Solution

DEC 22 MATHS FULL SYLLABUS

Class 10 - Mathematics
Section A
1.
(b) 5 × 23 × 3
Explanation:
We have,
120 = 5 × 23 × 3

2.
(c) 4
Explanation:
f(x) intersects the x -axis at 4 points. hence , f(x) has 4 zeroes.

3. (a) no solution
Explanation:
Since, we have y = 0 and y = -6 are two parallel lines.
therefore, no solution exists.

4. (a) b2 - 4ac > 0

Explanation:
A quadratic equation ax2 + bx + c = 0 has real and distinct roots, if b2 - 4ac > 0.
5.
(d) 37
Explanation:
a=3
d=3
an = 111

n=?
an = a + (n − 1)d

111 = 3 + (n − 1)3

108 = (n − 1)3
108
= (n − 1)
3

36 = n − 1

n = 37

6.
(c) 5
Explanation:
Distance from x axis = y
Here y = 5

7.
(b) (3, 5)
Explanation:
Point P divides the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2: 1
Let coordinates of P be (x, y), then
m1 x2 + m2 x1 2×4+1×1 8+1 9
x = = = = = 3
m1 + m2 2+1 3 3

1 / 10
 m1 y2 + m2 y1 2×6+1×3 12+3 15
y = = = = = 5
m1 + m2 2+1 3 3

∴ Coordinates of P are (3, 5)

8.
(d) 10
Explanation:
In △ADE and △ABC
∠ D = ∠ B {Corresponding angle}

∠ E = ∠ C {Corresponding angle}

∴ △ ADE and △ABC (by A A Similarity)

AD

AB
= DE

BC
2

5
= 4

X= 5×4

2
= 10
= 10 cm

9.
(d) 130o
Explanation:
OAPB is a quadrilateral
angle O+ angle A. + angle P + angle B = 360°
angle AOB + 90°+ 500° + 90° = 360°
angle AOB = 360° - 230°
angle AOB = 130°
OR
Sum of opposite angle is supplementary.
I.e. ∠P + ∠O = 180
50 + ∠O = 180

∠O = 180 − 50

∠O = 130

10.
(d) 0
Explanation:
we cannot draw parallel lines from a unique point, so we cannot draw any parallel tangents to PA through P.

–
11. (a) 3

2
(√2 − 1)

Explanation:
∘ ∘
sin 90 +cos 60
= ∘
sec 45 +tan 45
∘

1
1+

= 2

√2+1

= 2

√2+1

upon rationalization
√2−1
= 3

2
×
1
×
√2+1 √2−1

√2−1
= 3

2
×
√2−1

–
= 3

2
(√2 − 1)

12.
(b) − 17

Explanation:

2 / 10
 tanθ = 5

12
= P

b
−− −−−−−
h= 2
√5 + 12
2

= 13
Now, sinθ = 5

13

cosθ = 12

13
5 12
+
sinθ+cosθ
Now, =
13 13

sinθ−cosθ 5 12
−
13 13

=− 17

13.
(d) 5 m
Explanation:
Ratio of lengths of objects = ratio of lengths of their shadows.
Let the length of shadow of the tree be x m. Then,
⇒ 5x = 2 × 12.5 = 25
5 2
=
12.5 x

⇒ x=5

14.
(c) 22 cm
Explanation:
Arc length = 2πrθ

360
= (2 ×
22

7
× 21 ×
60

360
) cm = 22cm

15.
(c) 9.42 cm2
Explanation:
Radius of a circle = r = 6 cm
Central angle = θ = 30o
2

∴ Area of the sector = πr θ

360

cm2
∘
3.14×6×6×30
= ( ∘
)
360

= 9.42 cm2

16.
(d) 1

Explanation:
The number of possible outcomes when two dice are thrown is 36.
Now, the possible outcomes of getting a product of 12 are
{(2, 6),(3, 4),(4, 3),(6, 2)}, which means the number of favourable outcome is 4.
4 1
Required probability = 36
= 9

