Solutions to RSPL/2 (DS2)

1. Given that HCF of (306, 657) = 9

HCF (306, 657) × LCM (306, 657) = Ist number × IInd number

Product of the two numbers

or, LCM (306, 657) =
HCF (306, 657)
306 × 657
= = 22,338
9
2. Given: DE || BC, BC = 8 cm, DE = 6 cm, ar (DADE) = 36 cm­2
To find: ar DABC
A
Sol. As
DE || BC
\
∠ADE = ∠ABC (Corresponding angles)
D E
∠AED = ∠ACB (Corresponding angles) 6 cm

\
DADE ~ DABC (By AA similarity)
2
ar (∆ADE) DE B 8 cm C
\
=
ar( ∆ABC) BC2
36 (6 ) 2
⇒
=
ar (∆ABC) (8) 2
\
ar (DABC) = 64 cm2

3. We have 4x2 – 3px + 9 = 0

Comparing this equation with ax2 + bx + c = 0, we have
a = 4, b = – 3p and c = 9
For real and distinct roots D > 0
\
b2 – 4ac > 0
⇒
(–3p)2 – 4(4))(9) > 0
⇒
9p2 – 144 > 0
9p2 > 144  ⇒  p2 > 16  ⇒  – 4 < p < 4
4. The given equation is

2 sin 2q = 3
3
\
sin 2q =
2
⇒
sin 2q = sin 60°
So 2q = 60°
⇒
q = 30°

Mathematics—10

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 OR
Given A = 0°
So, sin 2A = sin 2(0°)
= sin 0° = 0
2 sin A = 2 sin 0°
= 2 × 0 = 0
Hence
sin 2A = 2 sin A
Yes, it is true.
5. Let C be the mid-point of AB. A(6, b – 2)
B(–2, 4) C(2, – 3)
6–2
\
2 =
2
4
2 = = 2 (true)
2
b–2+4
Also –3 =
2
b+2
⇒
–3 =
2
⇒
–6 = b + 2
⇒
b = – 8
6. Here a = 4, d = 7 – 4 = 3
Let nth term be 50.
So 50 = 4 + (n – 1)3
⇒
46 = 3n – 3
⇒
49 = 3n
49
⇒
n = (not possible)
3
50 is not the term of this AP because n is not natural number.
OR
Number divisible by 8 less than 100 are 8, 16, 24 ... 96.
Here a = 8, d = 8
an = a + (n – 1)d we get
88
96 = 8 + (n – 1) × 8 ⇒ = n – 1 ⇒ n = 12
8
Number of terms = 12
n (a + l) 12 (8 + 96)
So, sum of all terms, Sn = = = 624
2 2

Mathematics—10

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 357
7. The denominator of the rational number is 5000.
3 4
5000
Product of prime factors of 5000 = 2 × 5
357 357
\ =
5000 2 × 5 4
3

For writing the decimal expansion, without actual division, we multiply the numerator
and denominator by 2.
357 2 714
Hence × = = 0.0714
3
2 ×5 2 10000
4

OR
22
Suppose that p – is a rational number.
7
22 p
Hence p – can be written in the form , where p and q are positive co-primes and
7 q
q≠0

22 p
So π– =
7 q
p 22
⇒
p = +
q 7
⇒ p is rational, this contradict the fact that p is irrational

22
\ Our supposition is wrong. Hence, p –
is not rational.
7
8. The given quadratic polynomial is
f(x) = 6x2 – 3
f(x) = 3(2x2 – 1)
f(x) = 3 #^ 2 xh2 – (1) 2 -
f(x) = 3 ^ 2 x + 1h^ 2 x – 1h
The zeroes of f(x) are given by f(x) = 0
⇒ 3 ( 2 x + 1) ( 2 x – 1) = 0

⇒
2 x + 1 = 0 or 2x –1 = 0
1 1
⇒
x = – or x =
2 2
–1 1 coefficient of x 0 –b
Sum of zeroes = + =0= = =0=
2 2 coefficient of x 2 6 a
1 1 1 constant term –3 –1 c
Product of zeroes = – × =– = = = =
2 2 2 coefficient of x 2 6 2 a
9. The given quadratic equation is
2x2 + kx – 6 = 0
One of the two roots is 2.

