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Solution
PRE BOARD EXAM 1
Class 10 - Mathematics
Section A
1.
(d) 6
Explanation:
4 2
2800 = 2 × 5 × 7

On comparing
x = 4, y = 2
x+y=4+2=6
2.
(d) 2
Explanation:
Since 7 + 3 = 10
The least prime factor of a + b has to be 2; unless a + b is a prime number greater than
2.
Suppose a + b is a prime number greater than 2. Then a + b must be an odd number
and one of 'a' or 'b' must be an even number.
Suppose that 'a' is even. Then the least prime factor of a is 2; which is not 3 or 7. So 'a'
can not be an even number nor can b be an even number. Hence a + b can not be a
prime number greater than 2 if the least prime factor of a is 3 and b is 7.
Thus the answer is 2.
3.
(d) 1
Explanation:
Let α and β be the roots of quadratic equation 2x2 + kx + 4 = 0 in such a way that α = 2
Here, a = 2, b = k and c = 4
Then, according to question sum of the roots
−b
α + β =
a

2+β= −k

β = −k

2
-2
β = −k−4

And the product of the roots

c
α ⋅ β =
a

= 4

=2
Putting the value of β = −k−4

2
in above
=2
−k−4
2 ×
2

(-k - 4) = 2
k = -4 - 2
= -6
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Putting the value of k in β = −k−4

−(6)−4
β =
2

= 6−4

= 2

=1
β

Therefore, value of other root be β = 1

4. (a) ( – 6, 0) and (4, 0)
Explanation:
Here are the two solutions of each of the given equations. 3x + y − 12 = 0,
x 4 3 2
y 0 3 6
x − 3y + 6 = 0

x -6 0 -3
y 0 2 1

The triangle △PQR is formed by the given two lines and x-axis. Therefore, both lines
intersect the x-axis at (–6, 0) and at (4, 0).
5.
(b) No Real
Explanation:
The given equation is of the form: ax2 + bx + c = 0, where; a = 2, b = – 4 and c = 3.
Therefore, the discriminant (D) is given as D = b2 – 4ac
D = (– 4)2 – (4 × 2 × 3) = 16 – 24 = – 8
here D < 0
So, the given equation has no real roots.
6.
(d) 3:1
Explanation:
The point lies on y-axis
Its abscissa will be zero
Let the point divides the line segment joining the points (-3, -4) and (1, -2) in the ratio
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m:n
mx 2 +nx 1 m×1+n×(−3)
∴ 0 = ⇒ 0 =
m+n m+n

m−3n
⇒ = 0 ⇒ m − 3n = 0
m+n

m 3
⇒ m = 3n ⇒ =
n 1

∴ Ratio = 3:1
7.
(b) 4 : 7
Explanation:
4:7
8. (a) 30o
Explanation:
In triangle PQR, ∠P + ∠Q + ∠R = 180 ∘

∘ ∘ ∘
⇒ ∠P + 75 + 45 = 180

∘
⇒ ∠P = 60

In triangle PQR, if then as it is common.

PQ QM PQ QM PM
= , = =
PR MR PR MR PM

Therefore, ΔPQM ∼ ΔPRM (SSS)

Thus∠ QPM = ∠ RPM
∘
∠P = 60

⇒ ∠ QPM + ∠ RPM = 60o

⇒ 2∠ QPM = 60o
⇒ ∠ QPM = 30
o
9.
(c) 2 cm.
Explanation:
In right angled triangle ABC
AC2 = AB2 + BC2
AC2 =82 + 62 = 64 + 36 = 100
–
AC = √1 00 = 10cm

Here Area of ΔABC = ar (ΔAOB) + ar (ΔBOC) + ar (ΔCOA)

1 1 1 1
⇒ × 6 × 8 = × 8 × x + × 6 × x + × 10 × x
2 2 2 2

1
⇒24 = x [8 + 6 + 10]
2

1
⇒24 = x × 24
2

⇒x =
24×2

24
= 2 cm
10.
(d) cot4A
Explanation:
Given: cosec4A - 2 cosec2A + 1
= (cosec2A - 1)2
= (cot2A)2
= cot4A
–
11. (a) 50√3cm 2

Explanation:

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In a triangle PQR,
∘ PQ
cos 60 =
PR

