Michael's Numbers

We know that if we arrange $n$ numbers in increasing order then their median would be $(\frac{n}{2})^{th}$ term if $n$ is odd.

Given that $A<B<C<D<E$ , $C$ would be their median, so $C = 10$

Now, $\frac{A+B+C+D+E}{5} = 15 \Rightarrow A+B+C+D+E = 75 \Rightarrow A+B+D+E = 65$

To maximize $D$ , we need to minimize $A,B,E$ . Since they're are distinct positive integers, we let $A=1, B= 2$ . So, we are left with $D+E=62$ . Since $D<E \Rightarrow D<\frac{62}{2} = 31$

Hence, maximum value of $D = \boxed{30}$

C is the median since it is the middle-most number. Therefore, C = 10.

Since mean = 15, (A + B + 10 + D + E)/5 = 15

Simplifying: A + B + D + E = 65

We want to maximise D, but E is greater than D. Hence, substitute E = D + 1 A + B + D + D + 1 = 65 A + B + 2D = 64

Use the smallest numbers possible for A and B (1 and 2 respectively). But, 1 and 2 = 3, which will give a fraction for an answer for D.

Hence, substitute 1 and 3 for A and B: 2D = 60 D = 30

Since $C$ is the 3rd largest number out of the five, it is the median, so $C=10$ . Also, from the value of the 5 numbers mean, we can calculate that $A+B+C+D+E = 5 \times 15 = 75$ , and thus, $A+B+D+E=75-10=65$ .

As we want to maximize $D$ , the values of $A$ , $B$ and $E$ should be minimized. Thus, $A$ and $B$ should take the smallest possible positive values, 1 and 2 (ie. $A=1$ and $B=2$ ). This leaves us with $D+E=65-1-2 = 62$ . Noting the restriction that $D<E$ , the largest value of $D$ is therefore $\boxed {30}$ .

mean = 15 median =10 let A<B<C<D<E by using formula .... median = digit/member =10/5 =2 mean = product * meAN =2*15=30

First of all the meanings. When taking about a median here it is the middle value in anything, if you don't know about median then look it up. The median was given as 10, there are 5 values A, B, C, D, E which are arranged in order therefore C is the median whose value is 10. Now the question asks for the MAXIMUM value of D. Therefore let A=1, B=2, C=10 (given) and the two left are D and E. The mean is 15. A+B+C+D+E=75 . D+E= 75-13 D+E=62 Now since D is less than E and we need the MAXIMUM value of D therefore divide 62 by 2 and you get 31. D=30 because D<E

For maximizing value of D , we need to give least possible values to A & B, i.e., 1 and 2.

And , it is given that median (C) = 10.

Now,

1 + 2 + 10 + D + E = 5 * 15

D + E = 62

Maximum value which can be the value of D is 30 only because we are having a constraint D < E (i.e., 32), E can't be even 31 that will make D = E.

So, D = 30 (Ans)

For D to be maximum, we must find the smallest value for A and B. Not for C, because the median is 10, so the value of C has been determined.

The smallest positive integer is 1, hence $A = 1$

$A < B$ , do the smallest possible B is $1 + 1$ , $B = 2$

C is the median so, $C = 10$

Now, the mean is 5 and there are 5 numbers, the sum is $5 * 15 = 75$ $75 = A + B + C + D + E$

By the known variable above, we get $D + E = 62$

$D < E$ so,the maximum value of D $D = \frac {62}{2} - 1 = 30$

Firstly we need to find the lowest positive integer of A which is 1.

Secondly, we know that we are trying to find the maximum number for D therefore B must be the next smallest positive integer which is 2.

Thirdly we know C is 10 since it states that the median is 10

Finally finding D and E: Since the mean is 15, that means the sum of all letters (A to E) divided by 5 must equal to 15

(1+2+10+D+E) divided by 5 = 15

Times 5 on both sides we get: 13+D+E = 75

Then take away 13 from both sides we get: D + E = 62

Therefore D = 30 since D must be smaller than E.

+ +10+ _ _ _ _ _ _ _ _ _ _ _ = 15 5 i just think of a less two consecutive numbers to be filled in the first and second blank such as 1 and 2 so that the the number on the fourth and fifth blank will higher .. think of a number when u add to 13 the sum is 75 because 75 divided 5 is 15 .... the answer is 62 ...again think 0f a two number whose sum is 62 . the answer is 30 and 32