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Page No SA1.15:
Question 1:
Which of the following numbers has terminating decimal expansion?
(a)
(b)
(c)
(d)
Answer:
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).
Page No SA1.15:
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
Answer:
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 = 16x − px
So, to find the value of p we put the reminder is equal to zero.
Therefore
Hence option (d) is correct.
Page No SA1.15:
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°
Answer:
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
Page No SA1.15:
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
Answer:
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)
Page No SA1.15:
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
Answer:
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,
When k = 2,
and
Thus, for k = 2,
Hence, the given system of linear equations will have no solution when k = 2.
Page No SA1.16:
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 |
Answer:
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 |
It given that the mean of the distribution is 54.
Hence, the value of p is 11.
Page No SA1.16:
Question 7:
The [HCF × LCM] for the numbers 50 and 20 is
(a) 10
(b) 100
(c) 1000
(d) 50
Answer:
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
Page No SA1.16:
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
Answer:
Given that the pair of linear equation
It is given that the pair of equations represent parallel lines.
Hence the option (a) is correct.
Page No SA1.16:
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
Answer:
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.
Page No SA1.16:
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
Answer:
Here we have to evaluate the value of
Now we are going to solve this
Hence the option (c) is correct.
Page No SA1.16:
Question 11:
Find a quadratic polynomial with zeroes
Answer:
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
Page No SA1.16:
Question 12:
In Figure 2, ABCD is a parallelogram. Find the values of x and y.
Answer:
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
Page No SA1.16:
Question 13:
If sec 4A = cosec (A − 20°), where 4A is an acute angle, find the value of A.
Answer:
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°
Page No SA1.17:
Question 14:
In Figure 3, PQ || CD and PR || CB. Prove that .
Answer:
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.
Page No SA1.17:
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 â CE = BE â DE.
Answer:
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
Page No SA1.17:
Question 16:
Prove that :
Answer:
Here we have to prove that
Here we take the LHS and by using the trigonometric identities, we have
Hence
Page No SA1.17:
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.
Answer:
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,
Page No SA1.17:
Question 18:
Find the mode of the following data:
Class | 0−20 | 20−40 | 40−60 | 60−80 |
Frequency | 15 | 6 | 18 | 10 |
Answer:
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.
Page No SA1.17:
Question 19:
Prove that is an irrational number.
Answer:
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.
Page No SA1.17:
Question 20:
Use Euclid's division algorithm to find the HCF of 10224 and 9648.
Answer:
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
Page No SA1.17:
Question 21:
If α and β are zeroes of the quadratic polynomial x2 − 6x + a; find the value of 'a' if 3α + 2β = 20.
Answer:
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.
Page No SA1.18:
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 . Find the fraction.
Answer:
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
Page No SA1.18:
Question 23:
Prove that .
Answer:
Here we have to prove that
First we take LHS and use the identities
Now we take RHS
Hence proved.
Page No SA1.18:
Question 24:
In Figure 5, âABC is right angled at B, BC = 7 cm and AC − AB = 1 cm. Find the value of cos A − sin A.
Answer:
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
AC − AB = 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
Page No SA1.18:
Question 25:
In Figure 6, P and Q are the midpoints of the sides CA and CB respectively of âABC right angled at C. Prove that 4(AQ2 + BP2) = 5 AB2.
Answer:
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.
Page No SA1.18:
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.
Answer:
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,
Page No SA1.19:
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 |
Answer:
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
Page No SA1.19:
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 |
Answer:
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
Page No SA1.19:
Question 29:
Divide 30x4 + 11x3 − 82x2 − 12x + 48 by (3x2 + 2x − 4) and verify the result by division algorithm.
Answer:
Here we have to divide by.
According to division algorithm, Dividend = Divisor × Quotient + Remainder.
This can be verified as,
Divisor × Quotient + Remainder
Page No SA1.19:
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.
Answer:
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,
Page No SA1.19:
Question 31:
Without using trigonometric tables, evaluate the following:
Answer:
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.