17. (a) 1

26

Explanation:
black kings = club king + spade king = 2
Number of possible outcomes = 2
Number of Total outcomes = 52
∴ Required Probability =
2 1
=
52 26

18.
(d) 6
Explanation:

3 / 10
 Mean = 8.1
Σ fixi = 132 + 5k

∑ fi = 20
Σf xi 132+5k
∴ Mean = i

Σf
⇒ 8.1 = 20
i

⇒ 132 + 5k = 8.1 × 20 = 162

⇒ 5k = 162 - 132 = 30

⇒ k = =6 30

19.
(d) A is false but R is true.
Explanation:
A is false but R is true.

20. (a) Both A and R are true and R is the correct explanation of A.
Explanation:
Both A and R are true and R is the correct explanation of A.
Section B
21. 44 = 22 × 11
96 = 25 × 3
404 = 22 × 101
HCF = 22 = 4
LCM = 25 × 11 × 3 × 101
= 106656
22. If is is given that AB = 5.6 cm, BC = 6 cm and BD = 3.2 cm
In △ABC , AD is the bisector of ∠A, meeting side BC at D
AB BD
∴ =
AC DC
5.6cm 3.2cm

AC
=
2.8cm
[DC = BC - BD]
AC =
5.6×2.8

3.2
cm = 4.9
23. We know that the tangents from an external point to a circle are equal.
∴ AP = AS ......... (i)

BP = BQ ........ (ii)
CR = CQ ....... (iii)
DR = DS ....... (iv)
On adding eq. (i), (ii), (iii) and (iv), we get
(AP + BP) + (CR + DR) = (AS + BQ) + (CQ + DS)
⇒ AB + CD = (AS + DS) + (BQ + CQ)

⇒ AB + CD = AD + BC

24. We have,
LHS = (sinθ + cosecθ)2 + (cosθ + secθ)2
⇒ LHS = (sin2θ + cosec2θ + 2sinθ cosecθ) + (cos2θ + sec2θ + 2cosθ secθ)
2 2 1 2 2 1
⇒ LHS = (sin θ + cosec θ + 2 sin θ ) + (cos θ + sec θ + 2 cos θ )
sin θ cos θ

⇒ LHS = (sin2θ + cosec2θ + 2) + (cos2θ + sec2θ + 2)

⇒ LHS = sin2θ + cos2θ + cosec2θ + sec2θ + 4
⇒ LHS = 1 + (1 + cot2θ) + (1 + tan2θ) + 4 [∵ cosec2θ = 1 + cot2θ , sec2θ = 1 + tan2θ]
⇒ LHS = 7 + tan2θ + cot2θ = RHS
OR
2
3

2
(
1
) - 2(0) - 2 1

2
2
(2)
√3

3
=- 2

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25. Area of shaded region
∘
300
= [π(42) 2
− π(21) ]
2

360
∘

= 22

7
× 63 × 21 × 5

6
.
= 3465 cm2
OR
2 ∘
3.14×(6) × 60
Area of minor sector = ∘
360

= 18.84
Hence, area of minor sector is 18.84 cm2
Area of major sector = Area of circle - Area of minor sector
= 3.14 × (6)2 - 18.84
= 94.2
Hence, area of major sector is 94.2 cm2
Section C
– p
26. Let 5 + 3√2 is rational. It can be written in the form q
.
– p
(5 + 3√2) =
q
– p
3√2 = − 5
q

– p−5q
3√2 =
q

– p−5q
√2 =
3q

As p - 5q and 3q are integers .

p−5q
So, 3q
is rational number .
–
But is not rational number .
√2
–
Since a rational number cannot be equal to an irrational number. Our assumption that 5 + 3√2 is rational wrong.
–
Hence, 5 + 3√2 is irrational.
27. p(y) = 7y2 - 11

