PART A
SECTION – I
1. If A = {x: x is a factor of 24}, then find n(A).
Solution:
A = {1, 2, 3, 4, 6, 8, 12, 24}
n(A) = 8
2. Find the HCF of 24 and 33 by using a division algorithm.
Solution:
24 < 33
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3. Radha says “1, 1, 1,…. are in AP and also in GP”. Do you agree with Radha? Give reason.
Solution:
Given,
1, 1, 1,….
If a, b, c are in AP, then a + c = 2b
Thus, 1 + 1 = 2(1)
2 = 2
Therefore, 1, 1, 1 are in AP.
If a, b, a are in GP, then, ac = b2
(1)(1) = (1)2
1 = 1
Therefore, 1, 1, 1 are in GP.
Hence, we can say that 1,1,1 are in AP and also in GP.
4. If P(x) = x4 + 1, then find P(2) – P(-2).
Solution:
Given,
P(x) = x4 + 1
P(2) = (2)4 + 1 = 16 + 1 = 17
P(-2) = (-2)4 + 1 = 16 + 1 = 17
P(2) – P(-2) = 17 – 17 = 0
5. Find the roots of the quadratic equation x2 + 2x – 3 = 0.
Solution:
Given quadratic equation is: x2 + 2x – 3 = 0
x2 + 3x – x – 3 = 0
x(x + 3) – 1(x + 3) = 0
(x – 1)(x + 3) = 0
x – 1 = 0, x + 3 = 0
x = 1, x = -3
Hence, the roots of the given quadratic equation are 1 and -3.
6. Find the centroid of ΔPQR, whose vertices are P(1, 1), Q(2, 2), R(-3, -3).
Solution:
Let the given vertices of a triangle PQR are:
(x1, y1) = (1, 1)
(x2, y2) = (2, 2)
(x3, y3) = (-3, -3)
Centroid of a triangle = [(x1 + x2 + x3)/3, (y1 + y2 + y3)/3]
= [(1 + 2 – 3)/3, (1 + 2 – 3)/3]
= (0/3, 0/3)
= (0, 0)
7. For what value of ‘t’ the following pair of linear equations has a no solution?
2x – ty = 5 and 3x + 2y = 11
Solution:
Given pair of linear equations are:
2x – ty = 5
3x + 2y = 11
Comparing with the standard form,
a1 = 2, b1 = -t, c1 = -5
a2 = 3, b2 = 2, c2 = -11
Condition for no solution of linear equations is
a1/a2 = b1/b2 ≠ c1/c2
⅔ = -t/2
⇒ t = -4/3
SECTION – II
8. If µ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {2, 3, 5, 8} and B = {0, 3, 5, 7, 10}. Then represent A ⋂ B in the Venn diagram.
Solution:
Given,
µ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {2, 3, 5, 8}
B = {0, 3, 5, 7, 10}
A ⋂ B = {2, 3, 5, 8} ⋂ {0, 3, 5, 7, 10} = {3, 5}
9. Akhila says, “points A(1, 3), B(2, 2), C(5, 1) are collinear”. Do you agree with Akhila? Why?
Solution:
Given points are A(1, 3), B(2, 2) and C(5, 1).
If (x1, y1), (x2, y2) and (x3, y3) are collinear, then area of triangle = 0
i.e. ½ [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0
= ½ [1(2 – 1) + 2(1 – 3) + 5(3 – 2)]
= 1 + 2(-2) + 5(1)
= 1 – 4 + 5
= 2 ≠ 0
Thus, the given points are not collinear.
Hence, we do not agree with Akhila.
10. Write the quadratic equation, whose roots are 2 + √3 and 2 – √3.
Solution:
Let α and β be the roots of the quadratic equation.
Given,
2 + √3 and 2 – √3 are the roots of the quadratic equation.
