Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Basic with Solutions Set 4 are designed as per the revised syllabus.
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions
Time: 3 Hours
Maximum Marks: 80
General Instructions:
- This Question Paper has 5 Sections A, B, C, D and E.
- Section A has 20 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 value of \(\frac{\cot A+\tan B}{\cot B+\tan A}\) is: 1
(a) cot A tan B
(b) cot B tan A
(c) cot A
(d) tan B
Answer:
(a) cot A tan B
Explanation:
Question 2.
For any two positive integers ‘a’ and ‘b’, what is the value of HCF (a, b) × LCM (a, b)? 1
(a) ab
(b) a2b
(c) a × b
(d) a + b
Answer:
(c) a × b
Explanation: We know the relation,
HCF (a, b) × LCM (a, b) = Product of a and b.
Question 3.
Tours of the national capital and the white house begin at 8:30 am from tour agency. Tours for the national capital leave every 15 min. Tours for the white house leave every 20 min. How many minutes after do the tours leave at the same time? 1
(a) 50 min
(b) 60 min
(c) 35 min
(d) 40 min
Answer:
(b) 60 min
Explanation: Required time = LCM (15, 20)
By using prime factorisation method,
15 = 3 × 5
and, 20 = 2 × 2 × 5 = 22 × 5
∴ LCM (15, 20) = 22 × 3 × 5 = 60 mins
In every 60 mins, tour leaves at the same time.
Question 4.
Using prime factorisation method, the HCF and LCM of 210 and 175 is: 1
(a) 35,1000
(b) 34,1050
(c) 35,1050
(d) 35,1010
Answer:
(c) 35, 1050
Explanation: The prime factorisation of 210 and 175 are:
210 = 2 × 3 × 5 × 7
175 = 5 × 5 × 7
So, HCF (210, 175) = 5 × 7 = 35;
and LCM (210, 175)= 2 × 3 × 5 × 7 × 5 = 1050
Question 5.
If 2x, x + 10, 3x + 2 are in A.P., the value of x is: 1
(a) 6
(b) 7
(c) 9
(d) 10
Answer:
(a) 6
Explanation: Since 2x, x + 10, 3x + 2 are in A.P.,
∴ 2(x + 10) = 2x + (3x + 2)
⇒ 2x + 20 = 5x + 2
⇒ 3x = 18
⇒ x = 6
Question 6.
By taking A = 90° and B = 30°, sin A cos B – cos A sin B is: 1
(a) 0
(b) \(\frac{\sqrt{3}}{2}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{1}{\sqrt{2}}\)
Answer:
(b) \(\frac{\sqrt{3}}{2}\)
Explanation: sin A cos B – cos A sin B
= sin 90° cos 30° – cos 90° sin 30°
= 1 × \(\frac{\sqrt{3}}{2}\) – 0 × \(\frac{1}{2}\) = \(\frac{\sqrt{3}}{2}\)
Question 7.
If one of the zeroes of a quadratic polynomial of the form x2 + ax + b is the negative of the other, then it: 1
(a) has no linear term and the constant term is negative.
(b) has no linear term and the constant term is positive.
(c) can have a linear termbut the constant term is negative.
(d) can have a linear term but the constant term is positive.
Answer:
(a) has no linear term and the constant term is negative.
Explanation: f(x) = x2 + ax + b
Given: zeroes are α and – α
Sum of zeroes = α – α = 0
∴ f(x) = x2 + b, which is not linear Product of zeroes
= α(- α) = – α2 = \(\frac{b}{1}\)
⇒ – α2 = b
It is possible when, b < 0.
Hence, it has no linear term and the constant term is negative.
Question 8.
The ratio of the volume of a right circular cone to that of the volume of right circular cylinder, of equal diameter and height is: 1
(a) 4 : 3
(b) 1 : 4
(c) 2 : 3
(d) 1 : 3
Answer:
(d) 1 : 3
Explanation: Volume of a right circular cone = \(\frac{1}{3}\)πr2h
Volume of a right circular cylinder = πr2h
So, required ratio is 1: 3.
Question 9.
