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Give me the problem of calculating 300 crossings in the second volume of Grade Three, 1 hour is urgent! !
730- 190=

3700-370=

8500-5500=

3200+5 100=

96-42=

1200-400=

830-300=

800+900=

58-36=

7000+3000=

5×7=

72÷9=

36÷6=

8 1÷9=

42+ 13=

24÷3=

24+8=

27-3=

42+ 13=

42-6=

7+44=

42÷7=

64-8=

2 1÷3=

4×8=

54÷9=

49÷7=

42÷7=

35÷7=

19+4 1=

24-6×3=

4×8-5=

27÷3+6=

56÷7-3=

49÷9+7=

4×3÷2=

6÷3×7=

8÷4×3=

6×5+37=

12÷4+ 18=

16÷4×8=

8÷4+48=

25÷5+36=

6×8—42=

2×8÷4=

8×9÷9=

36÷6÷3=

9÷9×6=

42÷7×6=

96-80+7=

64-40-8=

40-4+20=

7×5-8=

14÷7+25=

15÷5+8=

8×9-8=

1×2+20=

16÷8+30

7×7+8=

54÷6+33=

5 10+200=

15-45÷9=

1 1-40÷8=

7+ 1×8=

130+80=

46-8=

8×5-7=

9×3-8=

30÷6=

36÷4=

4×5+44=

48÷8+7=

1800+200=

93-80=

6×5+37=

10-8÷ 1=

7×8=

73-4=

3×8÷4=

7×6+ 10=

72÷8=

64-8=

24×0=

7 1-7+30=

6+2×8

35-9=

70-5×9=

8×9+2 1=