Aimathic – Add and subtract decimals

Add and subtract decimals

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Difficulty: 1 – 5

5106645

Add or subtract. Line up the decimal points and use place value carefully. a) $9.42+15.8$ b) $14.6-6.75$ c) $0.734+3.59$ d) $30-12.841$

Hints

- Line up digits by place value, with decimal points directly under each other. - Add trailing zeros when they help show matching place values. - Keep the decimal point aligned in the answer.

Solution

1. For a), write $15.8$ as $15.80$. Then $9.42+15.80=25.22$. 2. For b), write $14.6$ as $14.60$. Then $14.60-6.75=7.85$. 3. For c), write $3.59$ as $3.590$. Then $0.734+3.590=4.324$. 4. For d), write $30$ as $30.000$. Then $30.000-12.841=17.159$.

Answer

a) $25.22$ b) $7.85$ c) $4.324$ d) $17.159$

5106705

Estimate each result first. Then calculate the exact value. a) $42.7+5.09+13$ b) $105.2-37.485$

Hints

- Round to nearby whole numbers for a quick estimate. - Align the decimal points when finding the exact value. - Add trailing zeros when they help show matching place values.

Solution

1. For a), an estimate is $43+5+13=61$. 2. The exact sum is $42.70+5.09+13.00=60.79$. 3. For b), an estimate is $105-37=68$. 4. The exact difference is $105.200-37.485=67.715$.

Answer

a) Estimate: $61$; exact: $60.79$ b) Estimate: $68$; exact: $67.715$

5106735

Two packages are being prepared for shipping. Package A contains items weighing $3.25$ pounds and $1.80$ pounds. Package B contains items weighing $2.65$ pounds and $2.45$ pounds. Which package is heavier, and by how many pounds?

Hints

- Find the total weight of each package first. - Align the decimal points when adding. - Subtract the smaller total from the larger total to find the difference.

Solution

1. Package A weighs $3.25+1.80=5.05$ pounds. 2. Package B weighs $2.65+2.45=5.10$ pounds. 3. Since $5.10>5.05$, Package B is heavier. 4. The difference is $5.10-5.05=0.05$ pound.

Answer

Package B is heavier. It weighs $5.10$ pounds, which is $0.05$ pound more than Package A.

5106745

In a two-leg swim relay, the times are: Blue Team: first leg $32.45\,\text{s}$, second leg $31.98\,\text{s}$ Red Team: first leg $32.12\,\text{s}$, second leg $32.26\,\text{s}$ Which team wins? Compare the total times and find the winning margin.

Hints

- Add the two leg times for each team. - In a timed race, the smaller total time wins. - Subtract the two total times to find the winning margin.

Solution

1. Blue Team's total time is $32.45+31.98=64.43\,\text{s}$. 2. Red Team's total time is $32.12+32.26=64.38\,\text{s}$. 3. The shorter total time wins, and $64.38<64.43$, so Red Team wins. 4. The winning margin is $64.43-64.38=0.05\,\text{s}$.

Answer

Red Team wins with a total time of $64.38\,\text{s}$. The winning margin is $0.05\,\text{s}$.

5106795

Calculate mentally. For each problem, describe a convenient strategy that avoids the standard written algorithm. a) $45.67-19.99$ b) $12.456+3.998$

Hints

- Look for a nearby whole number that is easier to add or subtract. - After using the nearby number, decide whether you added or subtracted too much. - Correct that small difference at the end.

Solution

1. For a), subtract $20$ instead of $19.99$, then add back $0.01$: $45.67-20+0.01=25.68$. 2. For b), add $4$ instead of $3.998$, then subtract the extra $0.002$: $12.456+4-0.002=16.454$.

Answer

a) $25.68$, using $45.67-20+0.01$ b) $16.454$, using $12.456+4-0.002$

5106855

Lucas buys school supplies: a ruler for $\$1.35$, three notebooks for $\$0.60$ each, and a pen for $\$9.49$. He pays with $\$20.00$. How much change should he receive?

Hints

- Find the cost of the three notebooks first. - Add all the item costs to find the total purchase price. - Subtract the total cost from the amount paid. - Align decimal points carefully.

Solution

1. The three notebooks cost $3\times\$0.60=\$1.80$. 2. The total cost is $\$1.35+\$1.80+\$9.49=\$12.64$. 3. The change is $\$20.00-\$12.64=\$7.36$.

Answer

$\$7.36$

5113895

An athlete records three training-route distances: $7.45$ miles, $5.7$ miles, and $9.333$ miles. a) Find the total distance. b) Round the total to the nearest hundredth of a mile.

