AP 10th Class Marking Scheme 2019 Mathematics - Paper 1 (English Medium)

A.P. 15E

S.S.C. PUBLIC EXAMINATIONS, MARCH-2019
MATHEMATICS
PAPER - I
PRINCIPLES OF VALUATION

General Instructions :

1. Any logical method is valid.

2. If a candidate exceeds choice all answers have to be valued. The
answer securing highest marks should be taken into account and
the others should be marked as “Excess”.

3. Wrong answer should be struck off and “0” (Zero) marks should
be awarded.

4. Marks should be awarded according to the logical sequential steps.

5. Final answer is not the criteria for awarding marks.

6. Page wise marks should be noted in brackets otherwise write nil.

7. While entering the total marks fraction should be rounded off to
the next integer.

– 1 –

A.P. 15E

SECTION - I

1. A = {2n / n ∈ N and n < 5} 1M

2. log5 625 = log5 25 ⎫
⎪ 1
⎬ /2 M
2 ⎪
= log5 5 ⎭

= 2log5 5 ⎫
⎪ 1
⎬ /2 M
⎪
= 2×1=2 ⎭

1M

3. Let the required supplementary angles be x and y. ⎫
⎪
⎪
∴ x + y = 180° ––––––– (1) ⎪
⎪ 1
⎬ /2 M
The larger angle exceeds the smaller by 58° ⎪
⎪
⎪
∴ x – y = 58° ––––––– (2) ⎪
⎭

For solving (1) and (2) ⎫
⎪
⎬ 1
/2 M
and Finding x = 119° and y = 61°. ⎪
⎭

1M

– 2 –

A.P. 15E
4. Radius of cylinder (r) = 7 cm

height (h) = 10 cm

Curved surface Area of cylinder = 2πrh ⎫ 1
/2 M
⎬
⎭
22
= 2× × 7 × 10 ⎫
7 ⎪ 1/ M
⎬ 2
⎪
= 440 cm2 ⎭

1M

SECTION - II
⎫
5. Let Rohan's age at present be 'x' years ⎪ 1
/2 M
⎬
⎪
His mother's age at present is (x + 26) years ⎭

⎫
After three years, Rohan's age = (x + 3) years ⎪
⎬ 1
/2 M
After three years, His mother's age = x + 26 + 3 = (x + 29) years ⎪⎭

The product of their ages after 3 years = (x + 3) (x + 29) ⎫ 1
⎪ /2 M
⎬
= x2 + 32x + 87 ⎪
⎭

By the sum, the product of their ages = 360 ⎫
⎪
2 ⎪
∴ x + 32x + 87 = 360 ⎪ 1/ M
⎬ 2
2 ⎪
⇒ x + 32x – 273 = 0
⎪
⎪
Which is the required quadratic equation. ⎭
2M

– 3 –

A.P. 15E

6. Given polynomial is x2 – x – 30

To find zeroes, x2 – x – 30 = 0 say ⎫ 1/ M
⎪ 2
⎬
2 ⎪
⇒ x – 6x + 5x – 30 = 0 ⎭

⇒ (x – 6) (x + 5) = 0 ⎫ 1/ M
⎪ 2
⎬
⇒ x = 6 or x = –5 ⎪
⎭

− (−1) − (Coefficient of x )
= 1
Sum of the zeroes = 6 + (–5) = 1 = 1 Coefficient of x 2
/2 M

−30 Constant term
Product of the zeroes = 6(–5) = –30 = = 1/ M
1 Coefficient of x 2 2

2M

7. Base radius of conical shape cap (r) = 7 cm.
height (h) = 24 cm.
⎫
Slant height (l) = r2 + h2 ⎪
⎪
⎪
= 7 2 + 242 ⎪
⎬ 1
/2 M
⎪
= 49 + 576 ⎪
⎪
⎪
= 625 = 25 cm. ⎭

– 4 –

A.P. 15E

Area of Sheet required to make a cap = Lateral surface area of the cap⎫
⎪1
⎬ /2 M
= πrl ⎪
⎭
22
= × 7 × 25 ⎫
7 ⎪ 1
⎬ /2 M
= 550 sq.cm. ⎪
⎭

