## RD Sharma Solutions For Class 7 Maths Chapter 1 Integers Free Online

Exercise 1.1 Page No: 1.5

**1. Determine each of the following products:**

**(i) 12 × 7**

**(ii) (-15) × 8**

**(iii) (-25) × (-9)**

**(iv) 125 × (-8)**

**Solution:**

(i) Given 12 × 7

Here we have to find the products of given numbers

12 ×7 = 84

Because the product of two integers of like signs is equal to the product of their absolute values.

(ii) Given (-15) × 8

Here we have to find the products of given numbers

(-15) ×8 = -120

Because the product of two integers of opposite signs is equal to the additive inverse of the product of their absolute values.

(iii) Given (-25) × (-9)

Here we have to find the products of given numbers

(-25) × (-9) = + (25 ×9) = +225

Because the product of two integers of opposite signs is equal to the additive inverse of the product of their absolute values.

(iv) Given 125 × (-8)

Here we have to find the products of given numbers

125 × (-8) = -1000

Because the product of two integers of opposite signs is equal to the additive inverse of the product of their absolute values.

**2. Find each of the following products:**

**(i) 3 × (-8) × 5**

**(ii) 9 × (-3) × (-6)**

**(iii) (-2) × 36 × (-5)**

**(iv) (-2) × (-4) × (-6) × (-8)**

**Solution:**

(i) Given 3 × (-8) ×5

Here we have to find the product of given number.

3 × (-8) × 5 = 3 × (-8 × 5)

=3 × -40 = -120

Since the product of two integers of opposite signs is equal to the additive inverse of the product of their absolute values.

(ii) Given 9 × (-3) × (-6)

Here we have to find the product of given number.

9 × (-3) × (-6) = 9 × (-3 × -6) [∵ the product of two integers of like signs is equal to

the product of their absolute values.]

=9 × +18 = +162

(iii) Given (-2) × 36 × (-5)

Here we have to find the product of given number.

(-2) × 36 × (-5) = (-2 × 36) × -5 [∵ the product of two integers of like signs is equal to

the product of their absolute values.]

=-72 × -5 = +360

(iv) Given (-2) × (-4) × (-6) × (-8)

Here we have to find the product of given number.

(-2) × (-4) × (-6) × (-8) = (-2 × -4) × (-6 × -8) [∵ the product of two integers of like signs is

equal to the product of their absolute values.]

=-8 × -48 = +384

**3. Find the value of:**

**(i) 1487 × 327 + (-487) × 327**

**(ii) 28945 × 99 – (-28945)**

**Solution:**

(i) Given 1487 × 327 + (-487) × 327

By using the rule of multiplication of integers, we have

1487 × 327 + (-487) × 327 = 486249 – 159249

Since the product of two integers of opposite signs is equal to the additive inverse of the product of their absolute values.

=327000

(ii) Given 28945 × 99 – (-28945)

By using the rule of multiplication of integers, we have

28945 × 99 – (-28945) = 2865555 + 28945

Since the product of two integers of like signs is equal to the product of their absolute values.

=2894500

**4. Complete the following multiplication table:**

**Second number**

First number | x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |

-4 | ||||||||||

-3 | ||||||||||

-2 | ||||||||||

-1 | ||||||||||

0 | ||||||||||

1 | ||||||||||

2 | ||||||||||

3 | ||||||||||

4 |

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

**Solution:**

Second number

First number | x | -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 |

From the table it is clear that, the table is symmetrical about the diagonal joining the upper left corner to the lower right corner.

**5. Determine the integer whose product with ‘-1’ is**

**(i) 58**

**(ii) 0**

**(iii) -225**

**Solution:**

(i) Given 58

Here we have to find the integer which is multiplied by -1

We get, 58 × -1 = -58

Since the product of two integers of opposite signs is equal to the additive inverse of the product of their absolute values.

(ii) Given 0

Here we have to find the integer which is multiplied by -1

We get, 0 × -1 = 0 [because anything multiplied with 0 we get 0 as their result]

(iii) Given -225

Here we have to find the integer which is multiplied by -1

We get, -225 × -1 = 225

Since the product of two integers of like signs is equal to the product of their absolute values.

