Madhya Pradesh Board Question Paper for Class 10th Maths 2016 In PDF

MP Board (MPBSE) Question Paper 2016 Class 10th Mathematics For English Medium with Solutions – Free Download

MP Board Question Paper 2016 Class 10th Maths with solutions are provided on this page by BYJU’S in downloadable pdf format and also in the text, so that the students are aware of the question paper pattern in the upcoming final exams. Maths Question Paper 2016 Class 10 is also added here, which helps in the exam preparation. Students are able to access all the Madhya Pradesh board previous year maths question papers here. In order to overcome the struggle of completing the paper on time, it is important for the students to practise different types of questions in the previous year papers and succeed in the final board examination.

Question 1: Choose the correct option and write it in your answer book.

(1 * 5 = 5)

Question i: The system of equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂ represents two parallel lines if,

(a) a₁ / a₂ = b₁ / b₂ = c₁ / c₂ (b) a₁ / a₂ ≠ b₁ / b₂

(c) a₁ / a₂ = b₁ / b₂ ≠ c₁ / c₂ (d) a₁ / a₂ ≠ b₁ / b₂ = c₁ / c₂

Answer: (c)

Question ii: The third proportional to 9, 12 is

(a) 6 √3 (b) 3√6 (c) 27 / 4 (d) 16

Answer: (d)

Question iii: The maximum degree of the variable used in the quadratic equation is.

(a) 1 (b) 2 (c) 3 (d) 4

Answer: (b)

Question iv: The measure of ∠PAQ in the figure given below is:

(a) 45° (b) 180° (c)90° (d) 60°

Answer: (c)

Question v: From a point 30 m away from the foot of the building the angle of elevation of the top of the building is 45°. The height of the building will be

(a) 25m (b) 30m (c) 25√2 m (d) 30√2 m

Answer: (b)

Question 2: Fill in the blanks: 1 * 5 = 5

(i) The zero of the linear polynomial ax + b is ____________ [-b / a]

(ii) The rate of the depreciation is __________ [negative]

(iii) If the sides of two triangles are proportional, then triangles will be ___________ [similar]

(iv) The radius of a circle is 7cm, then the area will be _________________ [154 cm²]

(v) The length of the diagonal of a cube is 12√3 cm. The length of the edge of the cube will be___________. [12 cm]

Question 3: Match the following. 1 * 5 = 5

Column – A Column – B

(i) sin² 25^o + cos² 25^o (a) cos θ

(ii) 1 + cot² θ (b) 1

(iii) sin (90° – θ) (c) cosec²θ

(iv) sec 60° (d) tan 41°

(v) tan 49° (e) 2

(f) 1 / 2

(g) cot 41°

Answers:

(i) – (b)

(ii) – (c)

(iii) – (a)

(iv) – (e)

(v) – (g)

Question 4: Write true/ false in the following: 1 * 5 = 5

(i) X + 2√x is not a polynomial. [True]

(ii) Income tax is a direct tax. [True]

(iii) Angles in the same segment of a circle are equal. [True]

(iv) The volume of a hemisphere is 3𝜋𝑟². [False]

(v) The probability of a definite event is always one. [True]

Question 5: Write the answer in one word / sentence of each: 1 * 5 = 5

(i) In equation x + 2y = 5, if y = 0 then write the value of x.

Answer:

x + 2y = 5

x = 5 – 2y

x = 5 – 0

x = 5

(ii) Write the formula of Hero to find the area of a triangle.

Answer:

√s (s – a) (s – b) (s – c)

(iii) Write the statement of Pythagoras theorem.

Answer: In a right-angle triangle, the square of the hypotenuse is equal to the sum of the square of the other two sides.

(iv) Write the number of circles passing through non-linear points.

Answer: 1

(v) Write the value of the mode of the following data: 2, 3, 4, 2, 12, 9, 7, 8, 9, 6, 9, 5, 9.

Answer: 9

Question 6: [i] Write the statement of basic proportionality (Thales Theorem).

