Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 3 are designed as per the revised syllabus.
CBSE Sample Papers for Class 10 Maths Standard Set 3 with Solutions
Time Allowed: 3 Hours
Maximum Marks: 80
General Instructions:
- This Question Paper has 5 Sections A-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 π =22/7 wherever required if not stated.
Section – A
(Section A consists of 20 questions of 1 mark each.)
Question 1.
The class interval of a given observation is 15 to 20, then the class mark for this interval will be: (1)
(a) 11.5
(b) 17.5
(c) 12
(d) 14
Answer:
(b) 17.5
Explanation:
Class mark = \(\frac{Upper limit + Lower limit}{2}\)
= \(\frac{20+15}{2}\)
= \(\frac{35}{2}\)
= 17.5
Question 2.
A bag contains 5 red, 8 green and 7 white balls. One ball is drawn at random from the bag. The probability of getting neither a green ball nor a red ball is: (1)
(a) \(\frac{5}{20}\)
(b) \(\frac{3}{20}\)
(c) \(\frac{7}{20}\)
(d) \(\frac{1}{20}\)
Answer:
(c) \(\frac{7}{20}\)
Explanation:
A ball which is neither a green ball nor a red ball is necessary a white ball.
So, required probability = \(\frac{7}{5+8+7}\) = \(\frac{7}{20}\)
Question 3.
The value of sin 30° cos 60° + sin 60° cos 30° is: (1)
(a) 0
(b) 1
(c) 2
(d) 4
Answer:
(b) 1
Explanation: sin 60° = \(\frac{\sqrt{3}}{2}\), sin 30° = \(\frac{1}{2}\),
cos 60° = \(\frac{1}{2}\) and cos 30° = \(\frac{\sqrt{3}}{2}\)
Therefore,
sin30°cos60° + sin60°cos30°
= \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{\sqrt{3}}{2}\) × \(\frac{\sqrt{3}}{2}\)
= \(\frac{1}{4}\) + \(\frac{3}{4}\)
= \(\frac{4}{4}\)
= 1
Question 4.
The outer and inner diameters of a circular ring are 34 cm and 32 cm respectively. The area of the ring is: (1)
(a) 27π cm2
(b) 30π cm2
(c) 33π cm2
(d) 31π cm2
Answer:
(c) 33π cm2
Explanation:
Area of the circlar ring
A = π[172 – 162] cm2 = 33π cm2
Question 5.
If 2 sin 2θ = \(\sqrt{3}\) , then fir 1 the value of θ. (1)
(a) 30°
(b) 60°
(c) 90°
(d) 45°
Answer:
(a) 30°
Explanation:
⇒ 2 sin 2θ = \(\sqrt{3}\)
sin 2θ = \(\frac{\sqrt{3}}{2}\)
= sin 60°
⇒ 2θ = 60°
⇒ θ = 30°
Question 6.
The degree of the polynomial, x5 – x4 + 2 is: (1)
(a) 2
(b) 5
(c) 1
(d) 0
Answer:
(b) 5
Explanation: Degree is the highest power of the variable in any polynomial.
Question 7.
If sin A = y , cos B = 1, 0 < A, B < π, then the value of cot (A + B) is: (1)
(a) \(\frac{\sqrt{3}}{2}\)
(b) \(\frac{1}{2}\)
(c) 0
(d) \(\sqrt{3}\)
Answer:
(d) \(\sqrt{3}\)
Explanation:
Question 8.
If r = 3 is a root of quadratic equation kr2 -kr- 3 = 0, then the value of k is: (1)
(a) \(\frac{3}{2}\)
(b) \(\frac{1}{2}\)
(c) 2
(d) \(\frac{5}{2}\)
Answer:
(b) \(\frac{1}{2}\)
Explanation:
As r = 3 is a root of kr² – kr – 3 = 0,
we have:
9k – 3k – 3 = 0 ⇒ 6k – 3 = 0
or k = \(\frac{1}{2}\)
Question 9.
The degree of the polynomial (x + 1) (x2 – x + x4 – 1) is: (1)
(a) 5
(b) 4
(c) 3
(d) 2
Answer:
(a) 5
Explanation: Given polynomial can be written as,
x5 + x4 + x3 – 2x -1
This is a polynomial of degree 5.
Question 10.
