Let the volumes of the three bottles be $5k$, $3k$, and $2k$.[cite: 10]
Bottle 1 (2:3): Phenyl = $5k \times \frac{2}{5} = 2k$. Water = $3k$. Taken $\frac{1}{3}$ part: Phenyl taken = $\frac{2k}{3}$, Water taken = $k$.[cite: 10]
Bottle 2 (1:2): Phenyl = $3k \times \frac{1}{3} = k$. Water = $2k$. Taken $\frac{1}{2}$ part: Phenyl taken = $0.5k$, Water taken = $k$.[cite: 10]
Bottle 3 (1:3): Phenyl = $2k \times \frac{1}{4} = 0.5k$. Water = $1.5k$. Taken $\frac{2}{3}$ part: Phenyl taken = $0.5k \times \frac{2}{3} = \frac{k}{3}$, Water taken = $1.5k \times \frac{2}{3} = k$.[cite: 10]
Total Phenyl mixed = $\frac{2k}{3} + \frac{k}{2} + \frac{k}{3} = k + 0.5k = 1.5k$.[cite: 10]
Total Water mixed = $k + k + k = 3k$.[cite: 10]
Ratio of Phenyl to Water = $1.5k : 3k = 1:2$.[cite: 10]
Answer: 1:2

Dividends paid = $3\frac{1}{3}\%$ of 864,000 = $\frac{10}{300} \times 864000 = 28,800$.[cite: 10]
Let Gross Receipts = $G$. Working Expenses = $0.40G$. Remainder = $0.60G$.[cite: 10]
Reserve = $0.40 \times 0.60G = 0.24G$. Balance (Dividends) = $0.60G - 0.24G = 0.36G$.[cite: 10]
$0.36G = 28800 \implies G = \frac{28800}{0.36} = 80,000$.[cite: 10]
Answer: Rs. 80,000

Let $x = a - \frac{1}{a}$. Squaring both sides, $x^2 = a^2 + \frac{1}{a^2} - 2 \implies a^2 + \frac{1}{a^2} = x^2 + 2$.[cite: 11]
Substitute into expression: $2(x^2 + 2) - x - 7 = 2x^2 + 4 - x - 7 = 2x^2 - x - 3$.[cite: 11]
Factorize quadratic: $2x^2 - 3x + 2x - 3 = x(2x - 3) + 1(2x - 3) = (2x - 3)(x + 1)$.[cite: 11]
Substitute back $x$: $(2(a - \frac{1}{a}) - 3)(a - \frac{1}{a} + 1) = (2a - \frac{2}{a} - 3)(a - \frac{1}{a} + 1)$.[cite: 11]
Answer: $(2a - \frac{2}{a} - 3)(a - \frac{1}{a} + 1)$

Factor Expression 1: $a^2 - (b^2 - 2bc + c^2) = a^2 - (b-c)^2 = (a-b+c)(a+b-c)$.[cite: 11]
Factor Expression 2: $b^2 - (a^2 - 2ac + c^2) = b^2 - (a-c)^2 = (b-a+c)(b+a-c)$.[cite: 11]
Factor Expression 3: $c^2 - (a^2 - 2ab + b^2) = c^2 - (a-b)^2 = (c-a+b)(c+a-b)$.[cite: 11]
There is no common polynomial factor present across all three expressions.[cite: 11]
Answer: 1

Rewrite numerator terms: $\frac{a}{a-x} = 1 + \frac{x}{a-x}$.[cite: 11]
Numerator becomes: $3 + x(\frac{1}{a-x} + \frac{1}{b-x} + \frac{1}{c-x})$.[cite: 11]
Factor out $x$: $x(\frac{3}{x} + \frac{1}{a-x} + \frac{1}{b-x} + \frac{1}{c-x})$.[cite: 11]
Denominator can be written identically: $\frac{3}{x} - \frac{1}{-(a-x)} \dots = \frac{3}{x} + \frac{1}{a-x} + \frac{1}{b-x} + \frac{1}{c-x}$.[cite: 11]
The entire bracket cancels out.[cite: 11]
Answer: $x$

Group terms in the numerator: $(\frac{a^2}{x-a} + a) + (\frac{b^2}{x-b} + b) + (\frac{c^2}{x-c} + c)$.[cite: 11]
Simplify each group: $\frac{a^2 + a(x-a)}{x-a} = \frac{ax}{x-a} = x(\frac{a}{x-a})$.[cite: 11]
Numerator becomes: $x(\frac{a}{x-a} + \frac{b}{x-b} + \frac{c}{x-c})$.[cite: 11]
The bracketed term is identical to the denominator and cancels out.[cite: 11]
Answer: $x$

Multiply the equation by $(1-x)(1-y)(1-z)$: $(1-y)(1-z) + (1-x)(1-z) + (1-x)(1-y) = 0$.[cite: 11]
Expand: $(1 - y - z + yz) + (1 - x - z + xz) + (1 - x - y + xy) = 0$.[cite: 11]
Combine terms: $3 - 2(x+y+z) + (xy+yz+zx) = 0$.[cite: 11]
Substitute $xy+yz+zx = 9$: $3 - 2(x+y+z) + 9 = 0 \implies 12 = 2(x+y+z) \implies x+y+z = 6$.[cite: 11]
Answer: 6