Sum of the roots = α + β = 2 + √3 + 2 – √3 = 4
Product of the roots = αβ = (2 + √3)(2 – √3)
= (2)2 – (√3)2
= 4 – 3
= 1
Hence, the quadratic equation is x2 – (α + β)x + αβ = 0
x2 – 4x + 1 = 0
11. Divide x3 – 4x2 + 5x – 2 by x – 2.
Solution:
Quotient = q(x) = x2 – 2x + 1
12. Write the formula of the nth term of GP and explain the terms in it.
Solution:
The formula for nth term of GP is
an = arn – 1
Here,
an = nth term of the sequence
a = first term
r = common ratio
13. Solve the pair of linear equations 2x + 3y = 8 and x + 2y = 5 by elimination method.
Solution:
Given,
2x + 3y = 8….(i)
x + 2y = 5….(ii)
(ii) × 2 – (i),
2x + 4y – (2x + 3y) = 10 – 8
y = 2
Substituting y = 2 in (ii),
x + 2(2) = 5
x = 5 – 4
x = 1
SECTION – III
14. (a) Draw the graph of the polynomial p(x) = x2 – 7x + 12, then find its zeroes from the graph.
Solution:
Given polynomial is p(x) = x2 – 7x + 12
| x | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| y | 12 | 6 | 2 | 0 | 0 | 2 | 6 | 12 |
The graph of the given polynomial cut the x-axis at (3, 0) and (4, 0)
Hence, x = 3 and x = 4 are the zeroes of the given polynomial.
OR
(b) Solve the equations graphically 3x + 4y = 10 and 4x – 3y = 5.
Solution:
Given,
3x + 4y = 10
4x – 3y = 5
Consider the first equation:
3x + 4y = 10
4y = -3x + 10
y = (-¾)x + (10/4)
y = (-¾)x + (5/2)
| x | -2 | 0 | 2 | 6 |
| y | 4 | 5/2 | 1 | -2 |
Now, consider the another equation,
4x – 3y = 5
3y = 4x – 5
y = (4/3)x – (5/3)
| x | -1 | 2 | 5 | 8 |
| y | -3 | 1 | 5 | 9 |
Graph:
The lines representing the given pair of equations intersect each other at (2, 1).
Hence, the solution of a given pair of linear equations is x = 2 and y = 1.
15. (a) Find the ratio in which X-axis divides the line segment joining the points (2, -3) and (5, 6). Then find the intersection point on X-axis.
Solution:
Let P(x, 0) divide the line segment joining the points A(2, -3) and B(5, 6) in the ratio m : n.
Here,
(x1, y1) = (2, -3)
(x2, y2) = (5, 6)
Using the section formula,
P(x, 0) = [(mx2 + nx1)/ (m + n), (my2 + ny1)/ (m + n)]
(x, 0) = [ (5m + 2n)/ (m + n), (6m – 3n)/ (m + n)]
⇒ (6m – 3n)/ (m + n) = 0
⇒ 6m = 3n
⇒ m/n = 3/6
⇒ m/n = 1/2
Therefore, the required ratio is 1 : 2.
Now,
x = (5m + 2n)/ (m + n)
x = [5 + 2(2)]/ (1 + 2)
x = 9/3
x = 3
Hence, the required point on the x-axis is (3, 0).
OR
(b) Find the sum of all two digit odd multiples of 3.
Solution:
Two digit odd multiples of 3 are 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99.
This is an AP with a = 15
Common difference = d = 6
n = 15
Sum of first n terms
Sn = n/2[2a + (n – 1)d]
S15 = (15/2) [2(15) + (15 – 1)6]
= (15/2) [30 + 14(6)]
= (15/2) [30 + 84]
= (15/2) × 114
= 855
Hence, the required sum is 855.
16. (a) If A = {x : 2x + 1, x ∈ N, x ≤ 5}, B = {x : x is a composite number, x ≤ 12}, then show that (A ⋃ B) – (A ⋂ B) = (A – B) ⋃ (B – A).
Solution:
A = {x : 2x + 1, x ∈ lN, x ≤ 5}
= {3, 5, 7, 9, 11}
B = {x : x is a composite number, x ≤ 12}
= {4, 6, 8, 9, 10, 12}
A ⋃ B = {3, 5, 7, 9, 11} ⋃ {4, 6, 8, 9, 10, 12}
= {3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
A ⋂ B = {3, 5, 7, 9, 11} ⋂ {4, 6, 8, 9, 10, 12}
= {9}
(A ⋃ B) – (A ⋂ B) = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12} – {9}
= {3, 4, 5, 6, 7, 8, 10, 11, 12}
A – B = {3, 5, 7, 9, 11} – {4, 6, 8, 9, 10, 12} = {3, 5, 7, 11}
B – A = {4, 6, 8, 9, 10, 12} – {3, 5, 7, 9, 11} = {4, 6, 8, 10, 12}
(A – B) ⋃ (B – A) = {3, 5, 7, 11} ⋃ {4, 6, 8, 10, 12} = {3, 4, 5, 6, 7, 8, 10, 11, 12}
Therefore, (A ⋃ B) – (A ⋂ B) = (A – B) ⋃ (B – A)
OR
(b) Prove that √2 + √7 is an irrational number.