Write the exponent of 3 in the prime factorisation of 1944. (1)
(a) 3
(b) 1
(c) 5
(d) 4
Answer:
(c) 5
Explanation: Prime factorisation of 1944
=2 × 2 × 2 × 3 × 3 × 3 × 3 × 3
= 23 × 35
Hence, the power of 3 is 5.
Question 11.
When placed along a wall, a ladder forms a 60° angle with the ground. If the ladder’s foot is 10 metres from the wall, the length of ladder is: (1)
(a) 4 m
(b) 8 m
(c) \(8 \sqrt{3}\) m
(d) 20 m
Answer:
(d) 20 m
Explanation:
Let AB be the wall AC be the length of the ladder.
In right angled triangle ABC,
cos 60° = \(\frac{BC}{AC}\)
\(\frac{1}{2}\) = \(\frac{10}{AC}\)
AC = 10 × 2 = 20
Therefore, the length of the ladder is 20 m.
Question 12.
The value of fc for which the pair of linear equations kx – 3y = k -2, 12x + ky = k has no solution. (1)
(a) 2
(b) 5
(c) 4
(d) None of these
Answer:
(d) None of these
Explanation:
Given, pair of equations are
kx – 3y = k – 2
and 12x + ky =k
Here, a1 = k, b1 = – 3, c1 = – (k – 2)
and a2 = 12, b2 = k, c2 = – k
For no solution,
\(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) ↑\(\frac{c_1}{c_2}\)
i.e,
\(\frac{k}{12}\) = \(\frac{-3}{k}\) ≠ latex]\frac{-(k-2)}{-k}[/latex]
Then,
k2 = -36(which is not possible)
-3k ≠ k2 – 2k
k2 ≠ -k
k ≠ – 1
Then, k ≠ – 1
Question 13.
How many multiples of 4 lie between 10 and 205? (1)
(a) 49
(b) 80
(c) 45
(d) 46
Answer:
(a) 49
Explanation:
Multiples of 4 between 10 and 205 are 12,16, ………, 204
The above series is an A.P. with a = 12, d = 4 and l = 204
Let, the number of terms in this A.P. be ‘n’.
Then, an = a + (n – 1) d
204 = 12 + (n – 1) × 4
204 = 12 + 4n – 4
4n = = 204 – 8
4n = 196
n = 49
Hence, number of multiples of 4 between 10 and 205 are 49.
Question 14.
The zeros of the polynomial x2 – 3x – m. (m + 3) is: (1)
(a) -2, (m + 1)
(b) (m + 3), 4
(c) (m + 1), 2
(d) – m, (m + 3)
Answer:
(d) -m, (m + 3)
Explanation:
Let, P(x) = x2 – 3x – m (m + 3) x2 – 3x – m(m + 3)
= x2 – [(m + 3) – m] x – m (m + 3)
= x2 – (m + 3) x + mx – m (m + 3)
= x[x – (m + 3)] + m [x – (m + 3)]
= (x + m) [x – (m + 3)]
Thus, zeroes of given polynomial are – m, (m + 3).
Question 15.
In the figure ∠ADC = 90°, BC = 38 cm, CD = 28 cm and BP = 25 cm
The radius of the circle is: (1)
(a) 20 cm
(b) 15 cm
(c) 16 cm
(d) 18 cm
Answer:
(b) 15 cm
Explanation:
Join OR and OS.
In quadrilateral RDSO, all the angles are right angles and adjacent sides are equal as OR = OS. So, RDSO is a square of side equal to the radius of the circle.
DS = DR = OR = OS
Now, BQ = BP gives CQ = BC – BQ
= (38 – 25) cm
= 13 cm
⇒ CR = 13 cm
Also, DR = CD – CR
= (28 – 13) cm
= 15 cm.
Thus, the radius of the circle is 15 cm.
Question 16.
In the figure, ∠D = ∠E and \(\frac{AD}{DB}\) = \(\frac{AE}{EC}\), that ABAC is a/an: (1)
(a) scalene
(b) equilateral
(c) isosceles
(d) right-angled
Answer:
(c) isosceles
Explanation:
Here, ∠E = ∠D
∴ AD = AE ……..(i)
(sides opposite to equal sides are equal)
and \(\frac{AD}{BD}\) = \(\frac{AE}{EC}\)
BD = EC
Adding (i) and (ii), we get
AD + DB = AE + EC
AB = AC
∴ ∆ABC is an isosoceles triangle.
