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Question 1:

Which of the following numbers has terminating decimal expansion?

(a) $\frac{37}{45}$

(b) $\frac{21}{{2}^{3}{5}^{6}}$

(c) $\frac{17}{49}$

(d) $\frac{89}{{2}^{2}{3}^{2}}$

Here we have to check terminating decimal expansion.

We know that if the numerator can be written in the form where m and n are non negative positive integer then the fraction will surely terminate. We proceed as follows to explain the above statement

Hence the correct option is (b).

Question 2:

The value of p for which the polynomial x3 + 4x2 − px + 8 is exactly divisible by (x−2) is

(a) 0

(b) 3

(c) 5

(d) 16

Here the given polynomial is

We have to find the value of p such that the polynomial is exactly divisible by

First we have to write equation in basic format of divisibility like this

After solving, we have seeing here the reminder is = 16xpx

So, to find the value of p we put the reminder is equal to zero.

Therefore

Hence option (d) is correct.

Question 3:

Δ ABC and Δ PQR are similar triangles such that ∠A = 32° and ∠R = 65°. Then, ∠B is

(a) 83°

(b) 32°

(c) 65°

(d) 97°

It is given that there are two similar triangles, ΔABC and in which A = 32° and R = 65°, then we have to find B

We have following two similar triangles.

We know the relation between angles in the two similar triangles and these are

In we have,

Hence the correct option is

Question 4:

In Fig. 1, the value of the median of the data using the graph of less than ogive and more than ogive is

(a) 5

(b) 40

(c) 80

(d) 15

Here we have to find the median from the given graph.

The following graph is given.

From the given graph we can easily see that the total cumulative frequency N = 80.

First we draw a line from the point on the less than ogive graph which is parallel to the x-axis and draw a line perpendicular to the x-axis from that point, which meet the x-axis at x = 15 which is the desired median.

Hence the correct option is (d)

Question 5:

For what value of k will the following system of linear equations has no solution:

3x + y = 1

(2k − 1)x + (k − 1)y = 2k + 1

The given system of equations is

3x + y = 1

(2k − 1)x + (k − 1)y = 2k + 1

Here, a1 = 3, b1 = 1, c1 = 1

a2 = 2k − 1, b2 = k − 1, c2 = 2k + 1

The given system of linear equations has no solution.

Now,

$\frac{3}{2k-1}=\frac{1}{k-1}\phantom{\rule{0ex}{0ex}}⇒3k-3=2k-1\phantom{\rule{0ex}{0ex}}⇒3k-2k=-1+3\phantom{\rule{0ex}{0ex}}⇒k=2$

When k = 2,

$\frac{1}{k-1}=\frac{1}{2-1}=1$ and $\frac{1}{2k+1}=\frac{1}{2×2+1}=\frac{1}{5}$

Thus, for k = 2, $\frac{1}{k-1}\ne \frac{1}{2k+1}$

Hence, the given system of linear equations will have no solution when k = 2.

Question 6:

If the mean of the following distribution is 54, find the value of p:

 Class              : 0−20 20−40 40−60 60−80 80−100 Frequency      : 7 p 10 9 13

Consider the table given below:

 Class Interval Frequency(fi) Class Mark(xi) fi xi 0–20 7 10 70 20–40 p 30 30p 40–60 10 50 500 60–80 9 70 630 80–100 13 90 1170 $\sum _{}^{}{f}_{i}=39+p$ $\sum _{}^{}{f}_{i}{x}_{i}=2370+30p$

It given that the mean of the distribution is 54.

$\therefore \mathrm{Mean}=\frac{\sum _{}^{}{f}_{i}{x}_{i}}{\sum _{}^{}{f}_{i}}\phantom{\rule{0ex}{0ex}}⇒\frac{2370+30p}{39+p}=54\phantom{\rule{0ex}{0ex}}⇒2370+30p=2106+54p\phantom{\rule{0ex}{0ex}}⇒54p-30p=2370-2106$

$⇒24p=264\phantom{\rule{0ex}{0ex}}⇒p=\frac{264}{24}=11$

Hence, the value of p is 11.

Question 7:

The [HCF × LCM] for the numbers 50 and 20 is

(a) 10

(b) 100

(c) 1000

(d) 50

Here we have to find of 50 and 20.

We know the product of HCF and LCM of two numbers a and b is the product of a and b. Therefore of 50 and 20 is as follow

Hence the correct option is

Question 8:

The value of k for which the pair of linear equation 4x + 6y − 1 = 0 and 2x + ky − 7 = 0 represent parallel lines is

(a) k = 3

(b) k = 2

(c) k = 4

(d) k = −2

Given that the pair of linear equation

It is given that the pair of equations represent parallel lines.

Hence the option (a) is correct.

Question 9:

If sin A + sin2 A = 1, then the value of cos2 A + cos4 A is

(a) 2

(b) 1

(c) −2

(d) 0

Here the given date is and

We have to find the value of

We know that the given relation is

……(1)

Now we are going to evaluate the value of

Here we are using the relation

This is same as the equation number (1)

Therefore

Hence the option (b) is correct.

Question 10:

The value of [(sec A + tan A) (1−sin A)] is equal to

(a) tan2 A

(b) sin2 A

(c) cos A

(d) sin A

Here we have to evaluate the value of

Now we are going to solve this

Hence the option (c) is correct.

Question 11:

Find a quadratic polynomial with zeroes

Given that the zeroes of the quadratic polynomial are and .

We have to find the quadratic polynomial from the given zeroes.

Let we assume that,

, then

Therefore the quadratic equation is given by

Hence the desire polynomial is

Question 12:

In Figure 2, ABCD is a parallelogram. Find the values of x and y.

The given parallelogram is

We have to find the value of x and y.

We know that if two diagonals are drawn in a parallelogram, then the intersection points is the mid-point of the two diagonals. Therefore, we have two equations as follow

x + y = 9…… (1)

x − y = 5…… (2)

Now, add the equation (1) and equation (2), we get

Now, we are going to put the value of x in equation (1), we get

Hence the value of x and y are

Question 13:

If sec 4A = cosec (A − 20°), where 4A is an acute angle, find the value of A.

Given that:

, then we have to find the value of A

Since it is given that the angle 4A is acute angle, therefore we can apply the identity

Therefore the given equation can be written as

Hence the value of A is 22°

Question 14:

In Figure 3, PQ || CD and PR || CB. Prove that $\frac{\mathrm{AQ}}{\mathrm{QD}}=\frac{\mathrm{AR}}{\mathrm{RB}}$.

Given that:

and , then we to prove that

The following diagram is given

We can easily see that, in the above figure are similar triangles, and also the are similar triangles.

Now, we have the following properties of similar triangles,

From equation (1) and Equation (2), we get

Hence proved.

Question 15:

In Figure 4, two triangles ABC and DBC are on the same base BC in which ∠A = ∠D = 90°. If CA and BD meet each other at E, show that AE âœ• CE = BE âœ• DE.

Given that, there are two triangles ABC and DBC are on the same base BC in which
A = D = 90°. If CA and BD meet each other at E, then we have to prove that AE × CE = BE × DE

The following figure is given

From the above figure, we can easily see thatare similar triangles, therefore we can use the property of similar triangle.

Hence

Question 16:

Prove that :

Here we have to prove that

Here we take the LHS and by using the trigonometric identities, we have

Hence

Question 17:

Prove that in a triangle if the square of one side is equal to the sum of the squares of the other two side then the angle opposite to the first side is a right angle.

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

Here, we are given a triangle ABC with. We need to prove that.

Now, we construct a triangle PQR right angled at Q such that and .

We have the following diagram.

Now, in, we have

…… (1)

But, it is given that

…… (2)

From equations (1) and (2), we get

…… (3)

From the above analysis in and, we have

Since, therefore

Hence,

Question 18:

Find the mode of the following data:

 Class 0−20 20−40 40−60 60−80 Frequency 15 6 18 10

We have to find the mode of the following distribution,

 Class 0-20 20-40 40-60 60-80 Frequency 15 6 18 10

The class (40-60) has the maximum frequency; therefore this is the modal class.

Lower limit of the modal class

Width of the class interval h = 20

Frequency of the modal class fk = 18

Frequency of the class preceding the modal class fk−1 = 6

Frequency of the class succeeding the modal class fk+1 = 10

Now, we have the following formula to find the value of mode.

Hence the value of mode is 42.

Question 19:

Prove that $\sqrt{7}$ is an irrational number.

