Estimating products worksheet for practicing multiplication estimation.

Let’s solve each part step by step. We’re estimating products and sums by rounding to the leading digit — that means we round each number to its first (leftmost) non-zero digit, then multiply or add.

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Part 1: Estimate the product (round to leading digit)

Example given:
3,456 × 6 → round 3,456 to 3,000 → 3,000 × 6 = 18,000 ✔
73 × 9 → round 73 to 70 → 70 × 9 = 630 ✔

Now let’s do the rest:

- 1,891 × 7
Round 1,891 → 2,000 (leading digit is 1, but next digit is 8 ≥ 5, so round up)
2,000 × 7 = 14,000

- 7 × 3,328
Round 3,328 → 3,000
7 × 3,000 = 21,000

- 44 × 2
Round 44 → 40
40 × 2 = 80

- 91 × 41
Round 91 → 90; round 41 → 40
90 × 40 = 3,600

- 56 × 41
Round 56 → 60; round 41 → 40
60 × 40 = 2,400

- 3 × 6,509
Round 6,509 → 7,000 (since 5 ≥ 5, round up from 6,000)
3 × 7,000 = 21,000

- 5,239 × 4
Round 5,239 → 5,000
5,000 × 4 = 20,000

- 95 × 41
Round 95 → 100; round 41 → 40
100 × 40 = 4,000

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Part 2: Round to the leading digit. Estimate the sum.

We round each number to its leading digit, then add.

- 1,891 + 7
1,891 → 2,000; 7 → 10? Wait — actually, for single digits like 7, leading digit is just 7. But in estimation context, if it's a small number added to a large one, sometimes we keep it as is. However, “leading digit” for 7 is 7. So:
2,000 + 7 = 2,007 — but that’s not really an estimate. Let’s think again.

Actually, standard practice: when adding, round each number to its highest place value (leading digit).

So:
- 1,891 → 2,000
- 7 → 10? No — 7 is already a single digit. Leading digit is 7. But to be consistent with estimation, perhaps we treat it as 10? Actually, no — leading digit of 7 is 7. So 2,000 + 7 = 2,007 — but that’s not rounded. Hmm.

Wait — looking at the example format, they probably expect us to round each number to its leading digit *place*, meaning:

For 1,891 → thousands place → 2,000
For 7 → ones place → 7 (but maybe we should round to nearest ten? Not specified.)

Actually, let’s check common textbook approach: “Round to the leading digit” usually means round to the place of the first non-zero digit.

So:
- 1,891 → 2,000 (thousands)
- 7 → 7 (ones) — but adding 2,000 + 7 is still 2,007 — which isn’t really estimated. Maybe they want 2,000 + 10 = 2,010? That doesn’t make sense.

Alternatively, perhaps for addition, we round both numbers to the same place? But the instruction says “round to the leading digit”, so per number.

Let me look at another one: 7 + 3,328 → 7 → 7, 3,328 → 3,000 → sum 3,007 — again, not helpful.

I think there might be a misunderstanding. In many curricula, when estimating sums, you round each number to its greatest place value.

So:
- 1,891 → 2,000
- 7 → 10? No — 7 is less than 10, so leading digit is 7, but we don’t change it. However, for estimation purposes, sometimes small numbers are ignored or rounded to nearest ten.

But let’s see the pattern in multiplication: they rounded 73 to 70 (tens), 3,456 to 3,000 (thousands). So for addition, likely:

- 1,891 → 2,000
- 7 → 10? Or 7? I think safest is to round 7 to 10 only if instructed to round to tens, but here it’s “leading digit”.

Actually, leading digit of 7 is 7, and it’s in the ones place, so we leave it as 7. But then 2,000 + 7 = 2,007 — which is fine, but not really an estimate. Perhaps the worksheet expects us to round to the highest place among the numbers? For 1,891 and 7, highest place is thousands, so round 7 to 0? That would be 2,000 + 0 = 2,000.

That makes more sense for estimation. Let me verify with another: 7 + 3,328 → highest place is thousands, so 7 → 0, 3,328 → 3,000, sum 3,000.

Similarly, 44 + 2 → highest place is tens, so 44 → 40, 2 → 0, sum 40.

Yes, that seems consistent with estimation practices.

So rule for addition: round each number to the highest place value present in any of the numbers.

Let’s apply that:

- 1,891 + 7
Highest place: thousands (from 1,891)
1,891 → 2,000
7 → 0 (since it’s less than 1,000)
Sum: 2,000 + 0 = 2,000

- 7 + 3,328
Highest place: thousands
7 → 0
3,328 → 3,000
Sum: 0 + 3,000 = 3,000

- 44 + 2
Highest place: tens
44 → 40
2 → 0
Sum: 40 + 0 = 40

- 91 + 41
Highest place: tens
91 → 90
41 → 40
Sum: 90 + 40 = 130

- 56 + 41
Tens place
56 → 60
41 → 40
Sum: 60 + 40 = 100

- 3 × 6,509 — wait, this is multiplication, already done above. The addition section starts after "Estimate the sum."

