A set consists of 5 distinct positive integers a, b, c, d, e, ........

given
b is least integer and d is highest no
and sum of a,b,c= 24
mean of b+24+d/5 =8.8
b+d=20
since d is the highest no it can be 19 and b is least can be 1
so 19-1= 18
IMO C

Hi Archit3110,

Please note that a, b, c, d and e are positive integers

Regards,
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EgmatQuantExpert

there is a confusion regarding usage of 0 ;
what i know is that 0 is neither +ve or -ve integer

if a statement says n is a set of +ve integers then we begin set n from (1,2,3...)

and if statement says n is a set of non negative integer then we begin set from ( 0,1,2,3,..)

will 0 be considered in set of +ve even integers? i.e (0,2,4,6,.) or not?

EgmatQuantExpert
in that case b =1 and d = 19
and d-b = 18
IMO C

Yes, 0 is neither a positive integer nor a negative integer.

It will be included in a set of non-negative and non-positive integers but it must be excluded from the set of positive or negative integers

Though, 0 is an even integer, it cannot be considered as a positive even integer. Since, 0 is not a positive integer

Regards,
_________________

EgmatQuantExpert

thanks

Solution

Given:

• A set of 5 distinct positive integers, {a, b, c, d, e}
• b is the least number and d is the highest number of the set
• a + c + e = 24
• Mean of the set = 8.8

To find:

• The maximum value of (d – b)

Approach and Working:

• a + c + e = 24 ………………………….………………………………… (1)
• \(\frac{(a + b + c + d + e)}{5} = 8.8\)

o Implies, a + b + c + d + e = 5 * 8.8 = 44 …………… (2)

(2) – (1), we get,

• b + d = 44 – 24 = 20
• For d – b to be maximum, b must be minimum

o The minimum value b can take is 1, since b is a positive integer

• If b = 1, then d = 20 – 1 = 19

Therefore, the maximum value of (d – b) = 19 – 1 = 18

Hence, the correct answer is option C.

Answer: C


_________________

The mean of all 5 numbers is 8.8
(a + b + c + d + e)/5 = 8.8
Multiply both sides by 5 to get: a + b + c + d + e = 44

The sum of a, c and e is 24
(a + c + e) = 24
So, take a + b + c + d + e = 44 and rewrite as: (a + c + e) + b + d = 44
We get: (24) + b + d = 44
So, b + d = 20

b is the least number and d is the highest number
Since b and d are POSITIVE INTEGERS, and since we're tying to MAXIMIZE the value of (d - b), we want to MAXIMIZE d and MINIMIZE b
This is accomplished when d = 19 and b = 1

What is the maximum value of (d – b)?
d - b = 19 - 1 = 18

Answer: C

Cheers,
Brent

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We know that a + c + e = 24. Because the mean of the 5 numbers is 8.8, we see that their sum must be 5 x 8.8 = 44. Thus, a + b + c + d + e = 44.

Subtracting the first equation from the second equation, we obtain:
b + d = 44 - 24 = 20.

To maximize the difference of b and d, we will choose b to be as small as possible and d to be as large as possible. Thus, if b = 1 then d = 19, and the maximum value of (d - b) is 18.

Answer: C
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