CBSE Class 10 - Maths

Solution:

The rational number given in the question is . It can also be written as

The above number is of type , in which x=2 and y=3.

Since x and y are positive integers, the rational number or will have a terminating decimal expansion.

Solution:

Let us assume that and are the zeroes of the polynomial.

The product of the zeroes = =

The sum of the zeroes = =

The required polynomial equation can be written as

=

=

If we take k=2, we will get the polynomial as .

One polynomial where the product of zeroes and sum of zeroes , is

Solution:

The equation x + y = 14 can also be written as x + y - 14 = 0

Similarly, x ? y = 4 can be written as x - y - 4 = 0

x + y -14 = 0 can be represented by where, = 1 and = - 14

x + y -14 = 0 can be represented by where , = - 1 and = - 4

=

=

We see from above that .

Therefore the pair of linear equations given in the question is consistent.

Solution:

The given equation can be represented by where a=4, b= and c=3.

The discriminant or D is given by . Substituting a=4, b= and c=3 in this, we get

= ()² - 4 = 48 - 48 =0.

From above, we see that there are two equal and real roots for the given quadratic equation.

Solution:

Let us assume that a, b and c are the three consecutive terms of the A.P. where a=x, b= 2x + k and c = 3x + 6.

For an A.P. , 2b= a + c

Substituting a=x, b= 2x + k and c = 3x + 6 in this, we get

2 = x + 3x + 6

4x + 2k = 4x + 6

2k = 6

k = 3

Thus x, 2x + k and 3x + 6 are three consecutive terms of an A.P. if k=3.

Solution:

?DAE = ?BAC as it is common angle

?ADE = ?ABC since they are corresponding angles

Hence according to the AA similarity condition, ?ADE ?ABC

For similar triangles, the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

= = =

Since DE = BC, =

=

= = .

area of ?ADE = area of ?ABC = 81 cm² = 36 cm².

Hence area of ?ADE = 36 cm².

Solution:

Since A + B = 90°, A = 90° - B

sec A = sec(90° - B)

As cosec = sec(90° - ),

sec A = cosec B

Given: sec A =

cosec B =

Hence value of cosec B is .

Solution:

Using mid-point formula, = - 3

= - 3

b + 2 = -6

b = -8

Hence we get the value of b as - 8.

Solution:

In 60 minutes, the angle that will be swept by the minute hand =

in 20 minutes, the angle that will be swept by the minute hand = 20 =

Area swept by the minute hand in 20 minutes = (7 cm)² .

=

⁼

Hence the area swept by the minute hand in 20 minutes is

Solution:

A modal class is the class interval having maximum frequency.

In the above frequency distribution, 27 is the maximum frequency and it comes in class interval 40 - 50.

Hence 40 - 50 is the modal class

40 is the lower limit of the modal class in the given frequency distribution.

Solution:

If and are two linear equations.

1. If , the above equations will represent intersecting lines

2. If , the equations will represent coincident lines

3. If , the equations will represent parallel lines

The equations given are :-

or

or

Here and

Similarly , and

and

Hence , which means that the given pair of equations represent coincident lines

Solution:

For an A.P. its n^th term or a_n is a_n = a + (n-1) d where a is the first term and d is the common difference between the terms.

The 17^th term of an A.P. or

The 10^th term of the A.P or

As given in the question, the difference between 17^th term and 10^th term of A.P. = 7.

7d = 7

Hence the common difference or d between the terms of the A.P. = 1

Solution:

and

=

=

= 5

The value of the given expression is 5.

Solution:

Using distance formula, we get

AB = = = =

BC = = = =

CA = = = =

As AB = CA, we see that the triangle is isosceles.

AB² + CA² = 106 + 106 =212 = BC² .

using Pythagoras theorem, we find that the triangle is right-angled at A.

Hence we have proved that the given points (?2, 5), (3, ?4) and (7, 10) are the vertices of a right-angled triangle.

