Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Basic with Solutions Set 5 are designed as per the revised syllabus.
CBSE Sample Papers for Class 10 Maths Basic Set 5 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 n = 22/7 wherever required if not stated.
Section – A (20 Marks)
Question 1.
If xy = 180 and HCF (x, y) = 3, then LCM (x, y) is: 1
(a) 60
(b) 70
(c) 80
(d) 90
Answer:
(a) 60
Explanation: By product formula,
LCM × HCF = Product of two numbers
⇒ LCM × 3 = 180
⇒ LCM = \(\frac{180}{3}\) = 60
Question 2.
If given A.P. is 11, 8, 5, 2, …………… then the sum of 10th term is: 1
(a) -23
(b) 28
(c) 24
(d) -25
Answer:
(d) -25
Explanation: Given AP is 11, 8, 5, 2, …
a = 11
d = 8 – 11 = -3
n = 10
S10 = \(\frac{10}{2}\)[2 × 11 + (10 – 1) × -3]
= 5(22 – 27)
= 5 × -5
= -25
Question 3.
The ideal times of year to have chilled shakes and ice creams are during the summer. During lockdown, Saumya was eager to try the watermelon sharbat she had recently learned to make from her online culinary classes. She cut a watermelon slice with a semi-circular cross section. If the perimeter of a semi-circular portion is 36 cm, then its radius is: 1
(a) 12 cm
(b) 15 cm
(c) 7 cm
(d) 14 cm
Answer:
(c) 7 cm
Explanation: Perimeter of semicircular portion
= 36 cm
πr + 2r = 36
⇒ (π + 2)r = 36
⇒ (\(\frac{22}{7}\) + 2) r = 6
⇒ \(\frac{36}{7}\)r = 36
⇒ r = \(\frac{36 \times 7}{36}\)
⇒ r = 7 cm
Question 4.
The solution of the pair of equations 2x + 3y = 9; 3x + 4y = 5 is: 1
(a) 21, -17
(b) 20, 14
(c) -21, 17
(d) 20, 3
Answer:
(c) -21, 17
Explanation: We have
2x + 3y = 9 ……….. (i)
3x + 4y = 5 …………. (ii)
Multiplying eq. (i) and (ii) by 3 and 2, respectively and then subtracting them, we get
From eq. (i),
2x + 51 =9
⇒ 2x = -42
⇒ x = -21
Thus, x = -21, y = 17 is the required solution.
Question 5.
Two vertices of a triangle are (4, -5) and (-5, -2). If the centroid of the triangle is the origin, the third vertex of the triangle is: 1
(a) (1, 5)
(b) (2, 4)
(c) (1, 4)
(d) (1, 7)
Answer:
(d) (1, 7)
Explanation: Let the third vertex be (x, y). Then,
= (\(\frac{4-5+x}{3}, \frac{-5-2+y}{3}\)) = (0, 0)
⇒ \(\frac{x-1}{3}\) = 0; \(\frac{y-7}{3}\) = 0
⇒ x = 1, y = 7
Thus, the third vertex is (1, 7).
Question 6.
In the adjoining figure, if PA = 10 cm, then the perimeter of ΔPCD is: 1
(a) 16 cm
(b) 21 cm
(c) 18 cm
(d) 20 cm
Answer:
(d) 20 cm
Explanation: We know lengths of tangents drawn from an external point to a circle are equal.
∴ From the figure, we have
PA = PB, CA = CE and DE = DB.
Now,
Perimeter of ΔPCD = PC + CE + ED + DP
= (PC + CE) + (ED + DP)
= (PC + CA) + (BD + DP)
= PA + PB
= 2 PA = 2 × 10 cm
= 20 cm
Question 7.
What is mid-point of the line segment AB where A (- 5, 0) and B (0, 5)? 1
(a) (-\(\frac{5}{2}\), \(\frac{5}{2}\))
(b) (3, 5)
(c) (\(\frac{5}{2}\), \(\frac{5}{2}\))
(d) (2, 4)
Answer:
(a) (-\(\frac{5}{2}\), \(\frac{5}{2}\))
Explanation: The mid-point of AB is
(\(\frac{-5+0}{2}, \frac{0+5}{2}\)), i.e.(-\(\frac{5}{2}\), \(\frac{5}{2}\))
Question 8.
