Exercise1.2 (Revised) - Chapter 1 - Integers - Ncert Solutions class 7 - Maths

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Chapter 1 - Integers | NCERT Solutions for Class 7 Maths

Ex 1.2 Question 1.

Find the each of the following products:
(a) $3 \times(-1)$
(b) $(-1) \times 225$
(c) $(-21) \times(-30)$
(d) $(-316) \times(-1)$
(e) $(-15) \times 0 \times(-18)$
(f) $(-12) \times(-11) \times(10)$
(g) $9 \times(-3) \times(-6)$
(h) $(-18) \times(-5) \times(-4)$
(i) $(-1) \times(-2) \times(-3) \times 4$
(j) $(-3) \times(-6) \times(2) \times(-1)$

Answer:

(a) $3 \times(-1)=-3$
(b) $(-1) \times 225=-225$
(c) $(-21) \times(-30)=630$
(d) $(-316) \times(-1)=316$
(e) $(-15) \times 0 \times(-18)=0$
(f) $(-12) \times(-11) \times(10)=132 \times 10=1320$
(g) $9 \times(-3) \times(-6)=9 \times 18=162$
(h) $(-18) \times(-5) \times(-4)=90 \times(-4)=-360$
(i) $(-1) \times(-2) \times(-3) \times 4=(-6 \times 4)=-24$
(j) $(-3) \times(-6) \times(2) \times(-1)=(-18) \times(-2)=36$

Ex 1.2 Question 2.

Verify the following:

(a) $18 \times[7+(-3)]=[18 \times 7]+[18 \times(-3)]$
(b) $(-21) \times[(-4)+(-6)]=[(-21) \times(-4)]+[(-21) \times(-6)]$

Answer:

(a) 18 x [7 + (-3)] = [18 x 7] + [18 x (-3)]

(a) $18 \times[7+(-3)]=[18 \times 7]+[18 \times(-3)]$
(b) $(-21) \times[(-4)+(-6)]=[(-21) \times(-4)]+[(-21) \times(-6)]$

Answer: (a) 18 x [7 + (-3)] = [18 x 7] + [18 x (-3)]

(c) $(-1) \times 0=0$

Ex 1.2 Question 4.

Starting from $(-1) \times 5$, write various products showing some patterns to show $(-1) \times(-1)=1$.

Answer:

$(-1) \times 5=-5,(-1) \times 4=-4$
$
\begin{aligned}
& (-1) \times 3=-3,(-1) \times 2=-2 \\
& (-1) \times 1=-1,(-1) \times 0=0 \\
& (-1) \times(-1)=1
\end{aligned}
$

Thus, we can conclude that this pattern shows the product of one negative integer and one positive integer whereas the product of two negative integers is a positive integer.