## Engage NY Eureka Math 2nd Grade Module 6 Lesson 15 Answer Key

### Eureka Math Grade 2 Module 6 Lesson 15 Sprint Answer Key

A. Subtract Crossing the Ten.  Question 1.
10 – 1 =
9

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 10 – 1. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 1 = 9. Hence the answer is 9.

Question 2.
10 – 2 =
8

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 10 – 8. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 2 = 8. Hence the answer is 8.

Question 3.
20 – 2 =
18

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 20 – 2. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 2 = 8. Hence the answer is 18.

Question 4.
40 – 2 =
38

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 40 – 2. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 2 = 8. Hence the answer is 38.

Question 5.
10 – 2 =
8

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 10 – 2. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 2 = 8. Hence the answer is 8.

Question 6.
11 – 2 =
9

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 11 – 2. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 2 = 9. Hence the answer is 9.

Question 7.
21 – 2 =
19

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 21 – 2. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 2 = 9. Hence the answer is 19.

Question 8.
51 – 2=
69

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 51 – 2. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 2 = 9. Hence the answer is 69.

Question 9.
10 – 3 =
7

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 10 – 3. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 3 = 7. Hence the answer is 7.

Question 10.
11 – 3 =
8

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 11 – 3. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 3 = 8. Hence the answer is 8.

Question 11.
21 – 3 =
18

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 21 – 3. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 3 = 8. Hence the answer is 18.

Question 12.
61 – 3 =
58

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 61 – 3. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 3 = 8. Hence the answer is 58.

Question 13.
10 – 4 =
6

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 10 – 4. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 4 = 6. Hence the answer is 6.

Question 14.
11 – 4 =
7

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 11 – 4. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 4 = 7. Hence the answer is 7.

Question 15.
21 – 4 =
17

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 21 – 4. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 4 = 7. Hence the answer is 7.

Question 16.
71 – 4 =
67

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 71 – 4. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 4 = 7. Hence the answer is 67.

Question 17.
10 – 5 =
5

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 10 – 5. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 5 = 5. Hence the answer is 5.

Question 18.
11 – 5 =
6

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 11 – 5. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 5 = 6. Hence the answer is 6.

Question 19.
21 – 5 =
16

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 21 – 5. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 5 = 6. Hence the answer is 16.

Question 20.
81 – 5 =
76

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 81 – 5. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 5 = 6. Hence the answer is 76.

Question 21.
10 – 6 =
4

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 10 – 4. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 6 = 4. Hence the answer is 4.

Question 22.
11 – 6 =
5

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 11 – 6. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 6 = 5. Hence the answer is 5.

Question 23.
21 – 6 =
15

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 21 – 6. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 6 = 5. Hence the answer is 15.

Question 24.
91 – 6 =
85

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 91 – 6. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 6 = 5. Hence the answer is 85.

Question 25.
10 – 7 =
3

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 10 – 7. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 7 = 3. Hence the answer is 3.

Question 26.
11 – 7 =
4

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 11 – 7. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 7 = 4. Hence the answer is 4.

Question 27.
31 – 7 =
24

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 31 – 7. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 7 = 4. Hence the answer is 24.

Question 28.
10 – 8 =
2

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 10 – 8. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 8 = 2. Hence the answer is 2.

Question 29.
11 – 8 =
3

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 11 – 8. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 11 – 8 = 3. Hence the answer is 3.

Question 30.
41 – 8 =
33

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 41 – 8. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 8 = 3. Hence the answer is 33.

Question 31.
10 – 9 =
1

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 10 – 9. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 9 = 1. Hence the answer is 1.

Question 32.
11 – 9 =
2

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 11 – 9. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 9 = 2. Hence the answer is 2.

Question 33.
51 – 9 =
42

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 51 – 9. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 9 = 2. Hence the answer is 42.

Question 34.
12 – 3 =
9

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 12 – 3. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 2 – 3 = 9. Hence the answer is 9.

Question 35.
82 – 3 =
79

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 82 – 3. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 2 – 3 = 9. Hence the answer is 79.

Question 36.
13 – 5 =
8

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 13 – 5. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 3 – 5 = 8. Hence the answer is 8.

Question 37.
73 – 5 =
68

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 73 – 5. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 3 – 5 = 8. Hence the answer is 68.

Question 38.
14 – 6 =
8

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 14 – 6. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 4 – 6 = 8. Hence the answer is 8.

Question 39.
84 – 6 =
78

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 84 – 6. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 4 – 6 = 8. Hence the answer is 78.

Question 40.
15 – 8 =
7

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 15 – 8. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 5 – 8 = 7. Hence the answer is 7.

Question 41.
95 – 8 =
87

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 95 – 8. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 5 – 8 = 7. Hence the answer is 7.

