Question

Is 7/8 or 11/10 closer to 1? How did you decide anybody help please?

Answer 1

Let

A-----> $\frac{7}{8}$

B-----> $\frac{11}{10}$

C-----> $1$

we know that

Point A-------> Multiply numerator and denominator by $10$

$\frac{7*10}{8*10} =\frac{70}{80}$

Point B-------> Multiply numerator and denominator by $8$

$\frac{11*8}{10*8} =\frac{88}{80}$

Point C-------> Multiply numerator and denominator by $80$

$\frac{1*80}{1*80} =\frac{80}{80}$

Find the distance Point A to Point C

$\frac{80}{80} -\frac{70}{80} =\frac{10}{80}$

Find the distance Point B to Point C

$\frac{88}{80} -\frac{80}{80} =\frac{8}{80}$

therefore

$\frac{8}{80} < \frac{10}{80}$

Point B is closer to Point C

$\frac{11}{10}$ is closer to $1$

the answer is

$\frac{11}{10}$ is closer to $1$

Answer 2

You can also figure it out by knowing that 7/8 is 1/8 away from 1 and 11/10 is 1/10 away from 1. since 1/10 is smaller than 1/8, you know that 11/10 is closer to 1 than 7/8.