These Sample papers are part of CBSE Sample Papers for Class 10 Maths. Here we have given CBSE Sample Papers for Class 10 Maths Paper 9. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.
CBSE Sample Papers for Class 10 Maths Paper 9
Board | CBSE |
Class | X |
Subject | Maths |
Sample Paper Set | Paper 9 |
Category | CBSE Sample Papers |
Students who are going to appear for CBSE Class 10 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 9 of Solved CBSE Sample Paper for Class 10 Maths is given below with free PDF download solutions.
Time Allowed: 3 hours
Maximum Marks: 80
General Instructions:
- All questions are compulsory.
- The question paper consists of 30 questions divided into four sections A, B, C and D.
- Section A contains 6 questions of 1 mark each. Section B contains 6 questions of 2 marks each. Section C contains 10 questions of 3 marks each. Section D contains 8 questions of 4 marks each.
- There is no overall choice. However, an internal choice has been provided in four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
- Use of calculators is not permitted.
SECTION-A
Question 1.
Given that HCF (54, 336) = 6, find LCM (54, 336).
Question 2.
Find the 10th term from the end of the AP: 5,10,15,………….. , 485.
Question 3.
If θ = 30°, verify that
\( cos\quad 2\theta =\cfrac { 1-{ tan }^{ 2 }\theta }{ 1+{ tan }^{ 2 }\theta } \)
Question 4.
In the given figure ΔABC and ΔPQR are similar. The area of ΔABC is 64 cm2 and the area of ΔPQR is 121 cm2. If QR – 15.4 cm, find BC
Question 5.
If a and b are the roots of the equation x2 + ax – b – 0, then find a and
Question 6.
The coordinates of the points P and Q are respectively (4, -3) and (-1, 7). Find the abscissa of a point R on the line segment PQ such that \(\cfrac { PR }{ PQ } \) = \(\cfrac { 3 }{ 5 } \)
SECTION—B
Question 7.
Find the ratio in which the line segment joining A(5, – 6) and B(-l, – 4) is divided by the y-axis. Also find the coordinates of the point of division.
Question 8.
Kartakay tosses two different coins simultaneously. What is the probability that he gets
(1) exactly one head?
(2) at least one head?
Question 9.
The decimal representation of \(\frac {6}{ 1250 } \) will terminate after how many places of decimal?
Question 10.
Find out whether the following pair of linear equations is consistent or not.
\(\cfrac {3}{ 2} \) x+ \(\cfrac {5}{3} y \) = 7
9x + 10y = 14
Question 11.
The first and the last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference.
Question 12.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
1. an orange flavoured candy?
2. a lemon flavoured candy?
SECTION-C
Question 13.
If p is a prime number, prove that \( \sqrt { p } \) is an irrational number.
OR
Use Euclid’s division lemma to show that the cube of any positive integer is of the form
9m, 9m + 1 or 9m + 8.
Question 14.
Draw the graphs of the equations x-y + 1 = 0 and 3x + 1y – 12 = 0. Find the area of the triangle formed by these lines and the x-axis.
Question 15.
Prove that: \(\cfrac { tan\quad \theta }{ 1-cot\quad \theta } +\cfrac { cot\quad \theta }{ 1-tan\quad \theta } =1+sec\theta +tan\theta \)= 1 + sec θ cosec θ
OR
Prove that:
\( \cfrac { sec\theta -cos\theta +1 }{ sin\theta +cos\theta -1 } =sec\theta +tan\theta \)
Question 16.
Find the point on the x-axis which is equidistant from the points (2, -5) and (-2, 9). Hence find the area of triangle formed by these points.
OR
Show that the points (7,10), (-2,5) and (3, -4) are the vertices of an isosceles right triangle.
Question 17.
In figure XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at Prove that ∠AOB = 90°.
Question 18.
In figure ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.
Question 19.
On dividing x3 – 3x2 + x+ 2 by a polynomial g(x), the quotient and remainder were (x – 2) and (4 – 2x), respectively. Find g(x).
Question 20.
In the given figure, Δ ABC and Δ DBC are on the same base BC. If AD intersects BC at O.
Prove that \( \cfrac { ar(\Delta ABC) }{ ar(\Delta ABC) } =\cfrac { AO }{ DO } \)
Question 21.
