18
Sufficient

(2) The middle two numbers are 5 and 7
Nothing new..
It has already been found in initial statement
Insufficient

A
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10 Jun 2018, 20:50
Hi Chetan,

I think C should be 7.
How did you deduce that c<7 because a scenario can be possible where c and d both can be 7 with the mean still 7.

I agree that a,b will be less than 7 and f>7.

I assume all c d e will have the same value I.e 7.

Correct me if I am wrong?

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10 Jun 2018, 21:09
prerakgoel03 wrote:
Hi Chetan,

I think C should be 7.
How did you deduce that c<7 because a scenario can be possible where c and d both can be 7 with the mean still 7.

I agree that a,b will be less than 7 and f>7.

I assume all c d e will have the same value I.e 7.

Correct me if I am wrong?

Sent from my iPhone using GMAT Club Forum mobile app

C and d would depend on median which is 6..
So c can be at the most 6 where even d is 6, otherwise d-6=6-c..
a,b,c,6,d,e,f

Let me further amplify the terms
1) median .. it is the middle value
If odd numbers the middle value in the set.. a,b,c..... B is the median
If the set contains an even number of elements, the average of two middle values..a,b,c,d....(b+c)/2
2) mode...
The number which shows up the maximum time in the set..
A,b,c,d,d,e...d is mode
3) mean..
It is the average of numbers..
a,b,c.....(a+b+c)/3

I am sure you would know this but are getting confused in mode and median
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10 Jun 2018, 21:15
prerakgoel03 wrote:
Hi Chetan,

I think C should be 7.
How did you deduce that c<7 because a scenario can be possible where c and d both can be 7 with the mean still 7.

I agree that a,b will be less than 7 and f>7.

I assume all c d e will have the same value I.e 7.

Correct me if I am wrong?

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Hi prerakgoel03,

If c=d=e=7 (as per your observation),

Is median of 6 numbers 7? we need to keep the median value as 6 as per question.

Here , we have even numbers of elements, n=6, For even numbers of elements, median=Average of $$(\frac{n}{2})$$th term and $$(\frac{n}{2} +1)$$th term

Hence , Median=Average of 3rd & 4th term

we have deduced that 4th term=d=7

So, 6=$$(\frac{c+7)}{2}$$
Or, c=12-7=5 which is less than 7.

Hope it helps.
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PKN

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10 Jun 2018, 21:20
I think I misjudged the value as 7.
Thanks for pointing out.
I am clear now .

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Re: Six numbers are randomly selected and placed within a set. If the set  [#permalink]

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10 Jun 2018, 21:22
chetan2u wrote:
Bunuel wrote:
Six numbers are randomly selected and placed within a set. If the set has a range of 16, a median of 6, a mean of 7 and a mode of 7, what is the greatest of the six numbers?

(1) The sum of the two smallest numbers is one-fifth of the sum of the two greatest numbers
(2) The middle two numbers are 5 and 7

Great question..
Let the numbers be a,b,c,d,e,f
Given info..
A) f-e=16
B) (c+d)/2=6
C) a,b,c are surely less than 7
D) mode means that 7 is atleast twice but the largest that is 'f' is >7 as it has to get the average to 7 and a,b,c are<7
E) so d and e are 7 each
F) now (c+d)/2=7 and d is 7 , so c is 5..
G) numbers become a,b,5,7,7,f

We have to find 'f'

Let's see what each statement tells us..

(1) The sum of the two smallest numbers is one-fifth of the sum of the two greatest numbers
This tells us that $$a+b=\frac{7+ f}{5}$$..(i)
Also $$\frac{a+b+5+7+7+f}{6}=7....a+b+f=42-(19)=23$$...(ii)
From (i) & (ii)..
$$\frac{(7+f)}{5}+f=23......7+f+5f=115.....6f=108...f=18$$
So largest number is 18
Rather all numbers can be found a=18-16=2 & b =3
2,3,5,7,7,18
Sufficient

(2) The middle two numbers are 5 and 7
Nothing new..
It has already been found in initial statement
Insufficient

A

Hi chetan2u,

Sir,

I have obtained the six numbers as:-

2, 3, 5, 7, 7, 18

I think the expression (A)highlighted above should be f-a=16.

Kindly correct me.
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Re: Six numbers are randomly selected and placed within a set. If the set  [#permalink]

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10 Jun 2018, 21:33
PKN wrote:
chetan2u wrote:
Bunuel wrote:
Six numbers are randomly selected and placed within a set. If the set has a range of 16, a median of 6, a mean of 7 and a mode of 7, what is the greatest of the six numbers?

