In the super market there are five shopkeepers A, B, C, D, E. if
shopkeeper A sells his 2/3 of products in a day, B sells 3/7 of products, C sells
4/5 of products in a day, D sells 5/6 of products in a day and E sells 7/8 of
products in a day, who sells more products in a day?​​ In the super market there are five shopkeepers A, B, C, D, E. if
shopkeeper A sells his 2/3 of products in a day, B sells 3/7 of products, C sells
4/5 of products in a day, D sells 5/6 of products in a day and E sells 7/8 of
products in a day, who sells more products in a day?

Dear Student,

Please find below the solution to the asked query:

Given : If shopkeeper A sells his $\frac{2}{3}$ of products in a day,

B sells $\frac{3}{7}$ of products,

C sells $\frac{4}{5}$ of products in a day,

D sells $\frac{5}{6}$ of products in a day

And E sells $\frac{7}{8}$ of products in a day .

To find who sells more products in a day , we find L.C.M. of denominators of all fractions .

3 =  1 $\times$ 3 ,

7 = 1 $\times$  7 ,

5 =  1 $\times$ 5 ,

6 = 2 $\times$ 3 ,

8 =  2 $\times$ 2 $\times$ 2

Then,

L.C.M. ( 3 , 7 , 5 , 6  and  8 ) = 2 $\times$ 2 $\times$ 2 $\times$ 3 $\times$ 7 $\times$ 5 = 840

Now we try to make each denominator = 840  , As :

A sells his $\frac{2}{3} = \frac{2}{3} \times \frac{280}{280} =\mathbf{ }\frac{\mathbf{560}}{\mathbf{840}}$ of products in a day,

B sells $\frac{3}{7} =\frac{3}{7}\times \frac{120}{120} = \frac{\mathbf{360}}{\mathbf{840}}$ of products,

C sells $\frac{4}{5} =\frac{4}{5} \times \frac{168}{168} = \frac{\mathbf{672}}{\mathbf{840}}$ of products in a day,

D sells $\frac{5}{6} =\frac{5}{6} \times \frac{140}{140} = \frac{\mathbf{700}}{\mathbf{840}}$ of products in a day

And E sells $\frac{7}{8} =\frac{7}{8} \times \frac{105}{105} = \frac{\mathbf{735}}{\mathbf{840}}$ of products in a day .

Now we can compare numerator of each fractions and observe that " 735 " is highest ,

Therefore,

E sells more products in a day .                                                                             ( Ans )


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