NCERT Solutions for Class 7th Maths Chapter 7 .1 Find the Unknowns — In-text Questions

Book page 1687 Updated on2026-09-19

Q1.

For the weighing scale problems in figures 7.6, 7.7, 7.8, 7.9, 7.10, and 7.11, frame equations by using letter-numbers to denote the unknown weight.

Answer

Name the unknown with a letter, write what each side weighs, and join the two sides with '=' because the scale is balanced.

Figure	Unknown	Left side	Right side
Fig. 7.6	e = weight of one fried egg	2 + 2 + 2 = 6	2e, so 2e = 6
Fig. 7.7	y = weight of one doughnut	4 stars = 16	4 + 2y, so 4 + 2y = 16
Fig. 7.8	b = weight of one banana	watermelon = 10	2b + 4, so 2b + 4 = 10
Fig. 7.9	s = weight of the sack	s + 2	12, so s + 2 = 12
Fig. 7.10	s = weight of one sack	2s	14 + s, so 2s = 14 + s
Fig. 7.11	s = weight of one sack	5s	21 + 2s, so 5s = 21 + 2s

Why it happens: a picture and an equation say the same thing in two languages. "Three slices of bread balance two eggs" and "6 = 2e" carry exactly the same information. Once it is written in symbols you no longer have to look at the picture — you can work on the symbols alone.

Tip: always write down in words what your letter stands for ("let e be the weight of one fried egg"). An equation with an unexplained letter is hard to check later.

Q2.

Solve the equations that you frame and check if you get the same value for the unknown weight as you got previously.

Answer

Yes — the equations give exactly the same answers as the pictures did.

Equation	Working	Answer	Matches the picture?
2e = 6	e = 6 ÷ 2	e = 3	Yes, Fig. 7.6
4 + 2y = 16	2y = 16 − 4 = 12; y = 12 ÷ 2	y = 6	Yes, Fig. 7.7
2b + 4 = 10	2b = 10 − 4 = 6; b = 6 ÷ 2	b = 3	Yes, Fig. 7.8
s + 2 = 12	s = 12 − 2	s = 10	Yes, Fig. 7.9
2s = 14 + s	2s − s = 14	s = 14	Yes, Fig. 7.10
5s = 21 + 2s	5s − 2s = 21; 3s = 21; s = 21 ÷ 3	s = 7	Yes, Fig. 7.11

Why it happens: every step you took with the picture had a twin in the algebra. "Lift the same weight off both pans" became "subtract the same number from both sides". "Share the weight equally among 3 sacks" became "divide both sides by 3". The pictures were never a different method — they were this method, drawn.

Check it yourself: put each answer back in. For 5s = 21 + 2s with s = 7: LHS = 35, RHS = 21 + 14 = 35. ✓

Q3.

Frame 5 equations. Find methods to solve them.

Answer

Here are five equations of five different shapes, each with the method that suits it. Yours may be different — compare them with your classmates'.

Equation	Shape	Method	Solution
x + 12 = 30	a term added	Remove 12: x = 30 − 12	x = 18
7m = 91	a factor	Remove the factor 7: m = 91 ÷ 7	m = 13
t ÷ 6 = 9	a divisor	Remove the divisor 6: t = 9 × 6	t = 54
4a − 5 = 23	term and factor	4a = 23 + 5 = 28, then a = 28 ÷ 4	a = 7
5p + 8 = 3p + 20	unknown on both sides	2p + 8 = 20, so 2p = 12, p = 12 ÷ 2	p = 6

Why it happens: there are really only three moves in the whole chapter — remove a term, remove a factor, remove a divisor — and every equation is some combination of them. Deal with the added or subtracted term first, and the multiplying factor last; that keeps the numbers whole for as long as possible.

Try This: make one equation whose answer is a negative number, such as 3x + 10 = 1 (giving x = −3). Swap equations with a friend and check each other's answers by substituting.

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