Problem Solving

Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?

78
77 1/5
66 1/7
55 1/7
52

I have the solution and I will provide it later, what I am concerned with is the probability a problem like this one (involving this number of steps in the solution) to appear on the real test. I read somewhere that the maximum length is 3-4 steps to solve a problem. Did anybody encounter more complex problems on the real GMAT ?

replayyyy wrote:Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?

78
77 1/5
66 1/7
55 1/7
52

I have the solution and I will provide it later, what I am concerned with is the probability a problem like this one (involving this number of steps in the solution) to appear on the real test. I read somewhere that the maximum length is 3-4 steps to solve a problem. Did anybody encounter more complex problems on the real GMAT ?

largest possible range is 66 1/7

largest possible range is 66 1/7[/quote]

number line: x, N2, 55, N4, 3x+20

x+55 < N2+ N4

x + N2 + 55 + N4 + (3x+20) = 275 (or 55*5)
4x + 75 + N2 + N4 = 275
4x + N2 + N4 = 200, N2 + N4 = 200 - 4x
x+55 < 200-4x
55 < 200 - 5x, 11 < 40 - x

Range (R) = (3x+20) - x, R = 20+2x

Now plug in answers for R max.:
78
77 1/5
66 1/7
55 1/7
52

Start from C (although we are given definite values, the solution is indefinite, Kaplan score 800 tip would be to start from middle, i.e. C)

R = 66 1/7, 66 1/7 = 20 + 2x, x = (66 1/7 - 20)/2, x = 23.07
11 < 40 - x, 11 < 40 - 23.07

answers C satisfies a condition of the largest possible R

IMO 78

From the question stem we know the following:

Mean = Median = 55
No of numbers = 5
Sum of numbers = 275

1) L = 3S + 20
2) 2S + 110 + L = 275

Equation 1) is from the question stem "largest number in the set is equal to 20 more than three times the smallest number".

Equation 2) maximizes the range. We have 5 numbers in the set. The median is 55. We need the other 4 numbers. To maximizes the range and maintain a median of 55, set 2 of the numbers to the smallest value (S), another one to the median (55), and the final number will be the largest number (L).

Solving these two equations for S results in S=29. Substituting 29 into the first equation results in L=107. Therefore, the maximum range is 78.

I can't answer your question about the frequency of a question like this. However, this one didn't seem to take as many steps as some of the other ones I have encountered. Does your solution use a different approach? For that matter, is my solution even correct?

rkanthilal wrote:IMO 78

From the question stem we know the following:

Mean = Median = 55
No of numbers = 5
Sum of numbers = 275

1) L = 3S + 20
2) 2S + 110 + L = 275

Equation 1) is from the question stem "largest number in the set is equal to 20 more than three times the smallest number".

Equation 2) maximizes the range. We have 5 numbers in the set. The median is 55. We need the other 4 numbers. To maximizes the range and maintain a median of 55, set 2 of the numbers to the smallest value (S), another one to the median (55), and the final number will be the largest number (L).

Solving these two equations for S results in S=29. Substituting 29 into the first equation results in L=107. Therefore, the maximum range is 78.

I can't answer your question about the frequency of a question like this. However, this one didn't seem to take as many steps as some of the other ones I have encountered. Does your solution use a different approach? For that matter, is my solution even correct?

I don't know why I brought into this problem an extra condition that the numbers are different. If N2=x and N4=Median then answer A is correct

rkanthilal wrote:IMO 78

From the question stem we know the following:

Mean = Median = 55
No of numbers = 5
Sum of numbers = 275

1) L = 3S + 20
2) 2S + 110 + L = 275

Equation 1) is from the question stem "largest number in the set is equal to 20 more than three times the smallest number".

Equation 2) maximizes the range. We have 5 numbers in the set. The median is 55. We need the other 4 numbers. To maximizes the range and maintain a median of 55, set 2 of the numbers to the smallest value (S), another one to the median (55), and the final number will be the largest number (L).

