Solution

SAMPLEQUESTION PAPER 7

Class 10 - Mathematics
Section A
1.
(b) 22 × 32 × 5
Explanation:
180 = 2 × 2 × 3 × 3 × 5
= 22 × 32 × 5.

ses
2. (a) 2
Explanation:

las
Since, The graph of P(x) intersect the x-asix at two distinct point.
hence, two zeroes are possible.
3. (a) (4, 4)

C
Explanation:
(4, 4)

(a) b2 - 4ac > 0

ion
4.

42
Explanation:
A quadratic equation ax2 + bx + c = 0 has real and distinct roots, if b2 - 4ac > 0.
5.
uit
74
(d) 4
Explanation:
4
01
T

6.
an

(c) 3 units
85

Explanation:
−−−−−−−−−−−−−−−−−−
2
−2
PQ = √( +
11
)
2
+ (5 − 5)
h

3 3
97

−−−−
= √9 + 0
au

= 3 units

7.
Ch

(b) 6 units
Explanation:
Distance of any Point form y-axis means the absolute value of its abscissa
∴ Distance of Point (6, 5) from y-axis is 6 units.

8.
(c) DE is not parallel to BC (DE || BC)
Explanation:
AD 1.6 1
= =
BD 1.8 3

AE
= 1.1
= 1

EC 2.2 2

AD AE
≠
BD EC

1 / 12
By Abhay Chauhan Mobile no. - 9785017442
 It means converse of BPT is not satisfied.
∴ DE is not parallel to BC (DE || BC)

9.
(b) 4 cm
Explanation:
Join OR & OQ
Now PQOR becomes a quadrilateral
∠QPR = 90o (given)
PQ = PR (∵ tangent from external point)

ses
OQ = OR = (radius of same circle)
∠ OQP = ∠ ORP = 90o (∵ tangents and radius are perpendicular)
∴ ∠ QOR = 360o - ∠ OQP - ∠ QPR - ∠ ORP
QOR = 360o - 90 - 90 - 90

las
∠

∠ QOR = 90o
∴ PQOR becomes a square

sides of a square are same

C
∴ PQ = 4 cm ion
10.

42
(d) 50 ∘

Explanation:
If the angle between two radii of a circle is 130 , the angle between tangents at ends of radii is∠ APB = 50 . Because the angle
o ∘
uit
74
between the two tangents drawn from an external point to a circle is supplementary of the angle between the radii of the circle
through the point of contact.
01
T
an
85
h
97

11.
au

(d) 1
Explanation:
1
cos θ =
Ch

√2

∘
cos θ = cos 45
∘
θ = 45
∘
Tan θ = Tan 45

=1

12.
(b) 3
Explanation:
Since, we know
Sec2 θ - tan2 θ = 1
(Secθ + tanθ) (Secθ - tanθ) = 1
(Secθ + tanθ) = 11

Secθ + tanθ = 3

2 / 12
By Abhay Chauhan Mobile no. - 9785017442
13.
–
(c) 30√3
Explanation:

height of tower = AB
∘ AB

ses
tan 60 =
BC
– AB
√3 =
30
–
AB = 30√3 m

las
14.
(d) 360
α
× πR
2

Explanation:

C
α 2
× πR
360

15.
ion
(c) cm

42
60

Explanation:
Given: Length of arc = 20 cm
θ
2π r= 20
uit

⇒
360
∘

∘
×
74
60
⇒
360
∘ × 2π r= 20
⇒
πr

3
= 20
01
nT

π
⇒ r(
3
) = 20
π
⇒ r(
3
) = 20
r= 60
cm
85

⇒
π
ha

16.
97

7
(b) 8
au

Explanation:
All possible outcomes are BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG.
Number of all possible outcomes = 8.
Ch

Let E be the event of having at least one boy.

Then, E contains GGB, GBG, BGG, BBG, BGB, GBB, BBB.
Number of cases favourable to E = 7.
7
Therefore,required probability = P( E) = 8

17.
(c) 1

Explanation:
Given,
Number of prizes = 5
Number of blanks in lottery = 20
Number of tickets in lottery = 5 + 20 = 25
Therefore, the probability of getting a prize,
Number of favorable outcomes
P(E) =
Total number of possible outcomes

3 / 12
By Abhay Chauhan Mobile no. - 9785017442
 = 5

25

= 1

18.
(c) decreases by 2
Explanation:
decreased by 2.

