Daily Practice Problems

📝 SolutionLet HCF $= h$. Then LCM $= 14h$.
Given: $h + 14h = 600 \Rightarrow 15h = 600 \Rightarrow h = 40$, so LCM $= 560$.
Now use: $\text{HCF} \times \text{LCM} = a \times b$
$40 \times 560 = 280 \times b \Rightarrow b = \dfrac{22400}{280} = \mathbf{80}$.

Key identity: product of two numbers = HCF × LCM. This holds only for two numbers.

📝 SolutionFrom the polynomial: $\alpha + \beta = 5$ and $\alpha\beta = k$.
Use the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$:
$1^2 = 5^2 - 4k \Rightarrow 1 = 25 - 4k \Rightarrow 4k = 24 \Rightarrow k = \mathbf{6}$.

Check: $x^2 - 5x + 6 = (x-2)(x-3)$; zeroes 3 and 2; difference = 1 ✓

📝 SolutionFor infinitely many solutions: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$
$\dfrac{2}{k-1} = \dfrac{3}{k+2} = \dfrac{7}{3k}$
From the first two: $2(k+2) = 3(k-1) \Rightarrow 2k + 4 = 3k - 3 \Rightarrow k = 7$.
Verify with the third ratio: $\dfrac{2}{6} = \dfrac13$ and $\dfrac{7}{21} = \dfrac13$ ✓
So $k = \mathbf{7}$.

Trap: You must check all three ratios. A value satisfying only two of them gives no solution, not infinite.

📝 SolutionLet the original speed be $x$ km/h. Time $= \dfrac{360}{x}$.
New speed $= (x+5)$, new time $= \dfrac{360}{x+5}$, which is 1 hour less:
$\dfrac{360}{x} - \dfrac{360}{x+5} = 1$
$360(x+5) - 360x = x(x+5)$
$1800 = x^2 + 5x \Rightarrow x^2 + 5x - 1800 = 0$
$D = 25 + 7200 = 7225,\ \sqrt{D} = 85$
$x = \dfrac{-5 + 85}{2} = \mathbf{40}$ km/h (reject the negative root).

Check: $360/40 = 9$ h; $360/45 = 8$ h ✓

📝 SolutionUse $a_n = S_n - S_{n-1}$:
$S_{n-1} = 3(n-1)^2 + 5(n-1) = 3n^2 - 6n + 3 + 5n - 5 = 3n^2 - n - 2$
$a_n = (3n^2 + 5n) - (3n^2 - n - 2) = \mathbf{6n + 2}$
Now set $a_k = 164$:
$6k + 2 = 164 \Rightarrow 6k = 162 \Rightarrow k = \mathbf{27}$.

Sanity check: $a_1 = S_1 = 3 + 5 = 8$, and $6(1)+2 = 8$ ✓