Question 17.
If a line segment AB of length 6 cm is divided internally by a point C in the ratio 3 : 2, then the length of AC is: (1)
(a) 4 cm
(b) 2.5 cm
(c) 2.6 cm
(d) 3.6 cm
Answer:
(d) 3.6 cm
Explanation:
Length of AC = \(\frac{3}{3+2}\) × 6
= 3.6 cm
Question 18.
Find the class marks of the class 45 – 60.
(a) 36.4
(b) 7.5
(c) 52.5
(d) 50
Answer:
(c) 52.5
Explanation:
Class mark of 45-60 = \(\frac{45+60}{2}\)
= \(\frac{105}{2}\)
= 52.5
DIRECTION: in the question number 19 and 20, a statement of assertion (A) is followed by a statement of reason (R).
Choose the correct option as:
(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): (cos4A – sin4 A) is equal to 2cos2 A – 1.
Statement R (Reason): The value of cose decreases as 6 increases. (1)
Answer:
(b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)
Explanation:
cos4 A – sin4 A = (cos2 A – sin2 A) (cos2 A + sin2 A)
= (cos2 A – sin2 A) .1
= cos2 A – (1 – cos2 A)
Question 20.
Statement A (Assertion): If the value of mode and mean is 60 and 66 respectively, then the value of median is 68.
Statement R (Reason): Median = \(\frac{1}{3}\) (mode + 2 mean) (1)
Answer:
(d) Assertion (A) is false but reason (R) is true.
Explanation:
Median = \(\frac{1}{3}\)(mode + 2 mean)
= \(\frac{1}{3}\)(60 + 2 x 66)
= 64
Section – B
(Section B consists of 5 questions of 2 marks each.)
Question 21.
The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q (2, -5) and R (-3,6), find the coordinates of P.
OR
Prove that the points (2, -2), (-2, 1) and (5, 2) are the vertices of a right-angled triangle. (2)
Answer:
Let the coordinates of point P be (2y, y). As P is equidistant from Q and R,
PQ = PR or PQ2 = PR2
i.e. (2 – 2y)2 + (-5 – y)2 = (-3 – 2y)2 + (6 – y)2
⇒ 4y2 – 8y + 4 + 25 + 10y + y2
⇒ 9 + 4y2 + 12y + 36 – 12y + y2
⇒ 2y = 16
⇒ y = 8
Thus, the coordinates of point P are (16, 8).
OR
Let the three given points be A (2, -2), B (-2, 1) and C (5, 2). Then, using distance formula, we have
AB = \(\sqrt{(-2 – 2)^2 + (1 + 2)^2}\) = \(\sqrt{16 + 9}\) = \(\sqrt{5}\)
BC = \(\sqrt{(5 + 2)^2 + (2 – 1)^2}\) = \(\sqrt{49 + 1}\)
= \(\sqrt{50}\) = \(5 \sqrt{2}\)
CA = \(\sqrt{(2 – 5)^2 + (-2 – 2)^2}\) = \(\sqrt{9 + 16}\)
= \(\sqrt{25}\) = 5
AB2 + CA2 = BC2
Hence, ABC is a right-triangle, right-angled at A.
Question 22.
If the HCF (210, 55) is expressible in the form 210 × 5 – 55y, then find y.
OR
Prove that the number 4n, n being a natural number, can never end with the digit 0. (2)
Answer:
Here,
210 = 2 × 3 × 5 × 7
55 = 5 × 11
So, HCF (210, 55) = 5
Thus, 5 = 210 × 5 – 55y
gives 55y = 209 × 5
⇒ 55y = 1045
⇒ y = 19
OR
If 4n ends with digit 0, then it must have 5, 2 as its factors.
But, (4)n = (22)n = 22n, i.e„ the only prime factor of 4n is 2.
Also, we know from the Fundamental Theorem of Arithmetic that the prime factorisation of each number is unique.
∴ 4n can never end with digit 0.
Question 23.
