**Maths**

**Section – A**

Q.1. If x = 3 is one root of the quadratic equation x^{2} – 2kx – 6 = 0, then find the value of k.

Solution:

Given quadratic equation is, x^{2} – 2kx – 6 = 0

x = 3 is a root of above equation, then

(3)^{2} – 2k (3) – 6 = 0

⇒ 9 – 6k – 6 = 0

⇒ 3 – 6k = 0

⇒ 3 = 6k

⇒ k = ^{3}/_{6}

⇒ k =½

Q.2. What is the HCF of the smallest prime number and the smallest composite number?

Solution:

Smallest prime number = 2

Smallest composite number = 4

Prime factorisation of 2 is 1 × 2

Prime factorisation of 4 is 1 × 2^{2}

HCF (2, 4) = 2

Q.3.Find the distance of a point P(x, y) from the origin.

Solution:

The given point is P (x, y).

The origin is O (0, 0)

The distance of point P from the origin,

Q.4. In an AP if the common difference (d) = -4 and the seventh term (a^{7}) is 4, then find the first term.

Solution:

Given,

d = -4, a^{7} = 4

a + 6d = 4

⇒ a + 6(-4) = 4

⇒ a – 24 = 4

⇒ a = 4 + 24

⇒ a = 28

Q.5. What is the value of (cos^{2} 67° – sin^{2} 23°) ?

Solution:

We have, cos^{2} 67° – sin^{2} 23°

= cos^{2} 67° – cos^{2} (90° – 23°) [∵ sin (90° – θ) = cos θ]

= cos^{2} 67° – cos^{2} 67°

= 0

Q.6. Given ΔABC ~ ΔPQR, if ^{AB}/_{PQ}=^{1}/_{3} then find ^{arΔABC}/_{arΔPQR}

Solution:

**Section – B**

Q.7.Given that √2 is irrational, prove that (5 + 3√2) is an irrational number.

Solution:

Given, √2 is an irrational number.

Let √2 = m

Suppose, 5 + 3√2 is a rational number.

But ^{a-5b}/_{3b} is rational number, so m is rational number which contradicts the fact that m = √2 is irrational number.

So, our supposition is wrong.

Hence, 5 + 3√2 is also irrational.

Hence Proved.

Q.8.In fig. 1, ABCD is a rectangle. Find the values of x and y.

Solution:

Given, ABCD is a rectangle.

AB = CD

⇒ 30 = x + y

or

⇒ x + y = 30 …(i)

Similarly, AD = BC

⇒ 14 = x – y

or

⇒ x – y = 14 …(ii)

On adding eq. (i) and (ii), we get

2x = 44

⇒ x = 22

Putting the value of x in eq. (i), we get

22 + y = 30

⇒ y = 30 – 22

⇒ y = 8

So, x = 22, y = 8.

Q.9.Find the sum of the first 8 multiples of 3.

Solution:First 8 multiples of 3 are 3, 6, 9,….. up to 8 terms

We can observe that the above series is an AP with

a = 3, d = 6 – 3 = 3, n = 8

Sum of n terms of an A.P is given by,

Q.10.Find the ratio in which P(4, m) divides the line segment joining the points A(2, 3) and B(6, -3). Hence find m.

Solution:Let P divides line segment AB in the ratio k : 1

Coordinates of P

Q.11. Two different dice are tossed together. Find the probability:

(i) of getting a doublet.

(ii) of getting a sum 10, of the numbers on the two dice.

Solution:

Total outcomes on tossing two different dice = 36

(i) A: getting a doublet

A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

Number of favourable outcomes of A = 6

(ii) B: getting a sum 10.

B = {(4, 6), (5, 5), (6, 4)}

Number of favourable outcomes of B = 3

Q.12. An integer is chosen at random between 1 and 100. Find the probability that it is:

(i) divisible by 8.

(ii) not divisible by 8.

Solution: The total number are 2, 3, 4, …….. 99

(i) Let E be the event of getting a number divisible by 8.

E = {8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96} = 12

(ii) Let E’ be the event of getting a number not divisible by 8.

Then, P(E’) = 1 – P(E) = 1 – 0.1224 = 0.8756

**Section – C**

Q.13. Find HCF and LCM of 404 and 96 and verify that HCF × LCM = Product of the two given numbers.

