Solution

PRE BOARD ( PAPER - 7 )

Class 10 - Mathematics
Section A
1.
(b) (18, 25)
Explanation:
The numbers that do not share any common factor other than 1 are called co-primes.
factors of 18 are: 1, 2, 3, 6, 9 and 18
factors of 25 are: 1, 5, 25
The two numbers do not share any common factor other than 1.
They are co-primes to each other.

2.
(c) 0
Explanation:
Here y = f(x) is not intersecting or touching the X-axis.
∴ Number of zeroes of f(x) = 0

3.
(d) 0
Explanation:
C
The number of solutions of two linear equations representing parallel lines is 0 because two linear equations representing
parallel lines has no solution and they are inconsistent.
JM

4.
(b) 16
Explanation:
2 is root equation x2 + ax + 12 = 0
∴(2)2 + a × 2 + 12 = 0 ⇒ 4 + 2a + 12 = 0
⇒ 2a + 16 = 0
−16
⇒ a= = −8
2

and given that roots of x2+ ax + q = 0 are equal.

∴ b2 - 4ac = 0
2 2
⇒ a − 4q = 0 ⇒ (−8) − 4q = 0

⇒ 64 − 4q = 0 ⇒ 4q = 64

⇒ q =
64

4
= 16
∴ q = 16

5.
(b) 9th
Explanation:
Let the nth term of 72, 63, 54,... be 0. Then,
Tn = 0 ⇒ 72 + (n -1) × (- 9) = 0 ⇒ 9n = 81 ⇒ n = 9.

6.
–
(d) √2 units
Explanation:

1 / 14
By - Apar Gupta
 A (5, -4) B (4, -5)
−−−−−−−−−−−−−−− −
AB = √(5 − 4) 2
+ (−4 + 5)
2

−−−−−−
= √1 + 1 2 2

–
AB = √2 units

7.
(d) 3 : 1
Explanation:
Let the Pt (0, y) intersect line segment joining (-6, 2) & (2, -6) in ratio m : n
using fiction formula,
2m−6n −6m+2n
(o, y) = [ m+n
,
m+n
]

on comparing obsc.
2m−6n
=0
m+n

2m - 6n = 0
2m = 6n
m

n
= 6

m:n=3:1

8.
(b) 7.5 cm
Explanation:
By BPT
AD AE
=
AB AC
4 AE
C
=
9 13.5
13.5×4
= AE
9

AE = 6 cm
JM

EC = AC - AE
= 13.5 - 6
EC = 7.5 cm

9.
(b) 5 cm
Explanation:
In the given figure, PA and PB are the tangents to the circle with centre O from P

PA = 5 cm, ∠ APB = 60°

PA = PB = 5 cm
In △APB, ∠ P = 60° and PA = PB
PAB is an equilateral triangle
AB = AP = BP = 5 cm

10.
(d) 10 cm
Explanation:

2 / 14
By - Apar Gupta
 We know that if two chords intersect each other at T outside the circle, then TP × TQ = TS × TR Let SR = x cm
⇒ (5 + 3) × 3 = (x + 2) × 2

⇒ x + 2 = 12

⇒x = 10 cm x = 10 cm
∴ SR = 10 cm

11.
(b) m2 + n2
Explanation:
Given,
a cos θ + b sin θ = m
a sin θ - b cos θ = n
Now, Squaring and adding, we have;
a2 cos2 θ + b2 sin2 θ + 2ab sin θ cos θ = m2
a2 sin2 θ + b2 cos2 θ - 2ab sin θ cos θ = n2
a2 (cos2 θ + sin2 θ) + b2 (sin2 θ + cos2 θ) = m2 + n2 { ∵ sin2 θ + cos2 θ= 1}
⇒ a2 + 1 + b2 × 1 = m2 + n2
⇒ a2 + b2 = m2 + n2
C
Hence a2 + b2 = m2 + n2
JM

12.
2
x −1
(d) 2x

Explanation:
Given, secθ + tanθ = x ...(i)
We know that
2 2
sec θ − tan θ= 1

⇒ (sec θ + tan θ)(sec θ − tan θ) = 1

⇒ x(sec θ − tan θ) = 1

⇒ sec θ − tan θ =
1

x
...(ii)
Subtracting (ii) from (i)
2
1 x −1
2 tan θ = x − =
x x
2
x −1
tan θ =
2x

13.
(b) equal
Explanation:
The angle of elevation and the angle of depression from an object on the ground to an object in the air are related as equal if the
height of objects is the same.

