12 Days of Christmas GMAT Competition - Day 5: A group of students

Question

12 Days of Christmas 2024 - 2025 Competition with $40,000 of Prizes

A group of students, with an average (arithmetic mean) weight of p pounds, is boarding a plane. The plane has space for only one more person: either Sally or Harry. If Sally, who weighs 105 pounds, joins the group, the average weight decreases by 6 pounds. If Harry, who weighs 195 pounds, joins instead, the average weight increases by 12 pounds. What is the value of p?

A. 129
B. 135
C. 141
D. 147
E. 153

Bunuel · Accepted Answer

GMAT Club's Official Explanation:

Assuming the initial group had n students, the total weight of these n students would be pn pounds.

If Sally joins, the total would become both pn + 105 and (p - 6)(n + 1). Thus, we have pn + 105 = (p - 6)(n + 1).
If Harry joins, the total would become both pn + 195 and (p + 12)(n + 1). Thus, we have pn + 195 = (p + 12)(n + 1).

Subtracting the first equation from the second:

90 = (p + 12)(n + 1) - (p - 6)(n + 1) 
90 = (n + 1)(p + 12 - p + 6)
90 = (n + 1)(18)
n + 1 = 5
n = 4

Substituting n = 4 into pn + 105 = (p - 6)(n + 1) gives p = 135.

Answer: B.

siddhantvarma · Answer

Let's say there are n students in the original group with an average weight p pounds
Sum of their weights = np

If Sally (105 lbs) joins:
New sum = np + 105
New average = (np + 105)/(n + 1) = p - 6
(np + 105)/(n + 1) = p - 6

If Harry (195 lbs) joins:
 New sum = np + 195
 New average = (np + 195)/(n + 1) = p + 12
(np + 195)/(n + 1) = p + 12

From Sally's equation:
np + 105 = (p - 6)(n + 1)
np + 105 = pn + p - 6n - 6
105 = p - 6n - 6
6n = p - 111 ...(1)

From Harry's equation:
np + 195 = (p + 12)(n + 1)
np + 195 = pn + p + 12n + 12
195 = p + 12n + 12
p = 183 - 12n ...(2)

Substituting equation (2) into equation (1):
6n = (183 - 12n) - 111
6n = 72 - 12n
18n = 72
n = 4

Now we can find p by substituting n = 4 back into either equation
Using equation (2):
p = 183 - 12(4)
p = 183 - 48
p = 135

Therefore p = 135 pounds.

The answer is B. 135.

Kinshook · Answer

A group of students, with an average (arithmetic mean) weight of p pounds, is boarding a plane. The plane has space for only one more person: either Sally or Harry. If Sally, who weighs 105 pounds, joins the group, the average weight decreases by 6 pounds. If Harry, who weighs 195 pounds, joins instead, the average weight increases by 12 pounds.

What is the value of p?

Let the number of students be x.
Total weight of x students = xp pounds

If Sally, who weighs 105 pounds, joins the group, the average weight decreases by 6 pounds.
New total weight = xp + 105
New average weight = (xp+105)/(x+1) = p - 6
xp + 105 = (x+1)(p-6) = xp - 6 + p - 6x
p - 6x = 111 (1)

If Harry, who weighs 195 pounds, joins instead, the average weight increases by 12 pounds.
New total weight = xp + 195
New average weight = (xp+195)/(x+1) = p + 12
xp + 195 = (x+1)(p+12) = xp + 12 + p + 12x
p + 12x = 183 (2)

(2) - (1)
18x = 72; x = 4
p - 6x = p - 24 = 111
p = 111 + 24 = 135

IMO B

ManifestDreamMBA · Answer

Very nice question, uses the concepts of extra/deficit in averages.

So, imagine if the avg money owned by 10 friends is $10, which means they have a total of $100. Amy joins the group with $21, then the new avg becomes (100+21)/(10+1) = 121/11=$11. now let's understand how this works.

Amy has an excess of $11 as compared to others, so if she generously distributes her $11, then everyone will have $11 (Extra money/total friends), which means the average is $11 now.

The reverse logic of deficit works if the avg goes down, everyone has to chip in to fill the deficit and everyone's share goes down.

Back to question.
Let's assume there are n students with average weight of p. Sally's weight is less (p>105), so everyone chips in and hence she takes away 6 from avg in a way. 
p - 105 = (n+1) 6 --- (1)
Now, if Harry joins, avg increases, means he generously shares with everyone.
195 - p = (n+1)12 --- (2)

Solving the 2 equations, gives p = 135

hr1212 · Answer

Let k be the number of students, then,

Total weight = k * p

(kp + 105)/(k + 1) = p - 6
kp + 105 = kp - 6k + p - 6
p - 6k = 111 --- (1)

(kp + 195)/(k + 1) = p + 12
kp + 195 = kp + 12k + p + 12
p + 12k = 183 --- (2)

Subtracting (1) from (2)

18k = 72
k = 4

Substituting value in (1)

p - 24 = 111
p = 135

Answer: B

Krunaal · Answer

Let S = the current no. of students in the group
T = Total weight of all S students
p = average weight of S students

T/S = p or pS = T

When Sally joins,

T + 105 / S + 1 = p - 6
T + 105 = pS + p - 6S - 6
105 = p - 6S - 6
p = 105 + 6S + 6 ...................(1)

When Harry joins,

T + 195 / S + 1 = p + 12
T + 195 = pS + 12S + p + 12
195 = p + 12S + 12
p = 195 - 12S - 12 ...................(2)

Solving eq (1) and (2), we get  p = 135

Answer B.