Hints

- Align the decimal points before adding. - Add trailing zeros if they help line up place values. - To round to the nearest hundredth, check the thousandths digit.

Solution

1. Add the distances: $7.45+5.7+9.333=22.483$ miles. 2. To round to the nearest hundredth, look at the thousandths digit, which is $3$. The rounded total is $22.48$ miles.

Answer

a) $22.483$ miles b) $22.48$ miles

5121815

Leon buys two books. The first costs $\$12.95$ and the second costs $\$8.98$. He adds the prices mentally by rounding each one up to the next whole dollar. Describe his steps and find the exact total.

Hints

- How much does each price increase when rounded up to the next whole dollar? - Add the rounded prices first. - Since the rounded prices are too high, how should the final total be adjusted?

Solution

1. Round $\$12.95$ up to $\$13.00$, an increase of $\$0.05$, and round $\$8.98$ up to $\$9.00$, an increase of $\$0.02$. 2. Add the rounded amounts: $\$13.00+\$9.00=\$22.00$. 3. The total rounding increase is $\$0.05+\$0.02=\$0.07$. 4. Subtract the extra amount: $\$22.00-\$0.07=\$21.93$.

Answer

Round to $\$13.00$ and $\$9.00$, add to get $\$22.00$, then subtract the $\$0.07$ total adjustment. The exact total is $\$21.93$.

5121825

Calculate $34.152-2.998$ mentally. Explain a strategy that first replaces the subtrahend with a nearby whole number.

Hints

- Which whole number is very close to $2.998$? - Decide whether replacing $2.998$ with that whole number subtracts too much or too little. - Correct the small difference at the end.

Solution

1. Replace $2.998$ with $3$. 2. Calculate the easier difference: $34.152-3=31.152$. 3. Subtracting $3$ removed $0.002$ too much because $3-2.998=0.002$. 4. Add that amount back: $31.152+0.002=31.154$.

Answer

$31.154$, using $34.152-3+0.002$.

5162985

Lukas wants to buy a board game that costs $\$45.00$. He has already saved $\$27.50$. How much more money does he need to save?

Hints

- Identify the full price and the amount already saved. - Find the difference between the two amounts. - Think about how much is needed to reach the next whole dollar first.

Solution

1. Subtract the amount already saved from the price: $\$45.00 - \$27.50 = \$17.50$.

Answer

Lukas needs to save $\$17.50$ more.

5163005

The Meyer family buys hiking gear: a backpack for $\$34.90$, a water bottle for $\$12.50$, and trekking poles for $\$45.00$. How much do the items cost altogether?

Hints

- Find the combined value of all three items. - Line up the decimal points when adding. - Add the cents and dollars carefully.

Solution

1. Add the three prices: $\$34.90 + \$12.50 + \$45.00 = \$92.40$.

Answer

The Meyer family pays $\$92.40$ altogether.

5200825

Calculate each sum. a) $\$3.45+\$6.55$ b) $\$12.80+\$4.35$ c) $\$0.99+\$7.05+\$1.10$ d) $85$ cents $+\$2.40$

Hints

- You can convert each amount to cents before adding. - One hundred cents equals $\$1.00$. - Align decimal points and add by place value. - In part d), write both amounts in the same unit first.

Solution

1. a) Add the dollars and cents: $\$3.45+\$6.55=\$10.00$. 2. b) $\$12.80+\$4.35=\$17.15$. 3. c) $\$0.99+\$7.05=\$8.04$, and $\$8.04+\$1.10=\$9.14$. 4. d) $85$ cents is $\$0.85$. Then $\$0.85+\$2.40=\$3.25$.

Answer

a) $\$10.00$ b) $\$17.15$ c) $\$9.14$ d) $\$3.25$

5201065

Subtract. Write each answer in dollars. a) $\$15.20-\$6.45$ b) $\$40.05-\$12.30$ c) $\$8.10-\$0.95$ d) $\$200.00-\$135.55$

Hints

- Align decimal points when subtracting. - You can convert each amount to cents to use whole-number subtraction. - Remember that $\$1.00$ is $100$ cents. - Preserve two decimal places for dollar amounts.

Solution

1. a) $\$15.20-\$6.45=\$8.75$. 2. b) $\$40.05-\$12.30=\$27.75$. 3. c) $\$8.10-\$0.95=\$7.15$. 4. d) $\$200.00-\$135.55=\$64.45$.

Answer

a) $\$8.75$ b) $\$27.75$ c) $\$7.15$ d) $\$64.45$

5205855

Calculate the value and write the result as a dollar amount: $\$8.45 + 215\text{ cents} - \$3.60$.