3
y- 2

3
= 1

3
(21y2 - 11y - 2)
= 1

3
(21y2 - 14y + 3y - 2)
= 1

3
[7y (3y - 2) + 1 (3y - 2)]
= 1

3
[(7y + 1)(3y - 2)]
∴ Zeroes are 2

3
,- 1

Sum of Zeroes = 2

3
- 1

7
= 11

21
−b

a
= 11

21
−b
∴ sum of zeroes = a
−1
Product of Zeroes = ( )(- 2

3 7
)=- 2

21
c

a
=- 2

3
( )=-
1

7 21
2

∴ Product = c

a
n
28. Given, Sn = 2
(3n + 5)
n−1
∴ Sn-1 = 2
[3(n - 1) + 5]
n−1
or Sn-1 = 2
(3n + 2)
Since, an = Sn - Sn-1
n n−1
= (3n + 5) − (3n + 2)
2 2
2 3n(n−1) 2(n−1)
3n 5n
= + − −
2 2 2 2
2 2
3n 5n 3n 3n
= + − + − n+ 1
2 2 2 2
8n
= 2
-n+1
= 4n - n + 1
= 3n + 1
Now, a25 = 3(25) + 1
or, a25 = 75 + 1 = 76

Thus, 25th term of AP. is 76

5 / 10
 OR
All the natural numbers less than 100 which are divisible by 6 are
6, 12, 18, 24,................., 96
Here, a1 = 6
a2 = 12
a3 = 18
a4 = 24
::
∴ a2 - a1 = 12 - 6 = 6
a3 - a2 = 18 - 12 = 6
a4 - a3 = 24 - 18 = 6
a2 − a1 = a3 − a2 = a4 − a3 = 6 ( =6 in each case)
∴ This sequence is an arithmetic progression whose difference is 6.

Here, a = 6
d=6
l = 96
Let the number of terms be n. Then,
l = a + (n - 1)d
⇒ 96 = 6 + (n - 1)6

⇒ 96 - 6 = (n - 1)6
⇒ 90 = (n - 1)6

⇒ (n - 1)6 = 90
90
⇒ n− 1 =
6

⇒ n - 1 = 15
⇒ n = 15 + 1

⇒ n = 16
n
∴ Sn = (a + l)
2

16
= ( ) (6 + 96)
2

= (8) (102)
= 816
29. In the given figure, two tangents RQ and RP are drawn from the external point R to the circle with centre O.
∠ PRQ = 120°

To prove: OR = PR + RQ
Construction: Join OP and OQ.
Also join OR.

Proof: OR bisects the ∠ PRQ

∘
120 ∘
∴ ∠PRO = ∠QRO = = 60
2

∵ OP and OQ are radii and RP and RQ are tangents.

∴ OP ⊥ PR and OQ ⊥ QR
In right △ OPR
∘ ∘ ∘
∠POR = 180 − (90 + 60 )

∘ ∘ ∘
= 180 − 150 = 30

6 / 10
 Similarly,
∘
∠QOR = 30

and cos θ =
PR

OR

∘ PR 1 PR
⇒ cos 60 = ⇒ =
OR 2 OR

⇒ 2PR=OR ......(i)
Similarly, in right △OQR
⇒ 2QR=OR .........(ii)

Adding (i) and (ii)

⇒ 2PR + 2QR = 2OR

⇒ OR = PR + RQ

Hence Proved.
OR
O is the center of a circle of radius 5cm,
PQ is a tangent to the given circle at P.
PQ meets the line OP passing through O & P.
OQ = 12cm
To find out PQ
We join OP.
Then OP is a radius of the given circle through the point of contact P of the tangent QP.
∴ OP ⊥ QP

i.e ∠ OPQ = 90o [∵The angle between a tangent and the radius through the point of contact is 90o]
∴ △OPQ is the right triangle with OQ as the hypotenuse.
−−−− −−− −−−
PQ = √O Q − O P [by applying Pythagoras theorem]
2 2