2M

OR

[ii] Write the property of angle-angle similarity.

Solution:

OR

Question 7: [i] The perimeters of two similar triangles are 30cm and 20cm, respectively. If one side of one triangle is 12cm, find the corresponding side of another triangle. 2M

OR

[ii] Triangle ACB is an isosceles triangle such that AC = BC. If AB² = 2 AC². Prove that ∆ACB is a right-angled triangle.

Solution:

⇒ Perimeter of △ABC = 20 cm.

⇒ Perimeter of △PQR = 30 cm.

⇒ QR = 12 cm, BC = ?

In the similarity case

Perimeter of △ABC / Perimeter of △PQR = AB / PQ = BC / QR = AC / PR

⇒ 20 / 30 = BC / 12

⇒ BC = 8 cm

OR

AB² = 2AC²

AB² = AC² + AC²

AB² = AC² + BC²

AB will be the largest side that is the hypotenuse.

By Pythagoras theorem,

H² = B² + P²

AB² = AC² + BC²

AB² = AC² + AC²

AB² = AB²

LHS = RHS

Since Pythagoras theorem is satisfied, ∆ACB is a right-angled triangle.

Question 8: [i] Triangle ABC and triangle PQR are two similar triangles. The areas of these are 64 cm² and 100 cm², respectively. If QR = 12cm. Then find the value of side BC. 2M

OR

[ii] In the figure given below DE║BC, if AD / DB = 3 / 5 and side AC = 6cm. Then, find the value of AE.

Solution:

Area of Δ ABC / Area of ΔPQR = (AB / PQ)² = (BC / QR)² = (AC / PR)²

Area of Δ ABC = 64 cm²

Area of Δ PQR = 100 cm²

QR = 12 cm

BC = ?

Area of ΔABC / Area of Δ PQR = (BC / QR)²

=> 64 / 100 = (BC / 12)²

=> 16 / 25 = BC² / 12²

=> 0.64 = BC² / 144

=> 92.16 = BC²

=> BC = 9.6 cm

OR

Using BPT theorem

AD / DB = AE / EC

EC = AC – AE

AD / DB = AE / AC – AE

3 / 5 = AE / 6 – AE

0.6 = AE / 6 – AE

3.6 – 0.6 AE = AE

3.6 = AE + 0.6 AE

3.6 = 1.6AE

AE = 2.25 cm

Question 9: [i] Write any two properties of Arithmetic Mean. 2M

OR

[ii] Find the Median of the following values:

5, 10, 3, 7, 1, 9, 6, 2, 11

Solution:

(b) The sum of the squared deviations of the items from Arithmetic Mean (A.M) is minimum, which is less than the sum of the squared deviations of the items from any other values.

OR

1, 2, 3, 5, 6, 7, 9, 10, 11

Median = {[n + 1] / 2}^th term

= {[9 + 1] / 2}^th term

= 5^th term

Median = 6

Question 10: [i] Write the probability of getting an odd number in a single throw of a die. 2M

OR

[ii] If two coins are tossed simultaneously, find the probability of getting two heads.

Solution:

Odd numbers = {1, 3, 5}

Probability (getting an odd number) = odd numbers / sample space

= 3 / 6

= 1 / 2

OR

S = {HT, TH, HH, TT}

⇒ n (S) = 4

Let E be the favourable outcomes of getting two heads, then E = {H, H}

⇒ n (E) = 1

P (getting two heads) = n (E) / n (S)

= 1 / 4

Question 11: [i] Solve the following system of equation 4M

3x – 2y = 4

y + 2x = 5

OR

[ii] Find the value of m, for which the system

2x+ my – 4 = 0

3x – 7y – 10 = 0 has

(i) a unique solution

(ii) no solution

Solution:

2x + y = 5 —- (2)

Multiply equation (1) by 2 and equation (2) by 3,

6x – 4y = 8

6x + 3y = 15

___________

-7y = -7

y = 1

Put y = 1 in (2),

2x + y = 5

2x + 1 = 5

2x = 4

x = 4 / 2

x = 2

OR

3x – 7y – 10 = 0

Comparing with ax + by + c = 0

a₁ = 2, b₁ = m, c₁ = – 4

a₂ = 3, b₂ = – 7, c₂ = – 10

(1) for unique solution,

a₁ / a₂ ≠ b₁ / b₂

2 / 3 = m / -7

-14 = 3m

m = -14 / 3

For all real values of m, [m ≠ -14 / 3] the system has a unique solution.