Hints

- Identify the two money units used in the expression. - How many cents equal one dollar? - Convert all amounts to the same unit before calculating. - Line up decimal points when adding and subtracting.

Solution

1. Convert cents to dollars: $215\text{ cents} = \$2.15$. 2. Add the first two amounts: $\$8.45 + \$2.15 = \$10.60$. 3. Subtract the third amount: $\$10.60 - \$3.60 = \$7.00$.

Answer

$\$7.00$

5214125

Find each difference. Pay close attention when you need to regroup $1$ dollar as $100$ cents. a) $\$74.25 - \$28.15$ b) $\$100.00 - \$63.45$ c) $\$12.05 - \$0.80$ d) $\$5.10 - \$2.55$

Hints

- You can rewrite each amount in cents before subtracting. - You can also subtract the dollars and cents by place value. - Remember that $\$1.00$ equals $100$ cents. - For a subtraction such as $\$100.00 - \$63.45$, try subtracting in parts.

Solution

1. For a), subtract the hundredths, tenths, ones, and tens: $\$74.25 - \$28.15 = \$46.10$. 2. For b), regroup from $\$100.00$ and subtract: $\$100.00 - \$63.45 = \$36.55$. 3. For c), regroup $\$12.05$ as $11$ dollars and $105$ cents. Then $105 - 80 = 25$, so the difference is $\$11.25$. 4. For d), regroup and subtract: $\$5.10 - \$2.55 = \$2.55$.

Answer

a) $\$46.10$ b) $\$36.55$ c) $\$11.25$ d) $\$2.55$

5351525

Find the missing decimal $x$ on each number line.

Hints

- Line up decimal places carefully when adding or subtracting. - A negative jump means subtracting, so it moves left on the number line. - If the starting value is missing, use the inverse operation to work backward.

Solution

1. For a), add the jump to the starting value: $4.5+2.7=7.2$, so $x=7.2$. 2. For b), the starting value is unknown. Undo the jump of $-3.5$: $6.7+3.5=10.2$, so $x=10.2$. 3. For c), subtract the jump size from the starting value: $12.3-4.6=7.7$, so $x=7.7$.

Answer

a) $x=7.2$ b) $x=10.2$ c) $x=7.7$

5352745

Find the two missing numbers in the jump sequence on the number line.

Hints

- Work step by step and find the middle number first. - Line up the decimal places carefully.

Solution

1. First jump: $12.4+3.8=16.2$. 2. Second jump: $16.2-2.5=13.7$.

Answer

The missing numbers are $16.2$ and $13.7$.

5106655

Estimate each result by rounding to whole numbers. Then calculate the exact value. a) $56.7+14+8.29$ b) $95.4-(24.38+31.7)$

Hints

- Round each decimal to a nearby whole number for the estimate. - In part b), evaluate the parentheses before subtracting. - Compare each exact result with its estimate to check whether it is reasonable.

Solution

1. For a), an estimate is $57+14+8=79$. The exact sum is $56.70+14.00+8.29=78.99$. 2. For b), an estimate is $95-(24+32)=39$. For the exact value, first calculate $24.38+31.70=56.08$, then $95.40-56.08=39.32$.

Answer

a) Estimate: $79$; exact: $78.99$ b) Estimate: $39$; exact: $39.32$

5106715

Sarah wrote: $15.8-2.45=13.45$ Explain Sarah's error. Then find the correct difference and explain why writing a trailing zero in the first number is helpful.

Hints

- Look carefully at the hundredths place in both numbers. - Write $15.8$ as $15.80$ and line up the decimal points. - Ask what must happen when subtracting $5$ hundredths from $0$ hundredths.

Solution

1. Sarah did not account correctly for the hundredths place. The first number has $0$ hundredths, so the subtraction requires regrouping. 2. Writing $15.8$ as $15.80$ makes the place values visible. Then $15.80-2.45=13.35$. 3. The trailing zero shows explicitly that the hundredths calculation begins with $0-5$, which requires regrouping from the tenths place.

Answer

Sarah ignored the regrouping needed in the hundredths place. The correct result is $13.35$. Writing $15.80$ makes the $0$ hundredths visible and helps keep the place values aligned.

5106755

A hiker plans an $11.5$-mile day hike. In the morning, the hiker walks $4.6$ miles. After a break, the hiker walks another $4.25$ miles. How many miles remain? Which of the three parts of the hike—morning, afternoon, or remaining distance—is the longest?

Hints

- Add the morning and afternoon distances first. - Subtract the distance already traveled from the total distance. - Compare all three segment lengths using place value.

