Back to the on-screen lesson ·
Working out what to do first, and checking the answer against an estimate.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson you solve problems that need more than one operation, deciding what has to be worked out first before the question can be answered. Then you check: round the numbers, redo it roughly, and see whether your exact answer has a reasonable size. An estimate can expose large mistakes; an exact relationship check and a rereading of the conditions are still needed.
You can calculate sums, differences, products and whole-number quotients. You also know that a remainder needs an interpretation. In this lesson the first challenge is choosing a relationship, not doing the arithmetic. Read a short situation and say what each number measures. Then name the quantity the question asks for. If that quantity depends on another unknown, work out the needed intermediate quantity first and label it before continuing.
| Term | What it means |
|---|---|
| Unknown | A quantity whose value you need to find. |
| Intermediate result | A quantity calculated on the way to the final answer. |
| Expression | Numbers or symbols connected by operations. |
| Equation | A statement that two expressions have the same value. |
| Constraint | A condition a valid answer must satisfy. |
| Reasonableness check | A comparison with the context, a bound or an estimate that can expose a mistake. |
A multistep problem contains relationships that depend on one another. If four boxes contain twenty-four pencils each and eighteen pencils are reserved, the shareable total cannot be read directly from the story. First the boxes supply 4 × 24 = 96 pencils. Removing the reserve leaves 96 - 18 = 78 pencils. Only then can you calculate an equal share among six groups: 78 divided by 6 = 13 pencils per group.
Each intermediate result has a job and a unit. Ninety-six is the total before reserving pencils; seventy-eight is the shareable total; thirteen is the amount per group. Writing only a row of unexplained answers makes it easy to stop early or use the wrong number in the next step. A good solution connects each operation to its relationship and ends by answering the exact question asked.
Another way: steps
Identify the unknown, model the relationships, calculate needed quantities in order, interpret any remainder, and check the complete answer.
For the pencil problem, draw four equal boxes and write twenty-four inside each. Under the four boxes, draw a single bar representing all ninety-six pencils. Split that bar into a reserved part of eighteen and a shareable part. Then split only the shareable part into six equal sections. The drawing shows why dividing the original total before removing the reserve would describe a different situation.
Your sketch does not have to show ninety-six individual pencil marks. It needs to show the relationships accurately. Equal boxes represent equal quantities, a split bar represents parts of one total, and six equal sections represent sharing. Label every part you know and place a question mark on the final unknown. If a diagram becomes cluttered, use a short chain of labeled equations instead. The purpose is to preserve the meaning of the problem while you calculate.
Sometimes it is easier to plan from the final question. Suppose the question asks how many full bags can be made after some items are removed. To answer, you need the number of available items and the number per bag. If the available number is itself unknown, identify how to calculate it from the starting amount and the removed part. Planning backward identifies dependencies even though the arithmetic usually proceeds forward.
For example, six cartons contain thirty-five cards each, and twenty-two damaged cards are discarded. Bags need eight usable cards. First find 210 cards in the cartons. Then find 188 usable cards. Finally divide 188 by eight to get twenty-three full bags and four loose cards. The number twenty-three answers the full-bag question. Reporting 188 would stop at an intermediate result, while reporting twenty-four would count a partial bag that does not meet the full-bag condition.
Let p represent the number of pencils each group receives in the earlier problem. A relationship that describes the whole situation is 6 × p + 18 = 4 × 24. Six equal shares together with the reserve account for all four boxes. You can solve it by calculating the total, subtracting the reserve and dividing the remainder among the six groups.
After finding p = 13, substitute it into the original relationship: 6 × 13 + 18 = 96 and 4 × 24 = 96. Both sides agree. This check tests more than a single multiplication fact; it tests whether the proposed share fits the original conditions. Letters stand for quantities, not object names. State 'p is pencils per group' so the symbol's unit remains clear. You do not need advanced equation rules to use a letter as a useful placeholder.
A story says, 'The blue team collected eighteen more cans than the red team. The blue team collected seventy-five cans.' How many cans did the red team collect? The word more appears, but the unknown is the smaller amount. Since red plus eighteen equals seventy-five, subtraction gives fifty-seven. If the red team's count were given instead, finding the blue team's count would require addition.
In a longer problem, the same care is needed at every stage. If both teams combine their cans after the comparison, you would add seventy-five and fifty-seven, giving 132. Choosing addition immediately because you saw more would produce an incorrect first result and contaminate the next step. Describe which quantity is the whole, which is a part, or which is the larger comparison amount before selecting an operation.
