Class 10 Maths Full Syllabus Practice Paper 01

Q1. This question paper contains 38 questions.

Answer: (d) an irrational number
Explanation:
If possible let a√b be rational.
p
Then a√b = q , where p and q are non-zero integers, having no common factor other than 1.
p
Now, a√b =
q p
⇒ √b = ... (i)
aq
But, p and aq are both rational and aq ≠ 0
p ∵ aq is rational.
Therefore, from eq. (i), it follows that √b is rational.
The contradiction arises by assuming that a√b is rational.
Hence, a√b is irrational.

Q2. This Question Paper is divided into 5 Sections A, B, C, D and E.

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

Q3. In Section A, Questions no. 1-18 are multiple choice questions (MCQs) and questions no. 19 and 20
are Assertion- Reason based questions of 1 mark each.

Answer: (b) one or many solutions
Explanation: A system of linear equations is said to be consistent if it has at least one solution or can have many solutions. If a consistent system has an infinite number of solutions, it is dependent. When you graph the equations, both equations represent the same line. If a system has no solution, it is said to be inconsistent. The graphs of the lines do not intersect, so the graphs are parallel and there is no solution.

Q4. In Section B, Questions no. 21-25 are very short answer (VSA) type questions, carrying 02 marks
each.

Answer: (a) 1 real root
Explanation:
Given: (x + 1)² - x² = 0
⇒ x² + 1 + 2x - x² = 0
⇒ 2x + 1 = 0
−1
⇒ x = ₂
Therefore, (x + 1)² - x² = 0 is a linear polynomial and has one real root.

Q5. In Section C, Questions no. 26-31 are short answer (SA) type questions, carrying 03 marks each.

Answer: (c) -320
Explanation:
I = T₁₆ = 10 + 15 × (-4) = -50.
n 16
.'. sum = ₂ (a + l) = ₂ (10 - 50) = 8 × (-40) = -320

Q6. In Section D, Questions no. 32-35 are long answer (LA) type questions, carrying 05 marks each.

Answer: –
(d) √2 units
Explanation: A (5, -4) B (4, -5) −−−−−−₂−−−−−−−−−−₂
AB = √(5 − 4) + (−4 + 5)
−−₂−−−−₂
= √1 + 1
–
AB = √2 units

Q7. In Section E, Questions no. 36-38 are case study-based questions carrying 4 marks each with sub-
parts of the values of 1,1 and 2 marks each respectively.

Answer: x₂+kx₁ y₂+ky₁
(c) ( , ) 1+k 1+k
Explanation:
Let coordinates of P be (x, y) which divides the line joining A(x₁, y₁) and B(x₂, y₂) in the ratio 1 : k
m₁ : m₂ = 1 : k
m₁x₂+m₂x₁
∴ x = m +m
1 2 1×x₂+k×x₁ x₂+kx₁
= =
1+k 1+k m1y₂+m2y₁
And y = m +m
1 2 1×y₂+k×y₁
=
1+k y₂+ky₁
=
1+k x₂+kx₁ y₂+ky₁
∴ P( , )
1+k 1+k

Q8. All Questions are compulsory. However, an internal choice in 2 Questions of Section B, 2 Questions of Section C and 2 Questions of Section D has been provided. An internal choice has been provided in all the 2 marks questions of Section E.

Answer: (d) 10
Explanation: In △ADE and △ABC
∠D = ∠B {Corresponding angle}
∠E = ∠C {Corresponding angle}
∴ △ADE and △ABC (by A A Similarity)
AD DE
=
AB BC 2 4
=
5 X
5×4
X = = 10
2
= 10 cm

Q9. Draw neat and clean figures wherever required.

Answer: (d) 18 cm
Explanation: In △DEF DF touches the circle at H and circle touches ED and EF Produced at K and M respectively
EK = 9 cm
EK and EM are the tangents to the circle
EM = EK = 9 cm
Similarly DH and DK are the tangent
DH = DK and FH and FM are tangents
FH = FM
Now, perimeter of ΔDEF
= ED + DF + EF
= ED + DH + FH + EF
= ED + DK + FM + EF
= EK + EM
= 9 + 9
= 18 cm

Q10. Take π = 22/7 wherever required if not stated.

