If the sum of the roots of the equation 3x2 – (3k – 2) x – (k – 6) = 0 is equal to the product of its roots then k = ?
A. 1
B. – 1
C. 0
D. 2
Let the roots of the given quadratic equation 3x2 – (3k – 2)x – (k – 6)=0 be α and β.
Now,
sum of roots = α + β = (3k – 2)/3 and,
product of roots = αβ = – (k – 6)/3
[∵ If α and β are the roots of quadratic equation ax2 + bx + c=0 then α + β = – b/a and αβ = c/a]
According to question –
sum of roots = product of roots
∴ α + β = αβ
⇒ (3k – 2)/3 = – (k – 6)/3
⇒ 3k – 2 = – k + 6
⇒ 4k = 8
∴ k = 2
Hence, The value of k is 2.
The number of all 2-digit numbers divisible by 6 is
A. 12
B. 15
C. 16
D. 18
All 2-digit numbers divisible by 6 are as follows: –
6, 12, ………., 96
The above series of numbers forms an arithmetic progression with
first term(a) = 6 and,
common difference(d) = (n + 1)th term – nth term = 12 – 6 = 6
last term or nth term(an) = 96
Let the number of terms in above series be n.
∵ an = a + (n – 1) × d
⇒ 96 = 6 + (n – 1) × 6
⇒ 90 = 6n – 6
⇒ 6n = 96
∴ n = 16
Thus, total no. of all 2-digit numbers divisible by 6.
A fair die is thrown once. The Probability of getting a composite number is
A. 1/3
B. 1/6
C. 2/3
D. 0
Let P be the event of getting a composite number while throwing a dice.
Total no. of outcomes when n number of die are thrown = 6n
∴ no. of total outcomes = n(S) = 6
Sample Space = {1, 2, 3, 4, 5, 6}
favourable elementary events = getting a composite number
= {4, 6}
∴ no. of favourable elementary events = n(P) = 2
Thus, the probability of getting a composite number = n(P)/n(S)
= 2/6
= 1/3
Which of the following statements is true?
A. The tangents drawn at the end points of a chord of a circle are parallel.
B. From a point P in the exterior of a circle, only two secants can be drown through P to the circle.
C. From a point P in the plane of a circle, two tangents can be drawn to the circle.
D. The perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.
A Tangent is a line that intersects a circle at exactly one point.
The tangents drawn at the end points of a chord of a circle can be parallel only if that chord is the diameter of the circle. This will be clear from the fig.1 and fig.2 shown below.
Thus, statement (a) is incorrect.
A secant is a segment that intersects a circle twice.
From a point P in the exterior of a circle, infinite no. of secants can be drown through P to the circle. This can be shown in the fig.3 drawn below.
Thus, statement (b) is incorrect.
A Tangent is a line that intersects a circle at exactly one point.
From a point P in the plane of a circle, two tangents can be drawn to the circle only if point P is exterior to the circle. This can be shown in the fig.4 drawn below.
Thus, statement (c) is incorrect.
In the above fig.5, we take a point Q on the tangent XY to the circle with centre O. Obviously, this point Q should lie outside to the circle otherwise XY will become secant. And, P is the point of contact. Clearly,
OQ > OP
Also, this is also true for all the points lying on the tangent XY except point P. And,
we know that perpendicular distance is always the shortest distance.
OP is shortest of all the distances b/w points O and any other points on XY i.e.
OP ⊥ XY
Hence, The perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.
Thus, statement (d) is correct.
In the given figure, PA and PB are tangents to a circle such that PA = 8 cm and ∠ APB = 60˚. The length of the chord AB is
A. 8 cm
B. 10 cm
C. 12 cm
D. 6 cm
In Δ PAB,
∠ APB = 60° and PA = 8 cm [given]
∴ PB = PA = 8 cm
[∵ tangents drawn from an exterior point to the circle are equal in length]
⇒ ∠ PAB = ∠ PBA = θ [LET]
Now, In Δ PAB
∠ APB + ∠ PAB + ∠ PBA = 180° [∵ Sum of all the angles of a Δ is 180°]
⇒ 60° + θ + θ = 180°
⇒ 60° + 2θ = 180°
⇒ 2θ = 120°
∴ θ = 60°
Thus, Δ PAB is an equilateral Triangle.
∴ Length of chord AB = 8 cm.
The angle of depression of an object from a 60-m-high tower is 30˚. The Distance of the object from the tower is
A. 20√ 3 m
B. 60√ 3 m
C. 40√ 3 m
D. 120 m
Let the Distance of the object from the tower be x meters.
∴ BC = x m
Given –
height of tower = AB = 60 m
Angle of depression = ∠ DAC = 30°
∴ ∠ BCA = ∠ DAC = 30°
[∵ When two‖lines are intersected by a third line then theAlternate interior angles will be equal.]
Now, In Δ ABC
tan 30° = AB/BC = 60/x [∵ tan θ = perpendicular/base]
⇒ 1/√3 = 60/x
∴ x = 60√3 meters
In what ratio does the point P (2, – 5) divide the line segment joining A (– 3,5) and B(4, – 9)?
A. 3:2
B. 2:1
C. 5:2
D. 5:3
Let the point P (2, – 5) divide the line segment joining A (– 3,5) and B(4, – 9) in the ratio m:n.
Let (x,y) ≡ (2, – 5)
(x1,y1) ≡ (– 3,5)
and (x2,y2) ≡ (4, – 9)
Using Section Formula,
⇒ 2 × (m + n) = 4m – 3n
⇒ 2m + 2n = 4m – 3n
⇒ 5n = 2m
∴ m:n = 5:2
Since the ratio is positive, Point P divides the line segment AB internally in the ratio 5:2.
Three solid spheres of radii 6 cm, 8 cm and 10 cm are melted to form a sphere. The radius of the sphere so formed is
A. 24cm
B. 16 cm
C. 14 cm
D. 12 cm
Let the radius of the sphere so formed be r cm.
Given –
Radius of 1st sphere(r1) = 6 cm
Radius of 2nd sphere(r2) = 8 cm
Radius of 3rd sphere(r3) = 10 cm
After Melting all these spheres, the volume will remain unchaged.
∴ Vol. of 1st sphere + Vol. of 2nd sphere + Vol. of 3rd sphere
= Vol. of new sphere so formed
⇒ (4/3)π(r1)3 + (4/3)π(r2)3 + (4/3)π(r3)3 = (4/3)π(r)3
Taking out (4/3)π from both sides, we get –
⇒ (r1)3 + (r2)3 + (r3)3 = (r)3
⇒ (6)3 + (8)3 + (10)3 = (r)3
⇒ 216 + 512 + 1000 = (r)3
⇒ (r)3 = 1728
∴ r = 12 cm
Thus, the radius of new sphere is 12 cm.
Find the value of p for which the quadratic equation
x2 – 2px + 1 = 0 has no real roots.
The given quadratic equation is x2 – 2px + 1 = 0.
And, Discriminant D of the quadratic equation ax2 + bx + c = 0 is given by –
D = b2 – 4ac
Comparing the equation ax2 + bx + c = 0 with given quadratic equation is x2 – 2px + 1 = 0, we get –
a = 1, b = – 2p and, c = 1
∴ D = (– 2p)2 – 4(1)(1) = 4p2 – 4 = 4(p2 – 1)
For no real roots,
D < 0
⇒ 4(p2 – 1) < 0
⇒ p2 – 1 < 0
⇒ (p + 1)(p – 1) < 0
∴ p ∈ (– 1,1)
Thus, p can take any values between – 1 and 1 for no real roots of given quadratic equation.
Find the 10th term form the end of the AP 4, 9, 14, .. , 254.
