## CBSE Previous Year Question Papers Class 10 Maths 2019 Outside Delhi

Time Allowed: 3 hours

Maximum Marks: 80

General Instructions:

- All questions are compulsory.
- This 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 and Section D contains 8 questions of 4 marks each.
- There is no overall choice. However, an internal choice has been provided in two questions of 1 mark each, two questions of 2 marks each, four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternative in all such questions.
- Use of calculators is not permitted.

CBSE Sample Papers Class 10 Maths

### CBSE Previous Year Question Papers Class 10 Maths 2019 Outside Delhi Set I

**Section – A**

Question 1.

If HCF (336, 54) = 6, find LCM (336, 54). [1]

Solution:

Given, HCF (336, 54) = 6

We know,

HCF Ã— LCM = one number Ã— other number

â‡’ 6 Ã— LCM = 336 Ã— 54

â‡’ LCM = \(\frac { 336\times 54 }{ 6 }\) = 336 Ã— 9 = 3024

Question 2.

Find the nature of roots of the quadratic equation 2x^{2} – 4x + 3 = 0. [1]

Solution:

Given, 2x^{2} – 4x + 3 = 0

Comparing it with quadratic equation ax^{2} + bx + c = 0

Here, a = 2, b = -4 and c = 3

D = b^{2} – 4ac = (-4)^{2} – 4 Ã— (2)(3) = 16 – 24 = -8 < 0

Hence, D < 0 this shows that roots will be imaginary.

Question 3.

Find the common difference of the Arithmetic Progression (A.P.) [1]

Solution:

Question 4.

Evaluate: sin^{2} 60Â° + 2 tan 45Â° – cos^{2} 30Â° [1]

OR

If sin A = \(\frac { 3 }{ 4 }\), calculate sec A.

Solution:

We know,

Question 5.

Write the coordinates of a point P on the x-axis which is equidistant from point A(-2, 0) and B(6, 0).

Solution:

Let coordinates of P on x-axis is (x, 0)

Given, A(-2, 0) and B(6, 0)

Here, PA = PB

On squaring both sides, we get

(x + 2)^{2} = (x – 6)^{2}

â‡’ x^{2} + 4 + 4x = x^{2} + 36 – 12x

â‡’ 4 + 4x = 36 – 12x

â‡’ 16x = 32

â‡’ x = 2

Co-ordinates of P are (2, 0)

Question 6.

In Figure 1, ABC is an isosceles triangle right angled at C with AC = 4 cm. Find the length of AB. [1]

OR

In Figure 2, DE || BC. Find the length of side AD, given that AE = 1.8 cm, BD = 7.2 cm and CE = 5.4 cm.

Solution:

Given, âˆ C = 90Â° and AC = 4 cm, AB = ?

âˆ†ABC is an isosceles triangle so, BC = AC = 4 cm

On applying Phythagoras theorem, we have

AB^{2} = AC^{2} + BC^{2}

â‡’ AB^{2} = AC^{2} + AC^{2} (âˆµ BC = AC)

â‡’ AB^{2} = 4^{2} + 4^{2} = 16 + 16 = 32

â‡’ AB = âˆš32 = 4âˆš2 cm

OR

Given, DE || BC

On applying, Thales theorem, we have

**Section – B**

Question 7.

Write the smallest number which is divisible by both 306 and 657. [2]

Solution:

Smallest number which is divisible by 306 and 657 is,

LCM (657, 306)

657 = 3 Ã— 3 Ã— 73

306 = 3 Ã— 3 Ã— 2 Ã— 17

LCM =3 Ã— 3 Ã— 73 Ã— 2 Ã— 17 = 22338

Question 8.

Find a relation between x and y if the points A(x, y), B(-4, 6) and C(-2, 3) are collinear. [2]

OR

Find the area of a triangle whose vertices are given as (1, -1) (-4, 6) and (-3, -5).

Solution:

Given, A(x, y), B(-4, 6), C(-2, 3)

x_{1} = x, y_{1} = y, x_{2} = -4, y_{2} = 6, x_{3} = -2, y_{3} = 3

If these points are collinear, then area of triangle made by these points is 0.