α, β are the zeros of the quadratic polynomial p(x) = x2 – (k + 6)x + 2 (2k – 1). Find the value of k, if α + β = \(\frac{1}{3}\) αβ. (2)
Answer:
Here,
α + β = \(\frac{k+6}{1}\)
and α × β = \(\frac{2(2k-1)}{1}\) = 2(2k – 1)
Since α + β = \(\frac{1}{3}\) αβ
⇒ k + 6= \(\frac{1}{3}\) [2(2k-l)]
⇒ 3k + 18 = 4k – 2
⇒ k = 20
Question 24.
Evaluate:
\(\frac{5 cos^2 60° + 4 sec^2 30° – tan^2 45°}{sin^2 30° + cos^2 30°}\) (2)
Answer:
Question 25.
A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. How much canvas cloth is required to just cover the heap? (2)
Answer:
Base diameter of cone = 24 m
Base radius of cone, r = 12 m
Height of cone, h = 3.5 m
Now, slant height, l = \(\sqrt{r^2 + h^2}\)
= \(\sqrt{12^2 + 3.5^2}\)
= \(\sqrt{144 + 12.25}\)
= \(\sqrt{156.25}\)
= 12.5 m
Area of canvas cloth = C.S.A. of cone
= πrl
= 3.14 × 12 × 12.5
= 471
Hence, area of canvas cloth required is 471 m.
Section – C
(Section C consists of 6 questions of 3 marks each.)
Question 26.
Find the greatest number of 6-digits exactly divisible by 15, 24 and 36.
OR
All the black face cards are removed from a pack of 52 playing cards. The remaining cards are well shuffled and then a card is drawn at random. Find the probability of getting a: (A) face card (B) red card. (2)
Answer:
The required greatest 6-digit number which is a multiple of 24, 15,36 is a multiple of LCM (24,15, 36).
Now, 24 = 23 × 3 ; 15 = 3 × 5 ; 36 = 22 x 32
So, LCM (24,15, 36) = 23 × 32 × 5 = 360
So, the required number is 2777 × 360, i.e. 999720, because 2778 × 360 is a 7-digit number.
OR
In a pack of 52 cards, black face cards are six.
So, the number of remaining cards in the pack is 46.
Thus, 46 cards consists of 26 red cards + 20 black cards (from 1 to 10 each)
Hence, total number of possible outcomes = 46.
Now,
(A) P (face card) = \(\frac{6}{46}\) = \(\frac{3}{23}\)
(B) P (red card) = \(\frac{26}{46}\) = \(\frac{13}{23}\)
Question 27.
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference. (2)
Answer:
Here, a = 5 , l = 45 and S = 400.
Let the AP contains n terms. Then, nth term is the last term.
⇒ a + (n – 1)d = 45
⇒ 5 + (n – l)d = 45
⇒ (n – l)d = 40 ……..(i)
also, Sn = \(\frac{n}{2}\) [2a + (n – 1)d] = 400
\(\frac{n}{2}\) [10 + 40] = 400
⇒ 50n = 800
⇒ n = 16
d = \(\frac{40}{15}\) i.e. \(\frac{8}{3}\)
Question 28.
Points A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of ∆ABC. Find the coordinates of the Q and R on median BE and CF respectively, such that BQ: QE = 2 :1 (2)
Answer:
The points A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of ∆ABC.
Median from B meets AC at E.
Coordinates of E are
\((\left(\frac{x_1+x_3}{2}, \frac{y_1+y_3}{2}\right)=\left(x_4, y_4\right))\)
Let, the coordinates of Q be (x, y)
Similarly, finding the coordinates of R dividing CF in the ratio of 2 :1, we get
R(p,q) = \((\left(\frac{x_1+x_3+x_3}{3}, \frac{y_1+y_3+y_3}{3}\right)\)
So, Q and R are the same points and their coordinates are \((\left(\frac{x_1+x_3+x_3}{3}, \frac{y_1+y_3+y_3}{3}\right)\)
Question 29.
In the given figure, DE || OQ and DF || OR, Prove that EF || OQ.
Let s denote the semi-perimeter of a triangle ABC in which BC = a, CA = b, AB = c. If a circle touches the side BC, CA, AB at D, E, F respectively, prove that BD = s – b. (2)
Answer:
In ∆OQP, DE || OQ
\(\frac{PE}{EQ}\) = \(\frac{PD}{DO}\)
In ∆OPR, DF || OR
\(\frac{PD}{DO}\) = \(\frac{PF}{FR}\)
From (i) and (ii), we get
\(\frac{PE}{EQ}\) = \(\frac{PF}{FR}\)
From ∆PQR,
EF || QR
OR
We know, lengths of tangents drawn from an external point to a circle are equal.