The above series of numbers forms an arithmetic progression with
first term(a) = 4 and,
common difference(d) = (n + 1)th term – nth term = 9 – 4 = 5
last term or nth term(an) = 254
Let the total no. of terms in above A.P be n.
∴ an = a + (n – 1) × d
⇒ 254 = 4 + (n – 1) × 5
⇒ 250 = 5n – 5
⇒ 5n = 255
∴ n = 51
∴ 10th term from the end of AP = 51 – 10 + 1 = 42th term from the beginning
∴ 42th term = a42 = a + (42 – 1)d
= 4 + 41 × 5
= 209
Hence, 10th term from the end of AP is 209.
Which term of the AP 24, 21 ,18, 15, … is the first negative term?
Let the nth term of the AP be the first negative term.
In the given AP –
first term(a) = 24 and,
common difference(d) = (n + 1)th term – nth term = 21 – 24 = – 3
According to question –
∴ an < 0
⇒ a + (n – 1) × d < 0
⇒ 24 + (n – 1) × (– 3) < 0
⇒ – 3n + 27 < 0
⇒ 3n > 27
∴ n > 9
Thus, the first negative term of given AP is 10th term.
A circle is touching the side BC of a Δ ABC at P and is touching AB and AC when produced at Q and R respectively.
Prove that AQ = 1/2 (perimeter of Δ ABC).
In the given figure,
AQ and AR are two tangents drawn from an exterior point A at contact points Q and R on the circle.
∴ AQ = AR
⇒ AQ = AC + CR…..(1)
Similarly,
BQ and BP are two tangents drawn from an exterior point B at contact points Q and P on the circle.
∴ BQ = BP…..(2)
And,
CR and CP are two tangents drawn from an exterior point C at contact points R and P on the circle.
∴ CR = CP…..(3)
Now, Equation (1) can be written as –
AQ = (AC + CR + AC + CR)/2
⇒ AQ = (AC + CP + AC + CR)/2 [using(3)]
⇒ AQ = (AC + CP + AR)/2
⇒ AQ = (AC + CP + AQ)/2
⇒ AQ = (AC + CP + AB + BQ)/2
⇒ AQ = (AC + CP + AB + BP)/2 [using(2)]
⇒ AQ = (AB + BC + AC)/2 [∵BP + CP=BC]
Thus, AQ = (1/2) × perimeter of Δ ABC
Two vertical of a ΔABC are given by A(6, 4) and B (– 2, 2) and its centroid is G(3, 4). Find the coordinates of its third vertex C.
Let the third vertex C ≡ (x3,y3)
In a Δ ABC,
Vertex A ≡ (x1,y1) ≡ (6,4)
Vertex B ≡ (x2,y2) ≡ (– 2,2)
Centroid(G) ≡ (x,y) ≡ (3,4)
Centroid of a Δ ABC is given by –
x = (x1 + x2 + x3)/3
⇒ 3 = (6 – 2 + x3)/3
⇒ 9 = 4 + x3
∴ x3 = 5
And,
y = (y1 + y2 + y3)/3
⇒ 4 = (4 + 2 + y3)/3
⇒ 12 = 6 + y3
∴ y3 = 6
Thus, the coordinates of third vertex C is (5,6).
A box contain 150 orange is taken out from the box at random and the probability of its being rotten is 0.06 then find the number of good orange in the box.
Total no. of Oranges = 150
Probability of rotten oranges =0.06
∴ Probability of good oranges = 1 – 0.06 = 0.94
⇒ (no. of good oranges)/(no. of total oranges) = 0.94
⇒ no. of good oranges = 0.94 × 150 = 141
Thus, the number of good orange in the box = 141
A toy is in the form of a cone mounted on a hemisphere of common base radius 7 cm. The total height of the toy is 31 cm. Find the total surface area of the toy.
fig.10
Given –
Total Height of cone = 31 cm
Radius of hemisphere(r) = Base Radius of Cone
= Height of hemisphere
= 7 cm
∴ Height of cone(h) = 31 – 7 = 24 cm
Slant Height of Cone(l) = √(h2 + r2) = √(242 + 72) = 25 cm
Now,
Total Surface Area of the Toy
= Curved Surface Area of Cone + Curved Surface Area of Hemisphere
= πrl + 2πr2
= π(rl + 2r2)
= π(7 × 25 + 2 × (7)2)
= π(175 + 98)
= π(273)
= 3.14 × 273
= 857.22 cm2
Solve: a2b2x2 – (4b4 – 3a4) x – 12a2b2 = 0.
The given quadratic equation is –
a2b2x2 –(4b4 – 3a4) x – 12a2b2 = 0
Discriminant D of the quadratic equation ax2 + bx + c = 0 is given by –
D = b2 – 4ac
Comparing the equation ax2 + bx + c = 0 with given quadratic equation is a2b2x2 –(4b4 –3a4) x – 12a2b2 = 0,we get –
a = a2b2 , b = –(4b4 – 3a4) and, c = – 12a2b2
∴ The roots of the given quadratic equation is given by –
Thus, the roots of the given quadratic equation are (4b2/a2) and (– 3a2/b2).
If the 8th term of an AP is 31 and its 15th term is 16 more than the 11th term, find the AP.
Let the first term and common difference of given AP be a and d respectively.
According to question –
8th term of AP = a8 = 31 [Given]
⇒ a + (8 – 1)d = 31
⇒ a + 7d = 31…..(1)
15th term of AP = a15 = 16 + a11
⇒ a + (15 – 1)d = 16 + a + (11 – 1)d
⇒ 14d = 16 + 10d
⇒ 4d = 16
∴ d = 4
Substituting the value of d in equation(1), we get –
a = 31 – 7 × 4 = 31 – 28 = 3
Thus, the required AP is 3,7,11,15,…….
Find the sum of all two-digit odd positive numbers.
All the two-digit odd positive numbers are –
11,13,15,17,……….,99
The above series of numbers forms an arithmetic progression with
first term(a) = 11 and,
common difference(d) = (n + 1)th term – nth term = 13 – 11 = 2
last term or nth term(an) = 99
Let the total no. of terms in above A.P be n.
∴ an = a + (n – 1) × d
⇒ 99 = 11 + (n – 1) × 2
⇒ 88 = 2n – 2
⇒ 2n = 90
∴ n = 45
Sum of all the 45 terms of the AP is given by –
S45 =(45/2)(11 + 99)
[∵Sn = (n/2)(a + l) =(n/2)[(2a + (n – 1)d]
=(45/2) × 110
=45 × 55
=2475
Thus, the sum of all two-digit odd positive numbers = 2475.
In the adjoining figure, PA and PB are tangents drawn from an external point P to a circle with centre O. Prove that
∠ APB = 2 ∠ OAB.
fig.11
Let ∠ APB = θ
In Δ APB,
PA = PB
[∵ Tangents drawn from an exterior point to the circle are equal in length]
⇒ Δ APB is an isoceles triangle.
∴ ∠ PAB = ∠ PBA = α [LET]
Now,
∠ APB + ∠ PAB + ∠ PBA = 180° [∵ sum of all the angles of Δ=180°]
⇒ θ + α + α = 180°
⇒ 2α = 180° – θ
∴ α = ∠ PAB = 90° – (θ/2)
Also, OA⊥AP
[∵ radius of a circle is always⊥to the tangent at the point of contact.]
∴ ∠ PAB + ∠ OAB = 90°
⇒ 90° – (θ/2) + ∠ OAB = 90°
⇒ ∠ OAB = (θ/2) = (1/2)∠ APB
∴ ∠ APB = 2 ∠ OAB
Hence, Proved.
In the adjoining figure, quadrilateral ABCD is circumscribed. If the radius of the in circle with centre O is 10 cm and AD ⊥ DC, find the value of x.
