CBSE Class 10 Mathematics Question Paper 2019 Set 3 Solutions

Strictly Confidential: (For Internal and Restricted use only)
Secondary School Examination
March 2019
MARKING SCHEME – MATHEMATICS (SUBJECT CODE -041 )
PAPER CODE: 30/3/1, 30/3/2, 30/3/3

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QUESTION PAPER CODE 30/3/1
EXPECTED ANSWER/VALUE POINTS
SECTION A

1
1. (x + 5)2 = 2(5x – 3) ⇒ x2 + 31 = 10
2

1
D = –124
2

27 3 1
2. 3 4 2 = 3 4
2 ⋅5 ⋅3 2 ⋅5 2

1
It will terminate after 4 decimal places
2

OR
429 = 3 × 11 × 13 1

10 1
3. S10 = [2 × 6 + 9 × 6]
2 2

1
= 330
2

4. AB = 5

1
⇒ (x – 0) 2 + (–4 – 0) 2 = 5 2

x2 + 16 = 25

1
x = ±3
2

5. Length of chord = 2 a 2 – b 2 1

1
6. PQ = 5 cm
2

PQ 5 1
tan θ = =
PR 9 2

OR

1
sec α = 1 + tan 2 α 2

25 13 1
= 1+ =
144 12 2

30/3/1 (1)

SECTION B
7. Diagonals of parallelogram bisect each other

3 + a 1+ b  5 + 4 1+ 3
∴  ,  =  ,  1
2 2 2 2 
3 + a = 9, 1 + b = 4

1 1
So a = 6, b = 3 +
2 2

OR

1:2
P divides AB in the ratio 1 : 2
A P Q B
(–2, 0) (0, 8)
 0 – 4 8 + 0   –4 8 
∴ Coordinates of P are 
 3
, = ,  1
2   3 3

Q divides AB in the ratio 2 : 1

 0 – 2 16 + 0   –2 16 
∴ Coordinates of Q are  ,  =  ,  1
 3 3  3 3

8. 3x – 5y = 4 ...(1)
9x – 2y = 7
9x – 15y = 12
9x – 2y = 7
– + –
–13y = 5 ⇒ y = –5/13 1

 9 –5 
From (1), x = 9/13 ∴ solution is  ,  1
 13 13 
9. HCF (65, 117) = 13 1

1
13 = 65n – 117
2

1
Solving, we get, n = 2
2

(2) 30/3/1

OR
Required minimum distance = LCM (30, 36, 40) 1
30 = 2 × 3 × 5 = 23 × 32 × 5
36 = 22 × 32 = 360 cm 1
40 = 23 × 5
10. Composite numbers on a die are 4 and 6

2 1
∴ P (composite number) = or 1
6 3
Prime numbers are 2, 3 and 5

3 1
∴ P(prime number) = or 1
6 2
11. x2 – 8x + 18 = 0
x2 – 8x + 16 + 2 = 0 1

1
(x – 4)2 = –2
2

Square of a number can’t be negative

1
∴ The equation has no solution.
2

1
12. Total number of possible outcomes = 34
2

Favourable number of outcomes is (7, 14, 21, 28 and 35) = 5 1

5 1
P(multiple of 7) =
34 2

SECTION C

1
13. AB2 = AD2 + BD2 Correct Figure
2
A
AC2 = AD2 + CD2 1

AB2 – AC2 = BD2 – CD2
C D B

= (3CD)2 – CD2

= 8 CD2 1

30/3/1 (3)

2
1 
= 8 ×  BC 
4 

⇒ 2AB2 – 2AC2 = BC2

1
or 2AB2 = 2AC2 + BC2
2

OR

A P 1
Correct Figure
2

∆ABC ~ ∆PQR
B D C Q M R
AB BC AC 1
∴ = =
PQ QR PR 2

AB 2BD BD
= or 1
PQ 2QM QM

Also ∠B = ∠Q

1
∴ ∆ABD ~ ∆PQM
2

AB AD 1
So =
PQ PM 2

14.

x3 – 3x + 1 x5 – 4x3 + x2 + 3x + 1 x2 – 1
x5 – 3x3 + x2
– + –
3
–x + 3x + 1 1
3 2
–x + 3x – 1 2
+ – +
2

