RD Sharma Solutions Class 7 Chapter 1 Integers

Read RD Sharma Solutions Class 7 Chapter 1 Integers below, students should study RD Sharma class 7 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 7 Mathematics have been prepared by teacher of Grade 7. These RD Sharma class 7 Solutions have been designed as per the latest NCERT syllabus for class 7 and if practiced thoroughly can help you to score good marks in standard 7 Mathematics class tests and examinations

Exercise 1.1


Question 1: Determine each of the following products:

(i) 12 × 7

(ii) (-15) × 8

(iii) (-25) × (-9)

(iv) 125 × (-8) 

Solution 1: 

(i) 12 × 7  

For find the products of following numbers:

RD Sharma Solutions Class 7 Chapter 1 Integers

As the product of two integer values of opposite signs is similar to the additive inverse of the product of their absolute values.

 

Question 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 2:  

(i) 3 × (-8) ×5 

For find the products of following numbers:

= 3 × (-8) × 5

= 3 × (-8 × 5)

= 3 × -40

= -120

The product of integers with opposite signs is equivalent to the additive inverse of the product of their absolute values.

 

(ii) 9 × (-3) × (-6)

For find the products of following numbers:

= 9 × (-3) × (-6)

= 9 × (-3 × -6)

= 9 × (+18)

= 162

[∵ The product of integers of like signs is equivalent to the product of their absolute values.]

 

(iii) (-2) × 36 × (-5)

For find the products of following numbers:

= (-2) × 36 × (-5)

= (-2 × 36) × (-5)

= -72 × -5 = +360

[∵ The product of integers of like signs is equivalent to the product of their absolute values.]

 

(iv) (-2) × (-4) × (-6) × (-8)

For find the products of following numbers:

= (-2) × (-4) × (-6) × (-8)

= (-2 × -4) × (-6 × -8)

= -8 × (-48) = +384

[∵ The product of integers of like signs is equivalent to the product of their absolute values.]

 

Question 3. Find the value of:

(i) 1487 × 327 + (-487) × 327

(ii) 28945 × 99 – (-28945) 

Solution 3:

(i) 1487 × 327 + (-487) × 327

According to the multiplication of integers,

1487 × 327 + (-487) × 327

= (1487 × 327) + (-487 × 327)

= 486249 – 159249

=327000

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

 

(ii) 28945 × 99 – (-28945)

According to the multiplication of integers,

28945 × 99 – (-28945)

= (28945 × 99) – (-28945)

= 2865555 + 28945

=2894500

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

 

Question 4 : Complete the following multiplication table:

RD Sharma Solutions Class 7 Chapter 1 Integers

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

Solution 4:

RD Sharma Solutions Class 7 Chapter 1 Integers


This is clear from the table that the table is symmetrical also with diagonal that links the upper left corner to the bottom right corner.

 

Question 5: Determine the integer whose product with ‘-1’ is

(i) 58

(ii) 0

(iii) -225 

Solution 5: 

(i) 58

For find the integer of which is multiplied by -1

= 58 × (-1) = -58

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

 

(ii) 0

For find the integer of which is multiplied by -1

= 0 × -1 = 0

[because when any digit  multiplied with 0 we get 0 as their result]

 

(iii) -225

For find the integer of which is multiplied by -1

= -225 × -1 = 225

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

 

Exercise 1.2

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 

Solution 1 :

RD Sharma Solutions Class 7 Chapter 1 Integers

RD Sharma Solutions Class 7 Chapter 1 Integers

 

Exercise 1.3 


Question 1: Find the value of 

1. 36 ÷ 6 + 3 

Solution 1: 

According to BODMAS rule 

we have to division first: 36 ÷ 6 = 6  

then we have to do addition: 6 + 3 = 9 

so, the answer is 9.

 

Question 2: 24 + 15 ÷ 3 

Solution 2: 

According to BODMAS rule 

we have to division first: 15 ÷ 3 = 5 

then we have to do addition: 5 + 24 = 29 

so, the answer is 29. 

