Students can access the CBSE Sample Papers for Class 10 Maths Standard with Solutions and marking scheme Set 4 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 10 Maths Standard Set 4 with Solutions
Time: 3 Hours
Maximum Marks: 80
General Instructions:
- This Question Paper has 5 Sections A, B, C, D and E.
- Section A has 20 MCQs carrying 1 mark each
- Section B has 5 questions carrying 02 marks each.
- Section C has 6 questions carrying 03 marks each.
- Section D has 4 questions carrying 05 marks each.
- Section E has 3 case based integrated units of assessment (04 marks each) with sub parts of the values of 1, 1 and 2 marks each respectively.
- All Questions are compulsory. However, an internal choice in 2. Qs of 5 marks, 2.Qs of 3 marks and 2. Questions of 2. marks has been provided. An internal choice has been provided in the 2 marks questions of Section E.
- Draw neat figures wherever required. Take n = 22/7 wherever required if not stated.
Section – A ( 20 Marks)
Section A Consists of Multiple Choice Type questions of 1 mark each
Question 1.
The HCF of k and 93 is 31, where k is a number.
Which of these CAN be true for SOME VALUES of k?
(i) k is a multiple of 31
(ii) k is a multiple of 93
(iii) k is a an even number
(iv) k is a an odd number
(A) Only (ii) and (iii)
(B) Only (i), (i) and (iii)
(C) Only (i), (iii) and (iv)
(D) All (i), (ii), (iii) and (iv)
Solution:
(C) Only (i), (iii) and (iv)
Explanation: HCF of k and 93 = 31
It means Factor of 93 = 31 × 3
Factor of k = 31 × a
Thus, It is a multiple of 31, which can be even or odd.
Hence (i), (iii) and (iv) are correct.
Question 2.
Given below is a pair of linear equations:
4x + y = 8
4x – 2y = 16
Is the given pair of equations:
(A) Consistent
(B) Inconsistent
(C) Consistent with unique solution
(D) Consistent with many solutions
Solution:
(C) Consistent with unique solution
∴ Equations have unique solution.
Hence they are consistent.
Question 3.
What is the value of k such that the following pair of equations have infinitely many solutions?
x – 2y = 3 and -3x + ky = -9.
(A) -6
(B) -3
(C) 3
(D) 6
Solution:
(D) 6
Explanation: Given equations are:
x – 2y – 3 = 0
and -3x + ky + 9 = 0
⇒ 3k = 18 ⇒ k = 6
Hence, the value of k is 6.
Question 4.
If the zeroes of quadratic polynomial are 1, 1; then the polynomial can be.
(A) x2 + x + 1
(B) x2 – 2x + 1
(C) x2 + 3x + 2
(D) x2 + 2x + 2
Solution:
(B) x2 – 2x + 1
Explanation: Given that
Two zeroes are 1 and 1
So the quadratic polynomial satisfying these roots is:
⇒ (x – 1)(x – 1)
⇒ x2 – 2x + 1
Question 5.
Which term of the progression 20, 19\(\frac{1}{4}\), 18\(\frac{1}{2}\), 17\(\frac{1}{3}\), is the first negative term:
(A) 27th term
(B) 28th term
(C) 26th term
(D) 25th term
Solution:
(B) 28th term
Explanation: Here, d = \(\frac{-3}{4}\)
Let the nth term be first negative term
∴ 20 + (n – 1)(\(\frac{-3}{4}\)) < 0
⇒ 3n = 83
⇒ n = 27\(\frac{2}{3}\)
Hence, 28th term is first negative term.
Question 6.
Points A(-1, y) and B(5, 7) lie on a circle with centre O(2, -3y). The values of y are:
(A) 1, -7
(B) -1, 7
(C) 2, 7
(D) -2, -7
Solution:
(B) -1, 7
Explanation: As points A and B lie on the circle and O is the centre.
AO and BO will be the radii of the circle.