BD = BF, CD = CE, AE = AF.
Now, from the figure, we have
BD = a – CD
= a- CE (∵ CD = CE)
= a – (b – AE)
= a (b – AF) (∵ AE = AF)
= a – b + (c – BF)
= a – b + c – BF
= a- b + c- BD (∵ BD = BF)
Thus, 2BD = a – b + c …(i)
Now, s = \(\frac{a+b+c}{2}\) ……(ii)
From (i) and (ii), we have :
s – b = \(\frac{a+b+c}{2}\) – b
= \(\frac{a-b+c}{2}\) = BD.
Hence Proved.
Question 30.
The rain water from a roof of dimensions 22 m × 20 m drains into a cylindrical vessel having the base of diameter 2 m and height 3.5 m. If the rain water collected from the roof just fill the cylindrical vessel, then find the rainfall in cm. (2)
Answer:
Let, the height of rainfall be ‘h’m.
Then, volume of roof = Volume of cylindrical vessel
l × b × h = πr2h
22 × 20 × h = \(\frac{22}{7}\) × 1 × 1 × 3.5
h = \(\frac{22×3.5}{22×7×20}\)
= \(\frac{0.5}{20}\) = \(\frac{5}{200}\) = \(\frac{1}{40}\)
= 0.025 m
= 2.5 cm
Question 31.
The median of the following frequency distribution is 24. Find the missing frequency f1 and f2. (2)
| Marks | 0-5 | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 | 35-40 | Total |
| Number of Students | 4 | 6 | f1 | 10 | 25 | 2 | 18 | 5 | 100 |
Answer:
| Marks | f | c.f. |
| 0-5 | 4 | 4 |
| 5-10 | 6 | 10 |
| 10-15 | f1 | 10 + f1 |
| 15-20 | 10 | 20 + f1 |
| 20-25 | 25 | 45 + f1 |
| 25-30 | f2 | 45 + f1 + f2 |
| 30-35 | 18 | 63 + f1+ f2 |
| 35-40 | 5 | 68 + f1 + f2 |
| Total | 100 |
Now, 68 + f1 + f2 = 100
f1 + f2 = 100 – 68 = 32
Here, median is 24, so median class is 20-25
Then, l = 20, b = 5, c.f. = 20 + f1
and f = 25, \(\frac{N}{2}\) = 50, Median = 24
Now, f2 = 32 – 10 = 22
Hence, the values of f1 and f2 are 10 and 22 respectively.
Section – D
(Section D consists of 4 questions of 5 marks each.)
Question 32.
The tangent at a point C of a circle with centre O and a diameter AB when extended intersect at P. If ∠PCA = 110°., find the measure of ∠CBA.
OR
The lower window of a house is at a height of 2 m above the ground and its upper window is 4 m above the lower window. At certain instant, the angles of elevation of a balloon from these windows are observed to be 60° and 30°, respectively. Find the height of the balloon from the ground. (5)
Answer:
Given, ∠PCA = 110°
A, O, B, P all are on the same line and P and C are points on the tangent.
AB is a diameter of a circle.
∴ ∠BCA = 90°
(∵ Angle inscribed in a semi-circle)
and ∠OCP = 90°
(∵ Tangent at any point on a circle is perpendicular to the radius)
Now, ∠PCA = ∠PCO + ∠OCA
⇒ 110° = 90° + ∠OCA
⇒ ∠OCA = 20°
Now, in ∆AOC
⇒ AO = OC (Radi of circle)
⇒ ∠OCA = ∠CAO = 20°
In ∆ABC,
⇒ ∠CAB + ∠CBA + ∠BCA = 180°
⇒ 20° + ∠CBA + 90° = 180°
⇒ ∠CBA = 70°
OR
In the figure, A and B represent two windows and E is the position of the balloon. If balloon is ‘h’ metres above the ground, then
in ∆ACE,
Hence, the height of the balloon above the ground is 8 metres.
Question 33.
ABCD is a trapezium in which AB || DC and P, Q are points on AD and BC respectively, such that PQ || DC. If PD = 18 cm, BQ = 35 cm and QC = 15 cm, then find the length of AD. (5)
Answer:
Question 34.