Question 9.

The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is \(\frac { 1 }{ 5 }\). The probability of selecting a black marble at random from the same jar is \(\frac { 1 }{ 4 }\). If the jar contains 11 green marbles, find the total number of marbles in the jar. [2]

Solution:

Let the probability of selecting a blue marble, black marble and green marble are P(x), P(y), P(z) respectively.

P(x) = \(\frac { 1 }{ 5 }\), P(y) = \(\frac { 1 }{ 4 }\) (Given)

We know,

P(x) + P(y) + P(z) = 1

Question 10.

Find the value(s) of k so that the pair of equations x + 2y = 5 and 3x + ky + 15 = 0 has a unique solution. [2]

Solution:

Given, x + 2y = 5, 3x + ky + 15 = 0

Comparing above equations with

a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0,

We get,

a_{1} = 1, b_{1} = 2, c_{1} = -5

a_{2} = 3, b_{2} = k, c_{2} = 15

Condition for the pair of equations to have unique solution is

k can have any value except 6.

Question 11.

The larger of the two supplementary angles exceeds the smaller by 18Â°. Find the angles. [2]

OR

Sumit is 3 times as old as his son. Five years later, he shall be two and a half times as old as his son. How old is Sumit at present?

Solution:

Let two angles A and B are supplementary.

A + B = 180Â° …(i)

Given, A = B + 18Â°

On putting A = B + 18Â° in equation (i),

we get B + 18Â° + B = 180Â°

â‡’ 2B + 18Â° = 180Â°

â‡’ 2B = 162Â°

â‡’ B = 81Â°

A = B + 18Â°

â‡’ A = 99Â°

OR

Let age of Sumit be x years and age of his son be y years.

Then, according to question we have, x = 3y …… (i)

Five years later, x + 5 = 2\(\frac { 1 }{ 2 }\)(y + 5) …….. (ii)

On putting x = 3y in equation (ii)

Question 12.

Find the mode of the following frequency distribution:

Solution:

Here, the maximum frequency is 50.

So, 35 – 40 will be the modal class.

l = 35, f_{0} = 34, f_{1} = 50, f_{2} = 42 and h = 5

**Section – C**

Question 13.

Prove that 2 + 5âˆš3 is an irrational number, given that âˆš3 is an irrational number. [3]

OR

Using Euclid’s Algorithm, find the HCF of 2048 and 960.

Solution:

Let 2 + 5âˆš3 = r, where, r is rational.

â‡’ (2 + 5âˆš3)^{2} = r^{2}

â‡’ 4 + 75 + 20âˆš3 = r^{2}

â‡’ 79 + 20âˆš3 = r^{2}

â‡’ 20âˆš3 = r^{2} – 79

â‡’ âˆš3 = \(\frac { { r }^{ 2 }-79 }{ 20 }\)

Now, \(\frac { { r }^{ 2 }-79 }{ 20 }\) is a rational number. So, âˆš3 must also be a rational number.

But âˆš3 is an irrational number (Given).

So, our assumption is wrong.

2 + 5âˆš3 is an irrational number.

Hence Proved.

OR

Step I:

Here 2048 > 960 so, On applying Euclid’s algorithm, we get

2048 = 960 Ã— 2 + 128

Step II:

Because remainder 128 â‰ 0, so, On applying Euclid’s algorithm between 960 and 128, we get

960 = 128 Ã— 7 + 64

Step III:

Again remainder 64 â‰ 0, so

128 = 64 Ã— 2 + 0

Here remainder is 0. So, the process ends here. And the dividend is 64 so, required HCF is 64.

Question 14.

Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same side of BC. If AC and BD intersect at P, prove that AP Ã— PC = BP Ã— DP. [3]

OR

Diagonals of a trapezium PQRS intersect each other at the point O, PQ || RS and PQ = 3RS. Find the ratio of the areas of triangles POQ and ROS.