We have to prove that is an irrational number

We will prove this by contradiction:

Let be an irrational number such that where x and y are co prime

So’

This means that:

From equation (1) and (2) we see that 7 is a common factor of x and y. This contradicts the fact that x and y have no common factor

Hence is an irrational number.

Question 20:

Use Euclid's division algorithm to find the HCF of 10224 and 9648.

Here we have to find the HCF of the numbers 10224 and 9648 by using Euclid’s division algorithm.

We know that If we divide a by b and r is the remainder and q is the quotient, Euclid’s Lemma says that

A = bq+r, where

And HCF of (a, b) = HCF of (b, r)

Here

Therefore, we have the following procedure,

Now, we apply the division algorithm on 9648 and 576.

Therefore the HCF of 432 and 144 is 144.

Hence the HCF of 10224 and 9648 is 144

Question 21:

If α and β are zeroes of the quadratic polynomial x2 − 6x + a; find the value of 'a' if 3α + 2β = 20.

Given that: α and β are the zeroes of the quadratic polynomial x2 − 6x + a and 3α + 2β = 20, then we have to find the value of a.

We have the following procedure.

Now, we are putting the value α in equation (1), we get

From equation (2), we have

Hence the value of a is −16.

Question 22:

The sum of the numerator and the denominator of a fraction is 8. If 3 is added to both the numerator and the denominator, the fraction becomes $\frac{3}{4}$. Find the fraction.

Here we are assuming that numerator and denominator are x and y respectively, then fraction will be and we have to find the value of .

From the given condition,

x + y = 8…… (1)

If 3 are added in numerator and denominator then fraction will be, from this we have

Now, multiply the equation (1) by 3, we get

3x + 3y = 24…… (3)

Now we take the addition of equations (2) and (3), we get

Put the value of x in the equation (1), we have

Hence the fraction is

Question 23:

Prove that .

Here we have to prove that

First we take LHS and use the identities

Now we take RHS

Hence proved.

Question 24:

In Figure 5, âˆ†ABC is right angled at B, BC = 7 cm and AC − AB = 1 cm. Find the value of cos A − sin A.

It is given that is right angled at B, BC = 7 cm and AC − AB = 1 cm then we have to find the value of

The following diagram is given

ACAB = 1…… (1)

Now, apply the Pythagoras theorem in, we get

Now add the equation (1) and (2), we get

Put the value of in equation (2), we have

Now,

Hence

Question 25:

In Figure 6, P and Q are the midpoints of the sides CA and CB respectively of âˆ†ABC right angled at C. Prove that 4(AQ2 + BP2) = 5 AB2.

In the given figure P and Q are the mid points of AC and BC, the we have to prove that

The following figure is given.

Using Pythagoras theorem in, we get

AB2 = AC2 + BC2…… (1)

Similarly, by using Pythagoras theorem in, we get

AQ2 = CQ2 + AC2…… (2)

BP2 = CP2 + BC2…… (3)

Now adding equation (2) and equation (3), we get

…… (4)

Now multiply equation (4) by 4, we get

Hence proved.

Question 26:

The diagonals of a trapezium ABCD with AB || DC intersect each other at point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.

Given that the diagonals of trapezium ABCD withintersect each other at point O and if, then we have to find the ratio of areas of triangle AOB and triangle COD.

We have the following diagram.

From the above figure, in triangles AOB and COD, we have

Therefore, we can apply the area property of similar triangles.

Hence,

Question 27:

The mean of the following frequency distribution is 50. Find the value of p.

 Classes 0−20 20−40 40−60 60−80 80−100 Frequency 17 28 32 p 19

Given that the mean value of the following distribution is 50.

 Class 0-20 20-40 40-60 60-80 80-100 Frequency 17 28 32 p 19

We have to find the value of p.

To find the value of p, we have following procedure

 Class Mid-value xi Frequency fi fixi 00-20 10 17 170 20-40 30 28 840 40-60 50 32 1600 60-80 70 p 70p 80-100 90 19 1710

Thus, we have

We know that the formula of mean is given by

Hence

Question 28:

Find the missing frequencies in the following frequency distribution table, if N = 100 and median is 32.

 Marks obtained 0−10 10−20 20−30 30−40 40−50 50−60 Total No. of Students 10 ? 25 30 ? 10 100

We have to find the missing term of the following distribution table if and median is 32.

 Marks obtained 0-10 10-20 20-30 30-40 40-50 50-60 Total No. of students 10 ? 25 30 ? 10 100

Suppose the missing term are x and y.

Now we have to find the cumulative frequency as

 Class 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 10 x 25 30 y 10 Cumulative Frequency 10

From the above distribution median class is 30-40 and

Therefore,

Now we are using the following relation

Now we are putting the value of x in the equation (1), we get

Hence the missing terms are

Question 29:

Divide 30x4 + 11x3 − 82x2 − 12x + 48 by (3x2 + 2x − 4) and verify the result by division algorithm.

Here we have to divide by.

According to division algorithm, Dividend = Divisor × Quotient + Remainder.

This can be verified as,

Divisor × Quotient + Remainder

Question 30:

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio.

Given that a in which a line parallel to BC is drawn which meet the other two sides at the point D and E, then we have to prove that

We have the following diagram with some additional construction.

In the above figure, we can see that EF is perpendicular to AB. Therefore EF is the height of the.

Now we are finding the area of and area of

Now take the ratio of the two equation, we have

Similarly, In , we have

But, the area of triangle DBE and triangle DEC are same, therefore equation (4) can be written as

From equation (3) and equation (5), we get

Hence,

Question 31:

Without using trigonometric tables, evaluate the following:

We have to evaluate the following expression

Here we are going to use following identities.

The given expression can be written as

Hence the value of the given expression is zero.

Question 32:

If 2 cos θ − sin θ = x and cos θ − 3 sin θ = y. Prove that 2x2 + y2 − 2xy = 5.

Given that:

Then we have to prove that

First we take the LHS and put the value of x and y, we get

Hence

Question 33:

Check graphically whether the pair of linear equation 4xy − 8 = 0 and 2x − 3y +  6 = 0 is consistent. Also, find the vertices of the triangle formed by these lines with the x-axis.

Here we have to draw the graph between two equations given by

Also we have to find the vertices of the triangle form by the x-axis and these lines.

The first equation can written as follow

Now we are going to find the value of y at different value of x

 x 2 3 y 0 4

Now mark the points (0,-8) and (2, 0) on xy −plane and we will draw a line which passes through these two points.

The second equation can be written as

 x 0 3 y 2 4

Now mark the points (0, 2) and (3, 4) on xy −plane and we will draw a line which passes through these two points.

From the above analysis, we have the following graph

From the above graph the vertices A, B and C are given as follow.

Hence the vertices are

Question 34:

The following table shows the ages of 100 persons of a locality.

 Age (yrs) 0−10 10−20 20−30 30−40 40−50 50−60 60−70 Number of persons 5 15 20 23 17 11 9

Represent the above as the less than type frequency distribution and draw an ogive for the same.

The given frequency distribution is

 Age in years 0-10 10-20 20-30 30-40 40-50 50-60 60-70 No. of Persons 5 15 20 23 17 11 9

We have to change the above distribution as less than type frequency distribution and also we have to draw its ogive.

We have the following procedure to change the given distribution in to less than type distribution.

 Age in years 0-10 10-20 20-30 30-40 40-50 50-60 60-70 No. of persons 5 5+15=20 20+20=40 40+23=63 63+17=80 80+11=91 91+9=100

=

 Age in years 0-10 10-20 20-30 30-40 40-50 50-60 60-70 No. of persons 5 20 40 63 80 91 100

To draw its ogive, take the number of persons on y-axis and age in years on x-axis. Mark the points (10, 5), (20, 20), (30, 40), (40, 63), (50, 80), (60, 91) and (70, 100) on the
xy-plane. Now these points are joined by free hand.

Question 1:

Evaluate:

Here we to evaluate the value of the expression given as follow

Now we are using the following identities

Therefore the given expression can be written as

Hence the value of the given expression is

Question 2:

In Fig. 1, the graph of a polynomial p (x) is shown. The number of zeroes of p (x) is

(a) 4

(b) 1

(c) 2

(d) 3

In the following graph we observe that it intersects x-axis at x = 1. So it has only one zero.

Hence the correct option is

Question 1:

In the following graph we observe that it intersects x-axis at x = 1. So it has only one zero.