Solution:

Prime factorization of 404 = 2 × 2 × 101

Prime factorization of 96 = 2 × 2 × 2 × 2 × 2 × 3

HCF = 2 × 2 = 4

And LCM = 2 × 2 × 2 × 2 × 2 × 3 × 101 = 9696

HCF = 4, LCM = 9696

Verification:

HCF × LCM = Product of the two given numbers

4 × 9696 = 404 × 96

38784 = 38784

Hence Verified.

Q.14. Find all zeroes of the polynomial (2x^{4} – 9x^{3} + 5x^{2} + 3x – 1) if two of its zeroes are (2 + √3) and (2 – √3).

Solution:

Here, p(x) = 2x^{4} – 9x^{3} + 5x^{2} + 3x – 1

And two of its zeroes are (2 + √3) and (2 – √3).

Quadratic polynomial with zeroes is given by,

{x – (2 + √3)}. {x – (2 – √3)}

⇒ (x – 2 – √3) (x – 2 + √3)

⇒ (x – 2)^{2} – (√3)^{2}

⇒ x^{2} – 4x + 4 – 3

⇒ x^{2} – 4x + 1 = g(x) (say)

Now, g(x) will be a factor of p(x) so g(x) will be divisible by p(x)

For other zeroes,

2x^{2} – x – 1 = 0

2x^{2} – 2x + x – 1 = 0

or

2x (x – 1) + 1 (x – 1) = 0

(x – 1) (2a + 1) = 0

x – 1 = 0 and 2x + 1 = 0

x = 1, x = -½

Zeroes of p(x) are

1, -½, 2 + √3 and 2 – √3.

Q.15. If A(-2, 1) and B(a, 0), C(4, b) and D( 1, 2) are the vertices of a parallelogram ABCD, find the values of a and b. Hence find the lengths of its sides.

OR

Q.15. If A(-5, 7), B(-4, -5), C(-1, -6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.

Solution:

Given ABCD is a parallelogram.

Q.16. A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.

Solution: Let the usual speed of plane be x km/h.

Increased speed = (x + 100) km/h.

Distance to cover = 1500 km.

Time taken by plane with usual speed = ^{1500 }/x hr

Time taken by plane with increased speed = ^{1500 }/_{100 + x}

According to the question,

x^{2} + 100x = 300000

x^{2} + 100x – 300000 = 0

x^{2} + 600x – 500x – 300000 = 0

x(x + 600) – 500(x + 600) = 0

(x + 600) (x – 500) = 0

Either x + 600 = 0 ⇒ x = -600 (Rejected)

or

x – 500 = 0 ⇒ x = 500

Usual speed of plane = 500 km/hr.

Q.17. Prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonal.

OR

Q.17. If the area of two similar triangles is equal, prove that they are congruent.

Solution: Let ABCD be a square with side ‘a’.

Q.18. Prove that the lengths of tangents drawn from an external point of a circle are equal.

Solution:

Given: A circle with centre O on which two tangents PM and PN are drawn from an external point P.

To Prove: PM = PN

Construction: Join OM, ON and OP

Proof: Since tangent and radius are perpendicular at point of contact,

∠OMP = ∠ONP = 90°

In ΔPOM and ΔPON,

OM = ON (Radii)

∠OMP = ∠ONP

PO = OP (Common)

ΔOMP = ΔONP (RHS cong.)

PM = PN (C.P.C.T)

Hence Proved.

Q.19. If 4 tanθ = 3, evaluate [(4sinθ – cosθ + 1)/ (4sinθ + cosθ – 1)]

OR

Q.19. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

Solution:

Given, 4 tan θ = 3

⇒ tan θ = ¾ (=P/B)

OR

Solution: Given, tan 2A = cot (A – 18°)

⇒ cot (90° – 2A) = cot (A – 18°)

[∵ tan θ = cot (90° – θ)]

⇒ 90° – 2A = A – 18°

⇒ 90° + 18° = A + 2A

⇒ 108° = 3A

⇒ A = 36°

Q.20. Find the area of the shaded region in Fig. 2, where arcs are drawn with centres A, B, C and D intersect in pairs at mid-points P, Q, R and S of the sides AB, BC, CD and DA respectively of a square ABCD of side 12 cm. [Use π = 3.14]

Solution:

Given, ABCD is a square of side 12 cm.

