Let the tower be at B

Initially man was at point C then the distance BC was 200m

After moving, man comes at position A distance AB=300m

AC= 100√13 (By Pythagoras theorem)

Shortest distance will be BD

Let DC = x then AD = 100√13 – x

In Triangle BDC, BD = (200)² – x²

In Triangle ABD, BD=(300)² – (100√13 – x)²

Equating the two values of BD we get, x=400/√13

Calculating BDwe get 600/√13

Let p and q be the lengths of the slow and faster trains respectively .

OPPOSITE DIRECTION-

When trains are travelling in the opposite directions ,their relative speed ` 

Distance covered= Sum of length of 2 trains = p + q

Then according to the given condition we have

SAME DIRECTION-

When trains are travelling in the same direction, since we are given the time noted by a person in the faster train as 32 seconds the distance covered is equal to the length of the slower train.

So, distance covered= q

The length of the slower train=150m.

Total people = 24 + 1 = 25

Average = 15

Sum of the age of 24 members and their gardener = 25 × 15 = 375

Now If the age of the gardener is left out their average age decreases by 1

New average = 15 – 1 = 14

Sum of age of 24 members = 24 × 14 = 336

So age of gardener = 375 – 336 = 39 years

Steps to draw wavy curve method-

The amounts at the end of a year, in compound interest, are in Geometric Progression with common ratio . Similarly the interest accrue in each year, in compound

interest, are also in Geometric Progression of the same common ratio, i.e. .

Also , i.e. the progressions are increasing.

Also the difference between any two consecutive terms, say interest for i^st year and interest for (i + 1)^st year will be aRⁱ⁺¹ - aRⁱ

Hence, the difference in the interest for the a^th and (a+1)^th year will be more than that for the b^th and (b+1)^th year if a>b due to the increasing accountability of the common ratio in the geometric progression formed by the interest rate.

The diagonal of the cube [BF] measures 5√3 cm.

Assume that one side length is, a cm.

∴ EF = DE = BD = a cm.

Applying Pythagoras theorem for triangle EDF [∠DEF = 90°] we get,

DF² = DE² + EF²

⇒ DF² = 2×a²

⇒ DF = √2 × a cm.

Now, consider ΔBDF where ∠BDF = 90°

Applying Pythagoras theorem for ΔBDF:

BF² = DF² + DB²

⇒ BF² = 2a² + a² = 3a²

As, BF = 5√3 cm.

∴ 25 × 3 = 3a²

⇒ a = 5cm.

The volume of the cube = a³ = 5³ = 125 cm³

We have

Let Shruthi’s score be 8x then, Tanuja’s score is 11x.