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

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Time allowed:3 Hours                                              Maximum Marks:80 

SECTION – A

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. 

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

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/AB = AE/AC 

4AD = AD +7.2 

3AD = 7.2

AD = 2.4 cm Ans. 

SECTION – B

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

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

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. 

SECTION – C

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

Sol. Let 2 + 5√3 = r, where, r is rational. 

∴ (2 + 5√3)² = r² 

4 + 75 + 20 √3 = r²  

79 + 20 √3 = r²  

20√3 = 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 + 5√3 is an irrational number.  Hence Proved.

                                                                       OR

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

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

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.** 

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)  

AD = AC

      (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. 

17. Evaluate :**                                                                     [3]

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 = 15√2 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. 

SECTION – D

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

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.** 

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  T_n = -82 

∴ T_n = a + (n – 1) d

– 82 = -7 + (n – 1)(-5) 

– 82 = -7 – 5n + 5 

– 80 = -5n 

n = 16

Therefore, 16^th term will be – 82. 

Let T_n = -100 

Again, T_n = 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  S_n = 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 7^th to 10^th 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.