Solution:

Given, âˆ†ABC, âˆ†DBC are right-angle triangles, right-angled at A and D, on the same side of BC.

AC & BD intersect at P.

In âˆ†APB and âˆ†PDC,

âˆ A = âˆ D = 90Â°

âˆ APB = âˆ DPC (Vertically opposite)

âˆ†APB ~ âˆ†PDC (By AA Similarity)

\(\frac { AP }{ BP }\) = \(\frac { PD }{ PC }\) (by c.s.s.t.)

â‡’ AP Ã— PC = BP Ã— PD.

Hence Proved.

OR

Given, PQRS is a trapezium where PQ || RS and diagonals intersect at O and PQ = 3RS

In âˆ†POQ and âˆ†ROS, we have

âˆ ROS = âˆ POQ (vertically opposite angles)

âˆ OQP = âˆ OSR (alternate angles)

Hence, âˆ†POQ ~ âˆ†ROS by AA similarity then,

If two triangles are similar, then ratio of areas is equal to the ratio of square of its corresponding sides. Then,

Question 15.

In Figure 3, PQ and RS are two parallel tangents to a circle with centre O and another tangent AB with the point of contact C intersecting PQ at A and RS at B. Prove that âˆ AOB = 90Â°. [3]

Solution:

Given, PQ || RS

To prove: âˆ AOB = 90Â°

Construction: Join O and C, D and E

In âˆ†ODA and âˆ†OCA

OD = OC (radii of circle)

OA = OA (common)

AD = AC (tangent drawn from the same point)

By SSS congruency

âˆ†ODA = âˆ†OCA

Then, âˆ DOA = âˆ AOC …(i)

Similarly, in âˆ†EOB and âˆ†BOC, we have

âˆ†EOB = âˆ†BOC

âˆ EOB = âˆ BOC …(ii)

EOD is a diameter of the circle, therefore it is a straight line.

Hence, âˆ DOA + âˆ AOC + âˆ EOB + âˆ BOC = 180Â°

â‡’ 2(âˆ AOC) + 2(âˆ BOC) = 180Â°

â‡’ âˆ AOC + âˆ BOC = 90Â°

â‡’ âˆ AOB = 90Â°.

Hence Proved.

Question 16.

Find the ratio in which the line x – 3y = 0 divides the line segment joining the points (-2, -5) and (6, 3). Find the coordinates of the point of intersection. [3]

Solution:

Let the required ratio be k : 1

By section formula, we have

Question 17.

Evaluate:

Solution:

Question 18.

In Figure 4, a square OABC is inscribed in a quadrant OPBQ. If OA = 15 cm, find the area of the shaded region. (Use Ï€ = 3.14)

In Figure 5, ABCD is a square with side 2âˆš2 cm and inscribed in a circle. Find the area of the shaded region. (Use Ï€ = 3.14)

Solution:

Given, OABC is a square with OA = 15 cm

OB = radius = r

Let side of square be a then,

a^{2} + a^{2} = r^{2}

â‡’ 2a^{2} = r^{2}

â‡’Â r = âˆš2 a

â‡’Â r = 15âˆš2 cm (âˆµ a = 15 cm)

Area of square = Side Ã— Side = 15 Ã— 15 = 225 cm^{2}

Area of shaded region = Area of quadrant OPBQ – Area of square

= 353.25 – 225 = 128.25 cm^{2}

OR

Given, ABCD is a square with side 2âˆš2 cm

BD = 2r

In âˆ†BDC,

BD^{2} = DC^{2} + BC^{2}

â‡’ 4r^{2} = 2(DC)^{2} (âˆµ DC = CB = Side = 2âˆš2 )

â‡’ 4r^{2} = 2 Ã— 2âˆš2 Ã— 2âˆš2

â‡’ 4r^{2} = 8 Ã— 2

â‡’ 4r^{2} = 16

â‡’ r^{2} = 4

â‡’ r = 2 cm

Area of square BCDA = Side Ã— Side = DC Ã— BC = 2âˆš2 Ã— 2âˆš2 = 8 cm^{2}

Area of circle = Ï€r2 = 3.14 Ã— 2 Ã— 2 = 12.56 cm^{2}

Area of shaded region = Area of circle – Area of square. = 12.56 – 8 = 4.56 cm^{2}

Question 19.