In the given figure,
DS and DR are the two tangents drawn from an external point D at the point of contacts S and R respectively. And,
OS ⊥ DS and OR ⊥ DR
[∵ radius of a circle is always⊥to the tangent at the point of contact.]
⇒ OSDR is a square [∵ AD ⊥ DC (Given)]
∴ DR = 10 cm
Similarly,
BP and BQ are the two tangents drawn from an external point B at the point of contacts A and Q respectively.
∴ BP = BQ = 27 cm
[∵ Tangents drawn from an exterior point to the circle are equal in length]
⇒ QC = BC – BQ = 38 – 27 = 11 cm
Also, CR and CQ are the two tangents drawn from an external point C at the point of contacts R and Q respectively.
∴ CR = CQ = 11 cm
[∵ Tangents drawn from an exterior point to the circle are equal in length]
∴ DC = x = DR + CR = 10 + 11 = 21 cm.
Thus, the value of x is 21 cm.
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct a pair of tangents to the circle. Measure the length of each of the tangent segments.
Steps of Construction:
1. Draw a circle with centre O with radius OL and a point P outside it. Join PO and bisect it. Let M be the midpoint of PO.
fig.13
2. Taking M as centre and MO as radius, we will draw a circle.
Let it intersect the given circle at the points Q and R.
fig.14
3. Join PQ and PR.
Then PQ and PR are the required two tangents.
fig.15
4. Join OQ. Then ∠ PQO is an angle in the semicircle and,
∴ ∠ PQO = 90°
fig.16
Since, OQ is a radius of the given circle, PQ has to be a tangent to the circle.
Similarly,
PR is also a tangent to the circle.
After measuring the lenghts of tangents using scale, we find that both the tangents are equal in length which concludes that all the measurements and steps done correctly.
Length of Each Tangent = 8 cm
The three vertices of a parallelogram ABCD, taken in order are A (1, – 2), B (3, 6) and C (5, 10). Find the coordinates of the fourth vertex D.
Let the coordinates of the fourth vertex D be (x4,y4).
We know that –
Diagonals of a∥gm bisect each other.
∴ Mid – point of diagonal AC ≡ Mid – point of diagonal BD
⇒ 6 = 3 + x4 and 8 = 6 + y4
⇒ x4 = 3 and y4 = 2
Thus, the coordinates of the fourth vertex D is (3,2).
Find the third vertex of a Δ ABC if two of its vertices are
B (– 3, 1) and C (0, – 2) and its centroid is at the origin.
Let the third vertex A ≡ (x1,y1)
In a Δ ABC,
Vertex B ≡ (x2,y2) ≡ (– 3,1)
Vertex C ≡ (x3,y3) ≡ (0, – 2)
Centroid(G) ≡ (x,y) ≡ (0,0)
Centroid of a Δ ABC is given by –
x = (x1 + x2 + x3)/3
⇒ 0 = (x1 – 3 + 0)/3
⇒ 0 = x1 – 3
∴ x1 = 3
And,
y = (y1 + y2 + y3)/3
⇒ 0 = (y1 + 1 – 2)/3
⇒ 0 = y1 – 1
∴ y1 = 1
Thus, the coordinates of third vertex A is (3,1).
Cards marked with all 2-digit numbers are placed in a box and are mixed thoroughly. One card is drawn at random. Find the probability that the number on the card is
(a) divisible by 10
(b) a perfect square number
(c) a prime number less than 25
Sample Space = Cards marked with 2-digit numbers
= {10,11,12,……..,99)
No. of Sample Space = n(S) = 90
(a) Let P be the event of getting a card marked with 2-digit numbers which is divisible by 10.
∴ favourable elementary events = {10,20,30,…………,90}
no. of favourable elementary events = n(P) = 9
Thus, Probability of getting a card marked with number divisible by 10 = n(P)/n(S) = 9/90 = 1/10
(b) Let P be the event of getting a card marked with 2-digit square numbers.
∴ favourable elementary events = {16,25,36,…….,81}
no. of favourable elementary events = n(P) = 6
Thus, Probability of getting a card marked with number divisible by 10 = n(P)/n(S) = 6/90 = 1/15
(c) Let P be the event of getting a card marked with 2-digit prime numbers less than 25.
∴ favourable elementary events = {11,13,17,19,23}
no. of favourable elementary events = n(P) = 5
Thus, Probability of getting a card marked with number divisible by 10 = n(P)/n(S) = 5/90 = 1/18
A road which is 7 m wide surrounds a circular park whose circumference is 352 m. Find the area of the road. [Take π = 22/7.]
fig.17
Let the radius of the circular park be r meters.
Given –
Circumference of circular park = 352 m
⇒ 2 × π × r = 352
⇒ 2 × (22/7) × r = 352
⇒ r = (7/44) × 352
∴ r = 7 × 8 = 56 m
⇒ outer radius = 56 + 7 = 63 m
∴ Area of the road = π(632 – 562)
= (22/7)(63 + 56)(63 – 56)
= (22/7)(119)(7)
= 22 × 119
= 2618 m2
A round table cover shown in the adjoining figure has six equal designs. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.50 per cm2. [Use √ 3 = 1.73.]
In the given figure, all the six desings covering equal area of the circle, therefore each design will subtend equal angles at the centre which is equal to (360°/6) i.e. 60°.
Also, the six triangles will be equal in area which is obtained by joining vertices of hexagon to the centre.
The triangle obtained will be equilateral because adjacent sides will be equal to the radius i.e. base angles will be equal and angle b/w them is 60° which concludes that other two angles will also be equal to 60° each.
∴ Area of six equilateral Δ = 6 × (√3/4) × (radius)2
= (3√3/2) × (28)2
= 1.5 × 1.73 × 784
= 2034.48 cm2
Area of Circle = π × (radius)2 = (22/7) × (28)2 = (22/7) × 784
= 22 × 112
= 2464 cm2
Area of the designs = Area of Circle – Area of six equilateral Δ
= (2464 – 2034.48) cm2
= 429.52 cm2
∴ Cost of making designs =Rs. (0.50 × 429.52) = Rs. 214.76
In an equilateral triangle of side 12 cm, a circle is inscribed touching its sides. Find the area of the portion of the portion of the triangle not included in the circle. [Take √ 3 = 1.73 and π = 3.14.]
fig.19
Let the radius of the circle be r cm.
In the fig.19,
AR and AQ are making a pair of tangents drawn from vertex A of Δ ABC on the circle.
∴ AR = AQ = x [LET]
BR and BP are making a pair of tangents drawn from vertex B of Δ ABC on the circle.
∴ BR = BP = y [LET]
CP and CQ are making a pair of tangents drawn from vertex C of Δ ABC on the circle.
∴ CP = CQ = z [LET]
Given –
Δ ABC is an equilateral triangle.
∴ AB = BC = AC = 12 cm
⇒ AR + BR = BP + CP = AQ + CQ = 12
⇒ x + y = y + z = x + z = 12…..(1)
Now,
(x + y + y + z + x + z) = (12 + 12 + 12)
⇒ 2 × (x + y + z) = 36
⇒ x + y + z = 18…..(2)
Subtracting equation(1) from equation(2), we get –
x = y = z = 6 cm
Also, the line joining the centre the circle to the vertices of Δ which circumscribes the circle bisects the angles of a Δ.
∴ ∠ OBP = 30°
In Δ BOP,
tan ∠ OBP = OP/BP = r/6
⇒ tan 30° = r/6
⇒ 1/√3 = r/6
∴ r = 6/√3 = 2√3 = 3.46 cm
Area of Δ ABC = (√3/4) × (side)2
=(1.73/4) × (12)2
= 1.73 × 36
= 62.28 cm2
Area of circle = π × (radius)2
= 3.14 × (3.46)2
= 37.59 cm2
Thus, Area of the triangle which is not included in the circle
= Area of Δ ABC – Area of circle
= (62.28 – 37.59) cm2
= 24.69 cm2
If a sphere has the same surface area as the total surface area of a circular cone of height 40 cm and radius 30 cm, find the radius of the sphere.
