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AP State Syllabus AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.5 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 2nd Lesson Linear Equations in One Variable Exercise 2.5

Question 1.
Solve the following equations.
i) \(\frac{n}{5}-\frac{5}{7}=\frac{2}{3}\)
Solution:

ii) \(\frac{x}{3}-\frac{x}{4}=14\)
⇒ \(\frac{4 x-3 x}{12}\) = 14
⇒ \(\frac{x}{12}\) = 14
⇒ x = 12 × 14 = 168
∴ x = 168

iii) \(\frac{z}{2}+\frac{z}{3}-\frac{z}{6}=8\)

iv) \(\frac{2 p}{3}-\frac{p}{5}=11 \frac{2}{3}\)

v) \(9 \frac{1}{4}=y-1 \frac{1}{3}\)

vi) \(\frac{x}{2}-\frac{4}{5}+\frac{x}{5}+\frac{3 x}{10}=\frac{1}{5}\)

vii) \(\frac{x}{2}-\frac{1}{4}=\frac{x}{3}+\frac{1}{2}\)

viii) \(\frac{2 x-3}{3 x+2}=\frac{-2}{3}\)
⇒ 3(2x – 3) = – 2(3x + 2)
⇒ 6x – 9 = -6x – 4
⇒ 6x + 6x = -4 + 9
⇒ 12x = 5
∴ x = \(\frac{5}{12}\)

ix) \(\frac{8 p-5}{7 p+1}=\frac{-2}{4}\)
Solution:
⇒ \(\frac{8 p-5}{7 p+1}=\frac{-2}{4}\)
⇒ 2(8p – 5) = – (7p + 1)
⇒ 16p – 10 = – 7p – 1
⇒ 16p + 7p = – 1 + 10
⇒ 23p = 9
∴ x = \(\frac{9}{23}\)

x) \(\frac{7 y+2}{5}=\frac{6 y-5}{11}\)
⇒ 11 (7y + 2) = 5 (6y-5)
⇒ 77y + 22 = 30y – 25
⇒ 77y – 30y = – 25 – 22
⇒ 47y = – 47
∴ y = \(\frac{-47}{47}\)
∴ y = -1

xi) \(\frac{x+5}{6}-\frac{x+1}{9}=\frac{x+3}{4}\)

⇒ 4(x + 13) = 18 (x + 3)
⇒ 4x + 52 = 18x + 54
⇒ 4x – 18x = 54-52
⇒ – 14x = 2
⇒ x = \(\frac{2}{-1}\) = \(\frac{-1}{7}\)
∴ x = \(\frac{-1}{7}\)

xii) \(\frac{3 t+1}{16}-\frac{2 t-3}{7}=\frac{t+3}{8}+\frac{3 t-1}{14}\)
Solution:

⇒ -11t + 55 = 2(19t + 17) = 38t + 34
⇒ -11t – 38t = 34 – 55
⇒ -49t = – 21
⇒ \(\frac{-21}{-49}\) = \(\frac{3}{7}\)
∴ t = \(\frac{3}{7}\)

Question 2.
What number is that of which the third part exceeds the fifth part by 4?
Solution:
Let the number be ‘x’ say.
\(\frac{1}{3}\) rd of a number = \(\frac{1}{3}\) x x = \(\frac{x}{3}\)
\(\frac{3}{7}\) th of a number = \(\frac{1}{5}\) x x = \(\frac{x}{5}\)
According to the sum

∴ The required number is 30.

Question 3.
The difference between two positive integers is 36. The quotient when one integer is
divided by other is 4. Find the integers.
(Hint: If one number is ‘X’, then the other number is ‘x – 36’)
Solution:
Let the two positive numbers be x, (x – 36) say.
If one number is divided by second tten the quotient is 4.
∴ \(\frac{x}{x-36}=4\)
⇒ x = 4(x – 36) = 4x – 144
⇒ 4x – x = 144
3x = 144
x = 48
∴ x – 36 = 48 – 36 = 12
∴ The required two positive intgers are 48, 12.

