Vardaan Learning Institute

Let the numerator be $x$. Then denominator is $x+8$.
Given: $\frac{x+17}{(x+8)-1} = \frac{3}{2} \Rightarrow \frac{x+17}{x+7} = \frac{3}{2}$
$2(x+17) = 3(x+7) \Rightarrow 2x + 34 = 3x + 21$
$34 - 21 = 3x - 2x \Rightarrow x = 13$
Numerator = 13, Denominator = $13+8 = 21$.

Let Shriya's age = $x$. Arjun's age = $2x$.
5 years ago: Shriya = $x-5$, Arjun = $2x-5$.
According to question: $2x-5 = 3(x-5)$
$2x - 5 = 3x - 15 \Rightarrow 15 - 5 = 3x - 2x \Rightarrow x = 10$.
Arjun's age = $2(10) = 20$.

Let unit digit $= y$, tens digit $= x$. Number $= 10x + y$.
Sum of digits: $x + y = 9 \Rightarrow y = 9 - x$.
Reversed number: $10y + x$.
Given: $(10y + x) - (10x + y) = 27 \Rightarrow 9y - 9x = 27 \Rightarrow y - x = 3$.
Substitute $y$: $(9-x) - x = 3 \Rightarrow 9 - 2x = 3 \Rightarrow 2x = 6 \Rightarrow x = 3$.
$y = 9 - 3 = 6$. Number is 36.

Let speed of steamer in still water $= u$ km/hr.
Downstream speed $= (u+2)$, Upstream speed $= (u-2)$.
Distance is constant. $D = \text{Speed} \times \text{Time}$.
$4(u+2) = 5(u-2) \Rightarrow 4u + 8 = 5u - 10 \Rightarrow u = 18$.

Let number of 5-rupee coins $= x$. Then 2-rupee coins $= 3x$.
Total Value $= 5(x) + 2(3x) = 5x + 6x = 11x$.
Given Total Value $= 77$. So, $11x = 77 \Rightarrow x = 7$.
5-rupee coins = 7. 2-rupee coins = $3 \times 7 = 21$.

Marked Price (MP) = ₹1200. Discount = 10%.
Selling Price (SP) = $1200 - 10\% \text{ of } 1200 = 1200 - 120 =$ ₹1080.
Profit = 20%. Let CP be Cost Price.
$SP = CP \times (1 + \frac{20}{100}) \Rightarrow 1080 = CP \times 1.2$
$CP = \frac{1080}{1.2} = 900$.

P = 20000, R = 8%, T = 2 years.
SI $= \frac{P \times R \times T}{100} = \frac{20000 \times 8 \times 2}{100} = 3200$.
CI $= P(1 + \frac{R}{100})^T - P = 20000(1.08)^2 - 20000 = 20000(1.1664) - 20000 = 23328 - 20000 = 3328$.
Difference $= 3328 - 3200 = 128$.

Population in 2000 (after 3 years) $= P(1 + \frac{R}{100})^n$
$= 20000(1 + \frac{5}{100})^3 = 20000(1.05)^3$
$= 20000 \times 1.157625 = 23152.5$.
Since population cannot be fractional, round to nearest integer.

Fabina (SI): $\frac{12500 \times 12 \times 3}{100} = 125 \times 36 =$ ₹4500.
Radha (CI): $12500(1 + \frac{10}{100})^3 - 12500 = 12500(1.331) - 12500 = 16637.5 - 12500 =$ ₹4137.5.
Difference: $4500 - 4137.5 = 362.5$.

SP of each = ₹20,000. Total SP = ₹40,000.
Buffalo 1 (5% Gain): $CP_1 = \frac{20000}{1.05} \approx 19047.62$.
Buffalo 2 (10% Loss): $CP_2 = \frac{20000}{0.90} \approx 22222.22$.
Total CP $= 19047.62 + 22222.22 = 41269.84$.
Loss $= CP - SP = 41269.84 - 40000 = 1269.84$.

More workers, less time (Inverse).
$M_1 \times D_1 = M_2 \times D_2 \Rightarrow 15 \times 48 = M_2 \times 30$.
$M_2 = \frac{15 \times 48}{30} = \frac{48}{2} = 24$.

Initially: 120 men for 30 days.
After 5 days: Remaining food is sufficient for 120 men for $30-5=25$ days.
New men count $= 120 + 5 = 125$.
$120 \times 25 = 125 \times D \Rightarrow D = \frac{120 \times 25}{125} = \frac{120}{5} = 24$.