A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 20 cm and the diameter of the cylinder is 7 cm. Find the total volume of the solid. (Use Ï€ = \(\frac { 22 }{ 7 }\)) [3]

Solution:

ABCD is a cylinder and BFC and AED are two hemisphere which has radius (r) = \(\frac { 7 }{ 2 }\) cm

Total volume of solid = Volume of two hemisphere + Volume of cylinder

= 179.67 + 500.5 = 680.17 cm^{3}

Question 20.

The marks obtained by 100 students in an examination are given below: [3]

Find the mean marks of the students.

Solution:

Question 21.

For what value of k, is the polynomial f(x) = 3x^{4} – 9x^{3} + x^{2} + 15x + k completely divisible by 3x^{2} – 5? [3]

OR

Find the zeroes of the quadratic polynomial 7y^{2} – \(\frac { 11 }{ 3 }\) y – \(\frac { 2 }{ 3 }\) and verify the relationship between the zeroes and the coefficients.

Solution:

Given,

f(x) = 3x^{4} – 9x^{3} + x^{2} + 15x + k

It is completely divisible by 3x^{2} – 5

Question 22.

Write all the values of p for which the quadratic equation x^{2} + px + 16 = 0 has equal roots. Find the roots of the equation so obtained. [3]

Solution:

Given, equation is x^{2} + px + 16 = 0

This is of the form ax^{2} + bx + c = 0

where, a = 1, b = p and c = 16

D = b^{2} – 4ac = p^{2} – 4 Ã— 1 Ã— 16 = p^{2} – 64

for equal roots, we have D = 0

p^{2} – 64 = 0

â‡’ p^{2} = 64

â‡’ p = Â±8

Putting p = 8 in given equation we have,

x^{2} + 8x + 16 = 0

â‡’ (x + 4)^{2} = 0

â‡’ x + 4 = 0

â‡’ x = -4

Now, putting p = -8 in the given equation, we get

x^{2} – 8x + 16 = 0

â‡’ (x – 4)^{2} = 0

â‡’ x = 4

Required roots are -4 and -4 or 4 and 4.

**Section – D**

Question 23.

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio. [4]

Solution:

Given, A âˆ†ABC in which DE || BC and DE intersect AB and AC at D and E respectively.

To prove: \(\frac { AD }{ DB }\) = \(\frac { AE }{ EC }\)

Construction: Join BE and CD

Draw EL âŠ¥ AB and DM âŠ¥ AC

Proof: we have

area (âˆ†ADE) = \(\frac { 1 }{ 2 }\) Ã— AD Ã— EL

and area (âˆ†DBE) = \(\frac { 1 }{ 2 }\) Ã— DB Ã— EL (âˆµ âˆ† = \(\frac { 1 }{ 2 }\) Ã— b Ã— h)

Now, âˆ†DBE and âˆ†ECD, being on same base DE and between the same parallels DE and BC, We have

area (âˆ†DBE) = area (âˆ†ECD) …..(iii)

from equations (i), (ii) and (iii), we have

\(\frac { AD }{ DB }\) = \(\frac { AE }{ EC }\)

Hence Proved.

Question 24.

Amit, standing on a horizontal plane, finds a bird flying at a distance of 200 m from him at an elevation of 30Â°. Deepak standing on the roof of a 50 m high building, finds the angle of elevation of the same bird to be 45Â°. Amit and Deepak are on opposite sides of the bird. Find the distance of the bird from Deepak. [4]

Solution:

Let Amit be at C point and the bird is at A point. Such that âˆ ACB = 30Â°. AB is the height of bird from point B on ground and Deepak is at D point, DE is the building of height 50 m.

Hence, the distance of bird from Deepak is 50âˆš2 m.

Question 25.

