Question

on what fruction must be subtracted from 4-11/12 to get 2-3/10​

Answer 1

Step-by-step explanation:

4-11/12 - ( 2-3/10)

= -7/12 -(-1/10)

= -7/12 + 1/10 [as( -) x (- )= (+)]

Taking LCM

- 35 + 6 /60 [ as the LCM of 12 and 10 is 60]

= -29/60

Hence, -29/60 should be subtracted from 4-11/12 to get 2-3/10.

Answer 2

Appropriate Question -

• What fraction must be subtracted from 4 -11/12 to get 2 -3/10?

Answer -

• The required fraction is -433/60.

To find -

• The fraction that must be subtracted from 4 -11/12 to get 2 -3/10.

Step-by-step explanation -

• Here, we are asked to find out the fraction which must be subtracted from 4 -11/12 to get 2 -3/10.

Let -

• The fraction be "y".

Then -

• The difference of 4 -11/12 and "y" will be 2 -3/10.

Therefore -

$\mapsto \sf y - 4 \: \dfrac{ - 11}{ \: \: \: 12} = 2 \: \dfrac{ - 3}{ \: 10}$

$\mapsto \sf{y - \dfrac{-59}{12} = \dfrac{-23}{10}}$

$\mapsto \sf{y = \dfrac{-23}{10} + \dfrac{-59}{12}}$

$\mapsto \sf{y = \dfrac{(-23 \times 6) + (-59 \times 5)}{60}}$

$\mapsto \sf{y = \dfrac{-138 - 295}{60}}$

$\mapsto \sf{y = \dfrac{-433}{60}}$

Hence -

• The required fraction is -433/60.

-----------------------------------------------------------

V E R I F I C A T I O N -

• To check our answer, let's subtract 4 -11/12 from -433/60 and see whether we get 2 - 3/10.

$\hookrightarrow\tt \dfrac{ - 433}{ \: \: \: \: 60} - 4 \: \dfrac{ - 11}{ \: \: 12}$

$\hookrightarrow\tt \dfrac{ - 433}{ \: \: \: \: 60} - \dfrac{ - 59}{ \: \: \: 12}$

$\hookrightarrow \tt\dfrac{( - 433 \times 1) - ( - 59 \times 5)}{60}$

$\hookrightarrow \tt \dfrac{ - 433 + 295}{60}$

$\hookrightarrow\tt \cancel{\dfrac{ - 138}{ \: \: \: \: 60} }$

$\hookrightarrow\tt\dfrac{ -23}{ \: \: 10}$

$\hookrightarrow\tt2 \: \dfrac{ - 3}{ \: 10}$

We get 2 -3/10 on subtracting 4 -11/12 from -433/60.

Hence verified!!

________________________________