Answer: (b) 24 cm
Explanation: We know that, a tangent to a circle is perpendicular to the radius at the point of contact.
So, △OCB is right a triangle, right angled at C.
Hence, by Pythagoras' theorem, we have:
BC² = OB² - OC²
⇒ BC² = 225 - 81 = 144
BC = 12 cm
We also know that, the tangents drawn from the same external point to a circle are equal. Since BC and BD are tangents drawn from the same external point, B, we have:
BC = BD = 12 cm.
So, BC + BD = 24 cm.
Hence, BC + BD = 24 cm.

Q11. Use of calculators is not allowed. Section A
1. If a is a non-zero rational and √b is irrational, then a√b is:
a) a natural number b) a rational number
c) an integer d) an irrational number
2. The graph of y = f(x) is shown in the figure for some polynomial f(x).
The number of zeroes of f(x) is
a) 2 b) 0
c) 3 d) 4
3. A system of linear equations is said to be consistent, if it has
a) exactly one solution b) one or many solutions
c) no solution d) two solutions
4. (x + 1)² - x² = 0 has
a) 1 real root b) 4 real roots
c) 2 real roots d) no real roots
5. The sum of first 16 terms of the AP 10, 6, 2, ... is
a) -400 b) 320
c) -320 d) 352
6. The distance between the points A(5, -4) and B(4, -5) is –
a) 9√2 units b) 1 unit
–
c) 2 units d) √2 units
7. If the line segment joining the points A (x₁, y₁) and B(x₂, y₂) is divided by a point P in the ratio 1 : k internally, then the coordinates of the point P are x₁+kx₂ y₁+ky₂ a⁾ ( , ) 1+k 1+k x₂+kx₁ y₂+ky₁ b⁾ ( , ) 1−k 1−k x₂+kx₁ y₂+ky₁ c⁾ ( , ) 1+k 1+k x₂−kx₁ y₂−ky₁ d⁾ ( , ) 1+k 1+k
8. In the given figure, DE || BC. The value of x is:
a) 8 b) 6
c) 12.5 d) 10
9. In Figure, a circle touches the side DF of △EDF at H and touches ED and EF produced at K and M
respectively. If EK = 9 cm, then the perimeter of △EDF is
a) 12 cm b) 13.5 cm
c) 9 cm d) 18 cm
10. In the given figure, BC and BD are tangents to the circle with centre O and radius 9 cm. If OB = 15
cm, then the length (BC + BD) is:
a) 36 cm b) 24 cm
c) 18 cm d) 12 cm
11. If x = a sec θ and y = b tan θ, then b²x² - a²y² =
a) a² - b² b) a² b²
c) a2 + b2 d) ab 2 tan 30∘

Answer: (b) a² b²
Explanation:
Here we have x = a sec θ and y = b tan θ
Therefore, b²x² - a²y² = b² (a sec θ)² - a² (b tan θ)²
= a²b² sec² θ - a²b² tan² θ
= a²b² (sec² θ - tan² θ)
= a²b² × 1
= a²b²

Q12. ( ) is equal to: 1+tan² 30∘
a) sin 30° b) cos 60o
c) sin 60o d) tan 60o

Answer: (c) sin 60o
Explanation: 2× ¹
2tan30o √³
2 o = 1 2
1+tan 30 1+( )
√3
2 2
√3 √3 2 3 √3
₁ = ₄ = × =
1+ √3 4 2
3 3
= sin60o

Q13. The angle of depression of a car parked on the road from the top of a 150 m high tower is 30°. The distance of the car from the tower (in metres) is –
a) 75 b) 150√3
– –
c) 150√2 d) 50√3

Answer: –
(b) 150√3
Explanation:
Let AB be the tower of height 150 m
C is car and angle of depression is 30o
Therefore, ∠ACB = 30o (alternate angle)
In right-angled triangle ABC, BC o
= cot 30
AB BC – –
⇒ = √3 ⇒ BC = 150 √3 m
150 –
That is, the distance of the car from the tower is 150√3 m.

