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Difference between revisions of "2021 AMC 10A Problems/Problem 2"

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===Solution 4.1 (Quick Inspection)===
 
===Solution 4.1 (Quick Inspection)===
 
The number of students in Portia's high school must be a multiple of <math>3.</math> This eliminates <math>\textbf{(B)}</math>, <math>\textbf{(D)}</math>, and <math>\textbf{(E)}</math>. Since <math>\textbf{(A)}</math> is too small (as <math>600+600/3<2600</math> is clearly true), we are left with <math>\boxed{\textbf{(C)} ~1950}.</math>
 
The number of students in Portia's high school must be a multiple of <math>3.</math> This eliminates <math>\textbf{(B)}</math>, <math>\textbf{(D)}</math>, and <math>\textbf{(E)}</math>. Since <math>\textbf{(A)}</math> is too small (as <math>600+600/3<2600</math> is clearly true), we are left with <math>\boxed{\textbf{(C)} ~1950}.</math>
 +
 +
~MRENTHUSIASM
 +
===Solution 4.1 (Plug in the Answer Choices)===
 +
For <math>\textbf{(A)},</math> we have <math>600+600/3=800\neq2600.</math> So, <math>\textbf{(A)}</math> is incorrect.
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 +
For <math>\textbf{(B)},</math> we have <math>650+650/3=866\frac{2}{3}\neq2600.</math> So, <math>\textbf{(B)}</math> is incorrect.
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For <math>\textbf{(C)},</math> we have <math>1950+1950/3=2600.</math> So, <math>\boxed{\textbf{(C)} ~1950}</math> is correct. For completeness, we will check choices <math>\textbf{(D)}</math> and <math>\textbf{(E)}.</math>
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For <math>\textbf{(D)},</math> we have <math>2000+2000/3=2666\frac{2}{3}\neq2600.</math> So, <math>\textbf{(D)}</math> is incorrect.
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For <math>\textbf{(E)},</math> we have <math>2050+2050/3=2733\frac{1}{3}\neq2600.</math> So, <math>\textbf{(E)}</math> is incorrect.
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 05:10, 14 February 2021

Problem 2

Portia's high school has $3$ times as many students as Lara's high school. The two high schools have a total of $2600$ students. How many students does Portia's high school have?

$\textbf{(A)} ~600 \qquad\textbf{(B)} ~650 \qquad\textbf{(C)} ~1950 \qquad\textbf{(D)} ~2000\qquad\textbf{(E)} ~2050$

Solution

The following system of equations can be formed with $p$ representing the number of students in Portia's high school and $l$ representing the number of students in Lara's high school. \[p=3q\] \[p+q=2600\] Substituting $p$ with $3q$ we get $4q=2600$. Solving for $q$, we get $q=650$. Since we need to find $p$ we multiply $650$ by 3 to get $p=1950$, which is $\boxed{\text{C}}$

-happykeeper

Solution 2 (One Variable)

Suppose Lara's high school has $x$ students. It follows that Portia's high school has $3x$ students. We know that $x+3x=2600,$ or $4x=2600.$ Our answer is \[3x=2600\left(\frac 34\right)=650(3)=\boxed{\textbf{(C)} ~1950}.\]

~MRENTHUSIASM

Solution 3 (Arithmetics)

Clearly, $2600$ students is $4$ times as many students as Lara's high school. Therefore, Lara's high school has $2600\div4=650$ students, and Portia's high school has $650\cdot3=\boxed{\textbf{(C)} ~1950}$ students.

~MRENTHUSIASM

Solution 4 (Answer Choices)

Solution 4.1 (Quick Inspection)

The number of students in Portia's high school must be a multiple of $3.$ This eliminates $\textbf{(B)}$, $\textbf{(D)}$, and $\textbf{(E)}$. Since $\textbf{(A)}$ is too small (as $600+600/3<2600$ is clearly true), we are left with $\boxed{\textbf{(C)} ~1950}.$

~MRENTHUSIASM

Solution 4.1 (Plug in the Answer Choices)

For $\textbf{(A)},$ we have $600+600/3=800\neq2600.$ So, $\textbf{(A)}$ is incorrect.

For $\textbf{(B)},$ we have $650+650/3=866\frac{2}{3}\neq2600.$ So, $\textbf{(B)}$ is incorrect.

For $\textbf{(C)},$ we have $1950+1950/3=2600.$ So, $\boxed{\textbf{(C)} ~1950}$ is correct. For completeness, we will check choices $\textbf{(D)}$ and $\textbf{(E)}.$

For $\textbf{(D)},$ we have $2000+2000/3=2666\frac{2}{3}\neq2600.$ So, $\textbf{(D)}$ is incorrect.

For $\textbf{(E)},$ we have $2050+2050/3=2733\frac{1}{3}\neq2600.$ So, $\textbf{(E)}$ is incorrect.

~MRENTHUSIASM

Video Solution #1(Setting Variables)

https://youtu.be/qNf6SiIpIsk?t=119 ~ThePuzzlr

Video Solution

https://youtu.be/xXx0iP1tn8k

- pi_is_3.14

See Also

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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