Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Basic with Solutions Set 3 are designed as per the revised syllabus.
CBSE Sample Papers for Class 10 Maths Basic Set 3 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 Multiple Choice Questions (MCQs) carrying 1 mark each.
- Section B has 5 Short Answer-I (SA-I) type questions carrying 2 marks each.
- Section C has 6 Short Answer-II (SA-II) type questions carrying 3 marks each.
- Section D has 4 Long Answer (LA) type questions carrying 5 marks each.
- Section E has 3 sourced based/Case Based/passage 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 Qs of 2 marks, 2 Qs of 3 marks and 2 Questions of 5 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 (20 Marks)
Question 1.
The points (a, a), (-a, -a) and (\(-\sqrt{3 a}\),\(-\sqrt{3 a}\)) are the vertices of: 1
(a) equilateral triangle
(b) scalene triangle
(c) isosceles triangle
(d) right-angled isosceles triangle
Answer:
(a) equilateral triangle
Explanation: Let name the points as A (a, a), B (-a, -a) and C (\(-\sqrt{3 a}, \sqrt{3 a}\)),
By distance formula, distance between points (x1, y1) and (x2, y2).
We get, AB = BC = AC = 2\(\sqrt{2 a}\) units
∴ ΔABC is an equilateral triangle.
Question 2.
The value of c for which pair of linear equations cx – y = 2 and 6x – 2y = 4 will have infinitely many solutions is: 1
(a) 3
(b) 5
(c) -1
(d) 0
Answer:
(a) 3
Explanation: We have
cx – y = 2 and
6x – 2y = 4
For infinitely many solutions
\(\frac{c}{6}=\frac{-1}{-2}=\frac{2}{4}\)
\(\frac{c}{6}=\frac{1}{2}\)
⇒ c = 3
Question 3.
The distance between (7, 0) and (1, – 8) is: 1
(a) 7 units
(b) 10 units
(c) 8 units
(d) 9 units
Answer:
(b) 10 units
Explanation: Distance between (7, 0) and (1.-8)
[By distance formula]
= \(\sqrt{(1-7)^2+(-8-0)^2}\)
= \(\sqrt{(-6)^2+(-8)^2}\)
= \(\sqrt{36+64}\)
= \(\sqrt{100}\) = 10 units
Question 4.
School divides its students into 5 houses A,B, C, D and E. Class X A has 23 students, 4 from house A, 8 from house B, 5 from house C, 2 from house D and rest from house E. A single student is chosen at random to become the monitor of the class. The probability that the chosen student is not from houses A, B and C is: 1
(a) \(\frac{4}{23}\)
(b) \(\frac{6}{23}\)
(c) \(\frac{8}{23}\)
(d) \(\frac{17}{23}\)
Answer:
(b) \(\frac{6}{23}\)
Explanation: Total students in class X = 23
Students from house A = 4
Students from house B = 8
Students from house C = 5
Students from house D = 2
Students from house E = 23 – (4 + 8 + 5 + 2) = 4
Total students in house D and E = 2 + 4
= 6
Required probability = \(\frac{6}{23}\)
Question 5.
A line of length 10 units has one end at the point (-3, 2). If the ordinate of the other end is 10, then the abscissa will be: 1
(a) 9 or 3
(b) 4 or -3
(c) -3 or 5
(d) -9 or 3
Answer:
(d) -9 or 3
Explanation: Let A and B denote the points (-3, 2) and (a, 10), where a is to be determined.
Here, AB = 10 units.
So, \(\sqrt{(a+3)^2+(10-2)^2}\) = 10
Squaring on both sides, we have
⇒ (a + 3)2 + 64 = 100
or (a + 3)2 = 36
⇒ a + 3 = ±6
⇒ a = -9 or 3.
Question 6.
If ΔABC ~ ΔDEF, such that ∠A = 47° and ∠E = 83°, then the value of ∠C is: 1
(a) 50°
(b) 60°
(c) 45°
(d) 90°
Answer:
(a) 50°
Explanation: Since ΔABC ~ ΔDEF.
∠A = ∠D, ∠B = ∠E and ∠C = ∠F
⇒ ∠A = 47° and ∠B = 83°
So, in ΔABC, ZC = 180° – (∠A + ∠B)
= 180° – (47° + 83°)
= 50°
Question 7.