14. (a) 7546 cm2

Explanation:
2πR = 616
(616×7)
R= (2×22)

R = 98 cm

3 / 14
By - Apar Gupta
 2
π r
Area of quadrant = 4

(22×98×98)
=
(7×4)

= 7546 cm2
15.
60
(c) cmπ

Explanation:
Given: Length of arc = 20 cm
× 2π r= 20
θ
⇒ ∘
360
∘
60
⇒
360
∘
× 2π r= 20
⇒
πr

3
= 20
⇒ r(
π

3
) = 20
π
⇒ r(
3
) = 20
60
⇒ r= π
cm

16.
(b) 2

Explanation:
Leap year contains 366 days = 364 days + 2 days = 364

7
weeks + 2 additional days = 52 weeks + 2 additional days
52 weeks contain 52 Fridays
We will get 53 Fridays if one of the remaining two additional days is a Friday
These additional days can be :
{(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday,
C
Saturday), (Saturday, Sunday)}
Number of total outcomes = 7
JM

Number of possible outcomes = 2

N umber of possible outcomes 2
∴ Required Probability of the event = N umber of total outcomes
=
7

17.
(b) 1

Explanation:
Number of composite numbers on a dice = {4, 6} = 2
Number of possible outcomes = 2
Number of Total outcomes = 6
∴ Required Probability =
2 1
=
6 3

23
18. (a) 3

Explanation:
We know that
3 Median = Mode + 2 Mean
3 Median = 7 + 2 × 8
= 7 + 16
3 Median = 23
Median = 23

19. (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.
20.
(b) Both A and R are true but R is not the correct explanation of A.
Explanation:

4 / 14
By - Apar Gupta
 Both A and R are true but R is not the correct explanation of A.

Section B
21. Clearly, the maximum number of columns = HCF (612, 48).

Now, 612 =2 × 2 × 3 × 3 × 17 = (2 2
× 3
2
× 17)

and 48 = 2 × 2 × 2 × 2 × 3 = (2 4
× 3) .
∴ HCF (612, 48) = (2 2
× 3) = (4 × 3)

∴ HCF (612, 48)= 12.

∴ Maximum number of columns in which they can march = 12.

22. In△ABC , AB∥DE.

CD CE
∴
DA
=
EB
...(i) [by Thales' theorem]
In △C DB, BD∥EF
∴
CF
=
FD
...(ii) [by Thales' theorem]
CE

EB

From (i) and (ii) we get

CD CF
=
DA FD

⇒
DA

DC
=
FD

CF
[taking reciprocals]
DA FD
⇒ + 1 = + 1
DC CF

DA+DC FD+CF
⇒ =
DC CF
AC DC
⇒ =
C
DC CF

2
⇒ DC = C F × AC

23. Join OA, OC and OB. Clearly, ∠OC A is the angle in a semi-circle
JM

∴ ∠OC A = 90o
In right triangles OCA and OCB, we have
OA = OB = r
∠OC A = ∠OC B = 90o
and, OC = OC
So, by RHS-criterion of congruence, we get
△OCA ≅ △OCB

⇒ AC = CB

24. Given: A triangle ABC in which ∠ B = 90 ∘

Let BC = 3k and AC = 4k
Then, Using Pythagoras theorem,
−−−−−−−−−−−− −−−−−−−−−−−
AB = √(AC ) 2
− (BC )
2
= √(4k) 2
− (3k)
2

−−−− −−−−−
=√16k 2
− 9k
2

–
=k√7

5 / 14
By - Apar Gupta
 k√7 √7
cos A = AB

AC
=
4k
=
4
Sin A BC 3k 3
tan A = Cos A
=
AB
= =
k√7 √7

OR
tan2θ + sin θ = cos2θ
Taking θ = 45o, we have
√2+1
L.H.S. = tan 2
45
∘
+ sin 45
∘
= (1)
2
+
1
= 1 +
1
=
√2 √2 √2