Q14. The area of a sector whose perimeter is four times its radius r units, is r²
a) 2r² sq. units b) sq units 2 r²
c) r² sq. units d) sq units 4

Answer: (c) r² sq. units
Explanation:
Radius of sector = r
Perimeter = 4r
and length of arc = 4r - 2r = 2r
∴ Let angle at the centre = θ
θ θ
Then, 2πr = ∘ and 2r = ∘
360 360 θ
⇒ π × ∘ = 1 ...(i)
360 2 θ 2 θ
Now area = πr × ∘ = r (π × ∘ )
360 360
= r² × 1 [from (i)]
= r²

Q15. A pendulum swings through an angle of 30o and describes an arc 8.8 cm in length. Find the length of the pendulum.
a) 17 cm b) 8.8 cm
c) 15.8 cm d) 16.8 cm

Answer: (d) 16.8 cm
Explanation:
Length of the pendulum = Radius of a sector of the circle
Arc length = 8.8
{tex}\frac{θ}{360} (2πr){/tex} = 8.8
30 22
₃₆₀ × 2 × × r = 8.8
7
r = 16.8 cm

Q16. Two dice are thrown together. The probability that they show different numbers is: 1 2
a) b) 3 3 1 5
c) d) 6 6

Answer: 5
(d) ₆
Explanation:
E = dice show different no. favourable case for same no.
= (1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)
= 6
6 1
P(same no.) = =
36 6 1
p(not same no.) = 1 −
6 5
=
6

Q17. The probability that it will rain on a particular day is 0.76. The probability that it will not rain on that day is
a) 0 b) 0.24
c) 0.76 d) 1

Answer: (b) 0.24
Explanation:
Given: P (It will rain on a particular day) = 0.76
∴ P (It will not rain on a particular day) = 1 - P (It will rain particular day)
= 1 - 0.76 = 0.24

Q18. In a data, if l = 60, h = 15, f₁ = 16, f₀ = 6, f₂ = 6, then the mode is
a) 60 b) 67.5
c) 62 d) 72

Answer: (b) 67.5
Explanation: f₁−f₀
Mode = l + ( ) × h
2f₁−f₀−f₂ 16−6
= 60 + × 15
2×16−6−6
10
= 60 + × 15
32−12 10
= 60 + ₂₀ × 15
= 60 + 7.5
= 67.5

Q19. Assertion (A): In a solid hemisphere of radius 10 cm, a right cone of same radius is removed out. ₃ –
The volume of the remaining solid is 523.33 cm [Take π = 3.14 and √2 = 1.4]
Reason (R): Expression used here to calculate volume of remaining solid = Volume of
hemisphere - Volume of cone
a) Both A and R are true and R is the b) Both A and R are true but R is not the correct explanation of A. correct explanation of A.
c) A is true but R is false. d) A is false but R is true.

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

Q20. Assertion (A): Common difference of an AP in which a₂₁ - a₇ = 84 is 14
Reason (R): nth term of AP is given by an = a + (n - 1)d
a) Both A and R are true and R is the b) Both A and R are true but R is not the correct explanation of A. correct explanation of A.
c) A is true but R is false. d) A is false but R is true. Section B

Answer: (d) A is false but R is true.
Explanation: We have,
an = a + (n - 1)d
a₂₁ - a₇ = {a + (21 - 1)d} - {a + (7 - 1)d} = 84
a + 20d - a - 6d = 84
14d = 84
18
d = = 6
14
d = 6
So, A is false but R is true.
Section B

Q21. Given that HCF (306,1314) = 18. Find LCM (306,1314).
OR –
Show that 5 + 3√2 is an irrational number.

Answer: HCF(306, 1314) = 18
LCM(306, 1314) =?
Let, a = 306
b = 1314
LCM (a, b) × HCF (a,b) = a ×b
or, LCM (a, b) × 18 = 306 ×1314
306×1314
or LCM(a,b) = ₁₈ =22338
Therefore, LCM(306, 1314) = 22338
OR –
Let us assume that 5 + 3√2, is a rational number.
Then there exist co primes a and b such that – a
5 + 3√2=
b – a
3√2= -5
b a−5b
=
b – a−5b
So √2= -------(i)
3b a−5b –
is rational so this shows that √2 is rational
3b –
But √2 is irrational.
∴(i) presents a contradiction.,
–
Hence 5 + 3√2 is an irrational number.

