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08 Jan 2025, 02:42
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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
A. 129
B. 135
C. 141
D. 147
E. 153
08 Jan 2025, 02:42
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Official Solution:
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
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
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
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
17 Feb 2025, 22:15
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pn + 105 = (p-6)*(n+1)
105 = p - 6n -6
111 = p - 6n --- (1)
pn + 195 = (p+12)*(n+1)
195 = p + 12n + 12
183 = p + 12n --- (2)
From, 2*(1) + (2)
222 + 183 = 3p
405 = 3p
p = 135
Ans: B
105 = p - 6n -6
111 = p - 6n --- (1)
pn + 195 = (p+12)*(n+1)
195 = p + 12n + 12
183 = p + 12n --- (2)
From, 2*(1) + (2)
222 + 183 = 3p
405 = 3p
p = 135
Ans: B
