RD Sharma 2018 Solutions for Class 7 Math Chapter 1 Integers are provided here with simple step-by-step explanations. These solutions for Integers are extremely popular among class 7 students for Math Integers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2018 Book of class 7 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s RD Sharma 2018 Solutions. All RD Sharma 2018 Solutions for class 7 Math are prepared by experts and are 100% accurate.

#### Question 11:

Find the value of
(−15) + 4 ÷ (5 − 3)

On applying the BODMAS rule, we get:

(− 15) + 4 ÷ (5 − 3)

= (− 15) + 4 ÷ 2  (On simplifying brackets)

= (− 15) + 2        (On performing division)

= − 13

#### Question 12:

Find the value of
(−40) × (−1) + (−28) ÷ 7

On applying the BODMAS rule, we get:

(− 40) × (− 1) + (− 28) ÷ 7

=  40 + (− 4)                           (On performing division and multiplication)

=  36

#### Question 13:

Find the value of
(−3) + (−8) ÷ (−4) −2 × (−2)

On applying the BODMAS rule, we get:

(− 3) + (− 8) ÷ (− 4) − 2 × (− 2)

=  (− 3) + 2 + 4                (On performing division and multiplication)

= (− 3) + 6                       (On performing addition)

= 3                                  (On performing subtraction)

#### Question 14:

Find the value of
(−3) × (−4) ÷ (−2) + (−1).

On applying the BODMAS rule, we get:

(−3) â€‹× (−4) ÷ (−2) + (−1)

= (−3) × 2 + (−1)                    (On performing division)

=  −6 − 1                                (On performing multiplication)

#### Question 1:

Simplify each of the following:
3 − (5 − 6 ÷ 3)

On applying the BODMAS rule, we get:

3 − (5 − 6 ÷ 3)

= 3 − (5 − 2)     (On performing division)

= 3 − 3              (On performing subtraction)

= 0

#### Question 2:

Simplify each of the following:
−25 + 14 ÷ (5 − 3)

On applying the BODMAS rule, we get:

−25 + 14 ÷ (5 − 3)

= −25 + 14 ÷ 2       (On simplifying brackets)

= −25 + 7               (On performing division)

= −18

#### Question 3:

Simplify each of the following:
$25-\frac{1}{2}\left\{5+4-\left(3+2-\overline{)1+3}\right)\right\}$

On applying the BODMAS rule, we get:

â€‹25 $-$ 1/2 {5+4-( 3 + 2 − $\overline{)1+3}$ )}

#### Question 4:

Simplify each of the following:
$27-\left[38-\left\{46-\left(15-\overline{)13-2}\right)\right\}\right]$

On applying the BODMAS rule, we get:

27 $-$ [38 $-$ {46 $-$ (15 $-$11)}]    (On simplifying vinculum)

= 27 $-$ [38$-$ {46 $-$ 4}]              (On simplifying parentheses)

= 27 $-$ [38 $-$ 42]                       (On simplifying braces)

= 27 $-$ ($-$4) = 31

#### Question 5:

Simplify each of the following:

On applying the BODMAS rule, we get:

36 $-$ [18 $-$ { 14 $-$ (15 $-$ 4 ÷ 2 â€‹× 2)}]
= 36 − [18 − {14 − (15 − 2 × 2)}]                (On performing division)
= 36 $-$ [18 $-$ {14 $-$ (15 $-$ 4)}]                   (On performing multiplication)
= 36 $-$ [18 $-$ {14 $-$ 11}]                            (On simplifying parentheses)
= 36 $-$ [18 $-$ 3]                                          (On simplifying braces)
= 36 $-$ 15
= 21

#### Question 6:

Simplify each of the following:

On applying the BODMAS rule, we get:

45 $-$ [38 $-$ { 60 ÷  3 $-$ (6 $-$ 9 ÷ 3) ÷ 3}]
=  45 $-$ [38 $-$ {60 ÷ 3 $-$ (6 $-$ 3) ÷ 3}]           (On performing division)
=  45 $-$ [38 $-$ {60 ÷ 3 $-$ 3 ÷ 3}]                    (On simplifying parentheses)
=  45 $-$ [38 $-$ {60 ÷ 3 $-$ 1}]                          (On performing division)
=  45 $-$ [38 $-$ {20 $-$ 1}]                                (On performing division)
=  45 $-$ [38 $-$ 19]                                          (On performing subtraction)
=  45 $-$ 19
= 26