(3)2 + (-4y)2 = (-3)2 + (-3y – 7)2
9 + 16y2 = 9 + 9y2 + 49 + 42y
16y2 – 9y2 – 42y – 49 = 0
7y2 – 42y – 49 = 0
7(y2 – 6y – 7) = 0
y2 – 7y + 1y – 7 = 0
y(y – 7) + 1(y – 7) = 0
(y – 7)(y + 1) = 0
Question 7.
If the vertices of a parallelogram PQRS taken in order are P(3, 4), Q(-2, 3) and R(-3, -2), then the co-ordinates of its fourth vertex S are
(A) (-2, -1)
(B) (-2, -3)
(C) (2, -1)
(D) (1, 2)
Solution:
(C) (2, -1)
Explanation: Given PQRS is a parallelogram and diagonal PR and QS bisect each other at O.
Let fourth vertex be S(x, y), then
⇒ x – 2 = 0
⇒ x = 2
and y + 3 = 2
⇒ y = -1
Hence, co-ordinates of fourth vertex S are (2, -1).
Question 8.
In the figure below, DE || AC and DF || AE. Which of these is equal to \(\frac{B F}{F E}\)?
(A) \(\frac{D F}{A E}\)
(B) \(\frac{B E}{E C}\)
(C) \(\frac{B A}{A C}\)
(D) \(\frac{E E}{E C}\)
Solution:
(B) \(\frac{B E}{E C}\)
Explanation: In ΔABC,
DE || AC (Given)
∴ \(\frac{B D}{D A}\) = \(\frac{B E}{E C}\) (From BPT) …(i)
In ΔABE,
DF || AE (Given)
∴ \(\frac{B D}{D A}\) = \(\frac{F B}{F E}\) (From BPT) …(ii)
From eqs. (i) and (ii), we get
\(\frac{B F}{F E}\) = \(\frac{B E}{E C}\)
Question 9.
Ankit joins the centre of the two pulleys and observes line segments P1S1 and P2S2 when extended meet at a point X. Which line segment is equal to the length of P1S1?
(A) OQ
(B) QX
(C) XS2
(D) P2S2
Solution:
(D) P2S2
Explanation: The lengths of a tangent drawn from an external point to a circle are equal.
∴ P2S2 is equal to the length of P1S1.
Question 10.
The area of the circle that can be inscribed in a square of 6 cm is
(A) 36π cm2
(B) 18π cm2
(C) 22π cm2
(D) 9π cm2
Solution:
(D) 9π cm2
Explanation:
ABCD is a square of side 6 cm. PQ is a diameter of given circle such that
PQ = AB = 6 cm
∴ Radius (r) = \(\frac{\text { Diameter }}{2}\)
= 6/2 = 3 cm
Area of the circle = πr2
= π(3)2 = 9πr cm2
Question 11.
If x tan 60° cos 60° = sin 60° cot 60°, then x =
(A) cos 30°
(B) tan 30°
(C) sin 30°
(D) cot 30°
Solution:
(B) tan 30°
Explanation:
x tan 60° cos 60° = sin 60° cot 60°
Question 12.
If cot θ = \(\frac{1}{\sqrt{3}}\), the value of sec2 θ + cosec2 θ is a
(A) 1
(B) \(\frac{40}{9}\)
(C) \(\frac{38}{9}\)
(D) 5\(\frac{1}{3}\)
Solution:
(D) 5\(\frac{1}{3}\)
Explanation: It is given that
Question 13.
In the figure below, what is the length of AB?
(A) 45√3 m
(B) \(\frac{45}{\sqrt{3}}\)m
(C) 45(√3-1)m
(D) 45(√3 +1)m
Solution:
(C) 45(√3-1)m
Explanation: According to figure
Question 14.