## Exercise 1.2 Page No: 1.8

**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**

**Solution:**

(i) Given 102 by 17

We can write given question as 102 ÷ 17

102 ÷ 17 = |102/17| = |102|/|17| [by applying the mod]

= 102/17 = 6

(ii) Given -85 by 5

We can write given question as -85 ÷ 5

-85 ÷ 5 = |-85/5| = |-85|/|5| [by applying the mod]

= -85/5 = -17

(iii) Given -161 by -23

We can write given question as -161 ÷ -23

-161 ÷ -23 = |-161/-23| = |-161|/|-23| [by applying the mod]

= 161/23 = 7

(iv) Given 76 by -19

We can write given question as 76 ÷ -19

76 ÷ -19 = |76/-19| = |76|/|-19| [by applying the mod]

= 76/-19 = -4

(v) Given 17654 by 17654

We can write given question as 17654 ÷ -17654

17654 ÷ -17654 = |17654/-17654| = |17654|/|-17654| [by applying the mod]

= 17654/-17654 = -1

(vi) Given (-729) by (-27)

We can write given question as (-729) ÷ (-27)

(-729) ÷ (-27) = |-729/-27| = |-729|/|-27| [by applying the mod]

= 729/27 = 27

(vii) Given 21590 by -10

We can write given question as 21590 ÷ -10

21590 ÷ -10 = |21590/-10| = |21590|/|-10| [by applying the mod]

= 21590/-10 = -2159

(viii) Given 0 by -135

We can write given question as 0 ÷ -135

0 ÷ -135 = 0 [because anything is divided by 0 we get the result as 0]

## Exercise 1.3 Page No: 1.9

**Find the value of**

**1. 36 ÷ 6 + 3**

**Solution:**

Given 36 ÷ 6 + 3

According to BODMAS rule we have operate division first then we have to do addition

Therefore 36 ÷ 6 + 3 = 6 + 3 = 9

**2. 24 + 15 ÷ 3**

**Solution:**

Given 24 + 15 ÷ 3

According to BODMAS rule we have operate division first then we have to do addition

Therefore 24 + 15 ÷ 3 = 24 + 5 = 29

**3. 120 – 20 ÷ 4**

**Solution:**

Given 120 – 20 ÷ 4

According to BODMAS rule we have operate division first then we have to do multiplication

Therefore 120 – 20 ÷ 4 = 120 – 5 = 115

**4. 32 – (3 × 5) + 4**

**Solution:**

Given 32 – (3 × 5) + 4

According to BODMAS rule we have operate in brackets first then we have move to addition and subtraction.

Therefore 32 – (3 × 5) + 4 = 32 – 15 + 4

= 32 – 11 = 21

**5. 3 – (5 – 6 ÷ 3)**

**Solution:**

Given 3 – (5 – 6 ÷ 3)

According to BODMAS rule we have operate in brackets first then we have move to subtraction.

Therefore 3 – (5 – 6 ÷ 3) = 3 – (5 – 2)

= 3 –3 = 0

**6. 21 – 12 ÷ 3 × 2**

**Solution:**

Given 21 – 12 ÷ 3 × 2

According to BODMAS rule we have perform division first then multiplication and addition.

Therefore, 21 – 12 ÷ 3 × 2 = 21 – 4 × 2

= 21 – 8 = 13

**7. 16 + 8 ÷ 4 – 2 × 3**

**Solution:**

Given 16 + 8 ÷ 4 – 2 × 3

According to BODMAS rule we have perform division first then followed by multiplication, addition and subtraction.

Therefore, 16 + 8 ÷ 4 – 2 × 3 = 16 + 2 – 2 × 3

= 16 + 2 – 6

= 18 -6

= 12

**8. 28 – 5 × 6 + 2**

**Solution:**

Given 28 – 5 × 6 + 2

According to BODMAS rule we have perform multiplication first then followed by addition and subtraction.

Therefore, 28 – 5 × 6 + 2 = 28 – 30 +2

= 28 – 28 = 0

**9. (-20) × (-1) + (-28) ÷ 7**

**Solution:**

Given (-20) × (-1) + (-28) ÷ 7

According to BODMAS rule we have perform division first then followed by multiplication, addition and subtraction.

Therefore, (-20) × (-1) + (-28) ÷ 7 = (-20) × (-1) – 6

= 20 – 6 = 16

**10. (-2) + (-8) ÷ (-4)**

**Solution:**

Given (-2) + (-8) ÷ (-4)

According to BODMAS rule we have perform division first then followed by addition and subtraction.

Therefore, (-2) + (-8) ÷ (-4) = (-2) + 2

=0

**11. (-15) + 4 ÷ (5 – 3)**

**Solution:**

Given (-15) + 4 ÷ (5 – 3)

According to BODMAS rule we have perform division first then followed by addition and subtraction.

Therefore, (-15) + 4 ÷ (5 – 3) = (-15) + 4 ÷ 2

= -15 + 2

= -13

**12. (-40) × (-1) + (-28) ÷ 7**

**Solution:**

Given (-40) × (-1) + (-28) ÷ 7

According to BODMAS rule we have perform division first then followed by multiplication, addition and subtraction.

