Students can access the CBSE Sample Papers for Class 10 Maths with Solutions and marking scheme Term 2 Set 3 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 10 Maths Standard Term 2 Set 3 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 20th term from the last term of the A.E: 3,8,13, 253.
OR
If 7 times the 7th term of an A.E is equal to 11 times its 11th term, then find its 18th term.
Answer:
20th term from the end = l – (n – 1)d
= 253 – 19 × 5
= 158
Detailed Solution:
Given, A.P.: 3, 8,13, ………… 253
Here, first term (a) = 3, common difference (d)
= 8 – 3 = 5 and last term (l) = 253
Then, 20th term from the end of the A.P.
= l – (n – 1)d
= 253 – (20 -1)5
= 253 – 95
= 158.
OR
7a7 = 11a11
⇒ 7 (a + 6d) = 11(a + 10 d)
⇒ a + 17d = 0
∴ a18 = 0
Detailed Solution:
Given, 7a7 = 11a11
∵ an = a + (n – 1 )d
Then, 7[a + (7 – 1)d] = 11[a + (11 – 1)d]
⇒ 7 (a + 6d) = 11(a + 10 d)
⇒ 7a + 42d = 11a + 110 d
⇒ 11a – 7a = 42d – 110d
⇒ 4a = – 68 d
⇒ a = – 17d
⇒ a + 17d = 0
i.e., a + (18 – 1)d = 0
Hence, a18 = 0.
Question 2.
A fast train takes 3 hours less than a slow train for a journey of 600 km. If the speed of the slow train is 10 km/h less than that of the fast train, find the speed of each train.
Answer:
Total distance of the journey = 600 km
Let speed of fast train = x km/h,
then speed of slow train = (x – 10) km/h
According to question,
\(\frac{600}{x-10}\) – \(\frac{600}{x}\) = 3
[∵ Time = \(\frac{\text { Distance }}{\text { Speed }}\)]
⇒ 600 \(\left[\frac{x-x+10}{(x-10) x}\right]\) = 3
⇒ \(\frac{6000}{x^{2}-10 x}\) = 3
⇒ x2 – 10x – 2000 = 0
⇒ x2 – 50x + 40x – 2000 = 0
⇒ x(x – 50) + 40(x – 50) = 0
⇒ (x – 50) (x + 40) = 0
Either, x = 50 or x = – 40
∵ speed can not be negative.
So, the speed of fast train = 50 km/h, and the speed of slow train = 50 – 10 = 40 km/h.
Commonly Made Error:
Some students do not know how to frame the equation. Some frame it correctly but fail to solve it.
Answering Tip:
Emphasis on solving on quadratic equation based application problems is necessary.
Question 3
In the given figure PQ is chord of length 6 cm of the circle of radius 6 cm. TP and TQ are tangents to the
circle at points P and Q respectively. Find ∠PTQ.
Answer:
Here, PQ = 6 cm, OP = OQ = 6 cm (radii)
∴ PQ = OP = OQ
∴ ∠POQ = 60°
(angle of equilateral ∆)
∠OPT = ∠OQT = 90°
(radius ⊥ tangent)
∴ ∠PTO + 90° + 90° + 60° = 360°
(angle sum property)
∠PTQ = 120°.
Question 4.
A well of diameter 4 m dug 21 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 3 m to form an embankment. Find the height of the embankment.
Answer:
Diameter of well = 4 m :
Radius of well, r = 2 m
Depth of well, h = 21 m,
Volume of earth dug out = πr2h
= \(\frac{22}{7}\) × 2 × 2 × 21
= 264 m3
Width of embankment = 3 m
Outer radius of ring = 2 + 3 = 5 m
Let the height of embankment be h
∴ Volume of embankment = Volume of earth dug out
π(R2 – r2)h = 264
\(\frac{22}{7}\) × (25 – 4) × h = 264
h = \(\frac{264 \times 7}{22 \times 21}\)
∴ Height of embankment = 4 m,
Question 5.
Find the mode of the following frequency distribution:
Answer:
Modal class is 30 – 40 Mode
∴ Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h
= 30 + \(\left(\frac{16-10}{32-10-12}\right)\) × 10
= 36.
