CBSE Sample Papers for Class 10 Maths Paper 6
CBSE Sample Papers for Class 10 Maths Paper 6
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 6
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 4 questions of 3 marks each and 3 questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
- Use of calculator is not permitted.
SECTION-A
Question 1.
Use prime factorization method to find the HCF of 300 and 1000.
Question 2.
Find the value of k for which the quadratic equation 2x² + 5x + k = 0 has equal roots.
Question 3.
Find the 101th term of the A.P. – 4, – 2, 0, 2, 4, …
Question 4.
∆ABC is similar to ∆DEF. If AB = 10 cm, DE = 12 cm, perimeter of ∆ABC = 25 cm, find the perimeter of ∆DEF.
Question 5.
Find the distance of A(9, – 40) from origin.
Question 6.
Given that tan A = \(\frac { 1 }{ \sqrt { 5 } } \) what is the value of \(\frac { { cos }^{ 2 }A-2{ sin }^{ 2 }A }{ { cos }^{ 2 }A+2{ sin }^{ 2 }A } \) ?
SECTION-B
Question 7.
Use Euclid’s algorithm to find the HCF of 612 and 408. Also find their LCM using their HCF.
Question 8.
Find the 10th term from the end of the A.P. 60, 130, 200, …, 2160.
Question 9.
Solve for x and y : ax + by = a² + b²; bx + ay = 2ab.
Question 10.
Find a point on the Y-axis which is equidistant from the points A( – 4, 3) and B(6, 5).
Question 11.
Mean and median of a slightly asymmetric data are 52 and 54.5 respectively. Find the mode of the data.
Question 12.
A card is drawn from a well-shuffled deck of 52 playing cards. Find the probability of getting :
(i) A black king or a red queen
(ii) A non-face card.
SECTION-C
Question 13.
Show √7 is irrational.
OR
Show n(n + 1) (n – 1) is divisible by 3 for any positive integer ‘n’.
Question 14.
On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder are x – 2 and – 2x + 4, respectively. Find g(x).
Question 15.
Solve graphically : 3x – 2y = 5, 2x + 3y = 12. Also find area bounded by these lines with X-axis.
Question 16.
ABCD is a quadrilateral. Diagonals AC and BD meet at point O. Also AO x DO = BO x CO. Show ABCD is a trapezium.
OR
Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ∆ABC ~ ∆PQR.
Question 17.
Find the area of the quadrilateral whose vertices taken in order are P( – 5, – 3), Q( – 4, – 6), R(2, – 1) and S(1, 2).
Question 18.
Prove that : \(\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } =\frac { 2 }{ { sin }^{ 2 }A-{ cos }^{ 2 }A } \)
OR
Prove that : \(\frac { 1+cosA }{ sinA } +\frac { sinA }{ 1+cosA } =2cosecA\)
Question 19.
Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre of the circle.
Question 20.
The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle. Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205)
Question 21.
A solid sphere made of copper has a diameter of 6 cm. It is melted and recast into small spherical balls of diameter 2 cm each. Assuming that there is no wastage in the process, find the number of small spherical balls obtained from the given sphere.
OR
A square field and an equilateral triangular park have equal perimeters. Cost of tilling the field at the rate of ₹ 5 per m² is ₹ 720. What is the cost of grassing the park at the rate of ₹ 10 per m² correct to nearest rupee ?
Question 22.
The following table shows the marks obtained by 100 students of Class X in a school during a particular academic session. Find the mode of this distribution.
SECTION-D
Question 23.
In a class test, the sum of the marks obtained by P in Mathematics and Science is 28. Had he got 3 more marks in maths and 4 marks less in science, the product of marks obtained in the two subjects would have been 180. Find the marks obtained in the two subjects separately.
Question 24.
180 logs are stacked in the following manner : 19 logs in the bottom row, 18 in the next row, 17 in the row next to it and so on. In how many rows are the 180 logs placed and how many logs are in top row ?
Question 25.
Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the square of the other two sides.