Solution

1. The hiker has completed $4.6+4.25=8.85$ miles. 2. The remaining distance is $11.5-8.85=2.65$ miles. 3. Compare $4.6$, $4.25$, and $2.65$. The morning distance, $4.6$ miles, is the longest.

Answer

The hiker has $2.65$ miles left. The morning segment, $4.6$ miles, is the longest.

5106815

Students often use compensation to calculate mentally with decimals. Check each statement. If it is incorrect, fix the strategy and the result. 1) “For $17.4-3.9$, I subtract $4$ and then subtract another $0.1$.” 2) “For $5.65+2.98$, I add $3$ and then subtract $0.02$.”

Hints

- Check each proposed strategy step by step. - If you subtract more than needed, should the compensation be added or subtracted? - Think about the direction of each adjustment on a number line.

Solution

1. Statement 1 is incorrect. Subtracting $4$ removes $0.1$ too much, so that amount must be added back: $17.4-4+0.1=13.5$. 2. Statement 2 is correct. Adding $3$ adds $0.02$ too much, so subtracting $0.02$ compensates: $5.65+3-0.02=8.63$.

Answer

1) Incorrect. The correct strategy is $17.4-4+0.1=13.5$. 2) Correct. $5.65+3-0.02=8.63$.

5106865

For a picnic, a family buys fruit for $\$4.58$, bread for $\$2.25$, sandwich spread for $\$3.12$, and two bottles of soda for $\$1.35$ each. They use a $\$1.50$ coupon. Afterward, they must also pay $\$3.50$ for parking. Estimate first whether $\$15.00$ will be enough. Then calculate the exact total.

Hints

- Round the prices to convenient nearby amounts for the estimate. - Remember that two bottles of soda are purchased. - Subtract the coupon from the purchase total. - Add the parking cost after applying the coupon.

Solution

1. A reasonable estimate is $\$4.60+\$2.30+\$3.10+\$2.70-\$1.50+\$3.50=\$14.70$, so $\$15.00$ should be enough. 2. The two sodas cost $2\times\$1.35=\$2.70$. 3. The purchases total $\$4.58+\$2.25+\$3.12+\$2.70=\$12.65$. 4. After the coupon, the cost is $\$12.65-\$1.50=\$11.15$. 5. Including parking, the exact total is $\$11.15+\$3.50=\$14.65$, so $\$15.00$ is enough.

Answer

The estimate shows that $\$15.00$ should be enough. The exact total is $\$14.65$, so the money is sufficient.

5106875

A sports club buys equipment: $5$ jump ropes at $\$4.25$ each, one medicine ball for $\$18.90$, and $10$ marker cones for $\$12.50$ total. The store gives the club a $\$5.75$ discount. The coach has a $\$50.00$ budget. After buying the equipment, is there enough money left to buy $3$ towels at $\$2.80$ each? Show your calculations.

Hints

- Find the equipment cost before and after the discount. - Subtract the equipment cost from the total budget. - Find the total cost of the three towels. - Compare the remaining budget with the towel cost.

Solution

1. The jump ropes cost $5\times\$4.25=\$21.25$. 2. Before the discount, the equipment costs $\$21.25+\$18.90+\$12.50=\$52.65$. 3. After the discount, the cost is $\$52.65-\$5.75=\$46.90$. 4. The remaining budget is $\$50.00-\$46.90=\$3.10$. 5. The towels cost $3\times\$2.80=\$8.40$. 6. Since $\$3.10<\$8.40$, there is not enough money. The budget is short by $\$8.40-\$3.10=\$5.30$.

Answer

No. After the equipment purchase, $\$3.10$ remains. The towels cost $\$8.40$, so the budget is short by $\$5.30$.

5121345

In each calculation, the letters $a$ and $b$ stand for digits from $0$ through $9$. Find the value of each digit. The values of $a$ and $b$ may be different in the two parts. a) $4.a2+3.7b=8.59$ b) $12.b-a.5=8.7$

Hints

- Work one place value at a time, as you would in the standard addition or subtraction algorithm. - Account for any regrouping from one place to the next. - Each letter represents one digit from $0$ through $9$.

Solution

1. In part a), the hundredths digits give $2+b=9$, so $b=7$. In the tenths place, $a+7=15$, so $a=8$ and $1$ is regrouped to the ones place. Then $1+4+3=8$, which confirms the sum. 2. In part b), regroup in the tenths place: $10+b-5=7$, so $b=2$. After that regrouping, the ones digit is $1$. Regroup again from the tens place: $11-a=8$, so $a=3$. The remaining tens digit is $0$, giving $8.7$.