The number of revolutions made by a circular wheel of radius 0.7 m in rolLing a distance of 176 m is: 1
(a) 22
(b) 24
(c) 75
(d) 40
Answer:
(d) 40
Explanation: Number of revolutions
= \(\frac{\text { total distance }}{\text { circumference }}\)
= \(\frac{176}{2 \times \frac{22}{7} \times 0.7}\)
= 40
Question 10.
The zeroes of the quadratic polynomial x2 + 99x + 127 are: 1
(a) both positive
(b) both negative
(c) one positive and one negative
(d) both equal
Answer:
(b) both negative
Explanation: a = 1, b = 99, c = 127
= \(\frac{-2.6}{2}\) or \(\frac{-195.4}{2}\)
= -1.3 or -97.7
Both are negative.
Question 11.
The value of ‘Id for which the pair of linear equations kx – y = 2 and 6x – 2y = 3 will have infinitely many solutions is: 1
(a) 4
(b) 3
(c) 2
(d) 1
Answer:
(d) None of these
Explanation: The given pair of equations will have infinitely many solutions when
\(\frac{k}{6}=\frac{-1}{-2}=\frac{2}{3}\)
No value of k satisfy the above relation.
So, no value of k exist.
Question 12.
The pair of linear equation is x + 3y – 6 = 0, 2x + 17y – 13 = 0 is: 1
(a) inconsistent
(b) parallel lines
(c) consistent
(d) data insufficient
Answer:
(c) consistent
Explanation: Here, a1 = 1, b1 = 3, c1 = -6
a2 = 2, b2 17, c2 = – 13
Now, \(\frac{a_1}{a_2}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{3}{17}\) and \(\frac{c_1}{c_2}=\frac{6}{13}\)
⇒ \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
∴ Given pair of linear equations have a unique solution. Hence, given pair of linear equations is consistent.
Question 13.
The mean of following distribution is: 1
(a) 15.6
(b) 17
(c) 14.8
(d) 16.4
Answer:
(d) 16.4
Explanation:
xi | fi | fixi |
11 | 3 | 33 |
14 | 6 | 84 |
17 | 8 | 136 |
20 | 7 | 140 |
Σfi = 24 | Σfixi = 393 |
xmean = \(\frac{\sum f_i x_i}{\sum f_i}\)
= \(\frac{393}{24}\)
= 16.4
Question 14.
Tony and Monu go to an ice-cream parlour where they find 5 different varieties of ice creams. If they order one ice-cream each. The probability that they both order the same variety of ice creams is: 1
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{1}{3}\)
(d) \(\frac{1}{5}\)
Answer:
(d) \(\frac{1}{5}\)
Explanation: Total number of possible outcomes = 5 × 5 = 25
Let the five different varieties of icecreams be a, b, c, d and e.
So, favourable outcomes are {(a, a), (b, b), (c, c), (d, d), (e, e)}i.e., 5
∴ Required probability = \(\frac{5}{25}\) = \(\frac{1}{5}\)
Question 15.
A solid cube is cut into two cuboids of equal volumes. The ratio of surface areas of the given cube and one of the resulting cuboid is: 1
(a) 2 : 3
(b) 1 : 3
(c) 3 : 2
(d) 3 : 1
Answer:
(c) 3 : 2
Explanation: Since the cube is cut into two cuboids of equal volumes, so the two cuboids are equal.
Let the edge of the cube be 2a units and the cube be cut along its length.
∴ Dimensions of each cuboid formed are a × 2a × 2a.
So, surface area of cuboid
= 2 (a × 2a + 2a × 2a + 2a × a)
= 16a2
Also, surface area of cube
= 6(2a)2 = 24a2
∴ Required ratio = 24a2 : 16a2 = 3 : 2.
Question 16.
A stopwatch was used to find the time that it took a group of students to run 100m. 1
Time (in sec) | Number of students |
0 – 20 | 8 |
20 – 40 | 10 |
40 – 60 | 13 |
60 – 80 | 6 |
80 – 100 | 3 |
What will be the upper limit of the modal class? 1
(a) 20
(b) 40
(c) 60
(d) 80
Answer:
(c) 60
Explanation: Modal class = 40 – 60, therefore upper limit = 60.