Area of Sheet required to make 10 such caps = 10 × 550 ⎫
⎪ 1
⎬ /2 M
= 5500 sq.cm. ⎪
⎭

2M

8. Given numbers are 1260, 1440

1440 = 1260 × 1+ 180 ⎫
⎪ 11/2 M
⎬
1260 = 180 × 7 + 0 ⎪
⎭
1
H.C.F. of 1260, 1440 is 180. /2 M

2M

– 5 –

A.P. 15E
9. First term of A.P. is a = 10

Let common difference = d

Sum of first 15 terms is S15 = 675 ⎫
⎪
⎪
15 ⎪
∴ [2a + 14d] = 675 ⎪
2 ⎪
⎪
⎪
⎪
675 × 2 ⎬ 1M
⇒ [2 × 10 + 14d] = = 90
15 ⎪
⎪
⎪
⇒ 14d = 90 – 20 = 70
⎪
⎪
⎪
⎪
70 ⎪
⇒ d= =5 ⎭
14

25th term in that A.P. is a25 = a + 24d ⎫
⎪
⎪
= 10 + 24 × 5 ⎪
⎪
⎬ 1M
⎪
= 10 + 120 ⎪
⎪
⎪
⎭
= 130

2M

– 6 –

A.P. 15E

SECTION - III
⎫
10 a) Let us assume, the contrary, ⎪
⎪⎪
that 2 + 5 3 is rational. ⎬ 1M
⎪
⎪
i.e., we can find co-primes 'a' and 'b', b ≠ 0 ⎪⎭

a
such that 2 + 5 3 =
b

a ⎫
⇒5 3 = −2 ⎪
b ⎪
⎪
⎪
a 2 ⎪
⇒ 3= − ⎬ 2M
5b 5 ⎪
⎪
⎪
a 2 a 2 ⎪
Since and ∈ Q ⇒ − ∈ Q ⎪
⎭
5b 5 5b 5

and so 3 is rational.

But this contradicts the fact that 3 is irrational. ⎫
⎪
⎬ 1
/2 M
⎪
So, Our assumption that 2 + 5 3 is rational is wrong. ⎭

1
So, we conclude that 2 + 5 3 is irrational. /2 M

4M

– 7 –

A.P. 15E
(OR)

10 b) From the given A.P. : 22, 15, 8, 1, ...... ⎫
⎪ 1M
⎬
a = 22, d = –7 ⎪
⎭

nth term of an A.P. = an = a + (n – 1) d ⎫
⎪ 1M
⎬
In this A.P. let nth term be –321 ⎪
⎭

⇒ a + (n – 1) d = – 321 ⎫
⎪
22 + (n – 1) (–7) = –321 ⎪
⇒ ⎪
⎪
(n – 1) (–7) = – 343 ⎪
⇒ ⎪
⎬
11/2 M
⎪
−343 ⎪
⇒ n–1= = 49 ⎪
−7 ⎪
⎪
n = 49 + 1 = 50 ⎪⎭
⇒

∴ –321 will be the 50th term in the given A.P. 1/ M
2

4M

– 8 –

A.P. 15E
11 a) Given that the sum of Moulika's marks ⎫
⎪
in mathematics and English is 30. ⎪
⎪
⎬
Let the marks of Moulika in Mathematics = x ⎪ 1
⎪ /2 M
and that in English = 30 – x. ⎪
⎭
If she got 2 marks more in Mathematics ⎫
⎪
then marks in Maths = x + 2 ⎪
⎪
⎪ 1/ M
and she got 3 marks less in English then ⎬ 2
⎪
the marks in English = 30 – x – 3 ⎪
⎪
⎪
= 27 – x ⎭

Product of these two = (x + 2) (27 – x) = 210 ⎫
⎪
2 ⎪
⇒ x – 25x + 156 = 0 ⎪ 2M
⎬
⇒ (x – 12) (x – 13) = 0 ⎪
⎪
⎪
⇒ x = 12, x = 13 ⎭
Case : (1) If x = 12, the marks of Moulika ⎫
⎪ 1
⎪ /2 M
in Mathematics = 12 ⎬
⎪
⎪⎭
in English = 30 – 12 = 18

Case : (2) If x = 13, the marks of Moulika ⎫
⎪
⎪ 1
in Mathematics = 13 ⎬ /2 M
⎪
⎪⎭
in English = 30 – 13 = 17