P, Q, R and S are the midpoints of sides AB, BC, CD and AD respectively.

Area of shaded region = Area of square – 4 × Area of quadrant

= a^{2} – 4 × ¼πr^{2}

= (12)^{2} – 3.14 × (6)^{2}

= 144 – 3.14 × 36

= 144 – 113.04

= 30.96 cm^{2}

Q.21. A wooden article was made by scooping out a hemisphere form each end of a solid cylinder, as shown in Fig. 3. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm. Find the total surface area of the article.

OR

Q.21. A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap?

Solution:

Given, Radius (r) of cylinder = Radius of hemisphere = 3.5 cm.

Total SA of article = CSA of cylinder + 2 × CSA of hemisphere

Height of cylinder, h = 10 cm

TSA = 2πrh + 2 × 2πr^{2}

= 2πrh + 4πr^{2}

= 2πrh (h + 2r)

= 2 × ²²⁄7× 3.5 (10 + 2 × 3.5)

= 2 × 22 × 0.5 × (10 + 7)

= 2 × 11 × 17

= 374 cm^{2}

OR

Solution: Base diameter of cone = 24 m.

Radius r = 12 m

Height of cone, h = 3.5 m

Volume of rice in conical heap = ⅓πr^{2}h

= ⅓ × ²²⁄7 × 12 × 12 × 3.5 = 528 cm^{3}

Q.22. The table below shows the salaries of 280 persons:

Calculate the median salary of the data.

Solution:

^{N}∕_{2}= ^{280}∕_{2} = 140

The cumulative frequency just greater than 140 is 182.

Median class is 10 -15.

l = 10, h = 5, N = 280, c.f. = 49 and f = 133

**Section – D**

Q.23. A motorboat whose speed is 18 km/hr in still water takes 1 hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

OR

Q.23. A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete the total journey, what is the original average speed?

Solution:

Given, speed of motorboat instil

water = 18 km/hr.

Let speed of stream = x km/hr.

Speed of boat downstream = (18 + x) km/hr.

And speed of boat upstream = (18 – x) km/hr.

Time of the upstream journey = ^{24}∕_{18-x}

Time of the downstream journey = ^{24}∕_{18+x}

According to the question,

⇒ x^{2} + 48x – 324 = 0

⇒ x^{2} + 54x – 6x – 324 = 0

⇒ x(x + 54) – 6(x + 54) = 0

⇒ (x + 54)(x – 6) = 0

Either x + 54 = 0 ⇒ x = -54

Rejected, as speed cannot be negative

Or

x – 6 = 0 ⇒ x = 6

Thus, the speed of the stream is 6 km/hr.

OR

Solution: Let the original average speed of train be x km/hr.

Increased speed of train = (x + 6) km/hr.

Time taken to cover 63 km with average speed = ^{64}∕_{x} hr.

Time taken to cover 72 km with increased speed = ^{72}∕_{x+6} hr.

According to the question,

⇒ 135x + 378 = 3(x^{2} + 6x)

⇒ 135x + 378 = 3x^{2} + 18x

⇒ 3x^{2} + 18x – 135x – 378 = 0

⇒ 3x^{2} – 117x – 378 = 0

⇒ 3(x^{2} – 39x – 126) = 0

⇒ x^{2} – 39x – 126 = 0

⇒ x^{2} – 42x + 3x – 126 – 0

⇒ x(x – 42) + 3(x – 42) = 0

⇒ (x – 42) (x + 3) = 0

Either x – 42 = 0 ⇒ x = 42

or

x + 3 = 0 ⇒ x = -3

Rejected (as speed cannot be negative)

Thus, average speed of train is 42 km/hr.

Q.24. The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7:15. Find the numbers.

Solution: Let the first term of AP be a and d be a common difference.

Let your consecutive term of an AP be a – 3d, a – d, a + d and a + 3d

According to the question,

a – 3d + a – d + a + d + a + 3d = 32

⇒ 4a = 32

⇒ a = 8 …(i)

Also,

(a – 3d) (a + 3d) : (a – d) (a + d) = 7 : 15

For d = 2, four terms of AP are,

a – 3d = 8 – 3 (2) = 2

a – d = 8 – 2 = 6

a + d = 8 + 2 = 10

a + 3d = 8 + 3(2) = 14

For d = -2, four term are

a – 3d = 8 – 3(-2) = 14

a – d = 8 – (-2) = 10

a + d = 8 + (-2) = 6

a + 3d = 8 + 3 (-2) = 2

Thus, the four terms of AP series are 2, 6, 10, 14 or 14, 10, 6, 2.

