NCERT Solutions for Class 6 Maths Chapter 6 (Ex 6.3)
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Access NCERT Solutions for Class 6 Maths Chapter 6 – Integers
Exercise 6.3
1. Subtract the following integers:
(a) $35  \left( {20} \right)$
Ans: Integer is a set of both positive and negative numbers including 0.
Integers are subtracted just like positive numbers. However, the answer carries the sign of the number that is greater in magnitude.
We are given an expression $35  \left( {20} \right)$, and we are required to subtract and solve it.
To do so, we will first subtract the numbers irrespective of the sign. On doing so we get,
$35  \left( {20} \right) = 15$
Since 35 is greater than 20 and the sign of 35 is positive, so the answer will also carry the positive sign.
Therefore, the value of the given expression $35  \left( {20} \right)$ is 15.
(b) $72  \left( {90} \right)$
Ans: Integer is a set of both positive and negative numbers including 0.
Integers are subtracted just like positive numbers. However, the answer carries the sign of the number that is greater in magnitude.
We are given an expression $72  \left( {90} \right)$, and we are required to subtract and solve it.
To do so, we will first subtract the numbers irrespective of the sign. On doing so we get,
$90  72 = 18$
Since 90 is greater than 72 and the sign of 90 is negative, so the answer will also carry the negative sign.
$72  \left( {90} \right) =  18$
Therefore, the value of the given expression $72  \left( {90} \right)$ is $  18$.
(c) $\left( {  15} \right)  \left( {  18} \right)$
Ans: Integer is a set of both positive and negative numbers including 0.
Integers are subtracted just like positive numbers. However, the answer carries the sign of the number that is greater in magnitude.
We are given an expression $\left( {  15} \right)  \left( {  18} \right)$, and we are required to subtract and solve it.
To do so, we will first simplify the signs.
$\left( {  15} \right)  \left( {  18} \right) = \left( {  15} \right) + 18$
We will then subtract the numbers irrespective of the sign. On doing so we get,
$\left( {  15} \right) + 18 = 3$
Since 18 is greater than 15 and the sign of 18 is positive, so the answer will also carry the positive sign.
Therefore, the value of the given expression $\left( {  15} \right)  \left( {  18} \right)$ is 3.
(d) $\left( {  20} \right)  \left( {13} \right)$
Ans: Integer is a set of both positive and negative numbers including 0.
Integers are subtracted just like positive numbers. When two integers have the same sign then they are added together and the sign that is common to both the integers is the sign of the answer.
We are given an expression $\left( {  20} \right)  \left( {13} \right)$, and we are required to subtract and solve it.
To do so, we will add both the integers as both of them have the same negative sign. Since 20and 13 both have the negative sign, so the answer will also carry the negative sign. On doing so we get,
$\left( {  20} \right)  \left( {13} \right) =  33$
Therefore, the value of the given expression $\left( {  20} \right)  \left( {13} \right)$ is $  33$.
(e) $23  \left( {  12} \right)$
Ans: Integer is a set of both positive and negative numbers including 0.
Integers are subtracted just like positive numbers. However, the answer carries the sign of the number that is greater in magnitude.
We are given an expression $23  \left( {  12} \right)$, and we are required to subtract and solve it.
To do so, we will first simplify the signs.
$23  \left( {  12} \right) = 23 + 12$
We will then add both numbers. On doing so we get,
$23 + 12 = 35$
Therefore, the value of the given expression $23  \left( {  12} \right)$ is 35.
(f) $\left( {  32} \right)  \left( {  40} \right)$
Ans: Integer is a set of both positive and negative numbers including 0.
Integers are subtracted just like positive numbers. However, the answer carries the sign of the number that is greater in magnitude.
We are given an expression $\left( {  32} \right)  \left( {  40} \right)$, and we are required to subtract and solve it.
To do so, we will first simplify the signs.
$\left( {  32} \right)  \left( {  40} \right) = \left( {  32} \right) + 40$
We will then subtract the numbers irrespective of the sign. On doing so we get,
$\left( {  32} \right) + 40 = 8$
Since 40 is greater than 32 and the sign of 40 is positive, so the answer will also carry the positive sign.
Therefore, the value of the given expression $\left( {  32} \right)  \left( {  40} \right)$ is 8.
2. Fill in the blanks with $ > $, $ < $ or $ = $ sign:
(a) $\left( {  3} \right) + \left( {  6} \right)$__________$\left( {  3} \right)  \left( {  6} \right)$
Ans: Integer is a set of both positive and negative numbers including 0.
