## CBSE Sample Papers for Class 10 Maths Paper 5

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 5

**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.**

AOBC is a rectangle whose three vertices are A(0, 3), 0(0, 0) and B(5, 0). Find the length of its diagonal CO.

**Question 2.**

In the given figure, if ∠AOB = 125°, then find the measure of ∠DOC.

**Question 3.**

A vessel has 136 lit. of oil. Another vessel has 92 lit. of oil. What is the capacity of the largest possible ladle which can measure out all the oil in each vessel in exact number of times ?

**Question 4.**

For what value of k, the system : kx + 3y = 11; 2x + 5y = 3, has unique solution.

**Question 5.**

If one root of 2x² + kx – 6 = 0 is 2, find the value of k.

**Question 6.**

Evaluate : sin 30° cos 60° + sin 60° cos 30°.

**SECTION-B**

**Question 7.**

If the HCF of 85 and 153 is expressible in the form 85m – 153, find the value of m.

**Question 8.**

Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle or not.

**Question 9.**

A bag contains 5 red balls, 8 white balls, 4 green balls and 7 black balls. If one ball is drawn at random, find the probability that it is :

(i) Black ball

(ii) Red ball

(iii) Not a green ball

(iv) Green or white ball

**Question 10.**

A two digit number is selected at random. What is the chance that will be :

(i) A composite number which is odd

(ii) Successor of a prime number

**Question 11.**

Can x – 1 be the remainder on division of a polynomial p(x) by 2x + 3 ? Justify your answer.

**Question 12.**

Find k if the sum of roots of equation x² – x + k(2x – 1) is Zero.

**SECTION-C**

**Question 13.**

Show that exactly one of the numbers n, n + 2 or n + 4 is divisible by 3.

**Question 14.**

If P(9a – 2, – b) divides line segment joining A(3a + 1,- 3) and B(8a, 5) in the ratio 3 : 1, find the value of a and b.

**Question 15.**

In the given figure, D and E trisect BC. Prove that 8AE² = 3AC² + 5AD²

**Question 16.**

In the given figure, AB is a chord of the circle and AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then find the measure of ∠BAT.

A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.

**Question 17.**

A path of 4 m width runs round a semi-circular grassy plot whose circumference is \(163 \frac { 3 }{ 7 } \) m.

Find :

(i) The area of the path.

(ii) The cost of gravelling the path at the rate of Rs 1.50 per m².

(iii) The cost of turfing the plot at the rate of 45 paise per m².

**OR**

All the vertices of a square lie on a circle. Find the area of the square, if area of the circle is 1256 cm². (Use π = 3.14)

**Question 18.**

500 persons are taking a dip into a cuboidal pond which is 80 m long and 50 m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is 0.04 m3 ?

**OR**

How many spherical lead shots of diameter 4 cm can be made out of a solid cube of lead whose edge measures 44 cm ?

**Question 19.**

Show that the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667.

**OR**

If sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.

**Question 20.**

Solve for x :\(2\left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) -9\left( x+\frac { 1 }{ x } \right) +14=0\)

**Question 21.**

If tan (A + B) = √3 and tan (A – B) = \(\frac { 1 }{ \sqrt { 3 } } \);0° < A + B ≤ 90°; A > B, find sin 2A and cos 6B.

**Question 22.**

Find the mean of the following data :

**SECTION-D**

**Question 23.**

The angle of elevation of a stationary cloud from a point 2500 m above a lake is 15° and the angle of depression of its reflection in the lake is 45°. What is the height of the cloud above the lake level ?

**OR**

The angle of elevation of the top of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40 m vertically above X, the angle of elevation of the top is 45°. Calculate the height of the tower.

**Question 24.**

Construct a right triangle in which the sides (other than hypotenuse) are of lengths 5 cm and 4 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.

**Question 25.**

In a class test, the sum of Kamal’s marks in Maths and English is 40. Had he got 3 marks more in Maths and 4 marks less in English, the product of their marks would have been 360. Find his marks in two subjects separately.

**Question 26.**

A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel of 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 pen stand made of wood is in the shape of a cuboid with four conical depressions and a cuboidal depression to hold the pens and pins, respectively. The dimensions of the cuboid are 10 cm, 5 cm and 4 cm. The radius of each of the conical depressions is 0.5 cm and the depth is 2.1 cm. The edge of the cubical depression is 3 cm. Find the volume of the wood in the entire stand.

**Question 27.**

Solve graphically x – y = 0, 2x + 3y – 30 = 0. Also find coordinates of points where 2x + 3y – 30 = 0 meets axis of X and Y.