Mathematics—10

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 Substituting x = 2 in the given equation, we get
2(2)2 + k.2 – 6 = 0
⇒
8 + 2k – 6 = 0
⇒
2 + 2k = 0
\
k = – 1
OR
n (n – 3)
As per question = 54 ⇒ n2 – 3n – 108 = 0
2
⇒ n2 – 12n + 9x – 108 = 0

⇒
n(n – 12) + 9(n – 12) = 0
⇒
(n – 12)(n + 9) = 0
⇒
n – 12 = 0 or n + 9 = 0
⇒
n = 12 or n = – 19 (not possible)
So, number of sides are 12.
10. Number of days in non leap year are 365 and 365 = 52 complete weeks and 1 day. This
one day can be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday.
Clearly there are 7 events associated with this random experiment.
1
So the probability that a non leap year has 53 Mondays = .
7
11. The given points are A(a + b, a – b) and B(a – b, – a – b)
The distance between any two points is given by the formula,

Distance = (x2 – x1) 2 + (y2 – y1) 2

\
AB = {(a – b) – (a + b)} 2 + {(– a – b) – (a – b)} 2

AB = (a – b – a – b) 2 + (– a – b – a + b) 2

= (–2b) 2 + (–2a) 2

= 4b2 + 4a2 = 2 a2 + b2 units

12. Number of playing cards = 52
Black cards = 13 + 13 = 26
Club cards = 13
Total possible outcomes = 26 + 13 = 39
39 3
\ Probability the drawn card is black or a club =
=
52 4
13. Let 2 and (– 3) are zeroes of the quadratic polynomial p(x) = x2 + (a + 1)x + b
\
x = 2 is a zero of p(x).

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 \
p(2) = 0
2
So (2) + (a + 1)2 + b = 0
⇒
4 + 2a + 2 + b = 0
⇒
2a + b + 6 = 0 ...(i)
Also (– 3) is zero of p(x).
\
p(–3) = 0
⇒
(– 3)2 + (a + 1)(–3) + b = 0
⇒
9 – 3a – 3 + b = 0
⇒
– 3a + b + 6 = 0 ...(ii)
On solving (i) and (ii), we get

a = 0 and b = – 6
 14. xi fi fixi

3 6 18
5 8 40
7 15 105
9 p 9p
11 8 88
13 4 52
Total Sfi = 41 + p Sfixi = 303 + 9p
Sfi xi 303 + 9p
Mean = ⇒ 7.5 =
Sfi 41 + p
⇒
7.5(41 + p) = 303 + 9p
⇒
307.5 + 7.5p = 303 + 9p
⇒
307.5 – 303 = 9p – 7.5p
⇒
4.5 = 1.5p
⇒
p = 3
OR
The given data is
Class interval Frequency (fi)

0–6 7
6 – 12 10
12 – 18 12
18 – 24 5
24 – 30 6

Maximum frequency = 12

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 \  Modal class = 12 – 18

f1 – f0
Mode = l + e o× h
2f1 – f0 – f2
Here, f1 = frequency of modal class = 12
f0 = frequency of the class preceding the modal class = 10
f2 = frequency of the class succeeding the modal class = 5
h = size of the class interval = 6
l = lower limit of the modal class = 12
12 – 10
\
Mode = 12 + d n×6
2 × 12 – 10 – 5
2
= 12 + d n × 6
9
4
= 12 +
3
= 13.33
15. Area of the plot = Area of rectangle ABCD + Area of semicircle on side BC as diameter
πr2
=l×b+
2
1 22 × 14 × 14 28
= 60 × 28 + d n =<r = m 14 mF
2 7 2
1 22 × 14 × 14
= 1680 + ×
2 7
= 1680 + 22 × 14 = 1680 + 308
= 1988 m2
OR