Detailed Solution:
| Class | Frequency |
| 0 -10 | 8 |
| 10-20 | 10 |
| 20-30 | 10 |
| 30-40 | 16 |
| 40-50 | 12 |
| 50-60 | 6 |
| 60-70 | 7 |
Modal-class = 30 – 40
⇒ l = 30, f0 = 10, f1 = 16, f2= 12, h = 10
Question 6.
For what values of k, the roots of the equation x2 + 4x + k = 0 are real ?
OR
Find the nature of roots of the quadratic equation 2x2 – 4x + 3 = 0.
Answer:
Since roots of the equation x2 + 4x + k = 0 are real
⇒ 16 – 4k ≥ 0
⇒ k ≤ 4
Detailed Solution:
Given x2 + 4x + k = 0.
Comparing the given equation with ax2 + bx + c = 0,
we get a = 1, b = 4, c = k 1
Since, given the equation has real roots
⇒ D ≥ 0
⇒ b2 – 4ac ≥ 0
⇒ 42 – 4 × 1 × k ≥ 0
⇒ 4k ≤ 16
⇒ k ≤ 4
OR
Given: 2x2 – 4x + 3 = 0
On comparing above with ax2 + bx + c = 0,
we get, a = 2, b = – 4, c = 3
Now, D = b2 – 4ac
So, D = (- 4)2 – 4(2) × (3)
= – 8 < 0 or (- ve)
Hence, the given equation has no real roots.
Commonly Made Error: Students often make mistakes in analyzing the nature of roots as they get confused with the conditions.
Answering Tip: Understand tire different conditions tor nature of roots.
Section – B [3 marks Each]
Question 7.
The table below show the salaries of 280 persons:
Calculate the median salary of the data.
Answer:
\(\frac{280}{2}\) = \(\frac{N}{2}\) = 140.
Median class = 10 – 15.
Median = 1 + \(\frac{h}{f}\)(\(\frac{N}{2}\) – c.f.)
= 10 + \(\frac{5}{133}\) (140 – 49)
= 10 + \(\frac{5 \times 91}{133}\)
= 13.42
Hence, median salary is ₹ 13.42 thousand or ₹ 13420 (approx).
Question 8.
Draw two tangents to a circle of radius 4 cm, which are inclined to each other at an angle of 60°.
Answer:
Steps of construction:
- Draw a circle of radius 4 cm with O as centre.
- Draw two radii OA and OB inclined to each other at an angle of 120°.
- Draw AP ⊥ OA at A and BP ⊥ OB at B. Which meet at P.
- PA and PB are the required tangents inclined to each other an angle of 60°.
Question 9.
Daily wages of 110 workers, obtained in a survey, are tabulated below:
Compute the mean daily wages of these workers.
Answer:
Mean, x = \(\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}\) = \(\frac{18720}{110}\) = 170.82
Hence, mean daily: wages are ₹ 170.182 (Approx)
Question 10.
Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point in between them on the road, the angles of elevation of the top of poles are 60° and 30° respectively. Find the height of the poles and the distances of the point P from the poles.
OR
From a point P on the ground, the angle of elevation of the top of a tower is 30° and that of the top of the flag-staff fixed on the top of the tower is 45°. If the length of the flag-staff is 5 m, find the height of the tower. (Use √3 = 1.732)
Answer:
Let two poles be AB and CD, then AB = CD = h
In ∆ ABP, ∠A = 90°
\(\frac{h}{x}\) = tan 60°= √3
In ∆ PCD, ∠C = 90°
\(\frac{h}{80-x}\) = tan 30° = \(\frac{1}{\sqrt{3}}\) =
Dividing (i) by (ii), we get
\(\frac{80-x}{x}\) = \(\frac{3}{1}\)
⇒ 3x = 80 – x
or 4x = 80
⇒ x = 20 m.
and h = 20 √3 m.
∴ Height of poles is 20 √3 m and the distances of P are 20 m and 60 m from poles.
OR
Let AB be a tower and BC be a flagstaff.