Q22. In the given figure, ∠CAB = 90∘ and AD ⊥ BC . Show that △BDA ∼ △BAC . If AC = 75 cm,
AB = 1 m and BC = 1.25m, Find AD.

Answer: Given, ∠CAB = 90∘ and AD ⊥ BC .
Also given, AC = 75 cm = 0.75 m , AB = 1 m and BC = 1.25 m.
In △BDA and △BAC, we have:
∠BDA = ∠BAC = 90⁰
∠DBA = ∠CBA (common)
∴ △BDA ∼ △BAC [By AA similarity theorem]
AD AB
⇒ = [By proportionality theorem]
AC BC AD 1
⇒ ₀.₇₅ = ₁.₂₅
0.75
⇒ AD =
1.25
= 0.6 m or 60 cm
∴AD = 60 cm.

Q23. Two tangents PQ and PR are drawn from an external point to a circle with centre O. Prove that QORP is a cyclic quadrilateral.
√3 2

Answer: Given: Tangents PR and PQ from an external point P to a circle with centre O.
To prove: Quadrilateral QORP is cyclic.
Proof: RO and RP are the radius and tangent respectively at contact point R.
Therefore, ∠PRO = 90°
Similarly ∠PQO = 90°
In quadrilateral QOPR, we have
∠P + ∠R + ∠O + ∠Q = 360°
⇒ ∠P + ∠90° + ∠O + ∠90° = 360°
⇒ ∠P + ∠O = 360° - 180° = 180°
These are opposite angles of quadrilateral QORP and are supplementary.
Therefore, Quadrilateral QORP is cyclic. hence, proved.
2 2 2 2

Q24. If sin A = , find the value of 2cot A -1.
2 OR – ₃
Prove that: (√3 + 1) (3 – cot 30°) = tan 60° – 2 tan 60°

Answer: 2cot A − 1= 2(cosec A − 1) − 1 (∵ cot θ = −1 + cosec θ)
2
= 2cosec A − 2 − 1
2 1
= ₂ − 3 (∵ cosecθ = )
sin A sinθ 2
= ₂ − 3
√3
( ₂ ) 2 8 8−9 −1
2cot A − 1= − 3= =
3 3 3 OR – ₃
We have to prove that: (√3 + 1) (3 – cot 30°) = tan 60° – 2 tan 60°
–
Here, LHS = (√3 + 1) (3 – cot 30°)
– –
= (√3 + 1)(3 − √3)
– – –
= √3(3 − √3) + 1(3 − √3)
– –
= 3√3 − 3 + 3 − √3
–
= 2√3
RHS = tan³ 60° – 2 sin 60°
– 3 √3
= (√3) − 2 ×
2 – –
= 3√3 − √3
–
= 2√3
⇒ LHS = RHS
Hence, proved.

Q25. In a circle of radius 10.5 cm, the minor arc is one-fifth of the major arc. Find the area of the sector corresponding to the major arc. Section C

Answer: Let the major arc be x cm long
1
Then, length of the minor arc = ₅ x cm
1
Circumference= (x + ₅ x) cm
6x
= ₅ cm
6x 22 21
5 = 2 × 7 × 2
⇒ x = 55cm
1 21 2 1
Required area = ( ₂ × 55 × ₂ ) cm [Area = ₂ r − 1]
= 288.75 cm²
Section C

Q26. 105 goats, 140 donkeys and 175 cows have to be taken across a river. There is only one boat which will have to make many trips in order to do so. The lazy boatman has his own conditions for transporting them. He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally like to take the largest possible number each time. Can you tell how many animals went in each trip?

Answer: Given: Number of goats for trip = 105
Number of donkey for trip = 140
Number of cows for trip = 175
Therefore, The largest number of animals in one trip = HCF of 105. 140 and 175.
First consider 105 and 140 By applying Euclid’s division lemma, we get
140 = 105 × 1 + 35
105 = 35 × 3 + 0
Therefore, HCF of 105 and 140 = 35
Now consider 35 and 175 Again applying Euclid’s division lemma, we get
175 = 35 × 5 +0
HCF of 105, 140 and 175 is 35.
So 35 animals of same kind can go for trip in a single trip and number of trip is 105/35 +140/35 + 175/35
= 12

Q27. If one root of the quadratic polynomial 2x² - 3x + p is 3, find the other root. Also, find the value of p.