Question 4.
The numerator of a fraction is 4 less than the denominator. If 1 is added to both its
numerator and denominator, it becomes 1/2 . Find the fraction.
Solution:
Let the denominator of a fractin be x.
The numerator of a fraction is 4 less than the denominator.
∴ The numerator = x – 4
∴ Fraction \(\frac{x-4}{x}\)
If ‘1’ is added to both, its numerator and denominator, it becomes \(\frac{1}{2}\)
∴ \(\frac{1+x-4}{1+x}=\frac{1}{2}\)
2 + 2x – 8 = 1 + x
2x – x = 1 + 6 = 7
x = 7
∴ The denominator = 7
The numerator = 7 – 4 = 3
∴ Fraction = \(\frac{3}{7}\)

Question 5.
Find three consecutive numbers such that if they are divided by 10, 17, and 26 respectively,
the sum of their quotients will be 10.
(Hint: Let the consecutive numbers = x, x+ 1, x+ 2, then \(\frac{x}{10}+\frac{x+1}{17}+\frac{x+2}{26}=10\))
Solution:
Let the three consecutive numbers be assume that x, (x + 1), (x + 2) respectively.
Given that x, (x + 1), (x + 2) are divided by 10, 17, 26 respectively, the sum of the quotients is 10. Then
⇒ \(\frac{x}{10}+\frac{x+1}{17}+\frac{x+2}{26}=10\)
⇒ \(\frac{x \times 221+130(x+1)+85(x+2)}{2210}=10\)
⇒ 221x + 130x + 85x + 130 + 170 = 22,100
⇒ 436x + 300 = 22,100
⇒ 436x = 22,100 – 300
⇒ 436x = 21,800
⇒ \(\frac{21800}{436}\)
∴ x = 50
∴ The required three consecutive num-bers are x = 50
x + 1 =50+ 1 = 51
x + 2 = 50 + 2 = 52

Question 6.
In class of 40 pupils the number of girls is three-fifths of the number of boys. Find the
number of boys in the class.
Solution:
Let the number of boys = x say.
Total number of students = 40
Number of girls = \(\frac{3}{5}\) × x = \(\frac{3x}{5}\)
According to the sum 3x

∴ x = 25
∴ Number of boys in the class room = 25

Question 7.
After 15 years , Mary’s age will be four times of her present age. Find her present age.
Solution:
Let the present age of Mary = x years say.
After 15 years Mary’s age = (x + 15) years
According to the sum
(x + 15) = 4 x x
⇒ x + 15 = 4x
⇒ 4x – x =15
⇒ 3x = 15
⇒ x = 5
∴ The present age of Mary = 5 years.

Question 8.
Aravind has a kiddy bank. It is full of one-rupee and fifty paise coins. It contains 3 times
as many fifty paise coins as one rupee coins. The total amount of the money in the bank is
₹ 35. How many coins of each kind are there in the bank?
Solution:
Number of 1 rupee coins = x say.
Number of 50 – paise coins = 3 x x = 3x
The value of total coins = \(\frac{3x}{2}\) + x
[∵50 paisa coins of 3x = ₹\(\frac{3x}{2}\)
According to the sum
⇒ \(\frac{3x}{2}\) + x = 35
⇒ \(\frac{3 x+2 x}{2}\) = 35
⇒ 5x = 2 × 35
⇒ x = 2 × \(\frac{35}{5}\)
∴ x = 14
∴ Number of 1 rupee coins = 14
Number of 50 paisa coins = 3 × x = 3 × 14 = 42

Question 9.
A and B together can finish a piece of work in 12 days. If ‘A’ alone can finish the same work in 20days , in how many days B alone can finish it?
Solution:
A, B can do a piece of work in 12 days.
(A + B)’s 1 day work = \(\frac{1}{12}\) th part.
A can complete the same work in 20 days.
Then his one day work = \(\frac{1}{20}\)
B’s one day work = (A+B)’s 1 day work – A’s 1 day work
\(=\frac{1}{12}-\frac{1}{20}=\frac{5-3}{60}=\frac{2}{60}=\frac{1}{30}\)
∴ Number of days to take B to com¬plete the whole work = 30.