Page No SA1.19:
Question 32:
If 2 cos θ − sin θ = x and cos θ − 3 sin θ = y. Prove that 2x2 + y2 − 2xy = 5.
Answer:
Given that:
Then we have to prove that
First we take the LHS and put the value of x and y, we get
Hence
Page No SA1.19:
Question 33:
Check graphically whether the pair of linear equation 4x − y − 8 = 0 and 2x − 3y + 6 = 0 is consistent. Also, find the vertices of the triangle formed by these lines with the x-axis.
Answer:
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
Page No SA1.19:
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.
Answer:
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.
Page No SA1.2:
Question 1:
Evaluate:
Answer:
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
Page No SA1.2:
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
Answer:
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
Page No SA1.21:
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
Answer:
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.
Page No SA1.21:
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.
Answer:
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
Page No SA1.21:
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
Answer:
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
Page No SA1.21:
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
Answer:
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
Page No SA1.21:
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
Answer:
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
Page No SA1.21:
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
Answer:
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
Page No SA1.21:
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
Answer:
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
Page No SA1.21:
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
Answer:
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
Page No SA1.21:
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
Answer:
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
Page No SA1.21:
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
Answer:
The given two equations are
Therefore, the graphical representation of the given pair of equations represents parallel lines.
Hence the correct option is
Page No SA1.22:
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
Answer:
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.
Page No SA1.22:
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.
Answer:
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
Page No SA1.22:
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
Answer:
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 |
Page No SA1.22:
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 |
Answer:
We have to divide the expression by .
Hence
Page No SA1.22:
Question 15:
We have to divide the expression by .
Hence
Answer:
Here we have the following two equations
The given two equations will have infinitely many solutions if .
Hence
Page No SA1.22:
Question 16:
Here we have the following two equations
The given two equations will have infinitely many solutions if .
Hence
Answer:
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.
Page No SA1.22:
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.
Answer:
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
Page No SA1.22:
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
Answer:
Given:and, then we have to find the value of A and B.
We know that is the value of tan60° and is the value of tan30°, 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
Page No SA1.22:
Question 19:
Given:and, then we have to find the value of A and B.
We know that is the value of tan60° and is the value of tan30°, 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
Answer:
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
Page No SA1.23:
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
Answer:
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
Page No SA1.23:
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
Answer:
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
Page No SA1.23:
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
Answer:
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
Page No SA1.23:
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
Answer:
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
Page No SA1.23:
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
Answer:
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.
Page No SA1.23:
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.
Answer:
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
Page No SA1.23:
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
Answer:
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.
Page No SA1.23:
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.
Answer:
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.
Page No SA1.23:
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.
Answer:
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
Page No SA1.23:
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
Answer:
Here we have to show that
First we take the LHS, we have
Hence
OR
Here we have to prove that
Hence
Page No SA1.24:
Question 30:
Here we have to show that
First we take the LHS, we have
Hence
OR
Here we have to prove that
Hence
Answer:
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
Page No SA1.24:
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
Answer:
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 âplane 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 âplane and we will draw a line which passes through these points.
From the above analysis we have the following graph
The shaded region is ABC.
From the above graph, the area of triangle ABC is given by
Hence area is
Page No SA1.24:
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 âplane 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 âplane and we will draw a line which passes through these points.
From the above analysis we have the following graph
The shaded region is ABC.
From the above graph, the area of triangle ABC is given by
Hence area is
Answer:
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
Page No SA1.24:
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
Answer:
Here we have to prove that,
Multiply in both numerator and denominator by, we get
Hence
Page No SA1.24:
Question 34:
Here we have to prove that,
Multiply in both numerator and denominator by, we get
Hence
Answer:
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
Page No SA1.3:
Question 3:
In Fig. 2, If DE || BC, then x equals
(a) 6 cm
(b) 8 cm
(c) 10 cm
(d) 12.5 cm
Answer:
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
Page No SA1.3:
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°
Answer:
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.