In the given figure, if PQR is the tangent to a circle at Q, whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to
(A) 20°
(B) 40°
(C) 35°
(D) 45°
Solution:
(B) 40°
Explanation: Given that AB || PQR
∠B = ∠BQR = 70° [Alternate interior angles]
∠OQR = ∠AMQ [Alternate interior angles]
As PQR and OQ are tangent and radius at contact point Q
∴ ∠OQR = 90°
⇒ ∠1 + ∠70° = 90°
⇒ ∠1 = 90° – 70° = 20°
∴ ∠AMO = 90°
∵ Perpendicular from centre to chord bisect the chord
MA = MB
∠QMA = ∠QMB = 90°
MQ = MQ [Common]
ΔQMA = ΔQMB [SAS congruence]
⇒ ∠A = ∠B
⇒ ∠A = 70° [∵ ∠B = 70°]
∴ ∠A + ∠AMQ + ∠2 = 180° [Angle sum property of a triangle]
⇒ 70° + 90° + ∠2 = 180°
⇒ ∠2 = 180° – 160°
⇒ ∠2 = 20°
∴ ∠AQB = ∠1 + ∠2
= 20° + 20° = 40°
Question 15.
The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm is:
(A) 31 cm
(B) 25 cm
(C) 62 cm
(D) 50 cm
Solution:
(D) 50 cm
Explanation: Let r1 = 24 cm and r2 = 7 cm
Area of first circle = πr12 = π(24)2 = 576π cm2
Area of second circle = πr22 = π(7)2 = 49π cm2
According to question,
Area of circle = area of first circle + area of second circle
πR2 = 576π + 49π [where, R be radius of circle]
R2 = 625 ⇒ R = 25 cm
Diameter of a circle = 2R = 2 × 25 = 50 cm.
Question 16.
The probability that the drawn card from a pack of 52 cards is neither an ace nor a spade is
(A) \(\frac{9}{13}\)
(B) \(\frac{35}{52}\)
(C) \(\frac{10}{13}\)
(D) \(\frac{19}{26}\)
Solution:
(A) \(\frac{9}{13}\)
Explanation: Total ace cards = 4 and total spade cards
= 13 – 1 = 12 (One card among aces is also a spade)
Cards which are neither ace or spade = 52 – 16 = 36
Required probability = \(\frac{36}{52}\) = \(\frac{9}{13}\)
Question 17.
This is a paper cup.
Jaya tossed this cup.
When the cup lands on the table, it can land in three possible positions, as shown below.
To calculate the probability of falling in each position, Jaya tosses the cup 60 times.
She records her observations in the table below.
Position of cup after toss | Frequency |
Inverted | 20 |
upright | 5 |
Rolling side | 35 |
Jaya tosses the cup one more time.
What would be the probability of the cup falling in the upright position?
(A) \(\frac{1}{3}\)
(B) \(\frac{1}{11}\)
(C) \(\frac{1}{12}\)
(D) \(\frac{1}{60}\)
Solution:
(C) \(\frac{1}{12}\)
Explanation: Total number of outcomes = 60
Possible outcomes of upright position = 5
∴ Required probability = \(\frac{5}{60}\) = \(\frac{1}{12}\)
Question 18.
Arti owns a manufacturing company. She hires 5 supervisors and 20 operators for a 6-month project. The table given below shows their salary breakup.
Position | Salary for the two months | Salary for the remaining four months |
Supervisor | Between ₹18,000 to ₹20,000 | Between ₹22,000 to ₹25,000 |
Operator | Between ₹8,000 to ₹10,000 | Between ₹13,000 to ₹15,000 |
Arti agrees to pay the maximum decided amount as salary to the operators.
What would be the total amount (in Rs) that Arti will have to pay the operators for the first two months?
(A) ₹10,000
(B) ₹200,000
(C) ₹300,000
(D) ₹500,000
Solution:
(B) ₹200,000
Explanation: Maximum decided amount for first two months paid by Arti as salary to operators
= ₹10,000
Number of operators = 20
Total amount = 20 × 10,000
= ₹200,000
DIRECTIONS: Two statements are given below – one labelled Assertion (A) and the other labelled Reason (R). Read the statements carefully and choose the option that correctly describes statements (A) and (R).
