Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Standard with Solutions Set 6 are designed as per the revised syllabus.
CBSE Sample Papers for Class 10 Maths Standard Set 6 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 2 marks each.
- Section C has 6 questions carrying 3 marks each.
- Section D has 4 questions carrying 5 marks each.
- Section E has 3 case based integrated units of assessment (4 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 question of 5 marks, 2question 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 π = 22/7 wherever required if not stated.
Section – A
Section A consists of 20 questions of 1 mark each.
Question 1.
The sum of exponents of prime factors in the prime-factorization of 196 is:
(a) 3
(b) 4
(c) 5
(d) 2
Answer:
(b) 4
Explanation: Prime factors of 196 = 22 × 72
So, sum of exponents = 2 + 2 = 4.
Question 2.
The centroid of APQR whose vertices are P(-8, 0), Q(5, 5) and R(-3, -2) is (1)
(a) (-2, 1)
(b) (1, -2)
(c) (2, 1)
(d) (1, 2)
Answer:
(a) (-2, 1)
Here, (x1, y1) = (-8, 0), (x2, y2) = (5, 5) and (x3, y3) = (-3, -2)
∴ Centroid of ΔPQR = (x, y)
= \(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)
= \(\left(\frac{-8+5-3}{3}, \frac{0+5-2}{3}\right)\)
= (-2, 1)
Question 3.
What is the ratio in which the line joining the points (6, 4) and (1, -7) is divided by the X-axis?
(a) 5 : 4
(b) 4 : 7
(c) 7 : 4
(d) 4 : 5
Answer:
(b) 4 : 7
Explanation: Let, K : 1 be the ratio in which X-axis divides the line joining (6, 4) and (1, -7). Let, P be the point on X-axis that divides the line. Then
P(x, y) = [latex]\frac{\mathrm{K}+6}{\mathrm{~K}+1}, \frac{-7 \mathrm{~K}+4}{\mathrm{~K}+1}[/latex]
As P (x, y) lies on X-axis P(x, 0) = (\(\frac{\mathrm{K}+6}{\mathrm{~K}+1}, \frac{-7 \mathrm{~K}+4}{\mathrm{~K}+1}\)) [∵ x-coordinate will be x, 0]
⇒ \(\frac{-7 \mathrm{~K}+4}{\mathrm{~K}+1}\) = 0
⇒ – 7K + 4 = 0
⇒ 7K = 4
K = \(\frac{4}{7}\)
So, the required ratio be K : 1 = \(\frac{4}{7}\) : 1 = 4 : 7
Question 4.
30th term of the AP 10, 7, 4, ……, is (1)
(a) 97
(b) 77
(c) -77
(d) -87
Answer:
(c) -77
Here, a = 10, n = 30
∴ Tn = a + (n – 1)d and d = 7 – 10 = -3
Then, T30 = 10 + (30 – 1) × (- 3)
⇒ T30 = 10 + 29 × (-3)
⇒ T30 = 10 – 87 = – 77
Question 5.
If a, (a -2) and 3a are in the arithmetic progression, then what is the value of a?
(a) -2
(b) 2
(c) 1
(d) -1
Answer:
(a) – 2
Explanation: As, a, a -2, 3a are AP.
Then, a – 2 – a = 3a – (a – 2)
⇒ – 2 = 2a + 2
⇒ 2a = – 2 – 2
⇒ 2a = – 4
⇒ a = – 2
Question 6.
If α and β are the zeroes of the polynomial f(x) = x2 -5x + k such that α – β = 1, then the value of 4k is (1)
(a) 12
(b) 24
(c) 10
(d) 20
Answer:
(b) 24
Given, α and β are the zeroes of the given polynomial f(x) = x2 – 5x + k
α + β = –\(\left(-\frac{5}{1}\right)\) = 5 and αβ = \(\frac{k}{1}\) = k
Since, α – β = 1 ⇒ (α – β)2 = 1
⇒ (α + β)2 – 4αβ = 1 ⇒ (5)2 – 4 × k = 1
⇒ 25 – 4k = 1 ⇒ 4k = 24
Question 7.