Answer: The given quadratic polynomial is p(x) = 2x² - 3x + p
Since, 3 is a root (zero) of p(x)
⇒ 2(3)² - 3 × 3 + p = 0
⇒18 - 9 + p = 0
⇒9 + p = 0
⇒ p = - 9
Now p(x) = 2x² - 3x - 9
= 2x²- 6x + 3x - 9
= 2x (x - 3) + 3 (x - 3)
= (x - 3) (2x + 3)
For roots of polynomial, p(x) = 0
⇒(x - 3) (2x + 3) = 0
3
⇒ x = 3 or x = -
2 3
Hence the other root is - .
2 No of heads per toss No of tosses fixi

Q28. Five coins were simultaneously tossed 1000 times and at each toss the number of heads were observed. The number of tosses during which 0,1,2,3,4 and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss. No. of heads per toss No. of tosses 0 38 1 144 2 342 3 287 4 164 5 25 Total 1000

Answer: 0 38 0 1 144 144 2 342 684 3 287 861 4 164 656 5 25 125
∑fi = 1000 ∑fixi = 2470
∑ fixi 2470
Mean number of heads per toss = = ₁₀₀₀ = 2.47
∑ fi
Therefore, Mean = 2.47

Q29. The area of a rectangle remains the same if the length is increased by 7 metres and the breadth is decreased by 3 metres. The area remains unaffected if the length is decreased by 7 metres and breadth is increased by 5 metres. Find the dimensions of the rectangle. OR 1 If 2 is added to the numerator of a fraction, it reduces to and if 1 is subtracted from the 2 1 denominator, it reduces to ₃ . Find the fraction.

Answer: Let us suppose that the length and breadth of the rectangle be x m and y m respectively.
Then, Area of rectangle = xy meter²
Now, according to question if length is increased by 7m and the breadth is decreased by 3m, the area
remains same
∴ xy = (x + 7)(y - 3)
⇒ xy = xy - 3x + 7y - 21
⇒ 3x - 7y = -21 ............(i)
Again, according to question when length is decreased by 7m and breadth is increased by 5m, then area remains unaffected
∴ xy = (x - 7)(y + 5)
⇒ xy = xy + 5x - 7y - 35
⇒ 35 = 5x - 7y
⇒ 5x - 7y = 35 ..........(ii)
Subtracting equation (i) from (ii), we get
5x - 7y -(3x- 7y) = 35 - (-21)
or, 5x - 7y -3- + 7y = 35 + 21
⇒ 2x = 56
56
⇒ x = = 28
2
Put the value of x = 28 in equation (ii), we get
5 × 28 - 7y = 35
⇒ 140 - 7y = 35
⇒ -7y = 35 - 140
⇒ -7y = -105
105
⇒ y = ₇ = 15
Therefore, dimensions of the rectangle are 28m and 15m respectively.
OR x
Let the fraction be y
According to question, x+2 1
y =
2
or 2x - y = -4 ....(i)
x 1
and y−1 = 3
or 3x - y = -1 ...(ii)
on solving eq (i) and (ii), we get,
x = 3, y = 10
3
∴ fraction is ₁₀

Q30. In an acute angled triangle ABC, if tan (A+B-C) = 1 and, sec (B+C-A)=2, find the values of A, B and
C.

Answer: According to the question,
tan (A+B-C) = 1
∘
⇒ tan (A+B-C) = tan45
⇒ A + B - C = 45° ........(1)
Also given, sec (B+C-A) = 2
⇒ sec (B + C - A ) = sec 60°
∘
∴B + C − A = 60 ...........(2)
Adding equation (1) & (2); ∘ ∘
(A + B − C) + (B + C − A) = 45 + 60
⇒ 2B = 105∘
1∘
⇒ B = 52
2 ∘ 1
Putting B = 52 in equation (2); we get :-
∘ 2 1 ∘
52 + C − A = 60
2 ∘ 1
⇒ C − A = 7 ......(3)
2 Also, in ΔABC, we have ∘
A + B + C = 180
1∘ ∘ 1∘
⇒ A + 52 ₂ + C = 180 [∵ B = 52 ₂ ]
1∘
⇒ C + A = 127 ₂ ......(4)
Adding and subtracting (3) and (4), we get
2C = 135o and 2A = 120o
1 ∘ o
⇒ C = 67 ₂ and A = 60
Hence, we get the values of A = 60o,
1∘
B = 52
2 ∘ 1
and C = 67 .
2