The zeros of 2x2 – x – 45 are respectively: 1
(a) 5, \(\frac{1}{2}\)
(b) 2, –\(\frac{9}{2}\)
(c) 5, –\(\frac{9}{2}\)
(d) 5, -9
Answer:
(c) 5, –\(\frac{9}{2}\)
Explanation: Given,
2x2 – x – 45 = 2x2 – 10x + 9x – 45
= 2x(x – 5) + 9(x – 5)
= (2x + 9)(x- 5)
2x + 9 = 0,x – 5 = 0
x = –\(\frac{9}{2}\), x = 5
So, its zeros are 5 and –\(\frac{9}{2}\)
Question 8.
If cot A + \(\frac{1}{\cot A}\) = 2, then the value of cot2 A + \(\frac{1}{\cot ^2 A}[latex], is: 1
(a) 2
(b) 4
(c) 1
(d) 5
Answer:
(a) 2
Explanation:
Given, cot A + [latex]\frac{1}{\cot A}\) = 2
On squaring both sides, we get
Question 9.
The pairs of equations x + 2y – 5 = 0 and – 4x – 8y + 20 = 0 have: 1
(a) unique solution
(b) exactly two solutions
(c) infinitely many solutions
(d) no solution
Answer:
(c) infinitely many solutions
Explanation:
\(\frac{a_1}{a_2}=\frac{1}{-4}\)
\(\frac{b_1}{b_2}=\frac{2}{-8}=\frac{1}{-4}\)
\(\frac{c_1}{c_2}=\frac{-5}{20}=-\frac{1}{4}\)
This shows:
\(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Therefore, the pair of equations has infinitely many solutions.
Question 10.
If sinA = \(\frac{1}{2}\),then the value of(cotA – cosA)is: 1
(a) √3
(b) \(\frac{1}{\sqrt{3}}[latex]
(c) [latex]\frac{3}{2}\)
(d) \(\frac{\sqrt{3}}{2}\)
Answer:
(d) \(\frac{\sqrt{3}}{2}\)
Explanation:
sin A = \(\frac{1}{2}\) gives A = 30°
So, cot A – cos A = cot 30° – cos 30°
= √3 – \(\frac{\sqrt{3}}{2}\) = \(\frac{\sqrt{3}}{2}\)
Question 11.
If the circumference of a circle and the perimeter of a square are equal, then 1
(a) area of the circle = area of the square
(b) area of the circle > area of the square ,
(c) area of the circle < area of the square (d) nothing definite can be said about the relation between the areas of the circle and square Answer: (b) area of the circle > area of the square
Explanation: Let circumference of a circle be c
2πr = c
⇒ r = \(\frac{c}{2 \pi}\)
Perimeter of a square = c
⇒ 4a = c
⇒ a = \(\frac{c}{4}\)
\(\frac{\text { Area of the circle }}{\text { Area of the square }}=\frac{\pi r^2}{a^2}=\frac{\pi\left(\frac{c}{2 \pi}\right)^2}{\left(\frac{c}{4}\right)^2}\)
= \(\frac{\pi \times c^2}{4 \pi^2} \times \frac{16}{c^2}\)
= \(\frac{4}{\pi}=\frac{28}{22}=\frac{14}{11}\)
Question 12.
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 45°, The height of the tower (in metres) is: 1
(a) 15
(b) 30
(c) 30√3
(d) 10√3
Answer:
(b) 30
Explanation: Let AB be the tower and P be the point on the ground.
In ΔABP, \(\frac{\mathrm{AB}}{\mathrm{BP}}\) = tan 45°
⇒ \(\frac{\mathrm{AB}}{30}\) = 1
⇒ AB = 30 m
Question 13.
The total surface area of the hemisphere of radius V is: 1
(a) 3πr2
(b) 5πr2
(c) 2πr2
(d) 4πr22
Answer:
(a) 3πr2
Explanation: Total surface area of hemisphere
= \(\frac{4 \pi r^2}{2}\) + πr2
= 2πr2 + πr2 = 3πr2
Question 14.
What is the upper limit of the median class for the given below distribution? 1
Class Interval | Frequency |
0 – 5 | 13 |
5 – 10 | 10 |
10 – 15 | 15 |
15 – 20 | 8 |
20 – 25 | 11 |
(a) 14
(b) 10
(c) 15
(d) 20
Answer:
(c) 15
Explanation:
Here, N = 57
So, \(\frac{N}{2}\) = 28.5
Cumulative frequency just greater than 28.5 is 38, which belongs to 10 – 15.
So, the median class is 10 – 15.
Thus, its upper limit is 15.