What is a natural number whose square diminished by 84 is thrice the 8 more of given number? (a) 13
(b) 11
(c) 10
(d) 12
Answer:
(d) 12
Explanation: Let x be the natural number then according to the question,
x – 82 = 3 (x + 8)
⇒ x2 – 84 = 3x + 24
⇒ x2 – 3x- 108 = 0
⇒ x2 – 12x + 9x – 108 = 0
⇒ x(x – 12) + 9(x -12) = 0
⇒ (x – 12) (x + 9) = 0
⇒ x – 12 = 0 or x + 9 = 0
⇒ x = 12 or x = -9
Since -9 is not a natural number. So, the required number is 12.
Question 8.
The HCF of 96 and 404 by prime factorization method is (1)
(a) 0
(b) 4
(c) 2
(d) 3
Answer:
(b) 4
We have, 96 = 25 × 3
and 404 = 22 × 101
∴ HCF = Common factor in both numbers = 22 = 4
Question 9.
Σfi = 15, Σfixi = 3p + 36 and mean of any distribution is 3, find p:
(a) 6
(b) 3
(c) 4
(d) 5
Answer:
(b) 3
Explanation: Given: Σfi = 15, Σfixi = 3p + 36 and Mean = 3
We know, Mean = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\)
⇒ 3 = \(\frac{3 p+36}{15}\)
⇒ 45 = 3p + 36
⇒ 3p = 45 – 36
⇒ 3p = 9
∴ p = 3.
Question 10.
ΔABC and ΔDEF are similar such that 2AB = DE and BC = 8 cm, then the value of EF is (1)
(a) 15 cm
(b) 16 cm
(c) 18 cm
(d) None of these
Answer:
(b) 16 cm
Since, ΔABC ~ ΔDEF
⇒ \(\frac{A B}{D E}\) = \(\frac{B C}{E F}\) = \(\frac{A C}{D F}\) ⇒ \(\frac{2 A B}{2 D E}\) = \(\frac{B C}{E F}\)
⇒ \(\frac{D E}{2 D E}\) = \(\frac{8}{E F}\) [∵ 2AB = DE]
∴ EF = 16 cm
Question 11.
Two dice are thrown together. The probability of getting the same number on both the dice is:
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{6}\)
(c) \(\frac{1}{3}\)
(d) \(\frac{1}{12}\)
Answer:
\(\frac{1}{6}\)
Explanation: Total number of possible outcomes =
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} = 36
Thus, the favourable outcomes are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).
Hence, the total number of favourable outcomes = 6
⇒ P(E) = \(\frac{6}{36}\) = \(\frac{1}{6}\)
Question 12.
If the first term of an AP is 2 and the common difference is 4, then the sum of its 40 terms is (1)
(a) 3000
(b) 2800
(c) 3200
(d) None of these
Answer:
(c) 3200
Given, a = 2 and d = 4
Then, Sn = \(\frac{n}{2}\)[2a + (n – 1)d]
⇒ S40 = \(\frac{40}{2}\)[2 × 2 + (40 – 1) × 4]
= 20[4 + 39 × 4]
= 20 × 160 = 3200
Question 13.
If the H.C.F. of two numbers is 1, then the two numbers are called:
(a) composite
(b) twin primes
(c) co-primes
(d) none of these
Answer:
(c) co-primes
Explanation: Two numbers are said to be co-primes if they have only one common factor, namely 1.
Question 14.
In the given figure, AT is tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. (1)
Then, the value of AT is
(a) 2\(\sqrt{3}\) cm
(b) 4 cm
(c) 2 cm
(d) None of these
Answer:
(a) 2\(\sqrt{3}\) cm
From the given figure,
∠OAT = 90°
[∵ tangent and radius are perpendicular to each other at the point of contact]
In right-angled ΔOAT,
\(\frac{A T}{O T}\) = cos 30°
⇒ \(\frac{A T}{4}\) = \(\frac{\sqrt{3}}{2}\) ⇒ AT = 2\(\sqrt{3}\) cm
Question 15.
If one root of the quadratic equation 6x2 – x – k = 0 is \(\frac{2}{3}\), then what is the value of k?
