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Given:

Simple interest on a sum for 10 years is equal to 5% of the principle.

Concept used:

SI = \(\frac{P \times R \times T}{{100}}\)

Where, P = Principle, R = Rate, T = Time 

Calculation:

Let the principal be Rs. P

According to the above concept

SI = \(\frac{P \times R \times 10}{{100}}\) = 5% of P

⇒ \(\frac{P \times R \times 10}{{100}}\) = \(\frac{5P}{{100}}\)

⇒ R = 0.5%

Now, let in t years the simple interest is equal to the principle

⇒ SI = \(\frac{P \times 0.5 \times t}{{100}}\)

⇒ P = \(\frac{P \times t}{{200}}\)

⇒ t = 200 years

∴ The required time is 200 years.

Shortcut Trick For a quadratic equation to have real roots, the discriminant (D) must be greater than or equal to zero (D ≥ 0). Equation: 4x2 − kx + 9 = 0 ⇒ a = 4, b = −k, c = 9. Apply D ≥ 0: (−k)2 − 4(4)(9) ≥ 0 k2 − 144 ≥ 0 ⇒ k2 ≥ 144. This implies k ≤ −12 or k ≥ 12. ∴ The correct answer is k ≤ −12 and k ≥ 12. Alternate Method Given: Quadratic equation 4x2 − kx + 9 = 0. Formula Used: For real roots of ax2 + bx + c = 0, Discriminant D = b2 − 4ac ≥ 0.

Comparing the given equation with ax2 + bx + c = 0: ⇒ a = 4, b = −k, c = 9 For roots to be real: ⇒ b2 − 4ac ≥ 0 ⇒ (−k)2 − 4 × 4 × 9 ≥ 0 ⇒ k2 − 144 ≥ 0 ⇒ (k − 12)(k + 12) ≥ 0 Using the method of intervals (Wavy Curve method): ⇒ k belongs to the region (−∞, −12] ∪ [12, ∞) ⇒ k ≤ −12 or k ≥ 12 ∴ The correct answer is k ≤ −12 and k ≥ 12. Additional Information Nature of Roots based on Discriminant (D) If D > 0, roots are real and unequal. If D = 0, roots are real and equal. If D < 0, roots are imaginary/complex. Quadratic Inequality Rule For an inequality x2 ≥ a2, the solution is always x ≤ −a or x ≥ a. For x2 ≤ a2, the solution is −a ≤ x ≤ a. Perfect Square Condition A quadratic equation is a perfect square if its discriminant D = 0, meaning it has two identical real roots.

Calculation:

(5/28) ÷ (28/35) ÷ (20/112)

⇒ [(5/28) × (35/28)] ÷ (20/112)

⇒ (25/112) ÷ (20/112)

⇒ (25/112) × (112/20)

⇒ 5/4

∴ The required value is 5/4

Shortcut Trick

Distance = Number of Revolutions × Circumference

44000 = 5000 × 2 × (22/7) × Radius

44 = 5 × 2 × (22/7) × Radius

Radius = (44 × 7) ÷ (10 × 22) = 1.4 m

∴ The correct answer is 1.4 m.

Alternate Method

Given:

Total Distance = 44 km = 44000 m

Number of Revolutions (n) = 5000

Formula Used:

Distance = n × 2πr

Calculations:

⇒ 44000 = 5000 × 2 × (22/7) × r

⇒ 44 = 5 × 2 × (22/7) × r

⇒ 44 = (220/7) × r

⇒ r = (44 × 7) ÷ 220

⇒ r = 308 ÷ 220

⇒ r = 1.4 m

∴ The correct answer is 1.4 m.

Additional Information

Circumference of a Circle

The distance around a circular wheel is its circumference, calculated as 2 × π × r or π × d.

Unit Conversion

Always ensure units are consistent; 1 km = 1000 m. If the distance is in km and radius is required in meters, convert km to m first.

Revolution Concept

In one complete revolution, a wheel covers a linear distance equal to its outer circumference.

Given:

Three cubes each of sides 7 cm are joined end to end

Concept:

When 3 cubes of same side are joined end to end, the resulting solid will be a cuboid

Formula used:

Surface area of Cuboid = 2 × ((l × b) + (l × h) + (b × h))

Calculation:

Length of the Cuboid, l = 7 + 7 + 7 = 21 cm

Breadth of the Cuboid, b = Same as the side of the given Cube = 7 cm

Height of the Cuboid, h = Same as the side of the given Cube = 7 cm

Now,

Surface area of Cuboid = 2 × ((l × b) + (l × h) + (b × h))

⇒ S.A = 2 × ((21 × 7) + (21 × 7) + (7 × 7))

⇒ S.A = 2 × (147 + 147 + 49) = 686

∴  The surface area of the resulting solid = 686 cm²

Given:

Speed of the train = 72km/hr

The train passes the platform in 48 sec and the passenger in 30 sec

Concept used:

Speed = distance/time

While a train crossing a man it actually crossing it's own length.

Calculation:

Speed of the train is 72 km/hr = 72 × (5/18) = 20 m/sec

Length of train = speed × time

⇒ 20 × 30 = 600 m

Now, accordingly

\(20 = ;\frac{{x + 600}}{{48}}\)

⇒ \(20 × 48 = x + 600\)

⇒ x = 960 - 600

⇒ x = 360

∴ The length of the platform is 360 meter.

Let the Cost price of an article be Rs. 100.

Marked price = 100 × 110/100 = 110

Required discount = Rs. 10

Required discount percentage = 10 × 100/110 = 100/11%

Given:

A and B working together can complete a job in 30 days. The ratio of their efficiencies is 3 : 2.

Calculation: 

Let total work to be done = 150 units

Let efficiency of A and B respectively be 3x and 2x units/day

⇒ (3x + 2x) × 30 = 150

⇒ x = 1

Time taken by faster person, i.e A = \(\frac{150}{3 × 1} = 50\) days

∴ The time taken by faster person is 50 days

 

Time is taken by A alone = 30 × \(\frac{3 + 2}{3} = 50 \) days

∴ The time taken by faster person is 50 days

Given:

A man covers a certain distance on a scooter driving at 64 km/h.

He returns to the starting point riding on a bicycle at 16 km/h.

Formula used:

Formula:

Average speed = 2s1s2/(s1+s2)

Solution:

Average speed = 2s1s2/(s1+s2)

(2 × 64 × 16)/(64 + 16)

25.6 km/h

∴ The answer is 25.6 km/h.

Given:

The number of days in which 20 workers finish the work = 30 days.

Formula used:

MDH/W = constant

Where M = number of men or machine

D = number of days

H = number of hours or minute

W = Total work 

Calculation:

When 20 workers finish the work in 30 days,

20 workers × 30 days = 600 workers days.

Let the number of days after which 5 workers leave the job be x, then

(20 workers × x days) + (15 workers × (35 - x) days) = 600 workers days

⇒ 20x + 525 - 15x = 600

⇒ 5x = 600 - 525

⇒ 5x = 75

⇒ x = 75 / 5

⇒ x = 15 days

∴ For the work to be completed in 35 days, 5 workers should leave after working for 15 days.