Solution:
Let √2 + √7 be a rational number.
√2 + √7 = a, where a is an integer.
Squaring on both sides,
(√2 + √7)2 = a2
(√2)2 + (√7)2 + 2(√2)(√7) = a2
2 + 7 + 2√14 = a2
9 + 2√14 = a2
2√14 = a2 – 9
√14 = (a2 – 9)/2
(a2 – 9)/2 is a rational number since a is an integer.
Therefore, √14 is also an integer.
We know that integers are not rational numbers.
Thus, our assumption that √2 + √7 is a rational number is wrong.
Hence, √2 + √7 is an irrational number.
17. (a) Sum of the areas of two squares is 850 m2. If the difference of their perimeters is 40 m. Find the sides of the two squares.
Solution:
Let x and y be the sides of two squares.
According to the given,
Sum of the areas of two squares = 850 m2
x2 + y2 = 850….(i)
Difference of their perimeters = 40 m
4x – 4y = 40
4(x – y) = 40
x – y = 10
x = y + 10….(ii)
Substituting (ii) in (i),
(y + 10)2 + y2 = 850
y2 + 100 + 20y + y2 – 850 = 0
2y2 + 20y – 750 = 0
2(y2 + 10y – 375) = 0
y2 + 10y – 375 = 0
y2 + 25y – 15y – 375 = 0
y(y + 25) – 15(y + 25) = 0
(y – 15)(y + 25) = 0
y – 15 = 0, y + 25 = 0
y = 15, y = -25
Measure cannot be negative.
Therefore, y = 15
Substitute y = 15 in (ii),
x = 15 + 10 = 25
Hence, the sides of two squares are 25 cm and 15 cm.
OR
(b) Sum of the present ages of two friends are 23 years, five years ago the product of their ages was 42. Find their ages 5 years hence.
Solution:
Let x and (23 – x) be the present ages (in years) of two friends.
According to the given
(x – 5)(23 – 5 – x) = 42
(x – 5)(18 – x) = 42
18x – x2 – 90 + 5x = 42
⇒ x2 – 5x – 18x + 90 + 42 = 0
⇒ x2 – 23x + 132 = 0
⇒ x2 – 11x – 12x + 132 = 0
⇒ x(x – 11) – 12(x – 11) = 0
⇒ (x – 11)(x – 12) = 0
⇒ x = 12, x = 11
If x = 12, then 23 – x = 23 – 12 = 11
If x = 11, then 23 – x = 23 – 11 = 12
Therefore, the present ages of two friends are 11 and 12 years.
Hence, their ages after 5 years will be 16 and 17 years.
PART A
SECTION – I
1. Evaluate cosec 39° sec 51° – tan 51° cot 39°.
Solution:
cosec 39° sec 51° – tan 51° cot 39°
= cosec (90° – 51°) sec 51° – tan 51° cot (90° – 51°)
= sec 51° sec 51° – tan 51° tan 51°
= sec251° – tan251°
= 1
2. Write the similarity criterion by which the given pair of triangles are similar.
Solution:
From the given,
OA/OB = 3/6 = ½
OC/OB = 2.5/5 = ½
∠AOC = ∠BOD (vertically opposite angles)
By SAS similarity criterion,
ΔAOC ~ ΔBOD
3. From English alphabet, if a letter is chosen at random, then find the probability that the letter is a consonant.
Solution:
Total number of outcomes = 26
i.e. English alphabet = 26
Number of consonants = 21
P(choosing a consonant) = 21/26
4. In a right triangle ABC, right angled at C in which AB = 13 cm, BC = 5 cm, determine the value of cos2B + sin2A.
Solution:
Given,
In a right triangle ABC, right angled at C in which AB = 13 cm, BC = 5 cm.
