Given:

15 men can remove 750 bushes in 25 hours

Formula used:

= 300

When 5 men leave a job 300 bushes can be removed in 15 hours

Answer is 300.

Detailed solution:-

Total students who played Basketball in Schools A and C together:

= Number of students playing Basketball in School A + Number

of students playing Basketball in School C = 125 + 145 = 270

Total students who played Football in Schools B and D together:

= Number of students playing Football in School B + Number

of students playing Football in School D = 50 + 40 = 90

Ratio = 270 / 90 = 3 /1

∴ The required answer is 3:1.

Given:

A boat can go 60 km downstream and 40 km upstream in 12 hours 30 minutes.

It can go 84 km downstream and 63 km upstream in 18 hours 54 minutes.

Concept used:

Upstream speed = Boat speed - speed of the current 

Downstream speed = Boat speed + speed of the current

Distance = speed × time

Calculation:

Downstream speed = x km/h

The upstream speed=  y km/h

As per the question,

60 /x + 40/y = 25/2 ...... (1)

Again, 84/x + 63/y = 189/10 ....... (2)

By solving 1 and 2 we get,

x = 40 / 3 and y = 5

So Still water boat's speed is

⇒ (13..33 + 5) / 2 = 9km/hr

∴ The correct option is 3

Alternate Method 

Let the speed of the boat = u

and

speed of current/river = v

So,

upstream speed (US) = u - v

downstream speed (DS) = u + v

according to the question,

60/DS  +  40/US  = 12.5

⇒ 3/DS + 2/US = 0.625 ....(1)

and

84/(u + v)  + 63/(u - v) = 18.9

⇒ 4/DS + 3/US = 0.9 ....(2)

let

a = 1/DS   and  b = 1/US

then eq(1) and eq(2) will be

⇒ 3a + 2b  = 0.625 ....(3)

⇒ 4a  + 3b = 0.9....(4)

So, multiply eq(3) with 3 and eq(4) with 2:-

⇒ 9a + 6b = 1.875  ...(5)

⇒ 8a + 6b = 1.8 ....(6)

now,  eq(5) - eq(6)

a = 0.075

then DS = 40/3

and from eq(6)

6b = 1.2

⇒ b = 0.2

⇒ US = 5

Boat speed = (DS + US)/2 = 55/6

Hence; u ≈ 9 km/hr

Calculation:

To find the largest number that has both colors, we need to identify the common multiples of 8 and 9 (1 to 100) and see which one is the largest.

LCM of (8, 9) = 72, which is largest between 1 to 100 

∴  the largest number in the rectangular array that has both yellow and green shading is 72

Given:

3sin²θ + 4cosθ - 4 = 0

Formula used:

sin²θ + cos²θ = 1

Hypotenuse(H)² = Perpendicular(P)² + Base(B)² 

Calculation:

3sin²θ + 4cosθ - 4 = 0

⇒ 3(1 - cos²θ) + 4cosθ - 4 = 0

⇒ 3 - 3cos²θ + 4cosθ - 4 = 0

⇒ - 3cos²θ + 4cosθ - 1 = 0

⇒ 3cos²θ - 4cosθ + 1 = 0

⇒ 3cos²θ - 3cosθ - cosθ  + 1 = 0

⇒ 3cosθ(cosθ - 1) - 1(cosθ  - 1) = 0

⇒ (cosθ - 1)(3cosθ - 1) = 0

⇒ cosθ = 1, 1/3

But cosθ = 1 is not possible because it is given 0° < θ  < 90° and cos0° = 1

So, cosθ = 1/3

Now,

Cosθ = B/H

So,

H² = B² + P²

⇒ 3² = 1² + P²

⇒ 9 = 1 + P²

⇒ 9 - 1 = P²

⇒ 8 = P²

⇒ 2√2 = P

So, cosec2θ + cot2θ = (3/2√2)² + (1/2√2)²

⇒ (9/8) + (1/8)

⇒ (9 + 1)/8

⇒ 10/8 = 5/4

∴ The value of (cosec² θ + cot² θ) is 5/4.