(a) 2
(b) 3
(c) 1
(d) 4
Answer:
(a) 2
Explanation: Given quadratic equation is,
6x2 – x – k = 0
Since, \(\frac{2}{3}\) is one of the root of given quadratic equation
∴ 6(\(\frac{2}{3}\))2 – \(\frac{2}{3}\) – k = 0
⇒ 6 × \(\frac{4}{9}\) – \(\frac{2}{3}\) – k = 0
⇒ \(\frac{8}{3}\) – \(\frac{2}{3}\) – k = 0
⇒ \(\frac{2}{3}\) – k = 0
⇒ 2 – k = 0
⇒ k = 2
Question 16.
A wire is in the shape of a circle of radius 21 cm. It is bent in the form of a square. The side of the square is ( take π = \(\frac{22}{7}\)) (1)
(a) 22 cm
(b) 33 cm
(c) 44 cm
(d) 66 cm
Answer:
(b) 33 cm
As length of wire is same, so both figures have same perimeter,
∴ Circumference of circle = Perimeter of the square.
Let r be the radius of the circle and a is the side of the square.
So, 2πr = 4a
⇒ 4a = 2 × \(\frac{22}{7}\) × 21 [∵ r = 21 cm]
⇒ 4a = 132
⇒ a = 33 cm
Question 17.
In ΔPQR, if PS is the internal bisector of ∠P meeting QR at S and PQ = 15 cm, QS = (3 + x) cm, SR = (x – 3) cm and PR = 7 cm, then find the value of x.
(a) 2.85 cm
(b) 8.25 cm
(c) 5.28 cm
Answer:
(b) 8.25 cm
Explanation: Since, PS is the internal bisector of ∠P and it meets QR at S.
∴ \(\frac{P Q}{Q S}=\frac{P R}{S R}\)
⇒ \(\frac{15}{3+x}=\frac{7}{x-3}\)
⇒ 7(3 + x) = 15(x – 3)
⇒ 21 + 7x = 15x – 45
⇒ 15x – 7x = 45 + 21
⇒ 8x = 66
⇒ 4x = 33
⇒ x = 8.25 cm
Question 18.
The area of a quadrant of circle whose circumference is 12 cm, is (1)
(a) 102.34 cm2
(b) 95.15 cm2
(c) 34 cm2
(d) 45.81 cm2
Answer:
(d) 45.81 cm2
Given, the circumference of a quadrant = 12 cm
⇒ \(\frac{2 \pi r}{4}\) = 12 ⇒ πr = 24 ⇒ r = \(\frac{24}{\pi}\)
Now, area of a quadrant = \(\frac{\pi r^2}{4}\)
DIRECTION: In question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R).
Choose the correct option:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true, but reason (R) is false.
(d) Assertion (A) is false, but reason (R) is true.
Question 19.
Statement A (Assertion): If in a circle, the radius of the circle is 3 cm and distance of point from the centre of a circle is 5 cm, then length of the tangent from that point will be 4 cm.
Statement R (Reason): The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Solution:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
Explanation: Let’s construct a circle whose radius is 3 cm and distance of a point from the centre of a circle is 5 cm
As per the above diagram, by joining the distant point A, we get ΔAOB and AB is the tangent to the given circle.
∵ tangent is perpendicular to the radius.
∴ using the Pythagoras theorem,
(Hypotenuse)2 = (Base)2 + (Perpendicular height)2
⇒ (OA)2 = (AB)2 + (OB)2
⇒ (AB)2 = (OA)2 – (OB)2
= (5)2 – (3)2
= 25 – 9 = 16
⇒ AB = \(\sqrt{16}[latex] = 4 cm
Question 20.
Assertion (A) 3y2 + 17y – 30 = 0 have distinct roots.
Reason (R) The quadratic equation ax2 + bx + c = 0 has distinct roots (real roots), if D > 0. (1)
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Answer:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Assertion True
3y2 + 17y – 30 = 0
D = b2 – 4ac = (17)2 – 4 × 3 (-30)
= 289+ 360 = 649 > 0
So, roots are real and distinct.
Reason is True and is the correct explanation of assertion.
Hence, Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Section – B
Section B consists of 5 questions of 2 marks each.
Question 21.
Explain why (17 × 5 × 11 × 3 × 2 + 2 × 11) is a composite number?.