(2) for no solution, the lines are parallel such that,

a₁ / a₂ = b₁ / b₂ ≠ c₁ / c₂

2 / 3 = m / -7 = -4 / -10

m ≠ [-7 * 4] / -10

m ≠ -14 / 5

So, m takes every value except [-14 / 5].

Question 12: [i] The sum of two numbers is 7. If the sum of these numbers is seven times its difference. Find the number. 4M

OR

[ii] If in ∆ABC ∠C = 2∠B = ∠A + ∠B + 20°, then find all the three angles of a triangle.

Solution:

x + y = 7 ……….. (1)

x + y = 7(x – y)……(2)

x + y = 7x – 7y

0 = 6x – 8y

0 = 3x – 4y………… (3)

Solving (1) and (3)

x + y = 7 × 4

3x – 4y = 0

4x + 4y = 28

__________

7x = 28

x = 4

y = 7 – x (from equation 1)

= 7 – 4

= 3

OR

Given that ∠C = 2∠B

∠C = 2y

Also,

∠C = ∠A + ∠B + 20°

∠C = x + y + 20

Since ABC is a triangle, by angle sum property,

∠A + ∠B + ∠C = 180^o

x + y + 2y = 180^o

x + 3y = 180^o —- (1)

∠A + ∠B + ∠C = 180^o

x + y + [x + y + 20] = 180^o

x + y + x + y + 20 = 180^o

2x + 2y = 160^o

x + y = 80^o —- (2)

Consider the equations (1) and (2),

x + 3y = 180^o

x + y = 80^o

_________

2y = 100

y = 100 / 2

y = 50^o

∠C = 2y

= 2 * 50^o

= 100^o

Put y = 50 in (1),

x + 3 * 50 = 180^o

x = 180^o – 150^o

x = 30^o

∠A = 30^o

∠B = 50^o

∠C = 100^o

Question 13: [i] If x / [b + c] = y / [c + a] = z / [a + b], then prove that (b – c) x + (c – a) y + (a – b) z = 0. 4M

OR

[ii] What should be subtracted from 11, 20, 26 and 50 so that the remaining numbers are in proportion?

Solution:

x = (b + c) k …(1)

y = (c + a) k …(2)

z = (a + b) k …(3)

To prove (b – c) x + (c – a) y + (a – b) z = 0

LHS = (b – c) x + (c – a) y + (a – b) z

= (b – c) (b + c) k + (c – a) (c + a) k + (a – b) (a + b) k

= k [(b – c) (b + c) + (c – a) (c + a) + (a – b) (a + b) ]

= k [b² – c² + c² – a² + a² – b²]

= k [0]

= 0

= RHS

OR

11 – x / 20 – x = 26 – x / 50 – x

(11 – x) (50 – x) = (26 – x) (20 – x)

550 – 11x – 50x + x² = 520 – 26x – 20x + x²

550 – 61x = 520 – 46 x

550 – 520 = – 46x + 61x

30 = 15x

x = 30 / 15

= 2

Question 14: [i] Solve the equation 3x – 1 / x = 2 using the formula method. 4M

OR

[ii] Find the nature of the roots of the following equation, 6x² – x – 2 = 0.

Solution:

3x² – 1 = 2x

3x² – 2x – 1 = 0

a = 3, b = -2, c = -1

x = -b ± √b² – 4ac / [2a]

= (2) ± √(-2)² – 4 * 3 * (-1) / [2 * 3]

= 2 ± √16 / 6

= 2 ± 4 / 6

x = 2 + 4 / 6

= 6 / 6

x = 1

x = 2 – 4 / 6

= -2 / 6

= -1 / 3

x = 1, -1 / 3

OR

• If discriminant > zero, then it has real and unequal roots.
• If discriminant = zero, then it has real and equal roots.
• If discriminant < zero, then it has unreal roots.