Question 15.
Two coins are tossed simultaneously. The probability of getting at least one head is: 1
(a) \(\frac{3}{4}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{2}{5}\)
(d) \(\frac{5}{4}\)
Answer:
(a) \(\frac{3}{4}\)
Explanation: Possible outcomes
= {HH, HT, TH, TT} = 4
Favourable outcomes = {HH, HT, TH} = 3
So, required probability = \(\frac{3}{4}\)
Question 16.
The common difference of the AP., √3, \(\sqrt{12}\), \(\sqrt{27}\) ……………….. is: 1
(a) \(\sqrt{12}\)
(b) √4
(c) √3
(d) 2√3
Answer:
(c) √3
Explanation: Here the terms of the A.P. are:
√3, 2√3, 3√3,…….
So, the common difference = (2√3 – √3),
= √3.
Question 17.
The nature of roots of the quadratic equation ax2 – 3bx – 4a = 0 (a ≠ 0) is: 1
(a) equal
(b) real and distinct
(c) unreal and equal
(d) unreal
Answer:
(b) real and distinct
Explanation: Here, given quadratic equation is:
ax2 – 3bx – 4a = 0
We have A = a, B = – 3b and C = -4a
Discriminant D,
B2 – 4AC = (-3b)2 – 4(a)(-4a)
= 9b2 + 16a2 >0
So, the roots are real and distinct.
Question 18.
The arithmetic mean of the following frequency distribution is: 1
Class interval | Frequency |
0 – 10 | 5 |
10 – 20 | 18 |
20 – 30 | 15 |
30 – 40 | 16 |
40 – 50 | 6 |
(a) 25
(b) 37
(c) 45
(d) 22
Answer:
(a) 25
Explanation: Now, the frequency distribution table from the given data can be drawn as:
Question 19.
Assertion (A): The volume of a hall, which is 5 times as high as it is broad and 8 times as long as it is high, is 12.8m3. The breadth of the hall is 25 cm.
Reason (R): The total surface area of a cuboid of length (l), breadth (b) and height (h) is 2 [lb + bh + lh]. 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:
(d) Assertion (A) is false but reason (R) is true
Explanation: Let breadth of a hall be x and height = 5x
Length = 8 × 5x
= 40x
∴ Volume of hall = x × 5x × 40x = 200x3
But, volume of hall = 12.8 m3
∴ 200x3 = 12.8 m3
⇒ x3 = \(\frac{12.8}{200}=\frac{8}{125}\)
⇒ x = 0.4 m = 40 cm
∴ Assertion (A) is false but reason (R) is true.
Question 20.
Assertion (A): The simplest form of \(\frac{1095}{1168}\) is \(\frac{15}{16}\)
Reason (R): For finding the simplest form of a fraction the numerator and denominator are divided by their HCF. 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)
Explanation: Here, HCF of 1095 and 1168 is 73.
\(\frac{1095 \div 73}{1168 \div 73}=\frac{15}{16}\), is the simplest form of fraction.
Section – B (10 Marks)
Question 21.
Write the prime factorisation of 8190.
OR
Find the HCF of (23 × 32 × 51), (22 × 33 × 52) and (24 × 31 × 52 × 7). 2
Answer:
The prime factorisation of 8190 is:
8190 = 2 × 3 × 3 × 5 × 7 × 13.
OR
The given factors are (23 × 32 × 51), (22 × 33 × 52) and (24 × 31 × 52 × 7)
Now, HCF = Product of each prime factors with smallest power.
HCF = 22 × 31 × 51 /.e. 60.
Question 22.
Find the area of the shaded region in the given figure, if ABCD is a square of side 20 m and APD and BPC are semicircles. (Use π = \(\frac{22}{7}\)) 2
Answer:
Area of semicircle BPC
\(\frac{1}{2}\)πr2 = \(\frac{1}{2}\)π × (10)2
= 50π m2
Similarly
Area of semicircle APD = 50π m2
Total area of semicircle BPC and APD
= (50π + 50π)m2
= 100π m2 = 100 × \(\frac{22}{7}\)
= \(\frac{2200}{7}\)m2
Side of square = 20 m
∴ Area of the square ABCD
= 20 × 20 = 400 m2
∴ Area of the shaded region = (Area of square Area of both semircLes BPC and APD)
= (400 – \(\frac{2200}{7}\))m2
= \(\frac{600}{7}\)m2
Question 23.
