✅ STEP-BY-STEP SOLUTION:

Step 1: Formulate equations
Let price of 1 pencil = ₹x, price of 1 chocolate = ₹y
Aarush: 2 pencils + 3 chocolates = ₹11 ⇒ \(2x + 3y = 11\) …(1)
Tanish: 1 pencil + 2 chocolates = ₹7 ⇒ \(x + 2y = 7\) …(2)

Step 2: Find points for equation (1): 2x + 3y = 11

When x = 1: 2(1) + 3y = 11 ⇒ 3y = 9 ⇒ y = 3 → point (1, 3)

When x = 4: 2(4) + 3y = 11 ⇒ 8 + 3y = 11 ⇒ 3y = 3 ⇒ y = 1 → point (4, 1)

When x = 7: 2(7) + 3y = 11 ⇒ 14 + 3y = 11 ⇒ 3y = -3 ⇒ y = -1 → point (7, -1)

Step 3: Find points for equation (2): x + 2y = 7

When x = 1: 1 + 2y = 7 ⇒ 2y = 6 ⇒ y = 3 → point (1, 3)

When x = 3: 3 + 2y = 7 ⇒ 2y = 4 ⇒ y = 2 → point (3, 2)

When x = 5: 5 + 2y = 7 ⇒ 2y = 2 ⇒ y = 1 → point (5, 1)

Step 4: Graphical representation

Step 5: Find intersection point
From the graph, both lines intersect at (1, 3)
Therefore, x = 1, y = 3

Step 6: Verify

Equation (1): 2(1) + 3(3) = 2 + 9 = 11 ✓

Equation (2): 1 + 2(3) = 1 + 6 = 7 ✓

✅ Final Answer:

Price of 1 pencil = $${₹1}$$

Price of 1 chocolate = $${₹3}$$

✅ STEP-BY-STEP SOLUTION:

Step 1: Define variables
Let original duration of tour = x days
Original daily expense = $$\frac{4200}{x} rupees$$

Step 2: New conditions

New duration = (x + 3) days

New daily expense = $$\frac{4200}{x+3}$$ rupees

Given: New daily expense = Original daily expense – ₹70

Step 3: Form equation

$$\frac{4200}{x+3} = \frac{4200}{x} – 70$$

Step 4: Solve for x

Multiply throughout by x(x+3):

$$4200x = 4200(x+3) – 70x(x+3)$$

$$4200x $$

$$= 4200x + 12600 – 70x^2 – 210x$$

$$0 = 12600 – 70x^2 – 210x$$

$$70x^2 + 210x – 12600 = 0$$

Step 5: Simplify (divide by 70)

$$x^2 + 3x – 180 = 0$$

Step 6: Factorize

$$x^2 + 15x – 12x – 180 = 0$$

x(x + 15) – 12(x + 15) = 0

(x + 15)(x – 12) = 0

x = -15 or x = 12

Step 7: Select valid solution

x = -15 is not possible (days cannot be negative)

Therefore, x = 12 days

Step 8: Verify

Original: 12 days, daily expense = 4200/12 = ₹350

Extended: 15 days, daily expense = 4200/15 = ₹280

Reduction = 350 – 280 = ₹70 ✓

✅ Final Answer: Original duration of tour = $${12 \text{ days}}$$

✅ STEP-BY-STEP SOLUTION:

Step 1: Define variables
Let altitude = x cm
Base = (x + 10) cm (given: base exceeds altitude by 10 cm)

Step 2: Area formula for right triangle

Area $$= \frac{1}{2} \times \text{base} \times \text{height}$$

600 $$= \frac{1}{2} \times (x + 10) \times x$$

Step 3: Form equation

1200 = x(x + 10)

$$x^2 + 10x – 1200 = 0$$

Step 4: Solve quadratic

Using factorization: $$x^2 + 40x – 30x – 1200 = 0$$

x(x + 40) – 30(x + 40) = 0

(x + 40)(x – 30) = 0

x = -40 or x = 30

Step 5: Select valid solution

x = -40 not possible (length cannot be negative)

Therefore, altitude = 30 cm

Base = 30 + 10 = 40 cm

Step 6: Find hypotenuse (Pythagoras theorem)

Hypotenuse = $$\sqrt{(\text{base})^2 + (\text{altitude})^2}$$

= $$\sqrt{40^2 + 30^2} $$

=$$ \sqrt{1600 + 900} $$

=$$\sqrt{2500} = 50 cm$$

✅ Final Answer:

Altitude = $${30 \text{ cm}}$$

Base = $${40 \text{ cm}}$$

Hypotenuse = $${50 \text{ cm}}$$

✅ BASIC PROPORTIONALITY THEOREM (THALES THEOREM):

Statement: If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

Given: In ΔABC, line DE ∥ BC, intersecting AB at D and AC at E.

To prove: $$\frac{AD}{DB} = \frac{AE}{EC}$$

Construction: Join BE and CD. Draw DM ⟂ AC and EN ⟂ AB.