Answer

a) $a=8$, $b=7$ b) $a=3$, $b=2$

5121835

Sophie's hiking app records three trail segments: $4.97$ miles, $3.05$ miles, and $2.98$ miles. She says, “I can find the exact total mentally by starting with $5+3+3$ and making one small adjustment.” Check Sophie's claim. What adjustment is needed, and what is the exact total distance?

Hints

- Find how far each distance is above or below $5$, $3$, and $3$. - Write each small adjustment with a positive or negative sign. - Add the three adjustments before changing the total of $11$.

Solution

1. Compare each distance with the nearby whole number: $4.97=5-0.03$, $3.05=3+0.05$, and $2.98=3-0.02$. 2. The nearby whole numbers sum to $5+3+3=11$. 3. The total adjustment is $-0.03+0.05-0.02=0$. 4. Therefore, the exact total is $11$ miles.

Answer

The adjustment is $0$ because $-0.03+0.05-0.02=0$. The exact total is $11$ miles.

5122715

A loaded backpack weighs $10.5\,\text{lb}$. It contains books weighing $4.25\,\text{lb}$, notebooks weighing $1.1\,\text{lb}$, a pencil case weighing $0.8\,\text{lb}$, and a water bottle weighing $2\,\text{lb}$. The rest of the total weight is the empty backpack. Write an expression for the weight of the empty backpack and find its weight.

Hints

- What is the total weight, and which parts of it are already known? - How can you find the part that remains after removing the known weights? - Add all the contents before subtracting from the total.

Solution

1. Add the weights of the contents: $4.25+1.1+0.8+2=8.15\,\text{lb}$. 2. Subtract the contents from the total: $10.5-8.15=2.35\,\text{lb}$.

Answer

Expression: $10.5-(4.25+1.1+0.8+2)$ The empty backpack weighs $2.35\,\text{lb}$.

5160995

Supplies were purchased for a school festival. The two receipts show these amounts: Receipt A: $\$12.00$; $\$8.00$; $\$5.50$; $\$4.50$ Receipt B: $\$13.47$; $\$22.89$; $\$5.12$; $\$17.04$ For each receipt, decide whether mental math or the standard addition algorithm is more efficient. Then find both totals.

Hints

- Do any pairs of cents combine to make a whole dollar? - For which receipt would you need to keep track of several decimal place values at once? - Can you pair amounts on Receipt A to make friendly totals?

Solution

1. For Receipt A, use mental math by pairing amounts that make whole-dollar totals: $\$12.00 + \$8.00 = \$20.00$ and $\$5.50 + \$4.50 = \$10.00$. The total is $\$20.00 + \$10.00 = \$30.00$. 2. For Receipt B, line up the decimal points and use the standard addition algorithm: $\$13.47 + \$22.89 + \$5.12 + \$17.04 = \$58.52$.

Answer

Receipt A (mental math): $\$30.00$ Receipt B (standard algorithm): $\$58.52$

5161595

Tim wants to buy a science kit for $\$44.50$ and an additional experiment set for $\$12.00$. He has saved $\$35.00$, and his parents give him $\$25.00$. How much money will Tim have left after buying both items?

Hints

- Find the combined cost of the two science items. - Find the total amount Tim has. - Subtract the cost from the amount he has. - Keep track of dollars and cents.

Solution

1. Find the total cost: $\$44.50 + \$12.00 = \$56.50$. 2. Find the total amount Tim has: $\$35.00 + \$25.00 = \$60.00$. 3. Find the amount left: $\$60.00 - \$56.50 = \$3.50$.

Answer

Tim will have $\$3.50$ left.

5179225

At the start of summer break, Maya has $\$50.00$ in her savings jar. During the first week, she records all the money she receives and spends. How much money is in the jar at the end of the week? <table> <tr><td>Gift from Grandma:</td><td>$\$15.00$</td></tr> <tr><td>Sold a used book:</td><td>$\$10.50$</td></tr> <tr><td>Movie ticket:</td><td>$\$8.40$</td></tr> <tr><td>Popcorn:</td><td>$\$3.25$</td></tr> <tr><td>New T-shirt:</td><td>$\$12.80$</td></tr> <tr><td>Gum:</td><td>$\$0.95$</td></tr> </table>

Hints

- Decide which amounts increase the money in the jar and which amounts decrease it. - Add all money received and all money spent separately. - Line up the decimal points when adding or subtracting.