#### Question 7:

Simplify each of the following:
$23-\left[23-\left\{23-\left(23-\overline{)23-23}\right)\right\}\right]$

On applying the BODMAS rule, we get:

23 $-$ [23 $-$ {23 $-$ (23 $-$ $\overline{)23-23}$)}]
= 23 $-$ [23 $-$ {23 $-$ (23 $-$ 0}]            (On simplifying vinculum)
= 23 $-$ [23 $-$ {23 $-$ 23}]                    (On simplifying parentheses)
= 23 $-$ [23 $-$ 0]                                  (On simplifying braces)
= 23 $-$ 23 = 0

#### Question 8:

Simplify each of the following:
$2550-\left[510-\left\{270-\left(90-\overline{)80+70}\right)\right\}\right]$

On applying the BODMAS rule, we get:

2550 $-$ [510 $-$ {270 $-$ (90 $-$ $\overline{)80+70}$)}]
= 2550 $-$ [510 $-$ {270 $-$ (90 $-$ 150)}]        (On simplifying vinculum)
= 2550 $-$ [510 $-$ { 270 $-$ ($-$ 60)}]              (On simplifying parentheses)
= 2550 $-$ [510 $-$ 330]                                 (On simplifying braces)
= 2550 $-$ 180
= 2370

#### Question 9:

Simplify each of the following:
$4+\frac{1}{5}\left[\left\{-10×\left(25-\overline{)13-3}\right)\right\}÷\left(-5\right)\right]$

On applying the BODMAS rule, we get:

4 + $\frac{1}{5}$[{ $-$ 10 × ( 25 $-$ $\overline{)13-3}$)} ÷ ($-$5)]
=  4 + $\frac{1}{5}$[{$-$ 10 × (25 $-$ 10)} ÷ ($-$ 5)]        (On simplifying vinculum)
=  4 + $\frac{1}{5}$[{$-$ 10 × 15} ÷ ($-$5 )]                   (On simplifying parentheses)
= 4 + $\frac{1}{5}$[30]                                               (On simplifying braces)
= 4 + 6
= 10

#### Question 10:

On applying the BODMAS rule, we get:

4 + $\frac{1}{5}$[{ $-$ 10 × ( 25 $-$ $\overline{)13-3}$)} ÷ ($-$5)]
=  4 + $\frac{1}{5}$[{$-$ 10 × (25 $-$ 10)} ÷ ($-$ 5)]        (On simplifying vinculum)
=  4 + $\frac{1}{5}$[{$-$ 10 × 15} ÷ ($-$5 )]                   (On simplifying parentheses)
= 4 + $\frac{1}{5}$[30]                                               (On simplifying braces)
= 4 + 6
= 10

On applying the BODMAS rule, we get:

#### Question 11:

Simplify each of the following:

$63-\left(-3\right)\left\{-2-\overline{)8-3}\right\}÷3\left\{5+\left(-2\right)\left(-1\right)\right\}$

On applying the BODMAS rule, we get:

63 $-$ ($-$ 3) {$-$ 2 $-$ $\overline{)8-3}$} ÷ 3 {5 + ($-$ 2) ($-$1)}
=  63 $-$ ($-$ 3) {$-$ 2 $-$ 5} ÷ 3 {5 + 2}       (On simplifying vinculum)
= 63 $-$ ($-$ 3) ($-$ 7 ) ÷ 3 × 7                     (On simplifying braces)
= 63 $-$ ($21÷21$)
= 63 $-$ 1
= 62

#### Question 12:

Simplify each of the following:
$\left[29-\left(-2\right)\left\{6-\left(7-3\right)\right\}\right]÷\left[3×\left\{5+\left(-3\right)×\left(-2\right)\right\}\right]$

On applying the BODMAS rule, we get:

[29 $-$ ($-$ 2) {6 $-$ (7 $-$ 3)}] ÷ [3 × { $-$ 3) × ($-$ 2)}]
= [29 $-$ ($-$ 2) {6 $-$ 4}] ÷ [3 × { 5 + 6}]                       (On simplifying parentheses)
= [29 $-$ ($-$ 2) (2)] ÷ [3 × 11]                                        (On performing subtraction and addition)
= [29 + 4] $÷$ 33                                                           (On performing multiplication)
= 33 $÷$ 33
= 1

#### Question 13:

Using brackets, write a mathematical expression for each of the following:
(i) Nine multiplied by the sum of two and five.
(ii) Twelve divided by the sum of one and three.
(iii) Twenty divided by the difference of seven and two.
(iv) Eight subtracted from the product of two and three.
(v) Forty divided by one more than the sum of nine and ten.
(vi) Two multiplied by one less than the difference of nineteen and six.

(i) 9 (2 + 5)
(ii) 12 ÷ (1 + 3)
(iii) 20 ÷ (7 $-$ 2)
(iv) (2 × 3 ) $-$ 8
(v) 40 ÷ {(9 + 10) + 1}
(vi) 2 × {(19 $-$ 6) $-$ 1}

#### Question 1:

Mark the correct alternatives in each of the following:

(−1) × (−1) × (−1) × (−1) × ... 500 times =

(a) −1                                  (b) 1                                  (c) 500                                  (d) −500

The number of integers in the given product is even.

∴ (−1) × (−1) × (−1) × (−1) × ... 500 times

= 1 × 1 × 1 × 1 × ... 500 times

= 1

Hence, the correct answer is option (b).

#### Question 2:

Mark the correct alternatives in each of the following:

(−1) + (−1) + (−1) + (−1) + ... 500 times =

(a) 500                                  (b) 1                                  (c) −1                                 (d) −500

(−1) + (−1) + (−1) + (−1) + ... 500 times

= −(1 + 1 + 1 + 1 + ... 500 times)

= −500

Hence, the correct answer is option (d).

#### Question 3:

Mark the correct alternatives in each of the following:

The additive inverse of −7 is

(a) −7                                  (b) $-\frac{1}{7}$                                  (c) 7                                 (d) $\frac{1}{7}$

We know that, for every integer a, there exists integer −a such that

a + (−a) = 0 = −a + a

Here, −a is the additive inverse of a and a is the additive inverse of −a.

Now, 7 + (−7) = 0 = −7 + 7

∴ 7 is the additive inverse of −7.

Hence, the correct answer is option (c).

#### Question 4:

Mark the correct alternatives in each of the following:

The modulus of an integer x is 9, then

(a) x = 9 only                                  (b) x = −9 only                                  (c) x = ± 9                                  (d) None of these

The modulus (or absolute value) of an integer is its numerical value regardless of its sign. The absolute value of an integer is always non-negative.

It is given that,

Modulus of x = | x | = 9

Now, | −9 | = 9 and | 9 | = 9

x = −9 or x = 9

x = ± 9

Hence, the correct answer is option (c).

#### Question 5:

Mark the correct alternatives in each of the following:

By how much does 5 exceed −4?

(a) 1                                  (b) −1                                  (c) 9                                  (d) −9

Difference between 5 and −4 = 5 − (−4) = 5 + 4 = 9

Thus, 5 exceed −4 by 9.

Hence, the correct answer is option (c).

#### Question 6:

Mark the correct alternatives in each of the following:

By how much less than −3 is −7?

(a) 4                                  (b) −4                                  (c) 10                                  (d) −10

Difference between −3 and −7 = (−3) − (−7) = −3 + 7 = 4

Thus, −7 is less than −3 by 4.

Hence, the correct answer is option (a).

#### Question 7:

Mark the correct alternatives in each of the following:

The sum of two integers is 24. If one of them is −19, then the other is

(a) 43                                  (b) −43                                  (c) 5                                (d) −5

Sum of two integers = 24

One of the integers = −19

∴ Other integer = Sum of two integers − One of the integers

= 24 − (−19)

= 24 + 19

= 43

Hence, the correct answer is option (a).

#### Question 8:

Mark the correct alternatives in each of the following:

What must be subtracted from −6 to obtain −14?