OR

The center of the circle is given as (. The point it passes through is given as (-3, -1)

Using distance formula, the radius of the circle or OP

=

=

=

Since the diameter of circle is given as 20 units and diameter =

=

=

When , then

When , then

Solution:

Let us assume that the coordinates of C are m and n. Point C can be represented as C(m, n).

Given: 3AC = CB

Using section formula,

=

The coordinates of point C are ()

Solution:

Let us represent a positive integer with 'a'

It can be represented as

where and is an integer

Hence a will be

Positive odd integer will be of the form

Let us consider the above 4 positive odd integer forms seperately

Case 1 :

is an integer

Case 2:

is an integer

Case 3:

is an integer

Case 4:

is an integer

From all these cases we see that the square of a positive odd integer is of the form where is any integer

OR

Let us assume that is a rational number

If so it can be represented using integers a and b ( as

a and b are rational numbers. So should also be rational and should also be a rational number. But we know that is an irrational number and so our assumption is wrong.

Since our assumption is false it is proved that is not a rational number.

Solution:

p(x) = 6x⁴ + 8x³ ? 5x² + ax + b

q(x) = 2x² - 5

Dividing p(x) by q(x)

We get the remainder as .

As p(x) is exactly divisible by q(x), remainder should be 0.

= 0

From and

From

Therefore the value of and

Solution:

Let a be the 1^st term and d be the common difference of the A.P.

n^th term of A.P. =

9^th term of A.P =

Given:

19^th term of A.P. =

29^th term of A.P. =

Hence the 29^th term of the A.P. is double the 19^th term of the A.P.

Solution:

With radius as 3 cm, draw a circle. Mark the center point as O. Take a point P at a distance of 6 cm from O and join O and P.

Draw a perpendicular bisector to the line OP. Mark the point at which it intersects the circle as point Q.

Using Q as center and QP as radius, draw a circle. This will intersect the first circle at 2 points. Mark the points as M and N.

Join M with P to obtain line MP and N with P to obtain the line NP. MP and NP are the tangents.

On measuring, the length of the tangents, MP and NP is 5.20 cm .

Solution:

Consider and

PQ || BA and hence according to the basic proportionality theorem, we get

--------------------- (1)

Consider and

PR || BD and hence according to the basic proportionality theorem, we get

-----------------------(2)

Taking (1) and (2) together we get

, from above, using converse of proportionality theorem, we get QR|| AD.

Solution:

Given: BC is trisected at D and E.

Consider . Using Pythagoras Theorem, we get

---------------------(1)

Consider . Using Pythagoras theorem, we get

-----------------------(2)

Consider . Using Pythagoras theorem, we get

-----------------------(3)

We have to prove that

From (2), we know that . Hence substituting in above, we get

Hence proved.

OR

AD and AF are tangents drawn from point A to the circle. Similarly BD and BE are tangents from point B to the circle and CE and CF are tangents from point C

The tangents drawn to the circle from any point external to the circle are equal.

AD=AF

BD=BE and

CE =CF

Let us take AD = a, BE = b and CF = c

-----(1)

-----(2)

------(3)

------(4)

Substituting (1) in (4), we get

Substituting (2) in 4, we get

Substituting (3) in (4), we get

Thus we get

Solution:

Solution:

Since the points (x, y), (1, 2) and (7, 0) are collinear, the area of the triangle formed with these three points as vertices will be zero.

Area of triangle with points (x₁,y₁) , (x₂,y₂) , (x₃,y₃) =

Thus the area of the triangle is . Since this area is zero, we get

is the equation which shows the relation between x and y if the points have to be collinear.

Solution:

(i) Area of table top

Area of a sector can be obtained by where is the angle of the sector and r is the radius.

Given : .