Question 17.
If the ratio of the length of a rod to its shadow is 1 : √3, then, angle of elevation of the sun is: 1
(a) 30°
(b) 60°
(c) 45°
(d) 90°
Answer:
(a) 30°
Explanation: Let θ be the angle of elevation of the sun.
Then, tanθ = \(\frac{\mathrm{BC}}{\mathrm{AB}}\)
= \(\frac{1}{\sqrt{3}}\)
θ = 30°
Question 18.
A rolling pin is made by joining three cylindrical pieces of wood, as shown in the figure:
Assuming that there is no wastage of wood, the volume of wood used in making the rolling pin is: 1
(a) 64π cm3
(b) 280π cm3
(c) 480π cm3
(d) 544π cm3
Answer:
(d) 544π cm3
Explanation: For the bigger cylinder:
r1 = 4 cm
h1 = 30 cm
For the two smaller cylinders:
r2 = \(\frac{4}{2}\) = 2 cm,
h2 = 8 cm
Now, Volume of wood used = Volume of the pin
= Volume of bigger cylinder + ( 2 × Volume of smaller cylinder)
= πr12h1 + 2 × πr22h2
= π × (4)2 × 30 + 2 × π × (2)2 × 8
= 480π + 64π
= 544π cm3
Question 19.
Assertion (A): The base radii of two right circular cylinders of the same height are in the ratio 3 : 5. The ratio of their curved surface area is 3 : 5.
Reason (R): CSA of right circular cylinder is 2πr2h. 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:
(c) Assertion (A) is true but reason (R.) is faLse.
Explanation: Let r1 h1 and r2, h2 be the radii and the heights of the first and second cylinders respectively.
Then, \(\frac{r_1}{r_2}=\frac{3}{5}\)
and h1 = h2 = h (say)
So, \(\frac{\text { CSA of first cylinder }}{\text { CSA of second cylinder }}=\frac{2 \pi r_1 h_1}{2 \pi r_2 h_2}\)
= \(\frac{r_1 h}{r_2 h}\)
= \(\frac{r_1}{r_2}=\frac{3}{5}\)
Question 20.
Assertion (A): If the probability of the occurence of an event is \(\frac{3}{7}\), then the probability of its non-occurence is \(\frac{5}{7}\).
Reason (R): P(A) + P(\(\overline{\mathbf{A}}\)) = 1, when P(\(\overline{\mathbf{A}}\)) is a complement of P(A). 1
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation
(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: For an event A, the
P(A) + P(\(\overline{\mathrm{A}}\)) = 1
Here, P(A) = \(\frac{3}{7}\)
Then, P(\(\overline{\mathrm{A}}\)) = 1 – \(\frac{3}{7}\)
= \(\frac{4}{7}\)
Section – B (10 Marks)
Question 21.
Determine the A.P. whose 3rd term is 5 and the 7th term is 9.
OR
Find the last term of an AP having 9 terms whose last term is 28 and sum of all the terms is 144. 2
Answer:
Here, a3 = a + 2d = 5
and a7 = a + 6d = 9
Solving the two equations, we get
a = 3, d = 1
So, required A.P. is 3, 4, 5, 6, …………..
OR
Let ‘a’ be the first term of A.P. and ‘d’ be the common difference.
Here, total number of terms of A.P. is 9, i.e. n = 9
nth term = last term = an = a + 8d = 28 …….. (i)
Also, Sn = S9 = \(\frac{9}{2}\)[2a + (9 – 1 )d] = 144
⇒ 9 (a + 4 d) = 144
or 9 a + 36d = 144 ……. (ii)
Solving (i) and (ii) simultaneously, we get
a = 4; d = 3
Thus, the required first term is 4.
Question 22.
Find the zeros of 3x2 – x – 4. 2
Answer:
Let p(x) = 3x2 – x – 4
= 3x2 – 4x + 3x – 4
= x(3x – 4) + 1 (3x – 4)
= (3x – 4) (x + 1)
3x – 4 = 0, x + 1 = 0, x = \(\frac{4}{3}\), x = -1
So, the two zeros of 3x2 – x – 4 are \(\frac{4}{3}\) and -1.