In a right angle triangle ABC, right-angled at B, if sin (A – C) = \(\frac{1}{2}\) find the measures of angles A and C.
If sin θ = \(\frac{2 m n}{m^2+n^2}\), find the value of \(\frac{\sin \theta \cot \theta}{\cos \theta}\). 2
Answer:
Since, sin (A – C) = \(\frac{1}{2}\)
A – C = 30° …..(i)
But, A + C = 90° ….. (ii) (as, A + B + C = 180°)
On solving equations (i) and (ii), we get
So, C = 30° and A = 60°
OR
Question 24.
A path of width 7 m runs around outside a circular park whose radius is 18 m. Find the area of the path. 2
Answer:
Given, radius = 18 m
Area of the path = Area of the outer circle – Area of the inner circle
= [π(18 + 7)2 – π(18)2] sq. m
= (625π – 324π) sq. m
= 301π sq. m, or 946 sq. m.
Question 25.
A die is thrown once. Find the probability of getting:
(A) a prime number greater than 3
(B) an even prime number greater than 3. 2
Answer:
(A) P (prime number greater than 3) = \(\frac{1}{6}\)
[∵ Only 5 is the prime number greater than 3]
(B) P (even prime number greater than 3) = \(\frac{0}{6}\) i.e., 0.
[∵ Only even prime number is 2, which is not greater than 3]
Section – C (18 Marks)
Question 26.
Amita, Suneha and Rajiv start preparing cards. In order to complete one card, they take 10, 16 and 20 minutes respectively, If all of them started together, after what time will they started preparing a new card together?
OR
Show that 5 + 2√7 is an irrational number, where √7 is given to be an irrational number. 3
Answer:
Given, Amita, Suneha and Rajiv takes 10, 16 and 20 minutes respectively to complete one card.
Prime factorisation of 10,16 and 20 is 10 = 2 × 5, 16 = 2 × 2 × 2 × 2 = 24 and 20 = 2 × 2 × 5 = 22 × 5
∴ Time after which they start preparing a new card together = LCM (10, 16 and 20)
= 2 × 2 × 2 × 2 × 5
= 24 × 5 = 80 minutes.
OR
Suppose 5 + 2√7 is a rational number.
∴ We can find two integers a, b ( b ≠ 0) such that 5 + 2√7 = \(\frac{a}{b}\), where a and b are co-prime integers.
⇒ 2√7 = \(\frac{a}{b}\) – 5
⇒ √7 = \(\frac{1}{2}\)[\(\frac{a}{b}\) – 5]
⇒ √7 is a rational number
[∵ a, b are integers, so \(\frac{1}{2}\)[\(\frac{a}{b}\) – 5] number] is a rational number
But this contradicts the fact that √7 is an irrational number. Hence, our assumption is wrong. Thus, 5 + 2√7 is an irrational number.
Question 27.
In the current scenario, government has policy to make a highway as soon as possible. So, the constructor used the heavy machine instead of using man labour.
A machine operator dug out a field in the shape of circular having radius 3 m and some depth 10 m.
(A) Find the volume of the soil.
(B) Find the total inner surface area of the digging field. 3
Answer:
(A) We have radius and height of cylindrical shape r = 3 m and h = 10 m.
Volume of the soil = Volume of cylinder
= πr2h
= 3.14 × (3)2 × 10
= 90 × 3.14
= 282.6 m3
(B) Total inner surface area of the digging field = CSA of digging field + Area of bottom surface
= 2πrh + πr2
= 2 × 3.14 × 3 × 140 + 3.14 × (3)2
= 188.4 + 28.26
= 216.66 m2
Question 28.
ABCD is a quadrilateral such that ∠D = 90°, A circle C(O, r) touches the sides AB, BC, CD and DA at P, Q, R and S respectively. If BC = 38 cm, CD = 25 cm and BP = 27 cm, then find the value of r.
OR
In the given figures, ∠DAB = 90°, AD = 40 cm, CD = 35 cm and CQ = 18 cm. Find the radius of the circle.
Answer:
Since, tangents drawn from an external point to a circle are equal.
∴ AP = AS, BP = BQ, CQ = CR,
DS = DR
BP = 27 cm
⇒ BQ = 27 cm
Now, CQ = BC – BQ
= 38 – 27
= 11 cm
⇒ CR = 11 cm
DR = DC – CR
= 25 – 11 = 14 cm
Now, OR ⊥ CD and OS ⊥ AD
[∴ Tangent at any point of a circle is perpendicular to the radius through the point of contact.] and OS = OR [Radii of same circle]
Also, given that ∠D = 90°
∴ ORDS is a square of side 14 cm.