Solution:
17 × 5 × 11 × 3 × 2 + 2 × 11 = 17 × 5 × 3 × 22 + 22
= 22 (17 × 5 × 3 + 1)
– 22 (255 + 1)
= 2 × 11 × 256 …… (i)
Now equation (i) is divisible by 2,11 and 256, which means it has more than 2 prime factors.
∴ (17 × 5 × 11 × 3 × 2 + 2 × 11) is a composite number.
Question 22.
A conical tent is to accommodate 77 persons. Each person must have 4m3 of air to breathe. Given, the radius of the tent as 7 m. Find the height of the tent. (2)
Answer:
Given, in the conical tent, each person requires 16 m3 of air to breathe.
∴ Volume of the conical tent = 77 × 4 = 308 m3
⇒ [latex]\frac{1}{3}\)πr2h = 308
⇒ \(\frac{1}{3}\) × \(\frac{22}{7}\) × (7)2 × h = 308
[given, radius of the tent = 7 m]
⇒ \(\frac{1}{3}\) × 22 × 7 × h = 308
⇒ h = \(\frac{308 \times 3}{22 \times 7}\) = 24m
Thus, height of the conical tent = 6 m.
Question 23.
If the distance between the points (3, 0) and (0, y) is 5 units and y is positive then what is the value of y?
Solution:
Given, the distance between the points (3, 0) and (0, y) is 5 units.
Distance = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
Thus, 5 = \(\sqrt{(3-0)^2+(0-y)^2}\)
= \(\sqrt{9+y^2}\)
⇒ 9 + y2 = 25 ⇒ y2 = 16
⇒ y2 = 42 ⇒ y = ±4
But as y is positive so the value of y = 4.
Question 24.
Find the value of θ in the following sin 2θ = sin 60°.cos 30° – cos 60° sin 30°. (2)
Answer:
We have, sin2θ = sin 60°.cos 30° – cos 60° sin 30°
⇒ sin2θ = \(\frac{\sqrt{3}}{2}\) × \(\frac{\sqrt{3}}{2}\) – \(\frac{1}{2}\) × \(\frac{1}{2}\)
[∵ sin 60° = \(\frac{\sqrt{3}}{2}\), cos 30° = \(\frac{\sqrt{3}}{2}\)]
⇒ sin2θ = \(\frac{3}{4}\) – \(\frac{1}{4}\) = \(\frac{2}{4}\)
⇒ sin2θ = \(\frac{1}{2}\)
⇒ sin2θ = sin30° [∵ sin 30° = \(\frac{1}{2}\)]
⇒ 2θ = 30°
⇒ θ = 10°
Question 25.
Savita and Hamida are friends. What is the probability that both will have (i) same birthday, (ii) different birthdays (ignoring leap year)?
OR
A box contains cards numbered 11 to 123. A card is drawn at random from the box. Find the probability that the number on the drawn card is :
(i) a square number (ii) a multiple of 8
Solution:
Savita may have any one of the 365 days of the year as her birthday. Similarly, Hamida may have any one of the 365 days of the year as her birthday.
Total number of ways in which Savita and Hamida may have their birthday = 365 x 365
(i) Savita and Hamida may have same birthday on any one of 365 days of the year.
∴ Number of ways in which Savita and Hamida will have birthday on same day = 365
Probability that Savita and Hamida will have birthday on same day = \(\frac{365}{365 \times 365}=\frac{1}{365}\)
(ii) Probability that Savita and Hamida will have different birthday
= 1 – Probability that Savita and Hamida will have birthday on same day
= 1 – \(\frac{1}{365}\) = \(\frac{364}{365}\)
OR
We have
Sample space = {11, 12, 13, …….. , 123}
⇒ n( S) = 113
(i) Let E1 be the event that the card drawn is a square number.
So, E1 = } 16, 25, 36,…., 121}
⇒ n(E1) = 8
∴p(E1) = \(\frac{8}{113}\)
(ii) Let E2 be the event that card drawn is a multiple of 8.
So, E2 = {16, 24, 32, , 120}
⇒ n(E2) = 14
∴p(E2) = \(\frac{14}{113}\)
Section – C
Section C consists of 6 questions of 3 marks each.
Question 26.