Ajay had some bananas and he divided them into two lots A and B. He sold lot A at the rate of f 2 for 3 bananas and lot B at the rate of ₹ 1 per banana and got a total of ₹ 400. If he had sold lot A at the rate of ₹ 1 per banana and lot B at the rate of f 4 for 5 bananas his total collection would have been ₹ 460. Find the total number of bananas he had. (5)
Answer:
Let, the number of bananas in first lot be x and that in second lot be y.
∴ Total number of bananas = x + y
In the first case, price of x bananas at the 2 per 3 bananas = \(\frac{2x}{3}\) and price of y bananas
at the rate of ₹ 1 per banana = y.
According to the given condition,
\(\frac{2x}{3}\) + y =400
⇒ 2x + 3 y = 1200 ……(i)
In second case, price of x bananas at the rate of ₹ 1 per bananas = x and price of y bananas
at the rate of ₹ 4 per 5 bananas = \(\frac{4}{5}\)y
According to given condition,
x + \(\frac{4}{5}\)y = 460
⇒ 5x + 4y = 2300 …….(ii)
Multiplying (i) by 5 are (ii) by 2, and then subtracting them.
7y = 1400
⇒ y = 200
Put y = 200, in equation (i) we get
2x + 3 × 200 = 1200
⇒ x = 300
x + y = 300 + 200 = 500
Thus, Ajay had total of 500 bananas.
Question 35.
Agam received an overall of 30 marks in math and English on a class quiz. The total of his marks in Mathematics and English would have been 210 if he had got 2 more in Mathematics and 3 less in English. Find out what he scored in the two subjects.
OR
There are 12 balls in a box, and x of them are black. What is the probability that one ball will be picked at random from the box , will be a black ball? If there are now six additional black balls in the box, the probability of drawing a black ball is now double of what it was before. Find x. (5)
Answer:
Let Agam’s marks in Mathematics = x
Let Agam’s marks in English = 30 – x
If, he had got 2 marks more in Mathematics, his marks would be = x + 2
If, he had got 3 marks less in English, his marks in English would be = 30 – x – 3
= 27 – x
According to given condition:
(x + 2) (2 7 – x) = 210
⇒ 27x – x2 + 54-2x = 210
=> x2 – 25x + 156 = 0
Comparing quadratic equation x2 – 25x + 156 = 0 with general form ax2 + bx + c = 0,
We get a = 1, b = -25 and c = 156
Therefore, Agam’s marks in Mathematics = 13 or 12
Agam’s marks in English = 30 – x
= 30-13
= 17
Or
Agam’s marks in English = 30 – x
= 30 – 12
= 18
Therefore, his marks in Mathematics and English are (13,17) or (12,18).
OR
There are 12 balls in the box.
Therefore, total number offavourable outcomes = 12
The number offavourable outcomes = x
Therefore, Pi = P (getting a black ball) = \(\frac{x}{12}\)
If 6 more balls put in the box, then
Total number offavourable outcomes = 12 + 6 = 18
And number offavourable outcomes = x + 6
P2 = P (getting a black ball) = \(\frac{x+6}{18}\)
According to question, P2 = 2P1
⇒ \(\frac{x+6}{18}\) = 2 × \(\frac{x}{12}\)
⇒ \(\frac{x+6}{18}\) = \(\frac{x}{6}\)
⇒ 6(x + 6) = 18x
⇒ 18x – 6x = 36
⇒ 12x = 36
⇒ x = \(\frac{36}{12}\)
⇒ x = 3
Section – E
(Case Study Based Questions)
(Section E consists of 3 questions. ALL are compulsory.)
Question 36.
The diagram shows a grid of squares. A button is placed on one of the squares. A fair dice is thrown. If 1,2,3 or 4 is thrown, the button is moved one square to the left.
If 5 or 6 is thrown, the button is moved one square to the right.
On the basis of the above information, answer the following questions:
(A) The button is placed on square X. the dice is thrown once. Then find, the probability that the button is moved to the right. (1)
(B) On the other occasion, the button is placed on square Y. The dice is thrown once and the button is moved. The die is thrown a second time and the button is moved again. Find the probability that the button is moved at P and button is moved at Y.
