The product of two improper fractions is less than both the fractions.
Answer
The product of two improper fractions are greater than both the fractions. $\frac{3}{2}\times\frac{7}{4}=\frac{21}{8}$
Hence, $\frac{21}{8}$ is, greater then both $\frac{3}{2}\text{and}\frac{7}{4}$
The reciprocal of an improper fraction is an improper fraction.
Answer
False. Solution: The reciprocal of an improper fraction is a proper fraction, e.g. $\frac{7}{6}$ → Improper fraction Its reciprocal is $\frac{7}{6}$ i.e. proper fraction.
The product of two proper fractions is _____ than each of the fractions that are multiplied.
Answer
The product of two proper fractions is less than each of the fractions that are multiplied.Solution:
The product of two proper fractions is less than each of the fractions that are multiplied. e.g. $\frac{1}{2}\times\frac{1}{3}=\frac{1}{6}$ $\therefore\frac{1}{6}<\frac{1}{2}\ \text{and}\ \frac{1}{6}<\frac{1}{3}$
The normal body temperature is $98.6^\circ F.$ When Savitri was ill her temperature rose to $103.1^\circ F.$ How many degrees above normal was that?
Answer
Given, Normal body temperature $= 98.6^\circ F$ and temperature rise to $= 103.1^\circ F$
$\therefore$ Rise in temperature $= (103.1 - 98.6)^\circ F = 4.5^\circ F.$
Renu completed $\frac{2}{3}$ part of her home work in $2$ hours. How much part of her home work had she completed in $1\frac{1}{4}$ hours$?$
Answer
The part of the work finished bt Renu in $2h =\frac{2}{3}$
So, the part of the work finished by renu in $1h =\frac{2}{3}\times\frac{1}{2}=\frac{1}{3}$
$\therefore$ The part of the work finished bt Renu in $1\frac{1}{4}\text{h}=\frac{1}{3}\times1\frac{1}{4}$
$=\frac{1}{3}\times\frac{(1\times4)+1}{4}$
$=\frac{1}{3}\times\frac{5}{4}=\frac{5}{12}\text{part}$
Hence, $\frac{5}{12}$ part of Renu's home work is cpmpleted by her in $1\frac{1}{4}\text{h}$