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Floor and Ceiling: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Advanced (beyond the core course) Advanced. This problem set goes beyond core Algebra I. You can skip it.

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Problem 1 of 10

Problem 1 The partial record

A real number has floor $-4$ and fractional part $0.35$ . Find the number.
1. Hint 1
  The fractional part measures the distance above the floor.
2. Hint 2
  Add the recorded floor and fractional part.
Answer

$-3.65$ .

Full solution

By definition,
$x-\lfloor x\rfloor=0.35.$
Substituting the floor gives
$x=-4+0.35$
Hence $x=-3.65$ .

This lies between $-4$ and $-3$ , so its floor is indeed $-4$ , and subtracting that floor leaves $0.35$ .

Answer

$-3.65$ .

Key idea

A number is reconstructed by adding its floor and its fractional part.
Problem 1: Mark complete
Problem 2 The average reading

Find the floor of the average of $-6.8$ and $-4.4$ .
1. Hint 1
  The average must be calculated before the integer rounding.
2. Hint 2
  Locate the average between neighboring integers, then choose the lower integer.
Answer

$-6$ .

Full solution

The average is
$\frac{(-6.8)+(-4.4)}{2}=-5.6$
It lies strictly between $-6$ and $-5$ .

The greatest integer at or below it is therefore
$\lfloor-5.6\rfloor=-6$

Answer

$-6$ .

Key idea

A floor is taken after the input is fully computed, and it returns the greatest integer at or below that input, which for a negative noninteger lies further from zero.
Problem 2: Mark complete
Problem 3 The upward adjustment

How much must be added to $-2.6$ to reach its ceiling?
1. Hint 1
  First locate the least integer at or above the given number.
2. Hint 2
  The needed increase is that integer minus the original number.
Answer

$0.6$ .

Full solution

The ceiling is $-2$ , since
$-3<-2.6<-2.$
The required increase is
$-2-(-2.6)=0.6$

Adding $0.6$ returns exactly $-2$ , checking both the direction and amount.

Answer

$0.6$ .

Key idea

The increase to a ceiling is the ceiling minus the original input.
Problem 3: Mark complete
Problem 4 The plotting window

Sketch $y=\lfloor2x\rfloor$ for $-1\le x<1.5$ on the axes in the figure. State the input interval for each height that appears and mark all open and closed endpoints.

Blank axes for the sketch, with $x$ from $-1.5$ to $2$ and $y$ from $-3$ to $3$ .

Text description of this figure

Blank coordinate axes drawn on a light grid, with nothing plotted on them. The horizontal x-axis carries a tick mark, a gridline and a number label at negative 1.5, negative 1, negative 0.5, 0.5, 1, 1.5 and 2, that is at every half unit across that range. The vertical y-axis carries a tick mark, a gridline and a number label at negative 3, negative 2, negative 1, 1, 2 and 3, that is at every whole number across that range. The origin is labeled 0, both axes end in arrowheads at both ends, and one unit is the same length on the x-axis as on the y-axis. No graph, no points, no open or closed endpoints and no vertical joining lines appear.
1. Hint 1
  The floor is constant while its input stays between one integer and the next.
2. Hint 2
  For height $n$ , solve $n\le2x<n+1$ for $x$ .
Answer

For $n=-2,-1,0,1,2$ : height $n$ on $n/2\le x<(n+1)/2$ ; left ends closed, right ends open.

Full solution

The condition for height $n$ is
$n\le2x<n+1.$
Dividing every part by the positive number $2$ gives
$\frac n2\le x<\frac{n+1}{2}$

Within the specified domain, the heights are $-2,-1,0,1,2$ , each on a half-unit interval.
Draw five horizontal segments, with closed left endpoints and open right endpoints.
Do not join the jumps with vertical segments.

The first segment includes $x=-1$ , and the last excludes $x=1.5$ , matching the given domain.

Answer

For $n=-2,-1,0,1,2$ : height $n$ on $n/2\le x<(n+1)/2$ ; left ends closed, right ends open.

Key idea

A scale factor inside a floor changes the widths of its steps while their open and closed endpoint rules remain intact.
Problem 4: Mark complete
Problem 5 The open carton

There are $73$ items still to pack. An already partly filled carton has room for $3$ more items and must be filled first. Every new carton holds $8$ items. How many new cartons are needed, and how many further items could arrive without increasing that number of new cartons?
1. Hint 1
  Remove the items that fit into the existing carton before counting new cartons.
2. Hint 2
  Once the count is rounded up, compare the space those cartons hold with the number of items being packed.
Answer

$9$ new cartons; $2$ further items could arrive without increasing that number.

Full solution

Filling the existing carton leaves
$73-3=70$
items.
All items need space, so
$\left\lceil\frac{70}{8}\right\rceil=9$
new cartons are needed.

Those nine cartons hold
$9(8)=72$
items, which is $2$ more than the $70$ being packed.
So two further items still fit, while a third would bring the count to $73$ and force a tenth carton.

Answer

$9$ new cartons; $2$ further items could arrive without increasing that number.