Solution

1. Add the money received: $\$15.00 + \$10.50 = \$25.50$. 2. Add the money spent: $\$8.40 + \$3.25 + \$12.80 + \$0.95 = \$25.40$. 3. Find the final amount: $\$50.00 + \$25.50 - \$25.40 = \$50.10$.

Answer

Maya has $\$50.10$ in her savings jar at the end of the week.

5179235

A fifth-grade class fund has $\$84.50$ on Monday morning. A bake sale adds $\$45.00$. The teacher then buys art supplies for $\$18.75$ and pays for bus tickets for a class trip. On Friday afternoon, $\$92.45$ remains. How much did the bus tickets cost altogether?

Hints

- Find the balance after each known transaction. - Compare the balance after the art supplies with the final balance. - The difference is the cost of the bus tickets.

Solution

1. Find the balance after the bake sale: $\$84.50 + \$45.00 = \$129.50$. 2. Subtract the cost of the art supplies: $\$129.50 - \$18.75 = \$110.75$. 3. The bus tickets account for the difference between this balance and the final balance: $\$110.75 - \$92.45 = \$18.30$.

Answer

The bus tickets cost $\$18.30$ altogether.

5180075

Mr. Carter pays a $\$412.60$ bill at a hardware store. After paying, he has two $\$100$ bills, one $\$50$ bill, three $\$10$ bills, eight $\$1$ bills, and ten quarters. He also has a $\$20$ gift card that can be used only at a bookstore. a) What is the total value of the cash Mr. Carter has left? b) How much cash did he have before paying the hardware-store bill? c) Can he immediately buy a drill that costs $\$295$ using the cash he has left? Explain.

Hints

- Keep cash and the store-specific gift card separate. - Use the value and number of each type of bill or coin to find the cash total. - The amount before the purchase equals the amount spent plus the cash remaining.

Solution

1. Find the cash remaining: $2 \times \$100 + 1 \times \$50 + 3 \times \$10 + 8 \times \$1 + 10 \times \$0.25 = \$200 + \$50 + \$30 + \$8 + \$2.50 = \$290.50$. 2. Find the cash he had before the purchase: $\$290.50 + \$412.60 = \$703.10$. 3. Compare the remaining cash with the drill price. Since $\$290.50 < \$295$, the cash is not enough. The bookstore gift card cannot be used at the hardware store.

Answer

a) $\$290.50$ b) $\$703.10$ c) No. He is $\$4.50$ short because $\$295 - \$290.50 = \$4.50$, and the gift card is valid only at the bookstore.

5180085

Anna is saving for a new bicycle. Her savings jar contains three $\$50$ bills, six $\$20$ bills, eight $\$5$ bills, and twenty-four $\$1$ bills. She first buys a helmet and a bell for a total of $\$47.35$. a) How much money was originally in Anna’s savings jar? b) How much money remains after she buys the accessories? c) The bicycle Anna wants costs $\$285$. Does she have enough money left to buy it?

Hints

- Multiply to find the total value of each type of bill, then add. - Subtract the cost of the accessories from the original amount. - Compare the remaining money with the price of the bicycle.

Solution

1. Find the original amount: $3 \times \$50 + 6 \times \$20 + 8 \times \$5 + 24 \times \$1 = \$150 + \$120 + \$40 + \$24 = \$334$. 2. Subtract the cost of the accessories: $\$334 - \$47.35 = \$286.65$. 3. Compare the remaining amount with the bicycle price. Since $\$286.65 > \$285$, she has enough money.

Answer

a) $\$334$ b) $\$286.65$ c) Yes. She has $\$1.65$ more than the bicycle costs.

5200835

Find each missing amount. a) $\$14.20+\_\_\_=\$20.00$ b) $\$6.75+\_\_\_=\$10.50$ c) $\$0.55+\$1.25+\_\_\_=\$5.00$

Hints

- Use subtraction as the inverse of addition. - Count up to the next whole dollar when that helps. - In part c), add the known amounts first. - Think of paying at a store and finding the amount still needed.

Solution

1. a) Subtract to find the missing addend: $\$20.00-\$14.20=\$5.80$. 2. b) $\$10.50-\$6.75=\$3.75$. 3. c) First add the known amounts: $\$0.55+\$1.25=\$1.80$. Then $\$5.00-\$1.80=\$3.20$.

Answer

a) $\$5.80$ b) $\$3.75$ c) $\$3.20$

5200975

Add the four amounts. Write the total in dollars and cents. $\$18.50$; $\$9.99$; $\$24.05$; $\$7.85$.

Hints

- Align the decimal points before adding. - Record each carry when a column totals more than $9$. - You can check by converting every amount to cents.