Let the radius of the sphere be r cm.
Given –
Height of cone(h) = 40 cm
Radius of cone(r) = 30 cm
∴ Slant height of cone(l) = √(h2 + r2) = √(402 + 302) = 50 cm
According to question –
Surface Area of Sphere = Total Surface Area of Circular Cone
⇒ 4 × π × r2 = π × r × (r + l)
⇒ 4r = r + l
⇒ 3r = l
∴ r = (l/3) = (50/3) cm
Thus, the radius of the Sphere = (50/3) cm
A two–digit number is such that the product of its digits is 35. If 18 is added to the number, the digit interchange their places. Find the number.
Let the two-digit number be xy(i.e. 10x + y).
After reversing the digits of the number xy, the new number becomes yx (i.e. 10y + x).
According to question –
xy = 35…..(1)
And,
(10x + y) + 18 = (10y + x)
⇒ 9x – 9y = – 18
⇒ x – y = – 2…..(2)
From equation(2), we get –
x = y – 2…..(3)
Substitute the value of x in equation(1), we get –
y(y – 2) = 35
⇒ y2 – 2y – 35 = 0
⇒ y2 – 7y + 5y – 35 = 0
⇒ y(y – 7) + 5(y – 7) = 0
⇒ (y – 7)(y + 5) = 0
∴ y = 7 [∵ y = – 5 is invalid because digit of a number can't be – ve.]
Substituting the value of y in equation (3), we get –
x = 5
Thus, the required number is 57.
Two water taps together can fill a tank in hours. The larger tap takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Let the tap of the smaller diameter and larger diameter fills the tank alone in x and (x – 10) hours respectively.
In 1 hour, the tap of the smaller diameter can fill 1/x part of the tank.
In 1 hour, the tap of the larger diameter can fill 1/(x – 10) part of the tank.
Two water taps together can fill a tank in hours = 75/ 8 hours.
But in 1 hour the taps fill 8/75 part of the tank.
⇒
⇒
⇒ 4x2 – 40x = 75x – 375
⇒ 4x2 – 115x + 375 = 0
⇒ 4x2 – 100x – 15x + 375 = 0
⇒ 4x(x – 25) – 15( x – 25) = 0
⇒ (4x -15)( x – 25) = 0
⇒ x = 25, 15/4
Taking x = 15 / 4
⇒ x – 10 = -25 /4 (But, time cannot be negative)
Now, taking x = 25
⇒ x – 10 = 15
Larger diameter of the tap can the tank 15 hours and smaller diameter of the tank can fill
the tank in 25 hours.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact to the centre.
fig.20
In the fig.20, PA and PB are the two tangents drawn from an external point P at the point of contacts A and B on the circle with centre O respectively.
∴ OA ⊥ PA and OB ⊥ PB
[∵ radius of a circle is always⊥to the tangent at the point of contact.]
∴ ∠ OAP = ∠ OBP = 90°
we know that –
Sum of all the angles of a quadrilateral = 360°
In quadrilateral OAPB,
∠ OAP + ∠ OBP + ∠ APB + ∠ AOB = 360°
⇒ 180° + ∠ APB + ∠ AOB = 360°
∴ ∠ APB + ∠ AOB = 180°
Hence, the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact to the centre.
From the top of a 7 – m high building, the angle of elevation of the top of a cable tower is 60˚ and the angle of depression of its foot is 45˚. Find the height of the tower. [Given √ 3 = 1.73.]
fig.21
Given –
Angle of Elevation = ∠ EAC = 60°
Angle of Depression = ∠ EAD = ∠ BDA = 45°
Height of Building = AB = ED = 7 m
In Δ ABD,
tan 45° = AB/BD
⇒ 1 = 7/BD
⇒ BD = 7 m
∴ AE = BD = 7 m [from fig.21]
And, In Δ ACE
tan ∠ CAE = CE/AE
⇒ tan 60° = CE/7
⇒ √3 = CE/7
⇒ CE = 7√3 m
Thus, Height of Tower = CE + ED = 7√3 + 7
= 7(1.73 + 1)
= 7 × 2.73
= 19.11 m
Puja works in a bank and she gets a monthly salary of ₹ 35000 with annual increment of ₹ 1500. What would be her monthly salary in the 10th year?
Given –
Monthly Salary = Rs. 35000
∴ Annual Salary = Rs. (12 × 35000) = Rs. 420000
Annual Increment = Rs. 1500
Let us consider this situation as an AP with
first term = a = Rs. 420000
and, Common Difference = d = Rs. 1500
∴ Salary in 10th year is given by –
a10 = a + (10 – 1)d = 420000 + 9 × 1500 = Rs. 433500
Thus, Monthly Salary in 10th year = Rs. (433500/12)
= Rs. 36125
In the given figure ABCD represent the quadrant of a circle of radius 7 cm with centre A. Calculate the area of the shaded region. [Take π = 22/7.]
Area of quadrant CAB = (π/4) × (radius)2
= (22/28) × (7)2
= 37.5 cm2
Area of Δ EAB = (1/2) × base × height
= (1/2) × 7 × 2
= 7 cm2
Thus, Area of shaded Region
= Area of quadrant CAB – Area of Δ EAB
= (37.5 – 7) cm2
= 30.5 cm2
The radii of the circular end of a solid frustum of a cone are 33 cm and 27 cm and its slant height is 10 cm. Find its capacity and total surface area. [Take π = 22/7.]
fig.23
Given –
MC = 27 cm, NE = 33 cm and CE = 10 cm
Let AM = h cm, AN = H cm and AC = l cm
∴ AE = AC + CE = (l + 10) cm
In the above fig.19,
Δ AMC and Δ ANE are similar triangles because their corresponding angles are equal.
…..(1)
On cross multiplying last two fractional parts of equation(1), we get –
33l = 27l + 270
⇒ 6l = 270
∴ l = 45 cm
∴ AE = 45 + 10 = 55 cm
In Δ ANE,
AN2 + NE2 = AE2 [by using pythagoras theorem]
⇒ H2 + (33)2 = (55)2
⇒ H2 + 1089 = 3025
⇒ H2 = 1936
∴ H = 44 cm
From first and last fractional parts of equation(1), we get –
h = (27/33) × 44 = 36 cm
∴ Height of frustum = H – h = 44 – 36 = 8 cm
Now,
Capacity of Frustum = Vol. of Cone ADE – Vol. of cone ABC
= (1/3)π × (NE)2 × (AN) – (1/3)π × (MC)2 × (AM)
= (1/3)π × [(33)2 × (44) – (27)2 × (36)]
= (22/21) × [47916 – 26244]
= (22/21) × 21672
= 22 × 1032
= 22704 cm3
Total Surface Area of Frustum
= Area of Curved Part(Trapezium)
+ Area of Upper Circular Part
+ Area of lower Circular Part
= [(1/2) × (sum of parallel sides) × (height of frustum)]
+ [π × (MC)2] + [π × (NE)2]
= [(1/2) × 2π(27 + 33) × 8] + [(22/7) × (27)2] + [(22/7) × (33)2]
= 480(22/7) + (22/7) × [(27)2 + (33)2]
= 480(22/7) + (22/7) × 1818
= (22/7) × 2298
= 22 × 328.28
= 7222.16 cm2
Thus, Capacity of Frustum = 22704 cm3
and, Total Surface Area of Frustum = 7222.16 cm2
From an external point p, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at the point E and PA = 14 cm, find the perimeter of ΔPCD.
In the given fig.,
CA and CE are the two tangents drawn from an external point C at the point of contacts A and E respectively.