6x² – x – 2 = 0

a = 6 ; b = -1 ; c = -2

Δ = b² – 4ac

= 1 – 4 * 6 * (-2)

= 1 + 48

= 49

So, Δ > 0

It has real and unequal roots.

Question 15: [i] From the top of the hill, the angle of depression of the top and the bottom of the 16m high building is 30° and 60° respectively. Find the height of the hill. 4M

OR

[ii] An aeroplane is flying at a height of 8,000m. The angle of depression of the control tower of the airport from the aeroplane is 30°. Find the horizontal distance between the control tower and the aeroplane.

Solution:

Height of the tall building = 16 m = BE

The angle of depression of top of the tall building from the multistoried building = 30°

The angle of depression of bottom of the tall building from the multistoried building = 60°

So, BC = x and CD = 16 m [As BCDE forms a rectangle]

AD = AC + CD

So, AC = (h – 16) m

In ΔBCA

tan 30^o = AC/ BC

1 / √3 = (h – 16)/x

x = √3(h – 16)…… (i)

In ΔADE

tan 60^o = AD / ED

1.732 = h / x

h = x

h = √3(h – 16) [using (i)]

h = √3h – 16√3

(√3 – 1)h = 16√3

h = 16√3/ (√3 -1)

Rationalising the denominator by (√3 + 1), we have

h = 16√3(√3 + 1) / (3 – 1)

h = 16√3(√3 + 1) / 2

h = 48 + 16 / 2

= 64 / 2

= 32m

The height of the building is 32m.

OR

tan 30^o = 8000 / x

x = 8000 / tan 30^o

x = 13856.4m

Question 16: [i] A quadrants shaped flower bed is made of radius 14m in a square garden of side 100m, in all the four corners. Find the area of the remaining part of the square garden. 4M

OR

[ii] A rocket is in the form of a closed cylinder from below, and the upper part surmounted by a cone of equal radius. The radius of the cylinder is 2 m and the height is 21 m. The height of the cone is 8.4m. Find the volume of the rocket.

Solution:

Radius of quadrant, r = 14cm

Angle measure, θ = 90° (angle of each corner of the square is 90°)

Area of square = (side)²

= (100)²

= 10000 cm²

Area of each quadrant = [1 / 4] * 𝛑 * r²

= 1 / 4 * 22 / 7 * 14 *14

= 154cm²

Area of remaining part of square = Area of the square – 4 * Area of each quadrant

= 10000 – 4 * 154

= 10000 – 616

= 9384 cm²

OR

= [1 / 3]𝛑r²h + 𝛑R²h

= [1 / 3] * [22 / 7] * 2² * (8.4) + [22 / 7] * 2² * 21

= 35.2 + 264

= 299.2m³

Question 17: [i] The area of three adjacent faces of a cuboid are x, y and z. If the volume of the cuboid is V, then prove that V² = xyz. 4M

OR

[ii] An iron sphere of radius 8 cm is melted then recast into small spheres each of radius 1 cm. Find the number of small spheres.

Solution:

Length = l

Breadth = b

Height = h

Given the area of three faces x, y and z are

lb = x –(1)

bh = y —(2)

lh = z —(3)

Multiply (1), (2) and (3)

lb × bh × lh = xyz

l² × b² × h² = xyz

(lbh)² = xyz

V² = xyz [Since the volume of the cuboid = V = lbh]

OR

Radius of small spheres (r) = 1 cm

Let the number of small spheres = n

n × volume of a small spheres = volume of bigger sphere

n * [4 / 3]𝛑r³ = [4 / 3] 𝛑R³

n = R³ / r³

= 8³ / 1³

= 512

Question 18: [i] Find cyclic factors ab (a – b) + bc (b – c) + ca (c – a). 5M

OR

[ii] Multiply the rational expressions x² − 7x + 10 / (x − 4)² and x² − 7x + 12 / x − 5 and express the product in its lowest terms.

