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Andhra Pradesh SSC Class 10th Maths Question Paper 1 With Solution 2019 QUESTION PAPER CODE 15E(A) (B)

Andhra Pradesh SSC Class 10th Maths

Question Paper 1 Solutions | Year: 2019 | Code: 15E(A)

SECTION – I (4 Marks)

1. Write A = {2, 4, 8, 16} in set-builder form.

Step 1 Observe the pattern: $2=2^1, 4=2^2, 8=2^3, 16=2^4$.
Answer $A = \{x : x = 2^n, n \in \mathbb{N}, n \le 4\}$

2. Find the value of $\log_5 \sqrt{625}$.

Solution $\log_5 (625)^{1/2} = \log_5 (5^4)^{1/2}$
$= \log_5 5^2 = 2 \log_5 5$
$= 2(1) = 2$

3. The larger of two supplementary angles exceeds the smaller by 58°, find the angles.

Solution $x + y = 180^\circ$ (Supplementary)
$x - y = 58^\circ$ (Given)
Adding: $2x = 238^\circ \implies x = 119^\circ$.
$y = 180^\circ - 119^\circ = 61^\circ$.
Angles: 119° and 61°.

4. Find the curved surface area of the cylinder, whose radius is 7cm and height is 10cm.

Solution $CSA = 2\pi rh = 2 \times \frac{22}{7} \times 7 \times 10 = 440 \text{ cm}^2$

SECTION – II (10 Marks)

5. Rohan’s mother is 26 years older than him. The product of their ages after 3 years will be 360. Write the quadratic equation.

Solution Rohan = $x$. Mother = $x + 26$.
After 3 years: $(x + 3)$ and $(x + 29)$.
Product: $(x + 3)(x + 29) = 360$
$x^2 + 32x + 87 = 360 \implies x^2 + 32x - 273 = 0$.

6. Find the zeroes of $x^2 – x – 30$ and verify the relation between zeroes and coefficients.

Solution $x^2 - 6x + 5x - 30 = 0 \implies (x - 6)(x + 5) = 0$
Zeroes: $\alpha = 6, \beta = -5$.
Sum: $6 + (-5) = 1$ (Verified with $-b/a$).
Product: $6(-5) = -30$ (Verified with $c/a$).

7. A joker’s cap is in the form of a right circular cone (r=7cm, h=24cm). Find the area of sheet for 10 caps.

Solution $l = \sqrt{7^2 + 24^2} = 25\text{ cm}$.
Area (1 cap) = $\pi rl = \frac{22}{7} \times 7 \times 25 = 550 \text{ cm}^2$.
Area (10 caps) = $5500 \text{ cm}^2$.

8. Find the HCF of 1260 and 1440 by using Euclid’s division lemma.

Solution $1440 = 1260(1) + 180$
$1260 = 180(7) + 0$
HCF is the last divisor: 180.

9. If the sum of the first 15 terms of an AP is 675 and its first term is 10, then find the 25th term.

Solution $S_{15} = \frac{15}{2}[2(10) + 14d] = 675 \implies 20 + 14d = 90 \implies d = 5$.
$a_{25} = 10 + 24(5) = 130$.

SECTION – III (16 Marks)

10. Internal Choice Question

[A] Show that $2 + 5\sqrt{3}$ is irrational.
Assume it is rational: $2 + 5\sqrt{3} = \frac{a}{b}$.
$\sqrt{3} = \frac{a-2b}{5b}$. RHS is rational, implying $\sqrt{3}$ is rational (Contradiction).
Thus, it is irrational.

[B] Check whether -321 is a term of the AP 22, 15, 8, 1...
$22 + (n-1)(-7) = -321 \implies (n-1)(-7) = -343 \implies n-1 = 49 \implies n=50$.
Yes, it is the 50th term.

11. Internal Choice Question

[A] Moulika's Marks (Maths & English)
Maths($x$), Eng($30-x$).
$(x+2)(27-x) = 210 \implies x^2 - 25x + 156 = 0$.
$x = 12$ or $13$. Marks: (12, 18) or (13, 17).

[B] Oil Drum Painting Cost
$r=1m, h=7m$.
TSA $= 2\pi r(r+h) = \frac{44}{7}(8) = 50.28 m^2$.
Cost $= 50.28 \times 10 \times 5 = \text{Rs. } 2514$.

12. Internal Choice Question

[A] Sets Operations
$A=\{1..5\}, B=\{2,3,5\}, C=\{1,3,5,7,9\}, D=\{2,4,6,8,12,14,16,24,48\}$
$A \cup B = \{1,2,3,4,5\}$
$B \cap C = \{3,5\}$
$A - D = \{1,3,5\}$
$D - B = \{4,6,8,12,16,24,48\}$

[B] Linear Equations Cost
$6x + 4y = 90$ and $8x + 3y = 85$.
Solving gives $x=5$ (Pencil), $y=15$ (Notebook).

13. Internal Choice: Graphs

[A] Find zeroes of $p(x) = x^2 + x – 20$ using graph.
Graph for x^2 + x - 20