Thus, radius of circle is 14 cm.
OR
Join OP and OS.
Since, length of tangents drawn from an external point to a circle are equal
∴ AP = AS [Tangents from A] …… (i)
CQ = CR [Tangents from C] ,…… (ii)
DR = DS [Tangents from D] ……..(iii)
Now,
CQ = CR
⇒ CR = 18 cm [∵ CQ = 18 cm (given)]
DR = DC – CR = 35 – 18 = 17 cm [∵ CD = 35 cm (given)]
∴ DS = 17 cm [Using (iii)]
AS = AD – DS = 40 – 17 = 23 cm [∵ AD = 40 cm (given)]
∴ AP = 23 cm [Using (i)]
Now, OP ⊥ AP and OS ⊥ AS
[∵ Tangent at any point of circle is perpendicular to the radius through the point of contact]
Also, ∠DAB = 90° [Given]
Since, all angles are of 90° and adjacent sides are equal in APOS, So APOS is a square.
OP = OS = AS = AP
= 23 cm.
Thus, radius of the circle is 23 cm.
Question 29.
A parallelogram, PQRS is inside a DABC in which AB || PS. Prove that OC || SR.
Answer:
Given: PQRS is Parallelogram in which AB || PS
To Prove: OC || SR
Proof: In ΔOAB and ΔOPS
PS || AB (given) …… (i)
∴ ∠1 = ∠2
[Corresponding pair of angles]
∠3 = ∠4
[Corresponding pair of angles]
∴ ΔOPS ~ ΔOAB
[by AA-similarity criterion]
\(\frac{O P}{O A}=\frac{O S}{O B}=\frac{P S}{A B}\) ………. (iii)
PQRS is a parallelogram, so PS || QR …….. (iii)
⇒ QR || AB [from (i) and (iii]
In ΔCQR and ΔCAB
∠5 = ∠CAB
[Corresponding angle]
∠6 = ∠CBA
[Coresponding angle]
∴ ΔCQR ~ ΔCAB [by AA similarity]
\(\frac{C Q}{C R}=\frac{C R}{C B}=\frac{Q R}{A B}\)
[Corresponding sides of similar triangles]
∴ PS = QR
∴ \(\frac{P S}{A B}=\frac{C R}{C B}=\frac{C Q}{C A}\) ….. (iv)
⇒ \(\frac{C R}{C B}=\frac{O S}{O B}\) [from (ii) and (iv)]
So, by converse of BPT, SR || OC.
Hence, proved.
Question 30.
If sin A = m sin B and tan A = n tan B then show that (n2 – 1) cos2 A = m2 – 1. 3
Answer:
Given: tan A = n tan B
and sin A = m sin B
To Prove:
(n2 – 1) cos2 A = m2 – 1
Proof sin A = m sin B (given) ……… (i)
tan A = n tan B
\(\frac{\sin A}{\cos A}\) = n\(\frac{\sin B}{\cos B}\) ………… (ii)
On substituting sin B from eq. (i), we get
⇒ cos B = \(\frac{n}{m}\) cos A ………. (iii)
and sin2A = m2 sin2B
⇒ (1 – cos2A) = m2 (1 – cos2B)
Substituting cos B from eq. (iii), we get
1 – cos2A = m2 (1 – \(\frac{n^2}{m^2}\)cos2A)
⇒ 1 – cos2A = m2 (\(\frac{m^2-n^2 \cos ^2 \mathrm{~A}}{m^2}\))
1 – cos2A = m2 – n2 cos2A
n2 cos2A – COS2A = m2 – 1
cos2A (n2 – 1) = m2 – 1
Hence, proved.
Question 31.
ABCD is a trapezium with AB || DC. If ΔAED ~ ABEC, then prove that AD = BC.
Answer:
For triangles AEB and CED, we have:
∠EAB = ∠ECD
and ∠EBA = ∠EDC
[Alternate angles as AB || DC]
∴ By AA similarty criterion, we have:
ΔAEB ~ ΔCED
⇒ \(\frac{A E}{C E}=\frac{A B}{C D}=\frac{E B}{E D}\)
⇒ \(\frac{\mathrm{AE}}{\mathrm{EB}}=\frac{\mathrm{CE}}{\mathrm{ED}}\) ……. (i)
It is also given that, ΔAED ~ ΔBEC
So, \(\frac{\mathrm{AE}}{\mathrm{BE}}=\frac{\mathrm{ED}}{\mathrm{EC}}=\frac{\mathrm{AD}}{\mathrm{BC}}\) …… (ii)
From (i) and (ii), we have:
\(\frac{C E}{E D}=\frac{E D}{E C}\)
⇒ EC2 = ED2
or EC = ED.