If x sec 45° = 2, then what is the value of x? 1
(a) \(\frac{\sqrt{3}}{2}\)
(b) 2√2
(c) √2
(d) \(\frac{1}{\sqrt{2}}\)
Answer:
(c) √2
Explanation: Given,
x sec 45° = 2
⇒ x(√2) = 2
⇒ x = √2
Question 9.
The frequency of the class succeeding the modal class in the following frequency distribution is: 1
Class | Frequency |
10 – 15 | 3 |
15 – 20 | 7 |
20 – 25 | 16 |
25 – 30 | 12 |
30 – 35 | 9 |
35 – 40 | 5 |
40 – 45 | 3 |
(a) 3
(b) 6
(c) 9
(d) 12
Answer:
(d) 12
Explanation: The modal class is 20 – 25 as, it has maximum frequency. So, the class succeeding the modal class is 25 – 30 with frequency 12.
Question 10.
In an A.P., if a = 3.5, d = 0, n = 101, then the value of an is: 1
(a) 2.5
(b) 3
(c) 4
(d) 3.5
Answer:
(d) 3.5
Explanation: In the given A.P., d = 0
So, all its terms are same as a = 3.5,
∴ a101 = 3.5
Question 11.
In the given figure, ΔACB ~ ΔAPQ. If AB = 6 cm, BC = 8 cm and PQ = 4 cm, then AQ is:
(a) 2 cm
(b) 2.5 cm
(c) 3 cm
(d) 3.5 cm
Answer:
(c) 3 cm
Explanation: Given, ΔACB ~ ΔAPQ
⇒ \(\frac{B C}{P Q}=\frac{A B}{A Q}\)
⇒ \(\frac{8}{4}=\frac{6}{A Q}\)
⇒ AQ = \(\frac{6 \times 4}{8}\)
AQ = 3 cm
Question 12.
A 6 faced cube has letters A, B, C, D, A and C on its six faces. This cube is rolled once. What is the probability of getting B or C?
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{1}{4}\)
Answer:
(a) \(\frac{1}{2}\)
Explanation: Total number of outcomes = 6
∴ Number of favourable outcomes = 3
P (B or C) = \(\frac{3}{6}\) i.e. \(\frac{1}{2}\).
Question 13.
Which criterion of similarity will be used in proving that ΔABD ~ ΔACE? 1
(a) SAS
(b) AA
(c) R.HS
(d) SSS
Answer:
(b) AA
Explanation: In Δs ABD and ACE, we have
AD = AE [given]
So, ∠D = ∠E [Angles opposite to equal sides are equal]
∠A = ∠A [Common]
So, by AA similarty criterion, ΔABD ~ ΔACE.
Hence, AA similarity criteria will be used.
Question 14.
In the following distribution, the class mark of the modal class is: 1
Class interval | Frequency |
40 – 50 | 10 |
50 – 60 | 25 |
60 – 70 | 28 |
70 – 80 | 12 |
80 – 90 | 10 |
90 – 100 | 15 |
(a) 45
(b) 55
(c) 65
(d) 63
Answer:
(c) 65
Explanation: The class 60 – 70 is a modal class as it has highest frequency.
∴ class mark = \(\frac{60+70}{2}\)
= 65
Question 15.
If a2 = \(\frac{23}{25}\), then a is: 1
(a) rational
(b) irrational
(c) whole number
(d) integer
Answer:
(b) irrational
Explanation: a2 = \(\frac{23}{25}\) then a = \(\frac{\sqrt{23}}{5}\), which is irrational
Question 16.
Which term of the A.P.: -2, -7, -12,… will be -77? 1
(a) 16th
(b) 10th
(c) 15th
(d) 12th
Answer:
(a) 16th
Explanation: Let, the nth term of the A.P. be -77.
Then, a + (n – 1 )d = -77
For the given A.P., a = -2 and d = -5.
So, a + (n – 1)d = (-2) + (n – 1)(-5) = -77
-5(n – 1) = -75
or n – 1 = 15 or n = 16
So, the 16th term of the A.P. is (-77).
Question 17.