Page No SA1.3:
Question 5:
Given that tan θ = , the value of is
(a) − 1
(b) 1
(c)
(d)
Answer:
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
Page No SA1.3:
Question 6:
In Fig. 3, AD = 4 cm, BD = 3 cm and CB = 12 cm, then cot θ equals
(a)
(b)
(c)
(d)
Answer:
In Fig 3, if we have
AD = 4 cm,
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
Page No SA1.3:
Question 7:
The decimal expansion of will terminate after how many places of decimal?
(a) 1
(b) 2
(c) 3
(d) will not terminate
Answer:
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.
Page No SA1.3:
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
Answer:
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
Page No SA1.3:
Question 9:
In an isosceles triangle ABC with AB = AC and BD ⊥ AC. Prove that BD2 − CD2 = 2CD.AD.
Answer:
Given: âABC is an isosceles triangle with AB = AC and BD ⊥ AC.
To prove: BD2 − CD2 = 2CD.AD
Proof: In right âADB,
(Hence Proved)
Page No SA1.3:
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
Answer:
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
Page No SA1.4:
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
Answer:
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,
When k = −1,
Thus, for k = −1,
Hence, the given system of equations has no solution when k = −1.
Page No SA1.4:
Question 12:
Can (x − 2) be the remainder on division of a polynomial p(x) by (2x + 3)? Justify your answer.
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
Page No SA1.4:
Question 13:
In Fig. 4, ABCD is a rectangle. Find the values of x and y.
Answer:
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
Page No SA1.4:
Question 14:
If 7 sin2 θ + 3 cos2θ = 4, show that tan θ =
Answer:
Given: and we have to prove that
We can write the given expression as
Therefore,
Hence proved
Page No SA1.4:
Question 15:
In Fig. 5, DE || AC and DF || AE. Prove that
Answer:
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
Page No SA1.4:
Question 16:
In Fig. 6, AD BC and BD = CD. Prove that 2CA2 = 2AB2 + BC2
Answer:
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
In triangle ADC, we have
…… (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.
Page No SA1.5:
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.
Answer:
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,
Page No SA1.5:
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 |
Answer:
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
Page No SA1.5:
Question 19:
Show that any positive odd integers is of the form 4q+1 or 4q+3 where q is a positive integer.
Answer:
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.
Page No SA1.5:
Question 20:
Prove that is irrational.
Answer:
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.
Page No SA1.5:
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.
Answer:
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.
Page No SA1.5:
Question 22:
If , are zeroes of the polynomial x2 − 2x − 15, then form a quadratic polynomial whose zeroes are (2) and (2).
Answer:
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
Page No SA1.5:
Question 23:
Prove that (cosec θ − sin θ) (sec θ − cos θ) = .
Answer:
Here we have to prove that
Now using the identity, we get
Hence proved.
Page No SA1.5:
Question 24:
If cos θ + sin θ = cos θ, show that cos θ − sin θ = sin θ.
Answer:
Given that if, then we have to prove that
We have,
Hence proved.
Page No SA1.5:
Question 25:
In Fig. 7, AB BC, FG BC and DE AC. Prove that Δ ADE ΔGCF
Answer:
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
Page No SA1.5:
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
Answer:
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.
Page No SA1.6:
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 |
Answer:
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.
Page No SA1.6:
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 |
Answer:
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
Page No SA1.6:
Question 29:
Find other zeroes of the polynomial p(x) = 2x4 + 7x3 − 19x2 − 14x + 30 if two of its zeroes are and .
Answer:
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
Page No SA1.6:
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.
Answer:
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
Page No SA1.6:
Question 31:
Prove that:
Answer:
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, .
Page No SA1.6:
Question 32:
If sec θ + tan θ = p, prove that sin θ =
Answer:
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
Page No SA1.6:
Question 33:
Draw the graphs of following equations:
2x − y = 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.