(a) 8                                  (b) 20                                  (c) −20                                 (d) −8

Let x be subtracted from −6 to obtain −14.

∴  −6 − x = −14

Putting x = 8, we get

LHS = −6 − 8 = −6 + (−8) = −14 = RHS

Thus, 8 must be subtracted from −6 to obtain −14.

Hence, the correct answer is option (a).

#### Question 9:

Mark the correct alternatives in each of the following:

What should be divided by 6 to get −18?

(a) −3                                  (b) 3                                  (c) −108                                  (d) 108

Let x be divided by 6 to get −18.

$\therefore x÷6=-18\phantom{\rule{0ex}{0ex}}⇒\frac{x}{6}=-18$

Putting x = −108, we get

LHS = $\frac{-108}{6}=-\frac{\left|-108\right|}{6}=-\frac{108}{6}=-18$ = RHS

Thus, −108 should be divided by 6 to get −18.

Hence, the correct answer is (c).

#### Question 10:

Mark the correct alternatives in each of the following:

Which of the following is correct?

(a) −12 > −9                                  (b) −12 < −9                                  (c) (−12) + 9 > 0                               (d) (−12) × 9 > 0

We know that if a and b are two negative integers, then the integer with greater absolute value is less than the integer with smaller absolute value.

Absolute value of −12 = | −12 | = 12

Absolute value of −9 = | −9 | = 9

∴ −12 < −9

Also,

(−12) + 9 = −3 < 0

and  (−12) × 9 = −(12 × 9) = −108 < 0

Hence, the correct answer is option (b).

#### Question 11:

Mark the correct alternative in each of the following:

The sum of two integers is −8. If one of the integers is 12, then the other is

(a) 20                                   (b) 4                                   (c) −4                                   (d) −20

Sum of two integers = −8

One of the integers = 12

∴ Other integer = Sum of two integers − One of the integers

= −8 − 12

= −8 + (−12)

= −20

Hence, the correct answer is option (d).

#### Question 12:

Mark the correct alternative in each of the following:

On subtracting −14 from −18, we get

(a) 4                                   (b) −4                                   (c) −32                                   (d) −32

−14 subtracted from −18

= −18 − (−14)

= −18 + 14

= −4

Hence, the correct answer is option (b).

#### Question 13:

Mark the correct alternative in each of the following:

(−35) × 2 + (−35) × 8 =

(a) −350                                   (b) −70                                   (c) −280                                    (d) 350

(−35) × 2 + (−35) × 8

= (−35) × (2 + 8)                              [a × ba × c = a × (b + c)]

= (−35) × 10

= −350

Hence, the correct answer is option (a).

#### Question 14:

Mark the correct alternative in each of the following:

If x ÷ 29 = 0, then x =

(a) 29                                   (b) −29                                   (c) 0                                  (d) None of these

We know that if a is a non-zero integer, then 0 ÷ a = 0.

∴ x ÷ 29 = 0

x = 0

Hence, the correct answer is option (c).

#### Question 15:

Mark the correct alternative in each of the following:

If x = (−10) + (−10) + ... 15 times and y = (−2) × (−2) × (−2) × (−2) × (−2), then x − y =

(a) 118                                   (b) −118                                   (c) −182                                  (d) 182

x = (−10) + (−10) + ... 15 times

= − (10 + 10 + ... 15 times)

= −150

y = (−2) × (−2) × (−2) × (−2) × (−2)

= −(2 × 2 × 2 × 2 × 2)                 (When the number of negative integers in a product is odd, the product is negative)

= −32

xy = −150 − (−32) = −150 + 32 = −118

Hence, the correct answer is option (b).

#### Question 16:

Mark the correct alternative in each of the following:

If a = (−1) × (−1) × (−1) × ... 100 times and b = (−1) × (−1) × (−1) × ... 95 times, then a + b

(a) −1                                   (b) −2                                   (c) 0                                   (d) 1

a = (−1) × (−1) × (−1) × ... 100 times

Here, the number of integers in the product is even.

a = (−1) × (−1) × (−1) × ... 100 times

= 1 × 1 × 1 × ... 100 times

= 1

b = (−1) × (−1) × (−1) × ... 95 times

Here, the number of integers in the product is odd.

b = (−1) × (−1) × (−1) × ... 95 times

= −(1 × 1 × 1 × ... 95 times)

= −1

So,

a + b = 1 + (−1) = 0

Hence, the correct answer is option (c).

