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805+ (Hard)|   Arithmetic|      
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devmillenium2k
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The average of five natural numbers is 150. A particular number among 13 Nov 2014, 21:04
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26% (02:46) correct 74% (02:58) wrong based on 117 sessions

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The average of five natural numbers is 150. A particular number among the five exceeds another by 100. The rest three numbers lie between these two numbers and they are equal. How many different values can the largest number among the five take?

(A) 59
(B) 19
(C) 21
(D) 18
(E) 42
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B
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KarishmaB
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The average of five natural numbers is 150. A particular number among 13 Nov 2014, 21:52
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devmillenium2k
The average of five natural numbers is 150. A particular number among the five exceeds another by 100. The rest three numbers lie between these two numbers and they are equal. How many different values can the largest number among the five take?

(A) 59
(B) 19
(C) 21
(D) 18
(E) 42
Show more

This is not a GMAT question. GMAT doesn't use the terminology of natural numbers.

Let the 5 natural numbers in increasing order be: a, b, b, b, a+100
Note that a < b < a+100

\(\frac{(a+b+b+b+a+100)}{5} = 150\)
\(2a + 3b = 650\)

Since a < b, let's find the point where a = b
2a + 3a = 650
a = 130 = b
But b must be greater than a. If we increase b by 1, we need to decrease a by 3 to keep the average same. But that decreases the largest number too, so increase b by another 1.
So a = 127, b = 132 give us the numbers as 127, 132, 132, 132, 227

Since b < a+100, let's find the point where b = a+100
2a + 3(a+100) = 650
a = 70, b = 170
But b must be less than a+100. If we decrease b by 1, we need to increase a by 3 to keep the average same. But that increases the largest number too, so decrease b by another 1.
So a = 73, b = 168 give us the numbers as 73, 168, 168, 168, 173

Values of a will be: 73, 76, 79, ....127 (Difference of 3 to make b a natural number)
This is an AP.
Last term = First term + (n - 1)*Common difference
127 = 73 + (n - 1)*3
n = 19
So the last term (a+100) will also take 19 distinct values.

Answer (B)

*Edited
General Discussion
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The average of five natural numbers is 150. A particular number among 14 Nov 2014, 00:17
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Hi Karishma,

Correct me if I'm wrong, 73 + 169 + 169 + 169 + 173 ≠ 750 :? :?
Also 127 + 131 + 131 + 131 + 227 ≠ 750 :?: :?:

≠ is not-equal

Does anyone even check these solutions :cry: :cry:
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The average of five natural numbers is 150. A particular number among 14 Nov 2014, 01:18
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spencer1999
Hi Karishma,

Correct me if I'm wrong, 73 + 169 + 169 + 169 + 173 ≠ 750 :? :?
Also 127 + 131 + 131 + 131 + 227 ≠ 750 :?: :?:

≠ is not-equal

Does anyone even check these solutions :cry: :cry:
Show more

Note that you are not bothered about the value of b above as long as it is in the range. You can adjust it according to a and a+100 to bring the total to 750. All we care about is the possible number of values of a (which will be the same as the possible number of values of b).
I have edited the post above to show the desired adjustment to the value of b to balance out the addition.
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The average of five natural numbers is 150. A particular number among 14 Nov 2014, 01:27
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devmillenium2k
The average of five natural numbers is 150. A particular number among the five exceeds another by 100. The rest three numbers lie between these two numbers and they are equal. How many different values can the largest number among the five take?

(A) 59
(B) 19
(C) 21
(D) 18
(E) 42
Show more

Tricky question! Not sure if I will get this under 2 mins but here it goes.
We basically have 5 positive numbers - x, y, y, y, x+100 with sum = 750
so 2x + 3y = 650 and x<y<x+100.

Take extreme cases of inequality and we get for x=y case, 5x=650 so x = 130.
Another case, y=x+100, 5x+300 = 650. x=70. BOTH THESE CASES are not allowed since y lies between x and x+100. But these are the bounds.

