## CBSE Sample Papers for Class 10 Maths Paper 4

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 4

**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 P(a/3, 4) is the mid-point of the line segment joining the points Q(- 6, 5) and R(- 2, 3), then find the value of ‘a’.

**Question 2.**

It is given that ∆ABC ~ ∆PQR, with = \(\frac { BC }{ QR } =\frac { 1 }{ 4 } \) Then, find \(\frac { ar(PRQ) }{ ar(BCA) } \).

**Question 3.**

HCF and LCM of a and b are 19 and 152 respectively. If a = 38 then find the value of b.

**Question 4.**

Find a quadratic polynomial, the sum and product of whose zeroes are 6 and 5, respectively.

**Question 5.**

If x = 2 and x = 3 are roots of 3x² – 2kx + 2m = 0, find the value of k and m.

**Question 6.**

Evaluate : 2 tan² 45° + 4 sin² 30° – 8 cos² 60°.

**SECTION-B**

**Question 7.**

Determine the least number that is divisible by all the numbers from 2 to 10 (both inclusive).

**Question 8.**

Find the coordinates of the point which is equidistant from the three vertices of the ∆AOB as shown in the given figure.

**Question 9.**

1000 tickets of a lottery were sold and there are 5 prizes on these tickets. If Saket has purchased a lottery ticket, what is the probability of winning a prize ?

**Question 10.**

A and B throw a pair of dice. If A throws 9, find B’s chance of throwing a higher number.

**Question 11.**

Determine the value of k for which the system : 2x – 3y = 7; (k + 2)x – (2k + 1 )y = 3(2k – 1), represents coincident lines on the graph.

**Question 12.**

Determine the sum of first 99 even natural numbers.

**SECTION-C**

**Question 13.**

Prove that exactly one out of any three consecutive positive integers is divisible by 3.

**Question 14.**

Find the area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b).

**Question 15.**

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

**OR**

O is an interior point of a rectangle ABCD. Prove that OB² + OD² = QA² + OC².

**Question 16.**

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

**Question 17.**

Find the area enclosed between two concentric circles of radii 3.5 cm and 7 cm. A third concentric circle is drawn outside the 7 cm circle such that the area enclosed between it and the 7 cm circle is same as that between the two inner circles. Find the radius of the third circle correct to one decimal place.

**OR**

A circular pond is 17.5 m in diameter. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate Rs 25 per m².

**Question 18.**

A solid iron cuboidal block of dimensions 4.4 m × 2.6 m × 1 m is recast into a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm. Find the length of the pipe.

**OR**

A cone of radius 8 cm and height 12 cm is divided into two parts by a plane through the mid-point of its axis parallel to its base. Find the ratio of the volumes of two parts.

**Question 19.**

If the remainder on division of x³ + 2x² + kx + 3 by x – 3 is 21, find the quotient and the value of k. Hence, find the zeroes of the cubic polynomial x³ + 2x² + kx – 18.

**Question 20.**

Solve graphically : 3x – y = 3; 2x – y = – 2.

**OR**

2 audio and 3 video cassettes together cost 426, whereas 3 audio and 2 video cassettes cost Rs 350. What are the prices of an audio and a video cassettes ?

**Question 21.**

If cos (A – B) = \(\frac { \sqrt { 3 } }{ 2 } \), cos (A + B) = \(\frac { 1 }{ 2 } \), 0°<A+B≤90° find cot (A+3B)

**Question 22.**

The following table gives the daily wage distribution of 50 casual workers in a locality. Find the modal daily wage of a worker.

**SECTION-D**

**Question 23.**

A tower subtends an angle a at a point A in the plane of its base and the angle of depression of the foot of the tower from a point which is b metres just above A is p. Prove that the height of the tower is b tan α cot β .

**OR**

An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°. Find the speed of the aeroplane in km/hr.

**Question 24.**

Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.

**Question 25.**

In figure, O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm and OT intersects circle at E. If AB is a tangent to the circle at E, find the length of AB, where TP and TQ are two tangents to the circle.

**Question 26.**

The rain water from a roof of dimensions 22 m x 20 m drains into a cylindrical vessel having diameter of base 2 m and height 3.5 m. If the rain water collected from the roof just fill the cylindrical vessel, then find the height of rainfall in cm.

**OR**

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 Rs 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm². (Take π = 3.14)

**Question 27.**

The sum of the squares of the two consecutive natural numbers is 313. Find the numbers.

**OR**

Sum of two numbers is 16 and the sum of their reciprocals is 16/63. Find the numbers.

**Question 28.**

There are 25 trees at equal distance of 5 metre in a line with a well, the distance of the well from the nearest tree being 10 metre. A gardener waters all the trees separately starting from the well ’ and he returns to the well after watering each tree to get water for the next. Find the total distance, the gardener will cover in order to water all the trees.

**Question 29.**

If x = tan A + sin A and y = tan A – sin A, then prove that: x² – y² = √xy.