#### Question 17:

Mark the correct alternative in each of the following:

|| 3 − 12| − 4| =

(a) −5                                   (b) 5                                   (c) 7                                   (d) −7

|| 3 − 12| − 4|

= || 3 + (−12)| − 4|

= || −9| − 4|

= |9 − 4|                 (Absolute value of an integer is its numerical value regardless of its sign)

= |5|

= 5

Hence, the correct answer is option (b).

#### Question 18:

Mark the correct alternative in each of the following:

If the difference of an integer a and (−9) is 5, then a =

(a) 4                                   (b) 5                                  (c) −4                                   (d) −9

− (−9) = 5            (Given)

a + 9 = 5

Putting a = −4, we get

LHS = −4 + 9 = 5 = RHS

a = −4

Hence, the correct answer is option (c).

#### Question 19:

Mark the correct alternative in each of the following:

The sum of two integers is 10. If one of them is negative, then the other has to be

(a) negative                                                                    (b) positive

(c) may be positive or negative                                     (d) None of these

It is given that the sum of two integers is 10, which is a positive integer.

But, we know that the sum of two negative integers is always a negative integer.

So, if the sum of two integers is positive and one of them is negative, then the other has to be positive.

For example,

−2 + 12 = 10

−5 + 15 = 10

Thus, the other integer has to be positive.

Hence, the correct answer is option (b).

#### Question 20:

Mark the correct alternative in each of the following:

If x = (−1) × (−1) × (−1) × (−1) × ... 25 times, y = (−3) × (−3) × (−3), then xy

(a) −27                                   (b) 27                                   (c) 26                                  (d) −26

x = (−1) × (−1) × (−1) × (−1) × ... 25 times

The number of integers in the given product is odd.

x = (−1) × (−1) × (−1) × (−1) × ... 25 times

= −(1 × 1 × 1 × ... 25 times)

= −1

y = (−3) × (−3) × (−3)

The number of integers in the given product is odd.

∴ y = (−3) × (−3) × (−3)

= −(3 × 3 × 3)

= −27

So,

xy = (−1) × (−27) = 27               (Product of two negative integers is always positive)

Hence, the correct answer is option (b).

#### Question 1:

Determine each of the following products:
(i) 12 â˜“ 7
(ii) (−15) â˜“ 8
(iii) (−25) â˜“ (−9)
(iv) 125 â˜“ (−8)

(i) 12 × 7 = 84

(ii) (15) × 8 =  120

(iii) (25) × (9) =  225

(iv) 125 × (8) =  1000

#### Question 2:

Find each of the following products:
(i) 3 â˜“ (−8) â˜“ 5
(ii) 9 â˜“ (−3) â˜“ (−6)
(iii) (−2) â˜“ 36 â˜“ (−5)
(iv) (−2) â˜“ (−4) â˜“ (−6) â˜“ (−8)

(i) 3 × (−8) × 5 = $-3×\left(8×5\right)$ = $-$120

(ii) 9 × (−3) × (−6) = = 162

(iii) (−2) × 36 × (−5) = = 360

(iv) (−2) × (−4) × (−6) × (−8) =  $\left(2×4×6×8\right)$ = 384

#### Question 3:

Find the value of:
(i) 1487 × 327 + (−487) × 327
(ii) 28945 × 99 − (−28945)

(i) 1487 × 327 + (−487) × 327 =

(ii) 28945 × 99 − (−28945) =

#### Question 4:

Complete the following multiplication table:

Is the multiplication table symmetrical about the diagonal joining the upper left corner to the lower right corner?