Question 23.
Find the coordinates of the point which divides the line joining (1, -2) and (4, 7) internally in the ratio 1 : 2.
OR
Find the third vertex of a triangle, if two of its vertices are at (-3,1) and (0, -2) and the centroid is at the origin. 2
Answer:
Let P(x, y) divide AB in the ratio 1 : 2. Then,
P(\(\frac{2+4}{3}, \frac{-4+7}{3}\))
= P(2, 1)
OR
Let the third vertex be (x, y). Then,
(\(\frac{x-3+0}{3}, \frac{y+1-2}{3}[latex]) = (0, 0)
⇒ [latex]\frac{x-3}{3}[latex] = 0 and [latex]\frac{y-1}{3}[latex] = 0
⇒ x = 3 and y = 1
Thus, the third vertex is (3, 1).
Question 24.
Find the area of the shaded region. 2
Answer:
Area of the shaded region = Area of semi-circle of radius 14 cm + 2 × Area of semi-circle of radius 7 cm.
= ([latex]\frac{\pi}{2}\)(14)2 + 2 × \(\frac{\pi}{2}\)(7)2)sq. cm
= (\(\frac{1}{2}\) × \(\frac{22}{7}\) × 14 × 14 + \(\frac{22}{7}\) × 7 × 7)sq. cm
= (308 + 154) sq. cm
= 462 sq. cm.
Question 25.
Amrish wakes up in the morning and notices that his digital clock reads 07 : 25 am. After noon, he looks at the clock again.
What is the probability that:
(A) the number in column A is 4?
(B) the number in column B is 8? 2
Answer:
(A) The number in column A can be 0, 1, 2, 3, 4 and 5.
So, P(4) = \(\frac{1}{6}\)
(B) The number in column B can be 0, 1, 2, 3, ……………. 9.
So. P(8) = \(\frac{1}{10}\)
Section – C (18 Marks)
Question 26.
On a morning walk, three girls step off together and their steps measure 40 cm, 42 cm and 45 cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps? 3
Answer:
Required minimum distance
= LCM (40, 42, 45)
∵ 40 = 2 × 2 × 2 × 5 i.e., 23 × 5
42 = 2 × 3 × 7
45 = 3 × 3 × 5 i.e., 33 × 5
∴ LCM (40, 42, 45) = 23 × 32 × 5 × 7
= 2520 cm
Question 27.
There is a circular park of radius 24 m and there is a pole at a distance of 26 m from the centre of the park as shown in the figure. It is planned to enclose the park by planting trees along line segments PQ and PR tangential to the park.
(A) Prove that the length of PQ and PR are equal.
(B) If six trees are to be planted along each tangential line segments at equal distances, find the distance between any two consecutive trees. 3
Answer:
(A) In right-angled triangle PRO,
We have PR = \(\sqrt{P O^2-R O^2}\)
= \(\sqrt{26^2-24^2}\)
= \(\sqrt{676-576}\)
= \(\sqrt{100}\)
= 10m
⇒ PR = PQ 10m
Hence, proved.
(B) As six trees are to be planted along PQ and PR each.
Let’s assume the distance between consecutive trees is x.
TotaL trees are at 5 equal distances.
Hence, 5x = 10
x = 2 m
Question 28.
In the figure, sectors of two concentric circles of radii 7 cm and 3.5 cm are shown.
Find the area of the shaded region. 3
Answer:
Area of the shaded region
= Area of the sector of radius 7 cm. – Area of the sector of radius 3.5 cm
= \(\frac{30^{\circ}}{360^{\circ}}\)π(7)2 – \(\frac{30^{\circ}}{360^{\circ}}\)π(3.5)2) sq .com
= \(\frac{30}{360}\) × \(\frac{22}{7}\) × (49 – 12.25) sq. cm
= 9.625 sq. cm
Question 29.
5 books and 7 pens together cost ₹ 79, whereas 7 books and 5 pens together cost ₹ 77. Represent this situation in the form of linear equation in two variables. Find the cost of one book and one pen. 3
OR
Solve for x and y:
7x – 4y = 49;
5x – 6y = 57 3
Answer:
Let the cost of 1 book be ₹ x and the cost of 1 . pen be ₹ y.