Hence the correct option is

In the given rational number , the denominator is in the form of , therefore we get

We know that any rational number which has the denominator in the form of terminate after the largest of m and n places.

Since, the rational number terminates after 4 places.

Question 2:

In the given rational number , the denominator is in the form of , therefore we get

We know that any rational number which has the denominator in the form of terminate after the largest of m and n places.

Since, the rational number terminates after 4 places.

Here we have to find the value of.

By using the trigonometric identities, the given expression can be written as

Now, we use the relation, we get

Hence the correct option is

Question 3:

Here we have to find the value of.

By using the trigonometric identities, the given expression can be written as

Now, we use the relation, we get

Hence the correct option is

Given: the sides of the two similar triangles are in the ratio 4:9, then we have to find the ratio of areas of the two triangles.

We know the following property of the similar triangles.

Hence the correct option is

Question 4:

Given: the sides of the two similar triangles are in the ratio 4:9, then we have to find the ratio of areas of the two triangles.

We know the following property of the similar triangles.

Hence the correct option is

Given that, then we have to evaluate.

By using the given expression can be written as

Now put the value, we get

Hence the correct option is

Question 5:

Given that, then we have to evaluate.

By using the given expression can be written as

Now put the value, we get

Hence the correct option is

Here we have to find the product of HCF and LCM of the smallest prime number and smallest composite number.

We know that any number that can express in product of prime number is called composite number. The smallest prime number is 2 and therefore the smallest composite number is.

HCF of 2 and 4 = 2

LCM of 2 and 4 = 4

Therefore the product of HCF and LCM of 2 and 4 is

Hence the correct option is

Question 6:

Here we have to find the product of HCF and LCM of the smallest prime number and smallest composite number.

We know that any number that can express in product of prime number is called composite number. The smallest prime number is 2 and therefore the smallest composite number is.

HCF of 2 and 4 = 2

LCM of 2 and 4 = 4

Therefore the product of HCF and LCM of 2 and 4 is

Hence the correct option is

We know that, when we draw more than ogive and less than ogive, the Â­x coordinate of intersection of the less than ogive curve and more than ogive curve point gives the median.

Hence the correct option is

Question 7:

We know that, when we draw more than ogive and less than ogive, the Â­x coordinate of intersection of the less than ogive curve and more than ogive curve point gives the median.

Hence the correct option is

The given condition is .

We have to find the value of.

We know that, therefore the given condition can be written as

Hence the correct option is

Question 8:

The given condition is .

We have to find the value of.

We know that, therefore the given condition can be written as

Hence the correct option is

Here we have to find the quadratic equation whose sum of zeroes is 3 and product of zeroes is -2.

Now, we assume that the two zeroes are, then

We know that the general polynomial is

Hence the correct option is

Question 9:

Here we have to find the quadratic equation whose sum of zeroes is 3 and product of zeroes is -2.

Now, we assume that the two zeroes are, then

We know that the general polynomial is

Hence the correct option is

Here we have to find the value of

In the above expression the term exist in the middle and the value ofis zero, therefore the expression can be written as

Hence the correct option is

Question 10:

Here we have to find the value of

In the above expression the term exist in the middle and the value ofis zero, therefore the expression can be written as

Hence the correct option is

The given two equations are

Therefore, the graphical representation of the given pair of equations represents parallel lines.

Hence the correct option is

Question 11:

The given two equations are

Therefore, the graphical representation of the given pair of equations represents parallel lines.

Hence the correct option is

In a given, is the point on BC such that then we have to prove that

We have the following diagram

Since and, therefore the and are similar triangles.

By using the property of similar triangle we have

Hence proved.

Question 12:

In a given, is the point on BC such that then we have to prove that

We have the following diagram

Since and, therefore the and are similar triangles.

By using the property of similar triangle we have

Hence proved.

Here we have to find the HCF of 26, 52 and 91 by using fundamental theorem of arithmetic.

We have the following procedure to find the HCF.

We can see that the common factor is 13.

Hence

Question 13:

Here we have to find the HCF of 26, 52 and 91 by using fundamental theorem of arithmetic.

We have the following procedure to find the HCF.

We can see that the common factor is 13.

Hence

We have the following distribution

 Class 50-55 55-60 60-65 65-70 70-75 75-80 Frequency 2 8 12 24 38 16

We have to change the above distribution as a less than type distribution

We have the following procedure

 Less than 55 60 65 70 75 80 Cumulative Frequency 2 2+8=10 10+12=22 22+24=46 46+38=84 84+16=100

=

 Less than 55 60 65 70 75 80 Cumulative Frequency 2 10 22 46 84 100

Question 14:

We have the following distribution

 Class 50-55 55-60 60-65 65-70 70-75 75-80 Frequency 2 8 12 24 38 16

We have to change the above distribution as a less than type distribution

We have the following procedure

 Less than 55 60 65 70 75 80 Cumulative Frequency 2 2+8=10 10+12=22 22+24=46 46+38=84 84+16=100

=

 Less than 55 60 65 70 75 80 Cumulative Frequency 2 10 22 46 84 100

We have to divide the expression by .

Hence

Question 15:

We have to divide the expression by .

Hence

Here we have the following two equations

The given two equations will have infinitely many solutions if .

Hence

Question 16:

Here we have the following two equations

The given two equations will have infinitely many solutions if .

Hence

Given that is an isosceles triangle with and, then we have to prove that the is a right angled triangle

We have the following diagram

We know that when square of one side is equal to sum of squares

Of other two sides then triangle is called right angled triangle.

Question 17:

Given that is an isosceles triangle with and, then we have to prove that the is a right angled triangle

We have the following diagram

We know that when square of one side is equal to sum of squares

Of other two sides then triangle is called right angled triangle.

Here we have to find mean of the following distribution.

 Class 1-3 3-5 5-7 7-9 9-11 Frequency 7 8 2 2 1

We have the following procedure to obtain mean of the above distribution with

 Class Mid-value Frequency 1-3 2 -4 7 -28 3-5 4 -2 8 -16 5-7 6 0 2 00 7-9 8 2 2 4 9-11 10 4 1 4

We know formula to find mean of a given distribution is

Hence

Question 18:

Here we have to find mean of the following distribution.

 Class 1-3 3-5 5-7 7-9 9-11 Frequency 7 8 2 2 1

We have the following procedure to obtain mean of the above distribution with

 Class Mid-value Frequency 1-3 2 -4 7 -28 3-5 4 -2 8 -16 5-7 6 0 2 00 7-9 8 2 2 4 9-11 10 4 1 4

We know formula to find mean of a given distribution is

Hence

Given:and, then we have to find the value of A and B.

We know that is the value of tan60Â° and is the value of tan30Â°, therefore the above equations can be written as

By adding the equations (1) and (2), we get

Now, put the value of A in the equation (1), we get

Hence

OR

Given that and is acute angle then we have to find the value of A.

Here we are using the identity

Hence the value of A is

Question 19:

Given:and, then we have to find the value of A and B.

We know that is the value of tan60Â° and is the value of tan30Â°, therefore the above equations can be written as

By adding the equations (1) and (2), we get

Now, put the value of A in the equation (1), we get

Hence

OR

Given that and is acute angle then we have to find the value of A.

Here we are using the identity

Hence the value of A is

In the given figure, and XY divides triangular region ABC into two parts equal in area. Then we have to find the ratio

We have the following diagram.

We can see that the triangle ABC and triangle BXY are similar triangles, therefore we have

The ratio of area of triangle BXY and triangle ABC is given by

Hence the ratio

Question 20:

In the given figure, and XY divides triangular region ABC into two parts equal in area. Then we have to find the ratio

We have the following diagram.

We can see that the triangle ABC and triangle BXY are similar triangles, therefore we have

The ratio of area of triangle BXY and triangle ABC is given by

Hence the ratio

Here we have to evaluate the expression given by

Here we are using the following identities.

The given expression can be written as

Hence the value of the given expression is

Question 21:

Here we have to evaluate the expression given by

Here we are using the following identities.

The given expression can be written as

Hence the value of the given expression is

We have to find the mode of the following distribution,

 Class 25-35 35-45 45-55 55-65 65-75 75-85 Frequency 7 31 33 17 11 1

The class (45-55) has the maximum frequency; therefore this is the modal class.

Lower limit of the modal class

Width of the class interval

Frequency of the modal class

Frequency of the class preceding the modal class

Frequency of the class succeeding the modal class

Now, we have the following formula to finding the value of mode.

