Students can access the CBSE Sample Papers for Class 10 Maths with Solutions and marking scheme Term 2 Set 1 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 1 with Solutions
Time: 2 Hours
Maximum Marks: 40
General Instructions:
- The question paper consists of 14 questions divided into 3 sections A, B, C.
- All questions are compulsory.
- Section A comprises of 6 questions of 2 marks each. Internal choice has been provided in two questions.
- Section B comprises of 4 questions of 3 marks each. Internal choice has been provided in one question.
- Section C comprises of 4 questions of 4 marks each. An internal choice has been provided in one question.
- It contains two case study based questions.
Section – A [2 Marks Each]
Question 1.
Find the value of a25 – a15 for the AP: 6,9,12,15,
OR
If 7 times the seventh term of the AP is equal to 5 times the fifth term, then find the value of its 12th term.
Answer:
a = 6, d = 3;
a25 = 6 + 24(3) = 78
a15 = 6 + 14(3) = 48
a25 – a15 = 78 – 48
a25 – a15 = 30
OR
7(a + 6 d) = 5 (a + 4d)
⇒ 2a + 22d = 0
⇒ a + 11d = 0
Detailed Solution:
6, 9, 12, 15, ……………. is the given A.P.
First term = a = 6
Common Difference = d = 9 – 6 = 3
nth term = an = a + (n – 1)d
25th term = a25
= 6 + (25 -1) × 3
= 6 + (24) × 3
= 6 + 72
= 78
15th term = a15
= 6 + (15 – 1) × 3
= 6 + (14) × 3
= 6 + 42 = 48
a25 – a15 = 78 – 48 = 30
OR
Let ‘a’ and ‘d’ be the first term and common difference of AP
nth term = an = a + (n – 1)d
7th term = a7 = a + (7 – 1)d = a + 6d
5th term = a5 = a + (5 – 1)d = a + 4d
According to the Questions,
7a7 = 7a5
7(a + 6d) = 5(a + 4d)
7a + 42d = 5a + 20d
7a – 5a + 42d – 20d = 0
2a + 22d = 0
2(a + 11 d) =0
a + 11d =0 …(i)
12th term = a12
= a + 11d
= 0 (From (i)
Hence, 12th term is zero.
Question 2.
Find the value of m so that the quadratic equation mx(5x – 6) + q = 0 has two equal roots.
Answer:
Given 5m2 – 6mx + 9 = 0
Since, b2 – 4ac = 0
for equal roots (-6 m)2 – 4 (5 m) (9) = 0
⇒ 36m(m – 5) = 0
⇒ m = 0,5; rejecting m = 0,
we get m = 5
Detailed Solution:
Since,mi (5x – 6) + 9 = 0 has equal roots
Discriminant = 0
⇒ b2 – 4ac = 0 …(i)
Quadratic equation can be written as
5m2 – 6mx + 9 = 0
Here, a = 5m, b = – 6m, c = 9
Put in (i),
(6 m)2 – 4(5 m)(9) = 0 m = 0
= 36 m2 – 180m = 0 or m = 5
= 36 m(m – 5) = 0
Either m = 0 or m = 5
Since, m = 0 is not possible
Hence, m = 5
Question 3.
From a point P, two tangents PA and PB are drawn to a circle C(0, r).
If OP = 2r, then find ∠APB. What type of triangle is APB?
Answer:
Let ∠APO = θ
sinθ = \(\frac{O A}{O P}=\frac{1}{2}\)
⇒ ∠APB = 2θ = 60°
Also ∠PAB = ∠PBA = 60° (∴ PA = PB)
⇒ ∆APB is equilateral
Detailed Solution:
Radius of given circle OA = r units
Let, ∠APO = θ
Radius is always perpendicular to tangent
= OA ⊥ AP
So, ∠OAP = 90°
∆OAP is a right angled triangle
In ∆OAP,
sin θ = \(\frac{OA}{OP}\)
sin θ = \(\frac{r}{2 r}=\frac{1}{2}\)
sin θ = sin 30°
θ = 30°
∠APO = 30°
∠APB = 2 × ∠APO = 2 × 30° = 60°
PA = PB [Length of tangents from an external point are equal in length]
Let, ∠ABP = ∠BAP = x
So, from ∆APB,
∠APB + ∠ABP + ∠BAP = 180° [Angle sum property of triangle]
60° + x + x = 180°
or x = 60°
∴ ∆APB is equilateral triangle.