According to question,
5x + 7y = 79 …… (i)
and 7x + 5y = 77 ……. (ii)
Solving (i) and (ii) for x and y
Adding (i) and (ii)
12x + 12y = 156
x + y = 13
x = 13 – y
Put in (i)
5(13 – y) + 7y = 79
65 – 5y + 7y = 79
2 y = 14
y = 7
⇒ x = 6
OR
Given equations are:
7x – 4y = 49 …….. (i)
5x – 6y = 57 ……. (ii)
Multiplying eq. (i) by 5 and eq. (ii) by 7, we get
35x – 20y = 245 …….. (iii)
35x – 42y = 399 …….. (iv)
Subtracting equation (iv) from equation (iii), we get
⇒ 22y = -154
⇒ y = -7.
Substituting y = -7 in eq. (i), we get
7x + 28 = 49
⇒ 7x = 21
⇒ x = 3
Thus, x = 3 and y = -7 is the required solution.
Question 30.
In the given figure, DE || AQ and DF || AR. Prove that EF || QR. 3
Answer:
Given: In ΔPQR, DE || AQ and DF || AR
To prove: EF || QR
Question 31.
A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.
OR
Find two numbers whose sum is 27 and product is 182. 3
Answer:
Let the natural number be x. Then,
A.T.Q. x + 12 = 160 × \(\frac{1}{x}\)
⇒ x2 + 12x – 160 = 0
⇒ x2 + 20x – 8x – 160 = 0
⇒ x(x + 20) – 8 (x + 20) = 0
⇒ (x + 20) (x – 8) = 0
⇒ x + 20 = 0
x = – 20
[Rejected, as x is a natural number]
or x – 8 = 0
⇒ x = 8
Thus, the required natural number is 8.
OR
Let first number be x and let second number be (27 – x)
According to given condition, the product of two numbers is 182.
Therefore,
x(27 – x) = 182
⇒ 27x – x2 = 182
⇒ x2 – 27x + 182 = 0
⇒ x2 – 14x – 13x + 182 = 0
⇒ x(x – 14) – 13(x – 14) = 0
⇒ (x – 14) (x – 13) = 0
⇒ x = 14, 13
Therefore, the first number is equal to 14 or 13
And, second number is = 27 – x = 27 – 14 = 13
or second number = 27 – 13 = 14
Therefore, two numbers are 13 and 14.
Section – D (20 Marks)
Question 32.
Mr. Sharma and Mr. Arora are family and they decided to go for a trip. For the trip they reserved their rail tickets. Mr. Arora has not taken a half ticket for his child who is 6 years old whereas Mr. Sharma has taken half tickets for his two children who are 6.5 years and 8 years old. A railway half ticket costs half of the full fare but the reservation charges are the same as on a full ticket. Mr. and Mrs. Arora paid ₹ 1700, while Mr. and Mrs. Sharma paid ₹ 2700. Find the full fare of one ticket and the reservation charges per ticket.
OR
Rajiv walks and cycles at uniform speeds. When he walks for 2 hrs and cycles for 1 hr, distance travelled is 24 km. When he walks for 1 hr and cycles for 2 hrs, distance travelled is 39 km. Find his speed of walking and cycling. If he walked and cycled for equal to time in 3 hrs how much distance does he cover? 5
Answer:
Let the full fare of one ticket for a trip be ₹ x, then the railway half fare of one ticket for a trip be ₹ \(\frac{x}{2}\).
Let the reservation charges per ticket be ₹ y.
