Question

46 what value of p makes the equation true?
$32 \times 125 = (32 \times 100) + (32 \times p)$
a 5
b 20
c 25
d 32

47 what is $\frac{3}{4} \times \frac{2}{7}$?
a $\frac{5}{11}$
b $\frac{6}{11}$
c $\frac{6}{27}$
d $\frac{3}{14}$

48 what is $\frac{2}{3} \times \frac{3}{4}$?
a $\frac{6}{7}$
b $\frac{5}{7}$
c $\frac{1}{2}$
d $\frac{6}{12}$

Explanation:

Question 46

Step1: Apply distributive property

The right - hand side of the equation \(32\times125=(32\times100)+(32\times p)\) can be factored using the distributive property \(a\times(b + c)=a\times b+a\times c\) (in reverse, \(a\times b+a\times c=a\times(b + c)\)). So, \((32\times100)+(32\times p)=32\times(100 + p)\). Then our equation becomes \(32\times125 = 32\times(100 + p)\).

Step2: Divide both sides by 32

Since both sides of the equation have a factor of 32 (and \(32
eq0\)), we can divide both sides by 32. We get \(125=100 + p\).

Step3: Solve for p

Subtract 100 from both sides of the equation \(125=100 + p\). So, \(p=125 - 100=25\)? Wait, no, wait. Wait, \(125=100 + p\), so \(p = 125-100 = 25\)? Wait, no, let's check again. Wait, the original equation is \(32\times125=(32\times100)+(32\times p)\). Let's calculate the left - hand side: \(32\times125 = 4000\). The right - hand side: \(32\times100=3200\), and \(32\times p\) is the other term. So, \(4000=3200 + 32p\). Subtract 3200 from both sides: \(4000 - 3200=32p\), \(800 = 32p\). Then \(p=\frac{800}{32}=25\)? Wait, no, 32×25 = 800, 32×100 = 3200, 3200+800 = 4000, and 32×125 = 4000. Wait, but the options have A.5, B.20, C.25, D.32. Wait, maybe I made a mistake in the distributive property. Wait, \(125=100 + 25\), so \(32\times125=32\times(100 + 25)=32\times100+32\times25\). So \(p = 25\)? But let's check the calculation again. Wait, 32×125: 32×100 = 3200, 32×25 = 800, 3200 + 800=4000. And 32×125 = 4000. So the value of \(p\) is 25? But wait, the options: A.5, B.20, C.25, D.32. So the answer is C.25? Wait, no, wait, maybe I misread the equation. Wait, the equation is \(32\times125=(32\times100)+(32\times p)\). Let's solve for \(p\):

\(32\times125-32\times100 = 32\times p\)

\(32\times(125 - 100)=32\times p\)

\(32\times25=32\times p\)

So \(p = 25\). But the option C is 25. Wait, but let's check the original problem again. Wait, maybe I made a mistake. Wait, 125-100 = 25, so \(p = 25\). So the answer is C.

Question 47

To multiply two fractions \(\frac{a}{b}\times\frac{c}{d}=\frac{a\times c}{b\times d}\) (where \(b
eq0\) and \(d
eq0\)).

For \(\frac{3}{4}\times\frac{2}{7}\), we have \(a = 3\), \(b = 4\), \(c = 2\), \(d = 7\).

So \(\frac{3}{4}\times\frac{2}{7}=\frac{3\times2}{4\times7}=\frac{6}{28}=\frac{3}{14}\) (after simplifying by dividing numerator and denominator by 2).

So the answer is D. \(\frac{3}{14}\).

Question 48

To multiply two fractions \(\frac{a}{b}\times\frac{c}{d}=\frac{a\times c}{b\times d}\) (where \(b
eq0\) and \(d
eq0\)).

For \(\frac{2}{3}\times\frac{3}{4}\), we have \(a = 2\), \(b = 3\), \(c = 3\), \(d = 4\).

So \(\frac{2}{3}\times\frac{3}{4}=\frac{2\times3}{3\times4}=\frac{6}{12}=\frac{1}{2}\) (after simplifying by dividing numerator and denominator by 6).

So the answer is C. \(\frac{1}{2}\).

Answer:

s:

1. C. 25
2. D. \(\frac{3}{14}\)
3. C. \(\frac{1}{2}\)