# CBSE Sample Papers for Class 10 Maths Paper 8

## CBSE Sample Papers for Class 10 Maths Paper 8

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 8

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.
If two integers a and b are written as a = x3y2 and b = xy4; x, y are prime numbers, then find H.C.F. (a, b).

Question 2.
Find the perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0).

Question 3.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then find the measure of ∠POA.

Question 4.
Determine the pair of linear equations from 2x + 3y = 5, y + $$\\ \frac { 2 }{ 3 }$$ x = 5, 4x = 6y + 10, which has infinite solution.

Question 5.
The first term of an A.P. is p and its common difference is q. Find its 10th term.

Question 6.
If 9 sec A = 41, find cos A and cot A.

SECTION-B

Question 7.
Find a point which is equidistant from the points A(- 5, 4) and B(- 1, 6). How many such points are there ?

Question 8.
Which term of A.P. 129, 125, 121, … is its first negative term ?

Question 9.
Write the denominator of the rational number $$\\ \frac { 257 }{ 5000 }$$ in the form 2m x 5n, where m, n are non – negatives. Hence, write its decimal expansion without actual division.

Question 10.
It is known that a box of 600 electric bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. What is the probability that it is a non-defective bulb ?

Question 11.
In a lottery of 50 tickets numbered 1 to 50, one ticket is drawn. Find the probability that the drawn ticket bears a prime number.

Question 12.
For what value of k the quadratic equation (2k + 3) x2 + 2x – 5 = 0 has equal roots ?

SECTION-C

Question 13.
Find the smallest number which when divided by 161, 207 and 184 leaves remainder 21 in each case.

Question 14.
Find the values of k if the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear.

OR

If the points A( 1, – 2), B(2, 3), C(a, 2) and D(- 4, – 3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base.

Question 15.
O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB || DC. Through O, a line segment PQ is drawn parallel to AB meeting AD in P and BC in Q. Prove that PO = QO.

Question 16.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Question 17.
The diameter of coin is 1 cm (fig.). If four such coins be placed on a table so that the rim of each touches that of the other two, find the area of the shaded region (Take π = 3.1416).

OR
Area of a sector of a circle of radius 36 cm is 54π cm2. Find the length of the corresponding arc of the sector.

Question 18.
A hemispherical tank of diameter 3 m, is full of water. It is being emptied by a pipe at the rate of $$3 \frac { 4 }{ 7 }$$ litres per second. How much time will it take to make the tank half empty ?

OR

A wall 24 m long, 0.4 m thick and 6 m high is constructed with the bricks each of dimensions 25 cm x 16 cm x 10 cm. If the mortar occupies 1/10 of the volume of the wall, then find the number of bricks used in constructing the wall.

Question 19.
The first term of an A.P. is 2 and the last term is 50. The sum of all these terms is 442. Find the common difference.

OR

Find the sum of first 17 terms of an A.P. whose 4th and 9th terms are – 15 and – 30 respectively.

Question 20.
Solve for x :
$${ \left( \frac { 2x+1 }{ x+1 } \right) }^{ 4 }-13{ \left( \frac { 2x+1 }{ x+1 } \right) }^{ 2 }+36=0$$

Question 21.
If cosec A + cot A = p, then prove that sec A = $$\frac { { p }^{ 2 }+1 }{ { p }^{ 2 }-1 }$$

Question 22.
Find mode of the following data :

SECTION-D

Question 23.
From the top of a lighthouse, the angles of depression of two ships on the opposite sides of it are observed to be a and p. If the height of the lighthouse is h metres and the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is $$\frac { h\left( tan\alpha +tan\beta \right) }{ tan\alpha \quad tan\beta }$$ meters

OR

A boy is standing on the ground and flying a kite with 100 m of string at an elevation of 30°. Another boy is standing on the roof of a 10 m high building and is flying his kite at an elevation of 45°. Both the boys are on opposite sides of both the kites. Find the length of the string that the second boy must have so that the two kites meet.

Question 24.
Construct a right triangle in which sides (other than the hypotenuse) are of lengths 8 cm and 6 cm. Then construct another triangle whose sides are 3/4 times the corresponding sides of the first triangle.

Question 25.
State and prove the converse of Pythagoras theorem.

Question 26.
A hemispherical bowl of internal radius 9 cm is full of liquid. The liquid is to be filled into cylindrical shaped bottles each of radius 1.5 cm and height 4 cm. How many bottles are needed to empty the bowl ?

OR

A gulab jamun, contains sugar syrup about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.