According to the given conditions:
x + y + x + y= 1700 (Amount paid by Mr. and Mrs. Arora)
⇒ 2x + 2y = 1700 …….. (i)
and (x + y) + (x + y) + (\(\frac{x}{2}\) + y) + (\(\frac{x}{2}\) + y) = 2700
(Amount paid by Mr. and Mrs. Sharma and their two children)
⇒ 3x + 4y = 2700 ….. (ii)
From (i), y = \(\frac{1700-2 x}{2}\) …………. (iii)
Substituting value of y from (iii) in (ii), we get
3x + 4(\(\frac{1700-2 x}{2}\)) = 2700
⇒ 3x + 3400 – 4x = 2700
⇒ x = 700
Now, putting x = 700 in (iii), we get
y = \(\frac{1700-2(700)}{2}\)
⇒ y = \(\frac{1700-1400}{2}\)
= \(\frac{300}{2}\) = 150
Hence, the full fare of one ticket = ₹ 700 and the reservation charges = ₹ 150 per ticket.
OR
Let the speed of walking be x km/hr and the speed of cycling be y km/hr.
According to the question,
2x + y = 24 ……….. (i)
and x + 2y = 39 ………. (ii)
Multiplying (i) by 2, we get 4x + 2y = 48 ……… (iii)
Subtracting (ii) from (iii), we get
3x = 9
⇒ x = 3
Putting the value of x in (i), we get
2(3) + y = 24
⇒ y = 24 – 6 = 18
∴ Speed of walking is 3 km/hr and speed of cycling is 18 km/hr.
When Rajiv walked and cycled for equal time.
Distance covered = 3 × 1.5 + 18 × 1.5
= 4.5 + 27 = 31.5 km
Question 33.
Amit and Prem were very good cricketers and also represented their school team at district and even state level. One day, after their match, they measured the height of the wickets and found it to be 28 inches. They marked a point P on the ground as shown in the figure below: 5
(A) If cot P = \(\frac{3}{4}\), then find the length of P.
(B) Find the value of cosec P.
(C) Find the value of \(\frac{1+\sin P}{1+\cos P}\).
(D) Find the value of sec R.
(E) Find the value of cosec2 R – cot2 R.
Answer:
(A) It is given that QR = 28 inches and cot P = \(\frac{3}{4}\)
We know that cot P = \(\frac{\text { Base }}{\text { Perpendicular }}\)
= \(\frac{P Q}{Q R}=\frac{P Q}{28}\)
Therefore, \(\frac{P Q}{28}=\frac{3}{4}\)
⇒ PQ = \(\frac{28 \times 3}{4}\)
= 21 inches.
(B) To evaluate cosec P, we will first find PR by applying Pythagoras theorem in ΔPQR.
PR2 = PQ2 + QR2
= 212 + 282
= 441 + 784
= 1225
⇒ PR = 35 inches
∴ cosec P = \(\frac{\text { Hypotenuse }}{\text { Perpendicular }}\)
= \(\frac{\mathrm{PR}}{\mathrm{QR}}=\frac{35}{28}=\frac{5}{4}\)
Hypotenuse Perpendicular
Question 34.
In the given figure, O is the centre of circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30°, prove that BA : AT = 2 :1. 5
Answer:
Given: O is the centre of circle and TP is the tangent to circle and ∠PBT = 30°.
To prove: \(\frac{\mathrm{BA}}{\mathrm{AT}}=\frac{2}{1}\)
Construction: Join OP
Proof: ∠BPA = 90° [Angle in a semi circle]
In ΔBPA, ∠BPA + ∠PBA + ∠BAP = 180° [Angle sum property]
⇒ 90° + 30° + ∠BAP = 180°
⇒ ∠BAP = 60°
Also, ∠OPT = 90°
and OP = OA
(Radii of same circle) ……. (i)
∠OAP = ∠OPA
= 60° ….. (ii)
⇒ ∠APT = 90° – 60°
= 30°
Now, ∠OAP + ∠PAT = 180° [Linear pair]
⇒ 60° + ∠PAT = 180° [Using (ii)]
⇒ ∠PAT = 120°
In ΔPAT, ∠PAT + ∠APT + ∠PTA = 180°
⇒ 120° + 30° + ∠PTA = 180°
⇒ ∠PTA = 30°
⇒ PA = AT …… (iii)
Also, in ΔOAP,
∠AOP = 180° – 60° – 60° = 60°
∴ ∠AOP = ∠OPA
⇒ PA = OA …… (iv)
Hence, PA = AT = OA = OP
[Using (i), (iii) and (iv)]
Now, BA = BO + OA = 20A [∵ OA = OB]
⇒ BA = 2AT [∵ OA = AT]
⇒ \(\frac{\mathrm{BA}}{\mathrm{AT}}=\frac{2}{1}\)
Question 35.