**OR**

Students of a class are made to stand in rows. If 4 students are extra in each row, there would be two rows less. If 4 students are less in each row, there would be 4 more rows. Find the number of students in the class.

**Question 28.**

Two pipes running together can fill a tank in \(5 \frac { 5 }{ 21 } \) minutes. If one pipe takes 1 minute more than the other, then find the time in which each pipe would individually fill the tank.

**Question 29.**

If sin α = a sin β and tan α = b tan β, then prove that cos² α = \(\frac { { a }^{ 2 }-1 }{ { b }^{ 2 }-1 } \)

**Question 30.**

Draw more than ogive’ and ‘less than ogive’ for the following distribution on the same graph. Also find the median from the graph.

**SOLUTIONS**

**SECTION-A**

**Solution 1:**

We know, diagonals of rectangle are equal.

OC = AB

=> \(OC=\left| \sqrt { { (5-0) }^{ 2 }+{ (0-3) }^{ 2 } } \right| \)

=> \(OC=\left| \sqrt { 25+9 } \right| =\left| \sqrt { 34 } \right| \)

=> \(OC=\sqrt { 34 } \) units

**Solution 2:**

Given

∠AOB = 125°

We know, opposite sides of a quadrilateral circumscribing a circle subtend supplements angles at the centre of a circle.

∴∠AOB +∠DOC = 180°

=> 125° + ∠DOC = 180°

=> ∠DOC = 180° – 125°

= 55° Ans.

**Solution 3:**

Largest capacity of the ladle which can measure out all the oil

= H.C.F. of 136 and 92

Now, 136 = 2 x 2 x 2 x 17

and 92 = 2 x 2 x 23

H.C.F. (136, 92) = 4

∴ 4 litre capacity ladle can measure out all the oil in each vessel in exact number of times. Ans.

**Solution 4:**

Given equations are,

kx + 3y – 11 = 0

and 2x + 5y – 3 =0

**Solution 5:**

Given equation is,

2x² + kx – 6 = 0

One root of the equation is 2.

x = 2

2(2)² + k(2) – 6 = 0

8 + 2k – 6 = 0

2k = – 2

k = – 1.

**Solution 6:**

We have,

sin 30° cos 60° + sin 60° cos 30°

=> \(\frac { 1 }{ 2 } .\frac { 1 }{ 2 } +\frac { \sqrt { 3 } }{ 2 } .\frac { \sqrt { 3 } }{ 2 } \)

=> \(\frac { 1 }{ 4 } +\frac { 3 }{ 4 } \)

=> 1

**SECTION-B**

**Solution 7:**

85 = 5 x 17

153 = 3 x 3 x 17

∴H.C.F. (85, 153) = 17

But

∴H.C.F. = 85m – 153

17 = 85m – 153

17 + 153 = 85m

\(m= \frac { 170 }{ 85 } \)

= 2

**Solution 8:**

Let the given points be A(5, – 2), B(6, 4), C(7, – 2).

Now,

Here, we observe that

AB = BC

and AB+BC≠AC

∴Points A,B,C forms an isosceles triangle with AB = BC

**Solution 9:**

We have,

Number of red balls = 5

Number of white balls = 8

Number of green balls = 4

Number of black balls = 7

Total balls = 24

**Solution 10:**

Total two-digit numbers 10 to 99 = 90

∴n(S) = 90

(i) Composite numbers which is odd = {15, 21, 25, 27, 33, 35, 39, 45, 49, 51, 55, 57, 63, 65, 69, 75, 77, 81, 85, 87, 91, 93, 95, 99}

∴ n(E) = 24

Probability of odd composite number

= \(\frac { n(E) }{ n(S) } =\frac { 24 }{ 90 } =\frac { 4 }{ 15 } \)

(ii) Successor of a prime number = {12, 14, 18, 20, 24, 30, 32, 38, 42, 44, 48, 54, 60, 62, 68, 72, 74, 80, 84, 90, 98}

∴ n(E) = 21

Probability of a successor of a prime number

= \(\frac { n(E) }{ n(S) } =\frac { 21 }{ 90 } =\frac { 7 }{ 30 } \)

**Solution 11:**

No, degree of the remainder is always less than degree of the divisor.

**Solution 12:**

Given equation is,

x² – x + k(2x – 1) = 0

=> x² – x + 2kx – k = 0

=> x² + (2k – 1)x – k = 0

**SECTION-C**

**Solution 13:**

Let q be the quotient and r be the remainder when n is divisible by 3.