1 PQ
⇒ =
2 20

⇒ PQ = 10 cm
QR
And sin 60 ∘
=
PR

√3 QR
⇒ =
2 20

–
⇒ QR = 10√3 cm
∴ ar (ΔPQR)
–
= 1

2
× 10√3 × 10

= 50√–
3 cm
2

12.
(b) 60 ∘

Explanation:

Given: 2AB = √–3AC

Let DC be θ
√3
⇒ AB = 2
AC
AB
∴ sin θ =
AC

√3
AC
2
⇒ sin θ =
AC

√3
⇒ sin θ =
2

∘ ∘
⇒ sin θ = sin 60 ⇒ θ = 60

13. (a)
2
πr θ

360

Explanation:
2
πr θ

360

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

360
= (2 ×
22

7
× 21 ×
60

360
) cm = 22cm

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15.
(b) 1

Explanation:
The probability of drawing a diamond-faced card from a pack of 52 playing cards is
easy to determine. Since there are 13 diamond-faced cards in the deck, the probability
becomes 13/52 = 1/4
16. (a) 0.5
Explanation:
Given data = 30, 34, 35, 36, 37, 38, 39, 40
Here n = 8 which is even
∴ Median = 1

2
[
n

2
th + (
n

2
+ 1) th] term = 1

2
(4th + 5th term)
= 1

2
(36 + 37) = 73

2
= 36.5
After removing 35, then n = 7
∴ New median = 7+1

2
th term = 4th term = 37
∴ Increase in median = 37 - 36.5 = 0.5
17.
(b) 4πr 2

Explanation:
Here, height of cylinder would be equal to diameter of sphere i.e. 2r
So, CSA of the cylinder is 2πrh
= 2πr(2r)
= 4π r2
18.
(d) 10
Explanation:
The first 10 natural odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
Sum of first 10 natural odd numbers
∴ Mean = 10

= 1+3+5+7+9+11+13+15+17+19

10

= 100

10

= 10
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.
(c) A is true but R is false.
Explanation:
–
Here reason is not true. √4 = ±2, which is not an irrational number.
Section B
21. The pair of linear equations are given as:
x + 2y – 4 = 0 ...(i)
2x + 4y – 12 = 0 ...(ii)
We express x in terms of y from equation (i), to get

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x = 4 – 2y
Now, we substitute this value of x in equation (ii), to get
2(4 – 2y) + 4y – 12 = 0
i.e., 8 – 12 = 0
i.e., – 4 = 0
Which is a false statement. Therefore, the equations do not have a common solution.
So, the two rails will not cross each other.
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

AC
=
3.2cm

2.8cm
[DC = BC - BD]
AC =
5.6×2.8

3.2
cm = 4.9
OR
Since G is the mid-point of PQ, then,
PG = GQ. According to the question, GH ∥ QR.Therefore by basic theorm of
proportionality,we have,
PG PH
=
GQ HR

1 =
PH

HR
[as PG = GQ]
PH = HR
Hence,H is the mid-point of PR.

23.

Given: PQ is a diameter of a circle with centre O.

The lines AB and CD are the tangents at P and Q respectively.
To Prove: AB ∥ CD
Proof: Since AB is a tangent to the circle at P and OP is the radius through the point of
contact.
∴ ∠ OPA = 90
o ........ (i)
[The tangent at any point of a circle is ⊥ to the radius through the point of contact]
∵ CD is a tangent to the circle at Q and OQ is the radius through the point of contact.

∴ ∠ OQD = 90
o ........ (ii)
[The tangent at any point of a circle is ⊥ to the radius through the point of contact]
From eq. (i) and (ii), ∠ OPA = ∠ OQD
But these form a pair of equal alternate angles also,
∴ AB ∥ CD

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

2 2 2
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2 2 2
⇒ LHS = sin θ + cos θ + cosec θ + sec θ + 4
2
2 2 2 2 2 2
⇒ LHS = 1 + (1 + cot θ ) + (1 + tan θ ) + 4 [∵ cosec θ = 1 + cot θ , sec θ = 1 + tan θ ]
2 2
⇒ LHS = 7 + tan θ + cot θ = RHS

25. Let the radius of the circle be r cm.

Then, circumference of the circle = 2πr cm
According to the question,
2πr = 22
22
⇒ 2 × × r = 22
7