Given:

The difference between the simple interest and compound interest (interest is compounded half yearly) on a sum at the rate of 25% per annum for one year is ₹ 4375

Formula used:

Simple Interest = (P × N × R)/100

Compound Interest = [P(1 + (r/200))^T] - P      (for compounded half yearly)

Calculation:

Let P be the Principal,

S.I = (P × 1 × 25)/100 = P/4

C.I = [P(1 + (25/200))²] - P        ( T = 2   ∵ compounded half yearly for 1 year)

⇒ C.I = 17P/64

Now, C.I - S.I = (17P/64) - (P/4) = P/64

⇒ P/64 = 4375

∴ P = 64 × 4375 = 280000

Shortcut Trick

Formula used:

CI - SI = P(R/100)²

Rate (R) = 25%/2 due to the compounded half-yearly.

⇒ 4375 = P (25/200)²

⇒ P = 4375 × 64

⇒ P = 280,000

∴ The sum is Rs. 280,000.

Given:

Tap P can fill a tank = 20 hours

Tap Q can fill a tank = 25 hours

Tap R can fill a tank = 40 hours

Calculation:

Let the total work be LCM of 20, 25, and 40 = 200 units

⇒ Efficiency of tap P = 200/20 = 10 units

⇒ Efficiency of tap Q = 200/25 = 8 units

⇒ Efficiency of tap R = 200/40 = 5 units

Since the tap Q is kept open for 10 hours,

Work done by tap Q = 10 × 8 = 80 units

∵ Tap R is closed 9 hours before the tank overflows

⇒ Tap P alone worked for 9 hours.

⇒ Work done by tap P alone = 9 × 10 = 90 units

Remaining work = 200 - (80 + 90) = 30 units

The remaining work was done by tap P and tap R together

Time taken by tap P and Tap R to complete the remaining work = 30/(10 + 5) = 30/15 = 2 hours

∴ The total time to fill the tank is (10 + 9 + 2) 21 hours.

Given:

Water in the original solution = 64%

So, Acid in the original solution = 36% 

Acid in the new solution = 30%

Water in the new solution = 70%

Calculation:

Let the total solution be at the beginning was 100x

So, Water = 64x and acid = 36x

Now, According to the question,

{64x - 4 × (64/100) + 4}/{36x - 4 × (36/100)} = 7/3

⇒ (64x + 1.44)/(36x - 1.44) = 7/3

⇒ 192x + 4.32 = 252x - 10.08

⇒ 60x = 14.4

⇒ x = 0.24

So, 64x = 64 × 0.24

⇒ 15.36

∴ Water in the solution, at the beginning in the vessel, was 15.36 liters.

Alternate Method:

Let ratio of water and acid  = 16x : 9x

Since, 4 litres of solution was taken out and same quantity of water added

And solution taken out will be in the ratio of water and acid

So,

⇒ (16x - 2.56 + 4)/(9x - 1.44) = 7/3

⇒ (16x + 1.44)/(9x - 1.44) = 7/3

⇒ 48x + 4.32 = 63x - 10.08

⇒ 15x = 14.4

⇒ x = 0.96

Water initially = 16x 

⇒ 16 × 0.96

⇒ 15.36

Given data:

SI for 3 years = Rs 11,250

CI for 2 years at the same rate = Rs 7650

Formula used:

P = Principal

SI = Simple Interest

R = Rate

T = Time

Calculation:

SI for 1 year = 11,250 ÷ 3 = Rs 3,750

SI for 2 year = 2 × 3750 = Rs 7500

Difference between CI and SI for 2 year = 7650 - 7500 = Rs 150

⇒ This difference between CI and SI was on the SI for the 1st year i.e., Rs 3750

∴ The Principal amount was Rs 93,750 and the rate of interest was 4%.