OR
State and prove the Basic Proportionality Theorem.
Question 26.
If tan A = n tan B and sin A = m sin B, prove that \({ cos }^{ 2 }A=\frac { { m }^{ 2 }-1 }{ { n }^{ 2 }-1 } \)
Question 27.
The angle of elevation of the top of a building at a point on the level ground is 30°. After walking a distance of 100 m towards the foot of the building along the horizontal line through the foot of the building on the same level ground, the angle of the elevation of the top of the building is 60°. Find the height of the building.
OR
From the top of a building 100 m high, the angles of depression of the top and bottom of a tree are observed to be 45° and 60° respectively. Find the height of the tree.
Question 28.
Draw ∆ABC with AB = 5 cm, ∠B = 60° and BC = 4 cm. Now construct a triangle whose sides are \(\\ \frac { 4 }{ 5 } \) times the corresponding sides of ∆ABC. Give the steps of construction.
Question 29.
A metallic right circular cone 20 cm high and whose semi-vertical angle is 30° is cut into two parts from the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.
OR
A solid consisting of a right circular cone of height 120 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. How many litres of water is left in the cylinder, if the radius of cylinder is 60 cm and its height 180 cm ? (Use π = \(\\ \frac { 22 }{ 7 } \))
Question 30.
The following table gives the yield per hectare of wheat of 60 farms in a village.
Draw a more than ogive for the above data.
SOLUTIONS
SECTION-A
Solution 1:
Given numbers are 300 and 1000.
∵ 300 = 3 x 2² x 5²
1000 = 2³ x 5³
∴ HCF (300, 1000) = 2² x 5² = 100. Ans.
Solution 2:
Given quadratic equation is,
2x² + 5x + k = 0
Here, A = 2, B = 5, C = k
Equation has equal roots, if
D = 0
=> B² – 4AC = 0
=> 25 – 4 x 2 x k = 0
=> 8k = 25
=> k = \(\\ \frac { 25 }{ 8 } \)
Solution 3:
Given A.P. is – 4, – 2, 0, 2, 4,…
Here, a = – 4, d = 2, n = 101
a101 = a + 100d = – 4 + 100 x 2
= 196.
Solution 4:
Given, AB = 10 cm, DE = 12 cm, pr. (∆ABC) = 25 cm
It is given that ∆ABC ~ ∆DEF
=> \(\frac { pr.\left( \triangle ABC \right) }{ pr.\left( \triangle DEF \right) } =\frac { AB }{ DE } \)
=> \(\frac { 25 }{ pr.\left( \triangle DEF \right) } =\frac { 10 }{ 12 } \)
=> pr.(∆DEF) = 30 cm
Solution 5:
Given points are A(9, – 40) and O(0, 0).
Using distance formula,
OA = \(\sqrt { { (9-0) }^{ 2 }+{ (-40-0) }^{ 2 } } \)
= \(\sqrt { { 9 }^{ 2 }+{ (-40) }^{ 2 } } \)
= \(\sqrt { 81+1600 } \)
= \(\sqrt { 1681 } \)
= 41 units.
Solution 6:
Given
tan A = \(\frac { 1 }{ \sqrt { 5 } } \)
We have,
\(\frac { { cos }^{ 2 }A-2{ sin }^{ 2 }A }{ { cos }^{ 2 }A+{ sin }^{ 2 }A } \)
Dividing Nr and Dr by cos² A, we get
SECTION-B
Solution 7:
Given numbers are 612 and 408.
Now, using Euclid’s division algorithm
612 = 408 x 1 + 204
408 = 204 x 2 + 0
HCF = 204
We know that HCF x LCM = a x b
=> 204 x LCM = 612 x 408
=> LCM = \(\\ \frac { 612X408 }{ 204 } \)
= 1224
Solution 8:
Given AP is 60, 130, 200,…., 2160.
Here, l = 2160, d = 130 – 60 = 70
∴ nth term from end = l – (n – l)d
10th term from end = 2160 – (10 – 1) x 70
= 2160 – 630 = 1530. Ans.