22×7 7
⇒ r = ⇒ r = cm
2×22 2

For a quadrant of a circle,

Area = 1

4
πr
2

2
1 22 7
= × × ( )
4 7 2

1 22 7 7 77 2
= × × × = cm
4 7 2 2 8

OR
i. The area of that part of the field in which the horse can graze if the length of the rope
is 5cm
1 2 1 2 1 2
= πr = × 3.14 × (5) = × 78.5 = 19.625m
4 4 4

ii. The area of that part of the field in which the horse can graze if the length of the rope
is 10 m
1 2 1 2 2
= πr = × 3.14 × (10) = 78.5m
4 4

∴ The increase in the grazing area

= 78.5 - 19.625 = 58.875 m2
Section C
26. We can prove √2
1
irrational by contradiction.
Let us suppose that 1

√2
is rational.
It means we have some co-prime integers a and b (b ≠ 0)
Such that
1

√2
= a

–
⇒ √2 =
b

a
..........(1)
–
R.H.S of (1) is rational but we know that is√2 irrational.
It is not possible which means our supposition is wrong.
Therefore, √2
1
can not be rational.
Hence, it is irrational.
27. We know that, if x = a is a zero of a polynomial then x - a is a factor of quadratic
polynomials.
Since and 1 are zeros of polynomial.
−1

Therefore (x + 1

4
) (x - 1)
2 1 1
= x + x − x −
4 4

2 1 4 1
= x + x − x −
4 4 4

2 1−4 1
= x + x −
4 4

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2 3 1
= x − x −
4 4

Hence, the family of quadratic polynomials is f(x) = k (x 2

−
3

4
x −
1

4
) , where k is any non-
zero real number.
28. Let the fraction be x

Then, according to the question,

x+1

y−1
= 1 ......(1)
x

y+1
=
1

2
.........(2)
⇒ x + 1 = y - 1 ...........(3)
2x = y + 1................(4)
⇒ x - y = - 2................(5)

2x - y = 1......................(^)
Substituting equation (5) from equation (6), we get x =3
Substituting this value of x in equation (5), we get
3 - y = -2
⇒ y = 3 + 2

⇒ y = 5

Hence, the required fraction is 3

Verification: Substituting the value of x = 3 and y = 5,

we find that both the equations(1) and ( 2) are satisfied as shown below:
x+1 3+1 4
= = = 1
y−1 5−1 4

x 3 3 1
= = =
y+1 5+1 6 2

Hence, the solution is correct.

OR
Let the fraction be x

According to question,
x+2 1
=
y 2

or 2x - y = -4 ....(i)
and x

y−1
=
1

or 3x - y = -1 ...(ii)
on solving eq (i) and (ii), we get,
x = 3, y = 10
∴ fraction is 10
3

29. According to the question,

O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm and OT
intersects circle at E.

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∵ OP ⊥ TP [Radius from point of contact of the tangent]

∴ ∠ OPT = 90°

In right △OPT *
OT2 = OP2 + PT2
⇒ (13)
2 = (5)2 + PT2 ⇒ PT = 12 cm
Let AP = x cm AE = AP ⇒ AE = x cm
and AT = (12 - x)cm
TE = OT - OE = 13 - 5 = 8 cm
∵ OE ⊥ AB [Radius from the point of contact]

∴ ∠ AEO = 90° ⇒ ∠ AET = 90°

In right △AET,
2 2 2
AT = AE + ET

2 2 2
(12 − x) = x + 8

2 2
⇒ 144 + x − 24x = x + 64

⇒ 24x = 80 ⇒ x =
80

24
=
10

3
cm
Also BE = AE = 10

3
cm
⇒ AB =
10

3
+
10

3
=
20

3
cm

30.

In △OP Q , by Pythagoras theorem

OQ2 = OP2 + PQ2
2 2 2
⇒ (P Q + 1) = OP + PQ [∵ OQ − P Q = 1 ⇒ OQ = 1 + P Q]

2 2 2
⇒ PQ + 2P Q + 1 = 7 + PQ

⇒ 2P Q + 1 = 49

⇒ 2P Q = 48

⇒ P Q = 24 cm

∴ OQ − P Q = 1cm ⇒ OQ - 24 = 1 ⇒ OQ = 25 cm
Now, sin Q = OP

OQ
=
7

25

and, cos Q =
PQ 24
=
OQ 25

OR
We have, p = sinθ + cosθ and q = secθ + cosecθ
∴ LHS = q(p
2 -1 ) = (secθ + cosecθ ) {(sinθ + cosθ )2 - 1 }

=( 1

cos θ
+
1

sin θ
) {sin
2
θ + cos
2
θ + 2 sin θ cos θ − 1}

=( sin θ+cos θ

cos θ sin θ
) (1 + 2 sin θ cos θ − 1)

=( sin θ+cos θ

cos θ sin θ
) (2 sin θ cos θ) = 2(sin θ + cos θ) = 2p = RHS
31. When king, queen and jack of clubs are removed, number of cards remaining = 52 - 3 =
49

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Total no. of outcomes = 49

i. Let H be the event of getting a heart card.
Thus, favorable outcomes = 13
P(H) = Favorable outcomes

Total no. of outcomes

=
13

49

ii. Let Q be the event of getting a queen card.

