# CBSE 10 Class MATHEMATICS (Maths) Solved Latest Question Paper(Set-2)

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Time allowed:3 Hours                                              Maximum Marks:80

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

Sol. Given, HCF (336, 54) = 6

We know,

HCF x LCM = one number x other number

⇒ 6 x LCM = 336 x 54

⇒ LCM = 336 x 54 / 6

= 336 x 9

= 3024 Ans.

## 2. Find the nature of roots of the quadratic equation   [1]

2x2 – 4x + 3 = 0.

Sol. Given, 2x2 – 4x + 3 = 0

ax2 + bx + c = 0

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

∴ D = b2 – 4ac

= (-4)2 – 4 x (2) (3)

= 16 – 24

= -8 < 0

D < 0 shows that roots will not be real. Hence, roots will be imaginary. Ans.

Ans. Given,

## 4. Evaluate : sin2 60° + 2 tan 45°- cos2 30°                            [1]

Sol. We know,

sin 60° = √3/2, tan 45° = 1 and cos 30° = √3/2

∴  sin2 60° +2 tan 45° – cos2 30°

= 2 Ans.

OR

## If sin A = 3/4, calculate sec A.

Sol. Given,  sin A = 3/4

## 5. Write the coordinates of a point P on x-axis which is equidistant from the point A(-2, 0) and B(6, 0).                [1]

Sol. 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

x2 + 4 + 4x = x2 + 36 – 12x

⇒ 4 + 4x = 36 – 12

⇒ 16x = 32

⇒ x = 32/16 = 2

Co-ordinates of P are (2, 0) Ans.

Sol. Given,

## 7. For what value of k, will the following pair of equations have infinitely many solutions: [2]

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

Sol. Given, The system of equations is

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

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

-6 (1 – k) = 3k + 6 and 3(5k + 1) = -21 (1 – k)

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

## 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

## 9. The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is 1/5 The probability of selecting a black marble at random from the same jar is 1/4. If the jar contains 11 green marbles, find the total number of marbles in the jar.                                         [2]

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

P(x) = 1/5, P(y) = 1/4

We know,

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

1/5 + 1/4 + P(z) = 1

9/20 + P(z) = 1

(∵ No. of green marbles = 11)

⇒ Total no. of marbles = 20

∴ There are 20 marbles in the jar. Ans.

## 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]

Sol. Given,  x + 2y = 5

3x + ky + 15 = 0

Comparing above equations with

a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0,

We get,

a1 = 1, b1 = 2, c1 = -5

a2 = 3, b2 = k, c2 = 15

Condition for the pair of equations to have unique solution is

k ≠ 6

k can have any value except 6. Ans.

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

Sol. Let two angles A and B are supplementary.

∴ A + B = 180°     …(i)

Given, A = B +18°

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

B + 18° + B = 180°

2B + 18° = 180°

⇒ 2B = 162°

⇒ B = 81°

⇒ A = B + 18°

⇒ A = 99° Ans.

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?

Sol. Let age of Sumit be x years and age of his son be y years. Then, according to question we have,

x = 3y     .…(1)

Five years later,

On putting x = 3y in equation (ii)

y = 15 years

Then, present age of sumit is

3 x y = 3 x 15

= 45 years Ans.

## 12. Find the mode of the following frequency distribution:

Sol.

Here, maximum frequency is 50.

So, 35-40 will be the modal class.

l = 35, f0 = 34, f1 = 50, f2 = 42 and h = 5

= 38.33 Ans.

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

Sol. Given,

5XA = 2XA + 2AY

3XA = 2AY

XA/AY = ⅔

XA : AY = 2 : 3

So A divides XY in ratio 2 : 3

Here, m = 2, n = 3, x1 = 6, y1 = -6, x2 = -4 and y2 = -1

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 x 2 + k(-4 + 1) = 0

6 – 3k = 0

3k = 6

k = 2      Ans.