(-40) × (-1) + (-28) ÷ 7 = (-40) × (-1) – 4

= 40 – 4

= 36

**13. (-3) + (-8) ÷ (-4) -2 × (-2)**

**Solution:**

Given (-3) + (-8) ÷ (-4) -2 × (-2)

(-3) + (-8) ÷ (-4) -2 × (-2) = -3 + 2 -2 × (-2)

= -3 + 2 + 4

= 6 – 3

=3

**14. (-3) × (-4) ÷ (-2) + (-1)**

**Solution:**

Given (-3) × (-4) ÷ (-2) + (-1)

(-3) × (-4) ÷ (-2) + (-1) = -3 × 2 -1

= – 6 – 1

= -1

## Exercise 1.4 Page No: 1.12

**Simplify each of the following:**

**1. 3 – (5 – 6 ÷ 3)**

**Solution:**

Given 3 – (5 – 6 ÷ 3)

According to removal of bracket rule firstly remove inner most bracket

We get 3 – (5 – 6 ÷ 3) = 3 – (5 – 2)

= 3 – 3

= 0

**2. -25 + 14 ÷ (5 – 3)**

**Solution:**

Given -25 + 14 ÷ (5 – 3)

According to removal of bracket rule firstly remove inner most bracket

We get -25 + 14 ÷ (5 – 3) = -25 + 14 ÷ 2

= -25 + 7

= -18

**Solution:**

According to removal of bracket rule first we have to remove vinculum we get

= 25 – ½ {5 + 4 – (5 – 4)}

Now by removing the innermost bracket we get

= 25 – ½ {5 + 4 – 1}

By removing the parentheses we get

= 25 – ½ (8)

Now simplifying we get

= 25 – 4

= 21

**Solution:**

According to removal of bracket rule first we have to remove vinculum we get

= 27 – [38 – {46 – (15 – 11)}]

Now by removing inner most bracket we get

= 27 – [38 – {46 – 4}]

By removing the parentheses we get

= 27 – [38 – 42]

Now by removing braces we get

= 27 – (-4)

= 27 + 4

= 31

**Solution:**

By removing innermost bracket we get

= 36 – [18 – {14 – (11 ÷ 2 × 2)}]

= 36 – [18 – {14 – 11}]

Now by removing the parentheses we get

= 36 – [18 – 3]

Now remove the braces we get

= 36 – 15

= 21

**6. 45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}]**

**Solution:**

Given 45 – [38 – {60 ÷ 3 – (6 – 9 ÷ 3) ÷ 3}]

First remove the inner most brackets

= 45 – [38 – {20 – (6 – 3) ÷ 3}]

= 45 – [38 – {20 – 3 ÷ 3}]

Now remove the parentheses w get

= 45 – [38 – 19]

Now remove the braces we get

= 45 – 19

= 26

**Solution:**

Now first remove the vinculum we get

= 23 – [23 – {23 – (23 – 0)}]

Now remove the innermost bracket we get,

= 23 – [23 – {23 – 23}]

By removing the parentheses we get,

= 23 – [23 -0]

Now we have to remove the braces and on simplifying we get,

= 23 – 23

= 0

**Solution:**

First we have to remove the vinculum from the given equation we get,

= 2550 – [510 – {270 – (90 – 150)}]

By removing innermost bracket we get,

= 2550 – [510 – {270 – (-60)}]

= 2550 – [510 – {270 + 60}]

Now remove the parentheses we get,

= 2550 – [510 – 330]

Now we have to remove parentheses

= 2550 – 180

= 2370

**Solution:**

First we have to remove vinculum from the given equation,

= 4 + 1/5 [{-10 × (25 – 10)} ÷ (-5)]

Now remove the innermost bracket, we get

= 4 + 1/5 [{-10 × 15} ÷ (-5)]

Now by removing the parenthesis we get,

= 4 + 1/5 [-150 ÷ -5]

By removing the braces we get,

= 4 + 1/5 (30)

On simplifying we get,

= 4 + 6

= 10

**Solution:**

Now we have to remove innermost bracket

= 22 – ¼ {-5 – (- 48 ÷ – 16)}

= 22 – ¼ {-5 – 3}

Now remove the parenthesis we get

= 22 – ¼ (-8)

On simplifying we get,

= 22 + 2

= 24

**Solution:**

First we have to remove vinculum from the given equation then we get,

= 63 – [(-3) {-2 – 5}] ÷ [3 {5 + 2}]

Now remove the parenthesis from the above equation

= 63 – [(-3) (-7)] ÷ [3 (7)]

= 63 – [21] ÷ [21]

= 63 – 1

= 62

**Solution:**

First we have to remove the innermost brackets then we get,

= [29 – (-2) {6 – 4}] ÷ [3 × {5 + 6}]

Now remove the parenthesis in the above equation,

= [29 + 2 (2)] ÷ [3 × 11]

Now remove all braces present in the above equation,

= 33 ÷ 33

= 1

**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.**

**Solution:**

(i) 9 (2 + 5)

(ii) 12 ÷ (2 + 3)

(iii) 20 ÷ (7 – 2)

(iv) 2 × 3 -8

(v) 40 ÷ [1 + (9 + 10)]

(vi) 2 × [(19 -6) -1]