(A) Both (A) and (R) are true and (R) is the correct explanation of the (A).
(B) Both (A) and (R) are true but (R) is not the correct explanation of the (A).
(C) (A) is true but (R) is false.
(D) (A) is false but (R) is true.
Question 19.
Assertion (A): If the height of the cone is 10 cm and the radius of the base is 7 cm, then the volume of the cone is 513.3 cm3.
Reason (R): According to assertion, if \(\sqrt{149}\) = 12.2, then curved surface area of a cone is 268.4 cm2.
Solution:
(B) Both (A) and (R) are true but (R) is not the correct explanation of the (A).
Explanation: For assertion:
Height (h) = 10 cm
and radius (r) = 7 cm
So, reason is also true.
Both A and R are true and R is not the correct
explanation of A.
Question 20.
Assertion (A): a, b, c are in A.P. if and only if 2b = a + c.
Reason (R): The sum of first n odd natural numbers is n2.
Solution:
(B) Both (A) and (R) are true but (R) is not the correct explanation of the (A).
Explanation: Assertion is true because
b – a = c – b (a, b, c are in A.P.)
⇒ 2b = a + c
Reason: Let 1 + 3 + 5 + 7 + 9 + … + n, are sum of n odd natural numbers.
Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
Sn = \(\frac{n}{2}\)[2(1) + (n – 1)2]
Sn = \(\frac{n}{2}\)(2n)
Sn = n2
Hence, the sum of the first n natural number is n2.
Section – B (10 Marks)
Section B consists of 5 questions of 2 marks each.
Question 21.
Given that 3 is irrational, prove that 5 + 2√3 is irrational.
Solution:
Let us assume 5 + 2√3 is rational, then it must be in the form of \(\frac{p}{q}\) where p and q are co-prime integers and q ≠ 0
i.e., 5 + 2√3 = \(\frac{p}{q}\)
So √3 = \(\frac{p-5 q}{2 q}\) … (i)
Since p, q, 5 and 2 are integers and q ≠ 0, RHS of equation (i) is rational. But LHS of (i) is √3 which is irrational. This is not possible.
This contradiction has arisen due to our wrong assumption that 5 + 2√3 is rational. So, 5 + 2√3 is irrational.
Question 22.
In the figure below, OPQ is a triangle with OP = OQ. RS is an arc of a circle with centre O.
(Note: The figure is not to scale.)
Triangle OSR is similar to triangle OPQ.
Is the above statement true or false? Justify your reason.
Solution:
Let ∠ROS = ∠POQ = x
(vertically opposite angles are equal)
In ΔORS, OR =OS (radii of circle)
∴ ∠ORS = ∠OSR
(Angles opposite to equal sides)
Now, ∠ORS + ∠OSR + ∠ROS = 180°
∠ORS + ∠ORS + x = 180°
2∠ORS = (180° – x)
∠ORS = \(\frac{\left(180^{\circ}-x\right)}{2}\)
Similarly, In ΔOPQ
OPQ = \(\frac{\left(180^{\circ}-x\right)}{2}\)
Thus, ∠ORS = ∠OSR = ∠OPQ
= ∠OQP = \(\frac{\left(180^{\circ}-x\right)}{2}\)
In ΔORS and ΔOPQ
∠ROS = ∠POQ (proved above)
∠ORS = ∠OQP (proved above)
∠OSR = ∠OPQ (proved above)
∴ ΔORS ∼ ΔOPQ (AAA Similarity)
Thus, It is true.
Question 23.
Shown below is a ΔPQR inscribed in a semicircle.
A circle is drawn such that QR is a tangent to it at the point R.
How many such circles can be drawn? Justify your answer.
Solution:
Infinite number of circles can be drawn.
Justification: ∠PRQ = 90° (Angle in semicircle)
Radius is perpendicular to the tangent at the point of contact.
∴ Infinite circles with radii lying on extended PR and R being a point on circumference of circle can be drawn.
Question 24.
