Complete Step-by-Step Solutions Key
(Exemplary Solutions for Maximum Marks in Board Evaluation)
Solution 1: Boat and Stream Problem
Ch 3: Linear Equations
Let speed of boat in still water = $x\text{ km/h}$ and speed of stream = $y\text{ km/h}$.
Speed upstream = $(x - y)\text{ km/h}$, Speed downstream = $(x + y)\text{ km/h}$.
Let $\frac{1}{x-y} = u$ and $\frac{1}{x+y} = v$.
Case 1: $30u + 44v = 10 \implies 15u + 22v = 5$ --- (i)
Case 2: $40u + 55v = 13$ --- (ii)
Multiplying (i) by $4$ and (ii) by $3$:
$60u + 88v = 20$
$60u + 165v = 39$
Subtracting gives: $77v = 19 \implies v = \frac{1}{11}$.
Substitute $v = \frac{1}{11}$ in (i): $15u + 2 = 5 \implies 15u = 3 \implies u = \frac{1}{5}$.
Thus: $x - y = 5$ and $x + y = 11$.
Adding both equations: $2x = 16 \implies x = 8\text{ km/h}$.
Subtracting: $2y = 6 \implies y = 3\text{ km/h}$.
Final Answer: Speed of boat in still water = $8\text{ km/h}$, Speed of stream = $3\text{ km/h}$.
Solution 2: Uniform Speed Train Problem
Ch 3: Linear Equations
Let uniform speed = $x\text{ km/h}$ and actual time taken = $y\text{ hours}$.
Total Distance $D = x \cdot y\text{ km}$.
Condition 1: $(x + 10)(y - 2) = xy \implies xy - 2x + 10y - 20 = xy \implies -2x + 10y = 20 \implies -x + 5y = 10$ --- (i)
Condition 2: $(x - 10)(y + 3) = xy \implies xy + 3x - 10y - 30 = xy \implies 3x - 10y = 30$ --- (ii)
Multiply (i) by $2$: $-2x + 10y = 20$.
Add to (ii): $(-2x + 10y) + (3x - 10y) = 20 + 30 \implies x = 50\text{ km/h}$.
From (i): $-50 + 5y = 10 \implies 5y = 60 \implies y = 12\text{ hours}$.
Distance $D = 50 \times 12 = 600\text{ km}$.
Final Answer: Distance covered by the train = $600\text{ km}$.
Solution 3: Women & Men Embroidery Work
Ch 3: Linear Equations
Let $1\text{ woman}$ complete work in $x\text{ days} \implies 1\text{ day work} = \frac{1}{x}$.
Let $1\text{ man}$ complete work in $y\text{ days} \implies 1\text{ day work} = \frac{1}{y}$.
Let $\frac{1}{x} = u$ and $\frac{1}{y} = v$.
From condition 1: $\frac{2}{x} + \frac{5}{y} = \frac{1}{4} \implies 2u + 5v = \frac{1}{4} \implies 8u + 20v = 1$ --- (i)
From condition 2: $\frac{3}{x} + \frac{6}{y} = \frac{1}{3} \implies 3u + 6v = \frac{1}{3} \implies 9u + 18v = 1$ --- (ii)
Solving (i) and (ii): $u = \frac{1}{18}, v = \frac{1}{36}$.
Final Answer: Time taken by $1\text{ woman} = 18\text{ days}$, Time taken by $1\text{ man} = 36\text{ days}$.
Solution 7: Infinite Solutions Condition ($a, b$)
Ch 3: Linear Equations
For infinite solutions: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
$\frac{2}{a-b} = \frac{3}{a+b} = \frac{7}{3a+b-2}$.
Equating first two ratios: $2(a+b) = 3(a-b) \implies 2a + 2b = 3a - 3b \implies a = 5b$ --- (i)
Equating last two ratios: $3(3a+b-2) = 7(a+b) \implies 9a + 3b - 6 = 7a + 7b \implies 2a - 4b = 6$ --- (ii)
Substitute $a = 5b$ into (ii): $2(5b) - 4b = 6 \implies 6b = 6 \implies b = 1$.
Then $a = 5(1) = 5$.
Final Answer: $a = 5, b = 1$.
Solution 9: Water Taps and Tank Problem
Ch 4: Quadratic Equations
Let time taken by smaller tap = $x\text{ hours}$.
Time taken by larger tap = $(x - 10)\text{ hours}$.
Combined rate: $\frac{1}{x} + \frac{1}{x-10} = \frac{8}{75}$.
$\frac{x - 10 + x}{x(x-10)} = \frac{8}{75} \implies \frac{2x - 10}{x^2 - 10x} = \frac{8}{75}$.