The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequencies f1 and f2. 5
Classes | Frequency |
0 – 20 | 5 |
20 – 40 | f1 |
40 – 60 | 10 |
60 – 80 | f2 |
80 – 100 | 7 |
100 – 120 | 8 |
OR
Calculate the median for the following data:
Class | Frequency |
More than or equal to 150 | 0 |
More than or equal to 140 | 12 |
More than or equal to 130 | 27 |
More than or equal to 120 | 60 |
More than or equal to 110 | 105 |
More than or equal to 100 | 124 |
More than or equal to 90 | 141 |
More than or equal to 80 | 150 |
Answer:
The frequency distribution for calculating the mean, for the given data is:
We know that,
30 + f1 + f2 = 50
⇒ f1 + f2 = 20
⇒ f2 = 20 – f1 ………… (i)
Now, mean = A + \(\frac{\Sigma f_i d_i}{\Sigma f_i}\)
⇒ 62.8 = 50 + \(\frac{560+20 f_2-20 f_1}{50}\)
⇒ 62.8 = 50 + \(\frac{560+20\left(20-f_1\right)-20 f_1}{50}\) [Using (i)]
⇒ 640 = 960 – 40 f1
⇒ 40 f1 = 320
f1 = 8
∴ f2 = 20 – 8 = 12
∴ f2 = 8, f2 = 12.
OR
Converting the given distribution into continuous distribution
Cl | Frequency | c.f. (Less than type) |
80 – 90 | 9 | 9 |
90 – 100 | 17 | 26 |
100 – 110 | 19 | 45 |
110 – 120 | 45 | 90 |
120 – 130 | 33 | 123 |
130 – 140 | 15 | 138 |
140 – 150 | 12 | 150 |
150 or more | 0 | 150 |
Total | 150 |
Here,
N = 150, \(\frac{N}{2}\) = 75
∴ Median class is 110 – 120
Here, l = 110, cf= 45, f= 45, h = 10
Median = l + \(\frac{\left(\frac{N}{2}-c f\right)}{f}\) × h
= 110 + \(\frac{(75-45)}{45}\) × 10
= 110 + \(\frac{300}{45}\)
= 110 + 6.67
= 116.67 (approx) Hence, the median is 116.67.
Section – E (12 Marks)
CASE STUDY 1
Question 36.
Rishu is riding in a hot air bailoon. After reaching a point P, he spots a car parked at B on the ground at an angle of depression of 30°. The balloon rises further by 50 metres and now he spots the same car at an angle , of depression of 45° and a lorry parked at B’ at an angle of depression of 30°. (Use √3 . = 1.73)
The measurement of Rishu facing vertically is the height. Distance is defined as the measurement of car/lorry from a point in a horizontal direction. If an imaginary line is drawn from the observation point to the top edge of the car/lorry, a triangle is formed by the vertical, horizontal and imaginary line.
Based on the above information, answer the following questions:
(A) If the height of the balloon at point P is ‘h’m and distance AB is V m, then find the relation between V and ‘h’. 1
(B) What is the relation between the height of the balloon at point P and distance AB? 1
(C) Find the height of the balloon at point P and the distance AB on the ground.
OR
Find the distance B’B on the ground. 2
Answer:
(A) In ΔAPB,
tan 30° = \(\frac{A P}{A B}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{h}{x}\)
⇒ x = \(\sqrt{3} h\)
(B) In ΔAP’B
tan 45° = \(\frac{A P^{\prime}}{A B}\)
⇒ AB = AP’
⇒ x = h + 50 …….. (ii)
(C) On solving equation obtained in (i) and (ii), we get
\(\sqrt{3} h\) = h + 50
⇒ h(√3 – 1) = 50
⇒ h = \(\frac{50}{0.732}\) = 68.5
In ΔAPB,
tan 30° = \(\frac{A P}{A B}\)
⇒ AB = \(\frac{\mathrm{AP}}{\tan 30^{\circ}}\)
= \(\frac{68.25}{\frac{1}{\sqrt{3}}}\)
= 68.25 × 1.732
= 118 m
OR
In ΔAP’B’
tan 30° = \(\frac{\mathrm{AP}^{\prime}}{\mathrm{AB}^{\prime}}\)
\(\frac{1}{\sqrt{3}}=\frac{68.25+50}{A B^{\prime}}\)
⇒ AB’ = 118.25 × 1.732
= 204.809
BB’ = AB’-AB
= 204.809 – 118
= 86.80 = 87 m
Question 37.