We are given two expressions $\left( {  3} \right) + \left( {  6} \right)$ and $\left( {  3} \right)  \left( {  6} \right)$. We are required to solve the expressions and then compare the values.
To do so, we will solve both the expressions separately and then compare their values.
Consider the expression $\left( {  3} \right) + \left( {  6} \right)$.
To solve this expression, simplify the signs first.
$\left( {  3} \right) + \left( {  6} \right) =  3  6$
As the signs of both the numbers are the same, so we will add them. The answer will carry the common sign. On doing this, we get,
$  3  6 =  9$
Hence, the value of the given expression $\left( {  3} \right) + \left( {  6} \right)$ is $  9$.
Now consider the expression $\left( {  3} \right)  \left( {  6} \right)$.
To solve this expression, simplify the signs first.
$\left( {  3} \right)  \left( {  6} \right) =  3 + 6$
As the signs of both the numbers are different, so we will subtract them. The answer will carry the sign of the number that is greater in magnitude. On doing this, we get,
$  3 + 6 = 3$
Hence, the value of the given expression $\left( {  3} \right)  \left( {  6} \right)$ is 3.
Now we will compare both the values,
$  9 < 3$
Therefore, $\left( {  3} \right) + \left( {  6} \right) < \left( {  3} \right)  \left( {  6} \right)$.
(b) $\left( {  21} \right)  \left( {  10} \right)$__________$\left( {  31} \right) + \left( {  11} \right)$
Ans: Integer is a set of both positive and negative numbers including 0.
We are given two expressions $\left( {  21} \right)  \left( {  10} \right)$ and $\left( {  31} \right) + \left( {  11} \right)$. We are required to solve the expressions and then compare the values.
To do so, we will solve both the expressions separately and then compare their values.
Consider the expression $\left( {  21} \right)  \left( {  10} \right)$.
To solve this expression, simplify the signs first.
$\left( {  21} \right)  \left( {  10} \right) =  21 + 10$
As the signs of both the numbers are different, so we will subtract them. The answer will carry the sign of the number that is greater in magnitude. On doing this, we get,
$  21 + 10 =  11$
Hence, the value of the given expression $\left( {  21} \right)  \left( {  10} \right)$ is $  11$.
Now consider the expression $\left( {  31} \right) + \left( {  11} \right)$.
To solve this expression, simplify the signs first.
$\left( {  31} \right) + \left( {  11} \right) =  31  11$
As the signs of both the numbers are the same, so we will add them. The answer will carry the common sign. On doing this, we get,
$  31  11 =  42$
Hence, the value of the given expression $\left( {  31} \right) + \left( {  11} \right)$ is $  42$.
Now we will compare both the values,
$  11 >  42$
Therefore, $\left( {  21} \right)  \left( {  10} \right) > \left( {  31} \right) + \left( {  11} \right)$.
(c) $45  \left( {  11} \right)$__________$57 + \left( {  4} \right)$
Ans: Integer is a set of both positive and negative numbers including 0.
We are given two expressions $45  \left( {  11} \right)$ and $57 + \left( {  4} \right)$. We are required to solve the expressions and then compare the values.
To do so, we will solve both the expressions separately and then compare their values.
Consider the expression $45  \left( {  11} \right)$.
To solve this expression, simplify the signs first.
$45  \left( {  11} \right) = 45 + 11$
As the signs of both the numbers are the same, so we will add them. The answer will carry the common sign. On doing this, we get,
$45 + 11 = 56$
Hence, the value of the given expression $45  \left( {  11} \right)$ is 56.
Now consider the expression $57 + \left( {  4} \right)$.
To solve this expression, simplify the signs first.
$57 + \left( {  4} \right) = 57  4$
As the signs of both the numbers are different, so we will subtract them. The answer will carry the sign of the number that is greater in magnitude. On doing this, we get,
$57  4 = 53$
Hence, the value of the given expression $57 + \left( {  4} \right)$ is 53.
Now we will compare both the values,
$56 > 53$
Therefore, $45  \left( {  11} \right) > 57 + \left( {  4} \right)$.
(d) $\left( {  25} \right)  \left( {  42} \right)$__________$\left( {  42} \right)  \left( {  25} \right)$
Ans: Integer is a set of both positive and negative numbers including 0.
We are given two expressions $\left( {  25} \right)  \left( {  42} \right)$ and $\left( {  42} \right)  \left( {  25} \right)$. We are required to solve the expressions and then compare the values.
To do so, we will solve both the expressions separately and then compare their values.