Key idea

A ceiling counts the containers needed, and the space they hold beyond the contents is the room left before one more container is required.
Problem 5: Mark complete
Problem 6 The two bracket records

Find all real $x$ for which $\lfloor x+0.4\rfloor=-2$ and $\lceil x-0.2\rceil=-2$ .
1. Hint 1
  Translate each bracket record into its complete interval of inputs.
2. Hint 2
  Keep the inclusive and exclusive endpoints when shifting, then take the overlap.
Answer

$-2.4\le x\le-1.8$ .

Full solution

The floor record means $-2\le x+0.4<-1$ , giving
$-2.4\le x<-1.4$

The ceiling record means $-3<x-0.2\le-2$ , giving
$-2.8<x\le-1.8$
Their overlap is
$-2.4\le x\le-1.8.$

At $x=-2.4$ the shifted floor input is $-2$ , and $-2.4$ lies strictly inside $-2.8<x\le-1.8$ ; at $x=-1.8$ the shifted ceiling input is $-2$ , and $-1.8$ lies strictly inside
$-2.4\le x<-1.4.$
So each endpoint satisfies both records.

Answer

$-2.4\le x\le-1.8$ .

Key idea

Several floor and ceiling records constrain an input to the overlap of intervals with carefully preserved endpoint inclusion.
Problem 6: Mark complete
Problem 7 The separate strips

Three separate strips are $28$ cm, $23$ cm and $19$ cm long. Each is cut into complete $4$ cm pieces, and leftover lengths cannot be joined. The resulting pieces are packed into bags holding at most $7$ pieces each. Find the number of complete pieces, total unused length, and number of bags needed.
1. Hint 1
  Count complete pieces from each strip separately before adding the counts.
2. Hint 2
  After the cutting count is known, round upward for the bags needed to hold all pieces.
Answer

$16$ pieces; $6$ cm unused; $3$ bags.

Full solution

The strips supply $\lfloor28/4\rfloor=7$ , $\lfloor23/4\rfloor=5$ and $\lfloor19/4\rfloor=4$ pieces, so
$7+5+4=16$
pieces are made.
Their leftovers are $0$ cm, $3$ cm and $3$ cm, so
$0+3+3=6$
cm remains unused.

Two bags hold $14$ pieces, so one further bag is needed for the last two.
Equivalently,
$\left\lceil\frac{16}{7}\right\rceil=3$
bags.

Answer

$16$ pieces; $6$ cm unused; $3$ bags.

Key idea

Lengths that cannot be joined force a separate rounding down on each strip, and only the pieces that result are grouped by a ceiling.
Problem 7: Mark complete
Problem 8 Two floors added

Zoe claims $\lfloor u\rfloor+\lfloor v\rfloor=\lfloor u+v\rfloor$ for every pair of real inputs. Decide whether she is correct and justify your decision.
1. Hint 1
  Adding fractional parts may carry into the next integer step.
2. Hint 2
  Write each input as its floor plus its fractional part, then try a negative noninteger pair whose fractional parts add to at least $1$ .
Answer

No; $u=v=-0.4$ give left side $-2$ and right side $-1$ .

Full solution

Take $u=v=-0.4$ .
Each has floor $-1$ , so the left side is
$-1+(-1)=-2$

The sum is $-0.8$ , which has floor $-1$ .
Thus the two sides differ, so the claim fails.

Answer

No; $u=v=-0.4$ give left side $-2$ and right side $-1$ .

Key idea

Rounding before addition can differ from rounding the completed sum.
Problem 8: Mark complete
Problem 9 The whole-number shift

For every real $x$ and every integer $k$ , Eli claims $\lfloor x+k\rfloor=\lfloor x\rfloor+k$ . Is he correct? Justify your decision.
1. Hint 1
  An integer shift moves both neighboring integer bounds by the same amount.
2. Hint 2
  Name $n=\lfloor x\rfloor$ and add $k$ to the inequality defining that floor.
Answer

Yes, for every real $x$ and integer $k$ .

Full solution

Let $n=\lfloor x\rfloor$ , so
$n\le x<n+1.$
Adding the integer $k$ gives
$n+k\le x+k<n+k+1$

The lower bound $n+k$ is an integer, so it is exactly the floor of $x+k$ .
Therefore $\lfloor x+k\rfloor=n+k$ , proving the claim.

Answer

Yes, for every real $x$ and integer $k$ .

Key idea

An integer shift moves a number and its floor together without changing their separation.
Problem 9: Mark complete
Problem 10 The doubled reading

A real number $x$ has fractional part $0.6$ . Arun says $2x$ must have fractional part $1.2$ . Decide whether he is right and find the fractional part of $2x$ .
1. Hint 1
  A fractional part is at least zero and less than one.
2. Hint 2
  Write $x$ as an integer plus $0.6$ and see whether doubling crosses another integer.
Answer

Arun is wrong; the fractional part is $0.2$ .

Full solution

Write $x=n+0.6$ , where $n$ is an integer.
Then
$2x=2n+1.2$
This is the integer $2n+1$ plus $0.2$ .

Thus the floor of $2x$ is $2n+1$ , and its fractional part is $0.2$ .
The extra whole unit in $1.2$ belongs to the floor, not the fractional part.

Answer

Arun is wrong; the fractional part is $0.2$ .

Key idea

Scaling a fractional part may create a whole-number contribution that must be removed to find the new fractional part.
Problem 10: Mark complete

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