Thus, favorable outcomes = 3 (1 queen of clubs is removed)
P(Q) = Favorable outcomes

Total no. of outcomes

=
3

49

iii. Let C be the event of getting a clubs card.

Thus out of 49 cards, there are 10 clubs cards, because king, queen and jack of clubs
are removed
Hence, favorable outcomes = 10
P(C) = Favorable outcomes

Total no. of outcomes

=
10

49

Section D
32. Let the original duration of the tour be x days.
Total expenses of the tour = ₹ 4200
So, expense of one day = ₹ 4200

If tour extends for 3 more days, total duration = (x + 3) days

Hence, expense of one day = 4200

x+3

Now, according to the question ;

4200 4200
− = 70
x (x+3)

(x+3)−x
1 1 70
⇒ 4200 × [ − ] = 70 ⇒ =
x (x+3) x(x+3) 4200

⇒ x(x +3) = 180 ⇒ x2 + 3x - 180 = 0

⇒ x
2 + 15x - 12x - 180 = 0 ⇒ x(x + 15) - 12(x + 15) = 0
⇒ (x + 15) ( x - 12) = 0 ⇒ x + 15 = 0 or x - 12 = 0

⇒ x = -15 or x = 12

⇒ x = 12 [∵ number of days cannot be negative.]

∴ original duration of the tour is 12 days.

OR
1 1 1 1
− = +
a+b+x x a b

x−a−b−x b+a
=
x(a+b+x) ab

-ab = x2 + (a + b)x
x2 + (a + b)x + ab = 0
(x + a) (x + b) = 0
x = -a, x = -b
33. Given: ABC is a triangle in which DE || BC.
To prove: AD

BD
=
AE

CE

Construction: Draw DN ⊥ AE and EM ⊥ AD ., Join BE and CD.

Proof :

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In △ADE,
Area of ΔADE = 1

2
× AE × DN ...(i)
In ΔDEC ,
Area of Δ DC E = 1

2
× C E × DN ...(ii)
Dividing equation (i) by equation (ii),
1
area (ΔADE) 2
×AE×DN
⇒ =
1
area (ΔDEC ) 2
×C E×DN

area (ΔADE)
⇒ =
AE
...(iii)
area (ΔDEC ) CE

Similarly, In ΔADE,
Area of Δ ADE = 1

2
× AD × EM ...(iv)
In ΔDEB ,
Area of Δ DEB = 1

2
× EM × BD ...(v)
Dividing equation (iv) by equation (v),
1
area (ΔADE) 2
×AD×EM
⇒ =
area (ΔDEB)
1
×BD×EM
2

area (ΔADE)
⇒ =
AD

BD
...(vi)
area (ΔDEB)

ΔDEB and ΔDEC lie on the same base DE and between two parallel lines DE and BC.
∴ Area (ΔDEB ) = Area (ΔDEC )

From equation (iii),

area (ΔADE)
⇒ =
AE
. ......(vii)
area (ΔDEB) CE

From equation (vi) and equation (vii),

AE AD
=
CE BD

If a line is drawn parallel to one side of a triangle to intersect the other two sides in
∴

two points, then the other two sides are divided in the same ratio.
34. Radius of hemisphere = radius of cylinder = 2 mm
Length of cylindrical part = 14 - 4 = 10 mm.
Surface area of the capsule = CSA of cylinder + 2(CSA of hemisphere)
22 22
= 2 × × 2 × 10 + 2 × 2 × × 2 × 2
7 7

= 176 mm2
Volume of the capsule = volume of cylinder + 2(volume of hemisphere)
22 2 22
= × 2 × 2 × 10 + 2 × × × 2 × 2 × 2
7 3 7