Question 10.
If a train runs at 40 kmph it reaches its destination late by 11 minutes . But if it runs at 50 kmph it is late by 5 minutes only. Find the distance to be covered by the train.
Solution:
Let the distance to be reached = x km say. Time taken to travel ‘x’ km with speed x
40 km/hr = \(\frac{x}{40}\) hr.
Time taken to travel ‘x’ km with speed 50 km/hr = \(\frac{x}{50}\) hr.
According to the sum the difference between the times

∴ The required distance to be trav¬elled by a train = 20 kms‘.

Question 11.
One fourth of a herd of deer has gone to the forest. One third of the total number is
grazing in a field and remaining 15 are drinking water on the bank of a river. Find the total
number of deer.
Solution:
Number of deer = x say.
Number of deer has gone to the forest
= \(\frac{1}{4}\) × x = \(\frac{x}{4}\)
Number of deer grazing in the field
= \(\frac{1}{3}\) × x =\(\frac{x}{3}\)
Number of remaining deer =15
According to the sum

∴ x = 36
∴ The total number of deer = 36

Question 12.
By selling a radio for ₹903, a shop keeper gains 5%. Find the cost price of the radio.
Solution:
The selling price of a radio = ₹ 903
Profit % = 5%
C.P = ?
C.P = \(\frac{\mathrm{S.P} \times 100}{(100+\mathrm{g})}\)
= \(\frac{903 \times 100}{(100+5)}\)
= \(\frac{903\times 100}{105}\)
C.P. = 8.6 × 100 = 860
∴ The cost price of the radio = ₹ 860

Question 13.
Sekhar gives a quarter of his sweets to Renu and then gives 5 sweets to Raji. He has 7 sweets left. How many did he have to start with?
Solution:
Number of sweets with Sekhar = x say.
Number of sweets given to Renu
= \(\frac{1}{4}\) × x = \(\frac{x}{4}\)
Number of sweets given to Raji = 5
Till he has 7 sweets left.
x – ( \(\frac{x}{4}\) + 5) = 7
⇒ x – \(\frac{x}{4}\) – 5 = 7
⇒ x – \(\frac{x}{4}\) = 7 + 5 = 12
⇒ \(\frac{4 x-x}{4}\) = 12
⇒ \(\frac{3x}{4}\) = 12
⇒ x = 12 × \(\frac{4}{3}\) = 16
∴ x = 4 × 4 = 16
∴ Number of sweets with Sekhar at the beginning = 16

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 1 Rational Numbers Ex 1.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 1st Lesson Rational Numbers Exercise 1.2

Question 1.
Represent these numbers on the number line.
(i) \(\frac{9}{7}\)
(ii) \(-\frac{7}{5}\)
Solution:
(i) \(\frac{9}{7}\)

(ii) \(-\frac{7}{5}\)

Question 2.
Represent \(-\frac{2}{13}, \frac{5}{13}, \frac{-9}{13}\) on the number line.
Solution:

Question 3.
Write five rational numbers which are smaller than \(\frac{5}{6}\)
Solution:
The rational number which are less than
\(\frac{5}{6}=\left\{\frac{4}{6}, \frac{3}{6}, \frac{2}{6}, \frac{1}{6}, \frac{0}{6}, \frac{-1}{6}, \frac{-2}{6} \ldots \ldots .\right\}\)

Question 4.
Find 12 rational numbers between -1 and 2.
Solution:

Question 5.
Find a rational number between \(\frac{2}{3}\) and \(\frac{3}{4}\)
[Hint : First write the rational numbers with equal denominators.]
Solution:
The given rational numbers are \(\frac{2}{3}\) and \(\frac{3}{4}\)
\(\frac{2}{3} \times \frac{4}{4}=\frac{8}{12}, \frac{3}{4} \times \frac{3}{3}=\frac{9}{12}\)
The rational numbers between \(\frac{8}{12}, \frac{9}{12}\) is
\(\frac{\left(\frac{8}{12}+\frac{9}{12}\right)}{2}=\frac{\frac{17}{12}}{2}=\frac{17}{24}\)
(∵ the rational number between a, b is \(\frac{a+b}{2}\) )
∴ the rational number between \(\frac{2}{3}\) and \(\frac{3}{4}\) is \(\frac{17}{24}\)

Question 6.
Find ten rational numbers between \(-\frac{3}{4}\) and \(\frac{5}{6}\)
Solution:

The 10 rational numbers between \(-\frac{9}{12}\) and \(\frac{10}{12}\) are

∴ We can select any 10 rational numbers from the above number line.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 1 Rational Numbers Ex 1.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 1st Lesson Rational Numbers Exercise 1.1

Question 1.
Name the properly Involved in the following examples.

vii) 7a + (-7) = 0
viii) x + \(\frac{1}{x}\) = 1(x ≠ 0)
ix) (2 x x) + (2 x 6) = 2 x (x + 6)
Solution:
i) Additive identity
ii) Distributive law
iii) Multiplicative identity
iv) Multiplicative identity
v) Commutative law of addition
vi) Closure law in multiplication
vii) Additive inverse
viii) Multiplicative inverse
ix) Distributive

Question 2.
Write the additive and the multiplicative inverses of the following.
i) \(\frac{-3}{5}\)
ii) 1
iii) 0
iv) \(\frac{7}{9}\)
v) -1
Solution:

Question 3.
Fill in the blanks



Solution:
i) \(\left(\frac{-12}{5}\right)\)
ii) \(\left(\frac{4}{3}\right)\)
iii) \(\left(\frac{9}{11}\right)\)
iv) \(\left(\frac{6}{7}\right)\)
v) \(\left(\frac{3}{4}, \frac{1}{3}\right)\)
vi) 0

Question 4.
Multiply \(\frac{2}{11}\) by the reciprocal of \(\frac{-5}{14}\)
Solution:
The reciprocal of \(\frac{-5}{14}\) is \(\frac{-14}{5}\)
( ∵ \(\left(\frac{-5}{14}\right) \times\left(\frac{-14}{5}\right)=1\) )
∴ The product of \(\frac{2}{11}\) and \(\frac{-14}{5}\) is
\(\frac{2}{11} \times\left(\frac{-14}{5}\right)=\frac{-28}{55}\)

Question 5.
Which properties can be used computing \(\frac{2}{5} \times\left(5 \times \frac{7}{6}\right)+\frac{1}{3} \times\left(3 \times \frac{4}{11}\right)\)
Solution:
The following properties are involved in the product of
\(\frac{2}{5} \times\left(5 \times \frac{7}{6}\right)+\frac{1}{3} \times\left(3 \times \frac{4}{11}\right)\)
i) Multiplicative associative property.
ii) Multiplicative inverse.
iii) Multiplicative identity.
iv) Closure with addition

Question 6.
Verify the following
\(\left(\frac{5}{4}+\frac{-1}{2}\right)+\frac{-3}{2}=\frac{5}{4}+\left(\frac{-1}{2}+\frac{-3}{2}\right)\)
Solution:

Question 7.
Evaluate \(\frac{3}{5}+\frac{7}{3}+\left(\frac{-2}{5}\right)+\left(\frac{-2}{3}\right)\) after rearrangement.
Solution:

Let x = \(1.2 \overline{4}\)
⇒ x = 1.244……. …………………(1)
Here periodicity of equation (1) is 1. So
it should be multiplied by 10 on both
sides.
⇒ 10 x x = 10 x 1.244
10x = 12.44 …………..(2)

Question 8.
Subtract
(i) \(\frac{3}{4}\) from \(\frac{1}{3}\)
(ii) \(\frac{-32}{13}\) from 2
(iii) -7 from \(\frac{-4}{7}\)
Solution:

Question 9.
What numbers should be added to \(\frac{-5}{8}\) so as to get \(\frac{-3}{2}\) ?
Solution:
Let the number to be add ‘x’ say

∴ \(\frac{-7}{8}\) should be added to \(\frac{-5}{8}\) then we will get \(\frac{-3}{2}\)

Question 10.
The sum of two rational numbers is 8 If one of the numbers is \(\frac{-5}{6}\) find the other.
Let the second number be ‘x’ say
⇒ \(x+\left(\frac{-5}{6}\right)=8\)
⇒\(8+\frac{5}{6}=\frac{48+5}{6}=\frac{53}{6}\)
∴ The other number (x) = \(\frac{53}{6}\)

Question 11.
Is subtraction associative in rational numbers? Explain with an example.
Solution:
Let \(\frac{1}{2}, \frac{3}{4}, \frac{-5}{4}\) are any 3 rational numbers.
Associative property under subtraction
a – (b – c) = (a – b) – c

∴ L.H.S. ≠ R.H.S.
∴ a – (b – c) ≠ (a – b) – c
∴ Subtraction is not an associative in rational numbers.

Question 12.
Verify that – (-x) = x for
(i) x = \(\frac{2}{15}\)
(ii) x = \(\frac{-13}{15}\)
Solution:

Question 13.
Write-
(i) The set of numbers which do not have any additive identity
(ii) The rational number that does not have any reciprocal
(iii) The reciprocal of a negative rational number.
Solution:
i) Set of natural numbers ’N’ doesn’t possesses the number ‘0’.
ii) The rational number ‘0’ has no multiplicative inverse.
[ ∵ 1/0 is not defined]
iii) The reciprocal of a negative rational number is a negative rational number.
Ex : Reciprocal of \(\frac{-2}{5}=\frac{-5}{2}\)

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 9 Statistics InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 9th Lesson Statistics InText Questions

Activity

Divide the students of your class into four groups. Allot each group the work of collecting one of the following kinds of data: [Page No. 195]
Solution:
i) Weights of all the students in your class
ii) Number of siblings that each student have.
iii) Day wise number of absentees in your class during last month.
iv) The distance between the school and home of every student of your class.

Do This

Which of the following are primary and secondary data ? (Page No. 195)
i) Collection of the data about enroll¬ment of students in your school for a period from 2001 to 2010.
Solution:
Secondary data.

ii) Height of students in your class re¬corded by physical education teacher.
Solution:
Primary data.

Activity

Make frequency distribution table of the initial letters of that denotes surnames of your classmates and answer the following questions.
i) Which initial letter occured mostly among your classmates?
ii) How many students names start with’ the alphabat T?
iii) Which letter occured least as an initial among your classmates? [Page No. 197]

Think, Discuss anil Write

Question 1.
Give 3 situations, where mean, median and mode are separately appropriate and counted. [Page No. 202]
Solution:
Mean :
a) Rice required for a certain number of students for a given period.
b) To compare the marks of students of a class.
c) To calculate the daily income of a shop during a month.

Median :
a) To study the salaries of staff of an institution.
b) To compare heights of boys and girls.

Mode :
a) To find the size of the shoe with heighest sale.

Try These

Question 1.
Find the median of the scores 75, 21, 56, 36, 81, 05, 42. [Page No. 207]
Solution:
Arranging the data in ascending order
05, 21, 36, 42, 56, 75, 81
Number of terms in the data = 7 – odd
∴ \(\left(\frac{n+1}{2}\right)^{t h}\) term = \(\frac{7+1}{2}=\frac{8}{2}\) = 4th term is the median = 42.