Question 4.
The curved surface area of a right circular cone is 12320 cm2. If the radius of its base is 56 cm, then find its height.
Answer:
CSA (cone) = πrl = 12320 cm2
\(\frac{22}{7}\) × 56 × l = 12320
l = 70 cm
h = \(\sqrt{70^{2}-56^{2}}\) = 42 cm
Detailed Solution:
Let Height of cone = h
Radius of cone = 56 cm
C.S.A of cone = 12320 cm2
Let slant height of cone = l cm
Curved surface area of cone = πrl
⇒ 12320 = \(\frac{22}{7}\) × 56 × l
⇒ \(\frac{12320 \times 7}{22 \times 56}\) = l
⇒ l = 70 cm
⇒ l2 = h2 + r2
⇒ (70)2 = h2 + (56)2
⇒ h2 = 4900 – 3136
⇒ h2 = 1764
⇒ h2 = 422
⇒ h = 42 cm
Question 5.
Mrs. Garg recorded the marks obtained by her students in the following table. She calculated the modal marks of the students of the class as 45. While printing the data, a blank was left. Find the missing frequency in the table given below.
Answer:
Modal class is 40 – 60,
l = 40,
h = 20, f1 = ?,
f0 = 10, f2 = 6
45 = 40 + 20 × \(\left[\frac{f_{1}-10}{2 f_{1}-10-6}\right]\)
⇒ \(\frac{1}{4}=\frac{f_{1}-10}{2 f_{1}-16}\)
⇒ 2f1 – 16 = 4f1 – 40
⇒ f1 =12
Detailed Solution:
Marks obtained Number of students
| Marks obtained | Number of students |
| 0-20 | 5 |
| 20-40 | 10 |
| 40-60 | x (Say) |
| 60-80 | 6 |
| 80 -100 | 3 |
Mode = 45
⇒ Modal class =40 – 60
Lower limit of modal class (l) = 40
Size of modal class (h) = 20
Frequency corresponding to modal class (F1) = x
Frequency preceding to modal class (F0) = 10
Frequency preceding to modal class (F2) = 6
Question 6.
If Ritu were younger by 5 years than what she really is, then the square of her age would have been 11 more than five times her present age. What is her present age?
OR
Solve for x : 9x2 – 6px + (p2 – q2) = 0
Answer:
Let the present age of Ritu be x years
(x – 5)2 = 5x + 11
x2 – 15x + 14 = 0
(x – 14)(x – 1) =0
⇒ x = 1 or 14
x = 14 years (rejecting x 1 as in that case Ritu’s age 5 years ago will be – ve)
OR
9x2 – 6px + (p2 – q2) = 0
a =9, b = -6p, c = p2 – q2
D = b2 – 4ac
= (-6p)2 – 4(9)(p2 – q2)
= 36q2
x = \(\frac{-b \pm \sqrt{D}}{2 a}=\frac{6 p \pm 6 q}{18}\)
= \(\frac{p+q}{3}\) or \(\frac{p-q}{3}\)
Detailed Solution:
Let present age of Ritu = x years
As per question given the following equation can be formed:
(x – 5)2 = 11 + 5x
⇒ x2 – 10x + 25 = 11 + 5x
⇒ x2 – 10x – 5x + 25 – 11 = 0
⇒ x2 – 15x + 14 = 0
⇒ x2 – 14x – 1x + 14 =0
⇒ x(x – 14) -1(x -14) = 0
⇒ (x – 14)(x – 1) = 0
⇒ x – 14 =0, x – 1 = 0
⇒ x = 14, x = 1
Since, her age can’t be 1, so her present age will be 14 years.
