Question

Answer 1

`1 2/3 + 3/51 = 5/3 + 3/51 = 88/51`

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`2 4/7 - 11/42 = 18/7 - 11/42 = 108/42 - 11/42 = 97/42`

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`2 1/2 xx 1/3 + 3 1/4 = 5/2 xx 1/3 + 13/4 = 5/6 + 13/4 = 49/12`

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`3 1/2 + 1/4 : 1/6 = 7/2 + 1/4 xx 6 = 7/2 + 3/2 = 10/2 = 5`

Answer 2

a: =5/3+3/51=88/51

b: =18/7-11/42=97/42

c: =5/2x1/3+13/4

=5/6+13/4=49/12

d: \(=\dfrac{7}{2}+\dfrac{3}{2}=\dfrac{10}{2}=5\)

Answer 3

a, = 5/3 + 3/51 = 85/51 + 3/51 = 88/51

b, = 18/7 - 11/42 = 108 /42 - 11/42 = 97/42

c, = 5/2 x 1/3 + 13/4 = 5/6 + 13/4 = 10/12 + 39/12 = 49/12

d, = 7/2 + 1/4 : 1/6 = 7/2 + 3/2 = 10/2 = 5

Answer 4

`a) 1 2/3 +3/51 = 5/3 + 3/51 = 85/51 +3/51 =88/51 `

`b)2 4/7 -11/42 = 18/7 -11/42 = 108/42 -11/42 = 97/42`

`c) 2 1/2 xx 1/3 + 3 1/4 `

`= 5/2 xx 1/3 + 13/3`

`= 5/6 + 13/3 = 5/6 + 26/6 = 31/6`

`d) 3 1/2 +1/4 :1/6`

`= 7/2 + 1/4 xx 6`

`= 7/2 + 6/4 = 7/2 +3/2 = 10/2`

`=5`

Answer 5

\(a,1\dfrac{2}{3}+\dfrac{3}{51}=\dfrac{5}{3}+\dfrac{3}{51}=\dfrac{85}{51}+\dfrac{3}{51}=\dfrac{88}{51}\)

\(b,2\dfrac{4}{7}-\dfrac{11}{42}=\dfrac{18}{7}-\dfrac{11}{42}=\dfrac{108}{42}-\dfrac{11}{42}=\dfrac{97}{42}\)

\(c,2\dfrac{1}{2}x\dfrac{1}{3}+3\dfrac{1}{4}=\dfrac{5}{2}x\dfrac{1}{3}+\dfrac{13}{4}=\dfrac{5}{6}+\dfrac{13}{4}=\dfrac{49}{12}\)

d,\(3\dfrac{1}{2}+\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{7}{2}+\dfrac{1}{4}x\dfrac{6}{1}=\dfrac{7}{2}+\dfrac{6}{4}=\dfrac{20}{4}=5\)

Answer 6

\(a,1\dfrac{2}{3}+\dfrac{3}{51}=\dfrac{5}{3}+\dfrac{3}{51}=\dfrac{85}{51}+\dfrac{3}{51}=\dfrac{88}{51}\)

\(b,2\dfrac{4}{7}-\dfrac{11}{42}=\dfrac{18}{7}-\dfrac{11}{42}=\dfrac{108}{42}-\dfrac{11}{42}=\dfrac{97}{42}\)

\(c,2\dfrac{1}{2}\times\dfrac{1}{3}+3\dfrac{1}{4}=\dfrac{5}{2}\times\dfrac{1}{3}+\dfrac{13}{4}=\dfrac{5}{6}+\dfrac{13}{4}=\dfrac{10}{12}+\dfrac{39}{12}=\dfrac{49}{12}\)

\(d,3\dfrac{1}{2}+\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{7}{2}+\dfrac{1}{4}\times\dfrac{6}{1}=\dfrac{7}{2}+\dfrac{3}{2}=\dfrac{10}{2}=5\)