(4) 30/3/1

1
Since remainder ≠ 0 ∴ g(x) is not a factor of p(x)
2

15. Coordinates of mid points are
A(0, –1)
D(1, 2)
E F

E (1, 0)
B(2, 1) D C(0, 3)
1
F(0 ,1) 1
2

1
Area of ∆DEF = [1(0 – 1) + 1(1 – 2) + 0] 1
2

1 1
Y = (–2) = 1sq. unit
(0, 6) 2 2
6 0
5 +1=
16. y Correct graph 2
3) x–
4 (2,
3
2 Solution is
(1,2)
(–1,0) 1 (0,1) (4, 0)
X 1 1
–2 –1 0 1 2 3 4 5 6 x = 2, y = 3 +
1 2 2
3x

2
+2
y–
12
=0

17. Let us assume that 3 be a rational number

p 1
3 = q where p and q are co-primes and q ≠ 0 2

⇒ p2 = 3q2 ...(1)

∴ 3 divides p2
i.e., 3 divides p also ...(2)
Let p = 3m, for some integer m 1
From (1), 9m2 = 3q2
⇒ q2 = 3m2
∴ 3 divides q2 i.e., 3 divides q also ...(3) 1

30/3/1 (5)

From (2) and (3), we get that 3 divides p and q both which is a contradiction to the fact that p and q
1
are co-primes.
2
Hence our assumption is wrong
∴ 3 is irrational
OR
1251 – 1 = 1250, 9377 – 2 = 9375, 15628 – 3 = 15625 1
Required largest number = HCF (1250, 9375, 15625)

1250 = 2 × 54 

9375 = 3 × 54  1
 1
2
6250 = 2 × 55 
1
∴ HCF (1250, 9375, 15625) = 54 = 625
2

18. A, B, C are interior angles of ∆ABC
1
∴ A + B + C = 180°
2

B+C  180° – A 
sin   sin 
(i)  2  =  2 

 A
= sin  90° – 
 2
A 1
= cos 1
2 2

B+ C  90° 
tan   tan 
(ii)  2  =  2  (∵ ∠ A = 90°)

= tan 45° 1
=1
OR
tan (A + B) = 1 ∴ A + B = 45° 1

1
tan (A – B) = ∴ A – B = 30° 1
3

1° 1
Solving, we get ∠A = 37 or 37.5°
2 2

1° 1
∠B = 7 or 7.5°
2 2

(6) 30/3/1

19. Let TR be x cm and TP be y cm
P
OT is ⊥ bisector of PQ
5cm
8cm
T
O R So PR = 4 cm

Q ln ∆OPR, OP2 = PR2 + OR2

∴ OR = 3 cm 1

1
ln ∆PRT, y2 = x2 + 42 ...(1)
2

1
ln ∆OPT, (x + 3)2 = 52 + y2
2

∴ (x + 3)2 = 52 + x2 + 16 [using (1)]

16 1
Solving we get x = cm
3 2

256 400 
From (1), y 2 = + 16 =
9 9 
 1
20 
So y = cm 2
3 

OR

1
∆ROC ≅ ∆QOC
2
D R C
8
S 7 1
2
∴ ∠1 = ∠ 2 
Q
6 O3 Similarly ∠ 4 = ∠3 
5 4  1
∠5 = ∠ 6 
A P B ∠8 = ∠7 

1
∠ROQ + ∠QOP + ∠POS + ∠SOR = 360°
2

∴ 2∠1 + 2∠4 + 2∠5 + 2∠8 = 360

⇒ ∠1 + ∠4 + ∠5 + ∠8 = 180°

30/3/1 (7)

So, ∠DOC + ∠AOB = 180°

and ∠AOD + ∠BOC = 180°. 1

20. Volume of water flowing through canal in 30 minutes

1
= 5000 × 6 × 1.5 = 45000 m3 1
2

8
Area = 45000 ÷
100
1
= 562500 m2 1
2

21.
Number of days Number of students (fi) xi fixi
0-6 10 3 30
6-12 11 9 99
12-18 7 15 105
18-24 4 21 84
24-30 4 27 108 Correct Table 2
30-36 3 33 99
36-42 1 39 39
Total 40 564

Σfi x i 564
x = =
Σf i 40
= 14.1 1
22. Total area cleaned = 2 × Area of sector

πr 2θ
= 2× 1
260°
22 120°
= 2× × 21 × 21 × 1
7 360°
= 924 cm2 1

(8) 30/3/1

SECTION D

1
23. Correct Figure
2
P (pole)
PB – PA = 7 m
B A 1
13 m Let AP be x m ∴ PB = (x + 7) m
2

