NCERT Solutions for Class 7th Maths Chapter 2 In-text Questions — A Quick Recap of Integers
Book page 26–27 Updated on2026-09-19
Q1.
To begin with, the coin is at point 0. If the coin is struck twice, with the first strike moving it by 4 units and the second strike moving it by 3 units, what will be the final position of the coin?
Answer
The coin ends at 7.
Start at 0 After the first strike: 0 + 4 = 4 After the second strike: 4 + 3 = 7
So the coin is 4 + 3 = 7 units from 0, to the right.
Two rightward strikes from 0: 4 units, then 3 more. The coin stops at 7.
Why it happens: the second strike does not care where the coin started. It simply pushes it 3 more units from wherever it is. So the two distances just pile up, and adding them gives the final position.
Q2.
If the coin is struck twice, and if the two movements are known, can you give a formula for the final position of the coin?
Answer
Yes. If the first strike moves the coin a units to the right and the second strike moves it b units to the right, then
P = a + b
where P is the distance of the coin from the starting point 0.
Why it happens: after the first strike the coin sits at a. The second strike adds another b units of travel to that. Nothing is lost or gained in between, so the total distance from 0 is a + b. This is why the formula holds no matter what a and b are.
Check it yourself: put a = 4 and b = 3 into P = a + b. You get 7 — exactly the answer found by walking the coin along the line.
Q3.
Now, suppose the coin can be struck to move it in either direction — left or right. The coin is at 0. If it is struck twice (the direction of the two strikes may be the same or different) can you give a formula for the final position of the coin?
Answer
The formula is the same one: P = a + b. Only the meaning of a and b widens.
Take a rightward movement as positive.
Take a leftward movement as negative.
Then a is the first movement (positive if the strike is to the right, negative if to the left), b is the second movement in the same sense, and
P = a + b
The four cases the book lists are all handled at once.
Strike 1
Strike 2
a
b
P = a + b
5 right
3 right
5
3
8
5 left
3 left
–5
–3
–8
5 right
3 left
5
–3
2
5 left
3 right
–5
3
–2
Why it happens: a signed number stores two things at once — the magnitude tells how far, the sign tells which way. Once the direction is packed into the number itself, the four separate cases stop being separate. One addition covers all of them, which is exactly why integers are worth inventing.
Q4.
Suppose the first strike moves the coin rightward by 5 units from 0, and the second strike leftward by 7 units, then we take the First Movement = 5 units, Second Movement = – 7 units. What is the final position of the coin?
Answer
The coin ends at –2.
P = a + b = 5 + (–7) = –2
So the coin is 2 units to the left of 0.
A strike of +5 followed by a strike of –7 leaves the coin at P = –2.
Why it happens: the coin first travels 5 units right, reaching 5. The second strike is stronger and points the other way, so it undoes those 5 units and carries the coin 2 units past 0. Adding a negative number is exactly this “undo, then overshoot” movement.
Q5.
If the first movement is – 4 and the final position is 5, what is the second movement?
Answer
The second movement is +9, that is, 9 units to the right.
P = a + b 5 = (–4) + b b = 5 – (–4) b = 5 + 4 b = 9
Check: (–4) + 9 = 5 ✓
Why it happens: the first strike carries the coin to –4, which is 4 units left of 0. To finish 5 units right of 0, the second strike must first bring it back across those 4 units and then push it 5 units further. That is 4 + 5 = 9 units, all rightward, so b = +9.
Q6.
If there are multiple strikes causing movements in the order 1, – 2, 3, – 4, …, – 10, what is the final position of the coin?
Answer
The coin ends at –5.
The ten movements are 1, –2, 3, –4, 5, –6, 7, –8, 9, –10. Pair each odd movement with the negative one just after it.
Why it happens: in every pair the leftward strike is exactly 1 unit stronger than the rightward strike before it. So each pair drags the coin 1 unit to the left. Five such pairs drag it 5 units left of 0.
Check it yourself: add them straight through — 1, –1, 2, –2, 3, –3, 4, –4, 5, –5. The running total after each strike lands on –5 at the end.