The pencil-sharing calculation can be recorded as (4 × 24 - 18) divided by 6. The parentheses show that the available total must be found before sharing. Another correct record uses separate equations: 4 × 24 = 96; 96 - 18 = 78; 78 divided by 6 = 13. Either record is useful if its meanings are clear.
Avoid writing a chain such as 4 × 24 = 96 - 18 = 78 divided by 6 = 13. The equals sign says the expressions on each side have the same value, but ninety-six and thirteen are not equal. Use a new line, a semicolon or explanatory words when moving to a different calculation. A mathematically correct result deserves a mathematically correct record. The notation should show which quantities are equal and which new operations change them.
A bus company has small vehicles with eight seats available for passengers. Three groups contain seventeen, nineteen and twenty-two passengers. Together they contain fifty-eight passengers. Seven vehicles would supply fifty-six seats, leaving two passengers without seats, so eight vehicles are needed under the assumption that groups may mix across vehicles.
If each group must travel separately, the answer changes. Seventeen passengers need three vehicles, nineteen need three, and twenty-two need three, for nine vehicles altogether. The combined-total calculation was not arithmetically wrong; it used a different grouping permission. A complete solution identifies the rule that allows or forbids combining remainders. This is why rereading the conditions after computing is essential. The story's organization can matter as much as its numbers.
A film lasts one hour and thirty-five minutes, followed by a twenty-minute discussion. You can convert the hour to sixty minutes, giving ninety-five minutes for the film and 115 minutes altogether. Alternatively, add twenty to the thirty-five minutes and keep the one hour, giving one hour fifty-five minutes. These are equivalent descriptions of the same duration.
You cannot add one hour directly to twenty minutes and call the result twenty-one minutes. Convert or keep separate unit labels until they match. If a total duration exceeds sixty minutes, regroup complete groups of sixty as hours. For 145 minutes, two hours account for 120 minutes and twenty-five minutes remain. The division remainder is a number of minutes, not a decimal digit to attach to the hour count. Two hours twenty-five minutes is not 2.25 hours.
Suppose five notebooks cost seven dollars each and you pay with fifty dollars. The cost is thirty-five dollars and the change is fifteen dollars. Before calculating exactly, you know the change must be less than fifty dollars and nonnegative if the payment is sufficient. You also know the cost lies between five times six and five times eight, or thirty and forty dollars, so the change lies between ten and twenty dollars.
These bounds reject an answer such as sixty-five dollars immediately. An estimate of about fifteen dollars supports the exact answer but cannot certify it: fourteen dollars would also fit the broad bounds. Check the exact relationship 35 + 15 = 50. Different checks have different strength. State what each one establishes instead of claiming that any reasonable-looking number must be correct.
A club has six boxes of markers and gives nine markers to each table. How many tables can receive markers? This cannot be determined unless you know the number of markers in each box or the total marker count. Multiplying six by nine would combine a box count with a per-table count without a relationship connecting them. Missing information cannot be repaired by choosing any available operation.
Other stories include information you do not need. If every box has twenty-four markers, its color is irrelevant to the total unless color matching is a condition. Identify each number's role, but do not force every number into a calculation. When you finish, explain why the data you used answer the question and name any assumption that matters. A short, justified plan is more reliable than a long string of operations that merely uses all the numbers.
A learner solves the four-box pencil problem by subtracting eighteen from twenty-four, multiplying by four and dividing by six. The arithmetic gives four pencils per group. Ask what the first subtraction means. It reserves eighteen pencils from every box, or seventy-two pencils altogether, but the story reserves eighteen only once. The mistake is a mismatch between the modeled relationship and the condition.
Repair the plan by combining the boxes before removing the single reserve. Then substitute the proposed thirteen pencils per group into the original story: six groups use seventy-eight, and the eighteen reserved pencils bring the total back to ninety-six. This review identifies a precise error and supplies an exact check. It also shows why checking arithmetic alone is insufficient when the wrong sequence of operations describes a different situation.
Three classes each need twenty-eight folders. A cupboard contains nineteen usable folders, and new folders are sold only in packs of ten. First calculate the demand: three times twenty-eight is eighty-four folders. Subtract the nineteen available to find sixty-five more needed. Six packs supply only sixty, so seven packs must be purchased. The school will then have eighty-nine folders altogether and five spares after meeting the demand. This spare count checks the decision and makes the consequence visible. Buying six packs would be cheaper but would fail the requirement. The plan assumes all cupboard folders are suitable and packs can be shared across classes. If each class must receive unopened packs, the conditions change and the calculation must be reconsidered.