Solving these two equations for S results in S=29. Substituting 29 into the first equation results in L=107. Therefore, the maximum range is 78.

I can't answer your question about the frequency of a question like this. However, this one didn't seem to take as many steps as some of the other ones I have encountered. Does your solution use a different approach? For that matter, is my solution even correct?

You are correct, it is A. May be you are dealing better with algebra than me. Here is the official explanation:

Whenever we're given partial information about the elements contained in a set, it's usually best to use an ordered set to try to organize the information we have and figure out what work we need to do to be able to answer the question. In this case, we know that the set contains 5 elements and that the median is 55. Therefore we know 55 is the middle number in the set:

{a, b, 55, c, d}

Where a is the smallest item in the set, b is the second smallest, c is the second largest, and d is the largest.

Furthermore, we know that the largest item is equal to 20 more than 3 times the smallest number:

{a, b, 55, c, 3a + 20}

We also know that the average number is equal to 55, so the sum of the 5 elements must be 55×5=275:

Sum of {a, b, 55, c, 3a + 20} = 275

The question asks us what the largest possible value for the range of the set is. The range is defined as the largest item minus the smallest item; in this case, the range is (3a + 20) - a = 2a + 20.

How do we maximize this range? Quite simply, we have to maximize a, because the range grows as a grows. This also entails maximizing the largest number (3a + 20). Thus we have to maximize the bolded elements here:

Sum of {a, b, 55, c, 3a + 20} = 275

How do we maximize these numbers? Since we know that a + b + 55 + c + (3a + 20) = 275, we have to MINIMIZE b and c in order to maximize a and 3a + 20. In order words, we should minimize the underlined numbers here:

Sum of {a, b, 55, c, 3a + 20} = 275

Because we organized the set as an ordered set, each number must be greater than or equal to the number to its left, and less than or equal to the number to its right. Thus, to make b as small as possible, set it equal to a:

Sum of {a, a, 55, c, 3a + 20} = 275

Similarly, to make c as small as possible, set it equal to 55:

Sum of {a, a, 55, 55, 3a + 20} = 275

Because we have narrowed it down to one variable, we can solve for a.

Sum of {a, a, 55, 55, 3a + 20} = 275
a + a + 55 + 55 + (3a +20) = 275
2a + 110 + 3a + 20 = 275
5a + 130 = 275
5a = 145
a = 29

Thus, the smallest number in the set is 29, and the largest number is 3a + 20 = 107. The range of the set is 2a + 20 = 78.

The correct answer is A.

replayyyy wrote:Set R contains five numbers that have an average value of 55. If the median of the set is equal to the mean, and the largest number in the set is equal to 20 more than three times the smallest number, what is the largest possible range for the numbers in the set?

78
77 1/5
66 1/7
55 1/7
52

A:78

Let x be the smallest number then largest is 3x+20

x,y,55,z,3x+20

we have mean=55
55X5=275
then
x+y+55+z+3x+20=275
4x=195-z-y

Range is depended on x....so smaller the value of z and y,higher the value of x and higher the range

so y is minimum when it is equal to x as 55 is median (it cannot be less than x-1 and smallest value will change....and we dont take x+1 as we are searching for least possible value,so that mean=median=55)

z is minimum when it is equal to 55 (it cannot be less than 55....and we dont take 56 as we are searching for least possible value so that mean=median=55)

so with that
x,x,55,55,3x+20
we have
5x+130=275
5x=145
x=29

so numbers are
29,29,55,55,29x3+20
(if you change any of the value(add max value,subtract from least value...relation will break)

max range=107-29=78
A

frank1

x+y+55+z+3x+20=275
4x=195-z-y

is not correct calculation on both sets of equation

x+y+55+z+3x+20=275 >>>>> x+3x=275-55-20-y-z>>>>>>4x=200-y-z