∴ CA = CE
[∵ Tangents drawn from an exterior point to the circle are equal in length]
Similarly, DE and DB are the two tangents drawn from an external point D at the point of contacts E and B respectively.
∴ DE = DB
[∵ Tangents drawn from an exterior point to the circle are equal in length]
Perimeter of Δ PCD = PC + CD + PD
= PC + CE + DE + PD
= PC + CA + BD + PD
= PA + PB [∵ PA = PC + CA and PB = PD + BD]
= 14 + 14 [∵ PA=PB]
= 28 cm
Construct a ΔABC in which BC = 5.4 cm, AB = 4.5 cm and ∠ ABC = 60˚.
Construct a triangle similar to this triangle, whose side are 3/4 of the corresponding sides of ΔABC.
Steps of Construction :
1. Draw a line Segment BC = 5.4 cm and draw an angle of 60°
at point B and mark a length of AB = 4.5 cm on the line passing through B. Then join AC.
fig.25
2. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
fig.26
3. Locate 4 points [the greater of 4 and 3 in (3/4)] B1, B2, B3, B4 on BX so that
BB1 = B1B2 = B2B3 = B3B4
fig.27
4. Join B3 [the 3rd point, 3 being smaller of 3 and 4 in (3/4)] to C and draw a line through B4 parallel to B3C, intersecting the extended line segment BC at C′.
fig.28
5. Draw a line through C′ parallel to CA intersecting the extended line segment BA at A′.
fig.29
Then A′BC′ is the required triangle.
A bag contain 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that of a red ball, find the number of blue balls in the bag.
Let the no. of blue balls in the bag be x.
Let B and R be the event of drawing a blue and red ball respectively.
∴ total no. of balls in the bag = x + 5
According to question –
Probability of drawing blue ball
= 3 × Probability of drawing blue ball
⇒ [no. of blue balls/total no. of balls]
= 3 × [no. of red balls/total no. of balls]
⇒ (x/x + 5) =3 × (5/x + 5)
∴ x = 15
Thus, the no. of blue balls in the bag is 15.
In what ratio is the line segment joining the points (– 2, – 3) and (3, 7) divided by the y-axis? Also, find the coordinates of the point of division.
Let the point on the y-axis which divides the line segment joining the points A(– 2, – 3) and B(3, 7) be C(0,y).
Let the ratio in which y-axis divides AB line segment be m:n.
Let (x,y) ≡ (0,y)
(x1,y1) ≡ (– 2, – 3)
and (x2,y2) ≡ (3,7)
Using Section Formula,
⇒ 3m = 2n
∴ m:n = 2:3
Now,
⇒ y = (5/5) = 1
Thus, the line segment joining the points (– 2, – 3) and (3, 7) divided by the y-axis in the ratio 2:3 internally and the coordinates of the point of division is (0,1).
The values of k for which the equation 2x2 + kx + 3 = 0 has two real equal root are
A. ± 2√ 3
B. ± 3√ 2
C. ± 2√ 6
D. ± √ 6
Any quadratic equation in the form ax2 + bx + c = 0 has equal roots if and only if Discriminant, D = 0
Where, D = b2 – 4ac
In the given equation,
a = 2
b = k
c = 3
Now, above equation will have equal roots if
D = 0
i.e.
(k)2 – 4(2)(3) = 0
⇒ k2 = 24
⇒ k= ±2√6
How many terms are there in the AP 7, 11,15, …, 139?
A. 31
B. 32
C. 33
D. 34
In the given AP,
First term, a = 7
Common difference, a2 – a1 = (11 – 7) = 4
Let the no of terms be n
Nth term, an = 139
We know that, For any AP
an = a + (n – 1)d
where,
an = nth term
d = common difference
n = no of terms
using the above formula for given AP, we have
139 = 7 + (n – 1)(4)
⇒ 4(n – 1) = 132
⇒ n – 1 = 33
⇒ n = 34
Hence, there are 34 terms in given AP.
One card is drawn from a well-shuffled deck of card. The probability of drawing a 10 of a black suit is
A. 1/13
B. 1/26
C. 1/52
D. 3/52
We know that,
Probability
Now,
No of total outcomes i.e. total no of cards = 52
No of favourable outcomes i.e. no of black suits of 10 = 2
Probability (Getting a 10 of black suit)
In a circle of radius 7 cm, tangent PT is drawn from a point P such that PT = 24 cm. If O is the centre of the circle then OP =?
A. 30 cm
B. 28 cm
C. 25 cm
D. 31 cm
Given, A circle with center O and radius, OT = 7 cm and PT = 24 cm
Now, we know that
Tangent at a point on the circle is perpendicular to the radius through the point of contact.
i.e.
OT ⏊ OP
By Pythagoras Theorem in ΔOTP [ i.e. Hypotenuse2 = Base2 + Height2]
(OP)2 = (OT)2 + (PT)2
⇒ (OP)2 = (7)2 + (24)2
⇒ (OP)2 = 49 + 576 = 625
⇒ OP = 25 cm
The ratio in which the line segment joining the points A(– 3/ 2) and B(6, 1) is divided by the y – axis is
A. 3:1
B. 1:3
C. 2:1
D. 1:2
We know that any point on y axis is in the form (0, x) where x is any real number, let y axis intersect the line segment AB at point P with coordinates (0, c)
And we have
Coordinates of A = (– 3, 2)
Coordinates of B = (6, 1)
Let P divides AB in K:1
Now, By using section formula i.e.
The coordinates of Point P which divides line AB in a ration m : n is
Where, (x1, y1) and (x2, y2) are the coordinates of points A and B respectively.
So,
Coordinates of P
So,
P divides AB in 2:1
The distance of the point P (6, – 6) from the origin is
A. 6 units
B. √ 6 units
C. 3√ 2 units
D. 6√ 2 units
Coordinates of Given Point (say P) = (6, – 6)
Coordinates of Origin (say O) = (0, 0)
By using distance formula i.e
Distance
Where, (x1, y1) and (x2, y2) are the coordinates of points A and B respectively.
So, we have
⇒ OP =√(36 + 36)
⇒ OP =6√2 units
A kite is flowing at a height of 75 cm from the level ground, attached to a string inclined at 60˚ to the horizontal. The length of string with no slack in it, is
A. 50√ 2 m
B. 25√ 3 m
C. 50√ 3 m
D. 37.5 m
Consider, the situation in the form of a triangle ABC where A is the kite and AC shows the height of kite i.e.
AC = 75 cm
And
AB be the string with angle of inclination i.e.
∠CAB = θ = 60°
We have to find length of string i.e. AB
Clearly, ABC is a right – angled triangle
So, we have
On cross multiplying we get,
On rationalizing we get,
A solid metal cone with radius of base 12 cm and height 24 cm is melted to form solid spherical balls of diameter 6 cm each. The number of balls formed is
A. 16
B. 32
C. 24
D. 28
For solid metal cone,
Height, h = 24 cm
Base radius, b = 12 cm
We know,
Where r is base radius and h is the height of cone.
Putting the values,
For a solid spherical ball,
Diameter = 6 cm
Radius, r = 3 cm
We know,
Where, r is radius of sphere.
Putting the values, we have
On solving, we get
No of balls = 32
If the roots of the equation (a – b) x2 + (b – c) + (c – a) = 0 are equal, prove that b + c = 2a.
As the equation is in the form Ax2 + Bx + C = 0 with non-zero A.