A workshop includes a ninety-five-minute making session, a twenty-minute break and a thirty-minute exhibition. Add the durations to obtain 145 minutes, then regroup sixty-minute units to report two hours twenty-five minutes. If the workshop begins at one o'clock in the afternoon with no gaps beyond the stated break, it ends at three twenty-five. A schedule check adds the parts in sequence: making ends at two thirty-five, the break at two fifty-five, and the exhibition at three twenty-five. These two routes agree. The end time depends on the stated start time, while the duration does not. Keeping that distinction clear prevents treating a clock reading as an amount of time that can be added without considering its reference point.
Label every intermediate result and check its role. Keywords do not determine an operation by themselves. An estimate can reject unreasonable results but cannot prove every nearby answer correct. Conditions about whole items, full packs and separate groups determine how a remainder is used.
Identify the final unknown.
Pencils per group
The question asks for an equal share after a reserve.
Find the starting total.
4 × 24 = 96 pencils
Four equal boxes supply the whole amount.
Remove the single reserve.
96 - 18 = 78 pencils
Only these pencils may be shared.
Share among six groups.
78 divided by 6 = 13
Every group receives the same amount.
Check the full relationship.
6 × 13 + 18 = 96
The shares and reserve account for every pencil.
Identify the purchase quantities.
5 notebooks at 7 dollars each; 50 dollars paid
The unknown is money returned.
Calculate the total cost.
5 × 7 = 35 dollars
Every notebook has the same price.
Check the payment is enough.
50 ≥ 35
Change is possible under these conditions.
Find the money returned.
50 - 35 = 15 dollars
Cost and change must equal the payment.
Verify and bound the result.
35 + 15 = 50; change is between 10 and 20 dollars
The exact check and rough bound agree.
Read the separation condition.
17, 19 and 22 passengers; 8 seats per vehicle; groups travel separately
Unused seats cannot be transferred between groups.
Allocate the first group.
17 divided by 8 is 2 remainder 1: 3 vehicles
One additional vehicle holds the remaining passenger.
Allocate the second group.
19 divided by 8 is 2 remainder 3: 3 vehicles
Two vehicles cannot seat this group.
Allocate the third group.
22 divided by 8 is 2 remainder 6: 3 vehicles
A third vehicle is again necessary.
Combine the vehicle counts.
3 + 3 + 3 = 9 vehicles
The independently required fleets are added.
Check why a combined calculation differs.
58 passengers could share 8 vehicles only if mixing were allowed
The separation condition rules out that smaller plan.
Find the starting quantity.
6 × 35 = 210 cards
Each carton contributes thirty-five.
Remove unusable cards.
210 - 22 = 188 cards
Damaged cards cannot enter bags.
Divide into full bags.
188 divided by 8 is 23 remainder 4
Each complete bag needs eight usable cards.
Interpret the requested output.
Check all parts of the record.
Find the number x for which 5 × x = 28 + 107. Calculate the total before dividing it into equal parts.
Answer:
5 boxes each contain 24 pencils; 18 pencils are reserved. Find the shareable total.
Calculate the combined starting amount.
5 × 24 = 120 pencils.
All boxes contribute before the reserve is taken.
Remove the reserve once.
The shareable amount is p pencils.
The reserve belongs to the entire collection, not each box.
Check both parts against the whole.
Shareable pencils plus 18 must equal 120.
The two parts account for all supplied pencils.
Find x in the equation 3 × 12 + x = 62.
Answer:
Find the number x for which 6 × x = 27 + 147. Calculate the total before dividing it into equal parts.
Answer:
Find x in the equation 4 × 8 + x = 39.
Answer:
Find the number x for which 4 × x = 32 + 80. Calculate the total before dividing it into equal parts.
Answer:
A maker opens 7 packs of 35 tiles. After reserving 18 tiles for repairs, how many tiles remain for the design?
Answer:
A session lasts 271 minutes and its discussion lasts 24 minutes. Report the combined duration in whole hours and remaining minutes.
Hours h; minutes m
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A session lasts 185 minutes and its discussion lasts 35 minutes. Report the combined duration in whole hours and remaining minutes.
Hours h; minutes m
You can solve a problem that takes two or more steps and check it. Without looking: what do you work out first, and how would an estimate tell you the answer was wrong?
21. Make full bags after removing damaged cards, step 4
23 full bags, with 4 cards left
The leftover cards cannot fill another complete bag.
21. Make full bags after removing damaged cards, step 5
23 × 8 + 4 + 22 = 210
Bagged, loose and damaged cards account for the total.