19. (a) Both A and R are true and R is the correct explanation of A.
Explanation:

ses
Both A and R are true and R is the correct explanation of A.
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.

las
Section B
–
21. Let us assume that 11 + 3 be a rational number.
√2
–
⇒ 11 + 3√2 = , where a and b are integers, b ≠ 0
a

C
b

– a−11 b
⇒ √2 = 3b

RHS is a rational number but LHS is irrational.

–
∴ Our assumption was wrong. Hence, 11 + 3√2 is an irrational number.
ion
22.
42
uit
74
01
nT

In △ADE and △ABC ,

∠ADE = ∠ABC [ ∵ given]

∠DAE = ∠BAC [∵ common angle]

85

So, △ADE ∼ △ABC [by AA similarity criterion]

ha

Then , = =
AD

AB
[since, corresponding sides of similar triangles are proportional]
AE

AC
DE

BC
AD DE
⇒ =
97

AE+EB 8.4

⇒
7.6
=
DE
[∵AB = AE + BE]
au

7.2+4.2 8.4

7.6
⇒ × 8.4 = DE
11.4

⇒ 0.66 × 8.4 = DE
Ch

⇒ DE = 5.6cm

23.

Given: AB and CD are two parallel tangents, another tangent BD intersects them at B and D respectively. The intercept BD
subtends ∠ BOD at center O.
To prove: ∠BOD = 90 ∘

Proof: In △BOP and △BOQ,

OP = OQ = radius
OB is common
BP = BQ (Tangents from one point B)

4 / 12
By Abhay Chauhan Mobile no. - 9785017442
 So △BOP ≅△ BOQ (By SSS criteria)
Hence ∠ OBP = ∠ OBQ
So ∠ QBP = 2∠ OBP........(1)
Similarly in △DOP and △DOR,
∠ ODP = ∠ ODR

and ∠ RDP = 2∠ ODP ......(2)

Now, AB ∥ C D and BD is a transversal line.
So ∠ QBP +∠ RDP = 180 (The interior angles formed on the same side of the transversal line)
∘

From eqn (1) and (2),

2 ∠ OBP + 2 ∠ ODP = 180 ∘

So ∠ OBP + ∠ ODP = 90 ∘

ses
Now in △BOD,
∠ BOD + ∠ OBP + ∠ ODP = 180
∘

∠ BOD + 90 = 180
∘

Therefore ∠ BOD = 90 ∘

las
Hence proved.
24. Given, acosθ - b sinθ = c
Squaring on both sides

C
(a cosθ - b sinθ)2 = c2
By Adding (a sinθ + b cosθ)2 on both sides, we get
(a cosθ - b sinθ)2 + (a sinθ + b cosθ)2 =c2 + (a sinθ + b cosθ)2
ion
42
(a2cos2θ + b2sin2θ - 2ab sinθ cosθ) + (a2sin2θ + b2cos2θ + 2ab sinθ cosθ) =c2 + (a sinθ + b cosθ)2
a2(cos2θ + sin2θ) + b2(sin2θ + cos2θ)=c2 + (a sinθ + b cosθ)2 ​
a2 + b2 = c2 + (a sinθ + b cosθ)2
uit
74
⇒ (a sinθ + b cosθ)2 = a2 + b2 - c2
−−− −− − −−−−
⇒ a sinθ + b cosθ = ± √a + b − c
2 2 2

OR
01
nT

L.H.S. = (cosec A − sin A)(sec A − cos A) = ( 1

sin A
− sin A) (
1

cos A
− cos A)

2 2
1− sin A 1− cos A
= ( )( )
85

sin A cos A

2 2
cos A sin A
= = sin A cos A
ha

sin A cos A

R.H.S. = 1

tan A+cot A

= 1
97

sin A c os A
+
au

c os A sin A

cos A sin A
= 2 2
sin A+ cos A

= sin A cos A
Hence, L.H.S. = R.H.S.
Ch

25. Radius (r) of circle = 21 cm

Angle subtended by the given arc = 60o
Length of an arc of a sector of angle θ = θ
∘
× 2πr
360

Area of sector OACB = 60

360
∘
× πr
2

1 22
=
6
×
7
× 21 × 21
= 231 cm2
∘

Length of arc ACB = 60

360
∘
× 2× 22

7
× 21

5 / 12
By Abhay Chauhan Mobile no. - 9785017442
 = 1

6
× 2 × 22 × 3
= 22 cm
OR
Area of circle = 3.14 × 10 × 10 = 314 cm2
cm2 or 78.5 cm2
3.14×10×10×90 157
Area of minor sector = 360
=
2