Substituting EC = ED in (ii), we have:
AD = BC
Hence, proved.
Section – D (20 Marks)
Question 32.
Solve the pair of equations graphically:
4x – y = 5; x + y = 5.
OR
Form a pair of linear equations for the following problems and find their ‘ solution by substitution method.
(A) The cost of a taxi in a city consists of a fixed charge and a charge for the distance travelled. The cost for a 10 km travel is ₹ 105, while for a 15 km journey, the cost is ₹ 155. What are the fixed charges and the km charged? How much will it cost someone to drive 25 km?
(B) For ₹ 3800, the cricket team’s coach purchases 6 balls and 7 bats. Then he spends 1750 for 3 bats and 5 balls. Find out how much each ball and bat costs. 5
Answer:
4x – y = 5
x | 1 | 2 |
y | -1 | 3 |
x + y = 5
x | 3 | 2 |
y | 2 | 3 |
The solution is x = 2, y = 3.
OR
(A) Let fixed charge = ₹ x and let charge for every km = ₹ y
According to given conditions, we have
x + y = 105 ………… (i)
x + 15y = 155 …………… (ii)
Using equation (i), we can say that
x = 105 – 10y
Putting this in equation (ii), we get
105 – 10y + 15y = 155
⇒ 5x = 50
⇒ y = 10
Putting value of y in equation (i), we get
x + 10(10) = 105
⇒ x = 105 – 100
⇒ x = 5
Therefore, fixed charge = ₹ 5
and charge per km = ₹ 10
To travel distance of 25 km, person will have to pay = ₹ (x + 25y)
= ₹ (5 + 25 × 10)
= ₹ (5 + 250)
= ₹ 255
(B) Let cost of each bat = ₹ x
and cost of each ball = ₹ y
According to given conditions, we have
7x + 6y = 3800 …………. (i)
And, 3x + 5 y = 1750 …………. (ii)
Using equation (i), we can say that
7x = 3800 – 6y
⇒ x = \(\frac{3800-6 y}{7}\)
Putting this in eq. (ii), we get “3800-6y’
3(\(\frac{3800-6 y}{7}\)) + 5y = 1750
⇒ (\(\frac{11400-18 y}{7}\)) + 5y = 1750
⇒ \(\frac{5 y}{1}-\frac{18 y}{7}=\frac{1750}{1}-\frac{11400}{7}\)
⇒ \(\frac{35 y-18 y}{7}=\frac{12250-11400}{7}\)
⇒ 17 y = 850
⇒ y = 50
Putting value of y in (ii), we get
3x + 250 = 1750
⇒ 3x = 1500
⇒ x = 500
Therefore, cost of each bat = ₹ 500 and
cost of each ball = ₹ 50.
Question 33.
Rahul got a total of 28 marks in math and science on a class quiz. The product of his scores would have been 180 if he had received 3 more in mathematics and 4 less in science. Find out what he scored in the two subjects.
OR
In a flight of 2800 km, an aircraft was slowed down due to bad weather, Its average speed is reduced by 100 km/h and time increased by 30 minutes, Find the original duration of the flight. 5
Answer:
Let Rahul has obtained ‘x’ marks in mathematics and ‘y’ marks in science.
Then by the problem,
x + y = 28 ……. (i)
If Rahul would have got 3 marks more in mathematics, then Rahul would have got (x + 3) in mathematics and if Rahul would have got 4 marks less in science then Rahul would have got (y – 4) marks in science.
A.T.Q.
(x + 3) (y – 4) = 180
(x + 3) (28 – x – 4) = 180
⇒ (x + 3) (24 – x) = 180
⇒ 72 + 21x – x2 = 180
⇒ x2 – 21x + 108 = 0
⇒ x2 – 12x – 9x + 108 = 0
⇒ x(x – 12) – 9(x – 12) = 0
⇒ (x – 12) (x – 9) = 0
⇒ x = 12, 9
Then, y = 16, 19
Hence, Rahul obtained 12 marks in mathematics and 16 marks in science.
Or Rahul obtained 9 marks in mathematics and 19 marks in science.