 

Question: 3. 120 – 20 ÷ 4 

Solution 3: 

According to BODMAS rule 

we have to division first: 20 ÷ 4 = 5 

then we have to do subtraction: 120 – 5 = 115 

so, the answer is 115 

 

Question: 4. 32 – (3 × 5) + 4 

Solution 4: 

According to BODMAS rule 

we have to multiplication first: 3 × 5 =15 

then addition: 15 ­+ 4 = 19 

then subtraction: 32 – 19 = 21 

so, the answer is 21 

 

Question: 5. 3 – (5 – 6 ÷ 3) 

Solution 5: 

According to BODMAS rule 

we have to division first: 6 ÷ 3 = 2 

then subtraction: (5 – 2) = 3 

then subtraction:3 – 3 = 0 

so, the answer is 0  

 

Question: 6. 21 – 12 ÷ 3 × 2 

Solution 6: 

According to BODMAS rule 

we have to division first: 12 ÷ 3= 4 

then multiplication: 4 × 2 =  

then subtraction: 21 – 8 = 13 

so, the answer is 13. 

 

Question: 7. 16 + 8 ÷ 4 – 2 × 3 

Solution 7: 

According to BODMAS rule 

we have to division first: 8 ÷ 4 = 2 

then multiplication: 2 × 3 = 6 

then addition: 16 + 2= 18 

then subtraction: 18 – 6 = 12 

so, the answer is 12. 

 

Question: 8. 28 – 5 × 6 + 2 

Solution 8: 

According to BODMAS rule 

we have to multiplication first: 5 × 6= 30 

then addition: -30 + 2 = -28 

then subtraction: 28 – 28 = 0 

So, the answer is 0. 

 

Question: 9. (-20) × (-1) + (-28) ÷ 7 

Solution 9: 

According to BODMAS rule 

we have to division first: 

then multiplication: 

then addition: 

then subtraction: 

Therefore, (-20) × (-1) + (-28) ÷ 7 = (-20) × (-1) – 4 

= 20 – 4 = 16 

 

Question: 10. (-2) + (-8) ÷ (-4) 

Solution 10: 

According to BODMAS rule 

we have to division first: (-8) ÷ (-4) = 2 

then addition: (-2) + 2 = 0 

so, the answer is 0. 

 

Question: 11. (-15) + 4 ÷ (5 – 3) 

Solution 11: 

we have to solve the bracket first: (5 – 3) = 2 

then division: 4 ÷ 2= 2 

then addition: ( -15) + 2 = -13 

so, the answer is -13 

 

Question: 12. (-40) × (-1) + (-28) ÷ 7 

Solution 12: 

According to BODMAS rule 

we have to division first: (-28) ÷ 7= -4 

then multiplication: (-40) × (-1)= 40 

then addition: 40 + (-4) = 36 

so, the answer is 36.

 

Question: 13. (-3) + (-8) ÷ (-4) -2 × (-2)

Solution 13:

According to BODMAS rule

we have to division first: (-8) ÷ (-4) = 2

then multiplication: -2 × (-2) = 4

then addition: 2 + 4 = 6

then subtraction: 6 -3 = 3

so, the answer is 3.

 

Question: 14. (-3) × (-4) ÷ (-2) + (-1)

Solution 14:

(-3) × (-4) ÷ (-2) + (-1)

According to BODMAS rule

We have to division first: (-4) ÷ (-2) = 2

Then multiplication: (-3) × 2 = -6

Then addition: (-6) + (-1) = -6 – 1 = -7 

Exercise 1.4

 

Question 1: Simplify each of the following:

1. 3 – (5 – 6 ÷ 3) 

Solution 1:

According to BODMAS rule firstly, solve the bracket:

3 – (5 – 6 ÷ 3)

= 3 – (5 – 2)

= 3 – 3

= 0

 

Question 2: -25 + 14 ÷ (5 – 3) 

Solution 2:

According to BODMAS rule firstly, solve the bracket:

-25 + 14 ÷ (5 – 3)

= -25 + 14 ÷ 2

= -25 + 7

= -18

 

 RD Sharma Solutions Class 7 Chapter 1 Integers

 

Question 5:  36 - [18 - {14- (15 - 4 ÷ 2 x 2)}] 

Solution 5:

According to BODMAS rule firstly, solve the inner most bracket first:

= 36 – [18 – {14 – (11 ÷ 2 × 2)}]

= 36 – [18 – {14 – 11}]

Now removing the parentheses we get

= 36 – [18 – 3]