$75(2x - 10) = 8(x^2 - 10x) \implies 150x - 750 = 8x^2 - 80x$.
$8x^2 - 230x + 750 = 0 \implies 4x^2 - 115x + 375 = 0$.
Factoring: $4x^2 - 100x - 15x + 375 = 0 \implies 4x(x - 25) - 15(x - 25) = 0$.
$(4x - 15)(x - 25) = 0 \implies x = 25$ or $x = 3.75$.
If $x = 3.75$, $x - 10 = -6.25$ (Negative, reject!).
Final Answer: Smaller tap = $25\text{ hours}$, Larger tap = $15\text{ hours}$.
Solution 10: Aircraft Delayed Weather Problem
Ch 4: Quadratic Equations
Let usual speed = $x\text{ km/h}$. Increased speed = $(x + 250)\text{ km/h}$.
Time difference = $30\text{ minutes} = \frac{1}{2}\text{ hour}$.
$\frac{1500}{x} - \frac{1500}{x+250} = \frac{1}{2}$.
$1500 \left[ \frac{x + 250 - x}{x(x + 250)} \right] = \frac{1}{2} \implies \frac{375000}{x^2 + 250x} = \frac{1}{2}$.
$x^2 + 250x - 750000 = 0$.
Factoring: $(x + 1000)(x - 750) = 0$.
Since speed cannot be negative, $x = 750\text{ km/h}$.
Final Answer: Usual speed of aircraft = $750\text{ km/h}$.
Solution 13: Solving Quadratic Equation with Parameters
Ch 4: Quadratic Equations
Given: $9x^2 - 9(a+b)x + (2a^2 + 5ab + 2b^2) = 0$.
Factorize constant term: $2a^2 + 5ab + 2b^2 = 2a^2 + 4ab + ab + 2b^2 = (2a + b)(a + 2b)$.
Now split middle coefficient $-9(a+b) = -3(2a+b) - 3(a+2b)$.
Rewrite equation: $9x^2 - 3(2a+b)x - 3(a+2b)x + (2a+b)(a+2b) = 0$.
$3x [3x - (2a+b)] - (a+2b) [3x - (2a+b)] = 0$.
$[3x - (2a+b)] [3x - (a+2b)] = 0$.
Final Answer: $x = \frac{2a+b}{3}$ or $x = \frac{a+2b}{3}$.
Solution 17: Equilateral Triangle $9AD^2 = 7AB^2$ Proof
Ch 6/7: Triangles
Given: Equilateral $\triangle ABC$ of side $a$. $BD = \frac{1}{3}a$.
To Prove: $9 AD^2 = 7 AB^2$.
Construction: Draw $AE \perp BC$. In equilateral triangle, altitude bisects the base $\implies BE = EC = \frac{a}{2}$.
Now, $DE = BE - BD = \frac{a}{2} - \frac{a}{3} = \frac{a}{6}$.
In right-angled $\triangle AEB$: $AE^2 = AB^2 - BE^2 = a^2 - \left(\frac{a}{2}\right)^2 = \frac{3a^2}{4}$.
In right-angled $\triangle AED$: $AD^2 = AE^2 + DE^2$.
$AD^2 = \frac{3a^2}{4} + \left(\frac{a}{6}\right)^2 = \frac{3a^2}{4} + \frac{a^2}{36}$.
LCM $= 36 \implies AD^2 = \frac{27a^2 + a^2}{36} = \frac{28a^2}{36} = \frac{7a^2}{9}$.
$9 AD^2 = 7 a^2$. Since $a = AB$, we get: $9 AD^2 = 7 AB^2$. (Hence Proved)
Solution 18: Medians Proof $4(BL^2 + CM^2) = 5 BC^2$
Ch 6/7: Triangles
In right $\triangle ABC$, $\angle A = 90^\circ$. $L$ is midpoint of $AC$, $M$ is midpoint of $AB$.
In right $\triangle BAL$: $BL^2 = AB^2 + AL^2 = AB^2 + \left(\frac{AC}{2}\right)^2 = AB^2 + \frac{AC^2}{4}$ --- (i)
In right $\triangle CAM$: $CM^2 = AC^2 + AM^2 = AC^2 + \left(\frac{AB}{2}\right)^2 = AC^2 + \frac{AB^2}{4}$ --- (ii)
Adding (i) and (ii):
$BL^2 + CM^2 = (AB^2 + AC^2) + \frac{AB^2 + AC^2}{4} = BC^2 + \frac{BC^2}{4} = \frac{5 BC^2}{4}$ (since $AB^2 + AC^2 = BC^2$).
Multiplying by $4$: $4(BL^2 + CM^2) = 5 BC^2$. (Hence Proved)