(A) If √3 sin θ – cos θ = 0 and 0° < θ < 90°, find the value of θ.
Solution:
Here √3 sin θ – cos θ = 0 and 0° < θ < 90°
or, √3 sin θ = cos θ
θ = 30°
OR
(B) Find an acute angle θ when \(\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}=\frac{1-\sqrt{3}}{1+\sqrt{3}}\).
Solution:
⇒ tan θ = √3
θ = 60°
Question 25.
Jaya drew this rangoli design during a competition.
Circles C1, C2, C3 and C4 have common centre P.
The radius of circle C1 is 6 cm.
The table given below shows the radii of circles in terms of the radius of circle C1.
Radius of circle | Terms of the radius of C1 |
C2 | 2 |
C3 | 2.5 |
C4 | 3.5 |
(A) Jaya says,” Since the radius of circle C4 is 3.5 times of the radius of circle C1, the area occupied by circle C4 is 3.5 times the area occupied by circle C1.”
Is Jaya correct? Give reason.
Solution:
Here, radius of circle C1 = 6 cm
Area of circle C1 = πr12
= \(\frac{22}{7}\) × 6 × 6
= 113.14 cm
Also, radius of C4 = 3.5 × radius of C1
= 3.5 × 6 = 21 cm
∴ Area of circle C4 = πr42
= \(\frac{22}{7}\) × 21 × 21
= 1386 cm2
No, Jaya is not correct as the area occupied by circle C1 is 113.14 cm2 whereas the area occupied by circle C4 is 1386 cm2 and 1386 is not 3.5 times 113.14.
OR
(B) If the area of a semi-circular field is 30800 sq.m, then find the perimeter of the field.
Solution:
Let the radius of the field be r.
We know that,
Area of semi-circle = \(\frac{1}{2}\)πr2
⇒ 30800 = \(\frac{1}{2}\)πr2
⇒ r2 = 30800 × 2 × \(\frac{7}{22}\) [∵ π = \(\frac{22}{7}\)]
⇒ r2 = 19600
⇒ r = 140 m [Taking positive square root]
How, perimeter of the field = πr + 2r
= \(\frac{22}{7}\) × 140 + 2 × 140
= 440 + 280
= 720 m
Section – C (18 Marks)
Section C Consists of 6 questions of 3 marks each
Question 26.
The LCM of 64, 82 and k is 124 where k is a positive integer. Find the smallest value of k. Show your steps.
Solution:
Prime factorization of 64 = 24 × 34
Prime factorization of 82 = 26
∴ LCM of 64, 82 and k = 26 × 34 × k … (i)
As, LCM of 64, 82 and k = 124 (given) …(ii)
From (i) and (ii) we get
26 × 34 × k = 124
k = 4
Thus smallest value of k = 4.
Question 27.
The roots a and B of the quadratic equation x2 – 5x + 3(k – 1) = 0 are such that α – β = 1. Find the value of k.
Solution:
We have,
α + β = 5 …(i)
α – β = 1 …(ii)
Solving (i) and (ii), we get
α = 3 and β = 2
also αβ = 5
or 3(k – 1) = 6
k – 1 = 2
k = 3
Question 28.
(A) Represent the following pair of linear equations graphically and hence comment on the condition of consistency of this pair.
x – 5y = 6, 2x – 10y = 12.
Solution:
Given, x – 5y = 6
⇒ y = \(\frac{x-6}{5}\)
x | 6 | 1 | -4 |
y | 0 | -1 | -2 |
and 2x – 10y = 12 ⇒ y = \(\frac{x-6}{5}\)
x | 6 | 1 | -4 |
y | 0 | -1 | -2 |
Since, the lines are coincident, so the system of linear equations is consistent with infinitely many solutions.
OR
(B) A fraction becomes \(\frac{1}{3}\) when 2 is subtracted from the numerator and it becomes \(\frac{1}{2}\) when 1 is subtracted from the denominator. Find the fraction.