R.H.S. = cos 2
45 = (
1
) =
1

2
√2

⇒ LH.S. ≠ R.H.S.
Hence given expression is not an identity.
Area of sector with θ=150∘
25. Part of the whole circle is the sector with central angle 150 = ∘

Area of the circle

2 150
π×(6) ×
360
=
2
π×(6)

150
=
360
5
=
12

Required ratio= (36π × 90

360
) : (36π ×
120

360
) : (36π ×
150

360
)

1 1 5
= : :
4 3 12

= 3 : 4 : 5

OR
Angle swept by minute hand in 60 minutes = 360o
Angle swept by minute hand in 5 minutes =30o
r = 14 cm
2 ∘

Area swept =
C
πr θ 30 22
∘
= ∘
× × 14 × 14
360 360 7

cm2 or 51.33 cm2

154
= 3
JM

Section C
26. This problem can be solved using H.C.F. because we are cutting or “dividing” the strips of cloth into smaller pieces of 36 and 24
and we are looking for the widest possible strips .
So,
H.C.F. of 36 and 24 is 12
So we can say that
Maya should cut each piece to be 12 inches wide.
–
27. Here, α + β = −2√3 and αβ = −9
f(x) = x2 – (α + β)x + αβ [Formula]
2 –
= x − (−2√3)x + (−9)

2 –
⇒ f (x) = x + 2√3x − 9

For zeroes of polynomial f(x), f(x) = 0

2
–
⇒ x + 2√3x − 9 = 0

2 – –
⇒ x + 3√3x − 1√3x − 9 = 0
– – –
⇒ x(x + 3√3) − √3(x + 3√3) = 0
– –
⇒ (x + 3√3)(x − √3) = 0
– –
⇒ x + 3√3 = 0 or (x − √3) = 0
– –
⇒ x = −3√3 or x = √3
– –
∴ α = −3√3 and β = √3
Hence the polynomial is x2 +2√3 x - 9 and its zeros are −3√3 and √3.
– – –

28. Let the present age of Nuri be x years and the present age of Sonu be y years.
After 10 years, Nuri’s age will be (x + 10) years and the age of Sonu will be (y + 10) years.
Thus using the given information, we have
x + 10 = 2(y + 10)
⇒ x + 10 = 2y + 20

⇒ x - 2y - 10 = 0

6 / 14
By - Apar Gupta
Before 5 years, the age of Nuri was (x – 5) years and the age of Sonu was (y – 5) years.
Thus using the given information, we have
x - 5 = 3(y - 5)
⇒ x - 5 = 3y - 15
⇒ x - 3y + 10 = 0

So, we have two equations

x - 2y - 10 = 0
x - 3y = 10 = 0
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
x −y 1
= =
(−2)×10−(−3)×(−10) 1×10−1×(−10) 1×(−3)−1×(−2)

x −y 1
⇒ = =
−20−30 10+10 −3+2

x −y 1
⇒ = =
−50 20 −1
x y
⇒ = = 1
50 20

⇒ x = 50, y = 20
Hence, the present age of Nuri is 50 years and the present age of Sonu is 20 years.
OR
The given pair of linear equations
2x + y = 4
and 2x-y = 4
Table for line 2x + y = 4
x 0 2

y 4 0
C
Points A B
and table for line 2x - y = 4
JM

x 0 2

y -4 0

Points C B

So the Graphical representation of both lines is as above.

Here, both lines and Y - axis from a △ABC .
Hence, the vertices of a △ABC are A (0,4), B(2,0) and C(0,- 4) where A and C are obtained by putting x = 0 in the given
equations abd B is obtained by solving them together.
∴Required area of △ABC = 2 × Area of △AOB
△ABC = 2 × ( × 4 × 2) = 8 sq. units.
1

Hence, the required area of the triangle is 8 sq units.

7 / 14
By - Apar Gupta
29.