**Question 30.**

The students of Class X of a school decided to donate their pocket money to purchase mineral water bottles for the people using contaminated water in a nearby village. They packed the mineral water bottles in different boxes. These boxes contained varying number of mineral water bottles. The following table shows the distribution of mineral water bottles according to the number of boxes :

Find the mean number of mineral water bottles kept in a packing box.

**SOLUTIONS**

**SECTION-A**

**Solution 1:**

Given : P is the mid-point of QR.

So by mid-point formula,

**Solution 2:**

Given

∆ABC ~ ∆PQR

and \(\frac { BC }{ QR } =\frac { 1 }{ 4 } \)

**Solution 3:**

We know,

Product of two numbers = H.C.F. × L.C.M.

⇒ a x b = H.C.F. x L.C.M.

⇒ 38 x b = 19 x 152

⇒ b = \(\frac { 19\times 152 }{ 38 } \) = 76.

∴ b = 76.

**Solution 4:**

Given : Sum of zeroes = 6

and product of zeroes = 5

We know, quadratic polynomial is given as,

= k[x² – (Sum of zeroes)x + Product of zeroes]

= k[x² – 6x + 5] where k ≠ 0.

**Solution 5:**

Given quadratic equation is,

x² – 2kx + 2m = 0

Since, 2 and 3 are roots of the given quadratic equation.

So, putting x = 2, 3 in given equation, we get

3(2)² – 2k(2) + 2m = 0

⇒ 12 – 4k + 2m = 0

and 3(3)² – 2k(3) + 2m=0

⇒ 27 – 6k + 2m = 0

Subtracting equation (i) and (ii), we get

k = \(\frac { 15 }{ 2 } \)

Substituting the value of k in equation (i), we get

12 – 4 x \(\frac { 15 }{ 2 } \) + 2m = 0

12 – 30 + 2m = 0

2m = 18 ⇒ m = 9

∴ k = \(\frac { 15 }{ 2 } \) and m = 9.

**Solution 6:**

We have,

**SECTION-B**

**Solution 7:**

Given numbers are 2, 3, 4, 5, 6, 7, 8, 9, 10

L.C.M. of 2, 4, 8 = 8

L.C.M. of 3, 9 = 9

L.C.M. of 5, 10 = 10

L.C.M. of 6, 2, 4, 8, 3, 9 = 8 x 9 = 72

L.C.M. of 2, 3, 4, 6, 8, 9, 5, 10 = 72 x 5 = 360

L.C.M. 2, 3, 4, 5, 6, 8, 9, 10, 7 = 360 x 7 = 2520

∴ 2520 is least number which is divisible by all the numbers from 2 to 10.

**Solution 8:**

Since ∠AOB = 90°

So, ∆AOB is right angle triangle.

We know that in a right angled triangle, mid-point of hypotenuse is equidistant from all three vertices of right angle triangle.

Mid-point of hypotenuse AB = \(\frac { 2x+0 }{ 2 } ,\frac { 0+2y }{ 2 } \) = (x, y)

Hence, C(x, y) is the equidistant from the three vertices of the ∆AOB.

**Solution 9:**

Total lottery tickets = 1000

∴ n(S) = 1000

Total winner tickets = 5

∴ n(E) = 5

Probability of winning the prize

P(E) = \(\frac { n(E) }{ n(S) } =\frac { 5 }{ 1000 } \) = 0.005.

**Solution 10:**

Let E be the event of B’s chance of throwing a higher number than 9.

⇒ E= {(4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)}

∴ Total chances, n(E) = 6

Hence, probability of throwing a higher number than 9

P(E) = \(\frac { n(E) }{ n(S) } =\frac { 6 }{ 36 } =\frac { 1 }{ 6 } \)

**Solution 11:**

2x – 3y – 7 = 0

and (k+2)x – (2k+1)y – 3(2k-1)=0

**Solution 12:**

Sum of first 99 even natural numbers

i.e., 2 + 4 + 6 + 8 …

These are in arithmetic progression with

a = 2, d = 4 – 2 = 2, n = 99

Sum of K terms of A.P. is given as

Hence, sum of first 99 even natural number = 9900.

**SECTION-C**

**Solution 13:**

Let the number be a = 3q + r where r = 0,1, 2

Case I: When r = 0,

Then a = 3q

It is divisible by 3.

Case II: When r = 1,

Then a = 3q + 1

It is not divisible by 3.

Case III: When r = 2,

Then a = 3q + 2

It is not divisible by 3

These are also three consecutive numbers hence exactly one out of any three consecutive number is divisible by 3.

**Solution 14:**

Given vertices of triangles are

A(a, b + c), B(b, c + a) and C(c, a + b)

We know, area of a triangle is given as

**Solution 15:**

Given

∆ABC and ∆DEF are two similar triangles, AP and DQ are medians of triangles respectively

**Solution 16:**

Given : ABCD is a quadrilateral circumscribing a circle.

To prove : ∠AOD + ∠BOC = 180°

∠AOB + ∠COD = 180°

Construction : Join OP, OQ, OR and OS.

**Solution 17:**

Let the radius of third circle be r cm.