Question 27.
A boat goes 12 km downstream and 26 km upstream in 8 hours. It can go 16 km upstream and 32 km downstream in same time. Find the speed of boat in still water and the speed of stream.

OR

Solve graphically x – 2y = 1; 2x + y = 7. Also find the coordinates of points where the lines meet X-axis.

Question 28.
A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each prize.

Question 29.
Show that :
$$\frac { sinA-cosA+1 }{ sinA+cosA-1 } =\frac { 1 }{ secA-tanA }$$

Question 30.
The median of the following data is 20.75. Find the missing frequencies ‘x’ and ‘y’, if the total frequency is 100.

SOLUTIONS

SECTION-A

Solution 1.
Given : a = x3y2
b = xy4
H.C.F.(a, b) = xy2.

Solution 2.
Given vertices are A(3, 0), B(0, 4) and 0(0, 0).
OA = 3 units
OB = 4 units
In ∆AOB, ∠O = 90°
∴From Pythagoras theorem,
AB2 = OA2 + OB2 = 32 + 42
AB = √9+16 = 5 unit
Perimeter of ∆OAB = OA + OB + AB
= 3 + 4 + 5 = 12 unit.

Solution 3.
Given :
∠APB = 80°
In ∆OAP and ∆OBP,
∠B = ∠A [each 90°]
OP = OP [common]
By RHS congruency rule

Solution 4.
Given equations are,
2x + 3y – 5 = 0
y + $$\\ \frac { 2 }{ 3 }$$x – 5=0
and 4x + 6y = 10
=> 2x + 3y – 5 = 0
2x + 3y – 15 = 0
and 4x + 6y – 10 =0

Solution 5.
Given : The first term of A.P. = a
and common difference = q
we know, an = a + (n – 1)d
∴ a10 = p + (10 – 1)d
=> a10 = P + 9d.

Solution 6.
Given : 9 sec A = 41
=> sec A = $$\\ \frac { 41 }{ 9 }$$
cos A = $$\\ \frac { 1 }{ sec A }$$
cos A = $$\\ \frac { 9 }{ 41 }$$

SECTION-B

Solution 7:
Let the point P(x, y) be equidistant from the points A (- 5, 4) and B(- 1, 6).

$$\left| \sqrt { { (x+5) }^{ 2 }+{ (y-4) }^{ 2 } } \right| =\left| \sqrt { { (x+1) }^{ 2 }+{ (y-6) }^{ 2 } } \right|$$
Squaring both sides
(x + 5)2 + (y – 4)2 = (x + 1)2 + (y – 6)2
x2 + 10x + 25 + y2 – 8y + 16 = x2 + 2x + 1 + y2 – 12y + 36
10x – 2x – 8y + 12y + 41 – 37 = 0
8x + 4y + 4 = 0
2x + y + 1 =0
The points lying on 2x + y + 1 = 0 are equidistant from A and B.

Solution 8:
Given A.P. is 129, 125, 121,…
Here, a = 129, d = 125 – 129 = 121 – 125 = – 4
Let nth term be the first negative term.
Then Tn < 0.
We know, nth term, Tn = a + (n – 1)d
= 129 + (n -1) (- 4)
= 129 – 4n + 4
= 133 – 4n
∴Tn< 0
∵133 – 4n < 0
=> 133 < 4n
=> 4n > 133
=> n =$$\\ \frac { 133 }{ 4 }$$
= $$33 \frac { 1 }{ 4 }$$
Hence, 34th term will be the first negative term.

Solution 9:
We have,

Solution 10:
Total bulbs = 600
Number of defective bulbs = 12
Number of non-defective bulbs = 600 – 12 = 588
probability of a non defective bulb = $$\frac { Number\quad of\quad non\quad defective\quad bulbs }{ total\quad bulbs }$$
= $$\\ \frac { 588 }{ 600 }$$
= $$\\ \frac { 147 }{ 150 }$$

Solution 11:
Total tickets 1, 2, 3, 4, …, 50
=> n(S) = 50
Prime number tickets = 2, 3, 5, 7,11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
n(E) = 15
Probability that the ticket bears a prime number
P(E) = $$\\ \frac { n(E) }{ n(S) }$$
= $$\\ \frac { 15 }{ 50 }$$
= $$\\ \frac { 3 }{ 10 }$$
= 0.3

Solution 12:
(2k + 3)x2 + 2x – 5 = 0
On comparing the equation with ax2 + bx + c = 0, we get
a = 2k + 3, b = 2, c = – 5
For equal roots,
b2 – 4ac = 0
(2)2 – 4(2k + 3) ( – 5) = 0
4 + 20(2k + 3) = 0
4 + 40k + 60 = 0
40k = – 64
k = $$\\ \frac { -64 }{ 40 }$$
k = $$\\ \frac { -8 }{ 5 }$$

SECTION-D

Solution 13:
Given numbers are 161, 207 and 184.
161 = 7 x 23
207 = 3 x 3 x 23
∴ 184 = 2 x 2 x 2 x 23
L.C.M. (161, 207, 1841) = 2 x 2 x 2 x 3 x 3 x 23 x 7
= 11592
Smallest number which when divided by 161, 207 and 184, leaves remainder 21 in each case
= 11592 + 21
= 11613. Ans.