Radius (OC) of the inner circle, r = 14 cm
and radius OA of the outer circle, R = 21 cm
Central angle = 90°
\  Area of the shaded region ABDC = Area of sector OAB – Area of sector OCD

θ θ
⇒
Required area = pR2 – pr2
360° 360°
π R2 90° πr2 90° π 2
= – = (R – r2)
360° 360° 4
22
= [(21)2 – (14)2]
7×4
22
= × (21 + 14)(21 – 14)
7×4
22
= × 35 × 7 = 192.5 cm2
7×4

Mathematics—10

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 270°
Area of major sector ODEC = πr2
360°
3 22
= × × 14 × 14 = 462 cm2
4 7
Total area = (192.5 + 462) cm2 = 654.5 cm2
16. Let ‘n’ be the number of small balls made.
⇒ Volume of all small balls = Volume of the big ball.

4 3 4 3
\
pr = pR
n×
3 3
R
According to the question, r =
4
4 R 3 4 3
\
n × π c m = pR
3 4 3
3
R
⇒
n × c m = R3
4
4 3
⇒
n = R3 × d n
R
⇒
n = 43
⇒
n = 64
Hence, 64 balls can be made.
17. Given points are A(4, 1), B(–5, 3) and (2, – 4).
There is one more point P(x, y).
We have to find the ratio of areas of DPBC and DABC.
1
|x (y – y3) + x2(y3 – y1) + x3(y1 – y2)|
We know that area of a triangle =
2 1 2
1
\ Area DPBC = |x(3 + 4) + (– 5)(– 4 – y) + 2(y – 3)|
2
1
= |7x + 20 + 5y + 2y – 6|
2
1
= |7x + 7y + 14|
2
7
= |x + y + 2| ...(i)
2
1
Also Area DABC = |4(3 + 4) + (– 5)(– 4 – 1) + 2(1 – 3)|
2
1
= |28 + 25 – 4|
2
1 49
= |49| = ...(ii)
2 2
7
|x+y+2| x+y+2
Area ∆ PBC
\ = 2 = Hence proved
Area ∆ ABC 49 7
2

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 OR
Coordinates of the vertices of the D are (t, t – 2), (t + 2, t + 2), (t + 3, t)

1
Area of triangle = |t(t + 2 – t) + (t + 2)(t – t + 2) + (t + 3)(t – 2 – t – 2)|
2
1
= |2t + 2(t + 2) + (t + 3)(– 4)|
2
1
= |2t + 2t + 4 – 4t – 12|
2
1 1
= |4 – 12| = |– 8| = 4 sq units
2 2
Hence, area of triangle formed by the given points is independent of t.
18. Given: Two non intersecting circles whose centres are A and B.
P is on the perpendicular bisector of AB. PQ and PR are tangents.
P

Q R

A B

To prove: PQ = PR

Construction: join OA, BR, AP and PB.

Proof: PA = PB
[Every point on the perpendicular bisector of a line
segment is equidistant from its end points]
Now in Ds PAQ and PBR
PA = PB (Proved above)
AQ = BR (Equal radii given)
∠PQA = ∠PRB (Each 90°)
⇒
DPQA @ DPRB(RHS)
\
PQ = PR (CPCT) Hence proved
19. The given equations are:

x = 3;  y = 5;  2x – y – 4 = 0

x = 3 is a line parallel to the y-axis at a distance of 3 units, on the right of the y-axis.

y = 5 is a line parallel to the x-axis at a distance of 5 units above the x-axis.
2x – y – 4 = 0
or y = 2x – 4
x 0 2 3
y – 4 0 2