DR = DS = 3 cm
∴ AR = AD - DR = 17 - 3 = 14 cm

⇒ AQ = AR = 14 cm

∴ QB = AB - AQ = 20 - 14 = 6 cm
Since QB = OP = r ∴ radius = 6 cm
OR

We know that tangent segments to a circle from the same external point are congruent
Now, we have
AE = AF,
BD = BE =6 cm and
C
CD = CF = 9 cm
Now,
JM

Area(△ABC )

= Area(△BOC ) + Area(△AOB) + Area(△AOC )

1 1 1
⇒ 54 = × BC × OD + × AB × OE + × AC × OF
2 2 2

⇒ 108 = 15 × 3 + (6 + x) × 3 + (9 + x) × 3

⇒ 36 = 15 + 6 + x + 9 + x
⇒ 36 = 30 + 2x
⇒ 2x = 6

⇒ x = 3cm

∴ AB = 6 + 3 = 9cm and

AC = 9 + 3 = 12cm.
30. Given,
x y
(
a
sin θ −
b
cos θ) = 1 .....(1)
y
and, ( x

a
cos θ + sin θ) = 1 ......(2)
b

Now, equation (1) is

x y
( sin θ − cos θ) = 1
a b

(Squaring both sides, we get)

2 2
x y 2xy

2
sin
2
θ+
2
cos
2
θ−
ab
sin θ cos θ = 1 ..(3)
a b

Equation (2) is
x y
( cos θ + sin θ) = 1
a b

(Squaring both sides, we get)

2 2
y 2xy
x

2
cos
2
θ+
2
sin
2
θ+
ab
sin θ cos θ = 1 ..(4)
a b

Adding (3) and (4), we get :-

8 / 14
By - Apar Gupta
 2 2
x y
2 2 2 2
(sin θ + cos θ) + (sin θ + cos θ) = 2
a2 b
2

2 2
y
⇒
x

a
+
b
= 2 .
Hence, Proved.
31. Take a = 0.14, h = 0.04

xi −0.14
Concentration of di = xi – 0.14 ui = fiui
Frequency (fi) Class Mark (xi) 0.04

SO2 (in ppm)

0.00-0.04 4 0.02 –0.12 –3 –12

0.04-0.08 9 0.06 –0.08 –2 –18
0.08-0.12 9 0.10 0.04 –1 –9
0.12-0.16 2 0.14 0 0 0
0.16-0.20 4 0.18 0.04 1 4
0.20-0.24 2 0.22 0.08 2 4

Total ∑ fi = 30 ∑ fiui = -31

Using the step-deviation method,

∑ f ui −31
¯¯
x
¯
= a +( i
)× h = 0.14 +( 30
)× (0.04)
∑ fi

= 0.14 – 0.041 = 0.0999 ppm.

Therefore, the mean concentration of SO2 in the air is 0.099
Section D
C
32.
JM

Let width of the pond be x m. Then,

The length of pond = (50 − 2x)m and the breadth of pond = (40 − 2x)m
Area of grass around the pond = 1184 m2
⇒ Area of Lawn - Area of Pond = 1184

⇒ 50 × 40 − (50 − 2x)(40 − 2x) = 1184

2
⇒ 2000 − (2000 − 100x − 80x + 4x ) − 1184 = 0

2
⇒ 2000 − (2000 − 180x + 4x ) − 1184 = 0

⇒ 2000 - 2000 + 180x - 4x2 - 1184 = 0

⇒ 4x2 - 180x + 1184 = 0
2
⇒ 4(x − 45x + 296) = 0

⇒ x2 - 45x + 296 = 0
Factorise now,
⇒ x2 - 37x - 8x + 296 = 0
⇒ x(x - 37) - 8(x - 37) = 0

⇒ (x - 37) (x - 8) = 0

⇒ x - 37 = 0 or x - 8 = 0

⇒ x = 37 or x = 8
When x = 37, then
The length of pond = 50 - 2 × 37
= 50 - 74
= -24 m (Length cannot be negative)
When x = 8, then
The length of pond = 50 – 2x