4M

– 9 –

A.P. 15E

(OR)

11 b) The diameter of the oil drum ⎫
⎪
⎪
which is in the shape of cylinder = d = 2m ⎪ 7m
⎪ 1/ M
⎬ 2
d ⎪
Radius of the drum = r = = 1 m 2m
2 ⎪
⎪
height = h = 7 m. ⎪
⎭
Total Surface Area of the drum which is ⎫
⎪ 1
⎬ /2 M
in the shape of a cylinder = 2πr(r + h) ⎪
⎭

22
= 2× × 1 × (1 + 7) ⎫
7
⎪
⎪
22 ⎪
= 2× ×8 ⎪
7 ⎪
⎬ 1 1/2 M
⎪
352 ⎪
=
7 ⎪
⎪
= 50.28 sq.m. ⎪
⎭

Charges to paint the drum per sq.m. = ` 5 ⎫
⎪
Total cost of painting 10 such type of drums = 50.28 × 5 × 10
⎪
⎬
1 1/2 M
⎪
= ` 2514 ⎪⎭

4M

– 10 –

A.P. 15E
1
12 a) A = {1, 2, 3, 4, 5} /2 M
1
B = {2, 3, 5} /2 M
1
C = {1, 3, 5, 7, 9} /2 M
1
D = {2, 4, 6, 8, 12, 16, 24, 48} /2 M
i) A ∪ B = {1, 2, 3, 4, 5} ∪ {2, 3, 5} = {1, 2, 3, 4, 5} 1/ M
2
ii) B ∩ C = {2, 3, 5} ∩ {1, 3, 5, 7, 9} = {3, 5} 1/ M
2
iii) A – D = {1, 2, 3, 4, 5} – {2, 4, 6, 8, 12, 16, 24, 48} ⎫ 1
⎬ /2 M
= {1, 3, 5} ⎭

iv) D – B = {2, 4, 6, 8, 12, 16, 24, 48} – {2, 3, 5} ⎫ 1/ M
⎬ 2
= {4, 6, 8, 12, 16, 24, 48} ⎭
4M
(OR)
⎫
12 b) Let the cost of one pencil = ` x ⎪ 1
⎬ /2 M
Cost of one Note Book = ` y ⎪
⎭
⎫
Total cost of 6 pencils and 4 Note Books = ` 90 ⎪
⎬ 1M
⇒ 6x + 4y = 90 ––––––––––– (1) ⎪
⎭
Total cost of 8 pencils and 3 Note Books = ` 85 ⎫
⎪
⇒ 8x + 3y = 85 ––––––––––– (2) ⎬ 1M
⎪
⎭
(1) : 6x + 4y = 90
⎫
(2) : 8x + 3y = 85 ⎪
⎪
Solving these equations to find x = 5, y = 15 ⎪
⎬ 1 1/2 M
and writing the cost of one pencil = ` 5 ⎪
⎪
the cost of one Note Book = ` 15 ⎪
⎭

4M

– 11 –

A.P. 15E

13 a) Let y = x2 + x – 20 1
/2 M
For finding the points of y = x2 + x – 20 11/2 M
For plotting the points and draw parabola 11/2 M
For writing zeroes of the polynomial are 4 and –5 1/ M
2
4M

Scale : on X-axis 1 cm = 1 unit
on Y-axis 1 cm = 2 units

– 12 –

A.P. 15E
(OR)
13b) For finding, the points and drawing ⎫
⎬ 11/2 M
the graph of 2x + y = 4 ⎭
For finding the points and drawing ⎫
the graph of 2x – 3y = 12 ⎬ 11/2 M
⎭
For finding the solution ⎫ 1M
⎬
x = 3 and y = –2 ⎭
4M

Scale : on X-axis 1 cm = 1 unit
on Y-axis 1 cm = 1 unit

– 13 –

A.P. 15E

PART - B 20 × 1/2 M = 10 M
SECTION - IV

14. C 24. C

15. A 25. A

16. B 26. A

17. C 27. B

18. C 28. A

19. B 29. D

20. B 30. D

21. D 31. D

22. A 32. A

23. D 33. B

– 14 –