 × −4 −3 −2 −1 0 1 2 3 4 −4 16 12 8 4 0 −4 −8 −12 −16 −3 12 9 6 3 0 −3 −6 −9 −12 −2 8 6 4 2 0 −2 −4 −6 −8 −1 4 3 2 1 0 −1 −2 −3 −4 0 0 0 0 0 0 0 0 0 0 1 −4 −3 −2 −1 0 1 2 3 4 2 −8 −6 −4 −2 0 2 4 6 8 3 −12 −9 −6 −3 0 3 6 9 12 4 −16 −12 −8 −4 0 4 8 12 16

Yes, the table is symmetrical along the diagonal joining the upper left corner to the lower right corner.

#### Question 5:

Determine the integer whose product with '−1' is
(i) 58
(ii) 0
(iii) −225

The integer, whose product with −1 is the given number, can be found by multiplying the given number by −1.

Thus, we have:

(i) 58 × (−1) =  −58

(ii) 0 â€‹× (−1) = = 0

(iii) (−225) â€‹× (−1) =  225

#### Question 6:

What will be the sign of the product if we multiply together
(i) 8 negative integers and 1 positive integer?
(ii) 21 negative integers and 3 positive integers?
(iii) 199 negative integers and 10 positive integers?

Negative numbers, when multiplied even number of times, give a positive number. However, when multiplied odd number of times, they give a negative number. Therefore, we have:

(i) (negative) 8 times × (positive)  1 time = = positive integer

(ii) (negative) 21 times  × (positive) 3 times = = negative integer

(iii) (negative) 199 times × (positive) 10 times = = negative integer

#### Question 7:

State which is greater:
(i) (8 + 9) × 10 and 8 + 9  × 10
(ii) (8 − 9) × 10 and 8 − 9 × 10
(iii) {(−2) − 5} × (−6) and (−2) −5 × (−6)

(i) ( 8 + 9) × 10 = 170   >   8 + 90 = 98

(ii) (8 − 9) â€‹× 10 = −10  >  8 − 90 = − 82

(iii) {(−2) − 5 } â€‹× (−6) = −7 â€‹× (−6) = 42  >   (−2) − 5 × (−6)  = ( −2 ) −  (−30)  = −2 + 30 = 28

#### Question 8:

(i) If a × (−1) = −30, is the integer a positive or negative?
(ii) If a × (−1) = 30, is the integer a positive or negative?

(i) × (−1) = −30

When multiplied by a negative integer, a gives a negative integer. Hence, a should be a positive integer.

a = 30

(ii) × (−1) = 30

â€‹When multiplied by a negative integer, a gives a positive integer. Hence, a should be a negative integerâ€‹.

a = −30

#### Question 9:

Verify the following:
(i) 19 × {7 + (−3)} = 19 × 7 + 19 × (−3)
(ii) (−23) {(−5) + (+19)} = (−23) × (−5) + (−23) × (+19)

(i)
LHS = 19 â€‹× { 7 + (−3) } = 19 × {4} =  76

RHS =  19 × 7 + 19 × (−3) = 133 + (−57) = 76

Because LHS is equal to RHS, the equation is verified.

(ii)
LHS = (−23) {(−5) + 19} = (−23) { 14} = −322

RHS = (−23) × (−5) + (−23) × 19 = 115 + (−437) = −322

Because LHS is equal to RHS, the equation is verified.

#### Question 10:

Which of the following statements are true?
(i) The product of a positive and a negative integer is negative.
(ii) The product of three negative integers is a negative integer.
(iii) Of the two integers, if one is negative, then their product must be positive.
(iv) For all non-zero integers a and b, a × b is always greater than either a or b.
(v) The product of a negative and a positive integer may be zero.
(vi) There does not exist an integer b such that for a> 1, a × b = b × a = b.

(i) True. Product of two integers with opposite signs give a negative integer.

(ii) True. Negative integers, when multiplied odd number of times, give a negative integer.

(iii) False. Product of two integers, one of them being a negative integer, is not necessarily positive. For example, (−1) × 2 = −2

(iv) False. For two non-zero integers a and b, their product is not necessarily greater than either a or b. For example, if a = 2 and  b = −2, then, × b = −4, which is less than both 2 and −2.

(v) False. Product of a negative integer and a positive integer can never be zero.