What type of lines are represented by the pair of equations lOx + 6y = 9 and 5x + 3y + 4 = 0? 1
(a) Straight lines
(b) Intersecting lines
(c) Parallel lines
(d) Data insufficient
Answer:
(c) Parallel lines
Explanation: The pair of equations satisfy the relation
\(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
as \(\frac{10}{5}=\frac{6}{3} \neq \frac{9}{-4}\)
Hence, these equations represent parallel lines.
Question 18.
A bridge across a river makes an angle of 45° with the river bank. If the length of the bridge across the river is 50 m, then what is the width of the river the river? 1
(a) 20√2 m
(b) 50√2 m
(c) 25√2 m
(d) 10√2 m
Answer:
(c) 25√2 m
Explanation: In right ΔABC, we have
sin 45° = \(\frac{\mathrm{BC}}{\mathrm{AC}}\)
⇒ \(\frac{1}{\sqrt{2}}=\frac{B C}{50}\)
⇒ BC = \(\frac{50}{\sqrt{2}}\)
= 25√2 m
Hence, the width of the river is 25√2 m
Question 19.
Assertion (A): The length of the tangent to a circle from a point P, which is 25 cm away from the centre is 24 cm. The radius of the circle is 17 cm.
Reason (R): Tangent is perpendicular to radius at the point of contact. 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: OQ is perpendicular to PQ.
∴ In ΔPOQ,
PQ2 + OQ2 = OP2
252 = OQ2 + 242
OQ2 = 625 – 576
= 49
OQ = 7 cm
Question 20.
Assertion (A): If circumference of two circles are equal, then their areas are also equal.
Reason (R): Two circles are congruent if their radii are equal. 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:
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)
Explanation: Let’s consider two circles of radii r1 and r2.
Then, 2πr1 = 2πr2
r1 = r2 = r
A1 = πr12 = πr2
Then, A2 = πr22 = πr2
∴ A1 = A2
Section – (B 10 Marks)
Question 21.
Assuming that √2 is irrational, show that 5√2 is an irrational number. 2
Answer:
Let 5√2 be rational. Then,
5√2 = \(\frac{p}{q}\), where p and q are co-prime and q ≠ 0.
⇒ √2 = \(\frac{p}{5 q}\)
Here, \(\frac{p}{5 q}\) is rational, which implies is
rational, which is a contradiction, as it is given that √2 is irrational.
⇒ 5√2 is irrational.
Question 22.
Find the greatest number that divides 45 and 210 completely. 2
Answer:
The greatest number that divides 45 and 210 completely is the HCF of 45 and 210.
Now, 45 = 3 × 3 × 5, or 32 × 51
210 = 2 × 3 × 5 × 7,
So, HCF (45, 210) = 31 × 51, i.e. 15
Hence, the required number is 15.
Question 23.
If x = a cos3 θ and y = b sin3 θ, then prove that:
\(\left(\frac{x}{a}\right)^{\frac{2}{3}}+\left(\frac{y}{b}\right)^{\frac{2}{3}}\) = 1
OR
Prove that \(\sqrt{\sec ^2 \theta+{cosec}^2 \theta}\) = tan θ + cot θ. 2
Answer:
OR
[∵ 2 = 2 tan θ cot θ]
= tan θ + cot θ
Question 24.
Mayank made a bird-bath for his garden in the shape of the cylinder with a hemispherical depression at one end (see figure). The height of the cylinder is 1.45m and its radius is 30 cm. Find the total surface area of the bird-bath.
OR
A toy is in the form of a cylinder with two hemispheres at two ends. If the height of the cylinder is 12 cm, and its base is of diameter 7 cm, find the total surface area of the toy. 2
Answer:
Let h be height of the cylinder, and r be the common radius of the cylinder and hemisphere. So, h = 1.45 m = 1.45 × 100 cm = 145 cm Then, the total surface area of the bird-bath = curved surface area of cylinder + curved
surface area of hemisphere
= 2πrh + 2πr2
= 2πr(h + r)
= 2 × \(\frac{22}{7}\) × 30(145 + 30)
= \(\frac{44}{7}\) × 30 × 175
= 33000 cm2
= \(\frac{33000}{10000}\)m2
= 3.3 m2
OR
Height of the cylinder = 12 cm
Diameter of the cylinder = 7 cm
∴ Radius of the cylinder = \(\frac{7}{2}\)cm
Curved surface area of cylinder
= 2πrh
= 2 × \(\frac{22}{7}\) × \(\frac{7}{2}\) × 12
= 264 cm2
Curved surface area of two hemispheres
= 4πr2
= 4 × \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\)
= 154 cm2
∴ Total surface area of toy = Curved surface area of the cylinder + Curved surface area of two hemispheres
= (264 + 154) cm2
= 418 cm2
Question 25.
In figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, Show that the area of the shaded region is 228 cm2. 2
Answer:
OB = \(\sqrt{O A^2+A B^2}=\sqrt{O A^2+O A^2}\)
= √2 OA = √2 × 20 = 20√2 cm
Area of shaded region = Area of quadrant OPBQ – Area of square OABC 90°
= \(\frac{90^{\circ}}{360^{\circ}}\) × 3.14(20√2)2 – 2o × 2o
= \(\frac{1}{4}\) × 3.14 × 800 – 400
= 200 × 3.14 – 400
= 228 cm2
Hence, proved.
Section – C (18 Marks)
Question 26.
Solve for x and y:
x + \(\frac{y}{4}\) = 11; \(\frac{5 x}{6}-\frac{y}{3}\) = 7
OR
A 2-digit number is such that the product of the digits is 20. If 9 is subtracted from the number, the digits interchange their places. Find the number. 3
Answer:
The given equations are rewritten as:
4x + y – 44 = 0
5x – 2y – 42 =0
From 4x + y – 44 = 0, we have
y = 44 – 4x ………… (i)
Substituting this value of y in 5x – 2y – 42 = 0,
we have:
5x – 2 (44 – 4x) – 42 = 0
⇒ 13x – 88 – 42 = 0
⇒ x = 10
From (i) y – 44 – 40 = 4
Thus, x = 10 and y = 4 is the required solution.
OR
Let ten’s digit and one’s digit of the two-digit number be a and b respectively. Then, the number is 10a + b.
Here, ab = 20 ………. (i)
and (10a + b) – 9 = 10b + a
i.e. 9a – 9b = 9
or a – b = 1 ……. (ii)
SoLving (i) and (ii) simultaneously, we get a = 5 and b = 4
Thus, the number is 54.
Question 27.
The diagonal BD of a parallelogram ABCD intersect the line segment AE at the point F, where E is any point on the side BC. Prove that DF × EF = FB × FA. 3
Answer:
In ΔFBE and ΔFDA, we have,
∠2 = ∠1 [Vertically opposite angles]
∠3 = ∠4 [Alternate angles]
∴ ΔFBE ~ ΔFDA [By AA similarity critrion]
⇒ \(\frac{F B}{F D}=\frac{F E}{F A}\)
⇒ DF × EF = FB × FA
Question 28.
Prove that the lengths of tangents drawn from an external point to a circle are equal. 3
Answer:
Let AP and AQ be the two tangents drawn to the circle from an external point A.
We need to show that AP = AQ.
Join OA, OP.and OQ.
We know that tangent is perpendicular to radius at the point of contact.
∴ OP ⊥ AP and OQ ⊥ QA.
Consider ΔOPA and ΔOQA.
Here, OQ = OP [radii of the circle]
OA = OA [common]
∠OPA = ∠OQA
So, ΔOPA ≅ ΔOQA [By SAS Criterion]
⇒ PA = QA or AP = AQ [By cpct]
Question 29.
A box contains 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Ramesh, a shopkeeper will buy only those shirts which are good but Suresh another shopkeeper will not buy shirts with major defects. A shirt is taken out of the box at random. What is the probability that:
(A) Ramesh will buy the selected shirt?
(B) Suresh will buy the selected shirt? 3
Answer:
Total number of shirts = 100
∴ Total number of possible outcomes = 100
(A) Let A be the event that shirt is good,
∵ Number of good shirts = 88
⇒ Favourable number of outcomes = 88
∴ p(A) = \(\frac{88}{100}\) = \(\frac{22}{25}\)
∴ Probability that Ramesh will buy the
selected shirt is \(\frac{22}{25}\).
(B) Let B be the event that shirt has no major defects.