OR
9x2 – 6px + (p2 – q2) = 0
In comparing with
ax2 + bx + c = 0
a = 9, b = -6p, c = p2 – q2
Solve by quadratic formula,
D = b2 – 4ac
= (- 6p)2 – 4(9)(p2 – q2)
= 36p2 – 36p2 + 36q2
= 36q2
Section – B [3 Marks Each]
Question 7.
Following is the distribution of the long jump competition in which 250 students participated. Find the median distance jumped by the students. Interpret the median
Answer:
\(\frac{n}{2}=\frac{250}{2}\)
⇒ 125 = median class is 2 – 3,
l = 2, h= 1, cf = 120, f = 62
Median = l + \(\frac{\frac{n}{2}-c f}{f}\) × h
= 2 + \(\frac{5}{62}\)
= \(\frac{129}{62}=2 \frac{5}{62}\)m or 2.08m
50% of students jumped below 2\(\frac{5}{62}\) m and 50% above it.
Detailed Solution:
| Distance(in m) | No. of students(Frequency) | Cumulative frequency |
| 0-1 | 40 | 40 |
| 1-2 | 80 | 120 |
| 2-3 | 62 | 182 |
| 3-4 | 38 | 220 |
| 4-5 | 30 | 250 |
| N = 250 |
Take \(\frac{N}{2}=\frac{250}{2}\)= 125
∴ Median dass =2-3
Lower limit of median class (1) = 2
Size of median class (h) = I
Frequency corresponding to median class (f) = 62
Total number of observations
Frequency (N) = 250
Cumulative frequency preceding
Median class (c.f.) = 120
∴ 50% of students jumped below 2.08 m and
50% of students jumped above 2.08 m.
Question 8.
Construct a pair of tangents to a circle of radius 4 cm, which are inclined to each other at an angle of 60°.
Answer:
Draw circle of radius 4cm
Draw OA and construct ∠AOB = 120″
Draw ∠OBP = ∠OBP = 90°
PA and PB are required tangents
Detailed Solution:
Steps of construction:
- Draw a circle of radius 4 cm.
- Draw two radii having an angle of 120°.
- Let the radii intersect circle at A and B.
- Draw angle of 90° on both A and B.
- The point where both rays of 90° intersect is P.
- PA and PB are the required tangents.
Question 9.
The distribution given below shows the runs scored by batsmen in one-day cricket matches. Find the mean number of runs.
Answer:
Mean(x̄) = \(\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{13280}{120}\) = 110.67 runs
Question 10.
Two vertical poles of different heights are standing 20 m away from each other on the level ground. The angle of elevation of the top of the first pole from the foot of the second pole is 60° and angle of elevation of the top of the second pole from the foot of the first pole is 30°. Find the difference between the heights of two poles. (Take √3 = 1.73)
OR
A boy 1.7 m tall is standing on a horizontal ground, 50 m away from a building. The angle of elevation of the top of the building from his eye is 60°. Calculate the height of the building. (Take √3 = 1.73)
Answer:
Let PRbe the building and AB be the boy
In ∆PQR,tan60°= PQ = 50√3m
Height of the building = (50√3 + 1.7)m
= 88.2m
Detailed Solution:
Let the heights of two pole be y and x.
Distance between the poles is QS = 20 m.
Difference between their heights
y – x = (20√3 – \(\frac{20}{3}\)√3)m
= \(\frac{20}{3}\)√3 × 2
= \(\frac{40}{3}\)√3
= 23.07 m
OR
Height of Boy = AB =1.7m
= QR = 1.7m
Distance between Boy and building
BR = 50 m
⇒ AQ = 50 m
In ∆PQA, ∠Q = 90°
tan 60° = \(\frac{P Q}{A Q}\)
√3 = \(\frac{P Q}{A Q}\)
50√3 = \(\frac{P Q}{50}\)
Total height of the building = PQ + QR
= 86.5 + 1.7
= 88.2m.
Section – C [4 Marks Each]
Question 11.