Mathematics—10

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 Y
x = 3
y = 5 V U
5

2 R(3, 2)

X′ Q (2, 0)
X
–3 –2 –1 O 1 2 3 4

–1

–2

–3

–4 P(0, –4)
=0
–4

Y′
–y
2x

PRUV is the quadrilateral formed by the three lines and the y-axis.
It is a trapezium; Height of the trapezium = UV = 3 – 0 = 3 units;
For parallel sides UR = 5 – 2 = 3 units
PV = 5 – (–4) = 5 + 4 = 9 units
1
\ Area of trapezium PRUV =
(sum of parallel sides) × height
2
1 1
= × ( 3 + 9) × 3 = × 12 × 3 = 18 sq units
2 2
4 2 2 2 2
20. 4(sin 30° + cos 60°) – (sin 45° – cos 90°) – tan 60°
1 4 1 2 1 2
= 4 )c m + c m 3 – *e o – (0 ) 2 4 – ( 3 )
2
2 2 2
1 1 1
= 4 d + n – c m – 3
16 4 2
1+4 1
= 4 d n– –3
16 2
5 1
= 4 × – –3
16 2
5 1 5 – 2 – 12 9
= – – 3 = =–
4 2 4 4

Mathematics—10

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 OR

sin2 A cos2 A 1
LHS = + –
2 2
cos A sin A sin A cos2 A
2

sin 4 A + cos 4 A – 1
=
sin2 A cos2 A
(sin2 A + cos2 A) 2 – 2 sin2 A cos2 A – 1
=
sin2 A cos2 A
1 – 2 sin2 A cos2 A – 1
=
sin2 A cos2 A
–2 sin2 A cos2 A
=
sin2 A cos2 A
= – 2
21. Given: DABC and PQR are equiangular
P

B L C Q M R

BC AL
To prove:
=
QR PM
Construction: Draw AL ^ BC and PM ^ QR

Proof: DABC ~ DPQR(Q Both Ds have same angles)

AB BC AC
\
= = ...(i)
PQ QR PR
Now in Ds ABL and PQM
∠B = ∠Q (Given)
∠ALB = ∠PMQ (Each 90°)
\
DALB ~ DPMQ (AA similarity)
AL LB AB
⇒
= = ...(ii)
PM MQ PQ
From (i) and (ii), we get
AL BC
=  Hence proved
PM QR

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 22. Let us asume to the contrary that 11 is a rational number
p
\ 11 = , p and q are co-prime integers having no common factor
q
except 1 and q ≠ 0
Squaring both the sides, we get
p2
11 =
q2
p2 = 11q2 ...(i)
2
i.e. p is a multiple of 11
⇒  p will also be a multiple of 11.

Let p = 11m, for some integer m.
Squaring both the sides, we get
p2 = 121m2 ...(ii)
From (i) and (ii), we get
11q2 = 121m2
or q2 = 11m2
i.e. q2 is a multiple of 11
⇒  q will also be a multiple of 11.

Hence it means that p and q have a common factor 11. Which contradicts the given fact
that p and q are co-prime.
Hence, our assumption is wrong. So 11 is an irrational number.
23. Here, a = 2750 (say) and h = 500
Expenditure (in `) Frequency Class marks di = xi – a di fi × u i
ui =
(fi) (xi) h
1000 – 1500 15 1250 –1500 –3 –45
1500 – 2000 40 1750 –1000 –2 –80
2000 – 2500 12 2250 –500 –1 –12
2500 – 3000 13 2750 0 0 0
3000 – 3500 7 3250 500 1 7
3500 – 4000 16 3750 1000 2 32
4000 – 4500 30 4250 1500 3 90
4500 – 5000 33 4750 2000 4 132
Total Sfi = 166 Sfiui = 124

Sfi ui
From formula, mean = a + × h, we get
Sfi
124
Mean = 2750 + × 500 = 2750 + 373.49 = 3123.49
166