∵ Number of shirts with no major defects
= 100 – 4 = 96
⇒ Favourable number of outcomes = 96
∴ P(B) = \(\frac{96}{100}\) = \(\frac{24}{25}\)
∴ Probability that Suresh will buy the selected shirts is \(\frac{24}{25}\).
Question 30.
Prove that:
\(\frac{5 \sin ^2 30^{\circ}+\cos ^2 45^{\circ}-4 \tan ^2 30^{\circ}}{2 \sin 30^{\circ} \cdot \cos 30^{\circ}+\tan 45^{\circ}}\) + cos0°
= \(\frac{17+6 \sqrt{3}}{6(\sqrt{3}+2)}\) 3
Answer:
Question 31.
150 workers were engaged to finish a work in a certain number of days. Four workers dropped the second day, four more workers dropeed the third day and so on. It takes 8 more days to finish the work now. Find the number of days in which the work was completed.
OR
Which term of the A.P. 3, 8,13, 18,……… is 78? 3
Answer:
Here, first day, 150 workers were engaged, second day, 150 – 4, i.e., 146 workers were engaged, third day 146 – 4, i.e., 142 workers were engaged and so on. Clearly, 150, 146, 142, 138,……. is an A.P., with 150 as first term and -4 as common difference.
Letn be the number of days in which work was completed.
∴ Sn = \(\frac{n}{2}\)[2(150) + (n – 1)(-4)]
= \(\frac{n}{2}\)[300 – 4n + 4]
\(\frac{n}{2}\)[304 – 4n]
= 152n – 2n2
According to the question,
150(n – 8) = 152n – 2n2
⇒ 2n2 – 152n + 150n – 1200 = 0
⇒ n2 – n – 600 = 0
⇒ n2 – 25n + 24n – 600 = 0
⇒ (n – 25) (n + 24) = 0
⇒ n = 25 or n = -24
Rejecting negative value of n, because number of days cannot be negative.
∴ n = 25
Hence, in 25 days the work was completed.
OR
Let nth term of the A.P. be 78. Then,
an = a + (n – l)d = 78
⇒ 3 + (n – 1)(5) = 78 (Here, a = 3, d = 5)
⇒ 5(n – 1) = 75
⇒ n- 1 = 15
⇒ n = 16
So, 16th term of the A.P. is 78.
Section – D
Question 32.
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see figure). Find the distance travelled by the ballon during the interval.
The shadow of a vertical tower on level ground increases by 16 m, when the altitude of the Sun changes from angles of elevation 60° to 45°. Find the height of the tower, correct to one place of decimal. (Use √3 = 1.73). 5
Answer:
In the figure, let C be the position of the girl. A and P are two positions of the balloon. CD is the horizontal line from the eyes of the girl.
Thus, the required distance between the two positions of the balloon is 58√3 m.
OR
Let AB represent the tower of height ‘h’ metres.
AS1, AS2 be, is shadows when sun’s attitude is 60° and 45° respectively.
∴ S1S2 = 16 m
From Δ S1AB,
\(\frac{\mathrm{AB}}{\mathrm{S}_1 \mathrm{~A}}\) = tan 60°
⇒ \(\frac{h}{\mathrm{~S}_1 \mathrm{~A}}=\sqrt{3}\)
⇒ S1A = \(\frac{h}{\sqrt{3}}\) …….. (i)
From Δ S2AB,
\(\frac{\mathrm{AB}}{\mathrm{S}_2 \mathrm{~A}}\) = tan 45°
⇒ \(\frac{h}{16+S_1 A}\) = 1
⇒ 16 + S1A = h ………… (ii)
Subtracting eqn. (i) from (ii), we have
16 = h – \(\frac{h}{\sqrt{3}}\)
= h(1 – \(\frac{1}{\sqrt{3}}\))
= 37.8 metres.
Question 33.
From an external point P, a tangent PT and a line segment PAB is drawn to a circle with centre O. ON perpendicular on the chord AB. Prove that:
(A) PA.PB = PN2 – AN2
(B) PN2 – AN2 = OP2 – OT2
(C) PA. PB = PT2
OR
In the given figure, a circle is inscribed in a ΔABC having sides BC = 8 cm, AB = 10 cm and AC = 12 cm. Find the lengths BL, CM and AN.