Consider the expression $\left( {  25} \right)  \left( {  42} \right)$.
To solve this expression, simplify the signs first.
$\left( {  25} \right)  \left( {  42} \right) =  25 + 42$
As the signs of both the numbers are different, so we will subtract them. The answer will carry the sign of the number that is greater in magnitude. On doing this, we get,
$  25 + 42 = 17$
Hence, the value of the given expression $\left( {  25} \right)  \left( {  42} \right)$ is 17.
Now consider the expression $\left( {  42} \right)  \left( {  25} \right)$.
To solve this expression, simplify the signs first.
$\left( {  42} \right)  \left( {  25} \right) =  42 + 25$
As the signs of both the numbers are different, so we will subtract them. The answer will carry the sign of the number that is greater in magnitude. On doing this, we get,
$  42 + 25 =  17$
Hence, the value of the given expression $\left( {  42} \right)  \left( {  25} \right)$ is $  17$.
Now we will compare both the values,
$17 >  17$
Therefore, $\left( {  25} \right)  \left( {  42} \right) > \left( {  42} \right)  \left( {  25} \right)$.
3. Fill in the blanks:
(a) $\left( {  8} \right) + \_\_\_\_\_\_\_\_\_\_ = 0$
Ans: Integer is a set of both positive and negative numbers including 0.
We are given an expression. We are required to fill the blank.
The sum of $\left( {  8} \right)$ and a number is 0. We know that when the sum of the same number with positive and negative signs is calculated, then it comes out to be 0.
This is also known as the additive identity of the numbers.
So, we will add 8 to $\left( {  8} \right)$ to obtain the sum 0.
Therefore, the completed sentence is as follows,
$\left( {  8} \right) + 8 = 0$
(b) $13 + \_\_\_\_\_\_\_\_\_\_ = 0$
Ans: Integer is a set of both positive and negative numbers including 0.
We are given an expression. We are required to fill the blank.
The sum of 13 and a number is 0. We know that when the sum of the same number with positive and negative signs is calculated, then it comes out to be 0.
This is also known as the additive identity of the numbers.
So, we will add $\left( {  13} \right)$ to 13 to obtain the sum 0.
Therefore, the completed sentence is as follows,
$13 + \left( {  13} \right) = 0$
(c) $12 + \left( {  12} \right) = \_\_\_\_\_\_\_\_\_\_$
Ans: Integer is a set of both positive and negative numbers including 0.
We are given an expression. We are required to fill the blank.
Consider the given expression $12 + \left( {  12} \right)$.
To solve this expression, simplify the signs first.
$12 + \left( {  12} \right) = 12  12$
As the signs of both the numbers are different, so we will subtract them. The answer will carry the sign of the number that is greater in magnitude. On doing this, we get,
$12  12 = 0$
Therefore, the completed sentence is as follows,
$12 + \left( {  12} \right) = 0$
(d) $\left( {  4} \right) + \_\_\_\_\_\_\_\_\_\_ =  12$
Ans: Integer is a set of both positive and negative numbers including 0.
We are given an expression. We are required to fill the blank.
The sum of $\left( {  4} \right)$ and a number is $  12$. Since the sign of the answer is negative, this means that the greater number is also negative.
Moreover, the operation is performed in an addition. If the other number will be positive then it will be greater than $  4$. So, in that case, the sign of the final answer will also be positive. But this is not the case. Hence, the second number cannot be positive.
Since the second integer is not positive so it is negative. On simplification, the addition sign will become negative. As both the numbers are negative so they will be added together. This means that the sum of $\left( {  4} \right)$ and a number is negative.
Now we know that when 8 is added to 4, then their sum is 12.
$4 + 8 = 12$
On multiplying the above equation with $  1$ on both the sides, we get,
$ 1\left( {4 + 8} \right) =  1\left( {12} \right)$
$ 4  8 =  12$
The above equation can also be written as,
$\left( {  4} \right) + \left( {  8} \right) =  12$
Therefore, the completed sentence is as follows,
$\left( {  4} \right) + \left( {  8} \right) =  12$
(e) $\_\_\_\_\_\_\_\_\_\_  15 =  10$
Ans: Integer is a set of both positive and negative numbers including 0.
We are given an expression. We are required to fill the blank.
The difference between a number and $15$ is $  10$. Since the sign of the answer is negative, this means that the greater number is also negative.
Moreover, the operation being performed is subtraction. If the other number will be negative then the sign of both the numbers will become the same. So, in that case, the numbers will have to be added. Since 15 is already greater than 10, hence it is not possible that something added to 15 can give 10.
