# CBSE 10 Class Latest MATHEMATICS (Maths) Solved Question Paper(Set-1)

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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

Comparing it with quadratic equation

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.

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

Sol. Given, ∠C=90° and AC = 4 cm

AB = ?

∵∆ABC is an isosceles triangle so,

BC = AC = 4 cm

On applying Phythagoras theorem, we have

AB2 = AC2 + BC2

= 42 + 42

⇒ AB² = 16 + 16 = 32

⇒ AB = √32

= 4√2 cm

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.

Sol. Given, DE || BC

On applying, Thales theorem, we have

AD = 2.4 cm Ans.

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

Sol. Smallest number which is divisible by 306 and 657 is LCM (657, 306)

657 = 3 x 3 x 73

306 = 3 x 3 x 2 x 17

LCM = 3 x 3 x 73 x 2 x 17

= 22338 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. Prove that 2 + 5√3 is an irrational number, given that √3 is an irrational number. [3]

Sol. Let 2 + 53 = r, where, r is rational.

∴ (2 + 53)2 = r2

4 + 75 + 20 3 = r2

79 + 20 3 = r2

203 = r2 -79

3 = r2 – 79 / 20

Now, r2 – 79 / 20 is a rational number. So, 3 must also be a rational number. But 3 is an irrational number (Given).

So, our assumption is wrong.

∴ 2 + 53 is an irrational number.  Hence Proved.

OR

## 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 x PC = BP x DP.            [3]

Sol. Given, ∆ABC, ∆DBC are right angle triangles, right angled at A and D, on same side of BC. AC & BD intersect at P.

In ∆APB and ∆PDC,

∠A = ∠D = 90°

∠APB = ∠DPC (Vertically opposite)

∴ ∆APB ~ ∆DPC (By AA Similarity)

⇒ AP x PC = BP x PD. Hence Proved.

OR

## 15. In Figure 3, PQ and RS are two parallel tangents to a circle with centreO and another tangent AB with point of contact C intersecting PO at A and RS at B. Prove that ∠AOB = 90°.   [3]

Sol. Given, PQ || RS

To prove: ∠AOB = 90°

Construction: Join O and C, D and E

In ∆ODA and ∆OCA

OD = OC       (radii of circle)

OA = OA       (common)

(tangent drawn from same point)

By SSS congruency

∆ODA ≅ ∆OCA

Then, ∠DOA = ∠AOC       ….(i)

Similarly, in ∆EOB and ∆BOC, we have

∆EOB ≅ ∆BOC

∠EOB = ∠BOC                 ….(ii)

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

Hence,

∠DOA + ∠AOC + ∠EOB + ∠BOC = 180°

2(∠AOC) + 2(∠BOC) = 180°

∠AOC + ∠BOC = 90°

∠AOB = 90°. Hence Proved.

## 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

OB = radius = r

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

Area of quadrant OPBQ

= 225 x 1.57

= 353.25 cm2

Area of shaded region

= 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 shaded region

= 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.

## 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]

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

To prove: AD/DB = AE/EC

Construction: Join BE and CD

Draw EL ⊥ AB and DM ⊥ AC

Proof: we have

Now, ∆DBE and ∆ECD, being on same base DE and between the same parallels DE and BC, we have

area (∆DBE) = area (∆ECD)           …..(iii)

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

AD/DB = AE/EC      Hence Proved.

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

Sol. Let Amit be at C point and bird is at A point. Such that ∠ACB = 30°. 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.

Now, In right triangle ABC, we have

sin 30° = P/H = AB/AC

1/2 = AB/200

AB = 100 m

In right ∆AFD, we have

sin 45° = P/H = AF/AD

(∵ AB = AF + BF

100 = AF + 50

AF = 50)

AD = 50√2 m

Hence, the distance of bird from Deepak is 50√2 m. Ans.

## 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 cm3 of iron has approximately 8 gm mass. (Use 𝛑 = 3.14) [4]

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

Volume of AB pole = 𝛑r12h1

= 3.14 x 12 x 12 x 220

= 99475.2 cm3

Volume of CD pole = 𝛑r22h2

= 3.14 x 8 x 8 x 60

= 12057.6 cm3

Total volume of the poles

= 99475.2 + 12057.6

= 111532.8 cm3

It is given that,

Mass of 1 cm3 of iron = 8 gm

Then mass of 111532.8 cm3 of iron

= 111532.8 x 8 gm

Then total mass of the pole is = 111532.8 x 8 gm

= 892262.4 gm

= 892.26 kg    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.