AC = 12 cm
cos B = 5/13
sin A = 5/13
cos2B + sin2A = (5/13)2 + (5/13)2
= (25 + 25)/ 1369
= 50/169
5. A point P is 25 cm from the center O of the circle. The length of the tangent drawn from P to the circle is 24 cm. Find the radius of the circle.
Solution:
Given,
We know that the radius is perpendicular to the tangent through the point of contact.
In right triangle OAP,
OP2 = AP2 + OA2
OA2 = OP2 – AP2
= (25)2 – (24)2
= 625 – 576
= 49
OA = 7 cm
Hence, the radius of the circle is 7 cm.
6. Find the median of the first seven composite numbers.
Solution:
The first seven composite numbers are 4, 6, 8, 9, 10, 12, 14.
These numbers are in ascending order.
n = 7
Median = (7 + 1)/2 th term
= 4th term
= 9
Hence, the median of the first seven composite numbers is 9.
7. In a hemispherical bowl of 2.1 cm radius ice-cream is there. Find the volume of the bowl.
Solution:
Given,
Radius of hemispherical bowl = r = 2.1 cm
Volume = (⅔)πr3
= (⅔) × (22/7) × 2.1 × 2.1 × 2.1
= 19.404 cm3
Therefore, the volume of the bowl is 19.404 cm3.
SECTION – II
8. Write the mode formula for grouped data and explain the terms in it.
Solution:
Mode for grouped data = l + [(f1 – f0)/ (2f1 – f0 – f2)] × h
Here,
l = Lower limit of the modal class
f1 = Frequency of the modal class
f0 = Frequency of the class preceding the modal class
f2 = Frequency of the class succeeding the modal class
h = class size (or class height)
9. In the given figure, TA and TB are tangents to the circle with centre O. If ∠ATB = 80°, then find the measure of ∠ABT.
Solution:
Given,
TA and TB are tangents to the circle.
∠ATB = 80°
In triangle TAB,
∠ATB + ∠TBA + ∠BAT = 180°
80° + ∠ABT + ∠ABT = 180° (TA and TB are tangents from an external point T)
2∠ATB = 180° – 80°
∠ATB = 100°/2
∠ATB = 50°
10. A bag contains balls which are numbered from 1 to 50. A ball is drawn at random from the bag, the probability that it bears a two digit number multiple of 7.
Solution:
Total number of outcomes = 50
i.e. numbers from 1 to 50
Two digit multiples of 7 are 14, 21, 28, 35, 42, 49.
Number of favourable outcomes = 6
Hence, the required probability = 6/50 = 3/25
11. From the top of the building the angle of elevation of the top of the cell tower is 60° and the angle of depression to its foot is 45°, if the distance of the building from the tower is 30 meters, draw the suitable diagram to the given data.
Solution:
Let AB be the building and CD be the cell tower.
BD is the distance of the building from the tower.
12. Find the value of (tan260° + cot230°)/ (sin230° + cos260°).
Solution:
(tan260° + cot230°)/ (sin230° + cos260°)
= [tan2(90° – 30°) + cot230°] / [sin230° + cos2(90° – 30°)]
= (cot230° + cot230°)/ (sin230° + cos230°)
= 2cot230°/1
= 2 (√3)2
= 2(3)
= 6
13. A right circular cylinder has a radius 3.5 cm and height 14 cm. Find the curved surface area.
Solution:
Given,
Radius of cylinder = r = 3.5 cm
Height of cylinder = h = 14 cm
Curved surface area = 2πrh
= 2 × (22/7) × 3.5 × 14
= 88 × 3.5
= 308 cm2
SECTION – III
14. Construct a triangle PQR, in which PQ = 4 cm, QR = 6 cm and ∠PQR = 70°. Construct a triangle such that each side of the new triangle is 3/4 of the triangle PQR.
Solution:
Therefore, ΔP’QR’ is the required triangle similar to the ΔPQR.