Now remove the braces we get

= 36 – 15

= 21

 

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

Solution 6:

According to BODMAS rule firstly, solve the inner most bracket first:

= 45 – [38 – {20 – (6 – 3) ÷ 3}]

= 45 – [38 – {20 – 3 ÷ 3}]

Now remove the parentheses:

= 45 – [38 – 19]

Now remove the braces:

= 45 – 19

= 26

 

RD Sharma Solutions Class 7 Chapter 1 Integers

RD Sharma Solutions Class 7 Chapter 1 Integers

RD Sharma Solutions Class 7 Chapter 1 Integers

 

Question 12:  [29 – (-2) { 6 – (7 – 3)}] ÷ [ 3 × {5 + (-3) × (-2)]} 

Solution 12:

According to BODMAS rule inner move bracket 

First we have to remove the innermost brackets,

= [29 – (-2) {6 – 4}] ÷ [3 × {5 + 6}]

Now remove the parentheses ,

= [29 + 2 (2)] ÷ [3 × 11]

Now remove all braces present,

= 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. 

Solution 13:

(i) 9 × (2 + 5)

(ii) 12 ÷ (1 + 3)

(iii) 20 ÷ (7 – 2)

(iv) 2 × 3 -8

(v) 40 ÷ [1 + (9 + 10)]

(vi) 2 × [(19 -6) -1]

 