CASE STUDY 2
The military base camps of neighbouring countries are at the ground level on either side of a mountain. Their corresponding sub camps are at certain points. Mora point and Lora point on moving a certain distance towards the top of the mountain as shown in the figure. The ratio between the distance from base camp B to Lora point and that of Lora Point to Mountain top is 2 : 5.
Based on the above information, answer the following questions:
(A) What is the ratio of the perimeters of the triangle formed by both base camps and Mountain top to triangle formed by Mora Point, Lora point and Mountain top? 1
(B) Find the distance between the Base camp A and Mora point.
OR
If the horizontal distance between the Mora point and Lora point is 6 miles, then what is the distance between the two base camps? 2
(C) If the distance from Mountain top to Mora point is 0.5 miles more than that of distance from base camp B to Mountain top, then what is the distance between Lora point and Mountain top? 1
Answer:
(A) Let ΔABC is the triangle formed by both base camps and mountain top (C) and ACDE is the triangle formed by Mora point (D), Lora point (E) and Mountain top (C) Clearly, DE || AB and so
ΔABC ~ ΔDEC [By AA-similarity criterion]
Now, require ratio corresponding sides = Ratio of their
= \(\frac{B C}{E C}=\frac{7}{5}\) fie., 7 : 5.
(B) Since, DE || AB, therefore
\(\frac{C D}{A D}=\frac{C E}{E B}\)
⇒ \(\frac{7.5}{A D}=\frac{5}{2}\)
⇒ 15 = 5AD
⇒ AD = 3 miles
OR
OR
Since ΔABC ~ ΔDEC
∴ \(\frac{B C}{E C}=\frac{A B}{D E}\)
[∵ Corresponding sides of similar triangle are proportional]
⇒ \(\frac{7}{5}=\frac{A B}{6}\)
⇒ AB = \(\frac{42}{5}\) = 8.4 miles
(C) Given, DC = 0.5 + BC and we need to find CE
Clearly, BC = 7.5 – 0.5 = 7 miles
Now, CE = \(\frac{5}{7}\) × BC
= \(\frac{5}{7}\) × 7 = 5 miles
Question 38.
CASE STUDY 3
DeLhi public school society, which has so many schools in different cities of India. One of the branch of Delhi Public School is in Ghaziabad. In that school thousand of students study in the classroom.
Out of them one of the boy is standing in the ground having coordinates (4, 1) facing ‘ towards east. He moves 4 units in straight line then take right and moves 3 units and stop. Now he is at his home.
The representation of the given situation on the coordinate axes is show below.
Based on the above information, answer the following questions:
(A) What is the shortest distance between his school and house?
OR
Suppose point D divide the line segment AB in the ratio 1: 2, then find the coordinate of D. 2
(B) If we draw perpendicular lines from
points A and B to the x-axis, find, the region covered by these perpendicular lines. 1
(C) Find the image of point C w.r.t. x-axis.l
Answer:
(A) From the given figure, shortest distance between school and house = AB
= \(\sqrt{(8-4)^2+(4-1)^2}\)
[from distance formula]
= \(\sqrt{4^2+3^2}=\sqrt{16+9}\)
= \(\sqrt{25}\) = 5 units
∴ Shortest distance = 5 km[l unit = 1 km]
OR
By using internal section formula.
(B) When, we draw perpendicular lines from the points, a rectangle is formed.
∴ Area of covered region
= l × b
= 4 × 1 = 4 sq units
(C) The image of a point C w.r.t. X-axis is (8,-4).