Class 7 Exam  >  Class 7 Notes  >  RD Sharma Solutions for Class 7 Mathematics  >  RD Sharma Solutions - Ex-1.4 Integers, Class 7, Math

Ex-1.4 Integers, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics PDF Download

QUESTION 1:

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

ANSWER 1:

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)

ANSWER 2:

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:

Ex-1.4 Integers, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics

ANSWER 3:

On applying the BODMAS rule, we get:

Ex-1.4 Integers, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics 

Ex-1.4 Integers, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics

Ex-1.4 Integers, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics


QUESTION 4:

Simplify each of the following:

Ex-1.4 Integers, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics

ANSWER 4:

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:
36[18{14(154 ÷2×2)}]36-18-{14-(15-4 ÷2×2)}

ANSWER 5:

On applying the BODMAS rule, we get:
36 - [18 - { 14 - (15 - 4 ÷ 2 × 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:
45[38{60÷3(69 ÷3)÷3}]45-38-{60÷3-(6-9 ÷3)÷3}

ANSWER 6:

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:

Ex-1.4 Integers, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics

ANSWER 7:

On applying the BODMAS rule, we get:

Ex-1.4 Integers, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics

= 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:

Ex-1.4 Integers, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics

Answer 8:

On applying the BODMAS rule, we get:

Ex-1.4 Integers, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics

= 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:

Ex-1.4 Integers, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics

Answer 9:

On applying the BODMAS rule, we get:

Ex-1.4 Integers, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics 


Ex-1.4 Integers, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics

= 4 + 6
= 10

QUESTION 10:

Ex-1.4 Integers, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics

Answer 10:

On applying the BODMAS rule, we get:

Ex-1.4 Integers, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics

Ex-1.4 Integers, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics

=22(2)  [Removing braces]

= 22+2 = 24 


Question 11:

Simplify each of the following:

Ex-1.4 Integers, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics

nswer 11:

On applying the BODMAS rule, we get:

Ex-1.4 Integers, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics 

=  63 - (- 3) {- 2 - 5} ÷ 3 {5 + 2}       (On simplifying vinculum)
= 63 - (- 3) (- 7 ) ÷ 3 × 7                     (On simplifying braces)
= 63 - (21÷2121÷21)
= 63 - 1
= 62


Question 12:

Simplify each of the following:
[29(2){6(73)}]÷[3×{5+(3)×(2)}]


Answer 12:

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.


Answer 13:

(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}

The document Ex-1.4 Integers, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
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FAQs on Ex-1.4 Integers, Class 7, Math RD Sharma Solutions - RD Sharma Solutions for Class 7 Mathematics

1. What are integers?
Ans. Integers are whole numbers that can be positive, negative, or zero. They do not have any fractional or decimal parts. Examples of integers include -3, 0, 5, and -10.
2. How to add integers?
Ans. To add integers, we need to follow the rules of addition. If the integers have the same sign, we add their absolute values and keep the common sign. If the integers have different signs, we subtract their absolute values and keep the sign of the integer with the greater absolute value.
3. Can we multiply two negative integers to get a positive integer?
Ans. Yes, when we multiply two negative integers, the result is a positive integer. For example, (-2) x (-3) = 6. This is because when we multiply two negative numbers, we are essentially multiplying their absolute values and then giving the result a positive sign.
4. How to subtract integers?
Ans. To subtract integers, we can convert the subtraction problem into an addition problem. We take the opposite of the integer that we want to subtract and then add it to the other integer using the rules of addition.
5. What are the properties of integers?
Ans. Integers have several properties, including closure property, commutative property, associative property, additive identity property, additive inverse property, and multiplicative identity property. These properties help us perform various operations with integers and solve mathematical problems.
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