In which,
A = a - b
B = b - c
C = c - a
And we know that if the roots of a equation are equal then we have
Discriminant, D = 0
Where, D = b2 - 4ac
⇒ b2 - 4ac = 0
⇒ (b - c)2 - 4(a - b)(c - a) = 0
⇒ (b - c)2 + 4(a - b)(a - c) = 0
⇒ b2 + c2 - 2bc + 4(a2 - ac - ab + bc) = 0
⇒ b2 + c2 - 2bc + 4a2 - 4ac - 4ab + 4bc = 0
⇒ 4a2+ b2 + c2 - 4ac + 2bc - 4ac = 0
⇒ (-2a)2+ b2 + c2 + 2(-2a)c + 2bc + 2(-2a)c = 0
⇒ (-2a + b + c)2 = 0
[using (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2za ]
⇒ -2a + b + c = 0
⇒ b + c = 2a
Hence Proved.
Find the 10th term form the end of the AP 4, 14, … , 254.
First term, a = 4
Common difference, a2 – a1 = 14 – (4) = 10
Let the no of terms be n
We know, that nth term of an AP is
an =a + (n – 1)d
where a is first term and d is common difference.
254 = 4 + (n – 1)10
⇒ 250 = (n – 1)10
⇒ n – 1 = 25
⇒ n = 26
10th term from last will be 17th term from starting
And a10 = a + 16d
= 4 + 16(10)
= 164
Or, which term of the AP 3, 15, 27, 34, … will be 132 more than its 54th term?
Which term of the AP 3, 15, 27, 39, … will be 132 more than its 54th term.
Given AP = 3, 15, 27, 39, …
First term, a = 3
Common difference, a2 - a1 = 15 - 3 = 12
And we know
Nth term of an AP, an = a + (n - 1)d
Where a is first term and d is common difference.
Now, let the mth term be 132 more than 54th term
In that case,
am = a54 + 132
⇒ a + (m - 1)d = a + 53d + 132
⇒ (m - 1)12 = 53(12) + 132
⇒ 12m - 12 = 636 + 132
⇒ 12m = 768 + 12
⇒ 12m = 780
⇒ m = 65
hence, 65th term will be 132 more than its 54th term.
Prove that the tangents drawn at the ends of a diameter of a circle are parallel
Let AB be the diameter of a circle with center O.
CD and EF are two tangents at ends A and B respectively.
To Prove : CD || EF
Proof :
OA ⏊ CD and OB ⏊ EF [Tangents drawn at a point on circle is perpendicular to the radius through point of contact]
∠OAD = ∠OBE = 90°
∠OAD + ∠OBE = 90° + 90° = 180°
Considering AB as a transversal
⇒ CD || EF
[Two sides are parallel, if any pair of the interior angles on the same sides of transversal is supplementary]
From an external point P tangents PA and PB are drawn to a circle with centre at the point E and PA = 14 cm, find the perimeter of ΔPCD.
Given: From an external point P, two tangents, PA and PB are drawn to a circle with center O. At a point E on the circle tangent is drawn which intersects PA and PB at C and D, respectively. And PA = 14 cm
To Find: Perimeter of ΔPCD
As we know that, Tangents drawn from an external point to a circle are equal.
So we have
AC = CE …[1] [Tangents from point C]
ED = DB …[2] [Tangents from point D]
Now Perimeter of Triangle PCD
= PC + CD + DP
= PC + CE + ED + DP
= PC + AC + DB + DP [From 1 and 2]
= PA + PB
Now,
PA = PB = 14 cm as tangents drawn from an external point to a circle are equal
So we have
Perimeter = PA + PB = 14 + 14 = 28 cm
The area of the circular base of a cone is 616 cm2 and its height is 48 cm. Find its whole surface area. [Take π = 22/7.]
Area of circular base = 616 cm2
We know that,
Area of circle = πr2
Where r is the radius of circle
Let the radius of circular base be r
We have,
πr2 = 616 cm2
⇒ r2 = 196
⇒ r = 14 cm
Now, Height = 48 cm [Given]
And we know,
Slant height, l =√(r2 + h2)
Where r is radius and h is the height of the cone
l=√(142 + 482)=√(196 + 2304)=√2500=50 cm
Now,
Total surface area of a cone = πr(l + r)
Where r is radius and l is slant height.
So, Putting values we have
Total surface of cone = π(14)(50 + 14)
In the adjoining figure, the area enclosed between two concentric circles is 770 cm2 and the radius of the outer circle is 21 cm. Find the radius of the inner circle.
Given,
Outer radius of circle, R = 21 cm
Area of enclosed region = 770 cm2
Let the radius of inner circle be r.
Area of enclosed region = Area of outer circle – Area of inner circle
⇒ 770 = πR2 – πr2
⇒
⇒ 35(7) = (21)2 – r2
⇒ r2 = 441 – 245
⇒ r2 = 196
⇒ r = 14 cm
Solve for x: 12abx2 – (9a2 – 8b2) x – 6ab = 0.
12abx2 – (9a2 – 8b2)x – 6ab = 0
12abx2 – 9a2x + 8b2x– 6ab = 0
3ax(4bx – 3a) + 2b(4bx – 3a) = 0
(3ax + 2b)(4bx – 3a) = 0
So, we have
3ax + 2b = 0 or 4bx – 3a = 0
If the 8th term of an AP is 31 and its 15th term is 16 more than the 11th term. Find the AP.
Let the a be first term and d be common difference
As we know
an = a + (n – 1)d
Given,
⇒ a8 = 31
⇒ a + 7d = 31
⇒ a = 31 – 7d …[1]
Also, As 15th term is 16 more than 11th term
⇒ a15 = a11 + 16
⇒ a + 14d = a + 10d + 16
⇒ 4d = 16
⇒ d = 4
Using this value in equation [1]
a = 31 – 7(4) = 3
So, AP is
a, a + d, a + 2d, …
3, 3 + 4, 3 + 2(4), …
3, 7, 11, …
Prove that the parallelogram circumscribing a circle is a rhombus.
Consider a circle circumscribed by a parallelogram ABCD, Let side AB, BC, CD and AD touch circles at P, Q, R and S respectively.
To Proof : ABCD is a rhombus.
As ABCD is a parallelogram
AB = CD and BC = AD [opposite sides of a parallelogram are equal] …[1]
Now, As tangents drawn from an external point are equal.
We have
AP = AS [tangents from point A]
BP = BQ [tangents from point B]
CR = CQ [tangents from point C]
DR = DS [tangents from point D]
Add the above equations
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ AB + CD = AS + DS + BQ + CQ
⇒ AB + CD = AD + BC
⇒ AB + AB = BC + BC [From 1]
⇒ AB = BC …[2]
From [1] and [2]
AB = BC = CD = AD
And we know,
A parallelogram with all sides equal is a rhombus
So, ABCD is a rhombus.
Hence Proved !
A ΔABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of length 8 cm and 6 cm respectively. Find AB and AC.
Given: ΔABC that is drawn to circumscribe a circle with radius r = 4 cm and BD = 6 cm DC = 8 cm
To Find: AB and AC
Now,
As we know tangents drawn from an external point to a circle are equal.
Then,
FB = BD = 6 cm [Tangents from same external point B]
DC = EC = 8 cm [Tangents from same external point C]
AF = EA = x (let) [Tangents from same external point A]
Using the above data we get
AB = AF + FB = x + 6 cm
AC = AE + EC = x + 8 cm
BC = BD + DC = 6 + 8 = 14 cm
Now we have heron's formula for area of triangles if its three sides a, b and c are given
ar = √(s(s – a)(s – b)(s – c))
Where,
So for ΔABC
a = AB = x + 6
b = AC = x + 8
c = BC = 14 cm
And
ar(ΔABC) = √((x + 14)(x + 14 – (x + 6))(x + 14 – (x + 8))(x + 14 – 14))
= √((x + 14)(8)(6)(x)) [1]
ar(ABC) = ar(AOB) + ar(BOC) + ar(AOC)
at, tangent at a point on the circle is perpendicular to the radius through point of contact,
So, we have
OF ⏊ AB, OE ⏊ AC and OD ⏊ BC
Therefore, AOB, BOC and AOC are right – angled triangles.