Q.25. In an equilateral ∆ABC, D is a point on side BC such that BD = ⅓ BC. Prove that 9(AD)^{2} = 7(AB)^{2}.

OR

Q.25. Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

Solution:

Given, ABC is an equilateral triangle and D is a point on BC such that BD = ⅓ BC.

To prove: 9AD^{2} = 7AB^{2}

Construction : Draw AE ⊥ BC

Proof: BD = ⅓ BC …(i) (Given)

AE ⊥ BC

We know that perpendicular from a vertex of equilateral triangle to the base divides base in two equal parts.

BE = EC = ½ BC …(ii)

In ∆AEB,

AD^{2} = AE^{2} + DE^{2} (Pythagoras theorem)

or

AE^{2} = AD^{2} – DE^{2} …(iii)

Similarly, In ∆AEB,

AB^{2} = AE^{2} + BE^{2}

OR

Solution: Given: ∆ABC is a right angle triangle, right-angled at A.

To prove : BC^{2} = AB^{2} + AC^{2}

Construction : Draw AD ⊥ BC.

Proof: In ∆ADB and ∆BAC,

∠B = ∠B (Common)

∠ADB = ∠BAC (Each 90°)

∆ADB ~ ∆BAC (By AA similarity axiom)

^{AB}∕_{BC}= ^{BD}∕_{AB} (CPCT)

AB^{2} = BC × BD

Similarly,

∆ADC ~ ∆CAB

^{AC}∕_{BC}= ^{DC}∕_{AC}

AC^{2} = BC × DC …(ii)

On adding equation (i) and (ii)

AB^{2} + AC^{2} = BC × BD + BC × CD = BC (BD + CD) = BC × BC

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

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

Hence Proved.

Q.26. Draw a triangle ABC with BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are ¾ of the corresponding sides of the ∆ABC.

Solution:

Draw a line segment BC = 6 cm.

Construct ∠XBC = 60°.

With B as centre and radius equal to 5 cm, draw an arc intersecting XB at A.

Join AC. Thus, ∆ABC is obtained.

Draw an acute angle ∠CBY below of B.

Mark 4-equal parts on BY as B_{1}, B_{2}, B_{3} and B_{4}

Join B_{4} to C.

From By draw a line parallel to B_{4}C intersecting BC at C’.

Draw another line parallel to CA from C’, intersecting AB at A’.

∆A’BC’ is required triangle which is similar to ∆ABC such that BC’ = ¾ BC.

Q.27.

Solution:

Q.28. The diameters of the lower and upper ends of a bucket in the form of a frustum of a cone are 10 cm and 30 cm respectively. If its height is 24 cm, find:

(i) The area of the metal sheet used to make the bucket.

(ii) Why we should avoid the bucket made by ordinary plastic? [Use π = 3.14]

Solution: Given, Height of frustum, h = 24 cm.

Diameter of lower end = 10 cm

Radius of lower end, r = 5 cm.

Diameter of upper end = 30 cm

Radius of upper end, R = 15 cm.

(i) Area of metal sheet used to make the bucket = CSA of frustum + Area of base

= πl(R + r) + πr^{2}

= π[26 (15 + 5) + (5)^{2}]

= 3.14 (26 × 20 + 25)

= 3.14 (520 + 25)

= 3.14 × 545

= 1711.3 cm^{2}

(ii) We should avoid the bucket made by ordinary plastic because plastic is harmful to the environment and to protect the environment its use should be avoided.

Q.29. As observed from the top of a 100 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. [Use √3 = 1.732]

Solution: Let AB be the lighthouse and two ships are at C and D.

Distance between two ships = y – x

= 100√3 – 100 [from equation (i) and (ii)]

= 100 (√3 – 1)

= 100(1.732 – 1)

= 100 (0.732)

= 73.2 m

Q.30. The mean of the following distribution is 18. Find the frequency f of the class 19-21.

OR

Q.30. The following distribution gave the daily income of 50 workers of a factory:

Convert the distribution above to a less than type cumulative frequency distribution and draw its give.

Solution:

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