Proof:

Step 1: Consider triangles ADE and BDE.
Area of ΔADE = \(\frac{1}{2} \times AD \times EN\)(taking AD as base, EN as height)
Area of ΔBDE = \(\frac{1}{2} \times DB \times EN\) (taking DB as base, EN as height)
\(\frac{ar(ADE)}{ar(BDE)}
= \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN}
= \frac{AD}{DB}\) …(1)

Step 2: Consider triangles ADE and CDE.

Area of ΔADE = $$\frac{1}{2} \times AE \times DM$$ (taking AE as base, DM as height)

Area of ΔCDE = $$\frac{1}{2} \times EC \times DM$$ (taking EC as base, DM as height)

$$\frac{ar(ADE)}{ar(CDE)}$$

$$= \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} $$

$$= \frac{AE}{EC}$$ …(2)

Step 3: Observe that triangles BDE and CDE are on the same base DE and between the same parallels DE and BC.

Therefore, ar(BDE) = ar(CDE) …(3)

Step 4: From (1) and (2):

$$\frac{AD}{DB}$$

=$$ \frac{ar(ADE)}{ar(BDE)} and \frac{AE}{EC}$$

= $$\frac{ar(ADE)}{ar(CDE)}$$

Step 5: Using (3), since ar(BDE) = ar(CDE), we get:

$$\frac{AD}{DB} = \frac{AE}{EC}$$

✅ Hence proved: $${\frac{AD}{DB} = \frac{AE}{EC}}$$

✅ STEP-BY-STEP SOLUTION:

Step 1: Convert units to same system
Girl’s height = 90 cm = 0.9 m
Lamp height = 3.6 m
Speed = 1.2 m/s
Time = 4 seconds

Step 2: Find distance walked in 4 seconds

Distance = Speed × Time = 1.2 × 4 = 4.8 m

After 4 seconds, girl is 4.8 m away from base of lamp post.

Step 3: Use similar triangles

Let length of shadow = x metres

In the figure, two similar triangles are formed:

· Large triangle: lamp post (height 3.6 m) and total distance from lamp to end of shadow (4.8 + x)

· Small triangle: girl (height 0.9 m) and her shadow length (x)

Step 4: Apply similarity property

$$\frac{\text{Lamp height}}{\text{Girl height}} $$
= $$\frac{\text{Total distance}}{\text{Shadow length}}$$

$$\frac{3.6}{0.9}$$
=$$ \frac{4.8 + x}{x}$$

Step 5: Solve for x

$$\frac{3.6}{0.9} = 4$$

4 = $$\frac{4.8 + x}{x}$$

4x = 4.8 + x

4x – x = 4.8

3x = 4.8

x = 1.6 m

✅ Final Answer: Length of shadow after 4 seconds = $${1.6 \text{ m}}$$ or $${160 \text{ cm}}$$

✅ STEP-BY-STEP SOLUTION:

PART A: FINDING MODAL AGE

Step 1: Identify modal class

Modal class is the class with highest frequency.

Highest frequency = 33 in class 35 – 40

So modal class = 35 – 40

Step 2: Note the values

Lower limit of modal class (l) = 35

Class size (h) = 5

Frequency of modal class (f₁) = 33

Frequency of class preceding modal class (f₀) = 21 (class 30-35)

Frequency of class succeeding modal class (f₂) = 11 (class 40-45)

Step 3: Apply mode formula

$$Mode = l + \frac{f₁ – f₀}{2f₁ – f₀ – f₂} \times h$$

$$= 35 + \frac{33 – 21}{2(33) – 21 – 11} \times 5$$

$$= 35 + \frac{12}{66 – 32} \times 5$$

$$= 35 + \frac{12}{34} \times 5$$

$$= 35 + \frac{60}{34}$$

= 35 + 1.7647…

$$= 36.7647… \approx 36.76 years$$

Modal age ≈ 36.76 years

PART B: FINDING MEDIAN AGE

Step 1: Create cumulative frequency table

Step 2: Find median class
Total frequency (N) = 100
\(\frac{N}{2} = 50\)
Cumulative frequency just greater than 50 is 78, corresponding to class 35-40.
So median class = 35 – 40

Step 3: Note values for median formula

Lower limit of median class (l) = 35

Class size (h) = 5

Cumulative frequency of class preceding median class (cf) = 45

Frequency of median class (f) = 33

$$\frac{N}{2} = 50$$

Step 4: Apply median formula

$$Median = l + \frac{\frac{N}{2} – cf}{f} \times h$$

$$= 35 + \frac{50 – 45}{33} \times 5$$

$$= 35 + \frac{5}{33} \times 5$$

$$= 35 + \frac{25}{33}$$

$$= 35 + 0.7575…$$

$$= 35.7575… \approx 35.76 years$$

✅ Final Answer:

Modal age = $${36.76 \text{ years}}$$

Median age = $${35.76 \text{ years}}$$