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Given:

Speed of Train X = 100 km/hr

Speed of Train Y = 60 km/hr

Relative speed of Train X with respect to Train Y = (100 - 60) km/hr = 40 km/hr

Time taken to cross = 9 seconds

Formula used:

Length of Train X = Relative Speed × Time

Note: To convert speed from km/hr to m/s, multiply by (5/18)

Calculations:

Relative Speed = 40 km/hr = 40 × (5/18) m/s = 100/9 m/s

⇒ Length of Train X = (Relative Speed) × Time

⇒ Length of Train X = (100/9) × 9

⇒ Length of Train X = 100 m

∴ The correct answer is option (2).

Given:

Speed of X = 4.8 km/hr

Speed of Y = 6 km/hr

Time difference = 10 minutes = $\frac{10}{60}$ hours = $\frac{1}{6}$ hours

Formula used:

Distance = Speed × Time

Relative Speed (when Y catches up with X) = Speed of Y - Speed of X

Calculation:

Relative Speed = 6 - 4.8 = 1.2 km/hr

Distance covered by X in $\frac{1}{6}$ hours:

Distance = Speed × Time

⇒ Distance = 4.8 × $\frac{1}{6}$ = 0.8 km

Now, time taken by Y to catch up with X:

Time = Distance / Relative Speed

⇒ Time = 0.8 / 1.2 = $\frac{2}{3}$ hours

Distance covered by X in 2/3 hour = 2/3 × 4.8 = 3.2 km

So, distance from P when both will meet = 0.8 + 3.2 = 4 km

∴ The correct answer is option (3).

Given:

X's salary is increased by 12% and then decreased by 10%.

Y's salary is increased by 10% and then decreased by 12%.

X and Y's current salaries are equal.

Formula used:

Net salary after increase and decrease = Initial salary × (1 + increase%/100) × (1 - decrease%/100)

Ratio of initial salaries = Final salary of X / Final salary of Y

Calculations:

Net salary of X: $\text{Initial salary of X} \times (1+\frac{12}{100}) \times (1-\frac{10}{100})$

⇒ Net salary of X = Initial salary of X × 1.12 × 0.9

⇒ Net salary of X = Initial salary of X × 1.008

Net salary of Y: $\text{Initial salary of Y} \times (1+\frac{10}{100}) \times (1-\frac{12}{100})$

⇒ Net salary of Y = Initial salary of Y × 1.1 × 0.88

⇒ Net salary of Y = Initial salary of Y × 0.968

Since the final salaries are equal:

⇒ Initial salary of X × 1.008 = Initial salary of Y × 0.968

⇒ Ratio of initial salaries = $\frac{\text{Initial salary of X}}{\text{Initial salary of Y}} = \frac{0.968}{1.008}$

⇒ Ratio = $\frac{968}{1008}$

⇒ Simplify the ratio:

⇒ Ratio = $\frac{121}{126}$

∴ The correct answer is option (3): 121 : 126.

Given:

Village population (p) = total number of persons

x% = percentage of total persons who can read and write

y% = percentage of males who can read and write

z% = percentage of females who can read and write

x > z and y > z

Formula used:

Number of males = $\dfrac{p(x - z)}{y - z}$

Calculation:

Given formula is:

⇒ Number of males = $\dfrac{p(x - z)}{y - z}$

∴ The correct answer is option (1).

Given:

Alloy X: Cu : Zn = 5 : 4

Alloy Y: Cu : Zn = 5 : 13

Equal quantities of X and Y are mixed to form alloy Z.

Formula used:

In equal quantities mixing, the average ratio of Cu and Zn is taken.

Cu in Z = (Cu in X + Cu in Y) ÷ 2

Zn in Z = (Zn in X + Zn in Y) ÷ 2

Calculation:

Let the quantity of each alloy (X and Y) be 9 units (LCM of 5+4 and 5+13).

For Alloy X:

Cu = (5/9) × 9 = 5 units

Zn = (4/9) × 9 = 4 units

For Alloy Y:

Cu = (5/18) × 9 = 2.5 units

Zn = (13/18) × 9 = 6.5 units

In Alloy Z:

Total Cu = 5 + 2.5 = 7.5 units

Total Zn = 4 + 6.5 = 10.5 units

Cu : Zn in Z = 7.5 : 10.5

Divide by 1.5:

⇒ Cu : Zn = 5 : 7

∴ The correct answer is option (2).

Given:

Score of X = Score of Y + 20

Score of X = 75% of (Score of X + Score of Y)

Formula used:

Let Score of Y = y

Score of X = y + 20

Score of X = 75% × (Score of X + Score of Y)

Calculations:

Let Score of Y = y

⇒ y + 20 = 75% × (y + 20 + y)

⇒ y + 20 = 75% × (2y + 20)

⇒ y + 20 = (75/100) × (2y + 20)

⇒ y + 20 = (3/4) × (2y + 20)

⇒ 4 × (y + 20) = 3 × (2y + 20)

⇒ 4y + 80 = 6y + 60

⇒ 80 - 60 = 6y - 4y

⇒ 20 = 2y

⇒ y = 10

Score of Y = 10

Score of X = y + 20 = 10 + 20 = 30

Ratio of Score of X to Score of Y = X : Y

⇒ 30 : 10

⇒ 3 : 1

∴ The correct answer is option (3).

Given:

Equation: 2x² − 5p x + 2p² = 0

One root exceeds the other by 4

Formula used:

Difference between roots = √(Discriminant) ÷ a

D = b² − 4ac

Calculations:

a = 2, b = −5p, c = 2p²

Difference of roots = 4

⇒ √D / a = 4

⇒ √D / 2 = 4

⇒ √D = 8

⇒ D = 64

Now compute Discriminant:

D = b² − 4ac

= (−5p)² − 4 × 2 × 2p²

= 25p² − 16p²

= 9p²

So, 9p² = 64

⇒ p² = 64/9

⇒ p = ±8/3

∴ The value of p = 8/3.

Given:

Principal (P) = ₹10,000

Rate of interest (r) = 10% per annum

Time (t) = 3 years

Installments = 3 equal annual payments

Formula used:

Installment Amount = $\dfrac{P \times r}{1-(1+r)^{-n}}$

Where, n = number of installments

Calculation:

Installment Amount = $\dfrac{10000 \times 0.10}{1-(1+0.10)^{-3}}$

⇒ Installment Amount = $\dfrac{1000}{1-(1.10)^{-3}}$

⇒ Installment Amount = $\dfrac{1000}{1 - 0.7513}$

⇒ Installment Amount = $\dfrac{1000}{0.2487}$

⇒ Installment Amount ≈ ₹4021

∴ The correct answer is option (3).

Given:

5x + 3 ≤ 8x - 9 and 2x + 20 > 5x + 2

Formula used:

Simplify inequalities to find the range of x.

Calculation:

5x + 3 < 8x - 9    and    2x + 20 > 5x + 2

⇒ 8x - 5x > 12    and    18 > 3x

⇒ x >4  and  6 > x

Option (1) is correct.

The correct answer is – 80% profit

Successive discounts

For discounts of 10%, 20%, and 25%, the net factor is 0.9 × 0.8 × 0.75 = 0.54.

This indicates that the final selling price is 54% of the marked price.

Application of given profit

After all discounts, the article is sold at a profit of 8% on cost price.

Assuming cost price = 100, the selling price becomes 108.

Thus, the marked price is calculated as 108 ÷ 0.54 = 200.

Profit after first discount

After only the  first discount of 10%, the selling price equals 90% of the marked price.

Selling price = 0.9 × 200 = 180.

Profit = 180 −  100 = 80, which is 80% profit.