The internal and external radii of a spherical shell are 3 cm and 5 cm respectively. It is melted and recast into a solid cylinder of diameter 14 cm, find the height of the cylinder. Also find the total surface area of the cylinder. (Take π = \(\frac{22}{7}\))
Answer:
Volume of shell = Volume of cylinder
⇒ \(\frac{4π}{2}\)[53 – 33] = π(7)2h
⇒ h = \(\frac{8}{3}=2 \frac{2}{3}\)
TSA of cylinder is = 2πr(r + h)
= 2 × \(\frac{22}{7}\) × 7 × (7 + \(\frac{8}{3}\))
= 44 × \(\frac{29}{3}\)
= \(\frac{1276}{3}\)cm2 or 425.33 cm2
Detailed Solution:
Internal radius (r) 3 cm
External radius (R) = 5 cm
Radius of cylinder (R1) = \(\frac{14}{2}\) = 7 cm
Let height of cylinder = h cm
According to the question,
[When one shape is reshaped into another shape the volumes are same of both shapes]
\(\frac{4}{3}\)π(R3 – r3) = π(R1)2h
\(\frac{4}{3}\)(53 – 33) = (7)2h
\(\frac{4}{3}\) × (125 – 27) = 49h
\(\frac{4}{3} \times \frac{98}{49}\) = h
h = \(\frac{8}{3}\) m or 2\(\frac{2}{3}\) cm
TSA of cylinder = 2πr(h + R1)
= 2 × \(\frac{22}{7}\) × 7(\(\frac{8}{3}\) + 7)
= 44 + \(\frac{29}{3}\)
= \(\frac{1276}{3}\) cm3
or
425.33 cm2
Question 12.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact to the centre.
OR
Two tangents TP and TQ are drawn to a circle with centre O from an external point T.
Prove that ∠PTQ = 2∠OPQ
Answer:
∠OAP + ∠OBP + ∠APB + ∠AOB = 360°
= 90° + 90° + ∠APB + ∠AOB = 360° (∵ Tangent ⊥ radius)
= ∠APB + ∠AOB = 180°
OR
Let ∠PTQ = θ
TPQ is an isosceles triangle
∠TPQ = ∠TPQ = \(\frac{1}{2}\)(180° – θ)
= 90° – \(\frac{\theta}{2}\)
∠OPT = 90°
∠OPT = ∠OPT – ∠OPT – ∠TPQ
= 90° – (90° – \(\frac{\theta}{2}\))
= \(\frac{\theta}{2}\)
∠OPT = \(\frac{1}{2}\)∠PTQ
2∠OPQ = ∠PTQ
Detailed Solution:
Given: PA and PB are two tangents to a circle with centre O.
To Prove:
∠APB + ∠AOB = 180°
Proof: OAPB is a quadrilateral, OA is radius and PA & PB are tangents.
∠OAP = ∠OBP = 90°
[Radius is always perpendicular to tangent] By angle sum property of quadrilateral.
.-. ∠AOB + ∠OBP + ∠APB + ∠OAP
= 360°
∠AOB + 90° + ∠APB + 90° = 360°
.-. ∠APB + ∠AOB = 180°
Hence Proved.
OR
Given: A circle with centre O two tangents
TP and TQ to the circle where P and Q are the point of contact.
To Prove: ∠PTQ = 2∠OPQ
Proof: TP = TQ
[length of tangents drawn from same external point to a circle are equal]
∴ ∠TQP = ∠TPQ …(i)
[Angles opposite to equal sides are equal] Now, PT is a tangent & OP is radius
∴ OP ⊥ TP[Radius and Tangent are perpendicular to each other]
⇒ ZOPT = 90°
∠OPQ + ∠TPQ = 90°
∠TPQ = 90° – ∠OPQ …(ii)
In APTQ,
By angle sum property of triangle,
∠TPQ + ∠TQP + ∠PTQ = 180° [From (i)]
∠TPQ + ∠TPQ + ∠PTQ = 180°
⇒ 2[TPQ + ∠PTQ = 180°
⇒ 2(90° – ∠OPQ) + ∠PTQ = 180° [From (ii)]
⇒ 180 – 2 ∠OPQ + ∠PTQ = 180°
∠PTQ = 2∠OPQ
Hence Proved.
Case Study-1
Question 13.
Trigonometry in the form of triangulation forms the basis of navigation, whether it is by land, sea or air. GPS a radio navigation system helps to locate our position on earth with the help of satellites.
