If the average (arithmetic mean) of four positive numbers is

Let's say these 4 positive numbers in ascending order are \(a\), \(b\), \(c\), and \(d\). Given: \(a+b+c+d=160\). Question: how many are less than 40? Obviously less than 40 can be only 0 (in case all numbers are 40), 1, 2, or 3 numbers (all 4 can not be less than 40 as in this case their sum won't be to 4*40=160).

(1) The two smallest numbers are identical --> \(a=b\) --> \(2a+c+d=160\) --> 0 is out as all numbers are not identical and 1 is also out as 2 smallest are equal and if 1 is less than 40 then another is also, but still two answers are possible: 2 or 3 numbers are less than 40. For example: {20, 20, 50, 50} or {20, 20, 30, 70}. Not sufficient.

(2) The average (arithmetic mean) of the two largest numbers is 50 --> \(c+d=100\) --> \(a+b=60\). Again 1, 2, or 3 numbers can be less than 40. For example: {10, 50, 50, 50} or {30, 30, 50, 50} or {30, 30, 30, 70}. Not sufficient.

(1)+(2) As from (1) \(a=b\) and from (2) \(a+b=60\) then \(a=b=30\), so two smallest numbers are 30 and 30: {30, 30, c, d}. But still 2 or 3 numbers can be less than 40: {30, 30, 50, 50} or {30, 30, 35, 65}. Not sufficient.

Answer: E.
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Let the #s be a,b,c,d.

Given a+b+c+d=160

note that the average of the 4 #s to be 40, some #s shud be < 40 and the rest shud be >40 OR all 4 shud be equal to 40

stmnt1: The two smallest numbers are identical.
let us say c and d are the smallest among 4. then, c=d
a+b+2c=160
if c and d are < 40 then 2c<80 then a+b shud be > 80 in which i can have both a and b > 40 or a>40 and b<40....NOT SUFF.

stmnt2: The average (arithmetic mean) of the two largest numbers is 50.

let a and b be the largest #s ==> a+b =100now 100+c+d = 160
==>c+d = 60
we can have both c and d < 40 or one < 40 and other > 40 for the SUM to be 60
NOT SUFF.

1&2
As you can see the above bold case points, both the statements, in a way, are pointing to a single piece of info. hence NOT SUFF.

Answer "E".

Regards,
Murali.
Kudos?

Well, I am little bit confused with the answer. Rather C seems more tempting to me. If we combine both statements then we come to know there two numbers greater than 40 and two numbers less than 40. We can come up with the answer using both the statements. I have interpreted smallest number as less than 40 whereas largest as more than 40. So, {30, 30, 35, 65} cant be present. Kindly rectify me where I went wrong.

How did you come up with the red part?

Also, why is set {30, 30, 35, 65} not valid? The average is 40, the two smallest numbers are identical and the average (arithmetic mean) of the two largest numbers is 50. All requirements are met.
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in questions like this...i try to plug in values right away...

x+y+w+z=160

1. x=y.
x and y can be 20, w can be 50, and z=70, or it can be x=y=20, w=30, z=90
2 outcomes, not sufficient.

2. y+w+z=150, or x=10. it can be x=10, y=20, w=30, z=100 or
x=10, y=50, w=50, z=50
so alone not sufficient.

1+2
x=10, y=10
we can have:
w=20, z=120
or
w=50, z=90

2 outcomes..the answer is E.

n=4; avg=40; sum=160

(1) The two smallest numbers are identical. insufic.
case 1: a,b=30; c,d=50; {30,30,50,50}; {a,b<40}
case 2: a,b,c,d=40; {40,40,40,40}; {0<40}

(2) The average (arithmetic mean) of the two largest numbers is 50. insufic.
case 1: a,b=30; c,d=50; {30,30,50,50}; {a,b<40}
case 2: a,b=30; c=30; d=70; {30,30,30,70}; {a,b,c<40}

(1 & 2) insufic.
case 1: a,b=30; c,d=50; {30,30,50,50}; {a,b<40}
case 2: a,b=30; c=30; d=70; {30,30,30,70}; {a,b,c<40}

Answer (E)

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