6. Salil wants to put a picture in a frame. The picture is 7*3/5 cm wi

1.	6. Salil wants to put a picture in a frame. The picture is 7*3/5 cm wide.
Answer» $\large{\underline{\rm{\green{\bf{Correct \: Question:-}}}}}$ Salil WANTS to put a PICTURE in a frame. The picture is $\sf 7\dfrac{3}{4} \: <klux>CM</klux>$ wide. To fit the frame in the picture cannot be more than $\sf 7 \dfrac{3}{10} \: cm$ wide. How much should be the picture be trimmed? $\large{\underline{\rm{\green{\bf{Given:-}}}}}$ Width of the picture = $\sf 7\dfrac{3}{4} \: cm = \dfrac{38}{5} \: cm$ Width of the frame = $\sf 7 \dfrac{3}{10} \: cm=\dfrac{73}{10} \: cm$ $\large{\underline{\rm{\green{\bf{To \: Find:-}}}}}$ Length of the picture that should be trimmed. $\large{\underline{\rm{\green{\bf{Solution:-}}}}}$ Given that, Width of the picture = $\sf 7\dfrac{3}{4} \: cm = \dfrac{38}{5} \: cm$ Width of the frame = $\sf 7 \dfrac{3}{10} \: cm=\dfrac{73}{10} \: cm$ ∴ The picture should be trimmed by = $\sf \bigg[ \bigg(\dfrac{38}{5}- \dfrac{73}{10} \bigg) \bigg]$ The LCM of 5, 10 = 10 So, LCM = 10 Now, change each of the given FRACTION into an equivalent fraction having 10 as the denominator. $\sf = \bigg[ \bigg( \dfrac{38}{5} \times \dfrac{2}{2} \bigg) \bigg] - \bigg[ \bigg( \dfrac{73}{10} \times \dfrac{1}{1} \bigg) \bigg]$ $\sf = \bigg( \dfrac{76}{10} \bigg) - \bigg( \dfrac{73}{10} \bigg)$ $\sf = \dfrac{(76-73)}{10}$ $\sf = \underline{\underline{\dfrac{3}{10} \: cm}}$ Thus, the picture should be trimmed by (3/10) cm

6. Salil wants to put a picture in a frame. The picture is 7*3/5 cm wide.

Answer»

$\large{\underline{\rm{\green{\bf{Correct \: Question:-}}}}}$

Salil WANTS to put a PICTURE in a frame. The picture is $\sf 7\dfrac{3}{4} \: <klux>CM</klux>$ wide. To fit the frame in the picture cannot be more than $\sf 7 \dfrac{3}{10} \: cm$ wide. How much should be the picture be trimmed?