Solution:
Let the fraction bc\(\frac{1}{3}\)
∴ \(\frac{x-2}{y}\) = \(\frac{1}{3}\) … (i)
and \(\frac{x}{y-1}\) \(\frac{1}{2}\) … (i)
Solving Eqs. (i) and (ii), we get
x = 7 and y = 15
∴ Required fraction is \(\frac{7}{15}\).
Question 29.
(A) ABDC is a quadrant of a circle of radius 28 cm and a semi-circle BEC is drawn with BC as diameter. Find the area of the shaded region. (use π = \(\frac{22}{7}\))
Solution:
As ABC is a quadrant of the drde, ∠BAC will be measured 90°.
In ΔABC, BC2 = AC2 + AB2
= (28)2 + (28)2
= 2(28)2
∴ BC = 28√2 cm
Radius of semi-circle drawn on BC = \(\frac{28 \sqrt{2}}{2}\)
= 14 √2 cm
Area of semi-circle = \(\frac{1}{2}\) × \(\frac{22}{7}\) × (14√2)
= 616 cm2
Area of ΔABC = \(\frac{1}{2}\) × 28 × 28
= 392 cm2
Area of of quadrant = \(\frac{1}{2}\) × \(\frac{22}{7}\) × 28 × 28
= 616 cm2
Area of the shaded region = Area of semi-circle + Area of Δ – Area of quadrant
= 616 + 392 – 616
= 392 cm2
OR
(B) The area of a circular playground is 22176 cm2. Find the cost of fencing this ground at the rate of ₹50 per metre.
Solution:
Area of a circular playground = 22176 cm2
i.e., πr2 = 22176 cm2
[where r is the radius of the play ground]
⇒ r2 = 22176 × \(\frac{7}{22}\) = 7056
⇒ r = 84 cm = 0.84 m
Cost of fencing this ground = ₹50 × 2πr
= ₹50 × 2 × \(\frac{22}{7}\) × 0.84
= ₹264
Question 30.
Prove that: \(\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\) = 1 + tan θ + cos θ
Solution:
= tan θ + 1 + cot θ
= RHS
Question 31.
The median of the following data is 16. Find the missing frequencies a and b, if the total of the frequencies is 70.
Solution:
Class | Frequency (f) | Cumulative frequency (c.f.) |
0-5 | 12 | 12 |
5-10 | a | 12 + a |
10-15 | 12 | 24 + a |
15-20 | 15 | 39 + a |
20-25 | b | 39 + a + b |
25-30 | 6 | 45 + a + b |
30-35 | 6 | 51 + a + b |
35-40 | 4 | 55 + a + b |
Total | 70 |
According to question,
55 + a + 5 = 70
a + 5 = 15 … (i)
a = 8
Substituting the value of o in equation (i), we get
8 + b = 15
⇒ b = 15 – 8
⇒ b = 7
Section – D (20 Marks)
Section D consists of 4 questions of 5 marks each
Question 32.
(A) The difference of the squares of two numbers is 180. The square of the smaller number is 8 times the greater number. Find the two numbers.
Solution:
Let the greater number be x.
The square of the smaller number is 8 times the greater number = 8x
Given, the difference of squares of two numbers is 180.
∴ x2 – 8x = 180
⇒ x2 – 8x – 180 = 0
⇒ x2 – 18x + 10x – 180 = 0
⇒ x(x – 18) + 10(x – 18) = 0
⇒ (x – 18)(x + 10) =0
Either (x – 18) = 0 or (x + 10) = 0
Either x = 18 or x = -10
Since, number cannot be negative. So, x = 18
Now, square of smaller number = 8x
= 8 × 18
= 144
∴ Smaller number = √144 = 12
Hence, smaller number is 12 and greater number is 18.
OR
(B) A shopkeeper buy certain number of books in ₹80. If he buy 4 more books then new cost price of each book is reduced by ₹1. Find the number of books initially he buy.