Q31. In the given figure, two tangents RQ and RP are drawn from an external point R to the circle with
centre O. If ∠PRQ = 120°, then prove that OR = PR + RQ.
OR A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ. Section D

Answer: 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 = ₂ = 60
∵ 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
Similarly,
∠QOR = 30∘
PR
and cos θ =
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
Given: In a circle a chord PQ and a tangent MRN at R such that QP || MRN
To prove: R bisects the arc PRQ.
Construction: Join RP and RQ.
Proof: Chord RP subtends ∠1 with tangent MN and ∠2 in alternates segment of circle so ∠1 = ∠2.
MRN || PQ
∴ ∠1 = ∠3 [Alternate interior angles]
⇒ ∠2 = ∠3
⇒ PR = RQ [Sides opp. to equal ∠s in Δ RPQ]
∵ Equal chords subtend equal arcs in a circle so
arc PR = arc RQ
or R bisect the arc PRQ. Hence, proved. Section D

Q32. Calculate the mode of the following frequency distribution table : Marks Number of students 25 or more than 25 52 35 or more than 35 47 45 or more than 45 37 55 or more than 55 17 65 or more than 65 8 75 or more than 75 2 85 or more than 85 0

Answer: The given data are: Marks Number of students 25 or more than 25 52 35 or more than 35 47 45 or more than 45 37 55 or more than 55 17 65 or more than 65 8 75 or more than 75 2 85 or more than 85 0 From above data we can calculate range data as following: Marks Number of students(f)
25 - 35 52 - 47 = 5
35 - 45 47 - 37 = 10
45 - 55 37 - 17 = 20
55 - 65 17 - 8 = 9
65 - 75 8 - 2 = 6
75 - 85 2 - 0 = 2
85 - 95 0
From table it is clear that maximum class frequency is 20 belonging to class interval 45 - 55
Modal class = 45 - 55
Lower limit (l) of modal class = 45
Class size (h) = 10
Frequency (f₁) of modal class = 20
Frequency (f₀) of class preceding modal class = 10
Frequency (f₂) of class succeeding the modal class = 9
f₁−f₀
Mode = l + ( ) × h
2f₁−f₀−f₂ 20−10
= 45 + ( ) × 10
2×20−10−9
10
= 45 + × 10
21
= 45 + 4.76
= 49.76
Therefore mode of data is 49.76

Q33. PQRS is a trapezium with PQ ∥ SR Diagonals PR and SQ intersect at M and ΔP MS ∼ ΔQMR .
Prove that PS = QR.

Answer: Given : ΔP MS ∼ ΔQMR and P Q||SR.
To show PS = QR
∵ △P MS ∼ △QMR P S P M MS
∴ = = ...(i)
QR QM MR [corresponding sides of similar triangles are proportional]
Now, consider △P MQ and △RMS
In these triangles, we have
∠P MQ = ∠RMS [ vertically opposite angles ]
∠MP Q = ∠MRS [ alternate angles ]
∵ △P MQ ∼ △RMS [AA criteria] P M MQ
∴ =
RM MS [corresponding sides of similar triangles are proportional] P M MR
⇒ = ...(ii)
QM MS From Eq (i) and Eq (ii), we get MS MR
⇒ =
MR MS
⇒ MS² = MR²
⇒ MS = MR
From Eq(i) , we get P S MS
∴ =
QR MR P S
= 1
QR
⇒ P S = QR Hence proved.

Q34. A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹ 750. We would like to find out the number of toys produced on that day. Represent the situations mathematically (quadratic equation). OR In a flight of 600 km, the speed of the aircraft was slowed down due to bad weather. The average speed of the trip was decreased by 200 km/hr and thus the time of flight increased by 30 minutes. Find the average speed of the aircraft originally.