Therefore, n = 3q + r, where r = 0, 1, 2

=> n = 3q or n = 3q + 1 or n = 3q + 2

Case I: If n = 3q, then n is divisible by 3 but n + 2 = 3q + 2 and n + 4 = 3a + 4 are not divisible by 3.

Case II: If n = 3q + 1, then n + 2 = 3q + 3 = 3(q + 1), which is divisible by 3 and n + 4 = 3q + 5, which is not divisible by 3.

So, only (n + 2) is divisible by 3.

Case III: If n = 3q + 2, then n + 2 = 3q + 4, which is not divisible by 3 and n + 4 = 3q + 6 = 3(q + 2) is divisible by 3.

So, only (n + 4) is divisible by 3.

Hence, one and only one out of n, (n + 2), (n + 4) is divisible by 3.

**Solution 14:**

By section formula,

**Solution 15:**

Here, D and E trisect BC.

=>BD = DE = EC = \(\\ \frac { 1 }{ 3 } \) BC

or BE = 2 BD,

BC = 3BD

In right ∆ABD,

we have

AD² = AB² + BD²….(i)

**Solution 16:**

In the given circle,

∠ABC = 90°

∠ACB + ∠CAB = 90°

50° + ∠CAB = 90°

∠CAB = 90° – 50°

∠CAB = 40°

Now, OA is radius of the circle and AT is the tangent of circle

∠OAT = 90°

∠OAB + ∠BAT = 90°

40° + ∠BAT = 90°

∠BAT = 90° – 40°

∠BAT = 50°

**Solution 17:**

We have

**Solution 18:**

Given

length of pond, l = 80m

breadth of pond, b = 50m

**Solution 19:**

Odd integers between 1 and 1000 divisible by 3 are 3, 9,15, 21, … 999

The above series form an A.P.

where a = 3, d = 9 – 3 = 6 and an = 999

We know

**Solution 20:**

We have

\(2\left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) -9\left( x+\frac { 1 }{ x } \right) +14=0\)

\(2\left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2 \right) -9\left( x+\frac { 1 }{ x } \right) +14-4=0\)

**Solution 21:**

We have

tan(A+B) = √3

=> tan(A+B) = tan 60°

**Solution 22:**

**SECTION-D**

**Solution 23:**

Let B be the point of oservation, P be the cloud and P’ be its reflection in the take.

AB = 2500 m

∠PBC = 15°

and ∠CBP = 45°

Let the height of cloud from lake be h m.

OP = OP’= h m

**Solution 24:**

Steps, of construction :

(1) Draw a line segment AB = 4 cm.

(2) Construct ∠XAB = 90°.

(3) With A as centre, draw an arc of radius 3 cm intersecting AX at C.

(4) Join BC. Thus, ∆ABC is obtained.

(5) Draw an acute angle ∠BAY below of AB.

(6) Mark points A1, A2, A3, A4, A5 on AY such that

AA1 = A1A2 = A2A3 = A3A4 = A4A5.

(7) Join A3B.

(8) Draw a line through A5, parallel to A3B intersecting extended line segment AB at B’.

(9) Through B’, draw a line parallel to BC intersecting AX at C’.

(10) Thus, ∆AB’C’ is obtained whose sides are \(\\ \frac { 5 }{ 3 } \) times the corresponding sides of ∆ABC.

**Solution 25:**

Let the marks obtained in Maths be x, then marks obtained in English = 40 – x

According to the question,

(x + 3) (40 – x – 4) = 360

=> (x + 3) (36 – x) = 360

=> – x² + 36x + 108 – 3x = 360

=> – x² + 33x – 252 = 0

=> x² – 21x – 12x + 252 = 0

=> x(x – 21) – 12(x – 21) = 0

(x – 21) (x – 12) = 0

Either x – 21 = 0 or x – 12 = 0

=> x = 21 or x = 12

If x = 21, then marks obtained in Maths = 21

and marks obtained in English = 40 – 21 = 19

If x = 12, then marks obtained in Maths = 12

and marks obtained in English = 40 – 12 = 28

**Solution 26:**

Given

Height of the cone = 20 cm

PQ (h) = 10 cm

**Solution 27:**

Given equations are

x – y = 0

y = x

=> x – y = 0

**Solution 28:**

Let one pipe fill the tank in x minute

then other will fill the tank in (x + 1) minutes.

According to the question,

**Solution 29:**

Given : sin α = a sin β

sin² α = > a² sin² β

**Solution 30:**

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