Solution:
Let the number of books bought be x. Then,
Cost of x books = ₹80
⇒ Cost of one book = ₹\(\frac{80}{x}\)
If the number of books bought is x + 4, then
Cost of one book = ₹\(\frac{80}{x+4}\)
It is given that the cost of one book is reduced by one rupee.
\(\frac{80}{x}\) – \(\frac{80}{x+4}\) = 1
80(\(\frac{1}{x}\) – \(\frac{1}{x+4}\)) = 1
80(\(\frac{x+4-x}{x(x+4)}\)) = 1
\(\frac{320}{x^2+4 x}\) = 1
⇒ x2 + 4x = 320
⇒ x2 + 4x – 320 = 0
⇒ x2 + 20x – 16x – 320 = 0
⇒ x(x + 20) – 16(x + 20) = 0
⇒ (x + 20)(x – 16) = 0
⇒ x = -20 or, x = 16
⇒ x = 16 [∵ x cannot be negative]
Hence, the number of books is 16.
Question 33.
In ΔABC, AD is a median and O is any point on AD. BO and CO on producing meet AC and AB at E and F respectively. Now AD is produced to X such that OD = DX as shown in figure.
Prove that:
(i) EF || BC
(ii) AO : AX = AF : AB
Solution:
Since, BC and OX bisect each other.
So, BXCO is a parallelogram then BE || XC and BX || CF.
In ΔABX, by B.PT.,
or, OA : AX = AF : AB Hence Proved.
Question 34.
(A) In a rain water harvesting system, the rain water from a roof 22 m × 20 m drains into a cylindrical tank having diameter of base 2 m and height 3.5 m. If the tank is full, find the rainfall in cm. Write your views on water conservation.
Solution:
Volume of rain water on the roof = volume of cylindrical tank
i.e., 22 × 20 × h = \(\frac{22}{7}\) × 1 × 3.5
⇒ h = \(\frac{1}{40}\)m
⇒ h = 2.5 cm
Water conservation must be encouraged.
Detailed Answer
Let length of roof, l = 22 m
breadth of roof, b = 22 m
and height of roof = h m
Also, given, height of cylinder = 3.5 m
Views on water conservation:
Water conservation reduces energy use and can even save household money. Water conservation must be encouraged.
OR
(B) Rohan wants to renovate his room. He calls an architect for this work to measure the room. The length, breadth and height of a room are 8 m 50 cm; 6m 25 cm and 4 m 75 cm respectively. He wants to put the longest rod that can measure the dimensions of the room exactly.
Solution:
Given Length = 8 m 50 cm = 850 cm
breadth = 6 m 25 cm = 625 cm
height = 4 m 75 cm = 475 cm
Since, the length of the longest rod is equal to
HCF of 850, 625 and 475.
Prime factor of 850 = 2 × 52 × 17
Prime factor of 625 = 54
Prime factor of 475 = 52 × 19
Hence, HCF (625, 850, 475) = 52 = 25
Thus, the longest rod that can measure the dimensions of the room exactly = 25 cm
Question 35.
The first term of an A.P. is 3, the last term is 83 and the sum of all its terms is 903. Find the number of terms and the common difference of the A.P.
Solution:
Here a = 3, an = 83 and Sn = 903
Therefore 83 = 3 + (n – 1)d
⇒ (n – 1)d = 80 …(i)
Also 903 = \(\frac{n}{2}\)[2a + (n – 1)d]
= \(\frac{n}{2}\)[6 + 80]
= 43n [using (i)]
⇒ n = 21
and d = 4
Section – E (12 Marks)
Case study based questions are compulsory.
Question 36.
Read the following text and answer the following questions:
Your elder brother wants to buy a car and plans to take loan from a bank for his car. He repays his total loan of ₹1,18,000 by paying every month starting with the first instalment of & 1000. If he increases the instalment by ₹100 every month, answer the following:
(i) Find the amount paid by him in 30th instalment.
(ii) If total instalments are 40 then amount paid in the last instalment?
(iii) How much the amount will he pay in 30 instalments?