Area of major sector = 314 - 78.5 = 235.5 cm2

Section C
–
26. Let us preassume that is a rational number.
3√2
– p
In that case, 3√2 can be writtin as , where p and q are co-prime integers and q is not zero.
q

p 3√2
So, =

ses
q 1

p √2
⇒ =
3q 1

Since, p is an integer and 3q is also an integer where 3q is not zero.

p –
So, is a rational number but the equalient number √2 should also be a rational number.
3q

las
–
But this contradicts the fact that √2 is an irrational number.
–
so, this assumption is wrong and 3√2 is an irrational number.
27. ∵ α and β are zeroes of given polynomial
So, x2 + 9x + 20 = 0

C
x2 + 4x + 5x + 20 = 0
x(x + 4) + 5(x + 4) = 0
ion
(x + 5)(x + 4) = 0

42
x = -5 and x = -4
∴ α = -5 and β = -4
Now, α + 1 = -4 and β + 1 = -3
uit
74
So, product of zeroes= (-4) × (-3) = 12
Sum of zeroes = -7
Now polynomial = x2 - (sum of zeroes)x + (product of zeroes)
01
nT

Polynomial = x2 + 7x + 12
28. It is given that the sum of seven cash prizes is equal to ₹ 700.
And, each prize is ₹ 20 less than its preceding term.
85

Let the value of first prize = ₹ a

ha

Let the value of second prize =₹ (a−20)

Let the value of third prize = ₹ (a−40)
97

So, we have a sequence of the form:

au

a, a−20, a−40, ...................

It is an arithmetic progression because the difference between consecutive terms is constant.
First term = a, Common difference = d = (a − 20) − a= −20
Ch

n = 7 (Because there are total of seven prizes)

S7 = ₹ 700 {given}
Applying formula, S n =
n

2
[2a + (n − 1)d] to find sum of n terms of AP, we get
7
S7 = [2a + (7 − 1)(−20)]
2
7
⇒ 700 = [2a − 120]
2

⇒ 200 = 2a− 120

⇒ 320 = 2a
⇒ a =160
Therefore, value of first prize = ₹ 160
Value of second prize = 160 - 20= ₹ 140
Value of third prize = 140 - 20= ₹ 120
Value of fourth prize = 120 - 20 = ₹ 100
Value of fifth prize = 100 - 20 = ₹ 80

6 / 12
By Abhay Chauhan Mobile no. - 9785017442
 Value of sixth prize = 80 - 20 = ₹ 60
Value of seventh prize = 60 - 20 = ₹ 40
OR
Let the first term of the Arithmetic progression be 'a'.
and the common difference be 'd'.
24th term of the Arithmetic progression, t24 = a + (24 - 1)d = a + 23d

10th term of the A.P., t10 = a + (10 - 1)d = a + 9d

72nd term of the A.P., t72 = a + (72 - 1)d = a + 71d

15th term of the A.P., t15 = a + (15 - 1)d = a + 14d

ses
t24 = 2×t10
⇒ a + 23d = 2(a + 9d)
⇒ a + 23d = 2a + 18d

⇒ 23d - 18d = 2a - a

las
⇒ 5d = a

And, t72 = a + 71d (substitute value of a)

= 5d + 71d

C
= 76d
= 20d + 56d
= 4 × 5d + 4 × 14d
ion
= 4(5d + 14d)

42
= 4(a + 14d)
= 4t15
Therefore, t72 = 4t15
uit
74
Hence proved.
29. Given, the radii of two concentric circles are 13 cm and 8 cm.
01
nT
85
ha

We have ∠AEB = 90° [angle in a semicircle].