Solution 14:
Given points are A(k + 1, 2k), B(3k, 2k + 3), C(5k – 1, 5k)
If the points A, B and C are collinear, then
ar (∆ABC) = 0

Solution 15:
Given, ABCD is trapezium in which
AB || CD || PQ

Solution 16:
Let O be the common centre of two concentric circles and let AB be a chord of large circle touching the smaller circle at P.
Construction : Join OP
Since OP is the radius of the smaller circle and AB is tangent to this circle at P,
∴ OP ⊥ AB
We know that the perpendicualr drawn from the centre of a circle to any chord of the circle bisects the chord.

Solution 17:
Given : Diameter of coin = 1 cm
Radius of coin, r = 0.5 cm
PQ = QR = RS = PS
= 2 x 0.5 = 1 cm

Solution 18:
Given : Diameter of hemispherical tank = 3 cm
Radius, r = $$\\ \frac { 3 }{ 2 }$$
Now, Volume of hemispherical tank = $$\frac { 2 }{ 3 } { \pi r }^{ 3 }$$

Solution 19:
Given, a = 2, an = 50, Sn = 442
We know that

Solution 20:
We have
$${ \left( \frac { 2x+1 }{ x+1 } \right) }^{ 4 }-13{ \left( \frac { 2x+1 }{ x+1 } \right) }^{ 2 }+36=0$$
let
$${ \left( \frac { 2x+1 }{ x+1 } \right) }^{ 2 }=a$$

Solution 21:
Given : cosec A + cot A = p ….(i)
We know,
cosec² A – cot² A = 1
=> (cosec A + cot A) (cosec A – cot A) = 1 [∵ a² – b² = (a+b)(a-b)]
=> (p) (cosec A – cot A) = 1

Solution 22:

SECTION-D

Solution 23:
Let PO be the lighthouse and A, B be the position of the two ships.
PO = h m
Angle of depression of ship at the point A = a°.
Angle of depression of ship at the point B = P°.
In ∆OPA, ∠P = 90°

Solution 24:
Steps of construction :
(1) Draw a line segment BC = 8 cm and make a right angle (∠DBC) at the point B.
(2) Taking B as centre and radius 6 cm, draw an arc which intersect BD at the point A.
(3) Join AC. Thus, ∆ABC is obtained.
(4) Draw a ray BX such that ∠CBX is an acute angle.
(5) Make four points B1 B2, B3 and B4 such that
BB1 = B1B2 = B2B3 = B3B4
(6) Join B4C.
(7) Through the point B3, draw a line B3C’ parallel to B4C which intersect the line segment BC at the point C’.
(8) Through the point C’, draw a parallel line C’A’ parallel to CA which intersect the line segment BA at the point A’.
(9) Thus, ∆A B’C’ is the required triangle whose sides are $$\\ \frac { 3 }{ 4 }$$ of corresponding sides of ∆ABC.

Solution 25:
Converse of Pythagoras Theorem : In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the side is a right angle.
Given : A triangle ABC such that AC² = AB² + BC²
Construction : Construct a triangle DEF such that
DE = AB, EF = BC and ∠E = 90°.
Proof : Since, ∆DEF is a right angled triangle with right angle at E, therefore, by Pythagoras theorem, we have

Solution 26:
Given :
Radius of bowl, r = 9 cm
For cylindrical bottle :
Height, H = 4 cm

Solution 27:
Let the speed of boat in still water be x km/h and speed of the stream be y km/hr.
.’. Speed of the boat in upstream = x – y
Speed of the boat in downstream = x + y

Solution 28:
Given :
Total amount = Rs 700
No. of cash prizes = 7
Each prize is Rs 20 less than preceding prize.
Therefore, the list of value of prizes form an A.P.
Let the first cash prize = Rs a
Difference between two consecutive prize = – Rs 20

Solution 29:
Given
L.H.S = $$\frac { sinA-cosA+1 }{ sinA+cosA-1 }$$

Solution 30:

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