(1) The sum of the two smallest numbers is one-fifth of the sum of the two greatest numbers
(2) The middle two numbers are 5 and 7

Great question..
Let the numbers be a,b,c,d,e,f
Given info..
A) f-e=16
B) (c+d)/2=6
C) a,b,c are surely less than 7
D) mode means that 7 is atleast twice but the largest that is 'f' is >7 as it has to get the average to 7 and a,b,c are<7
E) so d and e are 7 each
F) now (c+d)/2=7 and d is 7 , so c is 5..
G) numbers become a,b,5,7,7,f

We have to find 'f'

Let's see what each statement tells us..

(1) The sum of the two smallest numbers is one-fifth of the sum of the two greatest numbers
This tells us that $$a+b=\frac{7+ f}{5}$$..(i)
Also $$\frac{a+b+5+7+7+f}{6}=7....a+b+f=42-(19)=23$$...(ii)
From (i) & (ii)..
$$\frac{(7+f)}{5}+f=23......7+f+5f=115.....6f=108...f=18$$
So largest number is 18
Rather all numbers can be found a=18-16=2 & b =3
2,3,5,7,7,18
Sufficient

(2) The middle two numbers are 5 and 7
Nothing new..
It has already been found in initial statement
Insufficient

A

Hi chetan2u,

Sir,

I have obtained the six numbers as:-

2, 3, 5, 7, 7, 18

I think the expression (A)highlighted above should be f-a=16.

Kindly correct me.

Thanks a lot..
It is a typo..
Range is f-a
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11 Jun 2018, 18:05
Six numbers are randomly selected and placed within a set. If the set has a range of 16, a median of 6, a mean of 7 and a mode of 7, what is the greatest of the six numbers?

(1) The sum of the two smallest numbers is one-fifth of the sum of the two greatest numbers
(2) The middle two numbers are 5 and 7

let the number in asc. order be - a,b,c,d,e,f

1) Sufficient

5(a+b)=e+f
c+d=12 (from median)
f-a=16 (from range)
a+b+c+d+e+f=42 (from mean)

6(a+b)=30 so, a+b=5

e+f=25

now, e and f both cant be 7. so definitely e is 7 which also means d is 7.

now we can just fill in the series 2,3,5,7,7,18

2) No new info.

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12 Jun 2018, 01:14
Bunuel wrote:
Six numbers are randomly selected and placed within a set. If the set has a range of 16, a median of 6, a mean of 7 and a mode of 7, what is the greatest of the six numbers?

(1) The sum of the two smallest numbers is one-fifth of the sum of the two greatest numbers
(2) The middle two numbers are 5 and 7

Let the six numbers be a, b, c, d, e, f, in ascending order. a as smallest & f as greatest.

Given, Range f - a = 16,
Median (c + d)/2 = 6, c + d = 12, giving (c,d) = (5,7) or (4,8) or (3,9)...etc. for (c < d)
Mode 7, hence 7 is repeated at least twice.
Mean (a + b + c + d + e + f)/6 = 7
Hence a + b + c + d + e + f = 42.....(i)

Required is f = ?

Statement 1:
e + f = 5 (a + b)

we get, eq.(i) as,

1/5*(e + f) + 12 + e + f = 42

e + f = 25

Now, lets take (c,d) to satisfy (e,f), such that e + f = 25
If (c,d) = (3,9), we get the numbers, as a,b,3,9,e,f, for this we cant get Mode as 7. hence discard this (c,d) =(3,9)
From this we can see that, d has to be 7, or else the numbers don't fit the constraints.
Hence (c,d) = (5,7) & therefore e = 7, to satisfy that Mode = 7,

Substituting e = 7 in e + f = 25, we get f = 18.

Statement 1 is Sufficient.

Statement 2:
The middle two numbers are 5 and 7, hence we have c = 5 & d = 7, since c < d

So we have the numbers as, a, b, 5, 7, e, f

a + b +5 + 7 + e + f = 42

a + b + e + f = 30,

Now since mode is 7, the only possibility we have is e =7, we cannot take f = 7 as it will inflate a, b & the constraints will not be met.