OR
Let the original speed of the plane be x km/hr.
∴ Reduce speed of the plane = (x – 100) km/hr.
Time taken by plane to reach its destination at orignial speed = \(\frac{2800}{x}\) hr
Time taken by the plane to rech its destination at reduced speed \(\frac{2800}{x-100}\) hr
∴ According to question, \(\frac{2800}{x-100}-\frac{2800}{x}\) = \(\frac{1}{2}\)
⇒ \(\frac{2800 x-2800(x-100)}{x(x-100)}\) = \(\frac{1}{2}\)
⇒ 2800 × 100 × 2 = x2 – 100x
⇒ x2 – 100x – 560000 = 0
⇒ x2 – 800x + 700x – 560000 = 0
⇒ (x – 800) (x + 700) = 0
⇒ x = 800
or x = -700
∴ x = 800
[∵ Speed cannot be negative]
Original speed of flight = 800 km/hr
∴ Original duration of flight
= \(\frac{2800}{800}\)
= \(\frac{7}{2}\)
= 3\(\frac{1}{2}\) hrs
Question 34.
If the median of the distribution given below is 28.5, find the values of x and y.
Class Interval | Frequency |
0 – 10 | 5 |
10 – 20 | x |
20 – 30 | 20 |
30 – 40 | 15 |
40 – 50 | y |
50 – 60 | 5 |
Total | 60 |
Answer:
With the given frequency distribution table, we first prepare a cumulative frequency distribution table as given below:
Class interval | Frequency | Cum. frequency |
0 – 10 | 5 | 5 |
10 – 20 | x | 5 + x |
20 – 30 | 20 | 25 +x |
30 – 40 | 15 | 40 + x |
40 – 50 | y | 40 + x + y |
50 – 60 | 5 | 45 + x + y = 60 |
Sinc median given is 28.5, the median class is 20 – 30.
For this class,
l = 20, h = 10, \(\frac{n}{2}\) = 30, f = 20 and cf = 5 + x.
According to the formula,
Median = l + (\(\frac{\frac{n}{2}-c f}{f}\)) × h
28.5 = 20 + \(\frac{30-(5+x)}{20}\) × 10
⇒ \(\frac{25-x}{2}\) = 8.5
⇒ x = 8
Since, 45 + x + y=60
⇒ 45 + 8 + y = 60
⇒ y = 60 – 53
= 7
Thus, required values of x and y are 8 and 7 respectively.
Question 35.
A Ferris wheel, often known as a big wheel in the UK, is a type of amusement ride that consists of a rotating, upright wheel with several passenger-carrying units or passenger cars linked to the rim, so that they remain upright as the wheel rotates. AB is a chord of the outer wheel which touches the inner wheel at P. The radius of the inner wheel = 8 m and radius of outer wheel = 10 m.
(A) Find the length of the chord AB of the outer circle.
(B) The chord AB of the inner wheel is extended to a point C. If BC = 9 m, then find distance of the point C from the centre of the wheel. 5
Answer:
(A) Triangle OPB is right-angled at P Since, OP ⊥ AB.
OP = 8 m [radius of inner wheel] and OB = 10m [radius of outer wheel].
Therefore by Pythagoras theorem in
ΔOPB,
OB2 = OP2 + BP2
⇒ BP2 = OB2 – OP2
= 102 – 82
= 36 = 62
⇒ BP = 6 m
AP = BP = \(\frac{1}{2}\) AB
So, AB = 2PB
= 2 × 6
= 12 m
(B) As BC = 9 m
∴ PC = PB + BC
= 6 + 9 = 15 m
Triangle OPC is right angled at P.
Therefore, applying Pythagoras theorem, we get
OC2 = OP2 + PC2
= 82 + 152
= 64 + 225
= 289
⇒ OC = 17 m.
Section – E (12 Marks)
CASE STUDY 1
Question 36.
Satellite TV manufacturing businesses tend to have what economists call “economies of scale.” When economies of scale exist, bigness can be its own reward.
The more TV’s you manufacture in a single run, lower the costs per unit, which in turn increases your bottom-line margins.
Keeping that in mind, a T.V. manufacturing company increases its production uniformly by fixed number every year. The company produces 8000 sets in the 6th year and 11,300 sets in the 9th year.
Based on the above information, answer the following questions:
(A) Find the company’s production of the first year.