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank which is 10 m in diameter and 2 m deep. If the water flows through the pipe at the rate of 4 km per hour, in how much time will the tank be filled completely?
OR
The largest possible sphere is carved out of a wooden solid cube of side 7 cm. Find the volume of the wood left. [Use π = \(\cfrac { 22 }{ 7 } \) ]
Question 22.
If the mean of the following distribution is 6, find the value of p
SECTION-D
Question 23.
Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
OR
Two water taps together can fill a tank in 9\(\cfrac { 22 }{ 7 } \) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Question 24.
State and prove Pythagoras theorem. Using this theorem prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Question 25.
A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container at the rate of र 20 per liter. Also find the cost of metal sheet used to make the container, if it costs र 8 per 100 cm2. (Take π =3.14)
OR
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Question 26.
If the median of the distribution given below is 28.5, find the values of x and y
Question 27.
Find the sum of all three digit numbers which are divisible by 4 and 3.
OR
200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see figure). In how many rows are the 200 logs placed and how many logs are there in the top row?
Question 28.
Draw a triangle ABC with BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are \(\cfrac { 4 }{ 3 } \) times the corresponding sides of ΔABC.
Question 29.
Write all the other trigonometric ratios of ∠A in terms of sec A.
Question 30.
The angle of elevation of the top of a chimney from the foot of a tower is 60° and the angle of depression of the foot of the chimney from the top of the tower is 30°. If the height of the tower is 40 m, find the height of the chimney. According to pollution control norms, the minimum height of a smoke emitting chimney should be 100 m. State if the height of the above mentioned chimney meets the pollution norms. What value is discussed in this question?
Answers
SECTION-A
Answer 1.
LCM (54, 336) x HCF (54,336) = 54 x 336
⇒ \( LCM(54,336)\cfrac { 54\times 336 }{ HCF(54,336) } \)
= \( \cfrac { 54\times 336 }{ 6 } =3024 \)
Answer 2.
n th term from the end – l – (n -1 )d
Here, l= 485, n = 10, d = 10 – 5 = 5
∴ 10th term from the end = 485 – (10 – 1)5
= 485 – 45 = 440
Alternative method: 10th term from the end of the AP 5,10,15,………… 485 is equal to the 10th term from the beginning of the AP 485,480,……….. 10,5
Here, a = 485, d – 480 – 485 = – 5 n = 10
∴ an = a + (n – 1 )d
∴ a10 = 485 + (10 -1)(- 5)
= 485-45 = 440
Answer 3.
Answer 4.
Answer 5.
Sum of the roots = a + b = –\(\frac { B }{ A } \) = -a
Product of the roots = ab =\(\frac { C }{ A } \) = -b
⇒ a + b = -a and ab = -b
⇒ 2a = -b and a = -1
⇒ b = 2 and a =-1
Answer 6.
SECTION-B
Answer 7.
Let the required ratio be k : 1, then by section formula, the coordinates of the point which divides AB in the ratio k:1 are
\(\left( \cfrac { k+5 }{ k+1 } ,\cfrac { -4k-6 }{ k+1 } \right) \)
As this point lies on y-axis, therefore its x-coordinate
\(\cfrac { k+5 }{ k+1 }\)=0
⇒ K=5
Putting the value of k = 5, we get point of division as \(\left( \cfrac { -5+5 }{ 5+1 } ,\cfrac { -4\times 5-6 }{ 5+1 } \right) \)
i.e., \(0,\cfrac { -13 }{ 3 }\)
Answer 8.
Answer 9.
Answer 10.
Answer 11.
Answer 12.
(1) As the bag contains only lemon flavoured candies. So, the event related to the experiment of taking out an orange flavoured candy is an impossible event. So, its probability is 0.
(2) As the bag contains only lemon flavoured candies. So, the event related to the experiment of taking out lemon flavoured candies is certain event. So, its probability is 1.
SECTION-C
Answer 13.
Let us assume,to the contrary, that \( \sqrt { p } \) is a rational number. So, we can find integers r and s (≠0) such that \( \sqrt { p } \) \(\cfrac { r }{ s }\)
Suppose r and s have a common factor other than 1. Then, we divide by the common factor to get \( \sqrt { p } \) \( = [latex]\cfrac { a }{ b }\), where a and b are coprime.