Once we have that, it is only a matter of finding the allowed values of x. (650-2x) should be divisible by 3, so changes allowed in x should be only in factors of 3. Start writing numbers from 73, 76....upto 127 and we get 19!
Answer is B, but I didnt get this in under 2 minutes for sure!

Let me know if there is a faster way to solve. :D
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The average of five natural numbers is 150. A particular number among 10 Jul 2016, 05:45
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VeritasPrepKarishma
devmillenium2k
The average of five natural numbers is 150. A particular number among the five exceeds another by 100. The rest three numbers lie between these two numbers and they are equal. How many different values can the largest number among the five take?

(A) 59
(B) 19
(C) 21
(D) 18
(E) 42

This is not a GMAT question. GMAT doesn't use the terminology of natural numbers.

Let the 5 natural numbers in increasing order be: a, b, b, b, a+100
Note that a < b < a+100

\(\frac{(a+b+b+b+a+100)}{5} = 150\)
\(2a + 3b = 650\)

Since a < b, let's find the point where a = b
2a + 3a = 650
a = 130 = b
But b must be greater than a. If we increase b by 1, we need to decrease a by 3 to keep the average same. But that decreases the largest number too, so increase b by another 1.
So a = 127, b = 132 give us the numbers as 127, 132, 132, 132, 227

Since b < a+100, let's find the point where b = a+100
2a + 3(a+100) = 650
a = 70, b = 170
But b must be less than a+100. If we decrease b by 1, we need to increase a by 3 to keep the average same. But that increases the largest number too, so decrease b by another 1.
So a = 73, b = 168 give us the numbers as 73, 168, 168, 168, 173

Values of a will be: 73, 76, 79, ....127 (Difference of 3 to make b a natural number)
This is an AP.
Last term = First term + (n - 1)*Common difference
127 = 73 + (n - 1)*3
n = 19
So the last term (a+100) will also take 19 distinct values.

Answer (B)

*Edited
Show more

Quote:

Hi Karishma,
Do we have any shortcut method?
The entire method seems to be time-cosuming.

Please provide the logic behind the following:
“But that decreases the largest number too, so increase b by another 1.”
Thanks again!

Show more
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The average of five natural numbers is 150. A particular number among 11 Jul 2016, 22:53
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yosita18
VeritasPrepKarishma
devmillenium2k
The average of five natural numbers is 150. A particular number among the five exceeds another by 100. The rest three numbers lie between these two numbers and they are equal. How many different values can the largest number among the five take?

(A) 59
(B) 19
(C) 21
(D) 18
(E) 42

This is not a GMAT question. GMAT doesn't use the terminology of natural numbers.

Let the 5 natural numbers in increasing order be: a, b, b, b, a+100
Note that a < b < a+100

\(\frac{(a+b+b+b+a+100)}{5} = 150\)
\(2a + 3b = 650\)

Since a < b, let's find the point where a = b
2a + 3a = 650
a = 130 = b
But b must be greater than a. If we increase b by 1, we need to decrease a by 3 to keep the average same. But that decreases the largest number too, so increase b by another 1.
So a = 127, b = 132 give us the numbers as 127, 132, 132, 132, 227

Since b < a+100, let's find the point where b = a+100
2a + 3(a+100) = 650
a = 70, b = 170
But b must be less than a+100. If we decrease b by 1, we need to increase a by 3 to keep the average same. But that increases the largest number too, so decrease b by another 1.
So a = 73, b = 168 give us the numbers as 73, 168, 168, 168, 173

Values of a will be: 73, 76, 79, ....127 (Difference of 3 to make b a natural number)
This is an AP.
Last term = First term + (n - 1)*Common difference
127 = 73 + (n - 1)*3
n = 19
So the last term (a+100) will also take 19 distinct values.

