Question

Painter A finished colouring a picture in 7 12 hour. And Painter B finished colouring the same picture in 3 4 hour. Who worked longer? By what fraction was it longer?​

Answer 1

Answer:

painter A

Step-by-step explanation:

if we get them in fraction

=7/12and3/4

=7/4and1/4 by cancer cancellation method

so

painter a=0.58

painter b =0.25

Answer 2

GIVEN:

$\sf{Time\:taken\:by\:Painter\:A\:to\:colour\:the\:picture\:=\dfrac{7}{12}\:hour.}$

$\sf{Time\:taken\:by\:Painter\:B\:to\:colour\:the\:picture\:=\dfrac{3}{4}\:hour.}$

TO FIND:

• The painter who worked longer
• The fraction by which the work was longer

ANSWER:

We cannot compare their works like this since both of the fractions are unlike. Thus, we first make their denominators equal.

LCM of 12 and 4 = 12

So,

$\sf{Time\:taken\:by\:Painter\:A=\dfrac{7}{12}\:hour}$

$\sf{Time\:taken\:by\:Painter\:B=\dfrac{3\times3}{4\times3}=\dfrac{9}{12}hour}$

Now we have two like fractions to compare.

$\sf{9>7\:so\:\dfrac{9}{12}>\dfrac{7}{12}}$

$\sf{=\dfrac{3}{4}>\dfrac{7}{12}}$

Painter B worked longer than Painter A.

To find the fraction of time by which Painter B worked longer than Painter A, again, we take the like fractions and subtract the numerators.

$\sf{=\dfrac{9}{12}-\dfrac{7}{12}}$

$\sf{=\dfrac{9-7}{12}}$

$\sf{=\dfrac{2}{12}=\dfrac{1}{6}\:hour}$

Therefore Painter B worked longer than Painter A by 1/6 of an hour.