So, b\( \sqrt { p } \) = a
Squaring both the sides and rearranging,
We get pb2 – a2
⇒ p divides a2 [ ∵ p divides pb2]
⇒ p divides a [ ∵ v p is prime and p divides a2 ⇒ p divides a]
So, we can write a = pc, for some integer c,
Substituting for a, we get
pb2 – (pc)2, That is b2 = pc2
⇒p divides b2, So p divides b.
Therefore, a and b have at least p as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
So, our assumption is wrong and we conclude that \( \sqrt { p } \) is an irrational number.
OR
Let a be any positive integer. Then it is of the form 3q, 3q + 1 or 3q + 2, So we have the following cases.
Case (1) When a = 3q
a3 = (3 q)3 = 2 7q3
= 9(3q3) = 9m where m = 3q3
Case (2) When a = 3q+ 1,
a3 = (3 q + l)3
= (3q)3 + 3(3q)2 .1 + 3 (3q) .12 + 13
= 27 q3 + 27q2 + 9q + 1
= 9q (3q2 + 3q+ 1) + 1
= 9m + 1, where m = q (3q2 + 3q+ 1)
Case (3) When a = 3q + 2
a3 = (3q + 2)3 = (3q)3 + 3(3q)2 .2+3 (3q).22 + 23
= 27 q3 + 54 q2 + 36q + 8
= 9q(3q2 + 6q + 4) + 8
= 9m + 8, where m = q(3q2 + 6q + 4)
Hence, a3 is either of the form 9m or 9m + 1 or 9m+ 8
Answer 14.
We have, x – y + 1 = 0 and 3x + 2y – 12 = 0
i.e., y= x+1…(i)
Answer 15.
Answer 16.
Let P(x, 0) be the point on the x-axis, which is equidistant from points A( 2, -5) and B(-2, 9)
Answer 17.
Answer 18.
Answer 19.
Answer 20.
Answer 21.
Given, Diameter of tank = 10 m
Depth of tank (H) = 2m
Internal diameter of pipe = \(\cfrac {2}{ 10 }m \)
Answer 22.
SECTION-D
Answer 23.
Let the present age of Nuri be x years and present age of Sonu be y years. Now, five years ago
Nuri was (x – 5) years
and Sonu was (y – 5) years old.
∴ According to question, we have
(x – 5) = 3(y – 5)
⇒ x – 5 = 3y – 15
⇒ x-3y = -15 + 5 = -10
∴ x-3y = -10 …(i)
Again, Ten years later,
Nuri will be (x + 10) years
and Sonu will be (y + 10) years
So, according to question, (x + 10) = 2 (y + 10)
⇒ x + 10 = 2y + 20
x – 2y = 20 – 10
x – 2y = 10 …(ii)
Thus, we have system of equations (i) and (ii).
Hence, time taken by large pipe alone to fill the tank is 15 hours and time taken by smaller pipe alone to fill the tank is 25 hours.
Answer 24.
Given: A right triangle ABC right angled at B.
To Prove: AC2 = AB2 + BC2
Construction: Draw BD ⊥ AC
Proof: In ΔADB and ΔABC
Second part:
Let ABCD be a rhombus in which diagonals AC and BD bisect each other at O.
Since, the diagonals of rhombus bisect each other at right angles.
∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = 90° ”
and AO = CO, BO = OD
Since, AOB is a right triangle, right-angled at O.
∴ AB2 = OA2 + OB2 [Using Pythagoras theorem]
Answer 25.
Answer 26.
Answer 27.
Now ,if n=25,then number of logs in 25 th row is equal to 25th term of an AP with first term 20 and common difference – 1.
.’. Number of logs in 25th row = a + 24d = 20 – 24 = – 4
Clearly, this is not meaningful
Therefore, 22 = 16
Thus, logs are placed in 16 rows.
Now, Number of logs in top row = Number of logs in 16th row
⇒16th term of the AP = a + 15d
= 20 + 15 x -1 = 20 -15
= 5
Hence, there are 5 logs in top row.
Answer 28.
Answer 29.
Answer 30.
Let the height of the chimney AB be h. Height of tower CD = 40 m.
The distance between the tower and chimney be d.
The height of the chimney is 120 m which is more than the minimum requirement to meet the pollution norms.
Values: The values discussed here are:
- Environmental awareness
- Social concern
- Abiding the laws.
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