# Ex 1.2 Integers - NCERT Solution Class 7

Exercise 1.2

## Question 1

Write down a pair of integers, whose:

a) Sum is \(–7 \)

b) Difference is \(–10\)

c) Sum is \(0\)

### Solution

**Video Solution**

**What is known?**

Sum and difference of a pair of integers.

**What is the unknown?**

The two integers whose sum is given.

**Reasoning:**

Add or subtract the two values.

**Steps:**

a) sum is \(-7\)

Let us take a pair of integer \(-8\) and \(+1 \)

\[\begin{align}&={- 8}{+1}\\&=-7\end{align}\]

b) difference is \(-10 \)

Let us take a pair of integer \(-12\) and \(-2 \)

\[\begin{align}&={- 12\,- }{(-2)}\\&=-10\end{align}\]

c) sum is \(0\)

Let us take a pair of integer \(-5\) and \(+5\)

\[\begin{align}&={5\,+ }{(-5)}\\&=0\end{align}\]

## Question 2

a) Write the pair of negative integers whose difference gives \(8.\)

b) Write a negative integer and a positive integer whose sum is \(–5.\)

c) Write a negative integer and a positive integer whose difference is \(–3.\)

### Solution

**Video Solution**

**What is known?**

The two integers whose sum and difference is given.

**What is unknown?**

Sum and difference of a pair of integers.

**Reasoning:**

Add or subtract the two values.

**Steps:**

a) Let us have \(-2\) and \(-10\)

\[\begin{align}\therefore{\rm{Difference}}& = (-2)\,-(-10) \\&=- 2 + 10\\&= 8\end{align}\]

b) Let us have \(-10\) and \(5\)

\[\begin{align}\therefore\rm{Sum} &= -10\,+\,5 \\&=\,-5\end{align}\]

c) Let us have \(-2\) and \(1\)

\[\begin{align}\therefore\rm{Difference} &= -2\,-\,1\\&=\,-3\end{align}\]

## Question 3

In a quiz, team \(A\) scored \(–40,10,0\) and team \(B\) scored \(10,0, –40\) in \(3\) successive rounds. Which team scored more? Can we say that we can add integers in any order?

### Solution

**Video Solution**

**What is known?**

Score of team \(A\) and \(B\)

**What is unknown?**

The total score of team \(A\) and \(B\).

**Reasoning:**

Add the score

**Steps:**

Scores of team \[\begin{align}A = –40, 10, 0\end{align}\]

Total score of team \[\begin{align}A = – 40 + 10 + 0 = – 30\end{align}\]

Scores of team \[\begin{align}B= 10, 0, –40\end{align}\]

Total scores of team \[\begin{align}B= 10 +0 + (–40) = –30\end{align}\]

So,the scores of both the teams are equal.

Yes, we can add integers in any order.

We had already observed that scores obtained by both the teams are numerically equal.

But their order is different. So, the scores obtained by both the teams are equal.

## Question 4

Fill in the blanks to make the statements true:

i) \((–5) +(–8) = (–8) +(…..)\)

ii) \(–53+……= –53\)

iii) \(17+……= 0\)

iv) \(\begin{Bmatrix} [13+(–12)]+(…..)\\=13 + [(–12)+(–7)]\end{Bmatrix}\)

v) \(\begin{Bmatrix} (–4) + [15+(–3)] \\ = [–4 + 15] +…… \end{Bmatrix}\)

### Solution

**Video Solution**

**What is known?**

Different values

**What is unknown?**

Missing values

**Reasoning:**

We have to use different laws of addition.

**Steps:**

i)

\[\begin{align}&( {-5}){\rm{ }} + ( {-8}) = ( {-8}) +( {-5})\\&\text{[Commutative law of additions]}\end{align}\]

ii)

\(\begin{align}-53 + 0 =-53\text{[Additive Identity]}\end{align}\)

(adding to any integer, it gives the same value.)

iii)

\[\begin{align}{\rm{17 + }}\left( {{\rm{-17}}} \right)\,{\rm{ = 0}}\text{[Additive inverse]}\end{align}\]

iv)

\[\begin{align}&\left[ {13 \!+\! \left( -12 \right)} \right] \!+\! \left( {-7} \right) \!=\! 13 \!+\! \left[ {{\rm{ }}\left( {-12} \right){\rm{ }} \!+\! {\rm{ }}\left( {-7} \right)}\right]\\&\text{[Associative law of addition]}\end{align}\]

v)

\[\begin{align}& \left( {-4} \right) \!+ \!\left[ {15\!+\! \left( {-3} \right)} \right] \!=\! \left[ {\left( {-4} \right) \!+\! 15} \right] \!+\! \left( {-3} \right)\\&\text{[Associative law of addition]}\end{align}\]

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