And area of right angled triangle =1/2 × Base × Height
Using the formula,
Using [1] we have,
Squaring both side
⇒ 48x(x + 14) = (2x + 6 + 28 + 2x + 16)2
⇒ 48x2 + 672x = (56 + 4x)2
⇒ 48x2 + 672x = (4(14 + x))2
⇒ 48x2 + 672x = 16(196 + x2 + 28x)
⇒ 3x2 + 42x = 196 + x2 + 28x
⇒ 2x2 + 14x – 196 = 0
⇒ x2 + 7x – 98= 0
⇒ x2 + 14x – 7x – 98 = 0
⇒ x(x + 14) – 7(x + 14) = 0
⇒ (x – 7)(x + 14) = 0
⇒ x = 7 or x = – 14 cm
Negative value of x is not possible, as length can't be negative
Therefore,
x = 7 cm
⇒ AB = x + 6 = 7 + 6 = 13 cm
⇒ AC = x + 8 = 7 + 8 = 15 cm
Draw a circle of diameter 12 cm. From a point 10 cm away its centre, construct a pair of tangents to the circle. Measure the length of each tangent segment.
Steps of Construction:
1. Take a point O and draw a circle of radius 6 cm [ i.e. diameter 12 cm]
2. Mark a point P at a distance of 10 cm from O in any direction. Join OP
3. Draw right bisector of OP, intersecting OP at O'
4. Taking O' as center and O'O=O'P as radius, draw a circle to intersect the previous circle at T and T'.
5. Join PT and PT', which are required tangents.
6. Measured PT and PT' by a ruler and we get PT = PT' = 8 cm
Show that the point A (a, a), B (– a, – a) and C(– a√ 3, a√ 3) are the vertices of an equilateral triangle.
For the points A, B and C to be vertices of an equilateral triangle,
AB = BC = CA and we have distance formula,
For two point P(x1, y1) and Q(x2, y2)
PQ= √((x2 – x1)2 + (y2 – y1)2)
Using the above formula, and coordinates we have
AB=√((– a – a)2 + (– a – a)2)
⇒ AB= √(4a2 + 4a2 )=√8 a
As AB = BC = AC
ABC is an equilateral triangle.
Find the area of a rhombus if its vertices are A (3, 0), B (4, 5), C (– 1, 4) and D (– 2, – 1).
As the diagonal of rhombus divides it into two parts, it is sufficient to calculate the area of one part and double it.
Consider, the Diagonal AC,
Then,
ar(ABCD) = 2× ar(ΔABC)
Now,
A = (3,0); B = (4, 5); C = (– 1, 4)
As we know area of triangle formed by three points (x1, y1), (x2, y2) and (x3, y3)
=6 square units
⇒ ar(ABCD) = 2× ar(ΔABC) = 2(6) = 12 square units
Cards marked with numbers 13, 14, 15, …, 60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card drawn is
(a) Divisible by 5 (b) a number which is a perfect square
Total no of numbers = 60 – 13 + 1 = 48
[As total no's from a to b are (b – a + 1)]
(a) No Divisible by 5 in the given sequence = {15, 20, 25, 30, 35, 40, 45, 50, 55, 60}
So, we have
No of favourable outcomes = 10
No of total outcomes = 48
And,
Probability of an event
Therefore,
P(Getting a card having no divisible by 5)
(b) Perfect squares in the given sequence = {16, 25, 36, 49}
So, we have
No of favourable outcomes = 4
No of total outcomes = 48
And,
Probability of an event
Therefore,
P(Getting a card having a perfect square)
A window in a building is at a height of 10 m from the ground. The angle of depression of a point P on the ground from the window is 30˚. The angle of elevation of the top of the building from the point P is 60˚. Find the height of the building.
Let us consider this situation by a diagram as shown, in which AB is a building and C depicts the window and A be the top.
Now Given,
Height of window from the ground, BC = 10 m
Angle of depression of point P from window, ∠XCP = 30°
⇒ ∠XCP = ∠CPB = θ1 = 30° [Alternate Angles]
Angle of elevation of top of the building from point P, ∠APB = 60°
⇒ ∠ APB = θ2 = 60°
Now, In Δ BCP
Cross – Multiplying we get,
BP=10√3 meters
Now, In ΔABP
⇒ AB = 10√3 × √3 = 30 meters
So, Height of building is 30 meters.
In a violent storm, a tree got bent by the wind. The top of the tree meet the ground at an angle of 30˚, at a distance of 30 metres from the root. At what height from the bottom did the tree get bent? What was the original height of the tree?
Let AB be a tree, and P be the point of break,
And As tree falls, we can consider the situation as a right angled triangle at B
Given,
Angle of broken tree with ground, θ = 30°
Distance of top of broken tree from root, AB = 30 m
In ΔAPB
So, tree bents at a height ofmeters from the ground.
Also, In Δ APB
On cross – multiplying
AP = 10√3 × 2 = 20√3 meters
Original height of tree = AP + BP
= 20√3 + 10√3
= 30√3 meters
A wire bent in the form of a circle of radius 42 cm is cut and again bent in the form of a square. Find the ratio of the areas of the regions enclosed by the circle and the square.
Given,
Radius of circle made by wire, r = 42 cm
Circumference of circle of radius r = 2πr
Circumference of circle made by wire
As, the same wire is bent to make a square the perimeter of square will be equal to circumference of circle.
Let the side of square be a.
Perimeter of square of side 'a' = 4a
We have,
4a = 264
a = 66 cm
Now,
Ratio of areas
Area of circle of radius r = πr2
Area of square of radius a = a2
Putting value, we get
Ratio of areas
Required ratio is 14 : 11
A metallic sphere of radius 10.5 cm is melted and then recast into smaller cones, each of radius 3.5 cm and height 3 cm. How many cones are obtained?
We know volume of sphere of radius R is
And
Volume of cone of radius r and height h is
So, Given,
Radius of sphere, R = 10.5 cm
Radius of cone, r = 3.5 cm
Height of cone, h = 3 cm
No of cones can be made by melting sphere
Using formulas, and putting values
No of cones
Hence, 126 cones can be made.
In the given figure ΔABC is right angled at A. Semicircles are drawn on AB, AC and BC as diameter. It is given that AB = 3 cm and AC = 4 cm. Find the area of the shaded region.
Let semicircle I, II and III are semicircles with diameters AB, AC and BC respectively
Area of shaded region =
Area of semicircle I + Area of semicircle II + Area of triangle ABC – Area of semicircle III
As, ∠BAC is in semicircle,
∠BAC = 90° [Angle in a semicircle is right angle]
And ABC is a right – angled triangle at A
By Pythagoras Theorem
(Hypotenuse)2 = (Base)2 + (Perpendicular)2
(BC)2 = (AB)2 + (AC)2
⇒ (BC)2 = 32 + 42 = 9 + 16 = 25
⇒ BC = 5 cm
Now, For semicircle I
Diameter = AB = 3 cm
Radius
Area of semicircle of radius r
Area of semicircle I
For semicircle II
Diameter = AC = 4 cm
Radius
Area of semicircle of radius r
Area of semicircle II
For semicircle III
Diameter = BC = 5 cm
Radius,
Area of semicircle of radius r
Area of semicircle I
Area of a right – angled triangle
Area of ΔABC
Required area (From eqn [1])
₹ 250 is divided equally among a certain number of children. If there were 25 more children, each would have received 50paise less. Find the number of children.