OR
Draw less than Ogive for the following frequency distribution. Find the median from the obtained curve.
| IQ | 60 – 70 | 70 – 80 | 80 – 90 | 90 – 100 | 100 – 110 | 110 – 120 | 120 – 130 |
| No.of students | 2 | 5 | 12 | 31 | 39 | 10 | 4 |
Solution:
| Class | Cumulative Frequency |
| Less than 60 | 2 |
| Less than 70 | 2 + 5 = 7 |
| Less than 80 | 7 + 12 = 19 |
| Less than 90 | 19 + 31 = 50 |
| Less than 100 | 50 + 39 = 89 |
| Less than 110 | 89 + 10 = 99 |
| Less than 120 | 99 + 4 = 103 |
N = 103
n/2 = 103/2 = 51.5
Median = 92
15. Show that [cos θ/ (1 – sin θ)] + [(1 – sin θ)/ cos θ] = 2 sec θ
Solution:
LHS = [cos θ/ (1 – sin θ)] + [(1 – sin θ)/ cos θ]
= [cos2θ + (1 – sin θ)2]/ [cos θ (1 – sin θ)]
= [cos2θ + 1 + sin2θ – 2 sin θ]/ [cos θ (1 – sin θ)]
= (2 – 2 sin θ)/ [cos θ(1 – sin θ)]
= [2(1 – sin θ)]/ [cos θ(1 – sin θ)]
= 2/cos θ
= 2 sec θ
= RHS
Therefore, [cos θ/ (1 – sin θ)] + [(1 – sin θ)/ cos θ] = 2 sec θ
OR
In a right angle triangle, the hypotenuse is 10 cm more than the shortest side. If the third side is 6 cm less than the hypotenuse, find the sides of the right angle triangle.
Solution:
Let x cm be the shortest side of a right triangle.
Hypotenuse = (x + 10) cm
Length of the third side = (x + 10) – 6 = (x + 4) cm
By Pythagoras theorem,
(x + 10)2 = x2 + (x + 4)2
x2 + 20x + 100 = x2 + x2 + 8x + 16
⇒ x2 + 8x + 16 – 20x – 100 = 0
⇒ x2 – 12x – 84 = 0
⇒ x = [-(-12) ± √{(-12)2 – 4(1)(-84)}]/ 2(1)
= [12 ± √(144 + 336)]/ 2
x = 16.9 cm
Therefore, shortest side = 16.95 cm
Hypotenuse = 26.95 cm
Third side = 26.95 cm
16. Find the mean age of 100 residents of a colony from the following data.
| Age (in years) | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 |
| No. of Persons | 10 | 15 | 25 | 25 | 10 | 10 | 5 |
Solution:
| Age (in years) | Mid-point (x) | No. of persons (f) | fx |
| 0 – 10 | 5 | 10 | 50 |
| 10 – 20 | 15 | 15 | 225 |
| 20 – 30 | 25 | 25 | 625 |
| 30 – 40 | 35 | 25 | 875 |
| 40 – 50 | 45 | 10 | 450 |
| 50 – 60 | 55 | 10 | 550 |
| 60 – 70 | 65 | 5 | 325 |
| ∑f = 100 | ∑fx = 3100 |
Mean = ∑fx/∑f = 3100/100 = 31
Therefore, the mean age is 31 years.
OR
A toy is made with seven equal cubes of sides √7 cm. Six cubes are joined to six faces of a seventh cube. Find the total surface area of the toy.
Solution:
Given,
Side of cube = √7 cm
Surface area of one face of a cube = (√7)2 = 7 cm2
Six cubes are joined to six faces of a seventh cube.
Thus, only five sides of a cube are exposed and only a surface area of six cubes is effective.
Surface area of five faces of a cube = 5 × 7 = 35 cm2
Total surface area of the toy = 6 × Area of five faces of a cube
= 6 × 35
= 210 cm2
17. If two dice are thrown at the same time, find the probability of getting the sum of the dots on top is prime.
Solution:
Total number of outcomes = 62 = 36
Favourable outcomes of getting the sum of the dots on the top of dice is prime = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}
= 15
Hence, the required probability = 15/36
OR
The angle of elevation of the top of the hill from the foot of a tower is 60° and the angle of elevation of the top of the tower from the foot of the hill is 30°. If the tower is 50 m high. Find the height of the hill.
Solution:
Let AB be the hill and CD be the tower.
CD = 50 m
In right triangle ABC,
cot 60° = x/h
1/√3 = x/h
x = h/√3 ….(i)
In right triangle DBC,
cot 30° = x/50
√3 = x/50
x = 50√3….(ii)
From (i) and (ii),
h/√3 = 50√3
h = 50 × 3
h = 150 m
Hence, the height of the hill is 150 m.