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Cbse 10Th Questions 2E2915

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Cbse 10Th Questions 2E2915


1. Problem: Find the other number if LCM is 182, HCF is 13, and one number is 26. Formula: Product of two numbers = LCM × HCF Calculation: Let the other number be $x$. $26 \times x = 182 \times 13$ $x = \frac{182 \times 13}{26} = 91$ Answer: (d) 91 2. Problem: Find quadratic polynomial with sum of zeroes = 0 and product = $\sqrt{5}$. Formula: Polynomial is $x^2 - (sum) x + product = 0$ Here, sum = 0, product = $\sqrt{5}$ Polynomial: $x^2 + \sqrt{5} = 0$ Answer: (c) $x^2 - \sqrt{5}$ 3. Problem: System $kx - y = 2$ and $6x - 2y = 3$ has unique solution when? Rule: Unique solution if determinant $\neq 0$. Determinant: $\begin{vmatrix} k & -1 \\ 6 & -2 \end{vmatrix} = k(-2) - (-1)(6) = -2k + 6$ Set $\neq 0$: $-2k + 6 \neq 0 \Rightarrow k \neq 3$ Answer: (d) $k \neq 3$ 4. Problem: One root of $2x^2 + kx - 6 = 0$ is 2, find $k$. Substitute root: $2(2)^2 + k(2) - 6 = 0 \Rightarrow 8 + 2k - 6 = 0 \Rightarrow 2k + 2 = 0 \Rightarrow k = -1$ Answer: (b) -1 5. Problem: In AP, $d = -4$, $n=7$, $a_n=4$, find $a$. Formula: $a_n = a + (n-1)d$ $4 = a + 6(-4) \Rightarrow 4 = a - 24 \Rightarrow a = 28$ Answer: (d) 28 6. Problem: $k-1$, $k+3$, $3k-1$ in AP, find $k$. AP condition: $2 \times (middle) = first + last$ $2(k+3) = (k-1) + (3k-1)$ $2k + 6 = 4k - 2 \Rightarrow 6 + 2 = 4k - 2k \Rightarrow 8 = 2k \Rightarrow k = 4$ Answer: (a) 4 7. Problem: Find ratio in which x-axis divides join of $P(2,-3)$ and $Q(5,6)$. Let ratio be $m:n$. Since point lies on x-axis, y-coordinate = 0. Using section formula for y: $\frac{m \times 6 + n \times (-3)}{m + n} = 0$ $6m - 3n = 0 \Rightarrow 2m = n \Rightarrow m:n = 1:2$ Answer: (b) 1:2 8. Problem: Which is NOT similarity criterion? Options: AA, SAS, AAA, RHS AAA is not a valid similarity criterion (it is for congruence). Answer: (c) AAA 9. Problem: Find center of circle with diameter ends $(a,b)$ and $(-a,-b)$. Center is midpoint: $\left( \frac{a + (-a)}{2}, \frac{b + (-b)}{2} \right) = (0,0)$ Answer: (a) (0,0)