Speed = 75 km/hr.
(i) Time = 20 min = $\frac{20}{60} = \frac{1}{3}$ hr. Dist $= 75 \times \frac{1}{3} = 25$ km.
(ii) Dist = 250 km. Time $= \frac{250}{75} = \frac{10}{3}$ hours $= 3 \frac{1}{3}$ hours $= 3$ hrs 20 mins.

(A+B)'s 1 day work $= \frac{1}{12}$. B's 1 day work $= \frac{1}{30}$.
A's 1 day work $= \frac{1}{12} - \frac{1}{30}$. LCM of 12, 30 is 60.
$= \frac{5 - 2}{60} = \frac{3}{60} = \frac{1}{20}$.
A alone takes the reciprocal of $\frac{1}{20}$.

14 km in 25 min. Time = 5 hours = $5 \times 60 = 300$ min.
$\frac{14}{25} = \frac{x}{300} \Rightarrow x = \frac{14 \times 300}{25} = 14 \times 12 = 168$.

Area of 1 tile (parallelogram) $= \text{base} \times \text{height} = 24 \times 10 = 240 \text{ cm}^2$.
Floor Area $= 1080 \text{ m}^2 = 1080 \times 10000 \text{ cm}^2$ (since $1 \text{ m}^2 = 10000 \text{ cm}^2$).
Number of tiles $= \frac{10800000}{240} = 45000$.

Diameter = 84 cm $\Rightarrow$ Radius $r = 42$ cm = 0.42 m. Height $h = 1$ m.
CSA (Area covered in 1 rev) $= 2\pi rh = 2 \times \frac{22}{7} \times 0.42 \times 1 = 2 \times 22 \times 0.06 = 2.64 \text{ m}^2$.
Total Area $= 750 \times 2.64 = 1980 \text{ m}^2$.

Paper 11x4 folded to make height 4. The length 11 becomes the circumference.
$2\pi r = 11 \Rightarrow 2 \times \frac{22}{7} \times r = 11 \Rightarrow r = \frac{11 \times 7}{44} = \frac{7}{4} = 1.75$ cm.
Volume $= \pi r^2 h = \frac{22}{7} \times \frac{7}{4} \times \frac{7}{4} \times 4 = 22 \times \frac{1}{1} \times \frac{7}{4} \times 1 = \frac{154}{4} = 38.5$.

Volume of Godown $= 60 \times 40 \times 30 = 72000 \text{ m}^3$.
Volume of 1 box $= 0.8 \text{ m}^3$.
Number of boxes $= \frac{72000}{0.8} = \frac{720000}{8} = 90000$.

Draw a line parallel to the 6m side. This forms a parallelogram and a triangle.
Triangle sides are 6, 8, and $(30-20)=10$.
Check for right angle: $6^2 + 8^2 = 36 + 64 = 100 = 10^2$. It is a right-angled triangle!
Area of triangle $= \frac{1}{2} \times 6 \times 8 = 24$.
Also Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times h = 5h$.
$5h = 24 \Rightarrow h = 4.8$ m.
Area of Trapezium $= \frac{1}{2}(20+30) \times 4.8 = 25 \times 4.8 = 120$.

Plants $= 1000$. Rows = Columns $\Rightarrow$ Perfect Square needed.
$\sqrt{1000} \approx 31.6$. Next integer is 32.
$32^2 = 1024$.
Plants needed $= 1024 - 1000 = 24$.

$5^{2x+1} \div 25 = 125$
$5^{2x+1} \div 5^2 = 5^3$
$5^{(2x+1) - 2} = 5^3 \Rightarrow 5^{2x-1} = 5^3$
Equating powers: $2x - 1 = 3 \Rightarrow 2x = 4 \Rightarrow x = 2$.

Prime factorization of 675:
$675 = 25 \times 27 = 5^2 \times 3^3$.
To make it a perfect cube, we need powers to be multiples of 3.
We have $5^2$, we need one more 5. We have $3^3$ (perfect).
Smallest number to multiply $= 5$. Product $= 675 \times 5 = 3375$.
Cube root $= \sqrt[3]{5^3 \times 3^3} = 5 \times 3 = 15$.

Numbers: $2x, 3x, 4x$.
Sum of cubes: $(2x)^3 + (3x)^3 + (4x)^3 = 33957$
$8x^3 + 27x^3 + 64x^3 = 33957 \Rightarrow 99x^3 = 33957$
$x^3 = \frac{33957}{99} = 343$.
$x = \sqrt[3]{343} = 7$.
Numbers: $2(7), 3(7), 4(7)$.

Adjacent angles sum to $180^\circ$. Ratio 2:3.
$2x + 3x = 180 \Rightarrow 5x = 180 \Rightarrow x = 36$.
Angles are $2(36) = 72^\circ$ and $3(36) = 108^\circ$.
Opposite angles are equal.