Answer: Let the number of toys produced be x.
∴ Cost of production of each toy = Rs (55 − x)
It is given that, total production of the toys = Rs 750
∴ x(55 – x) = 750
⇒ x² – 55x + 750 = 0
Now to factorize this equation we have to find two numbers such that their product is 750 and sum is 55
⇒ x² – 25x – 30x + 750 = 0
⇒ x(x – 25 ) – 30(x – 25 ) = 0
⇒ (x – 25)(x – 30) = 0
Either x – 25 = 0 or x − 30 = 0
i.e., x = 25 or x = 30
Hence, the number of toys will be either 25 or 30.
OR
Let average speed of aircraft be x km/h
600 600 1
− =
x−200 x 2
x² - 200x – 240000 = 0
(x - 600) (x + 400) = 0
x = 600 km/h
∴ Original speed = 600 km/h

Q35. In Figure, from a solid cube of side 7 cm, a cylinder of radius 2.1 cm and height 7 cm is scooped out. Find the total surface area of the remaining solid. OR A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid. Section E

Answer: We have; A Cube, length
Cube's , a = 7 cm
Edge A Cylinder: 21
Cylinder's Radius, r = 2.1 cm or r = ₁₀ cm
Cylinder's Height, h = 7 cm
∵ A cylinder is scooped out from a cube,
∴ TSA of the resulting cuboid:
= TSA of whole Cube - 2 × (Area of upper circle or Area of lower circle) + CSA of the scooped out Cylinder
= 6a² + 2πrh - 2 × (πr²)
= 6 × (7)² + 2 × (22 ÷ 7 × 2.1 × 7) - 2 × [22 ÷ 7 × (2.1)²]
= 6 × 49 + (44 ÷ 7 × 14.7) - (44 ÷ 7 × 4.41)
= 294 + 92.4 - 27.72
= 294 + 64.68
= 358.68 cm²
Hence, the total surface area of the remaining solid is 358.68 cm²
OR According to the question,a hemispherical depression is cut from one face of the cubical block such that the diameter l of the hemisphere is equal to the edge of the cube.
Let the radius of hemisphere = r
l
∴ r = ₂
Now, the required surface area = Surface area of cubical block - Area of base of hemisphere + Curved
surface area of hemisphere.
= 6( side )² − πr² + 2πr²
2 l 2 l 2
= 6l − π( ) + 2π( )
2 2 2 πl² π 2
= 6l − + ₂ l
4 2 πl²
= 6l +
4 1 2
Surface area = (24 + π)l units.
4 1 22 2
= ₄ (24 + ₇ ) l
Section E

Q36. Read the following text carefully and answer the questions that follow: Akshat's father is planning some construction work in his terrace area. He ordered 360 bricks and instructed the supplier to keep the bricks in such as way that the bottom row has 30 bricks and next is one less than that and so on. The supplier stacked these 360 bricks in the following manner, 30 bricks in the bottom row, 29 bricks in the next row, 28 bricks in the row next to it, and so on.
i. In how many rows, 360 bricks are placed? (1)
ii. How many bricks are there in the top row? (1)
iii. How many bricks are there in 10th row? (2) OR If which row 26 bricks are there? (2)

