Question

Evaluate each of the following:

$\left(a\right)(-30)÷10$

$\left(b\right)50÷(-5)$

$\left(c\right)(-36)÷(-9)$

$\left(d\right)(-49)÷49$

$\left(e\right)13÷[-2+1]$

$\left(f\right)0÷(-12)$

$\left(g\right)(-31)÷\left[\right(-30)+(-1\left)\right]$

$\left(h\right)\left[\right(-36)÷12]÷3$

$\left(i\right)[-6+5]÷[-2+1]$

Answer 1

$\left(a\right)(-30)÷10=\frac{(-3)\times 10}{10}$
$\therefore (-30)÷10=-3$ [$10$ gets cancelled]

$\left(b\right)50÷(-5)=\frac{(-5)\times 10}{-5}$
$\therefore (-50)÷(-5)=10$ [$-5$ gets cancelled]

$\left(c\right)(-36)÷(-9)=\frac{(-9)\times 4}{-9}$
$\therefore (-36)÷(-9)=4$ [$-9$ gets cancelled]

$\left(d\right)(-49)÷\left(49\right)=\frac{(-49)\times 1}{49}$
$\therefore (-49)÷\left(49\right)=-1$ [$49$ gets cancelled]

$\left(e\right)13÷[-2+1]$
Since, $[-2+1]=-1$
Thus, $13÷[-2+1]=13÷[-1]=-13$

$\left(f\right)0÷(-12)=0$
As $0$ divide by any nonzero quantity gives us $0$ only.

$\left(g\right)(-31)÷\left[\right(-30)+(-1\left)\right]$
Since, $(-30)+(-1)=-30-1=-31$
Thus, $(-31)÷(-31)=1$

$\left(h\right)\left[\right(-36)÷12]÷3$
Since, $(-36)÷12=(-12\times 3)÷12=-3$
[$12$ gets cancelled]
Thus, $\left[\right(-36)÷12]÷3=[-3]÷3=-1$

$\left(i\right)[-6+5]÷[-2+1]$
Since $-6+5=-1$ and $[-2+1]=-1$
Thus, $[-6+5]÷[-2+1]=(-1)÷(-1)=1$