−−− −−−− −−−

= 2
√12 − 5
2
cm = √119 cm
tan θ cot θ cos θ+sin θ
30. To prove : 1−tan θ
−
1−cot θ
=
cos θ−sin θ
tan θ cot θ
L.H.S = 1−tan θ
−
1−cot θ
sin θ c os θ

sin θ
...[∵ tanθ = ]
c os θ sin θ
= −
sin θ c os θ cosθ
1− 1−
c os θ sin θ

sin θ cos θ
= +
cos θ−sin θ cos θ−sin θ

sin θ+cos θ
=
cos θ−sin θ

= R.H.S.
31. Calculation of mean by using step-deviation method.
xi −A
ui =
Class Interval Frequency(fi) Mid value xi h
(fi × ui )
xi −203.5
=
1

200 - 201 13 200.5 -3 -39

201 - 202 27 201.5 -2 -54

202 - 203 18 202.5 -1 -18

203 - 204 10 203.5 = A 0 0

204 - 205 1 204.5 1 1

205 - 206 1 205.5 2 2

Σfi = 70 Σ (fi × ui ) = -108

A = 203.5, h = 1
∑ f ui
Mean = A + {h × i
}
∑ fi

−108
= 203.5 + {1 × }
70

= 203.5 − 1.54 = 201.96

mean weight of packets is 201.96

Section D

7 / 10
32. Let the present age of Roohi be x years.
Then,
3 years ago, Roohi's age = (x - 3) years
5 years from now, Roohi's age = (x + 5) years
It is given that
1 1 1
+ =
x−3 x+5 3
x+5+x−3 1
⇒ =
(x−3)(x+5) 3

2x+2 1
⇒ =
2 3
x +2x−15

⇒ 6x + 6 = x2 + 2x - 15
⇒ x2 - 4x - 21 = 0
⇒ x2 - 7x + 3x - 21 = 0
⇒ x(x - 7) + 3(x - 7) = 0

⇒ x - 7 = 0 or x + 3 = 0

Since age cannot be negative, x ≠ -3

⇒ x=7
Hence, Roohi's present age is 7 years.
OR
Let number of books the shopkeeper buys = x
Price of each book = Rs 1200

cost of each book when x +10 books are bought = RS 1200

x+10

According to given question,

1200
−
x
= 20 x+10
1200

1 1
1200( x
−
x+10
) = 20
(
1

x
−
x+10
1
) = 1200
20

(x+10)−x
= 60
1

x(x+10)

2
x +10x
x +10- x = 60

600 = x + 10x 2

x + 10x - 600 = 0
2

Here, it is quadratic equation

x + 30x - 20x - 600 = 0
2

x (x +30) -20(x +30) = 0

(x +30) (x -20) = 0
either
(x +30) =0 or (x -20) = 0
x = -30 or x = 20

x = -30, is not possible because the number of books can't be negative.

so, number of books = x = 20.

33.

BC
In △ABC, AC
= sin 30
∘

2 1
=
AC 2

⇒ AC = 4 m

∴ length of slide meant for children below 6 years = 4m

8 / 10
 QR
In △P QR, PR
= sin 60
∘

4 √3
=
PR 2

8√3
⇒ PR =
3
m or 13.856

3
m or 4.62 m approx.
∴ length of slide meant for older children = 4.62 m

34.