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NCERT Exemplar Solutions Class 7 Maths Comparing Quantities
NCERT Exemplar Solutions Class 7 Maths Data Handling
NCERT Exemplar Solutions Class 7 Maths Exponents and Powers
NCERT Exemplar Solutions Class 7 Maths Fractions and Decimals
NCERT Exemplar Solutions Class 7 Maths Integers
NCERT Exemplar Solutions Class 7 Maths Lines and Angles
NCERT Exemplar Solutions Class 7 Maths Perimeter and Area
NCERT Exemplar Solutions Class 7 Maths Practical Geometry
NCERT Exemplar Solutions Class 7 Maths Rational Numbers
NCERT Exemplar Solutions Class 7 Maths Simple Equation
NCERT Exemplar Solutions Class 7 Maths Triangles and Its Properties
RS Aggarwal Class 7 Mathematics Solutions Chapter 1 Integers
RS Aggarwal Class 7 Mathematics Solutions Chapter 2 Fractions
RS Aggarwal Class 7 Mathematics Solutions Chapter 3 Decimals
RS Aggarwal Class 7 Mathematics Solutions Chapter 4 Rational Numbers
RS Aggarwal Class 7 Mathematics Solutions Chapter 5 Exponents
RS Aggarwal Class 7 Mathematics Solutions Chapter 6 Algebraic Expressions
RS Aggarwal Class 7 Mathematics Solutions Chapter 7 Linear Equations in One Variable
RS Aggarwal Class 7 Mathematics Solutions Chapter 8 Ratio and Proportion
RS Aggarwal Class 7 Mathematics Solutions Chapter 9 Unitary Method
RS Aggarwal Class 7 Mathematics Solutions Chapter 10 Percentage
RS Aggarwal Class 7 Mathematics Solutions Chapter 11 Profit and Loss
RS Aggarwal Class 7 Mathematics Solutions Chapter 12 Simple Interest
RS Aggarwal Class 7 Mathematics Solutions Chapter 13 Lines and Angles
RS Aggarwal Class 7 Mathematics Solutions Chapter 14 Properties of Parallel Lines
RS Aggarwal Class 7 Mathematics Solutions Chapter 15 Properties of Triangles
RS Aggarwal Class 7 Mathematics Solutions Chapter 16 Congruence
RS Aggarwal Class 7 Mathematics Solutions Chapter 17 Constructions
RS Aggarwal Class 7 Mathematics Solutions Chapter 18 Reflection and Rotational Symmetry
RS Aggarwal Class 7 Mathematics Solutions Chapter 19 Three-Dimensional Shapes
RS Aggarwal Class 7 Mathematics Solutions Chapter 20 Mensuration
RS Aggarwal Class 7 Mathematics Solutions Chapter 21 Collection and Organisation of Data
RS Aggarwal Class 7 Mathematics Solutions Chapter 22 Bar Graph
RS Aggarwal Class 7 Mathematics Solutions Chapter 23 Probability
RD Sharma Solutions Class 7 Maths
RD Sharma Solutions Class 7 Chapter 1 Integers
RD Sharma Solutions Class 7 Chapter 2 Fraction
RD Sharma Solutions Class 7 Chapter 3 Decimals
RD Sharma Solutions Class 7 Chapter 4 Rational Numbers
RD Sharma Solutions Class 7 Chapter 5 Operation on Decimal Numbers
RD Sharma Solutions Class 7 Chapter 6 Exponents
RD Sharma Solutions Class 7 Chapter 7 Algebraic Expressions
RD Sharma Solutions Class 7 Chapter 8 Linear Equations in One Variable
RD Sharma Solutions Class 7 Chapter 9 Ratio and Proportion
RD Sharma Solutions Class 7 Chapter 10 Unitary Method
RD Sharma Solutions Class 7 Chapter 11 Percentage
RD Sharma Solutions Class 7 Chapter 12 Profit and Loss
RD Sharma Solutions Class 7 Chapter 13 Simple Interest
RD Sharma Solutions Class 7 Chapter 14 Line and Angles
RD Sharma Solutions Class 7 Chapter 15 Properties of Triangle
RD Sharma Solutions Class 7 Chapter 16 Congruence
RD Sharma Solutions Class 7 Chapter 17 Constructions
RD Sharma Solutions Class 7 Chapter 18 Symmetry
RD Sharma Solutions Class 7 Chapter 19 Visualising Solid Shapes
RD Sharma Solutions Class 7 Chapter 20 Mensuration Perimeter and Area of Rectilinear Figures
RD Sharma Solutions Class 7 Chapter 21 Mensuration Area of Circle
RD Sharma Solutions Class 7 Chapter 22 Data Handling Collection and Organisation of Data
RD Sharma Solutions Class 7 Chapter 23 Data Handling Central Values
RD Sharma Solutions Class 7 Chapter 24 Data Handling Constructions of Bar Graph
RD Sharma Solutions Class 7 Chapter 25 Data Handling Probability
RS Aggarwal Class 7 Mathematics Solutions
RS Aggarwal Class 7 Mathematics Solutions Chapter 1 Integers
RS Aggarwal Class 7 Mathematics Solutions Chapter 2 Fractions
RS Aggarwal Class 7 Mathematics Solutions Chapter 3 Decimals
RS Aggarwal Class 7 Mathematics Solutions Chapter 4 Rational Numbers
RS Aggarwal Class 7 Mathematics Solutions Chapter 5 Exponents
RS Aggarwal Class 7 Mathematics Solutions Chapter 6 Algebraic Expressions
RS Aggarwal Class 7 Mathematics Solutions Chapter 7 Linear Equations in One Variable
RS Aggarwal Class 7 Mathematics Solutions Chapter 8 Ratio and Proportion
RS Aggarwal Class 7 Mathematics Solutions Chapter 9 Unitary Method
RS Aggarwal Class 7 Mathematics Solutions Chapter 10 Percentage
RS Aggarwal Class 7 Mathematics Solutions Chapter 11 Profit and Loss
RS Aggarwal Class 7 Mathematics Solutions Chapter 12 Simple Interest
RS Aggarwal Class 7 Mathematics Solutions Chapter 13 Lines and Angles
RS Aggarwal Class 7 Mathematics Solutions Chapter 14 Properties of Parallel Lines
RS Aggarwal Class 7 Mathematics Solutions Chapter 15 Properties of Triangles
RS Aggarwal Class 7 Mathematics Solutions Chapter 16 Congruence
RS Aggarwal Class 7 Mathematics Solutions Chapter 17 Constructions
RS Aggarwal Class 7 Mathematics Solutions Chapter 18 Reflection and Rotational Symmetry
RS Aggarwal Class 7 Mathematics Solutions Chapter 19 Three-Dimensional Shapes
RS Aggarwal Class 7 Mathematics Solutions Chapter 20 Mensuration
RS Aggarwal Class 7 Mathematics Solutions Chapter 21 Collection and Organisation of Data
RS Aggarwal Class 7 Mathematics Solutions Chapter 22 Bar Graph
RS Aggarwal Class 7 Mathematics Solutions Chapter 23 Probability