Solution 9:
Given equations are,
ax + by = a² + b² …(i)
and bx + ay = 2ab …(ii)
On multiplying equation (i) by a and equation (ii) by b, we get
Solution 10:
Given points are A( – 4, 3) and B(6, 5).
Let the point on Y-axis be P(0, b). (Given)
Now, PA = PB
=> PA² = PB²
=> (0 + 4)² + (b – 3)² = (0 – 6)² + (b – 5)²
=> 16 + b² – 6b + 9 = 36 + b² – 10b + 25
=> b² – 6b + 25 = b² – 10b + 61
=> 4b = 36
=> b = 9
The required point is (0, 9). Ans.
Solution 11:
Given : Mean = 52, Median = 54.5.
We know, Mode = 3 Median – 2 Mean
= 3 x 54.5 – 2 x 52
= 163.5 – 104
=> Mode = 59.5. Ans.
Solution 12:
Total number of cards = 52
No. of black king cards = 2
No. of red queen cards = 2
No. of face cards = 12
(i) P(black king or red queen) = P(black king) + P(red queen)
\(=\frac { 2 }{ 52 } +\frac { 2 }{ 52 } =\frac { 4 }{ 52 } =\frac { 1 }{ 13 } \)
(ii) P(non-face card) = 1 – P(face card)
\(=1-\frac { 12 }{ 52 } =\frac { 40 }{ 52 } =\frac { 10 }{ 13 } \)
SECTION-C
Solution 13:
Let √7 is a rational number.
Then, √7 can be written in the form
√7 = \(\\ \frac { p }{ q } \)
such that q ≠ 0
and p and q are co prime integers
On squaring both sides,
Solution 14:
We have,
Dividend = x³ – 3x² + x + 2
Quotient = x – 2
Remainder = – 2x + 4
We know,
Solution 15:
Given equations are,
3x – 2y = 5…(i)
and 2x + 3y = 12…(ii)
On solving, we get
l1 : 3x – 2y = 5
Solution 16:
Given : AO x DO = BO x CO
To prove : ABCD is a trapezium.
Proof : We have,
AO x DO = BO x CO
=> \(\\ \frac { AO }{ OC } \) = \(\\ \frac { BO }{ DO } \)
Also ∠1 = ∠2
∴ By SAS similarity axiom,
∆AOB ~ ∆COD
∠3 = ∠4
∠3 and ∠4 from a pair of alterate angles
AB || CD
Hence, ABCD is a trapezium.
Hence Proved.
OR
Given, ∆ABC and ∆PQR in which AD and PM are medians drawn on sides BC and QR respectively. It is also given that
\(\frac { AB }{ PQ } =\frac { BC }{ QR } =\frac { AD }{ PM } \)
Solution 17:
Given, vertices of quadrilateral are P(- 5, – 3), Q(- 4, – 6), R(2, – 1) and S(1, 2).
Construction : Join R.
Now, quadrilateral PQRS is divided into two triangles, ∆PSR and ∆PQR.
∴ar(\(\Box \) PQRS) = ar (∆PQR) + ar (∆PSR)
= \(\\ \frac { 1 }{ 2 } \) |- 5( – 6 + 1) – 4( – 1 + 3) + 2( – 3 + 6) | + \(\\ \frac { 1 }{ 2 } \) | – 5(2 + 1) + 1( – 1 + 3) + 2( – 3 – 2) |
= \(\\ \frac { 1 }{ 2 } \) | 25 – 8 + 6 | + \(\\ \frac { 1 }{ 2 } \) | – 15 + 2 – 10 |
= \(\\ \frac { 1 }{ 2 } \) | 23 | + \(\\ \frac { 1 }{ 2 } \) | – 23 |
= 23 sq. unit.