Answer: i. Number of bricks in the bottom row = 30. in the next row = 29, and so on.
Therefore, Number of bricks stacked in each row form a sequence 30, 29, 28, 27, ..., which is an AP
with first term, a = 30 and common difference, d = 29 - 30 = -1
Suppose number of rows is n, then sum of number of bricks in n rows should be 360.
i.e. Sn = 360
n n
⇒ ₂ [2 × 30 + (n − 1)(−1)] = 360 {Sn = ₂ (2a + (n − 1)d) }
⇒ 720 = n(60 − n + 1)
2
⇒ 720 = 60n - n + n
⇒ n² − 61n + 720 = 0
⇒ n² − 16n − 45n + 720 = 0 [by factorization]
⇒ n(n − 16) − 45(n − 16) = 0
⇒ (n − 16)(n − 45) = 0
⇒ (n − 16) = 0 or (n − 45) = 0
⇒ n = 16 or n = 45
Hence, number of rows is either 45 or 16.
n = 45 not possible so n = 16
a₄₅ = 30 + (45 − 1)(−1) {an = a + (n − 1)d }
= 30 − 44 = −14 [∵ The number of logs cannot be negative]
Hence the number of rows is 16.
ii. Number of bricks in the bottom row = 30. in the next row = 29, and so on.
Therefore, Number of bricks stacked in each row form a sequence 30, 29, 28, 27,..., which is an AP
with first term, a = 30 and common difference, d = 29 - 30 = -1
Suppose number of rows is n, then sum of number of bricks in n rows should be 360.
Number of bricks on top row are n = 16,
a₁₆ = 30 + (16 - 1) (-1) {an = a + (n - 1)d}
= 30 - 15 = 15
Hence, and number of bricks in the top row is 15.
iii. Number of bricks in the bottom row = 30. in the next row = 29, and so on.
therefore, Number of bricks stacked in each row form a sequence 30, 29, 28, 27, ..., which is an AP
with first term, a = 30 and common difference, d = 29 - 30 = -1.
Suppose number of rows is n, then sum of number of bricks in n rows should be 360
Number of bricks in 10th row a = 30, d = -1, n = 10
an = a + (n - 1)d
⇒ a₁₀ = 30 + 9 × -1
⇒ a₁₀ = 30 - 9 = 21
Therefore, number of bricks in 10th row are 21.
OR
Number of bricks in the bottom row = 30. in the next row = 29, and so on.
Therefore, Number of bricks stacked in each row form a sequence 30, 29, 28, 27,..., which is an AP
with first term, a = 30 and common difference, d = 29 - 30 = -1.
Suppose number of rows is n, then sum of number of bricks in n rows should be 360.
an = 26, a = 30, d = -1
an = a + (n - 1)d
⇒ 26 = 30 + (n - 1) × -1
⇒ 26 - 30 = -n + 1
⇒ n = 5
Hence 26 bricks are in 5th row.
−3+1 0+4

Q37. Read the following text carefully and answer the questions that follow: Ryan, from a very young age, was fascinated by the twinkling of stars and the vastness of space. He always dreamt of becoming an astronaut one day. So he started to sketch his own rocket designs on the graph sheet. One such design is given below: Based on the above, answer the following questions:
i. Find the mid-point of the segment joining F and G. (1)
ii. a. What is the distance between the points A and C? (2) OR
b. Find the coordinates of the point which divides the line segment joining the points A and B in the ratio 1 : 3 internally. (2)
iii. What are the coordinates of the point D? (1)

Answer: i. Mid point of FG is ( ₂ , ₂ ) = (-1, 2)
−−−−−−−−₂ −−−−−−−−−−₂
ii. a. AC = √(−1 − 3) + (−2 − 4)
−− −−
= √52 or 2√13
OR
1×3+3×3 1×2+3×4 7
b. The coordinates of required point are ( , )i.e. (3, )
1+3 1+3 2
iii. D(-2, -5)

Q38. Read the following text carefully and answer the questions that follow: Vijay lives in a flat in a multi-story building. Initially, his driving was rough so his father keeps eye on his driving. Once he drives from his house to Faridabad. His father was standing on the top of the building at point A as shown in the figure. At point C, the angle of depression of a car from the building was 60o. After accelerating 20 m from point C, Vijay stops at point D to buy ice cream and the angle of depression changed to 30o.
i. Find the value of x. (1)
ii. Find the height of the building AB. (1)
iii. Find the distance between top of the building and a car at position D? (2) OR Find the distance between top of the building and a car at position C? (2)

Answer: i. The above figure can be redrawn as shown below: From the figure,
let AB = h and BC = x
In △ABC, AB h
tan 60 = = x
BC – h
√3 = x
–
h = √3x ...(i)
In △ABD, AB h
tan 30 = BD = x+20
1 √3x
= [using (i)]
√3 x+20
x + 20 = 3x
x = 10 m
ii. The above figure can be redrawn as shown below: – –
Height of the building, h = √3x = 10√3 = 17.32 m
iii. The above figure can be redrawn as shown below: Distance from top of the building to point D. In △ABD o AB
sin 30 =
AD AB
⇒ AD = ₀
sin 30
10√3
⇒ AD = ₁
2 –
⇒ AD = 20√3m
OR The above figure can be redrawn as shown below: Distance from top of the building to point C is In △ABC o AB
sin 60 =
AC AB
⇒ AC = ₀
sin 60
10√3
⇒ AC = √₃
2
⇒ AD = 20 m