(vi) True. If a > 1, then,

#### Question 1:

Divide:
(i) 102 by 17
(ii) −85 by 5
(iii) −161 by −23
(iv) 76 by −19
(v) 17654 by −17654
(vi) (−729) by (−27)
(vii) 21590 by −10
(viii) 0 by −135

(i) 102  ÷ 17 = 6
(ii) −85  ÷  5 =  − 17
(iii) −161  ÷ ( − 23)  =  7
(iv) 76   ÷  − 19 =  − 4
(v) 17654   ÷  (− 17654) =  − 1
(vi) (−729)  ÷  (− 27) =   27
(vii) 21590   ÷  − 10 = − 2159
(viii) 0 â€‹ ÷ â€‹(−135) =

#### Question 2:

Fill in the blanks:
(i) 296 ÷ ... = −148
(ii) −88 ÷ ... = 11
(iii) 84 ÷ ... = 12
(iv) ....... ÷ −5 = 25
(v) ....... ÷ 156 = −2
(vi) ....... ÷ 567 = −1

(i) 296  ÷ $-$148 =
∴

(ii) − 88  ÷ 11 =

(iii) 84  ÷ 12 =

(iv)

(v)

∴

(vi)

#### Question 3:

Which of the following statements are true?
(i) 0 ÷ 4 = 0
(ii) 0 ÷ (−7) = 0
(iii) −15 ÷ 0 = 0
(iv) 0 ÷ 0 = 0
(v) (−8) ÷ (−1) = −8
(vi) −8 ÷ (−2) = 4

(i)

Because LHS is equal to RHS, the equation is true.

(ii)

Because LHS is equal to RHS, the equation is true.

(iii)

Because LHS is not equal to RHS, the equation is false.

(iv)

Because LHS is not equal to RHS, the equation is false.

(v)

Because LHS and RHS are not equal, the equation is false.

(vi)

Because LHS is equal to RHS, the equation is true.

#### Question 1:

Find the value of
36 ÷ 6 + 3

On applying the BODMAS rule, we get:

36  ÷  6  + 3

= 6 + 3  (On performing division)

= 9

#### Question 2:

Find the value of
24 + 15 ÷ 3

On applying the BODMAS rule, we get:

24 + 15 ÷ 3

= 24 + 5  (On performing division)

= 29

#### Question 3:

Find the value of
120 − 20 ÷ 4

On applying the DMAS rule, we get:

120 − 20 ÷ 4

= 120 − 5 (On performing division)

= 115

#### Question 4:

Find the value of
32 − (3 × 5) + 4

On applying the DMAS rule, we get:

32 − ( 3 â€‹× 5 ) + 4

= 32 − 15 + 4    (On performing multiplication)

=  36 − 15         (On performing addition)

= 21                  (On performing â€‹subtraction)

#### Question 5:

Find the value of
3 − (5 − 6 ÷ 3)

On applying the DMAS rule, we get:

3 $-$ ( 5 $-$ 6 ÷ 3)

= 3 $-$ ( 5 − 2)       (On performing division)

= 3 $-$ 3                (On performing subtraction)

= 0

#### Question 6:

Find the value of
21 − 12 ÷ 3 × 2

On applying the DMAS rule, we get:

21 − 12 ÷ 3 × 2

= 21 − 4 × 2         (On performing division)

= 21 − 8               (On performing multiplication)

= 13                      (On performing subtraction)

#### Question 7:

Find the value of
16 + 8 ÷ 4 − 2 × 3

On applying the DMAS rule, we get:

16 + 8 ÷ 4 − 2 â€‹× 3

= 16 + 2 − 6            (On performing division and multiplication)

= 18 − 6

= 12

#### Question 8:

Find the value of
28 − 5 × 6 + 2

On applying the DMAS rule, we get:

28 $-$ 5 × 6 + 2

= 28 $-$ 30 + 2 (On performing multiplication)

= 30 $-$ 30       (On performing addition)

= 0                 (On performing subtraction)

#### Question 9:

Find the value of
(−20) × (−1) + (−28) ÷ 7

On applying the DMAS rule, we get:

(− 20) â€‹× (− 1) + (−28) ÷ 7

=  20 + (− 4)   (On performing division and multiplication)

= 20 − 4

= 16

#### Question 10:

Find the value of
(−2) + (−8) ÷ (−4)