A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm^{3} of iron has approximately 8 gm mass. (Use Ï€ = 3.14) [4]

Solution:

Let AB be the iron pole of height 220 cm with base radius 12 cm and there is the other cylinder CD of height 60 cm whose base radius is 8 cm.

Volume of AB pole = Ï€r_{1}h_{1} = 3.14 Ã— 12 Ã— 12 Ã— 220 = 99475.2 cm^{3}

Volume of CD pole = Ï€r_{2}h_{2} = 3.14 Ã— 8 Ã— 8 Ã— 60 = 12057.6 cm^{3}

Total volume of the poles = 99475.2 + 12057.6 = 111532.8 cm^{3}

It is given that,

Mass of 1 cm^{3} of iron = 8 gm

Then mass of 111532.8 cm^{3} of iron = 111532.8 Ã— 8 gm

Then total mass of the pole is = 111532.8 Ã— 8 gm = 892262.4 gm = 892.2624 kg

Question 26.

Construct an equilateral âˆ†ABC with each side 5 cm. Then construct another triangle whose sides are \(\frac { 2 }{ 3 }\) times the corresponding sides. Draw two concentric circles of radii 2 cm and 5 cm. Take a point P on the outer circle and construct a pair of tangents PA and PB to the smaller circle. Measure PA.

Solution:

Steps for construction are as follows:

- Draw a line segment BC = 5 cm
- At B and C construct âˆ CBX = 60Â° and âˆ BCX = 60Â°
- With B as centre and radius 5 cm, draw an arc cutting ray BX at A. On graph paper, we take the scale.
- Join AC. Thus an equilateral âˆ†ABC is obtained.

- Below BC, make an acute angle âˆ CBY
- Along BY, mark off 3 points B
_{1}, B_{2}, B_{3}Such that BB_{1}, B_{1}B_{2}, B_{2}B_{3}are equal. - Join B
_{3}C - From B
_{2}draw B_{2}D || B_{3}C, meeting BC at D - From D, draw DE || CA, meeting AB at E.

Then âˆ†EBD is the required triangle, each of whose sides is \(\frac { 2 }{ 3 }\) of the corresponding side of âˆ†ABC.

Question 27.

Change the following data into ‘less than type’ distribution and draw its ogive:

Solution:

Question 28.

Prove that:

Solution:

Question 29.

Which term of the Arithmetic Progression -7, -12, -17, -22,…..will be -82? Is -100 any term of the A.P.? Give a reason for your answer. [4]

OR

How many terms of the Arithmetic Progression 45, 39, 33, …. must be taken so that their sum is 180? Explain the double answer.

Solution:

-7, -12, -17, -22, …….

Here a = -7, d = -12 – (-7) = -12 + 7 = -5

Let T_{n} = -82

T_{n} = a + (n – 1) d

â‡’ -82 = -7 + (n – 1) (-5)

â‡’ -82 = -7 – 5n + 5

â‡’ -82 = -2 – 5n

â‡’ -82 + 2 = -5n

â‡’ -80 = -5n

â‡’ n = 16

Therefore, 16th term will be -82.

Let T_{n} = -100

Again, T_{n} = a + (n -1) d

â‡’ -100 = -7 + (n – 1) (-5)

â‡’ -100 = -7 – 5n + 5

â‡’ -100 = – 2 – 5n

â‡’ -100 + 2 = -5n

â‡’ -98 = -5n

â‡’ n = \(\frac { 98 }{ 5 }\)

But the number of terms can not be in fraction.

So, -100 can not be the term of this A.P.

OR

45, 39, 33, …..