Question 2.
Median of a data arranged in ascending order 7, 10, 15, x, y, 27, 30 is 17 and when one more observation 50 is added to the data the median has become 18. Find x and y. [Page No. 207]
Solution:
The given data in ascending order is 7, 10, 15, x, y, 27, 30
Median = \(\left(\frac{n+1}{2}\right)^{t h}\) term = \(\left(\frac{7+1}{2}\right)\) = 4th term = x
∴ x = 17 by problem
When 50 is added, the data becomes 7, 10, 15, 17, y, 27, 30, 50
Median = \(\frac{\left(\left(\frac{n}{2}\right)+\left(\frac{n}{2}+1\right)\right)}{2}\) terms
18 = \(\frac{17+y}{2}\) (by problem)
17 + y = 36
y = 36 – 17 = 19

Question 3.
Find the median marks in the data.[Page No. 208]

Solution:

Median = \(\left(\frac{n+1}{2}\right)^{t h}\) term as N = 29 is odd
∴ \(\frac{29+1}{2}\) = 15th term.
15th observation is 15. (from the table)

Question 4.
In finding the median, the given data must be in order ? Why ? [Page No. 208]
Solution:
In finding the median, the observations must be in order. The data is to be arranged either in ascending/descending order to divide the data into two equal groups.

Think and Discuss

Question 1.
Classify your classmates according to their heights and find the mode of it.
(Page No. 208)
Solution:
Student to find the mode of the classmates according their heights.

Question 2.
If shopkeeper has to place an order for shoes, which number shoes should he order more ? [Page No. 208]
Solution:
Number 7 as it has highest sales.

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 8 Quadrilaterals InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 8th Lesson Quadrilaterals InText Questions

Try This

Question
Extend AB to E. Find ∠CBE. What do you notice? What kind of angles are ∠ABC and ∠CBE?
Solution:
Given that □ABCD is a parallelogram and∠A = 40°
∴ ∠ABC = 180°-40° = 140°
∠CBE = 40° ( ∵ ∠A and, ∠CBE are corresponding angles) And ∠CBE and ∠ABC are linear pair of angles.

Do This

Question
Cut out a parallelogram from a sheet of paper again and cut along one of its diagonal. What kind of shapes you obtain ? What can you say about these triangles 7 [Page No. 179]
Solution:
We get two triangles.

The two triangles are congruent to each other.

Think, Discuss and Write

Question 1.
Show that the diagonals of a square are equal and right bisectors of each other. (Page No. 185)
Solution:

Let □ABCD be a square.
Thus AB = BC = CD = DA
In ΔABC and ΔBAD
AB = AB (common base)
∠B =∠A (each 90°)
BC = AD (equal sides)
∴ ΔABC = ΔBAD (SAS congruence)
⇒ AC = BD (CPCT)
Also in ΔAOB and ΔCOD
∠OAB = ∠OCD [∵ alt. int. angles]
∠OBA = ∠ODC [∵ alt. int. angles]
AB = DC (sides of a square)
∴ ΔAOB = ΔCOD (ASA congruence)
Thus AO = OC (CPCT) ⇒ O is midpoint of AC
Also BO = OD (CPCT) ⇒ O is midpoint of BD
∴ O is midpoint of both AC and BD.
∴ Diagonals bisect each other.
In ΔAOB and ΔCOB
AB = BC (given)
OB = OB (common)
AO = OC (proved)
∴ ΔAOB ≅ ΔCOB (SSS congruence)
∠AOB = ∠COB (CPCT)
But ∠AOB + ∠COB = 180° (∵ linear pair of angles)
∴ ∠AOB = ∠COB = 180°/2 = 90°
Also ∠AOB = ∠COD (∵ vertically opposite angles)
∠BOC = ∠AOD (∵ vertically opposite angles)
∴ AC ⊥ BD
(i.e.,) In a square diagonals bisect at right angles.

Question 2.
Show that the diagonals of a rhombus divide it into four congruent triangles. (Page No. 185)
Solution:

□ABCD is a rhombus
Let AC and BD meet at ‘O’
In ΔAOB and ΔCOD
∠OAB = ∠ODC (alt.int.angles)
AB = CD (def. of rhombus)
∠OBA = ∠ODC (alt. mt, angles)
∴ ΔAOB ≅ΔCOD …………(1)
(ASA congruence)
Thus AO = OC (CPCT
Also ΔAOD ≅ ΔCOD …………..(2)
[ ∵AO = OC; AD = CD; OD = OD SSS congruence]
Similarly we can prove
ΔAOD ≅ ΔCOB …………..(3)
From (1), (2) and (3) we have
ΔAOB = ΔBOC = ΔCOD = ΔAOD
∴ Diagonals of a rhombus divide it into four congruent triangles.