## 14. Solve for x:                                                                 [3]

x2 + 5x – (a2 + a – 6) = 0

Sol.  Taking (a2 + a – 6)

= a2 + 3a – 2a – 6

= a(a + 3) – 2(a + 3)

= (a + 3) (a – 2)

∴ x2 + 5x – (a + 3) (a – 2) = 0

x2 +(a + 3)x- (a – 2)x – (a + 3) (a – 2) = 0

x[x + (a + 3)]- (a – 2) [x + (a + 3)] = 0

(x – a + 2) (x + a + 3) = 0

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

x = 4 – 2 and x = – (a + 3)

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

## 15. Find A and B if sin (A + 2B) = √3/2 and cos (A + 4B) = 0, where A and B are acute angles.                                         [3]

Sol. Given,

sin (A + 2B) = √3/2 and cos (A + 4B) = 0

sin (A + 2B) = sin 60°                                (∵ sin 60° = √3/2)

A + 2B = 60°                    …..(i)

and cos (A + 4B) = cos 90°                       (∵ cos 90° = 0)

A +4B = 90°                      …..(ii)

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

B = 15° and A = 30°     Ans.

## 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]

Sol. Let required ratio be k : 1

By section formula, we have

Here, x1 = -2, x2 = 6, y1 = -5, y2 = 3

m = k, n = 1

6k – 2 – 9k + 15 = 0

-3k + 13 = 0

k = 13/3

Hence required ratio is

Here, intersection point are,

= 3/2

∴ intersection point is (9/2, 3/2)  Ans.

## 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 𝛑 = 3.14)

Sol. Given, OABC is a square with OA = 15 cm

Let side of square be a then,

a2 + a2 = r2

2a2 = r2

r = 2a

r = 152 cm    (∵ a = 15 cm)

Area of square = Side x Side

= 15 x 15

= 225 cm2

= 225 x 1.57

= 353.25 cm2

= Area of quadrant OPBQ – Area of square OABC

=353.25 – 225

= 128.25 cm²

OR

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

Sol. Given, ABCD is a square with side 2√2 cm

∵ BD = 2r

In ∆BDC

BD2 = DC2 + BC2

4r2 = 2(DC)2

(∵ DC = CB = Side = 2√2 )

4r2 = 2 x 2√2 x 2√2

4r2 = 8 x 2

4r2 = 16

⇒ r2 = 4

r = 2 cm

Area of square BCDA = Side x Side

= DC x BC

= 2√2 x 2√2

= 8 cm2

Area of circle = 𝛑r2

= 3.14 x 2 x 2

= 12.56 cm2

= Area of circle – Area of square.

= 12.56 – 8

= 4.56 cm2  Ans.

## 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 𝛑 = 22/7)                                                                          [3]

Sol. ABCD is a cylinder and BFC and AED are two hemisphere which has radius (r) = 7/2 cm

= 1001/2

= 500.5 cm3

Volume of two hemisphere

= 539/3

= 179.67 cm3

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

= 179.67 + 500.5

= 680.17 cm3  Ans.

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

Find the mean marks of the students.

Sol.

= 44.82  Ans.

## 21. For what value of k, is the polynomial                      [3]

f(x) = 3x4 – 9x3 + x2 + 15x + k

completely divisible by 3x2 – 52**

OR

## Find the zeroes of the quadratic polynomial 7y2 – 11/3y – 2/3 and verify the relationship between the zeroes and the coefficients.

Sol. The given polynomial is

P(y) = 7y2 – 11/3y – 2/3

∵ P(y) = 0

7y2 – 11/3y – 2/3 = 0

⇒ 21y2 – 11y – 2 = 0

⇒ 21y2 – 14y + 3y – 2 = 0

⇒ 7y (3y – 2) + 1 (3y – 2) = 0

⇒ (3y – 2) (7y + 1) = 0

∴ y = 2/3, – 1/7

So zeroes of P(y) are 2/3, – 1/7  Ans.

Verification : On comparing 7y – 11/3y – 2/3

with ax2 + bx + c, we get

a = 7, b = 11/3, c = 2/3

Sum of zeroes =  -b/a

11/21 = 11/21

Product of zeroes = c/a

Hence Verified.