Combine first two terms: $\frac{1}{x-1} + \frac{1}{x+1} = \frac{x+1+x-1}{x^2-1} = \frac{2x}{x^2-1}$.[cite: 11]
Combine result with third term: $\frac{2x}{x^2-1} + \frac{2x}{x^2+1} = \frac{2x(x^2+1) + 2x(x^2-1)}{x^4-1} = \frac{4x^3}{x^4-1}$.[cite: 11]
Combine result with fourth term: $\frac{4x^3}{x^4-1} + \frac{4x^3}{x^4+1} = \frac{8x^7}{x^8-1}$.[cite: 11]
Combine result with final term: $\frac{8x^7}{x^8-1} - \frac{8x^7}{x^8+1} = \frac{8x^7(x^8+1 - (x^8-1))}{x^{16}-1} = \frac{16x^7}{x^{16}-1}$.[cite: 11]
Answer: $\frac{16x^7}{x^{16}-1}$

Let $\Delta ABC$ have $\angle B > \angle C$. We need to prove $AC > AB$.[cite: 11]
Assume $AC = AB$. Then $\angle B = \angle C$ (isosceles property), which contradicts $\angle B > \angle C$.[cite: 11]
Assume $AC < AB$. Then $\angle B < \angle C$ (angle opposite longer side is larger), contradicting $\angle B > \angle C$.[cite: 11]
Therefore, $AC > AB$ must be true.[cite: 11]

Let interior angles be $\angle A, \angle B, \angle C$. Their sum is $180^\circ$.[cite: 11]
When sides are extended, the exterior angles are $(180^\circ - \angle A)$, $(180^\circ - \angle B)$, and $(180^\circ - \angle C)$.[cite: 11]
Sum of exterior angles = $540^\circ - (\angle A + \angle B + \angle C) = 540^\circ - 180^\circ = 360^\circ$.[cite: 11]
Since $360^\circ$ is equal to 4 right angles ($4 \times 90^\circ$), the theorem is proven.[cite: 11]

Since $PQ \parallel AB$, $\angle BAP = \angle AQP$ (Alternate angles).[cite: 12]
AQ bisects $\angle A$, so $\angle BAQ = \angle QAC$. Therefore, $\angle PAQ = \angle PQA$, making $\Delta APQ$ isosceles with $AP = PQ$.[cite: 12]
Given P is the midpoint of AC, $AP = PC$, which means $PQ = PC$. Thus, $\Delta PQC$ is also isosceles.[cite: 12]
In $\Delta AQC$, the median QP equals half the side AC ($QP = AP = PC$). By geometric properties, if a median is half the base it bisects, the angle opposite the base is $90^\circ$.[cite: 12]
Proved: $\angle AQC = 90^\circ$

Let $\angle BAC = x$. Then $\angle ABC = \angle ACB = 2x$.[cite: 12]
Sum of angles: $x + 2x + 2x = 180^\circ \implies 5x = 180^\circ \implies x = 36^\circ$.[cite: 12]
Therefore, $\angle ABC = 72^\circ$ and $\angle ACB = 72^\circ$. BD bisects $\angle ABC$, so $\angle DBC = 36^\circ$.[cite: 12]
In $\Delta BCD$, $\angle BDC = 180^\circ - (\angle DBC + \angle ACB) = 180^\circ - (36^\circ + 72^\circ) = 72^\circ$.[cite: 12]
Since $\angle BDC = \angle BCD = 72^\circ$, the sides opposite them are equal ($BC = BD$), proving $\Delta BCD$ is isosceles.[cite: 12]

Since $\angle XYZ = 90^\circ$ and $\angle YXZ = 60^\circ$, the remaining angle $\angle XZY = 30^\circ$.[cite: 12]
In a $30^\circ-60^\circ-90^\circ$ right triangle, the side opposite the $30^\circ$ angle is exactly half the length of the hypotenuse.[cite: 12]
Here, XY is the side opposite $\angle XZY$ ($30^\circ$) and XZ is the hypotenuse.[cite: 12]
Proved: $XY = \frac{1}{2}XZ$

Each segment should measure exactly $10 \div 5 = 2\text{ cm}$.[cite: 12]
Draw a 10 cm line. Draw an acute angle from one end. Mark 5 equidistant arcs on the new ray.[cite: 12] Join the final arc to the end of the 10 cm line, and draw parallel lines from the other arcs to divide the primary line segment equally.[cite: 12]

Draw line PQ and mark point R above it.[cite: 12] Join R to any point A on PQ. Using a compass at A, draw an arc intersecting RA and PQ.[cite: 12] With the same radius, place the compass at R and draw an arc.[cite: 12] Measure the chord of the first arc and apply it to the arc at R. Draw a line through R and the intersection point to create the parallel line.[cite: 12]