Area enclosed by two concentric circles I and II of radii 3.5 cm and 7 cm, respectively

**Solution 18:**

Given : Dimensions of cuboidal block = 4.4 m x 2.6 m x 1 m

Internal radius of pipe, r = 30 cm = 0.30 m

Thickness of cylindrical pipe = 5 cm

External radius of pipe, R = 30 + 5 cm = 35 cm = 0.35 m

Now, volume of cuboidal block = 4.4 x 2.6 x 1 m³ = 11.44 m³

Volume of the hollow cylindrical pipe = π(R² – r²)h

= π(0.35² – 0.30²)h

**Solution 19:**

x^{3} + 2x^{2} -9x-18

it is divided by x – 3

x – 3 = 0 x = 3

From the given data, we get

f(3) = 21

⇒ (3)^{3} + 2(3)^{2} + k(3) = 21

⇒ 27 + 18 + 3k + 3 = 21

⇒ 48 + 3k = 21

⇒ 3k = – 27

⇒ k = – 9

∴ Polynomial becomes

x^{3} + 2x^{2} + (-9) x + 3

x^{3} + 2x^{2} – 9x + 3

We know,

Dividend = Quotient × Divisor + Remainder

⇒ x^{3} + 2x^{2} – 9x + 3 = Quotient × (x – 3) + 21

⇒ x^{3} + 2x^{2} – 9x – 18 = Quotient × (x – 3)

**Solution 20:**

Given equations are

3x – y = 3

and 2x – y = – 2

From Equations (i),

3x – y = 3

The point of intersection of these lines is (5, 12).

x = 5 and y = 12.

**OR**

Let the price of audio cassette be Rs x

and the price of video cassette be Rs y

According to the question,

2x + 3y = 426 …(i)

and 3x + 2y = 350 ….(ii)

Adding equations (i) and (ii), we get

5x + 5y = 776

⇒ x + y = 155.20 ….(iii)

Subtracting equation (ii) from equation (i), we get

– x + y =76 …(iv)

Adding equations (iii) and (iv), we get

2y = 231.20

⇒ y = \(\frac { 231.20 }{ 2 } \) = 115.60

Putting the value of y in equation (iii), we get

x + 115.60 = 125.20

⇒ x = 155.20 – 115.60

⇒ x = 39.60

Thus, Price of audio cassette = Rs 115.60

Price of video cassette = Rs 39.60.

**Solution 21:**

We have,

\(cos(A-B)=\frac { \sqrt { 3 } }{ 2 } \) …(i)

and \(cos(A+B)=\frac { 1 }{ 2 } \) ….(ii)

From equation (i)

**Solution 22:**

Modal class = Group with highest frequency

= 15000 – 20000

Now,

= 15000 + 3000

Mode = 18000

Modal daily wages = Rs 18000.

**SECTION-D**

**Solution 23:**

Let OP be the tower of height h m

Given AB = b m

Let distance between tower and point A(AO) = x m

In ∆AOP,

∠O = 90°

**Solution 24:**

**Steps of Construction :**

(1) Draw a circle with centre O and radius 3 cm.

(2) Draw a diameter of the circle. Extend it on both sides upto P and Q such that OP = OQ = 7 cm.

(3) Draw perpendicular bisectors of OP and OQ which meet them at M and N respectively.

(4) With M as centre and MP as radius and with N as centre and NQ as radius, draw circles which intersect the given circle at A, B and C, D respectively.

(5) Join PA, PB, QC, QD which are the required tangents.

**Solution 25:**

Given : A circle with centre of radius 5 cm and OT = 13 cm

Since, PT is a tangent at P and OP is a radius through P

OP ⊥ PT

In ∆OPT

**Solution 26:**

Given

Dimesions of roof = 22m x 20m

Diameter of cylinder = 2 m

Radius , r = \(\\ \frac { 2 }{ 2 }\) = 1m

**Solution 27:**

Let the two consecutive number be x and (x + 1).

Then, according to the question,

x² + (x + 1)² = 313

=> x² + x² + 2x + 1 – 313 = 0

=> 2x² + 2x – 312 = 0

=> x² + x – 156 = 0

=> x² + 13x – 12x – 156 = 0

=> x(x + 13) – 12(x + 13) = 0

=> (x + 13) (x – 12) = 0

Either x + 13 = 0 => x = – 13, rejected as it is not natural number.

or x – 12 = 0 => x = 12

∴ Consecutive numbers are 12 and 13.

**OR**

**Solution 28:**

Distance covered in watering the nearest tree = 2 x 10 = 20 m

Distance covered in watering the second tree = 15 x 2 = 30 m

Distance covered in watering third tree = 20 x 2 = 40 m

Sum of the distance covered = 20 + 30 + 40 + …

**Solution 29:**

Given

x = tanA + sin A

y = tanA – sinA

We know

**Solution 30:**

As this series is an inclusive one, we can make it exclusive by adding 0.5 to the upper limit and subtracting 0.5 from the lower limit of each class interval. Thus, we have

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