Since the second integer is not negative so it is positive. As both the numbers have different signs so they will be subtracted. Since the greater number is negative, the second number will be less than 15.
Now we know that 5 subtracted from 15 is 10.
$15  5 = 10$
On multiplying the above equation with $  1$ on both the sides, we get,
$ 1\left( {15  5} \right) =  1\left( {10} \right)$
$ 15 + 5 =  10$
The above equation can also be written as,
$5  15 =  10$
Therefore, the completed sentence is as follows,
$5  15 =  10$
4. Find the value of the following expressions:
(a) $\left( {  7} \right)  8  \left( {  25} \right)$
Ans: Integer is a set of both positive and negative numbers including 0.
We are given an expression $\left( {  7} \right)  8  \left( {  25} \right)$, and we are required to solve it.
We will first simplify the signs of the given expression.
$\left( {  7} \right)  8  \left( {  25} \right) = \left( {  7} \right)  8 + 25$
We will now add $  7$ and $  8$ together as they have the same sign.
$\left( {  7} \right)  8  \left( {  25} \right) =  15 + 25$
Now the signs of both the terms are different. So, the terms will be subtracted. The answer will carry the sign of the term which is greater in magnitude.
On subtracting the terms in the above equation, we get,
$\left( {  7} \right)  8  \left( {  25} \right) = 10$
Therefore, the value of the given expression $\left( {  7} \right)  8  \left( {  25} \right)$ is 10.
(b) $\left( {  13} \right) + 32  8  1$
Ans: Integer is a set of both positive and negative numbers including 0.
We are given an expression $\left( {  13} \right) + 32  8  1$, and we are required to solve it.
We will first add $  8$ and $  1$ together. Since both of them have the same sign, the answer will also carry that sign.
$\left( {  13} \right) + 32  8  1 = \left( {  13} \right) + 32  9$
We will now subtract 9 from 32 as both of them have different signs. The answer will carry the sign of the term which is greater in magnitude.
$\left( {  13} \right) + 32  8  1 = \left( {  13} \right) + 23$
Now the signs of both the terms are different. So, the terms will be subtracted. The answer will carry the sign of the term which is greater in magnitude.
On subtracting the terms in the above equation, we get,
$\left( {  13} \right) + 32  8  1 = 10$
Therefore, the value of the given expression $\left( {  13} \right) + 32  8  1$ is 10.
(c) $\left( {  7} \right) + \left( {  8} \right) + \left( {  90} \right)$
Ans: Integer is a set of both positive and negative numbers including 0.
We are given an expression $\left( {  7} \right) + \left( {  8} \right) + \left( {  90} \right)$, and we are required to solve it.
We will first simplify the signs of the given expression.
$\left( {  7} \right) + \left( {  8} \right) + \left( {  90} \right) = \left( {  7} \right)  8  90$
We will now add $  7$ and $  8$ together as they have the same sign.
$\left( {  7} \right) + \left( {  8} \right) + \left( {  90} \right) =  15  90$
Now we will add $  15$ and $  90$ together as they have the same sign.
$\left( {  7} \right) + \left( {  8} \right) + \left( {  90} \right) =  105$
Therefore, the value of the given expression $\left( {  7} \right) + \left( {  8} \right) + \left( {  90} \right)$ is $  105$.
(d) $50  \left( {  40} \right)  \left( {  2} \right)$
Ans: Integer is a set of both positive and negative numbers including 0.
We are given an expression $50  \left( {  40} \right)  \left( {  2} \right)$, and we are required to solve it.
We will first simplify the signs of the given expression.
$50  \left( {  40} \right)  \left( {  2} \right) = 50 + 40 + 2$
We will now add 50 and 40 together as they have the same sign.
$50  \left( {  40} \right)  \left( {  2} \right) = 90 + 2$
We will now add 90 and 2 together as they have the same sign.
$50  \left( {  40} \right)  \left( {  2} \right) = 92$
Therefore, the value of the given expression $50  \left( {  40} \right)  \left( {  2} \right)$ is 92.
NCERT Solutions for Class 6 Maths Chapter 6 Integers Exercise 6.3
Opting for the NCERT solutions for Ex 6.3 Class 6 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 6.3 Class 6 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.
Vedantu inhouse subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 6 students who are thorough with all the concepts from the Maths textbook and quite wellversed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 6 Maths Chapter 6 Exercise 6.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.
Besides these NCERT solutions for Class 6 Maths Chapter 6 Exercise 6.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our inhouse subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.
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