Let the no of children is x and amount given to each child is y
As the total amount is 250 ₹
We have,
xy = 250
…[1]
Also, given if no of children is increased by 25, the amount to each get less by 50 paise i.e. 0.5 ₹
So, we have
(x + 25)(y – 0.5) = 250
[By 1]
⇒ 500x – x2 + 12500 – 25x = 500x
⇒ x2 + 25x – 12500 = 0
⇒ x2 + 125x – 100x – 12500 = 0
⇒ x(x + 125) – 100(x + 125) = 0
⇒ (x – 100)(x + 125) = 0
so,
⇒ x – 100 = 0 or x + 125 = 0
⇒ x = 100 or – 125
However, no. of students can't be negative
Hence, x = 100
So, there were 100 students.
The hypotenuse of a right – angled triangle is 6 cm more than twice the shortest side. If the third side 2 cm less than the hypotenuse, find the sides of the triangle.
Let the shortest side be x cm [Let it be base]
Length of hypotenuse = 2x + 6 [in cm]
Length of other side = Length of hypotenuse – 2 = 2x + 6 – 2 = 2x + 4 [in cm] [Let it be perpendicular]
As we know, By Pythagoras Theorem
(hypotenuse)2 = (base)2 + (perpendicular)2
⇒ (2x + 6)2 = x2 + (2x + 4)2
⇒ 4x2 + 36 + 24x = x2 + 4x2 + 16 + 16x
[(a + b)2 = a2 + b2 + 2ab]
⇒ x2 – 8x – 20 = 0
⇒ x2 – 10x + 2x – 20 = 0
⇒ x(x – 10) + 2(x – 10) = 0
⇒ (x + 2)(x – 10) = 0
⇒ x = – 2 or x = 10 cm
However, Length can't be negative hence x = – 2 is not possible
Therefore,
x = 10 cm
we have,
Shortest Side = x = 10 cm
Hypotenuse = 2x + 6 = 2(10) + 6 = 26 cm
Third side = 2x + 4 = 2(10) + 4 = 24 cm
If the sum of first n, 2n and 3n term of an AP be S1 , S2 and S3 respectively, then prove that S3 = 3 (S2 – S1).
We know that sum of first n terms of an AP is
Where a is first term and d is common difference
So,
Sum of first 2n terms
Sum of first 3n terms
Now, Taking RHS
RHS = LHS
Hence Proved
The angle of elevation of a jet plane from point A on the ground is 60˚. After A flight of 15 seconds, the angle of elevation changes to 30˚. If the jet plane is flying at a constant height of 1500 √ 3 m, find the speed of the jet plane.
Let the jet plane goes from point P to point C and we have given,
Initially angle of elevation from point A, ∠PAQ = θ1 = 60°
After 15 seconds,
Angle of elevation from point A, ∠CAB = θ2 = 30°
As the plane is flying at a constant height,
BC = PQ = 1500√3 m
Now,
In ΔABC
In ΔAPQ
So, we have
QB = AB – AQ = 4500 – 1500 = 3000 m
And
QB = PC
So, jet plane travels 3000 m in 15 seconds
And we know,
Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Given: A circle with center O and P be any point on a circle and XY is a tangent on circle passing through point P.
To prove : OP⏊XY
Proof :
Take a point Q on XY other than P and join OQ .
The point Q must lie outside the circle. (because if Q lies inside the circle, XY will become a secant and not a tangent to the circle).
Therefore, OQ is longer than the radius OP of the circle. That is, OQ > OP.
Since this happens for every point on the line XY except the point P, OP is the shortest of all the distances of the point O to the points of XY.
So OP is perpendicular to XY.
[As Out of all the line segments, drawn from a point to points of a line not passing through the point, the smallest is the perpendicular to the line.]
A quadrilateral ABCD is drawn to circumscribe a circle, as shown in the given figure. Prove that: AB + CD = AD + BC
Given: A quadrilateral ABCD, And a circle is circumscribed by ABCD
Also, Sides AB, BC, CD and DA touch circle at P, Q, R and S respectively.
To Prove: AB + CD = AD + BC
Proof:
In the Figure,
As tangents drawn from an external point are equal.
We have
AP = AS [tangents from point A]
BP = BQ [tangents from point B]
CR = CQ [tangents from point C]
DR = DS [tangents from point D]
Add the above equations
⇒ AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ AB + CD = AS + DS + BQ + CQ
⇒ AB + CD = AD + BC
Hence Proved.
A solid is made up of a cube and a hemisphere attached on its top as shown in the figure. Each edge of the cube measures 5 cm and the hemisphere has a diameter of 4.2 cm. Find the total area to be painted.
Total area to be painted = TSA of cube + CSA of hemisphere – Base area of hemisphere
[TSA = Total surface area & CSA = Curved surface area]
Given,
Diameter of hemisphere = 4.2 cm
Radius of hemisphere, r = 2.1 cm
Side of cube, a = 5 cm
And
TSA of cube = 6a2, where a is the side of cube
CSA of hemisphere = 3πr2, where r is the base radius
Base area = πr2 [As base is circular]
Therefore,
Total area to be painted = 6a2 + 3πr2 – πr2 = 6a2+ 2πr2
= 6(25) + (2 × 22 × 0.3 × 2.1)
= 177.72 cm2
The diameter 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 capacity of the bucket
(ii) The area of the metal sheet used to make the bucket. [Take π = 3.14.]
Given,
The diameter of lower end = 10 cm
As Radius = Diameter/2
Radius of lower end, r2 = 5 cm
The diameter of upper end = 30 cm
Radius of upper end, r1 = 15 cm
Height of bucket, h = 24 cm
(i) As we know
volume of frustum of a cone =
Where, h = height, r1 and r2 are radii of two ends (r1 > r2)
Capacity of bucket
=3.14 × 8 × (25 + 225 + 75)
= 3.14 × 8 × 325 = 8164 cm3
(ii) Area of metal used to make bucket = CSA of frustum + base area
We know that,
Curved surface area of frustum = πl(r1 + r2)
Where, r1 and r2 are the radii of two ends (r1 > r2)
And l = slant height and
l = √(h2+ (r1 – r2 )2)
So, we have
Slant height, l=√(242 + (15 – 5)2)
⇒ l =√(576 + 100)
⇒ l =√676
⇒ l = 26 cm
And as the base has lower end,
Base area = πr22, where r2 is the radius of lower end
Therefore,
Area of metal sheet used = πl(r1 + r2) + πr22
= π(26)[15 + 5] + π(5)2
= 520π + 25π
= 545πFind the value of k for which the point A (– 1, 3), B (2, k) and C (5, – 1) are collinear.
Three points A, B and C are collinear if and only if
Area(ΔABC) = 0
As we know area of triangle formed by three points (x1, y1), (x2,y2) and (x3, y3)
⇒ 0= – 6k + 6
⇒ 6k = 6
⇒ k = 1
So, For k = 1, A, B and C are collinear.
Two dice are thrown at the same time. Find the probability that the sum of two numbers appearing on the top of the dice is more than 9.
When two dice are thrown, the 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)}
The outcomes in which sum of no's is 9 are ={(3,6) (4,5) (5,4) (6,3)}
No of Total possible outcomes = 36
No of favourable outcomes = 4
And, Probability of an event
Therefore,
P(Getting sum 9)
A circus tent is cylindrical to a height of 3 cm and conical above it. If its base radius is 52.5 m and the slant height of the conical portion is 53 m, find the area of canvas needed to make the tent. [Take π = 22/7.]
Total area of canvas required = CSA of cylindrical part + CSA of conical part [CSA = Curved surface area]
Now,
Radius of cone = Radius of cylinder = r = 52.5 m
Height of cylindrical part, h = 3 cm = 0.03 m [As 1 m = 100 cm]
Lateral height of conical part, l = 53 m
Now, we know
CSA of cylinder = 2πrh
Where, r is base radius and h is height of cylinder and
CSA of cone = πrl
Where, r is base radius and l is slant height.
Area of canvas required = 2πrh + πrl
= πr(2h + l)
= 8754.9 m2