=
3344

21
mm
3
or 159.24 mm 3

OR

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From the given figure,

Height (AB) of the cone = AC - BC (Radius of the hemisphere)
Thus, height of the cone = Total height - Radius of the hemisphere
= 9.5 - 3.5
= 6 cm
Volume of the solid = Volume of the cone + Volume of the hemisphere
1 2 2 3
= ( πr h) + ( πr )
3 3

1 2
= πr (h + 2r)
3

1 22
= × × 3.5 × 3.5(6 + 2 × 3.5)
3 7

1 22
= × × 3.5 × 3.5 × 13
3 7

= 166.83 cm3
Thus, total volume of the solid is 166.83 cm3.
35. We may observe from the given data that maximum class frequency is 40 belonging to
1500 - 2000 interval.
Class size (h) = 500
Mode = l + h
f −f
1
×
2f −f −f
1 2

Lower limit (l)of modal class = 1500

Frequency (f) of modal class = 40
Frequency (f1) of class preceding modal class = 24
Frequency (f2) of class succeeding modal class = 33
mode = 1500 + 40−24

2×40−24−33
× 500
= 1500 + 16

80−57
× 500
= 1500 + 347.826
= 1847.826 ≈ 1847.83
Expenditure (in ₹.) Number of families fi xi di = xi - 2750 ui ui f i
1000-1500 24 1250 -1500 -3 -72
1500-2000 40 1750 -1000 -2 -80
2000-2500 33 2250 -500 -1 -33
2500-3000 28 2750=a 0 0 0
3000-3500 30 3250 500 1 30
3500-4000 22 3750 1000 2 44
4000-4500 16 4250 1500 3 48
4500-5000 7 4750 2000 4 28

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Σfi = 200 Σfi di = - 35

Mean x̄ = a + h
Σf i di
¯¯
×
Σf
i

= 2750 + 500
−35
¯¯
x̄ ×
200

¯¯
¯
x = 2750 - 87.5
x = 2662.5
¯¯
¯

Section E
36. i. Let there be 'n' number of rows
Given 3, 5, 7... are in AP
First term a = 3 and common difference d = 2
n
Sn = [2a + (n − 1)d]
2

⇒ 360 = [2 × 3 + (n − 1) × 2]
n

⇒ 360 = n[3 + (n - 1) × 1]
2 + 2n - 360 = 0
⇒ n

⇒ (n + 20) (n - 18) = 0

⇒ n = -20 reject

n = 18 accept
ii. Since there are 18 rows number of candies placed in last row (18th row) is
an = a + (n - 1)d
⇒ a18 = 3 + (18 - 1)2

⇒ a18 = 3 + 17 × 2

⇒ a18 = 37

iii. If there are 15 rows with same arrangement

Sn = [2a + (n − 1)d]
n

⇒ S15 = 15

2
[2 × 3 + (15 − 1) × 2]

S15 = 15[3 + 14 × 1]
⇒

⇒ S15 = 255

There are 255 candies in 15 rows.

OR
The number of candies in 12th row.
an = a + (n - 1)d
⇒ a12 = 3 + (12 - 1)2

⇒ a12 = 3 + 11 × 2

⇒ a12 = 25

37. i. Q(x, y) is mid-point of B(-2, 4) and C(6, 4)

∴ (x, y) = ( −2+6

2
,
4+4

2
) = (
4

2
,
8

2
) = (2, 4)
ii. Since PQRS is a rhombus, therefore, PQ = QR = RS = PS.
−−−−−−−2
− −−−−−−−− −−−−− −−
∴ PQ = √(−2 − 2) + (1 − 4) = √16 + 9 = √25 = 5 units
2

Thus, length of each side of PQRS is 5 units.

iii. Length of route PQRS = 4 PQ
= 4 × 5 = 20 units
OR

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Length of CD = 4 + 2 = 6 units and length of AD = 6 + 2 = 8 units

∴ Length of route ABCD - 2(6 + 8) = 28 units

38. i. Let AB and CD be the 2 poles and M be a point somewhere between their bases in
the same line.

ii. tan 60o =

h
⇒ h = x√–3
x

o h
tan 30 = h=
(28−x)
⇒
28− x √3

–
∴ h= m 7√3

iii. BM = p and DM = q
o
sin 60 = ⇒ h = h p√3

p 2

o
sin 30 = ⇒ h = h q

q 2

= q = √–3 p
p√3 q
∴ ⇒
2 2

OR
o
tan 60 = x = 7m = AM
7√3
⇒
x

MC = 28 - x = 21m

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