Decimals: Chapter Test

Solution

Correct answer: C

You chose A: Skips the zero when counting columns after the point, so the $8$ is read as the second digit and lands in the hundredths.

You chose B: Gives the digit itself rather than what the column it sits in makes it worth.

You chose D: Counts the decimal point itself as the first column, which pushes every digit one place further right.

Solution

Correct answer: A

You chose B: Lines up the right-hand ends of the two digit strings instead of the points, which turns $0.46$ into $0.046$.

You chose C: Adds only the tenths column, the one place both numbers show without any padding, and stops there.

You chose D: Shortens the longer number to two places before adding, rounding $2.518$ to $2.52$ instead of padding the shorter number.

Solution

Correct answer: D

You chose A: Reads the digits over a hundred correctly but stops before reducing, so the fraction is equal in value but not in lowest terms.

You chose B: Divides both terms by $2$ and stops, though $22$ and $50$ still share a factor of $2$.

You chose C: Puts a two-digit decimal part over a thousand rather than a hundred, using three zeros for two digits.

Solution

Correct answer: B

You chose A: Rounds to the nearest hundredth instead of the place the question named.

You chose C: Reads the final digit $2$ as the deciding one rather than the digit immediately right of the tenths place, so it rounds down.

You chose D: Rounds to the nearest whole number and then writes a tenths digit, so the answer is not rounded to the named place at all.

Solution

Correct answer: C

You chose A: Gives the product only the two places of the longer factor instead of the places of both factors together.

You chose B: Adds the two factors instead of multiplying them.

You chose D: Counts an extra place for the decimal point itself on top of the places the two factors supply.

Solution

Correct answer: A

You chose B: Reads the digits after the point as whole numbers and orders those, so $7$ is treated as smaller than $68$.

You chose C: Compares most pairs place by place but settles $0.7$ against $0.699$ by digit count, letting the longer string win.

You chose D: Compares the four correctly but lists them from greatest to least.

Solution

Correct answer: B

You chose A: Writes the two nonzero digits side by side straight after the point, leaving no zero to hold the tenths column open.

You chose C: Puts the $7$ in the tenths column rather than the hundredths, so the hundredths term is read as a tenths term.

You chose D: Reads the decimal correctly but reports the fraction over the thousandths before reducing it.

Solution

Correct answer: D

You chose A: Moves the point two places in the divisor but only one in the dividend, so the division carried out is $28 \div 7$.

You chose B: Moves both points two places but adds an extra zero to the dividend as well, dividing $2800$ by $7$.

You chose C: Moves the point in the divisor and leaves the dividend untouched, dividing $2.8$ by $7$.

Solution

Correct answer: C

You chose A: Applies the prime-factor test to the denominator as written, where $60 = 2 \times 2 \times 3 \times 5$ carries a $3$, instead of reducing first to $\frac{7}{10}$.

You chose B: Sees the $7$ in $56 = 2 \times 2 \times 2 \times 7$ and stops, though reducing cancels that $7$ and leaves $\frac{3}{8}$.

You chose D: Reduces by dividing the numerator by $11$ and the denominator by $5$, using different numbers top and bottom, which lands on $\frac{2}{11}$ and so on a denominator that really does repeat.

Solution

Correct answer: B

You chose A: Multiplies correctly but lines up the right-hand ends for the subtraction, taking away $0.86$ instead of $8.6$.

You chose C: Borrows a ten into the tenths column but does not reduce the ones digit, so the ones subtraction starts from $7$ instead of $6$.

You chose D: Slides the point one place for the $100$ instead of two, so the product is $5.7$.

Solution

Correct answer: D

You chose A: Picks the decimal showing the most digits, reading a longer string as a smaller number once it begins with a zero.

You chose B: Miscounts the columns in this one, reading its $92$ as ten-thousandths instead of hundredths and thousandths, so it is compared as though it were $0.0092$.

You chose C: Reads the digits after the point as whole numbers, so $1$ looks smaller than $89$, $92$ and $925$.

Solution

Correct answer: A

You chose B: Takes the neighboring tenths themselves as the limits, rather than the two marks halfway to them.

You chose C: Allows only lengths at or above the recorded value, forgetting that a length just below it rounds up to it.

You chose D: Breaks both ties downwards, so an exact $8.95$ falls to $8.9$ and an exact $9.05$ falls to $9.0$, which is not the round-half-up convention this course uses.

Solution

Correct answer: C

You chose A: Treats any zero written after the decimal point as removable, though this one is holding the $6$ out of the tenths column.

You chose B: Judges by the zero contributing nothing to the expanded sum, which is true, and misses that deleting it slides the $3$ from the hundredths into the tenths.

You chose D: Reads a zero beside a nonzero digit as filler, though deleting it moves the $1$ from the tens column into the ones.

Solution

Correct answer: D

You chose A: Checks only the direction of the change, so any answer below $350$ passes, and never checks the size against an estimate.

You chose B: Gives the product one decimal place instead of the two that $0.62$ carries, so the point lands one column too far right.

You chose C: Counts the two zeros of $350$ as decimal places as well, giving the product four places instead of two.

Solution

Correct answer: A

You chose B: Applies the prime-factor test to the denominator as written, when the test holds only for a fraction in lowest terms, where the $13$ has already canceled.

You chose C: Divides the denominator by the numerator, which converts $\frac{52}{39} = \frac{4}{3}$ rather than the fraction asked about.

You chose D: Believes the prime-factor test can only predict repetition, so a prediction that a decimal ends still has to be confirmed by dividing.

Solution

Correct answer: B

You chose A: Moves the point in the divisor but not in the dividend, so the division carried out is $9.24 \div 4$.

You chose C: Borrows into each column of the subtraction but never reduces the column borrowed from, so the tenths start from $11$ and the ones still from $3$.

You chose D: Pads the quotient by writing the extra zero next to the point instead of at the right-hand end, subtracting from $23.01$.

Solution

Correct answer: D

You chose A: Requires the hundredths digit to beat $7$ outright, missing that a hundredths digit of $7$ still wins on the thousandths column.

You chose B: Counts a hundredths digit of $6$ as well, treating the thousandths $8$ as able to make up a whole hundredth.

You chose C: Reads the extra column in $0.4\square 8$ as making the reading larger whatever the missing digit is, so every digit qualifies.

Solution

Correct answer: C

You chose A: Stops the division after two places and reports the result as exact, although the remainder has not reached zero.

You chose B: Divides the denominator by the numerator, computing $11 \div 5$ instead of $5 \div 11$.

You chose D: Draws the bar over the last digit alone, which names $0.4555\ldots$ rather than the two-digit block that actually recurs.

Solution

Correct answer: A

You chose B: Reads the endless string of $4$s as eventually reaching a $5$ in the deciding column, so the hundredths digit is pushed up.

You chose C: Rounds to the nearest tenth instead of the place the question named.

You chose D: Moves the point in the divisor only, dividing $7.6$ by $9$ rather than $76$ by $9$.

Solution

Correct answer: B

You chose A: Slides the point two places for the thousand instead of three, so the first product is $2.6$.

You chose C: Counts $0.05$ as one decimal place rather than two, so the final product loses a place.

You chose D: Multiplies the two decimals together correctly but drops the factor of a thousand from the calculation entirely.