OR
In which year the company’s production is 9100 sets? 2
(B) Find the company’s production of the 8th year. 1
(C) Find the company’s total production of the first 6 years. 1
Answer:
(A) Given a6 = 8000
a9 = 11,300
Let, the first term be ‘a’ and common difference be ‘d’.
Then, a + (6 – l)d = 8000
⇒ a + 5d = 8,000 ……. (i)
and a + 8d = 11,300 ……. (ii)
On solving (i) and (ii) we get
d = 1,100
⇒ a = 8000 – 5 × 1100
= 2500
OR
Let, the year in which production is 9100 be ‘n’
Then, an = a + (n – 1)d
9100 = 2500 + (n – 1) × 1100
⇒ (n – 1) × 1100 = 6600
⇒ n – 1 = 6
⇒ n = 7
(B) Since, a = 2500
and d = 1100
∴ a8 = a + (8 – 1)d
= 2500 + 7 × 1100
= 2500 + 7700
= 10,200
(C) Production in 6th year = 8,000
∴ Sn = \(\frac{1}{2}\)[a + l]
= \(\frac{1}{2}\)[2500 + 8000]
= 3 × 10500
= 31,500
CASE STUDY 2
Question 37.
Ishani purchased a new building for her business. Being in the prime location, she decided to make some more money by putting up an advertisement sign for a rental ad income on the roof of the building.
From a point P on the ground level, the angle of elevation of the roof of the building is 30° and the angle of elevation of the top of the sign board is 45°. The point P is at a distance of 24 m from the base of the building.
Based on the above information, answer the following questions:
(A) Find the height of the building (without the sign board).
OR
Find the height of the building ( with the sign board). 2
(B) Find the height of the sign board. 1
(C) Find the distance of the point P from the top of the sign board. 1
Answer:
(A) Without the sign board, the height of the shop is AB.
In ΔPAB,
tan 30° = \(\frac{\mathrm{AB}}{\mathrm{PA}}\)
= \(\frac{1}{\sqrt{3}}=\frac{A B}{24}\)
AB = \(\frac{24}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)
= 8 √3 m
= 13.85 ≈ 14 m
OR
Considering, the diagram in the above question, AC as the new height of the shop including the sign-baard.
∴ In ΔAPC,
tan 45° = \(\frac{\mathrm{AC}}{\mathrm{AP}}\)
1 = \(\frac{\mathrm{AC}}{\mathrm{24}}\)
⇒ AC = 24 m
(B) Length of sign board,
BC = AC – AB
= 24 – 14
= 10 m
(C) In ΔAPC,
cos 45° = \(\frac{\mathrm{AP}}{\mathrm{PC}}\)
⇒ \(\frac{1}{\sqrt{2}}=\frac{24}{P C}\)
⇒ PC = 24√2 m
CASE STUDY 3
Question 38.
All of them know that smoking is injurious for health. So, college students decide to make a compaign.
To raise social awareness about hazards of smoking, a school decided to start “No SMOKING” compaign. 10 students are asked to prepare compaign banners in the shape of triangle (as shows in the figure)
Based on the above information, answer the following questions:
(A) If cost of per square centimetre of banner is ₹2, then find the overall cost incurred on such campaign.
OR
If we want to draw a circumscribed circle of given, then find the coordinate of the centre of circle. 2
(B) If we draw the image of figure about the line BC, then find the total area. 1
(C) Find the centroid of the given triangle. 1
Answer:
(A) Here from the figure,
Coordinates of A = (1, 1).
Coordinates of B = (6,1)
and Coordinates of C = (1, 5)
∴ Area of banner = Area of ΔABC
Now, area of one banner
\(\frac{1}{2}\) × AB × AC
\(\frac{1}{2}\) × 5 × 4 = 10
Then, area of 10 banners = 10 × Area of one banner
∴ Cost of 10 banners at the rate of ₹ 2 per cm2
= 2 × Area of 10 banners
= 2 × 10 × 10
= ₹ 200
OR
The centre of circumscribed circle of given triangle is the mid-point of hypotenuse.
Centre of circle = Mid-point of BC
= (\(\frac{1+6}{2}, \frac{5+1}{2}\))
= (\(\frac{7}{2}, \frac{6}{2}\))
= (3.5, 3)
(B) Total area of the required figure
= 2 × Area of ΔABC
= 2 × 10 = 20
(C) The centroid of the given triangle
= (\(\frac{1+1+6}{3}, \frac{5+1+1}{3}\))
= (\(\frac{8}{3}, \frac{7}{3}\))