OR
Find the probability that the button is moved at P, Q or Y. (2)
(C) Find the probability that the button is moved at Q. (1)
Answer:
(A) P(right) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
(B) getting (1, 2, 3 or 4 twice) = \(\frac{2}{6}\) × \(\frac{2}{6}\)
= \(\frac{1}{3}\) × \(\frac{1}{3}\)
= \(\frac{1}{9}\)
As per the situation given, if the button is at Y and it is moved twice, then it will reach to Y again if it is first move to left then to right or vice-versa.
P(Y) = \(\frac{2}{6} \times \frac{4}{6}+\frac{4}{6} \times \frac{2}{6}=\frac{16}{36}=\frac{4}{9}\)
By adding the probabilities obtained in (B), we get
Probability (P, Q or Y) = \(\frac{1}{9}\) + 0 + \(\frac{4}{9}\)
= \(\frac{5}{9}\)
(C) As per the situation given in part (B), if the button is at Y and it is moved twice, then there is no possibility that it reach to Q.
Question 37.
Earth is excavated to make a railway tunnel. The tunnel is a cylinder of radius 5 m and length 450 m.
A Level surface is Laid inside the tunnel to carry the railway lines. The Diagram 1 shows the circular cross – section of the tunnel. The level surface is represented by AB, the centre of the circle is O and ∠AOB = 90°. The space below AB is filled with rubble (debris from the demolition buildings).
Steel girders are erected above the tracks to strengthen the tunnel. Some of these are shown in Diagram 2. The girders are erected at 6 m intervals along the length of the tunnel, with one at each end.
On the basis of the above information, answer the following questions:
(A) Find the volume of earth removed to make the tunnel. (1)
(B) Find the area of AAOB shown in Diagram. (1)
(C) Find the length of each girder.
OR
Find total length of steel required in the 450 m length of tunnel and how many girders are errected? (2)
Answer:
(A) Volume of tunnel= πr2h
= π × 5 × 5 × 450
= 11250π cu cm
(B) Area of ∆AOB = \(\frac{1}{2}\) × AO × OB
= \(\frac{1}{2}\) × 5 × 5
= 12.5 sq. m
(C) As steel girders are in the shape of a circle or a sector of the circle with centra/angle 270°.
So, circumference of the circle with central angle 270°
= \(\frac{270°}{360°}\) × 2πr
= \(\frac{3}{4}\) × 2πr
= 1.5π × 5
= 7.5π cm
OR
Total length = 7.5 π × 76
= 570 π m
No. of girders = \(\frac{450}{6}\)
= 75
Total girder =75 + 1 (one at last)
= 76
Question 38.
A radio mast PQ , of height ‘h’ metres, is standing vertically on the horizontal ground. From A, the angle of elevation of the top of the mast is found to be 45° . On moving 50 m up a slope of 15°, the angle of elevation of P is found to be 75° from B, The horizontal through B is BC.
On the basis of the above information, answer the following questions:
(A) Find the measure of ∠APC. (1)
(B) Find the measure of ∠BPC. (1)
(C) Find the length BP.
OR
Find the length AP and also find the approximate value of h. (2)
Answer:
(A) In ∆APQ
∠A + ∠P + ∠Q = 180°
45° + ∠P + 90° = 180°
⇒ ∠P = 180° – 90° – 45°
= 180° – 135°
= 45°
(B) In ∆BPC
∠B + ∠BPC + ∠C = 180°
⇒ 75° + ∠BPC + 90° = 180°
⇒ ∠BPC = 180° – 165°
= 15°
(C) In ∆APQ
tan 45° = \(\frac{PQ}{AQ}\)
PQ = AQ = h
QQ’ = AQ – AQ’
= h – 49
BC = QQ’ = h- 49
Now, in ∆PBC
tan 75° = \(\frac{PC}{BC}\)
3.73 = \(\frac{h-13}{h-49}\)
3.73 h – 182.77 = h – 13
2.73 h = 169.77
h = 62.186
sin 75° = \(\frac{PC}{PB}\)
1 = \(\frac{62.2-13}{PB}\)
PB = 49.2 = 50 m
OR
In ∆APQ
sin 45° = \(\frac{h}{AP}\)
AP = \(\frac{h}{\frac{1}{\sqrt{2}}}\) = \(\sqrt{2} h\)
The approximate height h, as calculated in part (C) is 50 m.