Hence area of table top =

Hence area of table top =

(ii) Perimeter of table top

= length of the arc BD + OB + OD

Length of the arc BD =

Perimeter of the table top =

OR

Area of the shaded region = Area of the square ABCD - (area of semicircle APD + area of semicircle BPC)

Area of square ABCD

(since length of sides of the square = 14 cm)

Area of semicircle APD = where r = 7 cm (AD = 14 cm, )

Area of semicircle BPC = since r= 7 cm

Area of the shaded region =

Hence area of shaded region =

Solution:

The probability of an event =

Number of cards in box = 99 -14 + 1 = 86

(i)

Number of cards having odd number =

Hence probability of getting odd number =

(ii)

Number of cards having perfect squares

( Perfect squares between 14 and 9 are 16, 25, 36, 49, 64 and 81)

Hence probability of getting perfect square

(iii)

Number of cards with numbers divisible by 7

= 13

(Numbers divisible by 7 between 14 and 99 are 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91 and 98)

Hence probability of getting number divisible by 7 =

Solution:

Let the total number of articles bought be 'x'

Cost of x articles = Rs 900

Cost price of 1 article =Rs

Number of articles sold = x-5

Selling price of 1 article = Rs

Profit after selling x-5 articles = Rs 80

Since number of articles will not be negative, x= 75

Hence the number of articles bought by the trader = 75

OR

Let the son's age be 'x'

Son's present age =

Son's age 2 years ago =

Man's age 2 years ago =

Man's present age =

Son's age 3 years later from present =

Man's age 3 years later from present =

Man's present age =

Since age cannot be fraction, x = 5

Son's age = 5 years

Man's age = years

Solution:

Let the 2 similar triangles be .

To prove:

Let us draw AP ? BC and XQ ? YZ.

From above, we get

Using (4) and (5), we get

------------------------------------(6)

Substituting (6) in (3), we get

From (6), we get

To prove:

Let the sides of the square be of length 'a'

(since BC is diagonal)

All the angles of equilateral triangle are and all the sides of an equilateral triangle are equal.

Hence all equilateral triangles are similar.

Hence =

Therefore the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal.

Q28 The angle of elevation of the top of a building from the foot of a tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Solution:

Let the building be AB and the tower CD.

Length of CD = 50 m

Consider

Consider

Thus the height of the building is

Solution:

Let 'r' be the inner radius and 'R' be the external radius of the copper shell.

Given : External diameter of copper shell = 2R = 18 cm

Volume of the spherical copper shell =

The spherical copper shell is made into a solid cone

Hence volume of solid cone = volume of spherical shell

Volume of cone =

Hence

Therefore inner radius of the spherical copper shell = 8 cm and hence inner diameter = 16 cm

OR

Let the top radius of the bucket be 'r' and bottom radius of bucket be 'R'

r = 20 cm and R = 12 cm

Let the height of the bucket be 'h'

Volume of the bucket =

Let 's' be the slant height of the metal sheet

= = = 17 cm

The area of the metal sheet used in making the bucket

= CSA of the bucket + area of the bottom circular end of the bucket

=

Solution:

(a). Mode

The maximum frequency is 33 and the class having the maximum frequency is 45?55.

modal class = 45?55.

Lower class limit (l) of modal class = 45

Frequency (f₁) of modal class = 33

Frequency (f₀) of class preceding the modal class = 31

Frequency (f₂) of class succeeding the modal class = 17

Class size (h) = 10

= 46.11(approximate)

(B). Median

Let us calculate the cumulate frequency of the data in another column:-

We obtain n as 100

Cumulative frequency (cf) just greater than 50 = 71

71 comes under in the class interval 45-55. Hence median class = 45-55

Lower limit (l) of median class = 45

Class size (h) = 10

Frequency (f) of median class = 33

Cumulative frequency (cf) of class preceding median class = 38

Median =

(C). Mean

The class marks for each interval can be calculated by using the formula

Taking 60 as assured mean (a), d_i, u_i, and f_iu_i can be calculated as follows:

Mean =

mode = 46.11, median = 48.63 and mean = 49.7