So now we have, a + b + f = 23,

Hence we can try various combination for a, b & f to satisfy the constraints,

a = 3, b = 4, f = 16, we get the numbers as 3, 4, 5, 7, 7, 16, doesn't satisfy range constraint.

or a = 2, b = 3, f = 18, we get the numbers as 2, 3, 5, 7, 7, 18, satisfying all the constraints.

or a = 1, b = 2, f = 20, we get the numbers as 1, 2, 5, 7, 7, 20, again doesn't satisfy range constraint.

Hence we get, only f = 18 satisfying the constraints.

Statement 2 is Sufficient.

Thanks,
GyM

Ps - chetan2u, kindly review, i must have messed up somewhere, however cannot figure out where.
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Re: Six numbers are randomly selected and placed within a set. If the set  [#permalink]

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12 Jun 2018, 04:27
GyMrAT wrote:
Bunuel wrote:
Six numbers are randomly selected and placed within a set. If the set has a range of 16, a median of 6, a mean of 7 and a mode of 7, what is the greatest of the six numbers?

(1) The sum of the two smallest numbers is one-fifth of the sum of the two greatest numbers
(2) The middle two numbers are 5 and 7

Let the six numbers be a, b, c, d, e, f, in ascending order. a as smallest & f as greatest.

Given, Range f - a = 16,
Median (c + d)/2 = 6, c + d = 12, giving (c,d) = (5,7) or (4,8) or (3,9)...etc. for (c < d)
Mode 7, hence 7 is repeated at least twice.
Mean (a + b + c + d + e + f)/6 = 7
Hence a + b + c + d + e + f = 42.....(i)

Required is f = ?

Statement 1:
e + f = 5 (a + b)

we get, eq.(i) as,

1/5*(e + f) + 12 + e + f = 42

e + f = 25

Now, lets take (c,d) to satisfy (e,f), such that e + f = 25
If (c,d) = (3,9), we get the numbers, as a,b,3,9,e,f, for this we cant get Mode as 7. hence discard this (c,d) =(3,9)
From this we can see that, d has to be 7, or else the numbers don't fit the constraints.
Hence (c,d) = (5,7) & therefore e = 7, to satisfy that Mode = 7,

Substituting e = 7 in e + f = 25, we get f = 18.

Statement 1 is Sufficient.

Statement 2:
The middle two numbers are 5 and 7, hence we have c = 5 & d = 7, since c < d

So we have the numbers as, a, b, 5, 7, e, f

a + b +5 + 7 + e + f = 42

a + b + e + f = 30,

Now since mode is 7, the only possibility we have is e =7, we cannot take f = 7 as it will inflate a, b & the constraints will not be met.

So now we have, a + b + f = 23,

Hence we can try various combination for a, b & f to satisfy the constraints,

a = 3, b = 4, f = 16, we get the numbers as 3, 4, 5, 7, 7, 16, doesn't satisfy range constraint.

or a = 2, b = 3, f = 18, we get the numbers as 2, 3, 5, 7, 7, 18, satisfying all the constraints.

or a = 1, b = 2, f = 20, we get the numbers as 1, 2, 5, 7, 7, 20, again doesn't satisfy range constraint.

Hence we get, only f = 18 satisfying the constraints.

Statement 2 is Sufficient.

Thanks,
GyM

Ps - chetan2u, kindly review, i must have messed up somewhere, however cannot figure out where.

Hi
Yes, you will have two modes to satisfy then..
1,5,5,7,7,17 and few more but here mode will be 5 and 7..

But you don't require statement II at all because in the given conditions in the main statement, the middle two have to be 5 & 7..
So B really doesn't give you anything new
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Re: Six numbers are randomly selected and placed within a set. If the set  [#permalink]

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12 Jun 2018, 05:26
chetan2u wrote:
GyMrAT wrote:
Bunuel wrote:
Six numbers are randomly selected and placed within a set. If the set has a range of 16, a median of 6, a mean of 7 and a mode of 7, what is the greatest of the six numbers?

(1) The sum of the two smallest numbers is one-fifth of the sum of the two greatest numbers
(2) The middle two numbers are 5 and 7

Let the six numbers be a, b, c, d, e, f, in ascending order. a as smallest & f as greatest.

Given, Range f - a = 16,
Median (c + d)/2 = 6, c + d = 12, giving (c,d) = (5,7) or (4,8) or (3,9)...etc. for (c < d)
Mode 7, hence 7 is repeated at least twice.
Mean (a + b + c + d + e + f)/6 = 7
Hence a + b + c + d + e + f = 42.....(i)

Required is f = ?