Hence the value of mode is

Question 22:

We have to find the mode of the following distribution,

 Class 25-35 35-45 45-55 55-65 65-75 75-85 Frequency 7 31 33 17 11 1

The class (45-55) has the maximum frequency; therefore this is the modal class.

Lower limit of the modal class

Width of the class interval

Frequency of the modal class

Frequency of the class preceding the modal class

Frequency of the class succeeding the modal class

Now, we have the following formula to finding the value of mode.

Hence the value of mode is

Here we have to show that square of any positive odd integer is in the form of.

We know that any positive odd integer n is in the form of and, therefore

If, then

Where

If, then

Where

Hence the square of any positive odd integer is in the form of

Question 23:

Here we have to show that square of any positive odd integer is in the form of.

We know that any positive odd integer n is in the form of and, therefore

If, then

Where

If, then

Where

Hence the square of any positive odd integer is in the form of

We have following two equations.

We have to find the values of x and y.

The equation (2) can be written as

Now, subtract the equation (1) from equation (3), we get

Now put the value of x in equation (1), we get

Hence the solution of the given equations is

OR

Given that the sum of digits of two digit number is 12 and the number obtained by interchanging the two digits exceeds to the given number by 18, then we have to find the two digit number.

Let the two digit number whose unit place is x and tenth place is y, then by applying the given two conditions we have

Now add the equation (1) and equation (2), we get

Put the value of x in equation (1), we have

Hence the number is

Question 24:

We have following two equations.

We have to find the values of x and y.

The equation (2) can be written as

Now, subtract the equation (1) from equation (3), we get

Now put the value of x in equation (1), we get

Hence the solution of the given equations is

OR

Given that the sum of digits of two digit number is 12 and the number obtained by interchanging the two digits exceeds to the given number by 18, then we have to find the two digit number.

Let the two digit number whose unit place is x and tenth place is y, then by applying the given two conditions we have

Now add the equation (1) and equation (2), we get

Put the value of x in equation (1), we have

Hence the number is

Here we have to prove that the number is an irrational number.

Let us assume that the number is a rational number, therefore can be written as follow

Where a and b are positive co-prime integer numbers.

is an irrational number andis a rational number (Since a and b are integer).

This contradicts that is an irrational number. Therefore, our assumption is wrong.

Hence is an irrational number.

OR

Here we have to prove that is an irrational number.

Let us assume on the contrary thatis a rational number. Then, there exists positive integers a and b (co-prime numbers) such that

From (1), we have

From (2) and (3), we obtain that 2 is a common factor of a and b. But, this contradicts the fact that a and b have no common factor other than 1. This means our assumption is wrong.

Hence is an irrational number.

Question 25:

Here we have to prove that the number is an irrational number.

Let us assume that the number is a rational number, therefore can be written as follow

Where a and b are positive co-prime integer numbers.

is an irrational number andis a rational number (Since a and b are integer).

This contradicts that is an irrational number. Therefore, our assumption is wrong.

Hence is an irrational number.

OR

Here we have to prove that is an irrational number.

Let us assume on the contrary thatis a rational number. Then, there exists positive integers a and b (co-prime numbers) such that

From (1), we have

From (2) and (3), we obtain that 2 is a common factor of a and b. But, this contradicts the fact that a and b have no common factor other than 1. This means our assumption is wrong.

Hence is an irrational number.

Here we have to prove that

We are taking the LHS, we have

Hence

OR

Here we have to prove that

First we take LHS, we have

Now, we take RHS

Hence

Question 26:

Here we have to prove that

We are taking the LHS, we have

Hence

OR

Here we have to prove that

First we take LHS, we have

Now, we take RHS

Hence

We have an isosceles triangle with and, then we to prove that

We have the following diagram.

In right angle triangle ADB, we have

Hence proved.

Question 27:

We have an isosceles triangle with and, then we to prove that

We have the following diagram.

In right angle triangle ADB, we have

Hence proved.

Here we have to find the zeroes of the polynomial

To find the zeroes, we have to put the given polynomial zero

Now, we have or

Therefore the zeroes of the given polynomial are

The given polynomial is

The sum of zeroes of the above polynomial

And the product of zeroes f the above polynomial

The sum of zeroes

The product of zeroes

Hence verified.

Question 28:

Here we have to find the zeroes of the polynomial

To find the zeroes, we have to put the given polynomial zero

Now, we have or

Therefore the zeroes of the given polynomial are

The given polynomial is

The sum of zeroes of the above polynomial

And the product of zeroes f the above polynomial

The sum of zeroes

The product of zeroes

Hence verified.

The given distribution is

 Class 0-10 10-20 20-30 30-40 40-50 Total Frequency 8 16 36 34 6 100

We have to find the median of the above distribution.

Now we have to find the cumulative frequency as

 Class 0-10 10-20 20-30 30-40 40-50 Frequency 8 16 36 34 6 Cumulative Frequency 8 8+16=24 24+36=60 60+34=94 94+6=100

Sincelies in the cumulative frequency of the interval (20-30), therefore (20-30 belongs to the median class interval

Lower limit of median class interval

The frequency

Width of the class interval

Frequency of the median class

Cumulative frequency preceding median class frequency

Now we are using the following relation

Hence the median of the given distribution is

Question 29:

The given distribution is

 Class 0-10 10-20 20-30 30-40 40-50 Total Frequency 8 16 36 34 6 100

We have to find the median of the above distribution.

Now we have to find the cumulative frequency as

 Class 0-10 10-20 20-30 30-40 40-50 Frequency 8 16 36 34 6 Cumulative Frequency 8 8+16=24 24+36=60 60+34=94 94+6=100

Sincelies in the cumulative frequency of the interval (20-30), therefore (20-30 belongs to the median class interval

Lower limit of median class interval

The frequency

Width of the class interval

Frequency of the median class

Cumulative frequency preceding median class frequency

Now we are using the following relation

Hence the median of the given distribution is

Here we have to show that

First we take the LHS, we have

Hence

OR

Here we have to prove that

Hence

Question 30:

Here we have to show that

First we take the LHS, we have

Hence

OR

Here we have to prove that

Hence

The given polynomial is and two zeroes of this polynomial areand, then we have to find the two other zeroes of.

If two zeroes of the polynomial are and, then

is a factor of

Now we are going to by

Therefore

The other two zeroes are obtained from the polynomial

Hence the zeroes are

Question 31:

The given polynomial is and two zeroes of this polynomial areand, then we have to find the two other zeroes of.

If two zeroes of the polynomial are and, then

is a factor of

Now we are going to by

Therefore

The other two zeroes are obtained from the polynomial

Hence the zeroes are

Here we have to draw the graph between two equations given by

The first equation can written as follow

Now we are going to find the value of y at different values of x

 x 0 1 2 y 6 4 2

Now mark the points (0, 6), (1, 4) and (2, 2) on xy âˆ’plane and we will draw a line which passes through these points.

The second equation can be written as

 x 0 2 3 y 2 6 8

Now mark the points (0, 2), (2, 6) and (3, 8) on xy âˆ’plane and we will draw a line which passes through these points.

From the above analysis we have the following graph

From the above graph, the area of triangle ABC is given by

Hence area is

Question 32:

Here we have to draw the graph between two equations given by

The first equation can written as follow

Now we are going to find the value of y at different values of x

 x 0 1 2 y 6 4 2

Now mark the points (0, 6), (1, 4) and (2, 2) on xy âˆ’plane and we will draw a line which passes through these points.

The second equation can be written as

 x 0 2 3 y 2 6 8

Now mark the points (0, 2), (2, 6) and (3, 8) on xy âˆ’plane and we will draw a line which passes through these points.

From the above analysis we have the following graph

From the above graph, the area of triangle ABC is given by

Hence area is

Given that a in which a line parallel to BC is drawn which meet the other two sides at the point D and E, then we have to prove that

We have the following diagram with some additional construction.

In the above figure, we can see that EF is perpendicular to AB. Therefore EF is the height of the.

Now we are finding the area of and area of

Now take the ratio of the two equation, we have

Similarly, In , we have

But, the area of triangle DBE and triangle DEC are same, therefore equation (4) can be written as

From equation (3) and equation (5), we get

Hence proved.

OR

We have to prove that in a right angle triangle, the square of hypotenuse is the sum of the squares of the other two sides.

We have the following diagram.

Here we assuming that a right angle triangle ABC, right angled at B and BD is perpendicular to the hypotenuse AC.