Answer:
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.
Page No SA1.6:
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.
Answer:
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.
Page No SA2.2:
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
Answer:
Given the following quadratic equations
(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.
Page No SA2.2:
Question 2:
The sum of first five multiples of 3 is:
(a) 45
(b) 65
(c) 75
(d) 90
Answer:
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.
Page No SA2.2:
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
Answer:
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.
Page No SA2.2:
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°
Answer:
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
Therefore in Quadrilateral PQOR
Hence option (a) is correct.
Page No SA2.3:
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
Answer:
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.
Page No SA2.3:
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°
Answer:
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°
Therefore in Quadrilateral OBAC
Hence Option (c) is correct.
Page No SA2.3:
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 the volume of original cone. The height from the base at which the section is made is:
Answer:
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.
Page No SA2.3:
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
Answer:
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.
Page No SA2.3:
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.
Answer:
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.
Radius of glass sphere r
Volume of a glass sphere
Length of cuboidal box
Breadth of cuboidal box
Height of cuboidal box
Volume of water = Volume of cuboidal box − Volume of glass spheres
Page No SA2.3:
Question 10:
Find two consecutive odd positive integers, sum of whose squares is 290.
Answer:
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
Page No SA2.3:
Question 11:
Find the roots of the following quadratic equation:
Answer:
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 =
Page No SA2.3:
Question 12:
If the numbers x − 2, 4x − 1 and 5x + 2 are in A.P. find the value of x.
Answer:
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
Page No SA2.3:
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.
Answer:
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
In quadrilateral AOBP,
[Sum of angles of a quadrilateral = 360°]
We know that the sum of opposite angles of cyclic quadrilateral = 180°
Therefore from (1) and (2)
Quadrilateral AOBP is a cyclic quadrilateral.
Page No SA2.3:
Question 14:
In Fig. 2, a circle of radius 7 cm is inscribed in a square. Find the area of the shaded region
Answer:
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
Radius of circle cm
Given that radius of circle = 7 cm
Therefore
Page No SA2.3:
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?
Answer:
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
of 3cm which can be made from the cuboidal lead solid.
Let the length of cuboidal lead solid L = 9 cm
Let the breadth of cuboidal lead solid B = 11 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 =
Page No SA2.4:
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.
Answer:
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
Page No SA2.4:
Question 17:
Show that the point P (−4, 2) lies on the line segment joining the points A (−4, 6) and B (−4, −6).
Answer:
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.
Page No SA2.4:
Question 18:
Two dice are thrown at the same time. Find the probability of getting different numbers on both dice.
Answer:
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) =
Page No SA2.4:
Question 19:
A natural number, when increased by 12, becomes equal to 160 times its reciprocal. Find the number.
Answer:
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
Page No SA2.4:
Question 20:
Find the sum of the integers between 100 and 200 that are divisible by 9.
Answer:
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
Page No SA2.4:
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.
Answer:
Given a figure as shown. We have to prove that
Join OR
We know that sum of opposite angles of a cyclic quadrilateral
Therefore in quadrilateral PQOR,
…… (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.
Page No SA2.4:
Question 22:
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are time the corresponding sides of ΔABC.
Answer:
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.
Page No SA2.4:
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]
Answer:
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
Page No SA2.5:
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.
Answer:
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 )
Page No SA2.5:
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.
Answer:
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
Page No SA2.5:
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.
Answer:
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.
AB, BC, CD, AD
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.
Page No SA2.5:
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.
Answer:
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)
Page No SA2.5:
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.
Answer:
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
Page No SA2.5:
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?
Answer:
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
Page No SA2.5:
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.
Answer:
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
Page No SA2.5:
Question 31:
Prove that the lengths of tangents drawn from an external point to a circle are equal.
Answer:
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.
Page No SA2.5:
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.
Answer:
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,
=
Page No SA2.5:
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.
Answer:
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 )
Page No SA2.5:
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.
Answer:
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
Let AD is the building
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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