NCERT Solutions for Class 7th Maths Chapter 7 .1 Find the Unknowns — In-text Questions
Book page 164 – 1657 Updated on2026-09-19
Q1.
Find the unknown weights in the following cases:Fig. 7.1 — the scale is marked 16 and the beam hangs level. A leaf weighs 3.
Answer
Bud = 2 and flower = 8.
First read what the picture is telling us. The number in the circle at the top is the total weight hanging from the scale. The beam is level, so the two strings carry equal weight.
Total = 16, and the two sides are equal
So each side carries 16 ÷ 2 = 8
Left string: leaf + bud + leaf = 8
3 + b + 3 = 8
b + 6 = 8
b = 8 − 6 = 2
Right string: flower = 8, so flower = 8
Total 16 splits equally into 8 and 8, because the beam is level.
Why it happens: the sample pictures at the start of the section show the rule. A scale marked 4 with 2 and 2 below it hangs level; a scale marked 7 with 4 and 3 below it tilts. So the circled number is the sum of the two sides, and a level beam means the two sides are equal. Once you know one side is 8, the left string becomes a small puzzle: two leaves already use 6, so the bud must make up the last 2.
Check it yourself: 3 + 2 + 3 = 8 on the left, 8 on the right, and 8 + 8 = 16 — the number in the circle. Everything fits.
Q2.
Find the unknown weights in the following cases:Fig. 7.2 — the scale is marked 24 and the beam hangs level. A starfish weighs 2.
Answer
Striped fish = 8 and grey fish = 4.
Total = 24, beam is level, so each side = 24 ÷ 2 = 12
Left string: starfish + striped fish + starfish = 12
2 + f + 2 = 12
f + 4 = 12
f = 8 (the striped fish)
Right string: striped fish + grey fish = 12
8 + g = 12
g = 12 − 8 = 4 (the grey fish)
Why it happens: notice the order of work. The left string had only one unknown, so it could be solved straight away. That answer then unlocked the right string, which had two unknowns to start with. When a puzzle has several unknowns, always look first for the part that has just one.
Check it yourself: left = 2 + 8 + 2 = 12, right = 8 + 4 = 12, total = 24. ✓
Q3.
Find the unknown weights in the following cases:Fig. 7.3 — the scale is marked 8. A book hangs on the left; on the right a second beam carries two identical pencil boxes. Both beams hang level.
Answer
Book = 4 and each pencil box = 2.
Top beam is level, total = 8
So each side of the top beam = 8 ÷ 2 = 4
Left side is only the book, so book = 4
Right side = the small beam with its two boxes = 4
The small beam is level too, so its two sides are equal
box + box = 4
2 × box = 4
box = 4 ÷ 2 = 2
Why it happens: a scale hanging from a scale is just the same rule used twice. Whatever hangs from a beam is shared equally between its two strings when the beam is level. So 8 splits into 4 and 4, and then the 4 on the right splits again into 2 and 2.
Tip: work from the top downwards in these hanging puzzles. The top number is the biggest piece of information you have.
Q4.
Find the unknown weights in the following cases:Fig. 7.4 — the scale is marked 18 and the beam hangs level. The sun weighs 5.
Answer
Cloud = 1 and lightning bolt = 3.
Total = 18, beam is level, so each side = 18 ÷ 2 = 9
Left string: sun + 4 clouds = 9
5 + 4c = 9
4c = 9 − 5 = 4
c = 4 ÷ 4 = 1 (one cloud)
Right string: 3 lightning bolts = 9
3L = 9
L = 9 ÷ 3 = 3 (one bolt)
Why it happens: here the same object appears several times, so we can group. Four clouds weigh 4c together. Taking away the sun's 5 from both sides leaves 4c = 4, and sharing 4 equally among 4 clouds gives 1 each. This is exactly the two moves you will use again and again in this chapter: remove a term, then remove a factor.
Check it yourself: 5 + 1 + 1 + 1 + 1 = 9 and 3 + 3 + 3 = 9, and 9 + 9 = 18. ✓
Q5.
Find the unknown weights in the following cases:Fig. 7.5 — the scale is marked 40 and the beam hangs level.
Answer
Crown = 5 and gem = 3.
Total = 40, beam is level, so each side = 40 ÷ 2 = 20
Right string first — it has only one kind of object:
4 crowns = 20
4k = 20, so k = 20 ÷ 4 = 5 (one crown)
Why it happens: the right string is the easy door into this puzzle — four equal crowns sharing 20 must be 5 each. Neither string could be solved on its own if we had started on the left, because the left has two different unknowns. Choosing the right starting point saves all the work.
Check it yourself: left = 5 + 3 + 3 + 3 + 3 + 3 = 20, right = 5 + 5 + 5 + 5 = 20, total 40. ✓
Q6.
Find the unknown weights in the following cases:Fig. 7.6 — the balance hangs level. One slice of bread weighs 2.
Answer
One fried egg = 3.
This picture has no number at the top. It is a plain balance: the beam is level, so the two arms carry equal weight.
Left arm: 3 slices of bread = 2 + 2 + 2 = 6
Right arm: 2 fried eggs = e + e = 2e
The arms balance, so
2e = 6
e = 6 ÷ 2 = 3
Why it happens: a level beam is a statement of equality — exactly what an equation is. The book writes this same picture as 6 = e + e, or 2e = 6. From here on, every balance picture in this chapter can be turned into an equation in one line.
Check it yourself: two eggs weigh 3 + 3 = 6, the same as three slices of bread. The beam stays level. ✓
Q7.
Find the unknown weights in the following cases:Fig. 7.7 — the balance hangs level. One star weighs 4.
Answer
One doughnut = 6.
Left arm: 4 stars = 4 × 4 = 16
Right arm: doughnut + star + doughnut = y + 4 + y = 4 + 2y
The beam is level, so
4 + 2y = 16
2y = 16 − 4 (remove the term 4 from the left side)
2y = 12
y = 12 ÷ 2 = 6
Why it happens: there is a star on both arms. Taking that one star off both arms at the same time does not disturb the balance — it leaves 3 stars = 12 on the left and 2 doughnuts on the right, so 2y = 12. Removing equal weights from both pans is the whole idea of this chapter, and it is the same as the algebra step "subtract 4 from both sides".
Tip: the book frames this picture as the equation 4 + 2y = 16. Compare it with your own working — they are the same steps, written in symbols.
Q8.
Find the unknown weights in the following cases:Fig. 7.8 — the balance hangs level. The watermelon slice weighs 10 and the orange weighs 4.
Answer
One banana = 3.
Left arm: watermelon = 10
Right arm: banana + orange + banana = b + 4 + b = 2b + 4
The beam is level, so
2b + 4 = 10
2b = 10 − 4 (remove the orange from the right arm and 4 from the left)
2b = 6
b = 6 ÷ 2 = 3
Why it happens: think of it as taking the orange off the right pan. To keep the balance you must take away the same 4 units from the left pan too, leaving 6 units against two bananas. Each banana is therefore 3.
Check it yourself: 3 + 4 + 3 = 10, which is exactly the watermelon. ✓