AB2 = PB2 + AB2

∴ 132 = (x + 7)2 + x2

x2 + 7x – 60 = 0 1

= (x + 12) (x – 5) = 0 1

∴ x = 5, –12 Rejected

1
∴ Situation is possible
2

∴ Distance of pole from gate A = 5 m

1
and distance of pole from gate B = 12 m.
2

24. mam = nan
⇒ ma + m(m – 1)d = na + n(n – 1)d 1
⇒ (m – n)a + (m2 – m – n2 + n)d = 0 1
(m – n)a + [(m – n) (m + n) – (m – n)d] = 0 1
Dividing by (m – n)
So, a + (m + n – 1)d = 0
or am + n = 0 1
OR

1
Let first three terms be a –d, a and a + d
2

a – d + a + a + d = 18

1
So a=6
2

(a – d) (a + d) = 5d

30/3/1 (9)

⇒ 62 – d2 = 5d 1
or d2 + 5d – 36 = 0
(d + 9) (d – 4) = 0
so d = –9 or 4 1

1
For d = –9 three numbers are 15, 6 and –3
2

1
For d = 4 three numbers are 2, 6 and 10
2

25. Correct construction of ∆ABC 2
Correct construction of triangle similar to ∆ABC 2
26. (a) Total surface area of block
= TSA of cube + CSA of hemisphere – Base area of hemisphere 1
= 6a2 + 2πr2 – πr2
= 6a2 + πr2

 2 22  1
=  6 × 6 + × 2.1 × 2.1 cm 2
7  2

= (216 + 13.86) cm2

1
= 229.86 cm2
2

(b) Volume of block

3 2 22
= 6 + × × (2.1)3 1
3 7
= (216 + 19.40) cm3
= 235.40 cm3 1
OR
Volume of frustum = 12308.8 cm3

1
∴ πh(r12 + r22 + r1r2 ) = 12308.8
3
1
⇒ × 3.14 × h(202 + 122 + 20 × 12) = 12308.8 1
3
12308.8 × 3
h=
784 × 3.14
h = 15 cm 1
(10) 30/3/1

l= 152 + (20 – 12)2 = 17 cm. 1
Area of metal sheet used = πl (r1 + r2) + πr22
= 3.14[17 × 32 + 122]
= 3.14 × 688 cm2
= 2160.32 cm2 1

1
27. Correct figure, given, to prove and construction ×4=2
2

Correct proof. 2
OR

1
Correct figure, given, to prove and construction ×4=2
2

Correct proof. 2
28. 1 + sin2 θ = 3sin θ cos θ
Dividing by cos2 θ
sec2 θ + tan2 θ = 3tan θ 1
⇒ 1 + tan2 θ + tan2 θ = 3tan θ
⇒ 2 tan2 θ – 3tan θ + 1 = 0 1
(tan θ – 1) (2tan θ – 1) = 0 1

1 1 1
So tan θ = 1 or +
2 2 2

Alternate method
1 + sin2 θ = 3sin θ cos θ
sin2 θ + cos2 θ + sin2 θ – 3sin θ cos θ = 0 1
Dividing by cos2 θ
⇒ 2 tan2 θ – 3tan θ + 1 = 0 1
⇒ (tan θ – 1) (2 tan θ – 1) = 0 1

1 1 1
So tan θ = 1 or +
2 2 2

30/3/1 (11)

29. Class interval Cumulative Frequency
More than or equal to 20 100
More than or equal to 30 90
More than or equal to 40 82
More than or equal to 50 70
More than or equal to 60 46
More than or equal to 70 40
More than or equal to 80 15
1
Correct Table 1
2

1
Plotting of points (20, 100), (30, 90), (40, 82), (50, 70), (60, 46), (70, 40) and (80, 15) 1
2

Joining the points to get a curve 1
30. Correct Figure 1

Let AB = h be the height of tower
A

h
h ln ∆ABC, = tan 60°
x
60° 30°
B x C 40 m D h= x 3 1

h
ln ∆ABD, = tan 30°
x + 40

1
⇒ h 3 = x + 40
2

3x = x + 40

1
∴ x = 20
2

1
So, height of tower = h = 20 3 m
2

= 20 × 1.732 m
1
= 34.64 m
2

(12) 30/3/1