Here a = 45, d = 39 – 45 = -6

Let S_{n} = 180

â‡’ \(\frac { n }{ 2 }\) [ 2a + (n – 1) d] = 180

â‡’ \(\frac { n }{ 2 }\) [2 Ã— 45 + (n – 1) (-6)] = 180

â‡’ \(\frac { n }{ 2 }\) [90 – 6n + 6] = 180

â‡’ \(\frac { n }{ 2 }\) [96 – 6n] = 180

â‡’ n(96 – 6n) = 360

â‡’ 96n – 6n^{2} = 360

â‡’ 6n^{2} – 96n + 360 = 0

On dividing the above equation by 6

â‡’ n^{2} – 16n + 60 = 0

â‡’ n^{2} – 10n – 6n + 60 = 0

â‡’ n(n – 10) – 6 (n – 10) = 0

â‡’ (n – 10) (n – 6) = 0

â‡’ n = 10, 6

Sum of first 10 terms = Sum of first 6 terms = 180

This means that the sum of all terms from 7th to 10th is zero.

Question 30.

In a class test, the sum of Aran’s marks in Hindi and English is 30. Had he got 2 marks more in Hindi and 3 marks less in English, the product of the marks would have been 210. Find his marks in the two subjects. [4]

Solution:

Let Aran marks in Hindi be x and marks in English be y.

Then, according to question, we have

x + y = 30 …(i)

(x + 2)(y – 3) = 210 …(ii)

from equation (i) put x = 30 – y in equation (ii)

(30 – y + 2) (y – 3) = 210

â‡’ (32 – y) (y – 3) = 210

â‡’ 32y – 96 – y^{2} + 3y = 210

â‡’ y^{2} – 35y + 306 = 0

â‡’ y^{2} – 18y – 17y + 306 = 0

â‡’ y(y – 18) – 17(y – 18) = 0

â‡’ (y – 18) (y – 17) = o

â‡’ y = 18, 17

Put y = 18 and 17 in equation (i), we get x = 12, 13

Hence his marks in hindi can be 12 and 13 and in english his marks can be 18 and 17.

### CBSE Previous Year Question Papers Class 10 Maths 2019 Outside Delhi Set II

**Note:** Except for the following questions, all the remaining questions have been asked in Set I.

**Section – A**

Question 6.

Find the 21st term of the A.P. -4\(\frac { 1 }{ 2 }\), -3, -1\(\frac { 1 }{ 2 }\), … [1]

Solution:

**Section – B**

Question 7.

For what value of k, will the following pair of equations have infinitely many solutions:

2x + 3y = 7 and (k + 2)x – 3(1 – k)y = 5k + 1 [2]

Solution:

Given, The system of equations is

2x + 3y = 7 and (k + 2) x – 3 (1 – k) y = 5k +1

These equations are of the form a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0

where, a_{1} = 2, b_{1} = 3, c_{1} = -7

a_{2} = (k + 2), b_{2} = -3(1 – k), c_{2} = -(5k + 1)

Since, the given system of equations have infinitely many solutions.

Hence, the given system of equations has infinitely many solutions when k = 4.

**Section – C**

Question 13.

Point A lies on the line segment XY joining X(6, -6) and Y (-4, -1) in such a way that \(\frac { XA }{ XY }\) = \(\frac { 2 }{ 5 }\). If Point A also lies on the line 3x + k (y + 1) = 0, find the value of k. [3]

Solution:

Given,

Since, point A(2, -4) lies on line 3x + k (y + 1) = 0.

Therefore it will satisfy the equation.

On putting x = 2 and y = -4 in the equation, we get

3 Ã— 2 + k(-4 + 1) = 0

â‡’ 6 – 3k = 0

â‡’ 3k = 6

â‡’ k = 2

Question 14.

Solve for x: x^{2} + 5x – (a^{2} + a – 6) = 0 [3]

Solution:

Taking (a^{2} + a – 6)

= a^{2} + 3a – 2a – 6

= a(a + 3) – 2 (a + 3)

= (a + 3) (a – 2)

x^{2} + 5x – (a + 3) (a – 2) = 0

â‡’ x^{2} + (a + 3)x – (a – 2)x – (a + 3)(a – 2) = 0

â‡’Â x[x + (a + 3)] – (a – 2) [x + (a + 3)] = 0

â‡’ (x – a + 2)(x + a + 3) = 0

Hence, x – a + 2 = 0 and x + a + 3 = 0

x = a – 2 and x = -(a + 3)

Required values of x are (a – 2), -(a + 3)

Question 15.