Solution:
(i) Here, a = 1000
d = 100
a30 = a + (n – 1)d
= 1000 + (30 – 1)100
= 1000 + 2900 = 3900
(ii) Amount paid in 40th instalment,
a40 = a + (n – 1)d
= 1000 + (40 – 1)100
= 1000 + 3900
= 4900
(iii) Sum of 30 instalments
= \(\frac{n}{2}\)[2a + (n – 1)d]
= \(\frac{30}{2}\)[2 × 1000 + (30 – 1)100]
= 15[2000 × 4900 + 2900]
= 15 × 4900
= 73500
Total Amount paid in 30 instalments = ₹73500
OR
Find the ratio of the 1st instalment to the last instalment.
Solution:
Amount paid in last instalment = ₹4900 [From Part (ii)]
Principal for 1st instalment = ₹1000
Hence, Ratio of 1st instalment to the last instalment = \(\frac{1000}{4900}\)
= 10 : 49.
Question 37.
Read the following text and answer the questions given below it:
A tiling or tessellation of a flat surface is the covering of a plane using one or more geometric shapes, called tiles, with no overlaps and no gaps. Historically, tessellations were used in ancient Rome and in Islamic art. You may find tessellation patterns on floors, walls, paintings etc. Shown below is a tiled floor in the archaeological Museum of Seville, made using squares, triangles and hexagons.
A craftsman thought of making a floor pattern after being inspired by the above design. To ensure accuracy in his work, he made the pattern on the Cartesian plane. He used regular octagons, squares and triangles for his floor tessellation pattern.
(i) What is the length of the line segment joining points B and F?
(ii) What are the co-ordinates of the point on y-axis equidistant from A and G?
Solition:
(i) B(1, 2), F(-2, 9)
BF2 = (-2 – 1)2 + (9 – 2)2
= (-3)2 + (7)2
= 9 + 49
= 58
So BF = √58 units
(ii) A(-2, 2), G(-4, 7)
Let the point on y-axis be Z(0, y)
AZ2 = GZ2
(0 + 2)2 + ( y – 2)2 = ( 0 + 4)2 + ( y – 7)2
(2)2 + y2 + 4 – 4y = (4)2 + y2 + 49 – 14y
8 – 4y = 65 – 14y
10y = 57
So, y = 5.7
i.e., The required point is (0, 5.7)
OR
What is the area of Trapezium AFGH?
A(-2, 2), F(-2, 9), G(-4, 7), H(-4, 4)
Clearly GH = 7 – 4 = 3units
AF = 9 – 2 = 7 units
∴ So, height of the trapezium AFGH = 2 units
So, area of AFGH = \(\frac{1}{2}\)(AF + GH) × height
= \(\frac{1}{2}\)(7 + 3) × 2
= 10 sq. units
(iii) The centre ‘Z’ of the figure will be the point of intersection of the diagonals of quadrilateral WXOP. Then what are the co-ordinates of Z?
Solution:
W(~6, 2), X(-4, 0), O(5, 9), P(3, 11)
Clearly WXOP is a rectangle.
Point of intersection of diagonals of a rectangle is the mid point of the diagonals. So the required point is mid point of WO or XP.
= \(\left(\frac{-6+5}{2}, \frac{2+9}{2}\right)\)
= \(\left(\frac{-1}{2}, \frac{11}{2}\right)\)
Question 38.
Read the following and answer the following questions:
An electrician has to repair an electric fault on the pole of height 5 m. He needs to reach a point 1.3 m below the top of the pole to undertake the repair work (see figure).
(i) What is the length of BD?
(ii) What should be the length of ladder, when inclined at an angle of 60° to the horizontal?
(iii) How far from the foot of the pole should he place the foot of the ladder?
OR
If the horizontal angle is changed to 30°, then what should be the length of the ladder?
Solution:
(i) From figure, the electrician is required to reach at the point B on the pole AD.
So, BD = AD – AB
= (5 – 1.3) m = 3.7 m
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