CBSE Class 7 Mathematics — NCERT Solutions

Given: 1 year ≈ 365 days. We need to find y such that y × 365 is close to 1,00,000.

Let us choose y = 274.

Number of days = 274 × 365 = 1,00,010.

This is just 10 more than one lakh, so it is very close to one lakh.

Alternatively, choose y = 273:
Number of days = 273 × 365 = 99,645.
This is 1,00,000 − 99,645 = 355 less than one lakh.

So for y = 274, the number of days (1,00,010) is closest to one lakh — just 10 more than one lakh.

Given: Population = 75,000; One lakh = 1,00,000.

Concept: Subtraction.

$$1,00,000 - 75,000 = 25,000$$

∴ 75,000 is 25,000 less than one lakh.

Concept: Find the value of one side and work backwards to find the missing number.

(a) $13 + 4 = \_ + 6$

LHS $= 13 + 4 = 17$

So $\_ + 6 = 17 \Rightarrow \_ = 17 - 6 = 11$

$$\boxed{13 + 4 = 11 + 6}$$

(b) $22 + \_ = 6 \times 5$

RHS $= 6 \times 5 = 30$

So $22 + \_ = 30 \Rightarrow \_ = 30 - 22 = 8$

$$\boxed{22 + 8 = 6 \times 5}$$

(c) $8 \times \_ = 64 \div 2$

RHS $= 64 \div 2 = 32$

So $8 \times \_ = 32 \Rightarrow \_ = 32 \div 8 = 4$

$$\boxed{8 \times 4 = 64 \div 2}$$

(d) $34 - \_ = 25$

$\_ = 34 - 25 = 9$

$$\boxed{34 - 9 = 25}$$

Step 1: Evaluate each expression.

(a) $67 - 19 = 48$

(b) $67 - 20 = 47$

(c) $35 + 25 = 60$

(d) $5 \times 11 = 55$

(e) $120 \div 3 = 40$

Step 2: Arrange in ascending order: $47 < 48 < 40$...

Re-listing: $40, 47, 48, 55, 60$

Corresponding expressions:

$$\underbrace{120 \div 3}_{40} < \underbrace{67-20}_{47} < \underbrace{67-19}_{48} < \underbrace{5 \times 11}_{55} < \underbrace{35+25}_{60}$$

Ascending order: $(e),\ (b),\ (a),\ (d),\ (c)$

Given: $15 \frac{3}{10} \frac{4}{100} - 2 \frac{6}{10} \frac{8}{100}$

Step 1: Convert both numbers to hundredths notation (decimal form).

$15 \frac{3}{10} \frac{4}{100}$ means $15 + \frac{3}{10} + \frac{4}{100} = 15 + \frac{30}{100} + \frac{4}{100} = 15 + \frac{34}{100} = 15.34$

$2 \frac{6}{10} \frac{8}{100}$ means $2 + \frac{6}{10} + \frac{8}{100} = 2 + \frac{60}{100} + \frac{8}{100} = 2 + \frac{68}{100} = 2.68$

Step 2: Subtract.

$$15.34 - 2.68$$

Breaking it down by place value:
$$(10 + 5 + \tfrac{3}{10} + \tfrac{4}{100}) - (2 + \tfrac{6}{10} + \tfrac{8}{100})$$

$$= (15 - 2) + (\tfrac{3}{10} - \tfrac{6}{10}) + (\tfrac{4}{100} - \tfrac{8}{100})$$

Since $\frac{3}{10} < \frac{6}{10}$, regroup: borrow 1 unit from 15, converting it to $\frac{10}{10}$:

$$= (14 - 2) + (\tfrac{13}{10} - \tfrac{6}{10}) + (\tfrac{4}{100} - \tfrac{8}{100})$$

Since $\frac{4}{100} < \frac{8}{100}$, regroup: borrow $\frac{1}{10}$ from $\frac{13}{10}$, converting it to $\frac{10}{100}$:

$$= (14 - 2) + (\tfrac{12}{10} - \tfrac{6}{10}) + (\tfrac{14}{100} - \tfrac{8}{100})$$

$$= 12 + \tfrac{6}{10} + \tfrac{6}{100}$$

$$= 12.66$$

Answer: $15.34 - 2.68 = 12.66$

Convert to decimal form:

$\frac{3}{10} = 0.30$

$3\frac{4}{100} = 3 + \frac{4}{100} = 3.04$

Add:
$$0.30 + 3.04 = 3.34$$

Answer: $3.34$

Given: A triangle with all sides equal (equilateral triangle). Let each side have length $s$.

Concept: Perimeter = sum of all sides.

Working:
$$P = s + s + s = 3 \times s = 3s$$

Formula: $P = 3s$

Given: Two lines intersect at a point forming four angles: $\angle a$, $\angle b$, $\angle c$, $\angle d$ (going around the intersection).

Concept:

• A linear pair consists of two adjacent angles whose non-common arms form a straight line; they add up to $180°$.
• Vertically opposite angles are the angles across the intersection from each other; they are always equal.

Linear Pairs (each pair sums to $180°$):
$$\angle a \text{ and } \angle b, \quad \angle b \text{ and } \angle c, \quad \angle c \text{ and } \angle d, \quad \angle d \text{ and } \angle a$$

Pairs of Vertically Opposite Angles (each pair is equal):
$$\angle a = \angle c \quad \text{and} \quad \angle b = \angle d$$

Summary Table:

Since the actual figure is not visible, the method is as follows:

Given: Each child counts how many children standing in front of them (i.e., between them and the front of the line) are taller than them.

Step 1: Start from the child at the front of the line — they have no one in front, so they always say 0.

Step 2: For each subsequent child, count only those children who are positioned in front of them AND are taller than them.

Step 3: Write that count as the child's number.

Apply this rule to the given figure to obtain the sequence.

Given: Side lengths 4 cm, 4 cm, 6 cm.

Check Triangle Inequality:

• $4 + 4 = 8 > 6$ ✓
• $4 + 6 = 10 > 4$ ✓
• $4 + 6 = 10 > 4$ ✓

All conditions satisfied, so the triangle exists.

Steps of Construction:

Step 1: Draw base $AB = 6$ cm.

Step 2: With A as centre, draw an arc of radius 4 cm.

Step 3: With B as centre, draw an arc of radius 4 cm such that it intersects the first arc at point C.

Step 4: Join AC and BC.

$\triangle ABC$ is the required triangle with $AB = 6$ cm, $AC = 4$ cm, $BC = 4$ cm.

This is an isosceles triangle (two equal sides of 4 cm).

Given: Tenzin drinks $\frac{1}{2}$ glass of milk every day.

In a week (7 days):
$$7 \times \frac{1}{2} = \frac{7}{2} = 3\frac{1}{2} \text{ glasses}$$

In January (31 days):
$$31 \times \frac{1}{2} = \frac{31}{2} = 15\frac{1}{2} \text{ glasses}$$

Answer: Tenzin drinks $3\frac{1}{2}$ glasses in a week and $15\frac{1}{2}$ glasses in January.