Answer:
(A) L.H.S = PA-PB
= (PN – AN) (PN + BN)
= (PN – AN) (PN + AN) [∵ ON ⊥ AB .-. N is the mid-point of AB]
= PN2 – AN2 = R.H.S
(B) In right triangle PNO,
OP2 = ON2 + PN2 [By Pythagoras theorem ]
⇒ PN2 = OP2 – ON2
Now, LH.S = PN2 – AN2
= (OP2 – ON2) – AN2
= OP2 – (ON2 + AN2)
= OP2 – OA2 [Using Pythagoras theorem in ΔONA]
= OP2 – OT2 = R.H.S [Radii of the same circle]
(C) From (i) and (ii), we have
PA.PB = PN2 – AN2 and
PN2 – AN2 = OP2 – OT2
∴ PA.PB = OP2 – OT2 …….. (i)
In right angle ΔOTP,
OP2 = OT2 + PT2 [By Pythagoras theorem]
⇒ OP2 – OT2 = PT2
∴ PA.PB = PT2 …….. (ii)
∴ PA.PB = PT2 [Using (i) and (ii)]
OR
Let, BL = x
⇒ BN = x
[∵ Tangents drawn from an external point to the circle are equal in length]
∴ CL = CM = 8 – x [∵ BC = 8 cm]
and AN = AM = 10 = x [∵ AB = 10 cm]
Now, AC = 12 cm
∴ AM + MC = 12
⇒ 10 – x + 8 – x = 12
⇒ 18 – 2x = 12
⇒ 6 = 2x
⇒ x = 3
∴ Length of BL = 3 cm
Length of CM = 8 – 3 = 5 cm
Length of AN = 10 – 3 = 7 cm
Question 34.
Find the median marks for the following frequency distribution:
Marks | Number of Students |
0 – 20 | 7 |
20 – 40 | 12 |
40 – 60 | 23 |
60 – 80 | 18 |
80 – 100 | 10 |
Answer:
Marks | Frequency | Cumulative Frequency |
0 – 20 | 7 | 7 |
20 – 40 | 12 | 19 |
40 – 60 | 23 | 42 |
60 – 80 | 18 | 60 |
80 – 100 | 10 | 70 |
Here, N = 70,
∴ \(\frac{N}{2}\) = 35
Cumulative frequency just greater than 35 is 42 which belongs to class 40 – 60.
So, the median class is 40 – 60.
For this class,
l = 40, h = 20, c.f. = 19 and f = 23
So, Median = l + \(\frac{\frac{N}{2}-c . f}{f}\) × h
= 40 + \(\frac{35-19}{23}\) × 20
= 40 + 13.9 (approx.)
= 53.9 (approx.)
Question 35.
Between Mysore and Bangalore, 132 km apart, an express train travels in 1 hour less time than a passenger train (without taking into consideration the time they stop at intermediate stations). Find the average speed of the two trains if the express train’s average speed is 11 km/h higher than the passenger train’s. 5
Answer:
Let average speed of passenger train
= x km/h
Let average speed of express train
= (x + 11) km/h
Time taken by passenger train to cover 132 km
= \(\frac{132}{x}\) hours
Time taken by express train to cover 132 km
= (\(\frac{132}{x+11}\)) hours
According to the given conditions,
\(\frac{132}{x}=\frac{132}{x+11}\) + 1
⇒ 132 (\(\frac{1}{x}-\frac{1}{x+11}\)) = 1
⇒ 132 (\(\frac{x+11-x}{x(x+11)}\)) = 1
⇒ 132(11) = x(x + 11)
⇒ 1452 = x2 + 11x
⇒ x + 11x – 1452 =0
Comparing equation x2 + 11x – 1452 = 0 with general quadratic equation ax2 + bx + c = 0, we get a = 1, b = 11 and c = -1452
Applying quadratic formula
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
⇒ x = 33, -44
As speed cannot be in negative. Therefore, speed of passenger train = 33 km/h. And, speed of express train = x + 11
= 33 + 11 = 44 km/h
Section – E (12 marks)
CASE STUDY 1
Question 36.
Raghav and his family went for vacation to Rajasthan. There they had a stay in tent for a night. Raghav found that the tent in which they stayed is in the form of a cone surmounted on a cylinder. The total height of the tent is 35 m. Diameter of the base is 56 m and height of the cylinder is 14 m.