9 / 14
By - Apar Gupta
 = 50 - 2 × 8
= 50 - 16
= 34 m
And the breadth of the pond
= 40 - 2x
= 40 - 2 × 8
= 40 - 16
= 24 m
Therefore, the length and breadth of the pond are 34 m and 24 m respectively.
OR
Let breadth of the plot =,x m
∴ length of the plot = (2x + 1) m

(2x + 1)x = 528

2x2 + x - 528 = 0
(x - 16) (2x + 33) = 0
33
⇒ x = 16, x = − (not possible)
2

∴ length of the plot = 33 m

Breadth of the plot = 16 m
Cost of levelling the plot = 528 × 80 = ₹ 42,240

33.
C
Given: In △ABC and △PQR, AD is the median of △ABC, PM is the median of △PQR and △ABC ∼ △PQR.
JM

To Prove: = AB AD

PM
PQ

Proof:
Since AD is the median
BD = CD = BC 1

Similarly, PM is the median

QM = RM = QR 1

Now,
△ ABC ∼ △PQR. (∵ given)
AB

PQ
= =
BC

QR
(∵ Corresponding sides of similar triangle are proportional)
AC

PR

So,
AB BC

PQ
= QR

AB

PQ
= 2BD

2QM
(Since AD & PM are medians)
AB

PQ
= BD

QM
................(1)
Also, since △ABC ∼ △PQR.
∠ B = ∠ Q (∵ Corresponding angles of similar triangles are equal).............(2)

Now,
In △ABD & △PQM
∠ B = ∠ Q [∵ from (2)]
AB

PQ
= BD

QM
[∵ from (1)]
Hence by SAS similarly,
△ABD ∼ △PQM

Since corresponding sides of similar triangles are proportional,

AB AD

PQ
= PM

Hence proved.

10 / 14
By - Apar Gupta
34. Diameter of the cylinder = 7 cm
Therefore radius of the cylinder = 7

2
cm

Total height of the solid = 19 cm

Therefore, Height of the cylinder portion = 19 - 7 = 12 cm
Also, radius of hemisphere = cm 7

Let V be the volume and S be the surface area of the solid. Then,
V = Volume of the cylinder + Volume of two hemispheres
2 2 3 3
⇒ V = {πr h + 2 ( πr )} cm
3

2 4r 3
⇒ V = πr (h + ) cm
3

2
22 7 4 7 3 22 7 7 50 3 3
⇒ V = { × ( ) × (12 + × )} cm = × × × cm = 641.66cm
7 2 3 2 7 2 2 3

and,
C
S = Curved surface area of cylinder + Surface area of two hemispheres
2 2
⇒ S = (2πrh + 2 × 2πr ) cm
JM

2
⇒ S = 2πr(h + 2r)cm

22 7 7 2
⇒ S = 2 × × × (12 + 2 × ) cm
7 2 2

22 7 2
= (2 × × × 19) cm
7 2

= 418 cm2
OR
We have radius of cylinder = radius of cone = radius of hemisphere = 60 cm
Height of cone = 120 cm
∴ Height of cylindrical vessel = 120 + 60 =180 cm

∴ V = Volume of water that the cylinder contains = πr

2 2 3
h = {π × (60) × 180} cm

Let V1 be the volume of the conical part. Then,

1 2
V1 = πr h1
3
1 2 3 2 3
⇒ V1 = × π × 60 × 120cm = {π × 60 × 40} cm
3

11 / 14
By - Apar Gupta
 For hemispherical part r = Radius = 60 cm
Let V​2 be the volume of the hemisphere. Then,
2 3 3
V2 = { π × 60 } cm
3

2 3 2 3
⇒ V2 = {2π × 20 × 60 } cm = {40π ⋅ 60 } cm

Let V3 the the volume of the water left-out in the cylinder. Then,
V​3= V - V1 - V2
2 2 2 3
V3 = {π × 60 × 180 − π × 60 × 40 − 40π × 60 } cm

2 3
V3 = π × 60 × {180 − 40 − 40}cm
22 3
V3 = × 3600 × 100cm
7
22×360000 3 22×360000 3 22×36 3 3
⇒ V3 = cm = m = m = 1.1314m .
7 3 700
7×(100)