Also, OD ⊥ BE and OD bisects BE.
97

In right △OBD, we have

au

2
O B = O D + BD
2
[by Pythagoras' theorem]
2

−−−−−−− −−−
⇒ BD = √O B − O D
2 2

−−− −−−−
2 2
= √13 − 8 cm
Ch

−−−
= √105cm
−−−
BE = 2BD = 2√105cm [∵ D is the midpoint of BE]
In right △AEB, we have
AB2 = AE2 + BE2 [by Pythagoras' theorem]
−−−− −−−−−−
2 2
⇒ AE = √AB − BE
−−−−−−−−−−− −
2 −−− 2
= √26 − (2√105) cm

−−−
= √256cm

= 16 cm.
In right △AED, we have
AD2 = AE2 + DE2 [by Pythagoras' theorem]
−−−− −−−−−−
2 2
⇒ AD = √AE + DE
−−−−−−−−−− −
2
−−−
2
= √16 + (√105) cm

= 19 cm.
OR

7 / 12
By Abhay Chauhan Mobile no. - 9785017442
 Constructon: Join OR
It is given that, OP = 13 cm, OQ = OR = 5 cm ( both are radii of same circle )
Apply Pythagoras Theorem to the right-angled triangle OQP, we get
(OP)2 = (PQ)2 + (OQ)2
⇒ (13)2 = (PQ)2 + (5)2
⇒ 169 = 25 + (QP)2

ses
⇒ (QP)2 = 169 - 25
⇒ (QP)2 = 144
–
⇒ QP = √144
⇒ QP = 12 cm

las
Thus, QP = PR = 12 [lengths of tangents drawn from external point are equal]
30. We have
tan A + B = 1 = tan 45

C
tan A - B = = tan 30
1

√3

From the above we get,

A + B = 45o ...(1)
ion
42
A - B = 30o ..(2)
Adding (1) And (2)
2A = 75
uit

⇒ A = 37.5
74
Subtract (1) By (2)
⇒ 2B = 15
01
nT

⇒ B = 7.5

Upper class limit +Lower class limit

31. We know that class mark xi = 2

So xi and fixi can be calculated as follows:

85

Frequency (fi) xi fixi

ha

Class

10 - 30 15 20 300
97
au

30 - 50 18 40 720

50 - 70 25 60 1500

70 - 90 10 80 800
Ch

90 - 110 2 100 200

Total ∑ xi = 70 ∑ fi xi = 3520
Here from above, we get
∑ x = 70
i

∑ f x = 3520
i i

∑ fi xi
Hence x̄ =¯
¯
=
3520

70
= 50.28
∑ xi

Section D
32. Given, 1
+
1
=
2

3
(x−1)(x−2) (x−2)(x−3)

8 / 12
By Abhay Chauhan Mobile no. - 9785017442
 (x − 3 ) + (x − 1)
2
=
(x − 1)(x −2)( x − 3) 3

x− 3+ x −1 2
=
(x − 1)(x −2)( x − 3) 3

2x −4 2
=
(x − 1)(x −2)( x − 3) 3

2(x − 2) 2
=
(x − 1)(x − 2)( x − 3) 3

2 2
=
(x − 1)( x − 3) 3

(x - 1) (x - 3) = 3
x2 - 4x + 3 = 3
x2 -4x = 0

ses
x(x - 4) = 0
x = 0 , x - 4= 0
x = 0, x = 4

las
OR
Let the speed of the train be x km/hr for first 54 km and for next 63 km, speed is (x + 6) km/hr.
According to the question
54 63

C
+ = 3
x x+6

54(x+6)+63x
= 3
x(x+6)

or, 54x + 324 + 63x = 3x(x + 6)

ion
or, 117x + 324 = 3x + 18x 2

42
or, 3x − 99x − 324 = 0
2

or, x − 33x − 108 = 0

2

or, x − 36x + 3x − 108 = 0

2
uit

or, x(x - 36) + 3(x - 36) = 0

74
(x - 36)(x + 3) = 0
x = 36
01
nT

x = - 3 rejected.
(as speed is never negative)
Hence First speed of train = 36 km/h
85

33. Let us suppose that AB be the tower and suppose the angle of elevation of its top at C be 30∘
.
ha

Let us suppose that D be a point at a distance 150m from C such that the angle of elevation of the top to tower at D is 60
∘
.

Suppose h m be the height of the tower and let us suppose that AD = x meter
97
au
Ch

According to figure
In ΔC AB, we have
∘ AB
tan 30 =
AC

⇒
1
=
h

x+150
........(i)
√3

Now, In ΔDAB, we have

–
tan 60
∘
=
AB
⇒ √3 =
h

x
⇒ x =
h
.....(ii)
AD √3

9 / 12
By Abhay Chauhan Mobile no. - 9785017442
 Putting the value of x = h
in equation (i), we get
√3

1 h
=
√3 h
+150
√3

1 √3h
⇒ =
√3 h+150√3

– –
⇒ h + 150√3 = 3h ⇒ 3h − h = 150√3
–
2h = 150√3
150 – –
h = √3 = 75√3
2

h = (75 × 1.732)m

h = 129.9
Hence the height of tower is 129.9 m
34. The volume of the spherical vessel is

ses
calculated by the given formula
4
V = ​​ 3
π × r
3

Now,

las
V= 4

3
×
22

7
× 9 × 9 × 9

V = 3,054.85 cm3
The volume of the cylinder neck is calculated by the given formula.