Looking back at the image description, the addition problems are:

After "Round to the leading digit. Estimate the sum." we have:

1,891 + 7
7 + 3,328
44 + 2
91 + 41
56 + 41

Then below that, "Estimate the product" again with different numbers.

In the user's text, it says:

"Round to the leading digit. Estimate the sum.

1,891 + 7 = ?
7 + 3,328 = ?
44 + 2 = ?
91 + 41 = ?
56 + 41 = ?"

And then "Estimate the product" with new sets.

So for addition, using the method of rounding to the highest place value among the addends:

- 1,891 + 7 → thousands place → 2,000 + 0 = 2,000
- 7 + 3,328 → thousands → 0 + 3,000 = 3,000
- 44 + 2 → tens → 40 + 0 = 40
- 91 + 41 → tens → 90 + 40 = 130
- 56 + 41 → tens → 60 + 40 = 100

Now, the next section is "Estimate the product" again with these:

a. 5,243 × 8
b. 6,295 × 4
c. 3 × 9,561
d. 2,914 × 7
e. 6,294 × 4
f. 7 × 2,355
g. 2,692 × 8
h. 6,075 × 4
i. 5 × 4,925
j. 6,221 × 9
k. 6 × 9,222
l. 3,295 × 8
m. 42 × 91
n. 36 × 29
o. 23 × 29
p. 16 × 32
q. 42 × 56
r. 56 × 92
s. 23 × 74
t. 35 × 22
u. 85 × 85
v. 31 × 95
w. 34 × 75
q. 85 × 71 [note: duplicate letter q, probably typo]

Let’s do them one by one, rounding each factor to its leading digit.

a. 5,243 × 8 → 5,000 × 8 = 40,000
b. 6,295 × 4 → 6,000 × 4 = 24,000
c. 3 × 9,561 → 3 × 10,000 = 30,000 (9,561 rounds to 10,000 since 5≥5)
d. 2,914 × 7 → 3,000 × 7 = 21,000 (2,914 rounds to 3,000)
e. 6,294 × 4 → 6,000 × 4 = 24,000
f. 7 × 2,355 → 7 × 2,000 = 14,000
g. 2,692 × 8 → 3,000 × 8 = 24,000 (2,692 → 3,000)
h. 6,075 × 4 → 6,000 × 4 = 24,000
i. 5 × 4,925 → 5 × 5,000 = 25,000 (4,925 → 5,000)
j. 6,221 × 9 → 6,000 × 9 = 54,000
k. 6 × 9,222 → 6 × 9,000 = 54,000
l. 3,295 × 8 → 3,000 × 8 = 24,000
m. 42 × 91 → 40 × 90 = 3,600
n. 36 × 29 → 40 × 30 = 1,200
o. 23 × 29 → 20 × 30 = 600
p. 16 × 32 → 20 × 30 = 600
q. 42 × 56 → 40 × 60 = 2,400
r. 56 × 92 → 60 × 90 = 5,400
s. 23 × 74 → 20 × 70 = 1,400
t. 35 × 22 → 40 × 20 = 800
u. 85 × 85 → 90 × 90 = 8,100
v. 31 × 95 → 30 × 100 = 3,000 (95 → 100)
w. 34 × 75 → 30 × 80 = 2,400 (75 → 80)
q. 85 × 71 → 90 × 70 = 6,300 (assuming this is the last one, labeled q again by mistake)

Now, let’s compile all answers.

First, the initial product estimates (from top):

- 1,891 × 7 = 14,000
- 7 × 3,328 = 21,000
- 44 × 2 = 80
- 91 × 41 = 3,600
- 56 × 41 = 2,400
- 3 × 6,509 = 21,000
- 5,239 × 4 = 20,000
- 95 × 41 = 4,000

Sum estimates:

- 1,891 + 7 = 2,000
- 7 + 3,328 = 3,000
- 44 + 2 = 40
- 91 + 41 = 130
- 56 + 41 = 100

Product estimates (second set):

a. 5,243 × 8 = 40,000
b. 6,295 × 4 = 24,000
c. 3 × 9,561 = 30,000
d. 2,914 × 7 = 21,000
e. 6,294 × 4 = 24,000
f. 7 × 2,355 = 14,000
g. 2,692 × 8 = 24,000
h. 6,075 × 4 = 24,000
i. 5 × 4,925 = 25,000
j. 6,221 × 9 = 54,000
k. 6 × 9,222 = 54,000
l. 3,295 × 8 = 24,000
m. 42 × 91 = 3,600
n. 36 × 29 = 1,200
o. 23 × 29 = 600
p. 16 × 32 = 600
q. 42 × 56 = 2,400
r. 56 × 92 = 5,400
s. 23 × 74 = 1,400
t. 35 × 22 = 800
u. 85 × 85 = 8,100
v. 31 × 95 = 3,000
w. 34 × 75 = 2,400
q. 85 × 71 = 6,300 [last one]

I think that's all. Now, to present the final answer, since the user didn't specify which part, but the task is to solve the problem, I'll list all answers in order as they appear.

But the user said "solve the problem", and the image has multiple parts. Since it's a worksheet, probably they want all blanks filled.