## 22. Write all the values of p for which the quadratic equation x2 + px + 16 = 0 has equal roots. Find the roots of the equation so obtained.                                                 [3]

Sol. Given, equation is x2 + px + 16 = 0

This is of the form ax2 + bx + c = 0

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

∴ D = b2 – 4ac

= p2 – 4 x 1 x 16

= p2 – 64

for equal roots, we have

D = 0

p2 – 64 = 0

p2 = 64

p = ± 8

Putting p = 8 in given equation we have,

x2 + 8x + 16 = 0

(x – 4)2 = 0

x = 4, 4

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

## 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 60° and the angle of depression from the top of the other pole of point P is 30°. Find the heights of the poles and the distance of the point P from the poles.                [4]

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

From right ∆PAB, we have

AB/AP = tan 60° = √3

h/x = √3                        (∵ AP = x)

h = √3x          ….(i)

From right ∆DCP, we have

CD/PC =  tan 30° = 1/√3

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

⇒ 3x = 80 – x

⇒ 4x = 80

⇒ x = 20 m

On putting x = 20 in equation (1), we get

h = √3 x 20

= 20√3

h = 20√3 m

Thus, height of poles is 20√3 m and point P is at a distance of 20 m from left pole and (80 – 20) i.e., 60 m from right pole.

## 25. The total cost of a certain length of a piece of cloth is Rs 200. If the piece was 5 m longer and each metre of cloth costs Rs 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]

Sol. Let the original length of piece of cloth is xm and rate of cloth is Rs y per metre.

Then according to question, we have

x x y = 200                  …..(i)

and if length be 5 m longer and each meter of cloth be Rs 2 less then

(x + 5) (y – 2) = 200

(x + 5) (y – 2) = 200

xy – 2x + 5y – 10 = 200         ….(ii)

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

xy = xy – 2x + 5y – 10

⇒ 2x – 5y = -10                    …..(iii)

(y = 200/x) from equation (i)

⇒ 2x – 5 x 200/x = -10

⇒ 2x – 1000/x = -10

⇒ 2x2 – 1000 = -10x

⇒ 2x2 + 10x – 1000 = 0

⇒ x2 + 5x – 500 = 0

⇒ x2 + 25x – 20x – 500 = 0

⇒ x(x + 25) – 20 (x + 25) = 0

(x + 25) (x – 20) = 0

x = 20

(x ≠ -25 length of cloth can never be negative)

∴ x x y = 200

20 x y = 200

y = 10

Thus, length of the piece of cloth is 20 m and original price per metre is Rs 10.   Ans.

## 26. Construct an equilateral ∆ABC with each side 5 cm. Then construct another triangle whose sides are 2/3 times the corresponding sides of ∆ABC.**                               [4]

OR

## 27. Change the following data into less than type’ distribution and draw its ogive : [4]

Sol.

On graph paper, we take the scale.

## 28. Prove that:                                                      [4]

Sol.

Hence Proved.

OR

## Prove that :

Sol. LH.S.

R.H.S.

= 2 – (1 +  cos 𝜃)

= 1 – cos 𝜃                       …..(i)

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

L.H.S. = R.H.S.      Hence Proved.

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

Sol. -7, -12, -17, -22, ….

Here a = -7, d = -12 – (-7)

= -12 + 7

= -5

Let  Tn = -82

∴ Tn = a + (n – 1) d

– 82 = -7 + (n – 1)(-5)

– 82 = -7 – 5n + 5

– 80 = -5n

n = 16

Therefore, 16th term will be – 82.

Let Tn = -100

Again, Tn = a + (n – 1) d

– 100 = -7 + (n – 1) (-5)

– 100 = -7 -5n + 5

– 98 = -5n

n = 98/5

But the number of terms can not be in fraction. So, – 100 can not be a term of this A.P.   Ans.

OR

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

Sol. 45, 39, 33, ……..

Here a = 45, d = 39 – 45 = -6

Let  Sn = 180

n (96 – 6n) = 360

96n – 6n2 = 360

6n2 – 96n + 360 = 0

On dividing the above equation by 6

n2 – 16n + 60 = 0

n2 – 10n – 6n + 60 = 0

n(n – 10) – 6(n -10) = 0

(n – 10) (n – 6) = 0

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.    Ans.

## 30. In a class test, the sum of Arun’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]

Sol. Let Arun 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

(32 – y) (y – 3) = 210

32y – 96 – y2 + 3y = 210

y2 – 35y + 306 = 0

y2 -18y – 17y + 306 = 0

y (y – 18) – 17 (y – 18) = 0

(y – 18) (y – 17) = 0

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. Ans.