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Page No SA1.13:

Question 1:

Which of the following numbers has terminating decimal expansion?

(a) 3745

(b) 212356

(c) 1749

(d) 892232

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 SA1.13:

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 = 16xpx

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

Therefore

Hence option (d) is correct.

Page No SA1.13:

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 SA1.13:

Question 4:

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 SA1.13:

Question 5:

Find a quadratic polynomial with zeroes 3 + 2 and 3 - 2

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 SA1.13:

Question 6:

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 SA1.13:

Question 7:

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 SA1.14:

Question 8:

In Figure 3, PQ || CD and PR || CB. Prove that AQQD=ARRB.

 

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 SA1.14:

Question 9:

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 ✕ CE = BE ✕ 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 SA1.14:

Question 10:

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 SA1.14:

Question 11:

Prove that 7 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 SA1.14:

Question 12:

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 SA1.14:

Question 13:

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 SA1.14:

Question 14:

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 34. 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 SA1.14:

Question 15:

Prove that tan θ-cot θsin θ cos θ=tan2 θ-cot2 θ.

Answer:

Here we have to prove that

First we take LHS and use the identities

Now we take RHS

         

Hence proved.

Page No SA1.14:

Question 16:

In Figure 5, ∆ABC 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

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



Page No SA1.15:

Question 17:

In Figure 6, P and Q are the midpoints of the sides CA and CB respectively of ∆ABC 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 SA1.15:

Question 18:

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 SA1.15:

Question 19:

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 SA1.15:

Question 20:

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 SA1.15:

Question 21:

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 SA1.15:

Question 22:

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 SA1.15:

Question 23:

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 SA1.16:

Question 24:

Without using trigonometric tables, evaluate the following:

sec 37°cosec 53°+2 cot 15° cot 25° cot 45° cot 75° cot 65°-3sin2 18°+sin2 72°

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 SA1.16:

Question 25:

Prove that :

tan θ1-cot θ+cot θ1-tan θ=1+sec θ cosec θ

Answer:

Here we have to prove that

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

Hence

Page No SA1.16:

Question 26:

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 SA1.16:

Question 27:

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.

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 SA1.16:

Question 28:

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 SA1.16:

Question 29:

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.

a1a2=b1b2c1c232k-1=1k-112k+132k-1=1k-1 and 1k-112k+1

Now,

32k-1=1k-13k-3=2k-13k-2k=-1+3k=2

When k = 2,

1k-1=12-1=1 and 12k+1=12×2+1=15

Thus, for k = 2, 1k-112k+1

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

Page No SA1.16:

Question 30:

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
  fi=39+p   fixi=2370+30p

It given that the mean of the distribution is 54.

Mean=fixifi2370+30p39+p=542370+30p=2106+54p54p-30p=2370-2106

24p=264p=26424=11

Hence, the value of p is 11.

Page No SA1.16:

Question 31:

question

Answer:


Given: ∆ABC is an equilateral triangle. D is a point on BC such that BD=13BC.
Construction: Draw AEBC. Join AD.
To prove: 9AD2=7AB2
Proof: In ∆AEB and ∆AEC,
AB = AC
AEB=AEC=90°
AE = AE
Thus, ∆AEB ≅ ∆AEC     (RHS congruence criterion)
So, BE = EC
Thus, we have
 BD=13BC, DC=23BC and BE=EC=12BC
Since ∆ABC is an equilateral triangle, so C = 60º.
Therefore, ∆ADC is an acute triangle.
Now,
AD2=AC2+DC2-2DC×ECAD2=AC2+23BC2-2×23BC×12BCAD2=AC2+49BC2-23BC2AD2=AC2+49BC2-23AB2AD2=9AB2+4AB2-6AB29=79AB29AD2=7AB2
Hence Proved.



Page No SA1.2:

Question 1:

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 SA1.2:

Question 2:

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 3:

Given that tan θ = 13, the value of cosec2θ-sec2θcosec2θ+sec2θ is

(a) − 1

(b) 1

(c) 12

(d) -12

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 4:

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 SA1.3:

Question 5:

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 SA1.3:

Question 6:

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 SA1.3:

Question 7:

If 7 sin2 θ + 3 cos2 θ = 4, show that tan θ = 13

Answer:

Given: and we have to prove that

We can write the given expression as

Therefore,

Hence proved

Page No SA1.3:

Question 8:

In Fig. 5, DE || AC and DF || AE. Prove that EFBF=ECBE

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 SA1.3:

Question 9:

In Fig. 6, AD BC and BD = 13 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 SA1.4:

Question 10:

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 SA1.4:

Question 11:

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 SA1.4:

Question 12:

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 SA1.4:

Question 13:

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 SA1.4:

Question 14:

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 SA1.4:

Question 15:

Prove that (cosec θ − sin θ) (sec θ − cos θ) = 1tan θ+cot θ.

Answer:

Here we have to prove that

Now using the identity, we get

Hence proved.

Page No SA1.4:

Question 16:

If cos θ + sin θ = 2 cos θ, show that cos θ − sin θ = 2 sin θ.

Answer:

Given that if, then we have to prove that

We have,

Hence proved.

Page No SA1.4:

Question 17:

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 SA1.4:

Question 18:

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  area (Δ ABC)area (DBC) = AODO

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 SA1.4:

Question 19:

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 SA1.4:

Question 20:

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



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Question 21:

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

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

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Question 22:

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

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Question 23:

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,

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Question 24:

Prove that: sec θ+tan θ-1tan θ-sec θ+1=cos θ1-sin θ

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, .

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Question 25:

Evaluate:

sec θ cosec 90°-θ-tan θ cot 90°-θ+sin2 55°+sin2 35°tan 10° tan 20° tan 60° tan 70° tan 80°

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

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Question 26:

If sec θ + tan θ = p, prove that sin θ = p2-1p2+1

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

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Question 27:

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.

 

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.

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Question 28:

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.

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Question 29:

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

Answer:



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

To prove: BD2 − CD2 = 2CD.AD

Proof:  In right ∆ADB,

AB2=AD2+BD2                           Pythagoras TheoremAC2=AD2+BD2                      AB=ACAD+CD2=AD2+BD2AD2+CD2+2AD.CD=AD2+BD2

BD2-CD2=2AD.CD                  (Hence Proved)

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Question 30:

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.

a1a2=b1b2c1c23k+1k2+1=3k-2-2-53k+1k2+1=3k-2 and 3k-225

Now,

3k+1k2+1=3k-23k+1k-2=3k2+13k2-5k-2=3k2+3-5k=5
k=-1

When k = −1,

3k-2=3-1-2=3-3=-1

Thus, for k = −1, 3k-225

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

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Question 31:

question

Answer:

Let the number of students in each row be x and the number of rows be y.
Total number of students = xy
If one student is extra in a row, so
Number of students in each row = x + 1
Number of rows = y − 2
Total number of students = (x + 1)(y − 2)
(x + 1)(y − 2) = xy
xy − 2x + y − 2 = xy
⇒ 2x − y = −2             .....(i)
If one student is less in each row, so
Number of students in each row = x − 1
Number of rows = y + 3
Total number of students = (x − 1)(y + 3)
(x − 1)(y + 3) = xy
xy + 3xy − 3 = xy
⇒ 3xy = 3             .....(ii)
Subtracting (i) from (ii), we get
x = 5
Putting x = 5 in (i), we get
10 − y = −2
y = 12
So, the number of students in the class = xy = 60.



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