Solution:

a² b − ab² + b²c − bc² + ac² − a²c

= a²(b − c) − a (b − c) (b + c) + bc (b − c)

= (b − c) [a² − a (b + c) + bc]

= (b − c) [a² − ab − ac + bc]

= (b − c) [a (a − b) − c (a − b)]

= (b − c) (a − b) (a − c)

OR

= [x² – 5x – 2x – 10] / (x – 4)² * (x² – 4x – 3x – 12) / (x – 5)

= (x – 5) (x – 2) / (x – 4)² * (x – 4) (x – 3) / (x – 5)

= (x – 2) (x – 3) / (x – 4)

Question 19: [i] If α and β are the roots of quadratic equations 3x² – 5x – 7 = 0, then find the value of α / β + β / α. 5M

OR

[ii] The length of the side forming the right angle of a right angled triangle are x cm and (x + 1) cm. If the area of the triangle is 10 cm², then find the sides of the triangle.

Solution:

a = 3 ; b = -5 ; c = -7

α + β = -b / a = 5 / 3

αβ = c / a = -7 / 3

α / β + β / α = α² + β² / αβ

= (α + β)² – 2αβ / αβ

= [(5 / 3)² – 2 * (-7 / 3)] / [-7 / 3]

= [(25 / 9) + (14 / 3) / [-7 / 3]

= [67 / 9] / [-7 / 3]

= -67 / 21

OR

x² + x – 20 = 0

x² + x = 20

x² + 2 × x × 1 / 2 = 20

x² + 2 × x × 1 / 2 + (1 / 2)² = 20 + (1 / 2)²

( x + 1 / 2 )² = 20 + 1 / 4

( x + 1 / 2 )² = ( 80 + 1 ) / 4

( x + 1 / 2 )² = 81 / 4

x + 1 / 2 = ± √( 81 / 4 )

x + 1 / 2 = ± 9 / 2

x = -1 / 2 ± 9 / 2

x = ( -1 / 2 + 9 / 2 ) or x = ( -1 / 2 – 9 / 2 )

x = 8 / 2 or x = -10 / 2

x = 4 or x = -5

Hence, sides are 4 cm and 5 cm.

Question 20: [i] Find the compound interest on Rs, 2000 at the rate of interest 4% per annum for 2 years. 5M

OR

[ii] A sewing machine is available for Rs. 1600 Cash or for Rs. 1200 cash down payment and Rs. 460 to be paid after 6 months. Find the rate of interest charged under the instalment plan.

Solution:

Interest rate = 4% per annum = R

Time = 2 Years = n

Amount = P (1 + R / 100)ⁿ

=> Amount = 2000 ( 1 + 4 / 100)²

=> Amount = 2000 * 1.04²

=> Amount = 2,163.2

Compound Interest = Amount – Principal

= 2163.2 – 2000

= 163.2 Rs

163.2 Rs is the compound interest on rupees 2000 for 2 years at 4% per annum compounded annually.

OR

Down payment = 1200 Rs

Balance due = 1600 – 1200 = 400 Rs

No. of instalment = 1

Amount of instalment = 460 Rs

Interest paid = 460 – 400 = 60 Rs

SI = PRT / 100

60 = [1600 * R * 6] / [100 * 12]

R = 7.5 %

Question 21: [i] The side of the triangle is 4cm, 6cm and 8cm. Draw the circumcircle of the triangle and write the steps of construction. 5M

OR

[ii] Construct a cyclic quadrilateral in which AC = 6cm, ∠B = 90°, AB = 3cm and AD = 4cm. Write the steps for construction also.

Solution:

I) Construction of triangle

1) Let the base of the triangle be QR=8cm.

2) Taking Q as the centre and radius 4cm on the compass, draw an arc on the upper side of QR.