Prove that sin θ(1 + tan θ) + cos θ(1 + cotθ) = sec θ + cosec θ. (3)
Answer:
LHS = sinθ(1 + tanθ) + cos θ(1 + cotθ)
Or
Prove that \(\frac{\sin \theta-2 \sin ^3 \theta}{2 \cos ^3 \theta-\cos \theta}\) = tan θ
Answer:
Question 27.
State whether the given pairs of triangles are similar or not. In case of similarity, mention the criterion.
OR
In the given figure, DEFG is a square and ∠BAC = 90°. Prove that :
(i) ΔAGF ~ ΔDBG,
(ii) ΔAGF ~ ΔEFC,
(iii) ΔDBG ~ ΔEFC,
(iv) DE2 = BD × EC.
Solution:
(i) \(\frac{\mathrm{AO}}{\mathrm{DO}}\) = \(\frac{16}{9}\) and \(\frac{\mathrm{BO}}{\mathrm{CO}}\) = \(\frac{9}{5}\)
∵ \(\frac{\mathrm{AO}}{\mathrm{DO}}\) ≠ \(\frac{\mathrm{BO}}{\mathrm{CO}}\)
∵ Given triangles are not similar.
(ii) In ΔPQR, ∠P + ∠Q + ∠R = 180° [Angle-sum property of a triangle]
⇒ 45° + 78° + ∠R = 180°
⇒∠R = 180° – 45° – 78° = 57°
In ΔLMN, ∠L + ∠M +∠N = 180° [Angle-sum property of a triangle]
⇒ 57° + 45° + ∠N = 180°
∠N = 180° – 57° – 45°
= 78°
Now, in ΔPQR and ΔLMN ∠P = ∠M [Each 45°]
∠Q = ∠N [Each 78°]
∠R = ∠L [Each 57°]
By AAA similarity criterion ΔPQR ~ ΔMNL
OR
(i) In ΔAGF and ΔDBG, we have
∠GAF = ∠BDG = 90°
and ∠AGF = ∠GBD [Corresponding angles between parallel lines GF and BC with AB being the transversal as GDEF is a square]
Thus, by AA similarity criterion,
ΔAGF ~ ΔDBG. Hence proved.
(ii) In ΔAGF and ΔEFC, we have
∠GAF = ∠FEC = 90°
and ∠AFG = ∠FCE [Corresponding angles between parallel lines GF and BC with AC being the transversal]
Thus, ΔAGF ~ ΔEFC Hence proved.
(iii) From (i) and (ii), we have ΔAGF ~ ΔDBG
and ΔAGF ~ ΔEFC
Thus ΔDBG ~ AEFC Hence proved.
(iv) From (iii), we have ΔDBG ~ ΔEFC
Thus, \(\frac{B D}{F E}=\frac{D G}{E C}\) [C.P.C.T.]
⇒ \(\frac{\mathrm{BD}}{\mathrm{DE}}=\frac{\mathrm{DE}}{\mathrm{EC}}\) [DE = DG = FE = GF as they are the sides of a square]
⇒ DE2 = BD × EC. Hence proved.
Question 28.
Find the HCF and LCM of 84, 90 and 120 by prime factorisation method. (3)
Answer:
The prime factorisation of 84, 90 and 120 gives
84 = 22 × 3 × 7
90 = 2 × 32 × 5
and 120 = 23 × 3 × 5
To find the HCF, we list the common prime factors and their smallest exponents in 84, 90 and 192.
Here, 21 and 31 are the smallest exponents of the common factors 2 and 3, respectively.
So, HCF (84, 90, 120) = 21 × 31 = 2 × 3 = 6
To find the LCM, we list all prime factors of 84, 90 and 120 and their greatest exponents.
Here, 23, 32, 51 and 71 are the greatest exponents of the prime factors 2, 3, 5 and 7, respectively involved in three numbers.
So, LCM (84, 90, 120) = 23 × 32 × 51 × 71
= 8 × 9 × 5 × 7 = 2520
Question 29.
Prove that: 2 sec2 θ – sec4 θ – 2 cosec2 θ + cosec2 θ = cot2 θ – tan2 θ.