C
V=π×R ×h 2

Now,
22
V= 7
× 1 × 1 × 8
ion
V = 25.14 cm3

42
The total volume of the vessel is equal to the volume of the spherical shell and the volume of its cylindrical neck.
3054.85 + 25.14 = 3, 080 cm3
The total volume of the vessel is 3,080 cm3.
uit
74
As we know,
1 L = 1000 cm3
01
nT

3080

1000
= 3.080 L
Thus, the amount of water (in litres) it can hold is 3.080 L.
OR
85

Volume of solid = 1

3
×
22

7
× (7)2 × 3.5 + 2

3
×
22

7
× (7)3
ha

= 22

7
× (7)
2
× [
3.5

3
+
2

3
× 7]

or 898.33 cm3
97

= 898 1

3
au
Ch

CI fi xi di ui fiui
35.
0 - 10 5 5 -30 -3 -15

10 - 20 10 15 -20 -2 -20

20 - 30 18 25 -10 -1 -18

30 - 40 30 35 0 0 0

40 - 50 25 45 10 1 20

50 - 60 12 55 20 2 24

60 - 70 5 65 30 3 15

10 / 12
By Abhay Chauhan Mobile no. - 9785017442
 Total 100 6
Σfi ui
mean = A + Σfi
× h

= 35 + 6

100
× 10

= 356

10
or 35.6
Section E
36. i. x + y + 2 = 15
x + y = 13 ...(i)
Area of bedroom + Area of kitchen = 95
5 × x + 5 × x + 5 × y = 95
2x + y = 19 ...(ii)

ses
In △ABD
tan 60o =
120√3

BD
120√3
BD =
√3

las
BD = 120 m

C
ion
42
In △ABC
tan 30o = AB
uit

1
BC

120√3
74
=
√3 BC

BC = 360 m
CD = BC - BD
01
nT

= 360 - 120
= 240 m
ii. Length of outer boundary
85

= 12 + 15 + 12 + 15
ha

= 54 m
97

iii.
au

x=6
Ch

Area of bedroom 1 = 5 × x
= 5 × 6 = 30 m2
OR
Area of living room = (5 × 2) + (9 × 7)
= 10 + 63
= 73 m2
37. i. AA criterion

11 / 12
By Abhay Chauhan Mobile no. - 9785017442
 ii.

△ ABE ∼ △CDE (by AA criteria)

AB BE
=
CD DE
6×1.5
h= 1.8

ses
h=5
i.e., height of pole = 5 m.
iii. tan i = 6

OR

C las
ion
1.5 13−x
=

42
5 x

1.5x = 65 - 5x
6.5x = 65
x= 65
uit

6.5
74
= 10
∴ distance of Suresh from mirror

= 13 - x
01
nT

= 13 - 10
=3m
38. i. The distance between A and C
85

−−−−−−−−−−−−−− − −− −−−−
= √(8 − 4) 2
+ (5 + 3)
2
= √4 2
+ 8
2
ha

−−−−−− −− –
= √16 + 64 = √80 = 4√5 units
ii. Let the coordinates of I be (x, y)
97
au

Then, by section formula,

1×8+2×7 8+14 22
x= =
Ch

=
1+2 3 3

1×5+2×3 5+6 11
and y = 1+2
= 3
=
3

Thus, the coordinates of I is ( 22

3
,
11

3
)

iii. The mid-point of A and C

8+4 5−3
=( 2
,
2
) = (6, 1)
OR
8k+4 5k−3
Let B divides the line segment joining A and C in the ratio k : 1. Then, the coordinates of B will be ( k+1
,
k+1
) .
8k+4 5k−3
Thus, we have ( k+1
,
k+1
) = (7, 3)
8k+4 5k−3
⇒
k+1
= 7 and k+1
=3
8k+4
Consider, k+1
= 7 ⇒ 8k + 4 = 7k + 7 ⇒ k = 3
Hence, the required ratio is 3 : 1.

12 / 12
By Abhay Chauhan Mobile no. - 9785017442