**Maths**

**Section -A**

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

Answer : Given, HCF (336, 54) = 6

We know HCF × LCM = one number × other number

6 × LCM = 336 × 54

LCM = 336×54/6 = 336 × 9 = 3024

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

Answer: 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 × (2)(3) = 16 – 24 = -8 < 0

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

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

Answer :

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

OR

Q.4. If sin A = ¾, Calculate sec A.

Answer :

We know,

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

Answer: 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}

⇒ x^{2} + 4 + 4x = x^{2} + 36 – 12x

⇒ 4 + 4x = 36 – 12x

⇒ 16x = 32

⇒ x = 2

Coordinates of P are (2, 0)

Q.6. Find the 21st term of the A.P. -4½, – 3, -1½ ………

Answer :

**Section- B**

Q.7. For what value of k, will the following pair of equations have infinitely many Answers:

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

Answer: 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 Answers.

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

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

Q.8. Find the area of a triangle whose vertices are given as (1, -1) (-4, 6) and (-3, -5).

Answer: 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.

Q.9. The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is ⅕. The probability of selecting a black marble at random from the same jar is ¼. If the jar contains 11 green marbles, find the total number of marbles in the jar.

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

P(x) = ⅕ , P (y) = ¼ Given

We Know,

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

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 Answer.

Answer: 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 Answer is

k can have any value except 6.

Q.11. The larger of the two supplementary angles exceeds the smaller by 18°. Find the angles.

OR

Q.11. 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?

Answer: 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°

OR

Answer: 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 = 2½ (y + 5)……….(ii)

On putting x = 3y in equation (ii)

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

Answer :

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**

Q.13. Point A lies on the line segment XY joining X(6, -6) and Y (-4, -1) in such a way that XA/XY = ⅖.

If Point A also lies on the line 3x + k (y + 1) = 0, find the value of k.

Answer: 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 × 2 + k(-4 + 1) = 0

⇒ 6 – 3k = 0

⇒ 3k = 6

⇒ k = 2

Q.14. Solve for x: x^{2} + 5x – (a^{2} + a – 6) = 0 [3]

Answer: 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

⇒ x^{2} + (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 = a – 2 and x = -(a + 3)

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

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

Answer: Given

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

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

A + 2B =60 …(i)

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°

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.

Answer: Let the required ratio be k : 1

By section formula, we have

Q.17. Evaluate:

Answer :

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

Q.18. 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)

Answer: 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}

⇒ 2a^{2} = r^{2}

⇒ r = √2 a

⇒ r = 15√2 cm (∵ a = 15 cm)

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

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

= 353.25 – 225 = 128.25 cm

OR

Answer: Given, ABCD is a square with side 2√2 cm

BD = 2r

In ∆BDC,

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

⇒ 4r^{2} = 2(DC)^{2} (∵ DC = CB = Side = 2√2 )

⇒ 4r^{2} = 2 × 2√2 × 2√2

⇒ 4r^{2} = 8 × 2

⇒ 4r^{2} = 16

⇒ r^{2 }= 4

⇒ r = 2 cm

Area of square BCDA = Side × Side = DC × BC = 2√2 × 2√2 = 8 cm^{2}

Area of circle = πr^{2} = 3.14 × 2 × 2 = 12.56 cm^{2}

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

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).

Answer: ABCD is a cylinder and BFC and AED are two hemispheres which has radius (r) = 7/2 cm

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

= 179.67 + 500.5 = 680.17 cm

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

Find the mean marks of the students.

Answer :

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

Q.21. Find the zeroes of the quadratic polynomial 7y2 – 11/2y – ⅔ and verify the relationship between the zeroes and the coefficients.

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

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

Q.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.

Answer: 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 × 1 × 16 = p^{2} – 64

for equal roots, we have D = 0

p^{2} – 64 = 0

⇒ p^{2} = 64

⇒ p = ±8 Putting p = 8 in given equation we have,

x2 + 8x + 16 = 0

⇒ (x + 4)^{2} = 0

⇒ x + 4 = 0⇒ x = -4

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

x^{2 } > – 8x + 16 = 0

⇒ (x – 4)^{2} = 0

⇒ x = 4

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

**Section – D**

Q.23. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares on their corresponding sides.

Answer: Given, ΔABC ~ ΔDEF

Q.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.

Answer: 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.

h———–(i)

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

⇒ x√3 =

⇒ 3x = 80 – x

⇒ 4x = 80

⇒ x = 20m

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

h = √3 × 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.

Q.25. The total cost of a certain length of a piece of cloth is ₹ 200. If the piece was 5 m longer and each metre of cloth costs ₹ 2 less, the cost of the piece would have remained unchanged. How long is the piece and what is its original rate per metre?

Answer: Let the original length of the piece of cloth is x m and rate of cloth is ₹ y per metre.

Then according to question, we have

x × y = 200 …(i)

and if length be 5 m longer and each meter of cloth be ₹ 2 less than

(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 × 200/x = -10

= 2x – 1000/x -10

⇒ 2x^{2} – 1000 = -10x

⇒ 2x^{2} + 10x – 1000 = 0

⇒ x^{2} + 5x – 500 = 0

⇒ x^{2} + 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 × y = 200

20 × y = 200

y = 10

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

Q.26. Construct an equilateral ∆ABC with each side 5 cm. Then construct another triangle whose sides are ⅔ 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.

Answer: Steps for construction are as follows:

- Draw a line segment BC = 5 cm
- At B and C construct ∠CBX = 60° and ∠BCX = 60°
- 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 ∆ABC is obtained.

- Below BC, make an acute angle ∠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 ∆EBD is the required triangle, each of whose sides is ⅔ of the corresponding side of ∆ABC.

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

Answer :

Q.28. Prove that:

Answer :

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

Q.29. How many terms of the Arithmetic Progression 45, 39, 33, … must be taken so that their sum is 180? Explain the double answer.

Answer:-7, -12, -17, -22, …….

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

Let T_{n} = -82T_{n} = a + (n – 1) d

⇒ -82 = -7 + (n – 1) (-5)

⇒ -82 = -7 – 5n + 5

⇒ -82 = -2 – 5n

⇒ -82 + 2 = -5n

⇒ -80 = -5n

⇒ n = 16

Therefore, 16th term will be -82.

Let T_{n} = -100

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

⇒ -100 = -7 + (n – 1) (-5)

⇒ -100 = -7 – 5n + 5

⇒ -100 = – 2 – 5n

⇒ -100 + 2 = -5n

⇒ -98 = -5n

⇒ n = 98/5

But the number of terms can not be in fraction.

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

OR

Answer: 45, 39, 33, …..

Here a = 45, d = 39 – 45 = -6

Let S_{n} = 180

⇒ n/2 [ 2a + (n – 1) d] = 180

⇒ n/2 [2 × 45 + (n – 1) (-6)] = 180

⇒ n/2 [90 – 6n + 6] = 180

⇒ n/2 [96 – 6n] = 180

⇒ n(96 – 6n) = 360

⇒ 96n – 6n^{2} = 360

⇒ 6n^{2} – 96n + 360 = 0

On dividing the above equation by 6

⇒ n^{2} – 16n + 60 = 0

⇒ n^{2} – 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.

Q.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.

Answer: 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

⇒ (32 – y) (y – 3) = 210

⇒ 32y – 96 – y^{2} + 3y = 210

⇒ y^{2} – 35y + 306 = 0

⇒ y^{2} – 18y – 17y + 306 = 0

⇒ y(y – 18) – 17(y – 18) = 0

⇒ (y – 18) (y – 17) = o

⇒ 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.

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