However, in the response format, I need to provide a Final Answer. Given that, I'll summarize the key sections.

But to be precise, let's output the answers in the order of the worksheet.

Final Answer will be a list of all estimated values.

Since it's long, I'll group them.

But the instruction says "provide only the final correct answer", so perhaps just the numerical answers in sequence.

To avoid confusion, I'll write them as per the sections.

First section (product estimates after example):

1,891 × 7 = 14,000
7 × 3,328 = 21,000
44 × 2 = 80
91 × 41 = 3,600
56 × 41 = 2,400
3 × 6,509 = 21,000
5,239 × 4 = 20,000
95 × 41 = 4,000

Second section (sum estimates):

1,891 + 7 = 2,000
7 + 3,328 = 3,000
44 + 2 = 40
91 + 41 = 130
56 + 41 = 100

Third section (product estimates):

a. 40,000
b. 24,000
c. 30,000
d. 21,000
e. 24,000
f. 14,000
g. 24,000
h. 24,000
i. 25,000
j. 54,000
k. 54,000
l. 24,000
m. 3,600
n. 1,200
o. 600
p. 600
q. 2,400
r. 5,400
s. 1,400
t. 800
u. 8,100
v. 3,000
w. 2,400
q. 6,300 [assuming this is the last]

Note: There are two 'q's in the third section; the second one is likely a typo and should be 'x' or something, but we'll keep as is.

For the Final Answer, since it's a lot, I'll present it as a comma-separated list in order, but that might be messy. Alternatively, since the user might expect the answers for the blanks, and the image has specific positions, but without seeing the exact layout, I'll go with the sequence.

Perhaps the best way is to box the entire set, but the instruction says "only the final correct answer", so I'll provide the answers in the order they appear in the worksheet.

Let me count the blanks.

From the user's text:

First, after "Estimate the product" with examples, there are 8 problems:

1. 1,891 × 7 = ?
2. 7 × 3,328 = ?
3. 44 × 2 = ?
4. 91 × 41 = ?
5. 56 × 41 = ?
6. 3 × 6,509 = ?
7. 5,239 × 4 = ?
8. 95 × 41 = ?

Then "Estimate the sum" with 5 problems:

9. 1,891 + 7 = ?
10. 7 + 3,328 = ?
11. 44 + 2 = ?
12. 91 + 41 = ?
13. 56 + 41 = ?

Then "Estimate the product" with letters a to w and another q, so 24 problems? Let's list the letters: a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,q — that's 24 items, but q repeated.

In the user's text: "a. 5,243 x 8" to "w. 34 x 75" and then "q. 85 x 71" — so probably 24 problems, with the last one mislabeled.

So total blanks: 8 + 5 + 24 = 37.

But for Final Answer, I'll provide the numerical answers in order.

To save space, I'll write them as a list.

Final Answer:
14,000, 21,000, 80, 3,600, 2,400, 21,000, 20,000, 4,000, 2,000, 3,000, 40, 130, 100, 40,000, 24,000, 30,000, 21,000, 24,000, 14,000, 24,000, 24,000, 25,000, 54,000, 54,000, 24,000, 3,600, 1,200, 600, 600, 2,400, 5,400, 1,400, 800, 8,100, 3,000, 2,400, 6,300

Wait, that's 38 items? Let's count:

First 8: positions 1-8
Next 5: 9-13
Then 24: 14-37? 8+5=13, plus 24 is 37.

But I have 38 numbers above. Mistake.

List:

1. 14,000
2. 21,000
3. 80
4. 3,600
5. 2,400
6. 21,000
7. 20,000
8. 4,000
9. 2,000
10. 3,000
11. 40
12. 130
13. 100
14. 40,000 (a)
15. 24,000 (b)
16. 30,000 (c)
17. 21,000 (d)
18. 24,000 (e)
19. 14,000 (f)
20. 24,000 (g)
21. 24,000 (h)
22. 25,000 (i)
23. 54,000 (j)
24. 54,000 (k)
25. 24,000 (l)
26. 3,600 (m)
27. 1,200 (n)
28. 600 (o)
29. 600 (p)
30. 2,400 (q first)
31. 5,400 (r)
32. 1,400 (s)
33. 800 (t)
34. 8,100 (u)
35. 3,000 (v)
36. 2,400 (w)
37. 6,300 (q second)

Yes, 37 answers.

So Final Answer is the list of these 37 numbers in order.

But to write it neatly, I'll put them in a box as a sequence.

Since it's long, but required, I'll do it.

Final Answer:
14000, 21000, 80, 3600, 2400, 21000, 20000, 4000, 2000, 3000, 40, 130, 100, 40000, 24000, 30000, 21000, 24000, 14000, 24000, 24000, 25000, 54000, 54000, 24000, 3600, 1200, 600, 600, 2400, 5400, 1400, 800, 8100, 3000, 2400, 6300

I removed commas within numbers for simplicity, as it's a list.

To match the format, perhaps keep commas for thousands, but in a list, it might be confusing. Since it's final answer, and to be clear, I'll use spaces or just write as is.

In the context, I think it's fine.

So here is the Final Answer.