From the above figure we have, therefore we can apply the property of similar triangle.

Also the, from which we get

Now, we take addition of equation (1) and equation (2), to get

Question 33:

Given that a in which a line parallel to BC is drawn which meet the other two sides at the point D and E, then we have to prove that

We have the following diagram with some additional construction.

In the above figure, we can see that EF is perpendicular to AB. Therefore EF is the height of the.

Now we are finding the area of and area of

Now take the ratio of the two equation, we have

Similarly, In , we have

But, the area of triangle DBE and triangle DEC are same, therefore equation (4) can be written as

From equation (3) and equation (5), we get

Hence proved.

OR

We have to prove that in a right angle triangle, the square of hypotenuse is the sum of the squares of the other two sides.

We have the following diagram.

Here we assuming that a right angle triangle ABC, right angled at B and BD is perpendicular to the hypotenuse AC.

From the above figure we have, therefore we can apply the property of similar triangle.

Also the, from which we get

Now, we take addition of equation (1) and equation (2), to get

Here we have to prove that,

Multiply in both numerator and denominator by, we get

Hence

Question 34:

Here we have to prove that,

Multiply in both numerator and denominator by, we get

Hence

Here we can to change the following distribution in to more than type and also we have to draw its ogive.

 Class 100-120 120-140 140-160 160-180 180-200 Frequency 12 14 8 6 10

We have to find the median of the above distribution.

We have the following procedure to obtain more than type distribution.

 Class Frequency More than type cumulative frequency 100-120 12 12+14+8+6+10=50 120-140 14 14+8+6+10=38 140-160 8 8+6+10=24 160-180 6 6+10=16 180-200 10 10

=

 Class Frequency More than type cumulative frequency 100-120 12 50 120-140 14 38 140-160 8 24 160-180 6 16 180-200 10 10

To draw its ogive, the lower limits of the class interval are marked on x-axis and more than type cumulative frequencies are taken on y-axis. For drawing more than type curve, points (100, 50), (120, 38), (140, 24), (160, 16) and (180, 10) are marked on graph paper and these are joined by free hand to obtain more than type ogive curve. The curve drawn is given by

The median is the x coordinates corresponding to and therefore the median from the above curve is 139.

Hence

Question 3:

In Fig. 2, If DE || BC, then x equals

(a) 6 cm

(b) 8 cm

(c) 10 cm

(d) 12.5 cm

In the given figure we have DE||BC. We have to find x.

In the following given figure DE||BC, so triangle ADE and triangle ABC are similar triangle.

So we have the following relation

Hence the correct option is

Question 4:

If sin 3θ = cos (θ − 6°), where (3θ) and (θ − 6°) are both acute angles, then the value of θ is

(a) 18°

(b) 24°

(c) 36°

(d) 30°

Given that,

…… (1)

We have to find θ

Here the angles (3θ) and (θ − 6) are acute angles and we know that

Therefore we can rewrite the equation (1) as

Therefore option is correct.

Question 5:

Given that tan θ = $\frac{1}{\sqrt{3}}$, the value of $\frac{\mathrm{cos}e{c}^{2}\theta -se{c}^{2}\theta }{\mathrm{cos}e{c}^{2}\theta +se{c}^{2}\theta }$ is

(a) − 1

(b) 1

(c) $\frac{1}{2}$

(d) $-\frac{1}{2}$

Given:

We have to find the value of the following expression

Now, so

perpendicular =1

base =

and

Therefore

and

Put the above two values in the given expression, we have

Hence the correct option is

Question 6:

In Fig. 3, AD = 4 cm, BD = 3 cm and CB = 12 cm, then cot θ equals

(a) $\frac{3}{4}$

(b) $\frac{5}{12}$

(c) $\frac{4}{3}$

(d) $\frac{12}{5}$

In Fig 3, if we have

BD = 3 cm, and

CB = 12 cm, then

cot θ = ?

We have the following diagram

If we apply Pythagoras theorem in triangle ADB, we get

Now,

Hence the correct option is

Question 7:

The decimal expansion of $\frac{147}{120}$ will terminate after how many places of decimal?

(a) 1

(b) 2

(c) 3

(d) will not terminate

The given number is. We have to change the denominator in the form of and the given number terminates after largest of m and n.

Here,

Hence the given number terminates after 3 places.

Question 8:

The pair of linear equations 3x + 2y = 5; 2x − 3y = 7 have

(a) One solution

(b) Two solutions

(c) Many solutions

(d) No solution

The two equations are

3x + 2y = 5 …… (1)

2x − 3y = 7 …… (2)

Here,

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations. That is, there is only one solution.

Hence the correct option is

Question 9:

In an isosceles triangle ABC with AB = AC and BD ⊥ AC. Prove that BD2 − CD2 = 2CD.AD.

Given: âˆ†ABC is an isosceles triangle with AB = AC and BD ⊥ AC.

To prove: BD2 − CD2 = 2CD.AD

$⇒{\mathrm{BD}}^{2}-{\mathrm{CD}}^{2}=2\mathrm{AD}.\mathrm{CD}$                  (Hence Proved)

Question 10:

For a given data with 70 observations the 'less then ogive' and the 'more than ogive' intersect at (20.5, 35). The median of the data is

(a) 20

(b) 35

(c) 70

(d) 20.5

It is given that less than ogive and more than ogive intersects at (20.5, 35), then we have to find the median.

We know that for a given distribution, median is the x coordinates of intersection of less than ogive curve and more than ogive curve. Since the intersection point is (20.5, 35), therefore the median is 20.5

Hence

Hence the correct option is

Question 11:

Determine the value of k so that the following linear equations have no solution:

(3k + 1) x + 3y − 2 = 0

(k2 + 1) x + (k − 2) y − 5 = 0

The given system of equations is

(3k + 1) x + 3y − 2 = 0

(k2 + 1) x + (k − 2) y − 5 = 0

Here, a1 = 3k + 1, b1 = 3, c1 = −2

a2 = k2 + 1, b2 = k − 2, c2 = −5

The given system of equations has no solution.

Now,

$\frac{3k+1}{{k}^{2}+1}=\frac{3}{k-2}\phantom{\rule{0ex}{0ex}}⇒\left(3k+1\right)\left(k-2\right)=3\left({k}^{2}+1\right)\phantom{\rule{0ex}{0ex}}⇒3{k}^{2}-5k-2=3{k}^{2}+3\phantom{\rule{0ex}{0ex}}⇒-5k=5$
$⇒k=-1$

When k = −1,

$\frac{3}{k-2}=\frac{3}{-1-2}=\frac{3}{-3}=-1$

Thus, for k = −1, $\frac{3}{k-2}\ne \frac{2}{5}$

Hence, the given system of equations has no solution when k = −1.

Question 12:

Can (x − 2) be the remainder on division of a polynomial p(x) by (2x + 3)? Justify your answer.

If we divide any polynomial by another polynomial, then the degree of divisor is always greater than the degree of remainder.

In the given question the degrees of remainder (x−2) and divisor (2x+3) are same

Therefore if we divide the polynomial P(x) by (2x+3), then (x−2) cannot be the remainder

Question 13:

In Fig. 4, ABCD is a rectangle. Find the values of x and y.

The following diagram is given

The given diagram is rectangle and we know that in rectangle the face to face sides are same. Therefore

…… (1)

…… (2)

Now we are adding the equations (1) and (2) to have

Put the value of x in equation (1), we get

Hence, x = 10 and y = 2

Therefore the values of x and y are

Question 14:

If 7 sin2 θ + 3 cos2θ = 4, show that tan θ = $\frac{1}{\sqrt{3}}$

Given: and we have to prove that

We can write the given expression as

Therefore,

Hence proved

Question 15:

In Fig. 5, DE || AC and DF || AE. Prove that $\frac{EF}{BF}=\frac{EC}{BE}$

Given that:

If DE||AC and DF||AE, then we have to prove that

The following given figure is

We can easily see that in the given figure the triangle BDF and triangle BAE are similar triangles and also the triangle BDE and triangle BAC are similar triangles. Now we are applying the theorem of similar triangle in triangle BDF and triangle BAE, we get

…… (1)

Similarly in triangle BDE and triangle BAC, we get

…… (2)

Now we are comparing the equation (1) and (2), we have

Now take the reciprocal of the above equation, we have

Hence we have proved that

Question 16:

In Fig. 6, AD $\perp$ BC and BD = $\frac{1}{3}$ CD. Prove that 2CA2 = 2AB2 + BC2

In the given figure, we have

and, then we have to prove

The following given diagram is

Now, suppose the value of BD is x, then

…… (1)

And in triangle ADB, we have

…… (2)

Now add the equation (1) and equation (2), we get

…… (3)

Now we are putting the values of BD and CD, we get

Hence proved.