TSA of the article = 2π rh + 2(2π r2)

= 2π (3.5)(10) + 2[2π (3.5)2]
= 70π + 49π
= 119 π
= 119 × 22

= 374 cm 2
OR
Height of conical part = 10.2 - 4.2 = 6 cm
Volume of toy = Volume of conical part + Volume of hemispherical part
1 22 2 22
=( 3
×
7
× (4.2)
2
× 6) +( 3
×
7
3
× (4.2) )

= 266.112
Hence, Volume of toy is 266.112 cm3
−−−−−−−−− −
Slant height of conical part = √(4.2) + (6) ≈ 7.32 cm 2 2

TSA of the toy = CSA of hemispherical part + CSA of conical part

= (2 × 22

7
× (4.2) )
2
+( 22

7
× 4.2 × 7.32)

= 207.504
Hence, TSA of toy is 207.504 cm2
x−52⋅5

35. Marks x f u =
5
fu cf

40 - 45 42.5 8 -2 -16 8

45 - 50 47.5 9 -1 -9 17

50 - 55 52.5 10 0 0 27

55 - 60 57.5 9 1 9 36

60 - 65 62.5 5 2 10 41

65 - 70 67.5 4 3 12 45

45 6
6
Mean = 52.5 + 5 × 45
= 53.2 (approx)
5
Median = 50 + 10
(22.5 - 17)
= 52.75
Section E
36. i. Since, each poor child pays ₹ x
and each rich child pays ₹ y
∴ In batch I, 20 poor and 5 rich children pays ₹ 9000 can be represented as 20x + 5y = 9000

and in batch II, 5 poor and 25 rich children pays ₹ 26,000 can be represented as 5x + 25y = 26,000
ii. As we have 20x + 5y = 9,000 ...(i)
and 5x + 25y = 26,000
or x + 5y = 5,200 ...(ii)
On subtracting (ii) from (i), we get
19x = 3,800

9 / 10
 ⇒ x = 200
∴ Monthly fee paid by a poor child = ₹ 200
iii. As we have,
20x + 5y = 9000 ...(i)
and 5x + 25y = 26000
x + 5y = 5200 ...(ii)
On subtracting equation (ii) from (i), we have
19x = 3800
3800
x= 19

= 200
Put the value of x in equation (ii), we get
200 + 5y = 5200
5y = 5200 - 200
y = 1000
∴ y - x = 1000 - 200

= 800
Hence, difference in the monthly fee paid by a poor child and a rich child is ₹ 800.
OR
Total monthly fee = 10x + 20y
= 10(200) + 20(1,000)
= 2,000 + 20,000
= ₹ 22,000
37. i. Since ∠ D = ∠ C and ∠ B = ∠ A (Alternate interior angles)
∴ △OAC ∼ △OBD (By AA similarity)
ii. △OAC ∼ △OBD ⇒ = or OA

OB
AC

BD
OA

AC
=
OB

BD

OA OC
iii. a. △OAC ∼ △OBD ⇒ OB
=
OD
3x+4 3x+19
⇒
x
=
x+3
⇒ x=2
∴OC = 25
OR
OB OD BD
b. △OBD ∼ △OAC ⇒ OA
=
OC
=
AC
x+3
⇒
3x+4
x
=
3x+19
⇒ x=2
∴
BD

AC
= 2

10
or 1

38. i. The coordinates of point A are (9, 27), therefore its distance from x-axis = 27 units.
ii. Coordinates of B and C are (4, 19) and (14, 19)
−−−−−−−−−−−−−−−−− −
∴ Required distance = √(14 − 4) 2
+ (19 − 19)
2

−−
−
= √10 = 10 units
2

iii. Coordinates of F and G are (2, 6) and (16, 6) respectively.

−−−−−−−−−−−−−−− −
∴ Required distance = √(16 − 2) 2
+ (6 − 6)
2

−−
−
= √14 = 14 units
2

OR
Coordinates of L and N are (6, 4) and (7, 1) respectively.
−−−−−−−−−−−−−− −
Length of LN = √(7 − 6) + (1 − 4) 2 2

−−− − −−
= √1 + 9 = √10 units
−−
⇒ Length of MP = √10 units
Now, perimeter of LMPN = LN + LM + MP + NP
−− −− −−
= √10 + 6 + √10 + 4 = (2√10 + 10) units [∵ LM = 12 - 6 = 6 units and NP = 11 - 7 = 4 units]

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