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 24. In figure, OC is the radius of the base of the cone.
and OC = r
r2
Also
OB = h2 +
9 A B
Now in right triangle BAO
OB2 = OA2 + AB2 h
⇒
AB2 = OB2 – OA2
2
r2
= e h2 + o – (h) 2 O r C
9
r2
= h2 + – h2
9
r2
AB2 =
9
r
\
AB =
3
1 r 2 r
Volume of the frustum = πh =r2 + d n + r $ G
3 3 3
1 r2 r2
= πh e r2 + + o
3 9 3
1 9r 2 + r 2 + 3r 2
= πh e o
3 9
13πr2 h
= Hence proved.
27
sin θ sin θ
25. = 2 +
cot θ + cosec θ cot θ – cosec θ
sin q
Transposing from right to the left side we get
cot q – cosec q
sin θ sin q
– = 2
cot θ + cosec θ cot q – cosec q
1 1
LHS = sin θ < – F
cot θ + cosec θ cot θ – cosec θ
cot θ – cosec θ – cot θ – cosec θ
= sin θ < F
(cot θ + cosec θ) (cot θ – cosec θ)
–2 cosec q
= sin q = 2 G
cot q – cosec2 q
2
= sin q > sin q H
–
–1
2
= sin q ×
sin q
= 2
= RHS Hence proved

Mathematics—10

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 OR

LHS = l2m2(l2 + m2 + 3)
Putting the values of l and m, we get
LHS = (cosec q – sin q)2(sec q – cos q)2 [(cosec q – sin q)2 + (sec q – cos q)2 + 3]
2 2 2 2
1 – sin2 q 1 – cos2 q 1 – sin2 θ 1 – cos2 θ
=e o e o >e o +e o + 3H
sin q cos q sin θ cos θ
2 2 2 2
cos2 q sin2 q cos2 θ sin2 θ
=e o e o >e o +e o + 3H
sin q cos q sin θ cos θ
4 4
cos 4 q sin 4 q cos θ sin θ
= $ f 2
+
2
+ 3p
sin2 q cos2 q sin θ cos θ
cos6 θ + sin6 θ + 3 sin2 θ cos2 θ
= sin2q cos2q f p
sin2 θ cos2 θ
= cos6q + sin6q + 3sin2qcos2q
= {(cos2q + sin2q)3 – 3cos2qsin2q(cos2q + sin2q)} + 3sin2qcos2q
[Formula a3 + b3 = (a + b)3 – 3ab(a + b)]
= (1)3 – 3cos2q sin2q(1) + 3cos2qsin2q
= 1 = RHS
26. Converse of Pythagoras Theorem.
Statement: In a triangle, if the square of one side is equal to the sum of the squares of

the other two sides, then the angle opposite to the first side is a right angle.
A P

B C Q R
2 2 2
Given: A DABC, in which AC = AB + BC


To prove: ∠B = 90°
Construction: Construct a DPQR such that PQ = AB; QR = BC and ∠Q = 90°

Proof: In right triangle PQR

PR2 = PQ2 + QR2 [By Pythagoras Theorem]
or PR2 = AB2 + BC2 [By construction]
But AB2 + BC2 = AC2 [Given]
\
PR2 = AC2
Hence PR = AC

Mathematics—10

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 Now in DABC and DPQR
AB = PQ (By construction)
BC = QR (By construction)
AC = PR (As proved above)
\
DABC @ DPQR [SSS congruence]
\
∠B = ∠Q [CPCT]
But ∠Q = 90°
So ∠B = 90° Hence proved
27. A

3h m
E

hm
(90 – q) C q
B D
xm

In figure, heights of the two persons are DE = h m and AB = 3h m

Distance BD between their feet = x metres
Also ‘C’ is the mid-point of BD.
In right triangle ABC,
AB
tan(90° – q) =
BC
3h 3h × 2 6h
⇒ cot q = = =
x/2 x x
x
⇒ tan q = ...(i)
6h
Also in right triangle CDE,
DE h
tan q =
=
CD x/2
2h
⇒ tan q = ...(ii)
x
From (i) and (ii), we get
x 2h
=
6h x
⇒
x = 12h2
2

x2 x 3x
So h2 = ⇒h= , so 3h =
12 2 3 2
x 3x
\ Heights are
m and m.
2 3 2

Mathematics—10

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 OR
P
30° 45°

2 km

30° 45°
A x km C y km B

In figure aeroplane ‘P’ is flying at a height of 2 km above the ground, i.e. PC = 2 km

AB is the river whose width is to be found.