Question 17:

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

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

Here, we are given a triangle ABC with. We need to prove that

.

Now, we construct a triangle PQR right angled at Q such that and .

We have the following diagram.

Now, in, we have

…… (1)

But, it is given that

…… (2)

From equations (1) and (2), we get

…… (3)

From the above analysis in and, we have

Since, therefore

Hence,

Question 18:

Find the mode of the following distribution of marks obtained by 80 students:

 Marks obtained 0-10 10-20 20-30 30-40 40-50 Number of students 6 10 12 32 20

We have the following distribution

 Marks Obtained 0-10 10-20 20-30 30-40 40-50 Number of Students 6 10 12 32 20

We have to find the mode of the above distribution

From the given distribution, we can easily see that the class (20-30) has the maximum frequency i.e. 32.

Lower limit of the modal class

Class interval

Frequency of the modal class

Frequency of the class preceding the modal class

Frequency of the class succeeding the modal class

Therefore the mode value of the given distribution is 36.25

Question 19:

Show that any positive odd integers is of the form 4q+1 or 4q+3 where q is a positive integer.

Here we have to prove that for any positive integer q, the positive odd integer will be form of 4q+1 or 4q+3.

Now let us suppose that the positive odd integer is a then by Euclid’s division rule

a = 4q + r ……(1 )

Where q (quotient) and r (remainder) are positive integers, and

We are putting the values of r from 0 to 3 in equation (1), we get

But we can easily see that 4q and 4q+2 are both even numbers.

Therefore for any positive value q, the positive odd integer will be the form of 4q+1 and 4q+3.

Question 20:

Prove that is irrational.

Here we have to prove that the number is an irrational number.

Now let us suppose that, where is a rational number, then

As is rational number, therefore form equation (1), so is and is also a rational number which is a contradiction as is an irrational number.

Therefore is an irrational number.

Question 21:

A person can row a boat at the rate of 5 km/hour in still water. He takes thrice as much time in going 40 km upstream as in going 40 km downstream. Find the speed of the stream.

Given that,

Speed of boat in still water = 5 km/hour.

Distance covered by boat = 40 km

Now we are assuming that the speed of the stream is, then

Speed of the boat rowing upstream =, and

Speed of the boat rowing downstream =

According to the given condition, time taken to cover in upstream is three times the time taken to cover in downstream, therefore

Hence the speed of stream is 2.5 km/hour.

Question 22:

If , are zeroes of the polynomial x2 − 2x − 15, then form a quadratic polynomial whose zeroes are (2) and (2).

The given polynomial is andand are the zeroes of the polynomial , then we have to find another polynomial whose zeroes are,

Now we comparing the given polynomial with, we get

We know that

The polynomial whose zeroes are, is

Hence the polynomial is

Question 23:

Prove that (cosec θ − sin θ) (sec θ − cos θ) = .

Here we have to prove that

Now using the identity, we get

Hence proved.

Question 24:

If cos θ + sin θ = $\sqrt{2}$ cos θ, show that cos θ − sin θ = $\sqrt{2}$ sin θ.

Given that if, then we have to prove that

We have,

Hence proved.

Question 25:

In Fig. 7, AB $\perp$ BC, FG $\perp$ BC and DE $\perp$ AC. Prove that Δ ADE $~$ ΔGCF

In the given figure, we have

Then we have to prove that

The following diagram is given

In, we have

…… (1)

In, we have

…… (2)

From equation (1) and equation (2), we have

Similarly, we have

Since and are equiangular, therefore

Question 26:

In Fig. 8, and ΔDBC are on the same base BC and on opposite sides of BC ad Q is the point of intersection of AD and BC.

Prove that

We have given the diagram in which

We have to prove that

Firstly, we draw a line from A perpendicular to line BC and after that we draw a line from D perpendicular to BC.

From the above figure we can easily see that theand are similar

Therefore by the properties of similar triangle, we have

…… (1)

Now,

From equation (1), we get

Hence proved.

Question 27:

Find the mean of the following frequency distribution, using step-deviation method:

 Class 0-10 10-20 20-30 30-40 40-50 Frequency 7 12 13 10 8

The following frequency distribution is given

 Class 0-10 10-20 20-30 30-40 40-50 Frequency 7 12 13 10 8

We have to find the mean of the above frequency distribution using step deviation method.

By using step deviation method, we have

 Class Frequency Mid-value 00-10 7 5 10-20 12 15 20-30 13 25 0 0 30-40 10 35 1 10 40-50 8 45 2 16

From the above distribution, we have

We know the mean of a given distribution is given by

Hence the mean of the given distribution is 25.

Question 28:

Find the median of the following data

 Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 Frequency 5 3 4 3 3 4 7 9 7 8

The given distribution is

 Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 Frequency 5 3 4 3 3 4 7 9 7 8

We have to find the median of the above distribution.

Now we have to find the cumulative frequency as

 Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 Frequency 5 3 4 3 3 4 7 9 7 8 Cumulative Frequency 5 8 12 15 18 22 29 38 45 53

From the above distribution median class is 60-70 and

Now we are using the following relation

Hence the median of the given distribution is 66.43

Question 29:

Find other zeroes of the polynomial p(x) = 2x4 + 7x3 − 19x2 − 14x + 30 if two of its zeroes are  $\sqrt{2}$ and $\sqrt{-2}$.

The given polynomial is and two zeroes of this polynomial areand, then we have to find the two other zeroes of

If two zeroes of the polynomial are and, then

is a factor of

Therefore can be written as

The other two zeroes are obtained from the polynomial

Hence the other zeroes are and

Question 30:

Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Here we have to prove that the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

We have two triangles andin which

In the given figure AD is perpendicular to BC and PM is perpendicular to QR

The areas of triangle ABC and triangle PQR are given by

…… (1)

Since are same, therefore

…… (2)

By applying the property of similar triangles, we have

…… (3)

From (2) and (3), we get

…… (4)

From (1) and (4), we have

Hence

Question 31:

Prove that:

Here we have to prove that,

Firs we take the left hand side of the given equation

Now we are using the trigonometric identity

Therefore,

Hence, .

Question 32:

If sec θ + tan θ = p, prove that sin θ = $\frac{{p}^{2}-1}{{p}^{2}+1}$

Given that:

, then we have to prove that

We can rewrite the given data as

Now we take the right hand side

Now we are putting the value of p in the above expression, we get

Hence

Question 33:

Draw the graphs of following equations:

2xy = 1 and x + 2y = 13

(i) find the solution of the equations from the graph.

(ii) shade the triangular region formed by the lines and the y-axis.

Here we have to draw the graph between two equations given by

…… (1)

…… (2)

Also we have to find the solution of the given equations.

The first equation can written as follow

…… (3)

Now we are going to find the value of y at different value of x

 x 0 1 2 y − 1 1 3

Now mark the points (0,-1), (1,1) and (2,3) on xy-plane and we will draw a line which pass through these points.

The second equation can be written as

…… (4)

 x 0 2 3 y 6.5 5.5 5

Now mark the points (0, 6.5), (2, 5.5) and (3, 5) on xy-plane and we will draw a line which pass through these points.

From the above analysis we have the following graph

Form the above graph, the intersection point of the two lines is the solution.

And also the triangular region is shaded in the figure.

Question 34:

The following table gives the production yield per hectare of wheat of 100 farms of a village.

 Production yield in kg/hectare 50-55 55-60 60-65 65-70 70-75 75-80 Number of frames 2 8 12 24 38 16

Change the above distribution to more than type distribution and draw its ogive.

We have the following distribution

 Production yield 50-55 55-60 60-65 65-70 70-75 75-80 Number of frames 2 8 12 24 38 16

We have to change the above distribution in to the more than type distribution and we have to draw its ogive.

We have the following procedure to find the more than type distribution

 Class Frequency Cumulative frequency 50-55 2 2+8+12+24+38+16=100 55-60 8 8+12+24+38+16=98 60-65 12 12+24+38+16=90 65-70 24 24+38+16=78 70-75 38 38+16=54 75-80 16 16

To draw its ogive, take the number of frames on y-axis and production yield on x-axis. Mark the points (50,100), (55, 98), (60-90), (65, 78), (70, 54), and (75, 16) on the
xy-plane. Now these points are joined by free hand.