Solution 18:
Given
L.H.S = \(\frac { sinA+cosA }{ sinA-cosA } +\frac { sinA-cosA }{ sinA+cosA } \)
Solution 19:
Given : l and m are two parallel tangents, n is the intercept of tangents which intersect circle at point C.
To prove : ∠AOB = 90°.
Construction : Join OP and OC.
Proof : In ∆OPA and ∆OAC, we have
OP = OC [Radii]
AP = AC
[Tangents to a circle from an external point are equal in length]
AO = AO (common)
Solution 20:
Given ar(∆ABC) = 17320.5 cm²
let radius of circle be r
Then,AB = BC = CA = 2r
Solution 21:
Given,
Diameter of sphere of copper = 6 cm
Radius, R = \(\\ \frac { 1 }{ 2 } \) = 3 cm
Diameter of small spherical balls = 2 cm
Radius, r = \(\\ \frac { 2 }{ 2 } \) = 1 cm
Let number of small spheres be n.
Then,
n x Volume of small sphere = Volume of large sphere
Solution 22:
SECTION-D
Solution 23:
Let marks obtained in mathematics be x and marks obtained in science be y.
According to the question,
x + y = 28
=> y = 28 – x…(i)
Also (x + 3).(y – 4) = 180 ,
=> (x + 3) (28 – x – 4) = 180
=> (x + 3) (24 – x) = 180
=> 24x – x² + 72 – 3x = 180
=> x² – 21x + 108 = 0
=> x² – 12x – 9x + 108 = 0
=> x(x – 12) – 9(x – 12) = 0
=> (x – 12) (x – 9) = 0
=> x = 12 or x = 9.
∴Marks obtained in mathematics = 12 or 9.
If marks obtained in mathematics = 12 then marks obtained in science = 28 – 12 = 16
If marks obtained in mathematics = 9 then marks obtained in science = 28 – 9 = 19
Solution 24:
Let number of rows be ‘n’.
No. of logs in bottom row = 19
In next row = 18 and so on Also, total logs = 180
=> 19 + 18 + 17 + … (n terms) = 180
This is an A.P.
Here, a = 19, d = – 1, n = ?, Sn = 180
We know,
Solution 25:
Given : A right-angled triangle ABC in which ∠B = 90°.
To prove : (Hypotenuse)² = (Base)² + (Perpendicular)²
i.e„ AC² = AB² + BC²
Construction : From B, draw BD ⊥ AC.
Proof : In triangles ADB and ABC, we have
∠ADB = ∠ABC
and, ∠A = ∠A
Solution 26:
Given
sin A = m sin B,
tan A = n tan B
=> \(m=\frac { sinA }{ sinB } ,n=\frac { tanA }{ tanB } \)
Solution 27:
Let AB be the building, C and D be the points of observation.
Given : DC = 100 m
Let BC = x m and AB = h m
Solution 28:
Steps of construction :
(1) Draw a line segment BC = 6 cm.
(2) Construct ∠YBC = 60°.
(3) Taking B as centre and radius 5 cm, draw on arc, intersecting BY at A.
(4) Join AC. Thus, ∆ABC is obtained.
(5) Draw a ray BX such that ∠CBX is an acute angle and BX lies in exterior of ∆ABC.
(6) Mark B1 B2, B3 and B4 on BX such that BB1 = B1B2 = B2B3 = B3B4.
(7) Join B3 to C.
(8) From B4, draw a line parallel to B3C to meet BC produced at C’.
(9) From C’ draw a line parallel to CA to meet AB produced at A’.
Thus, ∆A’BC’ is the required triangle whose sides are \(\\ \frac { 4 }{ 3 } \) of corresponding sides of ∆ABC.
Solution 29:
Let VAD be the right circular cone of height AC = 20 cm. Suppose the cone is cut by a plane parallel to its base at a point B such that AB = BC
Solution 30:
We hope the CBSE Sample Papers for Class 10 Maths Paper 6 help you. If you have any query regarding CBSE Sample Papers for Class 10 Maths Paper 6, drop a comment below and we will get back to you at the earliest.