Let AC = x km and BC = y km

In right triangle PCA

In right DPCB,

PC PC
tan 30° = tan 45° =
AC BC
1 2 2
⇒
= ⇒ 1 =
3 x y
⇒
x = 2 3 ⇒ y = 2

Hence, width of the river, AB = 2 3 km + 2 km

2 ( 3 + 1) km
=
= 2(1.73 + 1) km
= 2 × 2.73 km = 5.46 km
28. The given quadratic equation is:
2(a2 + b2)x2 + 2(a + b)x + 1 = 0
To find the nature of roots of this quadratic equation we find discriminant.
Here A = 2(a2 + b2), B = 2(a + b), and C = 1
Discriminant, D = B2 – 4AC
⇒
D = {2(a + b)}2 – 4 × 2(a2 + b2).1
= 4(a2 + b2 + 2ab) – 8(a2 + b2)
= 4a2 + 4b2 + 8ab – 8a2 – 8b2
= – 4a2 – 4b2 + 8ab
= – (4a2 + 4b2 – 8ab)
= – (2a – 2b)2
It is given that a ≠ b
\
2a – 2b ≠ 0 ⇒ (2a – 2b)2 > 0 ⇒ – (2a – 2b)2 < 0
Hence, the given equation has no real roots. Hence proved.

Mathematics—10

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 OR

Given equation is (1 + m2)x2 + 2mcx + (c2 – a2) = 0

Here A = (1 + m2), B = 2mc and C = (c2 – a2)

Since the equation has equal roots.

So D = 0
2
\ B – 4AC = 0

⇒ (2mc)2 – 4(1 + m2)(c2 – a2) = 0

⇒ 4m2c2 – 4(c2 – a2 + m2c2 – m2a2) = 0

⇒
4m2c2 – 4c2 + 4a2 – 4m2c2 + 4m2a2 = 0
⇒
– 4c2 + 4a2 + 4m2a2 = 0
⇒ 4a2 + 4m2a2 = 4c2

⇒
a2(1 + m2) = c2

3n 2 5n
29. Given +
Sn =
2 2
Here n is a natural number. n = 1, 2, 3, ...

3 (1) 2 5 (1) 3 5
S1 = + = + =4
2 2 2 2
3 (2) 2 5 (2) 12 10
S2 = + = + = 6 + 5 = 11
2 2 2 2
Hence, the first term is 4.

Also, the sum of first two terms = 11

a25 = S25 – S24

3 (25) 2 5 (25) 3 (24) 2 5 (24)

= = + G–= + G
2 2 2 2
3 (25) 2 5 (25) 3 (24) 2 5 (24)
= + – –
2 2 2 2
3 (25) 2 3 (24) 2 5 (25) 5 (24)
= – + –
2 2 2 2
3 5
= (252 – 242) + (25 – 24)
2 2
3 5
= (49) + (1)
2 2
3 × 49 5 147 5
= + = +
2 2 2 2
152
= = 76
2

Mathematics—10

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 30. Steps of construction:
1. Draw concentric circles of radii 3.5 cm and 5.5 cm.
2. Take a point P on the outer circle.
3. Construct a circle on OP as diameter which intersects the inner circle at A and B.

3.5 cm

P O
5.5 cm

4. Join PA and PB.

5. Measure PA and PB.
On measurement we get PA = PB = 4.2 cm
Calculation: PA is tangent, OA is radius through the point of contact A ⇒ ∠OAP = 90°
\ In right triangle OAP,

OP2 = OA2 + PA2 [By Pythagoras Theorem]
2 2 2
⇒
PA = OP – OA
⇒
PA = (5.5) 2 – (3.5) 2
⇒
PA = (5.5 + 3.5) (5.5 – 3.5) = 9×2 = 18 = 4.24 cm

Mathematics—10

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