Statement 1:
e + f = 5 (a + b)

we get, eq.(i) as,

1/5*(e + f) + 12 + e + f = 42

e + f = 25

Now, lets take (c,d) to satisfy (e,f), such that e + f = 25
If (c,d) = (3,9), we get the numbers, as a,b,3,9,e,f, for this we cant get Mode as 7. hence discard this (c,d) =(3,9)
From this we can see that, d has to be 7, or else the numbers don't fit the constraints.
Hence (c,d) = (5,7) & therefore e = 7, to satisfy that Mode = 7,

Substituting e = 7 in e + f = 25, we get f = 18.

Statement 1 is Sufficient.

Statement 2:
The middle two numbers are 5 and 7, hence we have c = 5 & d = 7, since c < d

So we have the numbers as, a, b, 5, 7, e, f

a + b +5 + 7 + e + f = 42

a + b + e + f = 30,

Now since mode is 7, the only possibility we have is e =7, we cannot take f = 7 as it will inflate a, b & the constraints will not be met.

So now we have, a + b + f = 23,

Hence we can try various combination for a, b & f to satisfy the constraints,

a = 3, b = 4, f = 16, we get the numbers as 3, 4, 5, 7, 7, 16, doesn't satisfy range constraint.

or a = 2, b = 3, f = 18, we get the numbers as 2, 3, 5, 7, 7, 18, satisfying all the constraints.

or a = 1, b = 2, f = 20, we get the numbers as 1, 2, 5, 7, 7, 20, again doesn't satisfy range constraint.

Hence we get, only f = 18 satisfying the constraints.

Statement 2 is Sufficient.

Thanks,
GyM

Ps - chetan2u, kindly review, i must have messed up somewhere, however cannot figure out where.

Hi
Yes, you will have two modes to satisfy then..
1,5,5,7,7,17 and few more but here mode will be 5 and 7..

But you don't require statement II at all because in the given conditions in the main statement, the middle two have to be 5 & 7..
So B really doesn't give you anything new

Thanks for the reply!, however i beg to differ.

In my opinion, the given information, in the prompt, provides ambiguity for the middle 2 numbers. We cannot clearly deduce that the middle two numbers will be 5 & 7. Hence once the middle two terms are determined, the other numbers are to be found, such that they satisfy the constraints.

Statement 1 helps us to zone in on the middle two numbers & hence help in finding the greatest number, through the relationship between last two terms & first two terms.

Statement 2, on the other hand, provides us the middle two terms & hence help in finding the greatest number, by determining that there cannot be two modes & the mean as well as range constraints can be satisfied with only one set of numbers.

using statement 2, we cannot say the numbers are, 2.1, 3, 5, 7, 7, 18.1, which satisfies the range constraint but doesn't satisfy the constraint for mean. There will be only one unique set of numbers, containing 5 & 7 as middle terms & satisfying the other constraints.

Thanks,
GyM
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12 Jun 2018, 05:36
GyMrAT wrote:
determined, the other numbers are to be found, such that they satisfy the constraints.

Statement 1 helps us to zone in on the middle two numbers & hence help in finding the greatest number, through the relationship between last two terms & first two terms.

Statement 2, on the other hand, provides us the middle two terms & hence help in finding the greatest number, by determining that there cannot be two modes & the mean as well as range constraints can be satisfied with only one set of numbers.

using statement 2, we cannot say the numbers are, 2.1, 3, 5, 7, 7, 18.1, which satisfies the range constraint but doesn't satisfy the constraint for mean. There will be only one unique set of numbers, containing 5 & 7 as middle terms & satisfying the other constraints.

Thanks,
GyM

Hi..

It is good to differ...
But logic for why middle two numbers can be only 5 and 7 is..

Let the numbers be a,b,c,d,e,f
Given info..
A) f-a=16
B) (c+d)/2=6
C) a,b,c are surely less than 7
D) mode means that 7 is atleast twice but the largest that is 'f' is >7 as it has to get the average to 7 and a,b,c are<7
E) so d and e are 7 each
F) now (c+d)/2=7 and d is 7 , so c is 5..
G) numbers become a,b,5,7,7,f

You can see if there can be any other middle numbers...
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Re: Six numbers are randomly selected and placed within a set. If the set  [#permalink]

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04 Nov 2018, 12:29
Bunuel wrote:
Six numbers are randomly selected and placed within a set. If the set has a range of 16, a median of 6, a mean of 7 and a mode of 7, what is the greatest of the six numbers?

