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Question 1: Solve for $x$: $\frac{1}{a+b+x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}, [x \neq 0, -(a+b)]$.
Solution:
Step 1: Rearrange: $\frac{1}{a+b+x} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b}$.
Step 2: Simplify: $\frac{x - (a+b+x)}{x(a+b+x)} = \frac{a+b}{ab} \Rightarrow \frac{-(a+b)}{x^2+ax+bx} = \frac{a+b}{ab}$.
Step 3: Cross multiply: $-ab = x^2 + ax + bx \Rightarrow x^2 + ax + bx + ab = 0$.
Step 4: Factorize: $x(x+a) + b(x+a) = 0 \Rightarrow (x+a)(x+b) = 0$.
Answer: $x = -a, x = -b$. -
Question 2: Two water taps together can fill a tank in $9 \frac{3}{8}$ hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Solution:
Step 1: Let smaller tap take $x$ hrs. Larger takes $x-10$ hrs. Total time $= 75/8$ hrs.
Step 2: $\frac{1}{x} + \frac{1}{x-10} = \frac{8}{75} \Rightarrow \frac{2x-10}{x^2-10x} = \frac{8}{75}$.
Step 3: $75(2x-10) = 8(x^2-10x) \Rightarrow 150x - 750 = 8x^2 - 80x \Rightarrow 8x^2 - 230x + 750 = 0$.
Step 4: $4x^2 - 115x + 375 = 0$. Solving gives $x=25$ or $x=3.75$.
Step 5: If $x=3.75$, $x-10$ is negative. So $x=25$.
Answer: Smaller: 25 hrs, Larger: 15 hrs. -
Question 3: A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
Solution:
Step 1: Let stream speed $= x$. Upstream $= 18-x$, Downstream $= 18+x$.
Step 2: $\frac{24}{18-x} - \frac{24}{18+x} = 1$.
Step 3: $24(18+x - (18-x)) = (18-x)(18+x) \Rightarrow 48x = 324 - x^2$.
Step 4: $x^2 + 48x - 324 = 0 \Rightarrow (x+54)(x-6) = 0$.
Answer: Speed of stream is 6 km/h. -
Question 4: If the roots of the equation $(c^2 - ab)x^2 - 2(a^2 - bc)x + b^2 - ac = 0$ are equal, prove that either $a = 0$ or $a^3 + b^3 + c^3 = 3abc$.
Solution:
Step 1: For equal roots, $D = 0 \Rightarrow B^2 - 4AC = 0$.
Step 2: $4(a^2-bc)^2 - 4(c^2-ab)(b^2-ac) = 0$.
Step 3: Simplify: $(a^4 + b^2c^2 - 2a^2bc) - (b^2c^2 - ac^3 - ab^3 + a^2bc) = 0$.
Step 4: $a^4 - 3a^2bc + ac^3 + ab^3 = 0 \Rightarrow a(a^3 + b^3 + c^3 - 3abc) = 0$.
Answer: Hence $a=0$ or $a^3+b^3+c^3=3abc$. -
Question 5: A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km away in time, it has to increase its speed by 250 km/hr from its usual speed. Find its usual speed.
Solution:
Step 1: Let usual speed $= x$. Time diff $= 30$ min $= 1/2$ hr.
Step 2: $\frac{1500}{x} - \frac{1500}{x+250} = \frac{1}{2}$.
Step 3: $1500(\frac{250}{x^2+250x}) = \frac{1}{2} \Rightarrow 750000 = x^2 + 250x$.
Step 4: $x^2 + 250x - 750000 = 0 \Rightarrow (x+1000)(x-750) = 0$.
Answer: Usual speed is 750 km/hr. -
Question 6: Solve for $x$: $9x^2 - 9(a+b)x + (2a^2 + 5ab + 2b^2) = 0$.
Solution:
Step 1: Factorize constant term: $2a^2 + 5ab + 2b^2 = (2a+b)(a+2b)$.
Step 2: $D = 81(a+b)^2 - 36(2a+b)(a+2b) = 9(a-b)^2$.
Step 3: $x = \frac{9(a+b) \pm 3(a-b)}{18}$.
Answer: $x = \frac{2a+b}{3}, \frac{a+2b}{3}$. -
Question 7: Solve for $x$: $\frac{1}{x} - \frac{1}{x-2} = 3, x \neq 0, 2$.
Solution:
Step 1: $\frac{x-2-x}{x(x-2)} = 3 \Rightarrow \frac{-2}{x^2-2x} = 3$.
Step 2: $3x^2 - 6x + 2 = 0$.
Step 3: $x = \frac{6 \pm \sqrt{36-24}}{6} = \frac{6 \pm \sqrt{12}}{6}$.
Answer: $x = \frac{3 \pm \sqrt{3}}{3}$. -
Question 8: At t minutes past 2 pm, the time needed by the minutes hand of a clock to show 3 pm was found to be 3 minutes less than $\frac{t^2}{4}$ minutes. Find t.
Solution:
Step 1: Time needed to show 3 pm is $60 - t$ minutes.
Step 2: Equation: $60 - t = \frac{t^2}{4} - 3$.
Step 3: $63 - t = \frac{t^2}{4} \Rightarrow t^2 + 4t - 252 = 0$.
Step 4: $(t+18)(t-14) = 0 \Rightarrow t = 14$ (since time > 0).
Answer: $t = 14$ minutes. -
Question 9: Solve for $x$: $\sqrt{3}x^2 - 2\sqrt{2}x - 2\sqrt{3} = 0$.
Solution:
Step 1: $D = (-2\sqrt{2})^2 - 4(\sqrt{3})(-2\sqrt{3}) = 8 + 24 = 32$.
Step 2: $x = \frac{2\sqrt{2} \pm \sqrt{32}}{2\sqrt{3}} = \frac{2\sqrt{2} \pm 4\sqrt{2}}{2\sqrt{3}}$.
Step 3: $x_1 = \frac{6\sqrt{2}}{2\sqrt{3}} = \sqrt{6}$, $x_2 = \frac{-2\sqrt{2}}{2\sqrt{3}} = -\sqrt{\frac{2}{3}}$.
Answer: $x = \sqrt{6}, -\sqrt{2/3}$. -
Question 10: A shopkeeper buys a number of books for ₹80. If he had bought 4 more books for the same amount, each book would have cost ₹1 less. How many books did he buy?
Solution:
Step 1: Let books $= x$. Cost $= 80/x$. New cost $= 80/(x+4)$.
Step 2: $\frac{80}{x} - \frac{80}{x+4} = 1$.
Step 3: $80(4) = x(x+4) \Rightarrow x^2 + 4x - 320 = 0$.
Step 4: $(x+20)(x-16) = 0 \Rightarrow x = 16$.
Answer: 16 books.