A guard, stationed at the top of a 240 m tower, observed an unidentified boat coming towards it. A clinometer or inclinometer is an instrument used for measuring angles or slopes(tilt). The guard used the clinometer to measure the angle of depression of the boat coming towards the lighthouse and found it to be 30°.
(Lighthouse of Mumbai Harbour. Picture credits – Times of India Travel)
(i) Make a labelled figure on the basis of the given information and calculate the distance of the boat from the foot of the observation tower. [2]
(ii) After 10 minutes, the guard observed that the boat was approaching the tower and its distance from tower is reduced by 240(√3 – 1) m. He immediately raised the alarm. What was the new angle of depression of the boat from the top of the observation tower? [2]
Answer:
In ∆PTR, tan 30° = \(\frac{240}{x}\) ⇒ x = 240√3
(ii) Distance of boat from tower
= 240(√3 – 240(√3 – 1)
= 240 m
Let the angle of depression = θ
tan θ = \(\frac{240}{240}\) = 1
θ = 45°
Detailed Solution:
(i) Height of light house = 240 m
In ∆PRT, ∠R = 90°
tan 30° = \(\frac{2 A 0}{x}\)
\(\frac{1}{\sqrt{3}}=\frac{240}{x}\)
x = 240 √3 m.
Distance of foot from the foot of observation tower = 240 √3 m.
(ii) After 10 minutes,
Distance between boat and light house is reduced by
= 24(√3 – 1)m
Distance between boat and light house
= 240√3 – 240 (√3 – 1)
= 240√3 – 240√3 + 240
= 240 m
Let the point C is the new position of boat,
tan θ = \(\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{240}{240}\)
⇒ θ = 45°
Case Study-2
Question 14.
Push-ups are a fast and effective exercise for building strength. These are helpful in almost all sports including athletics. While the push-up primarily targets the muscles of the chest, arms, and shoulders, support required from other muscles helps in toning up the whole body.
Nitesh wants to participate in the push-up challenge. He can currently make 3000 push-ups in one hour. But he wants to achieve a target of 3900 push-ups in 1 hour for which he practices regularly. With each day of practice, he is able to make 5 more push-ups in one hour as compared to the previous day. If on first day of practice he makes 3000 push-ups and continues to practice regularly till his target is achieved. Keeping the above situation in mind answer the following questions:
(i) Form an A.P representing the number of push-ups per day and hence find the minimum number of days he needs to practice before the day his goal is accomplished? [2]
(ii) Find the total number of push-ups performed by Nitesh up to the day his goal is achieved. [2]
Answer:
(i) 3000,3005,3010,…, 3900
an = a + (n – 1 )d
3900 = 3000 + (n – 1)5
⇒ 900 = 5n – 5 ⇒ 5n = 905
⇒ n = 181
Minimum number of days of practice = n – 1 = 180 days
(ii) Sn = \(\frac{n}{2}\)(a + 1)
= \(\frac{181}{2}\) × (3000 + 3900)
= 624450 push-ups
Detailed Solution:
(i) By the given situation the A.P. to be formed is:
3000, 3005,3010,…………… 3900
First term = a = 3000
Common Difference = d = 3005 – 3000
d = 5
According to the problem nth term (an) = 3900
To Find n
nth term = an
= a + (n – 1)d
3900 = 3000 + (n -1)5
3900 – 3000 = (n -1)5
900 = (n -1)5
\(\frac{900}{5}\) = n -1
180 + 1 = n
or n = 181
Minimum number of days he needs to practice before his goal is accomplished
= 181 – 1 [Excluding the last day]
= 180
(ii) Total Number of push-ups performed means sum of all push-ups he did in 181 days.
Sn = \(\frac{n}{2}\) [a + l] …(i)
Sn = Sum of all push-ups in 181 days.
n = number of days = 181
a = first term of the A.P = 3000
l = last term = 3900
Put all values in (i)
⇒ Sn = \(\frac{181}{2}\) [3000 + 3900]
= \(\frac{181}{2}\) [6900]
= 624450
Total number of Push-ups performed in 181 days
= 624450.