Solution

1. Align the decimal points and add by place value. 2. Hundredths: $0+9+5+5=19$. Write $9$ and carry $1$. 3. Tenths: $5+9+0+8+1=23$. Write $3$ and carry $2$. 4. Ones: $8+9+4+7+2=30$. Write $0$ and carry $3$. 5. Tens: $1+0+2+0+3=6$. 6. The total is $\$60.39$.

Answer

$\$60.39$

5200985

A class raises $\$350.00$ for a school celebration. It spends $\$85.40$ on drinks. The buffet costs $\$42.75$ more than the drinks. How much money remains after both purchases?

Hints

- Find the buffet cost first. - Add the drink and buffet costs. - Subtract the total spent from the amount raised.

Solution

1. The buffet costs $\$85.40 + \$42.75 = \$128.15$. 2. The total spent is $\$85.40 + \$128.15 = \$213.55$. 3. The amount remaining is $\$350.00 - \$213.55 = \$136.45$.

Answer

$\$136.45$ remains.

5200995

A sports club receives a $\$1500.00$ donation. It spends $\$645.80$ on new uniforms. It spends $\$212.30$ less on new balls than on the uniforms. How much of the donation remains after both purchases?

Hints

- Use “less than” to find the cost of the balls. - Add the two purchase amounts. - Subtract the total spent from the donation.

Solution

1. The balls cost $\$645.80 - \$212.30 = \$433.50$. 2. The two purchases cost $\$645.80 + \$433.50 = \$1079.30$. 3. The amount remaining is $\$1500.00 - \$1079.30 = \$420.70$.

Answer

$\$420.70$ remains.

5201075

Consider the calculation $\$100.00 - \$25.50 - \$14.75$. 1. Find the final amount by subtracting from left to right. 2. Compare your result with $\$100.00 - (\$25.50 + \$14.75)$. What do you notice? Briefly explain why.

Hints

- In the first calculation, work from left to right. - In the second calculation, evaluate the parentheses first. - Compare the total amount subtracted in the two methods.

Solution

1. Subtract the first amount: $\$100.00 - \$25.50 = \$74.50$. 2. Subtract the second amount: $\$74.50 - \$14.75 = \$59.75$. 3. For the comparison, first add the two amounts being subtracted: $\$25.50 + \$14.75 = \$40.25$. Then $\$100.00 - \$40.25 = \$59.75$. 4. Both methods give the same result because both subtract a total of $\$40.25$ from $\$100.00$.

Answer

1. The final amount is $\$59.75$. 2. Both calculations equal $\$59.75$ because the same total amount is subtracted in each calculation.

5201085

Ms. Meyer buys fruit for $\$6.75$ and vegetables for $\$11.40$. She pays with a $\$50.00$ bill. How much change does she receive?

Hints

- Add the two purchase amounts first. - Subtract the total cost from the amount paid. - Align the decimal points.

Solution

1. The total cost is $\$6.75 + \$11.40 = \$18.15$. 2. The change is $\$50.00 - \$18.15 = \$31.85$.

Answer

Ms. Meyer receives $\$31.85$ in change.

5201095

Mila is saving for a remote-control car that costs $\$45.00$. She has already saved $\$27.65$, and her grandmother gives her $\$10.50$. How much more does Mila need to save?

Hints

- Add the money Mila has saved and received. - Subtract that total from the car’s price. - Align the decimal points.

Solution

1. Mila now has $\$27.65 + \$10.50 = \$38.15$. 2. She still needs $\$45.00 - \$38.15 = \$6.85$.

Answer

Mila still needs to save $\$6.85$.

5201115

Calculate each difference. a) $\$100.00-\$45.67$ b) $\$23.08-\$14.50$ c) $\$200.00-\$123.45$ d) $\$11.02-\$9.95$

Hints

- Think about subtracting from a whole-dollar amount such as $\$100.00$. - Pay close attention to zeros in the cents places. - Counting up to the next whole dollar can simplify a calculation. - Check by adding the difference to the amount subtracted.

Solution

1. a) $\$100.00-\$45.67=\$54.33$. 2. b) $\$23.08-\$14.50=\$8.58$. 3. c) $\$200.00-\$123.45=\$76.55$. 4. d) $\$11.02-\$9.95=\$1.07$.

Answer

a) $\$54.33$ b) $\$8.58$ c) $\$76.55$ d) $\$1.07$

5207985

Find the missing amounts. a) $\$100.00-\Box=\$37.55$ b) $\$73.20-\Box=\$28.45$ c) $\Box-\$19.99=\$45.50$

Hints

- Which inverse operation finds a missing value in a subtraction equation? - Rewrite each equation as an addition or subtraction fact. - Align decimal points and calculate by place value. - Check by substituting each answer into the original equation.