(1) The sum of the two smallest numbers is one-fifth of the sum of the two greatest numbers
(2) The middle two numbers are 5 and 7

Very nice problem (in which a "structured candidate" has considerable advantage)!

$${a_1} \le {a_2} \le \ldots \le {a_6}\,\,\,\,\,\left\{ \matrix{ \,{a_6} = {a_1} + 16\, \hfill \cr \,{a_3} + {a_4} = 12 \hfill \cr \,\sum\nolimits_6 {\, = } \,\,\,42 \hfill \cr \,7{\rm{s}}\,\,{\rm{appear}}\,\,{\rm{most}}\, \hfill \cr} \right.$$

$$? = {a_6}$$

We "MUST" start with statement (2): it is easier and it will help us "gain knowledge and sensitivity" to what is really going on!

$$\left( 2 \right)\,\,\,\,{a_1}\,\,\,{a_2}\,\,\,5\,\,\,7\,\,\,7\,\,\,{a_6}\mathop \ne \limits^{\left( {**} \right)} 7\,\,\,\,\,\left\{ \matrix{ \,{a_6} = {a_1} + 16\, \hfill \cr \,{a_3} + {a_4} = 12\,\,\,\,\left( {{\rm{sure}}} \right) \hfill \cr \,\sum\nolimits_6 {\, = } \,\,\,42\,\,\,\left( * \right) \hfill \cr \,7{\rm{s}}\,\,{\rm{appear}}\,\,{\rm{most}}\,\,\,\,\left( {\,\left( {**} \right)\,\,{a_6} = 7\,\,\, \Rightarrow \,\,\,{a_1} = 7 - 16 = - 9\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,{a_2} = 25 > {a_3}\,\,{\rm{impossible}}\,} \right)\, \hfill \cr} \right.$$

$$\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {{a_1},{a_6},{a_2}} \right) = \left( {1,17,5} \right)\,\,\,{\rm{viable}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{a_6} = \,\,17 \hfill \cr \,{\rm{Take}}\,\,\left( {{a_1},{a_6},{a_2}} \right) = \left( {2,18,3} \right)\,\,\,{\rm{viable}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{a_6} = \,\,18\,\, \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{INSUFF}}.$$

$$\left( 1 \right)\,\,\,5\left( {{a_1} + {a_2}} \right) = {a_5} + {a_6}\,\,\,\,\left\{ \matrix{ \,{a_6} = {a_1} + 16\, \hfill \cr \,{a_3} + {a_4} = 12 \hfill \cr \,\sum\nolimits_6 {\, = } \,\,\,42 \hfill \cr \,7{\rm{s}}\,\,{\rm{appear}}\,\,{\rm{most}}\, \hfill \cr} \right.$$

$${a_3} = 7\,\,\,\, \Rightarrow \,\,{a_4} = 12 - 7 = 5 < {a_3}\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{impossible}}$$

$${a_{`5}} = {a_6} = 7\,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{ \,{a_1} = 7 - 16 = - 9 \hfill \cr \,{a_2} = \sum\nolimits_6 {\, - \left[ {{a_1} + \left( {{a_3} + {a_4}} \right) + {a_5} + {a_6}} \right] = } \,\,42 - \left( { - 9 + 12 + 2 \cdot 7} \right) = 25\,\,\, > \,\,\,{a_5} \hfill \cr} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{impossible}}$$

$${\rm{Hence}}:\,\,\,\,{a_1}\,\,\,{a_2}\,\,\,{a_3}\,\,\,7\,\,\,7\,\,\,{a_6} \ne 7\,\,\,\,\,\left\{ \matrix{ \,{a_6} = {a_1} + 16\, \hfill \cr \,{a_3} + {a_4} = 12\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{a_3} = 12 - 7 = 5 \hfill \cr \,\sum\nolimits_6 {\, = } \,\,\,42\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{a_1} + {a_2} + {a_6} = 42 - {a_3} - \left( {{a_4} + {a_5}} \right) = 42 - 5 - 14 = 23 \hfill \cr \,5\left( {{a_1} + {a_2}} \right) = {a_5} + {a_6}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,5\left( {23 - {a_6}} \right) = 7 + {a_6}\,\,\,\,\, \Rightarrow \,\,\,\,? = {a_6}\,\,\,{\rm{unique}} \hfill \cr} \right.$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Re: Six numbers are randomly selected and placed within a set. If the set   [#permalink] 04 Nov 2018, 12:29
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