Question 42.
16 – 7 =
9

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 16 – 7. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 6 – 7 = 9. Hence the answer is 9.

Question 43.
46 – 7 =
39

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 46 – 7. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 6 – 7 = 9. Hence the answer is 39.

Question 44.
68 – 9 =
59

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 68 – 9. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 8 – 9 = 9. Hence the answer is 59.

B. Subtract Crossing the Ten  Question 1.
10 – 2 =
8

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 10 – 2. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 2 = 8. Hence the answer is 8.

Question 2.
20 – 2 =
18

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 20 – 2. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 2 = 8. Hence the answer is 18.

Question 3.
30 – 2 =
28

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 30 – 2. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 2 = 8. Hence the answer is 28.

Question 4.
50 – 2 =
48

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 50 – 2. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 2 = 8. Hence the answer is 48.

Question 5.
10 – 2 =
8

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 10 – 2. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 2 = 8. Hence the answer is 8.

Question 6.
11 – 2 =
9

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 11 – 2. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 2 = 9. Hence the answer is 9.

Question 7.
21 – 2 =
19

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 21 – 2. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 2 = 9. Hence the answer is 19.

Question 8.
61 – 2 =
59

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 61 – 2. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 2 = 9. Hence the answer is 59.

Question 9.
10 – 3 =
7

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 10 – 3. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 3 = 7. Hence the answer is 7.

Question 10.
11 – 3 =
8

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 11 – 3. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 3 = 8. Hence the answer is 8.

Question 11.
21 – 3 =
19

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 10 – 3. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 3 = 7. Hence the answer is 7.

Question 12.
71 – 3 =
68

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 71 – 3. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 3 = 8. Hence the answer is 68.

Question 13.
10 – 4 =
6

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 10 – 4. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 4 = 6. Hence the answer is 6.

Question 14.
11 – 4 =
7

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 11 – 4. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 4 = 7. Hence the answer is 7.

Question 15.
21 – 4 =
17

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 21 – 4. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 4 = 7. Hence the answer is 17.

Question 16.
81 – 4 =
77

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 81 – 4. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 4 = 7. Hence the answer is 7.

Question 17.
10 – 5 =
5

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 10 – 5. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 5 = 5. Hence the answer is 5.

Question 18.
11 – 5 =
6

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 11 – 5. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 5 = 6. Hence the answer is 6.

Question 19.
21 – 5 =
16

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 21 – 5. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 5 = 6. Hence the answer is 16.

Question 20.
91 – 5 =
86

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 91 – 5. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 5 = 6. Hence the answer is 86.

Question 21.
10 – 6 =
4

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 10 – 6. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 6 = 4. Hence the answer is 4.

Question 22.
11 – 6 =
5

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 11 – 6. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 6 = 5. Hence the answer is 5.

Question 23.
21 – 6 =
15

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 21 – 6. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 6 = 5. Hence the answer is 15.

Question 24.
41 – 6 =
35

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 41 – 6. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 6 = 5. Hence the answer is 35.

Question 25.
10 – 7 =
3

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 10 – 7. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 7 = 3. Hence the answer is 3.

Question 26.
11 – 7 =
4

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 11 – 7. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 7 = 4. Hence the answer is 4.

Question 27.
51 – 7 =
44

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 51 – 7. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 7 = 4. Hence the answer is 44.

Question 28.
10 – 8 =
2

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 10 – 8. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 8 = 2. Hence the answer is 2.

Question 29.
11 – 8 =
3

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 11 – 8. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 8 = 3. Hence the answer is 3.

Question 30.
61 – 8 =
53

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 61 – 8. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 8 = 3. Hence the answer is 53.

Question 31.
10 – 9 =
1

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 10 – 9. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 – 9 = 1. Hence the answer is 1.

Question 32.
11 – 9 =
2

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 11 – 9. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 9 = 2. Hence the answer is 2.

Question 33.
31 – 9 =
22

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 31 – 9. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 1 – 9 = 2. Hence the answer is 22.

Question 34.
12 – 3 =
9

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 12 – 3. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 2 – 3 = 9. Hence the answer is 9.

Question 35.
92 – 3 =
89

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 92 – 3. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 2 – 3 = 9. Hence the answer is 89.

Question 36.
13 – 5 =
8

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 13 – 5. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 3 – 5 = 8. Hence the answer is 8.

Question 37.
43 – 5 =
38

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 43 – 5. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 3 – 5 = 8. Hence the answer is 38.

Question 38.
14 – 6 =
8

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 14 – 6. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 4 – 6 = 8. Hence the answer is 8.

Question 39.
64 – 6 =
58

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 64 – 6. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 4 – 6 = 8. Hence the answer is 58.