Question 1:

Which of the following equations has the sum of its roots as 3?

(a) x2 + 3x − 5 = 0

(b) −x2 + 3x + 3 = 0

(c)

(d) 3x2 − 3x − 3 = 0

(a) x2 + 3x − 5 = 0 (b) −x2 + 3x + 3 = 0

(c) (d) 3x2 − 3x − 3 = 0

We are to find out which of the above equations has sum of roots = 3.

The sum of the roots of the quadratic equation ax2 + bx + c = 0 is given by .

For given equation (a)

a = 1, b = 3, c = −5

For given equation (b)

a = −1, b = 3, c = 3

For given equation (c)

For given equation (d)

a = 3, b = −3, c = −3

Hence option (b) is correct.

Question 2:

The sum of first five multiples of 3 is:

(a) 45

(b) 65

(c) 75

(d) 90

We have to find the sum of first five multiples of 3

First five multiples of 3 are 3, 6, 9, 12 and 15

It forms an Arithmetic Progression (A.P) with

First term

We know that the sum of the n terms of an arithmetic progression

Here n = 5

Therefore

Hence option (a) is correct.

Question 3:

If radii of the two concentric circles are 15 cm and 17 cm, then the length of each chord of one circle which is tangent to other is:

(a) 8 cm

(b) 16 cm

(c) 30 cm

(d) 17 cm

We are given that radii of two concentric circles are 15 cm and 17 cm

We have to find the length of each chord of one circle which is tangent to other

Let A be the centre of the two concentric circles

Let BC be the chord of bigger circle tangent to smaller at F

Therefore AF is the radius of smaller circle

AB is the radius of bigger circle

We know that radius of a circle is perpendicular to its tangent

Therefore

We know that a perpendicular from centre of circle to chord of circle bisects the chord

Therefore BF = FC = 8 cm

Hence option (b) is correct.

Question 4:

In Fig. 1, PQ and PR are tangents to the circle with centre O such that ∠QPR = 50°,

(a) 25°

(b) 30°

(c) 40°

(d) 50°

We are given the below figure in which

PQ and PR are tangents to the circle with centre O and

We have to find

PQ is the tangent to circle

Therefore [Since Radius of a circle is perpendicular to tangent]

Similarly

We know that sum of angles of a quadrilateral

Hence option (a) is correct.

Question 5:

Two tangents making an angle of 120° with each other are drawn to a circle of radius 6 cm, then the length of each tangent is equal to

(a)

(b) 6

(c)

(d) 2

We are given two tangents to a circle making an angle of 120° with each other. The radius of circle is 6 cm

We have to find the length of each tangent.

Let O be the center of the given circle

Let AB and AC be the two tangents to the given circle drawn from point A

Therefore

Now OB and OC represent the radii of the circle

Therefore

[Since Radius of a circle is perpendicular to tangent]

Therefore by SSS rule

Hence [Corresponding angles of congruent triangles]

In right

Hence option (d) is correct.

Question 6:

To draw a pair f tangents to a circle which are inclined to each other at an angle of 100°, It is required to draw tangents at end points of those two radii of the circle, the angle between which should be:

(a) 100°

(b) 50°

(c) 80°

(d) 200°

Given a pair of tangents to a circle inclined to each other at angle of 100°

We have to find the angle between two radii of circle joining the end points of tangents that is we have to find in below figure.

Let O be the center of the given circle

Let AB and AC be the two tangents to the given circle drawn from point A

Therefore BAC = 100°

Now OB and OC represent the radii of the circle

Therefore

[Since Radius of a circle is perpendicular to tangent]

We know that sum of angles of a quadrilateral = 360°

Hence Option (c) is correct.

Question 7:

The height of a cone is 60 cm. A small cone is cut off at the top by a plane parallel to the base and its volume is $\frac{1}{64}th$ the volume of original cone. The height from the base at which the section is made is:

It is given that the height of a cone = 60 cm

The volume of a small cone cut on the top by plane parallel to base of given cone
= volume of given cone

We have to find the height from the base at which section is made.

Let R be the radius of base of given cone

Let H be the height of given cone

H = 60 cm

Let r be the radius of base of small cone

Let h be the height of small cone

In ΔAPC and ΔAQE

PC || QE

Therefore

Hence ΔAPC ~ ΔAQE

…… (1)

[From (1)]

Hence Option (c) is correct.

Question 8:

If the circumference of a circle is equal to the perimeter of a square then the ratio their areas is:

(a) 22 : 7

(b) 14 :11

(c) 7 :22

(d) 7 :11

It is given that circumference of a circle = perimeter of a square

We have to find the ratio of their areas

Let the radius of circle = r

Let the circumference of circle =

Let the area of circle =

Let the side of the square = b

Let the perimeter of square =

Let the area of square =

Therefore

…… (1)

Area of Circle

[From Equation (1)]

Hence Option (b) is correct.

Question 9:

21 glass spheres, each of radius 2 cm are packed in a cuboidal box of internal dimensions 16 cm × 8 cm × 8 cm and, then the box is filled with water. Find the volume of water filled in the box.

It is given that 21 glass spheres each of radius 2 cm are packed in a cuboidal box of inner dimensions and then filled with water.

We have to find the volume of water.

Volume of a glass sphere

Length of cuboidal box

Height of cuboidal box

Volume of water = Volume of cuboidal box − Volume of glass spheres

Question 10:

Find two consecutive  odd positive integers, sum of whose squares is 290.

We have to find two consecutive integers sum of whose squares is 290.

Let the two consecutive integers be x and x+2

According to the question

Dividing both sides by 2

Therefore two consecutive integers are

Question 11:

Find the roots of the following quadratic equation:

$\frac{2}{5}{x}^{2}-x-\frac{3}{5}=0.$

It is given that:

We have to find the roots of above equation.

Multiplying both sides by 5

2x2 − 5x − 3 = 20

2x2 − 6x + x − 3 = 0

2x(x − 3) + 1 (x − 3) = 0

(x − 3) (2x + 1) = 0

x = 3, x =

Question 12:

If the numbers x − 2, 4x − 1 and 5x + 2 are in A.P. find the value of x.

It is given that the numbers

We have to find the value of x

We know that if x, y and z are in A.P, then

Therefore for the given numbers

Hence

Question 13:

Two tangents PA and PB are drawn from an external point P to a circle with centre O. Prove that AOBP is a cyclic quadrilateral.

We are given two tangents PA and PB drawn to a circle with centre O from external point P

We are to prove that quadrilateral AOBP is cyclic

We know that tangent at a point to a circle is perpendicular to the radius through that point.

Therefore from figure

That is

[Sum of angles of a quadrilateral = 360°]

We know that the sum of opposite angles of cyclic quadrilateral = 180°

Therefore from (1) and (2)

Question 14:

In Fig. 2, a circle of radius 7 cm is inscribed in a square. Find the area of the shaded region

It is given that a circle of radius 7 cm is inscribed in a square

We have to find the area of shaded region shown in figure

We are given the following figure

Let the side of the square = a cm

Since the circle in inscribed in the square

Diameter of the circle = a cm

Given that radius of circle = 7 cm

Therefore

Question 15:

How many spherical lead shots each having diameter 3 cm can be made from a cuboidal lead solid of dimensions 9 cm × 11 cm × 12 cm?

Given a cuboidal lead solid with dimensions 9 cm × 11 cm × 12 cm

We have to find the number of spherical lead shots each having a diameter

Let the length of cuboidal lead solid L = 9 cm

Let the length of cuboidal lead solid H = 9 cm

Let the number of spherical lead shots = x

Volume of a spherical lead shot =

Volume of the cubical lead shot =

Question 16:

Point P(5, −3) is one of the two points of trisection of the line segment joining the points A (7, −2) and B (1, − 5) near to A. Find the coordinates of the other point of trisection.

We are given a line segment joining points A (7, −2) and B (1, −5)

P (5, −3) is one of the two points of trisection of line segment AB

P is near to A

We are to find the coordinates of other points of trisection

Let the other point of trisection is Q

Therefore

AP = PQ = QB

That is Q is the mid point of line segment PB

We know that the coordinates of mid point of line segment with coordinates of end points

Question 17:

Show that the point P (−4, 2) lies on the line segment joining the points A (−4, 6) and B (−4, −6).