Based on the above information, answer the following questions:
(A) If each person need 176 m of floor, then how many persons can be accommodated in the tent? 1
(B) Find total surface area of cone. 1
(C) How much canvas is needed to make the tent?
OR
Find the volume of the tent. 2
Answer:
(A) Area of floor = πr2 = \(\frac{22}{7}\) × 28 × 28
= 2464 m2
Number of persons that can be accommodated in the tent = \(\frac{2464}{176}\) = 14
(B) Total surface area of a cone = curved surface area of a cone + area of circular base
= πrl + πr2 = πr(l + r)
[Since, slant height of cone,
l = \(\sqrt{r^2+h^2}=\sqrt{(28)^2+(21)^2}\)
= \(\sqrt{784+441}=\sqrt{1225}\) = 35 m
= \(\frac{22}{7}\) × 28 (35 + 28) = 5544 m2
(C) Required area of canvas = curved surface area of cone + curved surface area of cylinder
= πrl + 2πrh
= πr(l + 2 h)
= \(\frac{22}{7}\) × 28 (35 + 28)
= 5544 m2
OR
Volume of tent = Volume of cone + Volume of cylinder
= \(\frac{1}{3}\) × π × r2 × 21 + πr2 × 14
= πr2(7 + 14)
= \(\frac{22}{7}\) × 28 × 28 × 21
= 51744 m3
CASE STUDY 2
Question 37.
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 though of amking 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.
Based on the above information, answer the following questions:
(A) What is the length of the line segment joining points B and F? 1
(B) The centre ‘Z’ of the figure will be the point of intersection of the diagonals of quadrilateral WXOP. Then what are the coordinates of Z? 1
(C) What are the coordinates of the point on y-axis equidistant from A and G?
OR
What is the area of trapezium AFGH? 2
Answer:
(A) From the given figure, B(1, 2), F(-2,9)
The distance between the points B and F is:
BF = \(\sqrt{(-2-1)^2+(9-2)^2}\)
= \(\sqrt{(-3)^2+(7)^2}\)
= \(\sqrt{9+49}\) = \(\sqrt{58}\)
So, BF = \(\sqrt{58}\) units.
W(-6, 2), X(-4, 0), 0(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.
= (\(\frac{-6+5}{2}, \frac{2+9}{2}\))
= (\(\frac{-1}{2}, \frac{11}{2}\))
(C) 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
The required point is (0, 5.7).
OR
From the given figure,
A(-2, 2), F(—2, 9), G(-4, 7), H(-4, 4)
Clearly,GH =7-4 = 3 units
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.
CASE STUDY 3
Question 38.
Shweta is a baker. She has chocolates and pancakes which together are 40 in number. If she has 50 more pancakes and 50 less pancakes, then number of pancakes would become 4 times the number of chocolates. Denote original number of chocolates and pancakes by x and y respectively.
Based on the above information, answer the following questions:
(A) Represent algebraically the above situations. 1
(B) Which type of solution, the given system of equation has?
OR
Find out the number of chocolates and pancakes that Shweta is having. 2
(C) If cost of 1 chocolates is ₹ 20 and cost of 1 pancake is ₹ 3, then find the total cost of chocolate and pancake that
Shweta had originally.
Answer:
(A) Since, every student get one chocolate.
So, number of chocolates Rohit has = Number of students in the class = 54
Let, number of milk chocolates Rohit has = x
Probability of distributing milk chocolates = \(\frac{1}{3}\)
⇒ \(\frac{x}{54}=\frac{1}{3}\)
⇒ x = \(\frac{54}{3}\) = 18
(B) Let number of dark chocolates Rohit has = y
Probability of distributing dark chocolates = \(\frac{4}{9}\)
⇒ \(\frac{y}{54}=\frac{4}{9}\)
⇒ y = \(\frac{4 \times 54}{9}\) = 24
OR
Number of white chocolates Rohit has
= 54 – (18 + 24) = 12
∴ Required probability = \(\frac{12}{54}\) = \(\frac{2}{9}\)
(C) Total number of milk and white chocolates
= 18 + 12 = 30
∴ Required probability = \(\frac{30}{54}\) = \(\frac{5}{9}\)