Question 40.
15 – 8 =
7

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 15 – 8. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 5 – 8 = 7. Hence the answer is 7.

Question 41.
85 – 8 =
77

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 85 – 8. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 5 – 8 = 7. Hence the answer is 77.

Question 42.
16 – 7 =
9

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 16 – 7. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 6 – 7 = 9. Hence the answer is 9.

Question 43.
76 – 7 =
69

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 76 – 7. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 6 – 7 = 9. Hence the answer is 69.

Question 44.
58 – 9 =
49

Explanation:
To solve the problem we must put the largest number on top and the smallest number at the bottom 58 – 9. Now we need to subtract the one’s digit. In this case, the subtrahend one’s digit is larger than the minuend one’s digit. Therefore, we need to “borrow” 10 from the minuend ten’s digit which gives you 10 + 8 – 9 = 9. Hence the answer is 49.

### Eureka Math Grade 2 Module 6 Lesson 15 Problem Set Answer Key

Question 1.
Shade in an array with 2 rows of 3. Write a repeated addition equation for the array.
3 + 3 = 6 Explanation:
I have shaded an array with 2 rows and then 3 columns. Hence the repeated addition equation for the new array is 3 + 3 = 6.

Question 2.
Shade in an array with 4 rows of 3. Write a repeated addition equation for the array.
3 + 3 + 3 + 3 =12 Explanation:
I have shaded an array with 4 rows of 3 as given in the question. Hence the repeated addition equation for the new array is 4 + 4 + 4 + 4 + 4 =20

Question 3.
Shade in an array with 5 columns of 4. Write a repeated addition equation for the array.
4 + 4 + 4 + 4 + 4 =20 Explanation:
I have shaded an array with 5 columns of 4 as asked in the question. Hence the repeated addition equation for the new array is 4 + 4 + 4 + 4 + 4 =20.

Question 4.
Draw one more column of 2 to make a new array. Write a repeated addition equation for the new array.
5 + 5 =10 Explanation:
I have drawn one more column of 2 to make an array. Hence the repeated addition equation for the new array is 5 + 5 = 10

Question 5.
Draw one more row of 4 and then one more column to make a new array. Write a repeated addition equation for the new array.
5 + 5 + 5 =15 Explanation:
I have drawn one more row of 4 and then one more column to make an array. Hence the repeated addition equation for the new array is 5 + 5 + 5 = 15

Question 6.
Draw one more row and then two more columns to make a new array. Write a repeated addition equation for the new array.
5 + 5 + 5 + 5 + 5 =25 Explanation:
I have drawn one more row and then two more columns to make an array. Hence the repeated addition equation for the new array is 5 + 5 + 5 + 5 + 5 =25.

### Eureka Math Grade 2 Module 6 Lesson 15 Exit Ticket Answer Key

Question 1.
Shade in an array with 3 rows of 5. Write a repeated addition equation for the array.
5 + 5 + 5 + 5 =20 Explanation:
I have shaded in an array with 3 rows of 5 as asked in the question. Hence the repeated addition equation for the new array is 5 + 5 + 5 + 5=20

### Eureka Math Grade 2 Module 6 Lesson 15 Homework Answer Key

Question 1.
Shade in an array with 3 rows of 2. Write a repeated addition equation for the array.
2 + 2 + 2 =6 Explanation:
I have shaded in with 3 rows of 2 to make an array as asked in the question. Hence the repeated addition equation for the new array is 2 + 2 + 2 = 6

Question 2.
Shade in an array with 2 rows of 4. Write a repeated addition equation for the array.
4 + 4 = 8 Explanation:
I have shaded in with 2 rows of 4 to make an array as asked in the question. Hence the repeated addition equation for the new array is 4 + 4 = 8

Question 3.
Shade in an array with 4 columns of 5. Write a repeated addition equation for the array.
5 + 5 + 5 + 5 =20 Explanation:
I have shaded in with 4 columns of 5 to make an array as asked in the question. Hence the repeated addition equation for the new array is 5 + 5 + 5 + 5=20

Question 4.
Draw one more column of 2 to make a new array. Write a repeated addition equation for the new array.
4 + 4 = 8 Explanation:
I have drawn one more column of 2 to make an array. Hence the repeated addition equation for the new array is 4 + 4 = 8

Question 5.
Draw one more row of 3 and then one more column to make a new array. Write a repeated addition equation for the new array.
5 + 5 + 5 =15 Explanation:
I have drawn one more row of 3 and then one column to make an array. Hence the repeated addition equation for the new array is 5 + 5 + 5 =15

Question 6.
Draw one more row and then two more columns to make a new array. Write a repeated addition equation for the new array. 