35. Let the missing frequency be f the assumed mean be A = 47 and h = 3.

Calculation of Mean
Class-Intervals mid-values xi fi di = xi - 47.5 ui =
xi −47.5
fiui
3

40-43 41.5 31 -6 -2 -62

43-46 44.5 58 -3 -1 -58

46-49 47.5 60 0 0 0

49-52 50.5 f 3 1 f

52-55 53.5 27 6 2 54

N = ∑ f = 176 + f i Σfi ui = f - 66
We have,
C
¯¯¯
¯
X = 47.2 , A = 47.5 and h = 3
¯¯¯
¯
∴ X =A+h{ 1

N
Σfi ui }
JM

f −66
⇒ 47.2 = 47.5 + 3 × { 176+f
} ​​​​​​​

f −66
- 0.3 = 3 × { 176+f
}

−1 f −66
⇒
10
=
176+f
⇒ - 176 - f = 10 f - 660 ⇒ 11f = 484 ⇒ f = 44
Hence, the missing frequency is 44
Section E
36. i. Let production in a 1st year be 'a' unit and increase in production (every year) be 'd' units.
Increase in production is constant, therefore unit produced every year forms an AP.
Now, a3 = 6000
a + 2d = 6000 ⇒ a = 6000 - 2d ...(i)
and a7 = 7000 ⇒ a + 6d = 7000
⇒ (6000 - 2d) + 6d = 7000 ⇒ 4d = 1000 [using eq. (i)]
⇒ d = 250

When d = 250, eq. (i) becomes

a = 6000 - 2(250) = 5500
∴ Production in 1st year = 5500

ii. Production in fifth year

a5 = a + 4d = 5500 + 4(250) = 6500
iii. Total production in 7 years = 7

2
(5500 + 7000) = 43750
OR
an = 1000 units
an = 1000
⇒ 10000 = a + (n - 1)d
⇒ 1000 = 5500 + 250n - 250
⇒ 10000 - 5500 + 250 = 250n

12 / 14
By - Apar Gupta
 ⇒ 4750 = 250n
⇒ n= = 19
4750

250

37. i. A(1, 9) and B(5, 13)

ii. C(9, 13) and D(13, 9)
Mid-point of CD is (11, 11)
iii. a. M(5, 11) and Q(9, 3)
−−−−−−−−−−−−−−−− −− –
MQ = √(9 − 5) + (3 − 11) = √80 or 4√5
2 2

OR
b. M(5, 11) and N(9, 11)

1×9+3×5 1×11+3×11
Z( , )
1+3 1+3

Z(6, 11)

38. i.

Suppose AB and CD are the two towers of equal height h m. BC be the 80 m wide road. P is any point on the road. Let CP be
x m, therefore BP = (80 – x).
Also, ∠ APB = 60o and ∠ DPC = 30o
In right angled triangle DCP,
C
tan 30o = CD

CP

⇒ h 1
JM

=
x √3

⇒h= x
.......(i)
√3

In right angled triangle ABP,

tan 60o = AB

AP
h –
⇒ = √3
80−x
–
⇒ h = √3(80 − x)
x –
⇒ = √3(80 − x)
√3

⇒ x = 3(80 – x)
⇒ x = 240 – 3x

⇒ x + 3x = 240
⇒ 4x = 240
⇒ x = 60

Thus, the position of the point P is 60 m from C.

ii.

x 60 –
Height of the tower, h = = = 20√3
√3 √3
–
The height of each tower is 20√3 m.

13 / 14
By - Apar Gupta
iii.

The distance between Neeta and top of tower AB.

In △ABP
sin 60o =
AB

AP
AB
⇒ AP =
0
sin 60

20√3
⇒ AP =
√3

⇒ AP = 40 m
OR

The distance between Neeta and top of tower CD.

In △CDP
C
sin 30o = CD

PD
CD
⇒ PD =
JM

0
sin 30

20√3 –
⇒ PD = = 40√3
1

2
–
⇒ P D = 40√3

14 / 14
By - Apar Gupta