A_Nishith · Answer

Step 1 : Express the new averages when Sally or Harry joins.
When Sally (105 pounds) joins, the new average decreases by 6:

Total weight of group+105 =p−6.
Multiplying through by n+1:
Total weight of group+105=(n+1)(p−6).
When Harry (195 pounds) joins, the new average increases by 12:

Total weight of group+195 = p+12.
Multiplying through by n+1 
Total weight of group+195=(n+1)(p+12).

Step 2:  Express the total weight of the group in terms of n and p.
The total weight of the group is np. 
Substituting Total weight of group= np into both equations:

From Sally joining: np+105=(n+1)(p−6)

Expanding and rearranging: np+105=np−6n+p−6 ⇒ 105=−6n+p−6.

Simplify: p−6n=111 ------eq(1)

From Harry joining: np+195=(n+1)(p+12).

Expanding and rearranging: np+195=np+12n+p+12 ⇒ 195=12n+p+12.

Simplify:p+12n=183 ----- eq(2)

Step 3:  Solve the system of equations.
From equation (1): p=6n+111.

Substitute p=6n+111 into equation (2): 6n+111+12n=183.

Combine terms: 18n+111=183.

Solve for n: 18n=72⇒n=4.

Substitute n=4 into p=6n+111:

p=6(4)+111 = 135

Final Answer:
The value of  p is 135

Navya442001 · Answer

Let  n  be the number of students in the group. The total weight of the students is  np .

If Sally joins the group, the new average weight is p - 6. The total weight is now np + 105, and the number of people is n + 1. So, we have the equation:
(np + 105) / (n + 1) = p - 6
np + 105 = (p - 6)(n + 1) np + 105 = pn + p - 6n - 6 105 = p - 6n - 6 p - 6n = 111 --- (1)

If Harry joins the group, the new average weight is p + 12. The total weight is now np + 195, and the number of people is n + 1. So, we have the equation:
(np + 195) / (n + 1) = p + 12
np + 195 = (p + 12)(n + 1) np + 195 = pn + p + 12n + 12 195 = p + 12n + 12 p + 12n = 183 --- (2)

Now we have a system of two linear equations:
p - 6n = 111 --- (1) p + 12n = 183 --- (2)

Subtract equation (1) from equation (2):
(p + 12n) - (p - 6n) = 183 - 111 18n = 72 n = 4

Substitute n = 4 into equation (1):
p - 6(4) = 111 p - 24 = 111 p = 135
Therefore, the average weight of the students is 135 pounds.

Karanjotsingh · Answer

Solve this step by step:

Let's say there are n students in the original group
 
 Original total weight = n × p  
 
 If Sally joins:
 
 New average = (original total + 105)/(n + 1) = p - 6 
 (n × p + 105)/(n + 1) = p - 6  
 
 If Harry joins:
 
 New average = (original total + 195)/(n + 1) = p + 12 
 (n × p + 195)/(n + 1) = p + 12  
 
 From Sally's equation:
 
 n × p + 105 = (n + 1)(p - 6) 
 n × p + 105 = n × p - 6n + p - 6 
 105 = -6n + p - 6 
 111 = -6n + p  
 
 From Harry's equation:
 
 n × p + 195 = (n + 1)(p + 12) 
 n × p + 195 = n × p + 12n + p + 12 
 195 = 12n + p + 12 
 183 = 12n + p  
 
 Now we have two equations:
 
 111 = -6n + p 
 183 = 12n + p  
 
 Subtracting them:
 
 -72 = -18n 
 n = 4  
 
 Plug n = 4 back:
 
 111 = -6(4) + p 
 111 = -24 + p 
 p = 135  
  
Therefore, the average weight is 135 pounds.

The answer is B.

Mardee · Answer

Number of students in the group = n
Total weight of the group = X
The average weight of the group = p

=> X = n.p
Now,
Sally, who weighs 105 pounds, joins the group, the new average weight decreases by 6 pounds
=> (X + 105)/n+1 = p - 6
Harry, who weighs 195 pounds, joins the group, the new average weight increases by 12 pounds
=> (X+195)/ n+1 = p + 12

X + 105 = (p - 6)* (n+1) Now substitute X = np and simplify we get,
=> p = 6n + 111

Now, 
(X+195) = (p + 12) * (n+1)
Now here too substitute X = np and simplify we get,
p = 183 - 12n

Now simplify the 2 equations
p = 6n + 111
p = 183 - 12n

=> n = 4
p = 6n + 111 = 135

Solution B. 135

Lizaza · Answer

Let's say there're n students of total weight w. Then:  \frac{w}{n}=p  
With Sally:  \frac{w+105}{n+1}=p-6 
With Harry:  \frac{w+195}{n+1}=p+12

p= \frac{w+105}{n+1} + 6 
then  \frac{w+195}{n+1} = \frac{w+105}{n+1} + 6 + 12

Here we can iterate:  \frac{w+195}{n+1} = \frac{w+105}{n+1} + \frac{90}{n+1} 
So,  \frac{w+105}{n+1} +\frac{ 90}{n+1} = \frac{w+105}{n+1} + 18 
And  \frac{90}{n+1}=18 
Then,  n+1=5  and  n=4

From this, we can easily get  w=pn=4p 
Then,  \frac{4p+105}{5}=p-6 
And  4p+105=5p-30 
Therefore,  p=135 , and   the answer is B) 135.

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