We have to show that point P (−4, 2) lies on line segment AB with points A (−4, 6) and

B (−4, −6)

If P (−4, 2) lies on the line segment joining A (−4, 6) and B (−4, −6), then the three points

must be collinear.

Let the three points be not collinear and form a triangle PAB

We know that area of a triangle with coordinates of vertices

Since area of the triangle is 0, no triangle exists.

Therefore points P (−4, 2), A (−4, 6) and B (−4, −6) are collinear.

Question 18:

Two dice are thrown at the same time. Find the probability of getting different numbers on both dice.

It is given that two dice are thrown at the same time.

We have to find the probability of getting different numbers on both dice.

Total number of possible choices in rolling a dice = 6

Total number of possible choices in rolling two dice [Using multiplication rule]

Probability of getting same number if two dice are thrown

Probability of getting different number if two dice are thrown

OR

It is given that a coin is tossed two times.

Therefore, sample space is given by,

{HH, HT, TH, TT}

Let E be the event of getting two heads. I.e., E = {HH}

Then, probability of getting atmost one head is given by,

P (E) = P (HT or TH or TT) = 1 − P (E) = 1 − P (HH) =

Question 19:

A natural number, when increased by 12, becomes equal to 160 times its reciprocal. Find the number.

It is given that a number when increased by 12 becomes 160 times its reciprocal.

We have to find the number.

Let the number be x

According to the question

Question 20:

Find the sum of the integers between 100 and 200 that are divisible by 9.

We have to find the sum of integers between 100 and 200 that are divisible by 9.

Integers divisible by 9 between 100 and 200 are

108, 117, 126, … 198

The above equation forms an Arithmetic Progression (A.P)

Let there be n such integers in the above A.P

We know that the nth term of an A.P

Where = First term of A.P

= Common difference of successive members

Therefore total number of integers in the A.P = 11

We know that the sum of the n terms of an arithmetic progression

Question 21:

In Fig. 3, two tangents PQ are PR are drawn to a circle with centre O from an external point P. Prove that ∠QPR = 2 ∠OQR.

Given a figure as shown. We have to prove that

Join OR

We know that sum of opposite angles of a cyclic quadrilateral

…… (1)

In

[Since OQ, OR are radii of the circle]

Therefore is an isosceles triangle.

Hence [Angles opposite to equal side of isosceles triangle]

Now, [since,]

[From (1)]

Hence proved.

Question 22:

Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are $\frac{3}{4}$ time the corresponding sides of ΔABC.

We have to draw a triangle

Then we have to construct a similar triangle with side of

Steps of construction

1. Draw a line segment

2. At B draw so that a ray BX is made

3. Keep compass at B and mark an arc of 5 cm at ray BX and name that point as A

4. Join AC to make

5. Draw a ray making an acute angle with BC.

6. Mark 4 points, B1, B2, B3, and B4 along BY such that BB1 = B1B2 = B2B3 = B3B4

7. Join CB4

8. Through the point B3, draw a line parallel to B4C by making an angle equal to ∠BB4C, intersecting BC at C′.

9. Through the point C′, draw a line parallel to AC, intersecting BA at A′.

Thus, ΔA′BC′ is the required triangle.

Question 23:

In Fig. 4, OABC is a square inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of shaded region [Use π = 3.14]

It is given that OABC is a square in the quadrant OPBQ.OA is 20 cm.

We have to find the area of the shaded region.

OABC is a square , therefore sides of the OABC must be equal

Hence OA, AB, BC, OC = 20 cm.

Join the points O and B to form a line segment OB.

Since OB is the diagonal of OABC, is a right angled triangle.

Applying Pythagoras Theorem in

It can be seen from the figure that Quadrant OPBQ is of a circle with radius OB( r )

Therefore area of OPBQ

Now, area of the shaded region

Question 24:

A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter 'l' of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

It is given that a hemisphere is cut from a cubical box with edge l such that

diameter of hemisphere is also l.

We have to find the surface area of the remaining solid.

Surface area of the cubical box with side l

Let r be the radius of hemisphere

Surface area of the hemisphere

( since )

Question 25:

A tower stands vertically on the ground. From a point on the ground which is 20 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.

Given a tower at the ground such that at a distance of 20 m away from foot of tower,

the top of tower makes an angle of with the ground.

We have to find the height of the tower.

Let the tower be AB

Height of the tower = h

Distance of the tower from point C where top of the tower makes an angle of = 20 m

Since B is the foot of the tower, CB = 20 m

Height of the tower

Question 26:

Prove that the points A (4, 3), B (6, 4), C (5, −6) and D (3, −7) in that order are the vertices of a parallelogram.

Given points A (4, 3), B (6, 4), C (5, −6) and D (3, −7).

We have to prove that these points form vertices of a parallelogram.

The above four points will form 4 line segments.

We know that length of a line segment having coordinates

Therefore

It can be seen that AB = CD , BC = AD

Since opposite sides are equal, ABCD is a parallelogram

Hence proved.

Question 27:

The points A (2, 9), B (a, 5), C (5, 5) are the vertices of a triangle ABC right angled at B. Find the value of 'a' and hence the area of ΔABC.

It is given that ABC is a right angled triangle..Vertices of triangle are

A( 2,9 ), B( a,5 ), C( 5, 5 ).

We have to find the value of a and area of the triangle.

is a right angled triangle.

We know that length of a line with coordinates of end points

Hence in

…… (1)

…… (2)

Therefore, applying Pythagoras Theorem to right

Putting a = 2 in (1) and (2)

Question 28:

Cards with numbers 2 to 101 are placed in a box. A card is selected at random from the box. Find the probability that the card which is selected has a number which is a perfect square.

It is given that cards with numbers 2 to 101 are placed in a box. A card is picked at random.

We have to find the probability that the card selected has a number which is a perfect square.

Perfect squares between 2 and 101 are 4, 9, 16, 25, 36, 48, 64, 81, 100

Total number of perfect squares

Total number of cards

Probability (perfect square)

Hence probability for the selected card number to be a perfect square

Question 29:

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

It is given that a train travels a distance of 63 km with a certain average speed and 72 km with a speed which is 6 km/hr more than the original average speed.

Time taken for the whole journey is 3hr.

We have to find out the original average speed of the train.

Let the original average speed of the train = x km/hr

While covering the distance of 72 km, the speed of train

Total time taken by the train

Therefore

Dividing both sides by3

Speed cannot be negative

Question 30:

A sum of Rs. 1400 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs. 40 less than the preceding price, find the value of each of the prizes.

It is given that total prize money is Rs 1400 /-. There are a total of 7 prizes distributed in a way that each prize is less than the previous prize by Rs 40/-

We have to find the value of the prizes.

Let a is the value of a prize

The difference between the consecutive prizes d

Total number of prizes n

Now it can be seen that the value of prizes forms an Arithmetic Progression (A.P)

Therefore

We know that for an A.P

Substituting the values

Therefore the value of prizes

Question 31:

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

We have to prove that the lengths of tangents drawn from an external point to a circle are equal.

Draw a circle with centre O and tangents PA and PB, where P is the external point and A and B are the points of contact of the tangents.
Join OA, OB and OP.
Now

Hence proved.

Question 32:

A well of diameter 3 m ad 14 m deep is dug. The earth, taken out of it, has been evenly spread all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

Given a well with diameter 3 m and height 14 m. The earth dug out from well is used to make a circular embankment of 4m width.

We have to find the height of the embankment.

Let R be the radius of well

Let H be the height of well

Let r be the radius of embankment

Let h be the height of embankment

H

Width of the circular embankment

According to the question

Volume of the earth dug = Volume of the circular embankment

Therefore,

=

Question 33:

The slant height of the frustum of a cone is 4 cm and the circumferences of its circular ends are 18 cm and 6 cm. Find curved surface area of the frustum.

It is given that slant height of frustum of a cone is 4 cm. Circumferences of its ends are

18 cm and 6 cm.

We have to find the curved surface area of the frustum.

Let l be the slant height

Let be the radii of two circular ends of the cone

Circumference of one end

Circumference of other end

Now,

Curved surface area of frustum

(since )

Question 34:

From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find height of the tower.

It is given that tower is placed at a 20 m high building. The top and bottom of the tower makes an angle of   respectively with the ground.

We have to find the height of the tower.

Let DB is the tower

Height of the building = 20 m

Height of the tower = x m

According to the figure,

…… (1)

…… (2)

Since (1) = (2)

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