Prove that: \(\frac{{cosec} A}{{cosec} A-1}+\frac{{cosec} A}{{cosec} A+1}\) = 2 sec2 θ
Solution:
L.H.S. = 2 sec2 θ – sec4 θ – 2 cosec2 θ + cosec4 θ
= 2 sec2 θ – (sec2 θ)2 – 2 cosec2 θ + (cosec2 θ)2
= 2(1 + tan2 θ) – (1 + tan2 θ)2 – 2(1 + cot2 θ) + (1 + cot2 θ)2
= 2 + 2 tan2 θ – 1 – tan4 θ – 2 tan2 θ – 2 – 2 cot2 θ + 1 + cot4 θ + 2 cot2 θ
= cot4 θ – tan4 θ
= R.H.S. Hence Proved.
OR
= 2 cosec2 A tan2 A
= 2(1 + cot2 A) tan2 A
= 2 tan2 A + 2 tan2 A cot2 A
= 2 + 2 tan2 A = 2(1 + tan2 A)
= 2 sec2 A
= R.H.S. Hense proved.
Question 30.
If the pth, qth and rth terms of an AP are a, b and c respectively, then show that a(q – r) + b(r – p) + c(p – q) = 0. (3)
Answer:
Let m be the first term and d the common difference of the given AR Then,
pth term = m + (p – 1)d = a ……..(i)
qth term = m + (q – 1)d = b ……..(ii)
and rth term = m + (r – 1)d = c …….(iii)
On multiplying (q – r), (r – p) and (p – q) in Eqs. (i), (ii) and (iii), respectively and then adding, we get
a(q – r) + b(r – p) + c(p – q)
= (q – r){m + (p – 1)d} + (r – p){m + (q – 1 )d} + (p – q){m + (r – 1)d}
= m(q – r + r – p + p – q) + d[(p – 1 )(q – r) + (q – 1)(r – p) + (r – 1)(p – q)]
= m × 0 + d[pq – pr – q + r + qr – qp – r + p + rp – qr – p + q]
= 0 + d (0) = 0
⇒ a(q – r) + b(r – p) + c(p – q ) = 0
Hence proved.
Question 31.
The largest possible sphere is carved out of a solid wooden cube of side 7 cm. Find the volume of the wood left. [Use π = \(\frac{22}{7}\)]
Solution:
∴ Length of each side of the cube = 7 cm
∴ Largest possible diameter of the sphere = 7 cm
Radius of the sphere = 3.5 cm
Thus, Volume of cube = (7)3 cm3 = 343 cm3
and Volume of sphere = \(\frac{22}{7}\) × \(\frac{22}{7}\) (3.5)3 cm3
= \(\frac{539}{3}\) cm3
∴ Remaining volume of wood = (343 – \(\frac{539}{3}\))cm3 = \(\frac{1029-539}{3}\))cm3
= 163.33 cm3.
Section – D
Section D consists of 4 questions of 5 marks each.
Question 32.
A girl of height 80 cm is walking away from the base of a lamp-post at a speed of 2.1 m/sec. If the lamp is 4 m above the ground, then find the length of her shadow after 4 sec. (5)
Answer:
Let AB be the lamp-post, CD be the girl and D be the position of girl after 4 sec.
Again, let DE = x m be the length of shadow of the girl.
Given, CD = 80 cm = 0.8 m, AS = 4m and speed of the girl = 2.1 m/sec
∴ Distance of the girl from lamp-post after 4 sec,
BD = 2.1 × 4 = 8.4 m
[∵ distance = speed × time]
In ΔABE and ΔCDE,
∠B = ∠D [each 90°]
∠E = ∠E [common angle]
ΔABE ~ ΔCDE [by AA similarity criterion]
⇒ \(\frac{B E}{D E}\) = \(\frac{A B}{C D}\) …… (i)
[since, corresponding sides of similar triangles are proportional]
On substituting all the values in Eq, (i), we get
\(\frac{8.4+x}{x}\) = \(\frac{4}{0.8}\) [∵ BE = BD + DE = 8.4 + x]
⇒ \(\frac{8.4+x}{x}\) = \(\frac{40}{8}\)
⇒ \(\frac{8.4+x}{x}\) = 5
⇒ 8.4 + x = 5x
⇒ 4x = 8.4 ⇒ x = 2.1 m
Hence, the length of her shadow after 4 sec is 2.1 m.