Find A and B if sin (A + 2B) = \(\frac { \surd 3 }{ 2 }\) and cos (A + 4B) = 0, where A and B are acute angles. [3]

Solution:

Given,

sin (A + 2B) = \(\frac { \surd 3 }{ 2 }\) and cos (A + 4B) = 0

â‡’ sin (A + 2B) = 60Â° (âˆµ sin 60Â° = \(\frac { \surd 3 }{ 2 }\))

A + 2B =60 …(i)

cos (A + 4B) = cos 90Â° (âˆµ cos 90Â° = 0)

â‡’ A + 4B = 90Â° …(ii)

On solving equation (i) and (ii), we get

B = 15Â° and A = 30Â°

**Section – D**

Question 23.

Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares on their corresponding sides.

Solution:

Given, Î”ABC ~ Î”DEF

Question 24.

Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point P between them on the road, the angle of elevation of the top of a pole is 60Â° and the angle of depression from the top of the other pole of point P is 30Â°. Find the heights of the poles and the distance of the point P from the poles. [4]

Solution:

Let AC is the road of 80 m width. P is the point on road AC and height of poles AB and CD is h m.

â‡’ h = \(\frac { 80-x }{ \surd 3 }\) …… (ii)

Equating the values of h from equation (i) and (ii) we get

â‡’ xâˆš3 = \(\frac { 80-x }{ \surd 3 }\)

â‡’ 3x = 80 – x

â‡’ 4x = 80

â‡’ x = 20m

On putting x = 20 in equation (i), we get

h = âˆš3 Ã— 20 = 20âˆš3

h = 20âˆš3 m

Thus, height of poles is 20âˆš3 m and point P is at a distance of 20 m from left pole and (80 – 20) i.e., 60 m from right pole.

Question 25.

The total cost of a certain length of a piece of cloth is â‚¹ 200. If the piece was 5 m longer and each metre of cloth costs â‚¹ 2 less, the cost of the piece would have remained unchanged. How long is the piece and what is its original rate per metre? [4]

Solution:

Let the original length of the piece of cloth is x m and rate of cloth is â‚¹ y per metre.

Then according to question, we have

x Ã— y = 200 …(i)

and if length be 5 m longer and each meter of cloth be â‚¹ 2 less then

(x + 5) (y – 2) = 200

â‡’ (x + 5) (y – 2) = 200

â‡’ xy – 2x + 5y – 10 = 200 …(ii)

On equating equation (i) and (ii), we have

xy = xy – 2x + 5y – 10

â‡’ 2x – 5y = -10 …… (iii)

â‡’ y = \(\frac { 200 }{ x }\) from equation (i)

â‡’ 2x – 5 Ã— \(\frac { 200 }{ x }\) = -10

â‡’ 2x – \(\frac { 1000 }{ x }\) = -10

â‡’ 2x^{2} – 1000 = -10x

â‡’ 2x^{2} + 10x – 1000 = 0

â‡’ x^{2} + 5x – 500 = 0

â‡’ x^{2} + 25x – 20x – 500 = 0

â‡’ x(x + 25) – 20 (x + 25) = 0

â‡’ (x + 25) (x – 20) = 0

â‡’ x = 20 (x â‰ -25 length of cloth can never be negative)

âˆ´ x Ã— y = 200

20 Ã— y = 200

y = 10

Thus, length of the piece of cloth is 20 m and original price per metre is â‚¹ 10.

### CBSE Previous Year Question Papers Class 10 Maths 2019 Outside Delhi Set III

**Note:** Except for the following questions, all the remaining questions have been asked in previous sets.

**Section – B**

Question 7.

A die is thrown twice. Find the probability that

(i) 5 will come up at least once. [2]

(ii) 5 will not come up either time.

Solution:

When two dice are thrown simultaneously, all possible outcomes are

(1.1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

Total number of outcomes = 36

Total outcomes where 5 comes up at least once = 11

**Section – C**

Question 13.