Solution

1. a) Subtract the difference from the minuend: $\$100.00-\$37.55=\$62.45$. 2. b) $\$73.20-\$28.45=\$44.75$. 3. c) Add the subtrahend and difference to find the minuend: $\$45.50+\$19.99=\$65.49$.

Answer

a) $\$62.45$ b) $\$44.75$ c) $\$65.49$

5319265

Decimal addition and subtraction can be shown as jumps on a number line. For number lines a), b), and c), find the decimals represented by the red question marks. Give the missing values in jump order.

Hints

- Line up decimal places carefully when adding or subtracting. - A positive jump moves to the right; a negative jump moves to the left. - When working backward, use the inverse operation for each jump. - Check your answers by following the jumps forward in the order shown.

Solution

1. For a), start at $2.8$: $2.8+1.5=4.3$, then $4.3+0.9=5.2$. 2. For b), start at $7.3$: $7.3-2.6=4.7$, then $4.7-1.8=2.9$. 3. For c), work backward from $5.5$. Undo the final jump of $-1.6$: $5.5+1.6=7.1$. Then undo the jump of $+3.4$: $7.1-3.4=3.7$.

Answer

a) $4.3$, $5.2$ b) $4.7$, $2.9$ c) $3.7$, $7.1$

5351395

Three metal rails, $A$, $B$, and $C$, are shown above centimeter scales. a) Find the length of rail $A$. b) Find the length of rail $B$. c) Which rail is exactly $1.25\,\text{cm}$ longer than rail $C$? Show the subtraction. d) What is the total length of rails $A$ and $C$?

Hints

- Subtract the starting value from the ending value to find a rail’s length. - Each small interval represents $0.25\,\text{cm}$. - For part c), find rail C’s length before comparing. - For part d), add the two rail lengths.

Solution

1. a) Rail A runs from $0.5$ to $3.25$, so its length is $3.25-0.5=2.75\,\text{cm}$. 2. b) Rail B runs from $1.25$ to $5.0$, so its length is $5.0-1.25=3.75\,\text{cm}$. 3. Rail C runs from $2.75$ to $5.25$, so its length is $5.25-2.75=2.5\,\text{cm}$. 4. c) $3.75-2.5=1.25\,\text{cm}$, so rail B is $1.25\,\text{cm}$ longer than rail C. 5. d) $2.75+2.5=5.25\,\text{cm}$.

Answer

a) $2.75\,\text{cm}$ b) $3.75\,\text{cm}$ c) Rail $B$, because $3.75\,\text{cm}-2.5\,\text{cm}=1.25\,\text{cm}$ d) $5.25\,\text{cm}$

5106945

Lucas shops for a class party: $3$ packs of napkins at $\$1.49$ each, one tablecloth for $\$6.95$, and $2$ packs of paper cups at $\$2.25$ each. He has store credits worth $\$5.00$, $\$2.20$, and $\$3.50$. a) Estimate by rounding to whole dollars whether the store credits will cover the entire purchase. b) Calculate the exact amount Lucas still owes, or the unused store-credit balance if the credits exceed the cost. c) After all items and credits are entered, the register shows that Lucas owes $\$3.27$. What likely data-entry error involving the $\$6.95$ tablecloth would explain that amount?

Hints

- Round the item prices and store credits to whole dollars for part a). - Find the exact purchase total and exact credit total separately. - Compare the correct amount owed with $\$3.27$. - See whether that difference matches a plausible error in the tablecloth price.

Solution

1. For a), estimate the purchase as $3\times\$1+\$7+2\times\$2=\$14$. Estimate the store credits as $\$5+\$2+\$4=\$11$. The credits should not cover the full purchase. 2. For b), the exact purchase total is $3\times\$1.49+\$6.95+2\times\$2.25=\$15.92$. 3. The store credits total $\$5.00+\$2.20+\$3.50=\$10.70$. 4. Lucas still owes $\$15.92-\$10.70=\$5.22$. 5. For c), the difference between the correct amount and the register amount is $\$5.22-\$3.27=\$1.95$. 6. Entering the tablecloth as $\$5.00$ instead of $\$6.95$ lowers the purchase total by exactly $\$1.95$, which explains the displayed $\$3.27$.

Answer

a) No. The purchase is about $\$14$, while the credits total about $\$11$. b) Lucas still owes $\$5.22$. c) The tablecloth was likely entered as $\$5.00$ instead of $\$6.95$.

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