Question 33.
Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution:
Given, a circle with centre O touches the opposite sides of a quadrilateral at points P, Q, R and S.
To prove the above, the following need to be satisfied :
∠AOB + ∠COD = 180° and ∠AOD + ZBOC = 180°
Join OS, OP, OQ and OR.
In ΔOQC and ΔOCR,
QC = CR [Tangents from external point C]
OC = OC [Common]
OQ = OR [Radii of circle]
Thus ΔOQC ≅ ΔORC [by S.S.S. congruency]
Hence ∠QOC = ∠COR [Corresponding angles of congruent triangles]
Similarly, ∠ROB = ∠SOB, ∠SOA = ∠POA
and ∠POD = ∠QOD
Now ∠QOC + ∠COR + ∠ROB + ∠SOB + ∠SOA + ∠POA + ∠POD + ∠QOD = 360° (∵ O is point)
⇒ 2(∠QOC + ∠ROB + ∠SOA + ∠POD) = 360°
and 2(∠COR + ∠SOB + ∠POA + ∠QOD) = 360°
⇒∠QOC + ∠ROB + ∠ZSOA + ∠POD = 180°
and ∠COR + ∠SOB + ∠POA + ∠QOD = 180°
⇒∠QOC + ∠SOB + ∠SOA + ∠QOD = 180° (∵ ∠ROB = ∠SOB, ∠POD = ∠QOD)
and ∠COR + ∠ROB + ∠POA + ∠POD = 180°
⇒∠QOC + ∠QOD + ∠SOA + ∠SOB = 180°
and ∠COR + ∠ROB + ∠POA + ∠POD = 180°
⇒∠AOB + ∠COD = 180°
and ∠AOD + ∠BOC = 180° Hence Proved.
Question 34.
The height of a cone is 40 cm. A small cone is cut at the top by a plane parallel to the base. If the volume of the small cone be \(\frac{1}{27}\) of the volume of the given cone, at what height above the base are the sections made? (5)
Answer:
Let R be the radius of the given cone, r the radius of the small cone, h be the height of the frustum and be the height of the small cone. In figure, ∆ONC and ∆OMA are similar by AA similarity
⇒ \(\left(\frac{r}{R}\right)^3\) = \(\frac{1}{27}\) = \(\left(\frac{1}{3}\right)^3\)
⇒ \(\frac{r}{R}\) = \(\frac{1}{3}\) ….. (ii)
From Eqs. (i) and (ii) h1 = \(\frac{1}{3}\) × 40 = \(\frac{40}{3}\) cm
= 13.33 cm
Therefore, h = 40 – h1 = 40 – \(\frac{40}{3}\) = 26.67 cm
Hence, the section is made at a height of 26.67 cm above the base of the cone.
Question 35.
Find the mean, median and mode of the following data:
Class | Frequency |
0 – 10 | 5 |
10 – 20 | 10 |
20 – 30 | 18 |
30 – 40 | 30 |
40 – 50 | 20 |
50 – 60 | 12 |
60 – 70 | 5 |
OR
Find the Arithmetic mean for the following frequency distribution:
Class | Frequency |
25 – 29 | 14 |
30 – 34 | 22 |
35 – 39 | 16 |
40 – 44 | 6 |
45 – 49 | 5 |
50 – 54 | 3 |
55 – 59 | 4 |
Solution:
Mean = \(\frac{\Sigma\left(f_i \times x_i\right)}{\Sigma f_i}\)
= \(\frac{3560}{100}\) = 35.6
Now, N = Σfi = 100
So, \(\frac{N}{2}\) = 50
The cumulative frequency just above 50 is 63.
Hence, the median class is 30 – 40.
So, l = 30, h = 10, f = 30, F = 33 and \(\frac{N}{2}\) = 50
Median = l + \(\frac{\frac{N}{2}-F}{f}\) × h
30 + \(\frac{50-33}{30}\) × 10 = 30 + 5.67
= 35.67
We know, Mode = 3 Median – 2 Mean
⇒ Mode = 3(35.67) – 2(35.6)
1 =107.01 – 71.2
= 35.81.
OR
As this series is an inclusive one, we can make it exclusive by adding 0.5 to the upper limit and subtracting 0.5 from the lower limit of each class interval. Thus, we have
Section – E
Question 36.
