Question

Hiram's algebra notes are $50$ pages long and are printed on $25$ sheets of paper; the first sheet contains pages $1$ and $2,$ the second sheet contains pages $3$ and $4,$ and so on. One day he leaves his notes on the table before leaving for lunch, and his roommate decides to borrow some pages from the middle of the notes. When Hiram comes back, he discovers that his roommate has taken a consecutive set of sheets from the notes and that the average (mean) of the page numbers on all remaining sheets is exactly $19.$ How many sheets were borrowed?

$10$

$13$

$15$

$17$

$20$

Video solution:

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Written solution:

Suppose the borrowed sheets are sheets $a$ through $b$ . Then the borrowed pages run from $2a-1$ through $2b$ , so there are $2(b-a+1)$ borrowed pages.

The total sum of all page numbers is $50\cdot51/2=1275$ . If the remaining pages have mean $19$ , then

$1275-\frac{(2a-1+2b)\cdot 2(b-a+1)}{2}=19\bigl(50-2(b-a+1)\bigr).$

Simplifying gives

$(2a+2b-39)(b-a+1)=325=25\cdot13.$

The positive solution with consecutive sheets in the middle is

$2a+2b-39=25,\qquad b-a+1=13.$

Thus $a+b=32$ and $b-a=12$ , so $a=10$ and $b=22$ . Therefore $13$ sheets were borrowed.

Thus, B is the correct answer.

2021 AMC 10A Spring Problem 22

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Problem 22 in Other Years