Answer (B)

*Edited

Quote:

Hi Karishma,
Do we have any shortcut method?
The entire method seems to be time-cosuming.

Please provide the logic behind the following:
“But that decreases the largest number too, so increase b by another 1.”
Thanks again!

Show more

This comes from the concept of integer solutions for equations.

Check it here: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/06 ... -of-thumb/

You have 2a + 3b = 650
One solution: a = 70, b = 170
Next integer solution for this equation will be a = 73, b = 168
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The average of five natural numbers is 150. A particular number among 08 Nov 2019, 05:33
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devmillenium2k
The average of five natural numbers is 150. A particular number among the five exceeds another by 100. The rest three numbers lie between these two numbers and they are equal. How many different values can the largest number among the five take?

(A) 59
(B) 19
(C) 21
(D) 18
(E) 42
Show more

Natural numbers are positive integers;
n=5; mean=150; sum=750; set={a,b,b,b,tn};
tn-a=100, tn=100+a;
a<b<tn.

a+3b+tn=750…a+3b+(100+a)=750…2a+3b=650
2a+3b=650…2a=650-3b…a=(650-3b)/2
a=(650-3b)/2=integer;
if 3b=odd: E-O=O…O/E≠integer; so, 3b=even…b=even;

a<b<tn: (650-3b)/2<b<(650-3b)/2+100
(650-3b)/2<b…650<2b+3b…130<b;
b<(650-3b)/2+100…2b<850-3b…b<170;
a<b<tn: 130<b<170

The different values that (a,tn) can take is the number of evens in b's range, since b=even;
So find then evens between 130 and 170:
num multiples between range = largest m(2) - smallest m(2) / 2 + 1
num multiples between range = 168 - 132 / 2 + 1 = 18 + 1 = 19

Ans (B)
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The average of five natural numbers is 150. A particular number among 01 Jun 2024, 16:47
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KarishmaB

devmillenium2k
The average of five natural numbers is 150. A particular number among the five exceeds another by 100. The rest three numbers lie between these two numbers and they are equal. How many different values can the largest number among the five take?

(A) 59
(B) 19
(C) 21
(D) 18
(E) 42
This is not a GMAT question. GMAT doesn't use the terminology of natural numbers.

Let the 5 natural numbers in increasing order be: a, b, b, b, a+100
Note that a < b < a+100

\(\frac{(a+b+b+b+a+100)}{5} = 150\)
\(2a + 3b = 650\)

Since a < b, let's find the point where a = b
2a + 3a = 650
a = 130 = b
But b must be greater than a. If we increase b by 1, we need to decrease a by 3 to keep the average same. But that decreases the largest number too, so increase b by another 1.
So a = 127, b = 132 give us the numbers as 127, 132, 132, 132, 227

Since b < a+100, let's find the point where b = a+100
2a + 3(a+100) = 650
a = 70, b = 170
But b must be less than a+100. If we decrease b by 1, we need to increase a by 3 to keep the average same. But that increases the largest number too, so decrease b by another 1.
So a = 73, b = 168 give us the numbers as 73, 168, 168, 168, 173

Values of a will be: 73, 76, 79, ....127 (Difference of 3 to make b a natural number)
This is an AP.
Last term = First term + (n - 1)*Common difference
127 = 73 + (n - 1)*3
n = 19
So the last term (a+100) will also take 19 distinct values.

Answer (B)

*Edited
Show more
­hey I understood the explanation. however i didnt understand  "Values of a will be: 73, 76, 79, ....127 (Difference of 3 to make b a natural number)" why difference of 3 has been taken, like how did you decide that
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The average of five natural numbers is 150. A particular number among Updated on: 03 Jun 2024, 07:03
Originally posted by samarpan.g28 on 02 Jun 2024, 09:51.
Last edited by samarpan.g28 on 03 Jun 2024, 07:03, edited 1 time in total.
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033

KarishmaB

devmillenium2k
The average of five natural numbers is 150. A particular number among the five exceeds another by 100. The rest three numbers lie between these two numbers and they are equal. How many different values can the largest number among the five take?