Fun Game
In summer vacation Sachin and Rahul invite some friends in a park and they want to play some fun game. So, they consider block in the shapes of a cube with one letter/number written on each face as shown below
While through the cube, they want to know the change of getting some particular number or alphabet.
Based on above information, answer the following questions.
(i) Find the probability of getting an alphabet. (2)
Answer:
(i) When a cube is thrown once, all possible outcomes are
Total number of possible outcomes = 6
Let E1 be the event of getting an alphabet.
Then, the favourable outcomes are A and B. Number of favourable outcomes = 2
∴ P (getting an alphabet) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
Or
Find the probability of getting a prime number. (2)
(ii) Find the probability of getting a consonant. (1)
(iii) When we have no reason to believe that one is more likely to occur than the other, then what is it said? (1)
Answer:
Let E2 be the event of getting a prime number. Then, the favourable outcomes are 2, 3, 5, 7
Number of favourable outcomes = 4
∴ P (getting a prime number) = \(\frac{4}{6}\) = \(\frac{2}{3}\)
(ii) Let E3 be the event of getting a consonant.
Then, the favourable outcome is B.
Number of favourable outcomes = 1
∴ P (getting a consonant) = \(\frac{1}{6}\)
(iii) When we have no reason to believe that one is more likely to occur, then it is said to be the equally likely outcome.
Question 37.
A company wants to make frames as part of a new product that they are launching. This frame is rectangular and cut out of a sheet of steel. In this frame a smaller rectangle will be cut out from the bigger sheet of steel, to lower its weight. The inner dimensions of the steel frame are 6 cm × 11 cm and the width of the frame is cm. The final area of the frame is 28 cm2.
Answer the following questions:
(i) What is the dimensions of outer frame in terms of ‘x’?
(ii) Form a quadratic equation in terms of x, if the final area of the frame is 28 cm2.
(iii) What is the value of ‘k’, if the given equation (9x2 + 6kx + 4 = 0) has equal roots?
OR
What is the width of the frame?
Solution:
(i) Length of outer frame = (11 + 2x) cm
Breadth of outer frame = (6 + 2x) cm
(ii) Here, Area of frame = Area of outer rectangle – Area of Inner rectangle
= (6 + 2x) (11 + 2x) – 6 × 11
= 66 + 22x + 12x + 4x2 – 66
= 34x + 4x2
⇒ 34x + 4x2 = 28
⇒ 2x2 + 17x – 14 = 0
(iii) Here, 9x2 + 6kx + 4 = 0
For equal roots, D = b2 – 4ac = 0
= (6k)2 – 4 × 9 × 4 = 0
⇒ 36k2 – 144 = 0
⇒ k2 = 4
⇒ k = ±2
OR
Based on the equation obtained in part (ii), we get 2x2 + 17x – 14 = 0
.. x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}=\frac{-17 \pm \sqrt{289+112}}{2 \times 2}\)
= \(\frac{-17 \pm \sqrt{401}}{4}=\frac{-17 \pm 20.02}{4}\)
= 0.8 or – 9.3
≈ 1 cm
Question 38.
Sattelite Towers in the Himalayas
The satellite image of Himalaya Mountain is shown below. In this image there are many signal towers are standing. The angle of elevation of the top of a hill from the foot of a tower is 60° and the angle of elevation of the top of the tower with a height 50 m from the foot of the hill is 30°.
Based on above information, answer the following questions.
(i) Find the horizontal distance between hill and tower. (2)
Answer:
(i) Given, AB = 50 m is the height of the tower.
Let CD = h m be the height of the hill.
Then, ∠ACB = 30° and ∠CAD = 60°
In right angled ΔBAC,
cot 30° = \(\frac{A C}{A B}\) [∵ cotθ = \(\frac{\text { base }}{\text { perpendicular }}\)]
Or
Find the height of the hill, if the distance between bottom of hill and tower is 50\(\sqrt{3}\) m. (2)
(ii) Find the distance from foot of tower to the top of the hill. (1)
(iii) Find the distance from foot of the hill to the top of the tower. (1)
Answer:
(ii) ∴ Required distance,
(ii) ∴ Required distance,