Integers - Exercise 1(C)

Find the product :

18 x 4

Answer

We have

18 x 4 = 72

Find the product :

(-25) x 6

Answer

We have

(-25) x 6 = -150

[∵ Negative x Positive = Negative]

Find the product :

(-30) x 7

Answer

We have

(-30) x 7 = -(30 x 7) = -210

[∵ Negative x Positive = Negative]

Find the product :

8 x (-15)

Answer

We have

8 x (-15) = -(8 x 15) = -120

[∵ Positive x Negative = Negative]

Find the product :

20 x (-10)

Answer

We have

20 x (-10) = -(20 x 10) = -200

[∵ Positive x Negative = Negative]

Find the product :

(-12) x (-15)

Answer

We have

(-12) x (-15) = +(12 x 15) = 180

[∵ Negative x Negative = Positive]

Find the product :

(-8) x (-13)

Answer

We have

(-8) x (-13) = +(8 x 13) = 104

[∵ Negative x Negative = Positive]

Find the product :

(-20) x (-1)

Answer

We have

(-20) x (-1) = +(20 x 1) = 20

[∵ Negative x Negative = Positive]

Find the product :

(-9) x 0

Answer

We have

(-9) x 0 = -(9 x 0) = 0

[∵ Anything multiplied with 0 becomes 0]

Find the product :

0 x (-11)

Answer

We have

0 x (-11) = -(0 x 11) = 0

[∵ Anything multiplied with 0 becomes 0]

Find the product :

{(-9) x 8} x (-5)

Answer

We have

{(-9) x 8} x (-5)

First multiply inside the brace:
(-9) x 8 = -72

Now multiply the result with last number:
(-72) x (-5) = +(72 x 5) = 360

∴ {(-9) x 8} x (-5) = 360

Find the product :

{(-10) x (-5)} x 6

Answer

We have

{(-10) x (-5)} x 6

First multiply inside the brace:
(-10) x (-5) = 50

Now multiply the result with last number:
50 x 6 = 300

∴ {(-10) x (-5) x 6} = 300

Find the product :

{(-12) x (-15)} x (-2)

Answer

We have

{(-12) x (-15)} x (-2)

First multiply inside the brace:
(-12) x (-15) = 180

Now multiply:
180 x (-2) = -(180 x 2) = -360

∴ {(-12) x (-15)} x (-2) = -360

Find the product :

(-8) x {(-5) x (-3)}

Answer

We have

(-8) x {(-5) x (-3)}

First multiply inside the brace:
(-5) x (-3) = 15

Now multiply the result with the first number:
(-8) x 15 = -(8 x 15) = -120

∴ (-8) x {(-5) x (-3)} = -120

Find the product :

(-11) x {(-8) x 5}

Answer

We have

(-11) x {(-8) x 5}

First multiply inside the brace:
(-8) x 5 = -40

Now multiply the result with the first number:
(-11) x (-40) = +(11 x 40) = 440

∴ (-11) x {(-8) x 5} = 440

Verify the following :

(-14) x (-8) = (-8) x (-14)

Answer

We have

(-14) x (-8) = (-8) x (-14)

LHS = (-14) x (-8) = 112
RHS = (-8) x (-14) = 112

Since LHS = RHS,

∴ (-14) x (-8) = (-8) x (-14) $\hspace{1.5cm}$[a x b = b x a ⇒ Commutative Property of multiplication.]

Verify the following :

{(-7) x 5} x (-6) = (-7) x {5 x (-6)}

Answer

We have

{(-7) x 5} x (-6) = (-7) x {5 x (-6)}

LHS = {(-7) x 5} x (-6)
= (-35) x (-6)
= 210

RHS:(-7) x {5 x (-6)}
= (-7) x (-30)
= 210

Since LHS = RHS,

∴ {(-7) x 5} x (-6) = (-7) x {5 x (-6)} $\hspace{1.5cm}$[{a x b} x c = a x {b x c} ⇒ Associative Property of multiplication.]

Verify the following :

(-10) x {(-7) + (-9)} = {(-10) x (-7)} + {(-10) x (-9)}

Answer

We have

(-10) x {(-7) + (-9)} = {(-10) x (-7)} + {(-10) x (-9)}

LHS = (-10) x {(-7) + (-9)}
=(-10) x {-16}
= 160
RHS = {(-10) x (-7)} + {(-10) x (-9)}
= 70 + 90
= 160

Since LHS = RHS,

∴ (-10) x {(-7) + (-9)} = {(-10) x (-7)} + {(-10) x (-9)} $\hspace{1.5cm}$[a x {b + c} = {a x b} + {a x c} ⇒ Distributive Property of multiplication over addition.]

Find the quotient :

28 ÷ (-7)

Answer

We have

28 ÷ (-7)

= $\dfrac {+28}{-7}$

= -4

Find the quotient :

(-65) ÷ 13

Answer

We have

(-65) ÷ 13

= $\dfrac {-65}{+13}$

= -5

Find the quotient :

(-66) ÷ (-6)

Answer

We have

(-66) ÷ (-6)

= $\dfrac {-66}{-6}$

= 11

Find the quotient :

(-9) ÷ (-1)

Answer

We have

(-9) ÷ (-1)

= $\dfrac {-9}{-1}$

= 9

Find the quotient :

0 ÷ (-2)

Answer

We have

0 ÷ (-2)

= $\dfrac {0}{-2}$

= 0

Find the quotient :

(-12) ÷ (-12)

Answer

We have

(-12) ÷ (-12)

= $\dfrac {-12}{-12}$

= 1

Write all even integers between

(i) (-4) and 11

(ii) (-13) and (-7)

Answer

We have

(i) All even integers between -4 and 11 are:
-2, 0, 2, 4, 6, 8 and 10

(ii) All even integers between -13 and -7 are:
-12, -10 and -8

Write all odd integers between

(i) (-1) and 7

(ii) (-20) and (-14)

Answer

(i) All odd integers between -1 and 7 are:
1, 3 and 5

(ii) All odd integers between -20 and -14 are:
-19, -17 and -15

Write five consecutive even integers succeeding -21.

Answer

The required consecutive even integers succeeding -21 are:
-20, -18, -16, -14 and -12

Write five consecutive odd integers preceding -36.

Answer

The required consecutive odd integers preceding -36 are:
-37, -39, -41, -43 and -45

The product of two integers is -120. If one number is 15, find the other.

Answer

Let the two integers be p and q.

Given that p x q = -120 and p = 15

Substitute p in the below equation:

p x q = -120

15 x q = −120 $\hspace{2cm}$(Substitute p)

q = $\dfrac{-120}{15}$ $\hspace{2.7cm}$(Solve for q)

q = -8

∴ The other number is -8.