Question 7 asks us to factorize the expression $6pq - 3rs - 3ps + 6qr$. Question 8 asks us to find the number that should be subtracted from the sum of $2\frac{1}{6}$ and $2\frac{7}{12}$ to give $3\frac{1}{4}$.

1. Problem Description

Question 7 asks us to factorize the expression $6pq - 3rs - 3ps + 6qr$.

Question 8 asks us to find the number that should be subtracted from the sum of $2\frac{1}{6}$ and $2\frac{7}{12}$ to give $3\frac{1}{4}$.

2. Solution Steps

Question 7:

We need to factorize the expression $6pq - 3rs - 3ps + 6qr$.

Rearrange the terms: $6pq + 6qr - 3ps - 3rs$.

Factor by grouping:

$6q(p+r) - 3s(p+r)$.

Factor out the common term $(p+r)$:

$(p+r)(6q-3s)$.

Factor out 3 from $(6q-3s)$:

$3(p+r)(2q-s)$.

This matches option B.

Question 8:

Let the number to be subtracted be $x$.

The problem can be written as:

$(2\frac{1}{6} + 2\frac{7}{12}) - x = 3\frac{1}{4}$.

First, convert the mixed fractions to improper fractions:

$2\frac{1}{6} = \frac{2\times 6 + 1}{6} = \frac{13}{6}$.

$2\frac{7}{12} = \frac{2\times 12 + 7}{12} = \frac{31}{12}$.

$3\frac{1}{4} = \frac{3\times 4 + 1}{4} = \frac{13}{4}$.

So the equation becomes:

$(\frac{13}{6} + \frac{31}{12}) - x = \frac{13}{4}$.

To add the fractions, we need a common denominator, which is

1

2. $\frac{13}{6} = \frac{13 \times 2}{6 \times 2} = \frac{26}{12}$.

Thus, $\frac{26}{12} + \frac{31}{12} = \frac{26+31}{12} = \frac{57}{12}$.

The equation is now:

$\frac{57}{12} - x = \frac{13}{4}$.

We want to isolate $x$:

$x = \frac{57}{12} - \frac{13}{4}$.

We need a common denominator, which is

1

2. $\frac{13}{4} = \frac{13 \times 3}{4 \times 3} = \frac{39}{12}$.

Thus, $x = \frac{57}{12} - \frac{39}{12} = \frac{57-39}{12} = \frac{18}{12}$.

Simplify the fraction:

$x = \frac{18}{12} = \frac{6 \times 3}{6 \times 2} = \frac{3}{2}$.

Convert to a mixed fraction:

$x = \frac{3}{2} = 1\frac{1}{2}$.

This matches option B.

3. Final Answer

Question 7: B.

Question 8: B.