(A) 59
(B) 19
(C) 21
(D) 18
(E) 42
This is not a GMAT question. GMAT doesn't use the terminology of natural numbers.

Let the 5 natural numbers in increasing order be: a, b, b, b, a+100
Note that a < b < a+100

\(\frac{(a+b+b+b+a+100)}{5} = 150\)
\(2a + 3b = 650\)

Since a < b, let's find the point where a = b
2a + 3a = 650
a = 130 = b
But b must be greater than a. If we increase b by 1, we need to decrease a by 3 to keep the average same. But that decreases the largest number too, so increase b by another 1.
So a = 127, b = 132 give us the numbers as 127, 132, 132, 132, 227

Since b < a+100, let's find the point where b = a+100
2a + 3(a+100) = 650
a = 70, b = 170
But b must be less than a+100. If we decrease b by 1, we need to increase a by 3 to keep the average same. But that increases the largest number too, so decrease b by another 1.
So a = 73, b = 168 give us the numbers as 73, 168, 168, 168, 173

Values of a will be: 73, 76, 79, ....127 (Difference of 3 to make b a natural number)
This is an AP.
Last term = First term + (n - 1)*Common difference
127 = 73 + (n - 1)*3
n = 19
So the last term (a+100) will also take 19 distinct values.

Answer (B)

*Edited
­hey I understood the explanation. however i didnt understand  "Values of a will be: 73, 76, 79, ....127 (Difference of 3 to make b a natural number)" why difference of 3 has been taken, like how did you decide that
Show more
­Hi 033, I am not Karishma but felt like helping you :)
You find the values of a concerning b, or vice versa. We know, 2a+3b=350 or, (650-3b)/2=a. Now, if you start for b=130, a will also be 130. But, b<a. Moreover, a<b<e where e=(a+100).

1. Let us now increase the value of b by 2 i.e. 132 (because the denominator is 2) then a becomes 127. This is our first set of (a,b)=(127,132).
2. If b=134, then a is 124. Therefore, (a,b)=(124,134).
3. Let's now jump to b=140, then a is 115.
4. Let's make a big jump to b=160, a will be 85.
5. Let's try for b=170, here a will be 70. Then 'e' will be 170. But we know b<e.
6. If you take b=168, then a=73. We must stop here i.e. at (a,b)=(73,168).

Did you see the pattern in the values of 'a' in lines 3 and 4? The value of a decreases from 127 to 124 i.e. by 3. Finally, it settles at 73.
Here apply AP. 127=73+(n-1)*3 or n=19.
We could have used values of b as well. What would be the series then? 132,134,....168 i.e. increase by 2. 168=132+(n-1)*2 or, n=19.
Option (B) is correct.
 ­
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The average of five natural numbers is 150. A particular number among 03 Jun 2024, 05:33
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Call the least number X, so the greatest number is:

X+100

Call the middle three equal numbers Y. So the statement becomes:

X+3Y+X+100 = 5*150 = 750, or:


2X+3Y = 650


The conditions are:

X<Y<X+100

If X and Y were equal then both X and Y would equal:

5X=5Y=650 and X=Y=130

So we know X<130 and Y>130

We can see that Y must be even because both 650 and 2X are even.


Try Y=132. Solving for X:

2X=650-3*132 = 254 and:

X=127. So X+100 = 227 and the difference between Y and X+100 is:

227-132 = 95


Try Y=134. Using above math:

X=124 and X+100 = 224 and the difference between Y and X+100 is:

90

We can see that the difference between Y and X+100 diminishes by 5 for every 2 point increase in Y.

Since the difference starts at 95 and can't be permitted to go to 0, it must stop at 5:

(95-5)/5 = 18. Add 1 because the position at 95 isn't counted:

18+1 = 19

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