Find the ratio in which the y-axis divides the line segment joining the points (-1, -4) and (5, -6). Also, find the coordinates of the point of intersection. [3]

Solution:

Let the y-axis cut the line joining point A(-1, -4) and point B(5, -6) in the ratio k : 1 at the point P(0, y)

Then, by section formula, we have

Question 14.

Find the value of: [3]

Solution:

Question 15.

Two spheres of same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the new sphere. [3]

Solution:

Given, a radius of small sphere be r = 3 cm

Both spheres are made by same metal, then their densities will be same.

Let radius of bigger sphere = r’ then,

Then according to question, we have,

Volume of bigger sphere + Volume of smaller shpere = Volume of new sphere.

\(\frac { 4 }{ 3 }\) (r’)^{3} + \(\frac { 4 }{ 3 }\) (r)^{3} = \(\frac { 4 }{ 3 }\) (R)^{3}

â‡’ r’^{3} + r^{3} = R^{3}

â‡’ 189 + 27 = R^{3}

â‡’ 216 = R^{3}

â‡’ R = 6

D = 6 Ã— 2 = 12

Radius of new sphere is 6 cm.

So, the diameter is 12 cm.

**Section – D**

Question 23.

In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then prove that the angle opposite the first side is a right angle. [4]

Solution:

Given, âˆ†ABC in which AC^{2} = AB^{2} + BC^{2}

To prove: âˆ B = 90Â°

Consturction : Draw a âˆ†DEF such that

DE = AB, EF = BC and âˆ E = 90Â°.

Proof: In âˆ†DEF we have âˆ E = 90Â°

So, by Pythagoras theorem, we have

DF^{2} = DE^{2} + EF^{2}

â‡’ DF^{2} = AB^{2} + BC^{2} …(i)

(âˆµ DE = AB and EF = BC)

AC^{2} = AB^{2} + BC^{2} …(ii) (Given)

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

AC^{2} = DF^{2} â‡’ AC = DF.

Now, in âˆ†ABC and âˆ†DEF, we have

AB = DE, BC = EF and AC = DF.

âˆ†ABC = âˆ†DEF.

Hence, âˆ B = âˆ E = 90Â°.

Hence Proved.

Question 24.

From a point P on the ground, the angle of elevation of the top of a tower is 30Â° and that of the top of the flag-staff fixed on the top of the tower is 45Â°. If the length of the flag-staff is 5 m, find the height of the tower. (Use âˆš3 = 1.732) [4]

Solution:

Let AB be the tower and BC be the flag-staff.

Let P be a point on the ground such that

âˆ APB = 30Â° and âˆ APC = 45Â°, BC = 5 m

Let AB = h m and PA = x metres

From right âˆ†PAB, we have

Hence, the height of the tower is 6.83 m

Question 25.

A right cylindrical container of radius 6 cm and height 15 cm is full of ice-cream, which has to be distributed to 10 children in equal cones having a hemispherical shape on the top. If the height of the conical portion is four times its base radius, find the radius of the ice-cream cone. [4]

Solution:

Let R and H be the radius and height of the cylinder.

Given, R = 6 cm, H = 15 cm.

Volume of ice-cream in the cylinder = Ï€R^{2}H = Ï€ Ã— 36 Ã— 15 = 540Ï€ cm^{3}

Let the radius of cone be r cm

Height of the cone (h) = 4r

Radius of hemispherical portion = r cm.

Volume of ice-cream in cone = Volume of cone + Volume of the hemisphere

Number of ice cream cones distributed to the children = 10

â‡’ 10 Ã— Volume of ice-cream in each cone = Volume of ice-cream in cylindrical container

â‡’ 10 Ã— 2Ï€r^{3} = 540Ï€

â‡’ 20r^{3} = 540

â‡’ r^{3} = 27

â‡’ r = 3

Thus, the radius of the ice-cream cone is 3 cm.

CBSE Previous Year Question Papers CBSE Previous Year Question Papers Class 10 Maths