## 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 = x^{3}y^{2} and b = xy^{4}; 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 2^{m} x 5^{n}, 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) x^{2} + 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 cm^{2}. 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 = x^{3}y^{2}

b = xy^{4}

H.C.F.(a, b) = xy^{2}.

**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,

AB^{2} = OA^{2} + OB^{2} = 3^{2} + 4^{2}

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°]

OA = OB [radii]

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, a_{n} = a + (n – 1)d

∴ a_{10} = p + (10 – 1)d

=> a_{10} = 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}

x^{2} + 10x + 25 + y^{2} – 8y + 16 = x^{2} + 2x + 1 + y^{2} – 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 T_{n} < 0.

We know, nth term, T_{n} = a + (n – 1)d

= 129 + (n -1) (- 4)

= 129 – 4n + 4

= 133 – 4n

∴T_{n}< 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:**

Given quadratic equation is,

(2k + 3)x^{2} + 2x – 5 = 0

On comparing the equation with ax^{2} + bx + c = 0, we get

a = 2k + 3, b = 2, c = – 5

For equal roots,

b^{2} – 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, a_{n} = 50, S_{n} = 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 B_{1} B_{2}, B_{3} and B_{4} such that

BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}

(6) Join B_{4}C.

(7) Through the point B_{3}, draw a line B_{3}C’ parallel to B_{4}C 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 :

Radius, R = 1.5 cm

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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