Question

Michael finished colouring a picture in 7/12 hour. Vaibhav finished colouring the same picture in 3/4 hour. Who worked longer? By what fraction was it longer?​

Answer 1

Step-by-step explanation:

Time taken by Michal in coloring = 7/12 hours.

Time taken by Vaibhav in coloring = 3/4 hours.

The LCM of 12 and 4 is 12. So making the denominator of both fraction equal to 12, we get,

=> 7/12 and 3/4 × 3/3

=> 7/12 and 9/12

As

9/12 > 7/12

For calculating by how much, we do Subtraction

9/12 - 7/12 = 9-7/12 = 2/12 = 1/6

Vaibhav worked longer by the fraction 1/6.

Thank you..

Hope it helps you...

Answer 2

Answer:

So , the Vaibhav worked longer .

$By,\frac{1}{6}\:he\:worked\:longer$

Step by step explanation:

》》Given:

Michael finished colouring the picture in :

$=\frac{7}{12}hour$

Vaibhav finished colouring the picture in :

$=\frac{3}{4}hour$

》》To find:

(i) Who worked longer

(ii) By what fraction was it longer

》》Concept:

First find the L.C.M .

Then multiply the denominator till coming the L.C.M at denominator.

And to find the fraction of time who worked longer .

Subtract who worked longer with the who worked less .

》》Solution:

First find the L.C.M of the denominator of both the fraction is :

$L.C.M = 12$

(You can see in the attachement)

Now , find who worked longer :

$\Rightarrow{\frac{7}{12},\frac{3}{4}}$

$\Rightarrow{\frac{7}{12}×\frac{1}{1}=\frac{7}{12}}$

$\Rightarrow{\frac{3}{4}×\frac{3}{3}=\frac{9}{12}}$

$\Rightarrow{\frac{7}{12}<\frac{9}{12}}$

So , the Vaibhav worked longer .

At last , find the fraction of time he worked longer :

$\mapsto{\frac{9}{12}-\frac{7}{12}}$

$\mapsto{\frac{9-7}{12}}$

$\mapsto{\frac{2}{12}}$

Make it in simplest form :

$\mapsto{\frac{2}{12}=\frac{1}{6}}$

》》Extra information:

To find Acceleration:

$A=\frac{V1-V}{T}$

Where,

A = Acceleration

V1 = Final velocity

V = Initial velocity

T = Time

To find the ratio :

$\frac{Number}{Number}$

Divide till it come to the simplest form than make it in ratio .

Ex = 2:3

To find velocity :

$V=\frac{D}{T}$

Where,

V = Velocity

D = Displacement

T = Time