3) Taking R as the centre and radius 6cm on the compass, draw an arc on the upper side of QR, intersecting the previous arc. Mark this point of intersection as point P.

4) Join points P to Q and P to Q to complete the △PQR.

II) Construction of perpendicular bisectors

1) Taking Q as the centre and radius more than half of QR, mark arcs below and above the line.

2) Now, with R as the centre and same radius, draw arcs above and below the line to intersect the already drawn arcs. Name the new points as X and Y.

3) Join points X and Y. This line XY is the required perpendicular bisector of side QR.

4) Similarly, draw perpendicular bisectors of sides PQ and PR.

III) Construction of circumcircle

1) All the perpendicular bisectors of △PQR will intersect at one point in the interior of the triangle.

Name that point as O.

2) Taking point O as the centre and OP as the radius draw a circle. Points Q and R should also lie on this circle.

3) This is the required circumcircle.

OR

(i) Draw AB = 3 cm

(ii) Draw ∠B + ∠D = 180^o

(iii) From A, draw an arc of 6 cm cutting BC.

(iv) Draw ∠A = 90^o.

(v) Cut an arc of 4 cm from A.

(vi) ABCD is the required cyclic quadrilateral.

Question 22: [i] Prove that sin²θ + cos² θ = 1. 5M

OR

[ii] Show whether the following is the identity or not: tan θ + sin θ / tan θ − sin θ = sec θ + 1 / sec θ − 1.

Solution:

sin θ = L / H

cos θ = B / H

L² + B² = H²

LHS = sin² θ + cos² θ

= [L / H]² + [B / H]²

= L² + B² / H²

= 1

= RHS

OR

= (sin θ / cos θ + sin θ) / (sin θ / cos θ – sin θ)

= (sin θ + sin θ . cos θ) / (sin θ – sin θ . cos θ)

= sin θ (1 + cos θ) / sin θ (1 – cos θ)

= (1 + cos θ) / (1 – cos θ)

= (1 + 1 / sec θ) / (1 – 1 / sec θ)

= (sec θ + 1) / (sec θ – 1)

= RHS

Question 23: [i] Prove that the length of two tangents drawn from an external point to a circle is equal. 6M

OR

[ii] Write the definition of a cyclic quadrilateral. Prove that the sum of pairs of opposite angles of a cyclic quadrilateral is 180°.

Solution:

Construction: Draw a line segment from centre O to external point T {touching point of two tangents}.

The tangents make a right angle with the radius of the circle.

PO and QO are radii.

So, ∠OPT = ∠OQT = 90°

Both the triangles ∆POT and QOT are right-angled triangles with a common hypotenuse OT.

∆POT and ∆QOT

∠OPT = OQT = 90°

Common hypotenuse OT and OP = OQ [OP and OQ are radii]

So, RHS rule of similarity,

∆POT ~ ∆QOT

Hence, OP / OQ = PT / QT = OT / OT

PT / QT = 1

PT = QT

OR

Construction: Draw AC and DB

Proof: ∠ACB = ∠ADB and ∠BAC = ∠BDC [Angles in the same segment]

∴ ∠ACB + ∠BAC = ∠ADB + ∠BDC = ∠ADC

Adding ∠ABC on both the sides,

∠ACB + ∠BAC + ∠ABC = ∠ADC + ∠ABC

But ∠ACB + ∠BAC + ∠ABC = 180° [Sum of the angles of a triangle]

∴ ∠ADC + ∠ABC = 180°

∴ ∠BAD + ∠BCD = 360° – (∠ADC + ∠ABC) = 180°.

Question 24: Compute the mean by the short cut method of the following frequency distribution: 6M

OR

Calculate the cost of living index number for the year 1999 on the basis of 1996 of a medium family from the following information.

Solution:

Mean = ∑f_ix_i / ∑f_i

= 1430 / 40

= 35.75

OR

Cost of living index in 1999 = [Total Expense in 1999 / Total Expense in 1996] × 100

= [1405 / 1171] * 100

= 119.98

= 120