According to question,
In ∆ BAC, ∠A = 90°
\(\frac{A C}{A P}\) = tan 45° = 1
Commonly Made Error:
Many candidates were not able to understand the language of question.
Answering Tip:
Do sufficient practice of drawing correct diagrams for problems based on Heights and Distances.
Section – C [4 Marks Each]
Question 11.
The given figure is a decorative block, made up of two solids : a cube and a hemisphere. The base of the block is a cube of side 6 cm and the hemisphere fixed on the top has a diameter of 3.5 cm. Find the total surface area of the block. [Use π = \(\frac{22}{7}\)].
Answer:
Surface area of block
= 216 – \(\frac{22}{7}\) × \(\frac{3.5}{2}\) × \(\frac{3.5}{2}\) + 2 × \(\frac{22}{7}\) × \(\frac{3.5}{2}\) × \(\frac{3.5}{2}\) lYz+lYz
= 225.625 cm2.
Detailed Solution:
Given, Side of cube = 6 cm
Diameter of hemisphere = 3.5 cm
Radius of hemisphere = \(\frac{3.5}{2}\)
Total surface area of cube = 6a2
= 6 × (6)2 = 216 cm2
Total surface area of solid = TSA of cube – Area of circle + CSA of hemisphere.
Question 12.
In given figure, AB is the diameter of a circle with centre O and AT is a tangent. If ∠AOQ = 58°, find ∠ATQ.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Answer:
∠ AOQ = 58° (Given)
∠ABQ = \(\frac{1}{2}\) ∠AOQ
(Angle on the circumference of the circle by the same arc)
= \(\frac{1}{2}\) × 58°
= 29°
∠BAT = 90° (∵ OA ⊥ AT)
∴ ∠ATQ = 90° – 29
= 61°
OR
Given: A circle with centre O is inscribed in a quadrilateral ABCD.
In ∆AEO and ∆AFO,
OE = OF (radii of circle)
∠OEA = ∠OFA = 90°
(radius is ⊥r to tangent)
The point of contact is perpendicular to the tangent.
OA = OA (common side)
∆AEO ≅ ∆AFO (R.H.S. congruency)
∠7 = ∠8(By cpct) …(i)
Similarly,
∠1 = ∠2 …(ii)
∠3 = ∠4 …(iii)
∠5 = ∠6 …(iv)
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
(angle around a point is 360°)
2 ∠1 + 2 ∠8 + 2 ∠4 + 2∠5 = 360°
∠1 + ∠8 + ∠4 + ∠5 = 180°
(∠1 + ∠8) + (∠4 + ∠5) = 180°
∠AOB + ∠COD = 180°
Hence Proved.
Case Study-1
Question 13.
A circus artist is climbing from the ground along a rope stretched from the top of a vertical pole and tied at the ground. The height of the pole is 12 m and the angle made by the rope with ground level is 30°.
Give answer of the following questions:
(i) Find the distance covered by the artist in climbing the top of the pole.
Answer:
Clearly, distance covered by the artist is equal to the length of the rope AC.
Let AB be the vertical pole of height 12 m.
It is given that ∠ACB = 30°
Thus, in right-angled triangle ABC,
sin 30° = \(\frac{A B}{A C}\)
⇒ \(\frac{1}{2}\) = \(\frac{12}{A C}\)
∴ AC = 24 m.
(ii) Calculate the length of BC.
Answer:
In ∆ABC, ∠B = 90°
tan 30° = \(\frac{A B}{B C}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{12}{B C}\)
⇒ BC = 12√3 m.
Case Study-2
Question 14.
Jaspal Singh takes a loan from a bank for his car.
Jaspal Singh repays his total loan of₹ 118000 by paying every month starting with the first installment of ₹ 1000. If he increases the installment by ₹ 100 every month.
(i) Calculate the amount paid by him in 25th installment.
Answer:
The amount paid by him in 25th installment
T25 = a + 24d
= 1000 + 24 × 100
= 1000 + 2400
= ₹ 3400
(ii) Calculate the amount paid by him in 30th installment.
Answer:
The amount paid by him in 30th installment,
T30 = a + 29 d
= 1000 + 29 × 100
= 1000 + 2900
= ₹ 3900.