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

OR

Q.6. 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.

Answer:

Given, ∠C = 90° and AC = 4 cm, AB = ?

∆ABC is an isosceles triangle so, BC = AC = 4 cm

On applying Pythagoras theorem, we have

AB^{2} = AC^{2} + BC^{2}

⇒ AB^{2} = AC^{2} + AC^{2} (∵ BC = AC)

⇒ AB^{2} = 42 + 42 = 16 + 16 = 32

⇒ AB = √32 = 4√2 cm

OR

Answer: Given, DE || BCOn applying, Thales theorem, we have

**Section – B**

Q.7. A die is thrown twice. Find the probability that

(i) 5 will come up at least once.

(ii) 5 will not come up either time.

Answer: When two dice are thrown simultaneously, all possible outcomes are

(1.1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

Total number of outcomes = 36Total outcomes where 5 comes up at least once = 11

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. Find the ratio in which the y-axis divides the line segment joining the points (-1, -4) and (5, -6). Also, find the coordinates of the point of intersection.

Answer: Let the y-axis cut the line joining point A(-1, -4) and point B(5, -6) in the ratio k : 1 at the point P(0, y)

Then, by section formula, we have

Q.14. Find the value of:

Answer :

Q.15. Two spheres of same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the new sphere.

Answer: Given, a radius of small sphere be r = 3 cm

Both spheres are made by the same metal, then their densities will be same.

Let radius of bigger sphere = r’ then,

Then according to question, we have,

Volume of bigger sphere + Volume of smaller sphere = Volume of new sphere.

4/3(r)^{3} + 4/3 (r)^{3 }= 4/3 (R)^{3}

⇒ r’^{3} + r^{3} = R^{3}

⇒ 189 + 27 = R^{3}

⇒ 216 = R^{3}

⇒ R = 6

D = 6 × 2 = 12

Radius of new sphere is 6 cm.

So, the diameter is 12 cm.

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. In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then prove that the angle opposite the first side is a right angle.

Answer: Given, ∆ABC in which AC^{2} = AB^{2} + BC^{2}

To prove: ∠B = 90°Construction : Draw a ∆DEF such that

DE = AB, EF = BC and ∠E = 90°.

Proof: In ∆DEF we have ∠E = 90°

So, by Pythagoras theorem, we have

DF^{2} = DE^{2} + EF^{2}

⇒ DF^{2} = AB^{2} + BC^{2} …(i)

(∵ DE = AB and EF = BC)

AC^{2} = AB^{2} + BC^{2} …(ii) (Given)

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

AC^{2} = DF^{2} ⇒ AC = DF.

Now, in ∆ABC and ∆DEF, we have

AB = DE, BC = EF and AC = DF.

∆ABC = ∆DEF.

Hence, ∠B = ∠E = 90°.

Hence Proved.

Q.24. From a point P on the ground, the angle of elevation of the top of the tower is 30° and that of the top of the flag-staff fixed on the top of the tower is 45°. If the length of the flag-staff is 5 m, find the height of the tower. (Use √3 = 1.732) [4]

Answer: Let AB be the tower and BC be the flag-staff.

Let P be a point on the ground such that

∠APB = 30° and ∠APC = 45°, BC = 5 m

Let AB = h m and PA = x metres

From right ∆PAB, we have

Hence, the height of the tower is 6.83 m

Q.25. A right cylindrical container of radius 6 cm and height 15 cm is full of ice-cream, which has to be distributed to 10 children in equal cones having a hemispherical shape on the top. If the height of the conical portion is four times its base radius, find the radius of the ice-cream cone.

Answer: Let R and H be the radius and height of the cylinder.

Given, R = 6 cm, H = 15 cm.

Volume of ice-cream in the cylinder = πR^{2}H = π × 36 × 15 = 540π cm^{3}

Let the radius of cone be r cm

Height of the cone (h) = 4r

Radius of hemispherical portion = r cm.

Volume of ice-cream in cone = Volume of cone + Volume of the hemisphere

Number of ice cream cones distributed to the children = 10

⇒ 10 × Volume of ice-cream in each cone = Volume of ice-cream in cylindrical container

⇒ 10 × 2πr =540π

⇒ 20r^{3} = 540

⇒ r^{3} = 27

⇒ r = 3

Thus, the radius of the ice-cream cone is 3 cm.

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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