The document Math Past Year Paper with Solution - 2019, Class 10 Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.

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**Section A**

**Q 1. If HCF (336, 54) = 6, find LCM (336, 54).**

= 336 × 9 = 3024

**Q 2. Find the nature of roots of the quadratic equation 2x ^{2} – 4x + 3 = 0.**

2x

^{2}– 4x + 3 = 0 ⇒ D = 16 – 24 = –8

∴ Equation has NO real roots

**Q 3. Find the common difference of the Arithmetic Progression (A.P.)**

**Q 4. Evaluate : sin ^{2} 60^{o} + 2 tan 45^{o} – cos^{2} 30^{o}**

**OrIf sin A = 3/4 , calculate sec A.**

sin

^{2}60° + 2 tan 45° – cos^{2}30

[For any two correct values]= 2

oR

**Q 5. Write the coordinates of a point P on the x-axis which is equidistant from ****points A(– 2, 0) and B(6, 0).**

The point on x-axis is (2, 0)

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

**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.**

∆ABC: Isosceles ∆ ⇒ AC = BC = 4 cm.

**Section B**

**Q 7. Write the smallest number which is divisible by both 306 and 657.**

Smallest number divisible by 306 and 657 = LCM (306, 657)

LCM (306, 657) = 22338

**Q 8. Find a relation between x and y if the points A(x, y), B(– 4, 6) and**

**C(– 2, 3) are collinear.**

**OR**

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

**(– 3, – 5).**

A, B, C are collinear ⇒ ar. (∆ABC) = 0

OR

**Q 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.**

P(blue marble) = 1/5 , P(black marble)= 1/4

**Q 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.**

For unique solution

⇒ k ≠ 6

**Q 11. The larger of two supplementary angles exceeds the smaller by 18 ^{o}. Find**

**the angles.**

**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?**

Let larger angle be x°

∴ Smaller angle = 180° – x°∴ (x) – (180 – x) = 18

2x = 180 + 18 = 198 ⇒ x = 99

∴ The two angles are 99°, 81°

OR

Let Son’s present age be x years

Then Sumit’s present age = 3x years.

**Q 12. Find the mode of the following frequency distribution :**

Maximum frequency = 50, class (modal) = 35 – 40.

**Section C**

**Q 13. Prove that 2 + 5 √3 is an irrational number, given that √3 is an irrational number**

**OR**

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

Let 2+5√3 = a , where ‘a’ is a rational number.

than

Which is a contradiction as LHS is irrational and RHS is rational

∴ 2+5√3 can not be rational

Hence 2+5√3 is irrational.

Alternate method:Let 2+5√3 be rationalLHS is irrational and RHS is rational

which is a contradiction.∴ 2+5√3 is irrational.

OR

**Q 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.**

**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.**

Correct Figure

∆APB ~ ∆DPC [AA similarity]

ORCorrect Figure

In ∆POQ and ∆ROS∴ ∆POQ ~ ∆ROS [AA similarity]

**Q 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 ∠ AOB = 90 ^{o}.**

Correct Figure

∆AOD ≅ AOC [SAS]

Alternate method:Correct Figure

∆OAD ≅ ∆AOC [SAS]

**Q 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.**

Let the line x – 3y = 0 intersect the segment

joining A(–2, –5) and B(6, 3) in the ratio k : 1

**Q 17. Evaluate :**

**Q 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)**

**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)**

Radius of quadrant

Shaded area = Area of quadrant – Area of square

OR

∴ Radius of circle = 2 cm

∴ Shaded area = Area of circle – Area of square

**Q 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)**

Height of cylinder = 20 – 7 = 13 cm.

**Q 20. The marks obtained by 100 students in an examination are given below :**

**Find the mean marks of the students.**

Mean =Note: If N is taken as 100, Ans. 44.55

**Q 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 ?**

**OR**

**Find the zeroes of the quadratic polynomial and verify the relationship between the zeroes and the coefficients.**

OR

∴ Zeroes are 2/3, –1/7

Sum of zeroes =

**Q 22. Write all the values of p for which the quadratic equation x ^{2} + px + 16 = 0 **

x

^{2}+ px + 16 = 0 have equal roots if D = p^{2}– 4(16)(1) = 0p

^{2}= 64 ⇒ p = ±8∴ x

^{2}± 8x + 16 = 0 ⇒ (x ± 4)^{2}= 0

x ± 4 = 0∴ Roots are x = –4 and x = 4

**Section D**

**Q 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.**

Given: DE ∣∣ BC

To prove that:

EC / AE = BD / AD

Proof: ∠AED=∠ACB Corresponding angles

∠ADE = ∠ABC Corresponding angles

∠EAD is common to both the triangles

⇒ ΔAED ∼ ΔACB by AAA similarity

⇒ AE / AC = AD / AB

⇒ AE + EC / AE = AD + BD / AD

⇒ EC / AE = BD / AD

Hence proved

**Q 24. Amit, standing on a horizontal plane, finds a bird flying at a distance of 200 m from him at an elevation of 30 ^{o}. Deepak standing on the roof of a 50 m high building, finds the angle of elevation of the same bird to be 45^{o}. Amit and Deepak are on opposite sides of the bird. Find the distance of the bird from Deepak.**

In ∆APQ

**Q 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)**

Total volume = 3.14 (12)

^{2}(220) + 3.14(8)^{2}(60) cm^{3}

= 99475.2 + 12057.6 = 111532.8 cm^{3}

**Q 26. Construct an equilateral ΔABC with each side 5 cm. Then construct another triangle whose sides are 2/3 times the corresponding sides of ΔABC**

**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.**

Constructing an equilateral triangle of side 5 cm

Constructing another similar ∆ with scale factor 2/3OR

Constructing two concentric circles of radii 2 cm and 5 cm

Drawing two tangents PA and PB

PA = 4.5 cm (approx)

**Q 27. ****Change the following data into ‘less than type’ distribution and draw its ogive :**

Plotting of points (40, 7), (50, 12), (60, 20), (70, 30), (80, 36), (90, 42) and (100, 50)

Joining the points to get the curve

**Q 28. Prove that :**

**OR**

**Prove that :**

ORConsider

**Q 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.**

**OR**

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

Let –82 = a

_{n}

∴ –82 = –7 + (n – 1) (–5)

⇒ 15 = n – 1 or n = 16

Again –100 = am = –7 + (m – 1) (–5)

⇒ (m – 1)(–5) = –93

OR

**Q 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.**

Let marks in Hindi be x

Then marks in Eng = 30 – x

∴ (x + 2) (30 – x – 3) = 210

⇒ x^{2}– 25x + 156 = 0

or (x – 13) (x – 12) = 0

⇒ x = 13 or x = 12

∴ 30 – 13 = 17 or 30 – 12 = 18

∴ Marks in Hindi & English are

(13, 17) or (12, 18)

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