Try This

Question
Draw a triangle ABC and mark the mid points E and F of
two sides of triangle. \(\overline{\mathbf{A B}}\) and \(\overline{\mathbf{A C}}\) respectively. Join the point E and F as shown in the figure. Measure EF and the third side BC of triangle. Also measure ∠AEF and ∠ABC.
We find ∠AEF = ∠ABC and \(\overline{\mathrm{EF}}=\frac{1}{2} \overline{\mathrm{BC}}\)

As these are corresponding angles made by the transversal AB with lines EF and BC, we say EF//BC.
Repeat this activity with some more triangles. (Page No. 188)

Solution:

P, Q are mid points of XY and XZ
PQ // YZ
PQ = \(\frac{1}{2}\)YZ

X, Y are mid points of PQ and PR
XY // QR
XY = \(\frac{1}{2}\)QR

A, B are midpoints of DE and DF
AB // EF
AB = \(\frac{1}{2}\)EF
( ∵ In all cases the pairs of respective corresponding angles are equal.)

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 7 Triangles InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 7th Lesson Triangles InText Questions

Do This

Question 1.
There are some statements given below. Write whether they are true or false. [Page No. 150]
Solution:
i) Two circles are always congruent. – False
ii) Two line segments of same length are always congruent. – True
iii) Two right angle triangles are some times congruent. – True
iv) Two equilateral triangles with their sides equal are always congruent. – True

Question 2.
Which minimum measurements do you require to check if the given figures are congruent? [Page No. 150]

i) Two rectangles.
Solution:
Length and breadth are required.

ii) Two rhombuses.
Solution:
One side and one interior angle are required.

Do These

Question 1.
State whether the following triangles are congruent or not. Give reasons for your answer. [Page No. 153]

Solution:
i) ΔABC ≅ ΔDEF
∵ ∠B = ∠E (∵ Angle sum property Z E = 180° – (70° + 60°) = 50°)
BC = EF
∠C = ∠F
∴ By SAS congruence
ΔABC ≅ ΔDEF

ii) In ΔMNL and ΔTSR
MN = ST
NL = TR
∠ M = ∠ T (∵ ΔMNL is flipped to get ΔTSR)
∴ ΔMNL ≅ ΔTSR

Question 2.
In the given figure, the point P bisects AB and DC. Prove that ΔAPC ≅ ΔBPD. [Page No. 153]

Solution:
Given that P bisects AB and DC.
Now in ΔAPC and ΔBPD
AP = BP (∵ P bisects AB)
CP = DP (∵ P bisects CD)
∠ APC = ∠ BPD
∴ ΔAPC ≅ ΔRPP (∵ SAS congruence)

Activity

1. On a tracing paper draw a line segment BC of length 6 cm.
2. From vertices B and C draw rays with angle 600 each. Name the point A where they meet
3. Fold the paper so that B and C fit precisely on top of each other. What do you observe? Is AB = AC?
[Page No. 160]

Do This

Question 1.
In the figure given below ΔABC and ΔDBC are two triangles such that \(\overline{\mathbf{A B}}=\overline{\mathbf{B D}}\) and \(\overline{\mathbf{A C}}=\overline{\mathbf{C D}}\) . Show that ΔABC ≅ ΔDBC [Page No. 164]

Solution:
Given that \(\overline{\mathbf{A B}}=\overline{\mathbf{B D}}\) and \(\overline{\mathbf{A B}}=\overline{\mathbf{B D}}\)

In ΔABC and ΔDBC
AB = BD (∵ given)
AC = DC (∵ given)
BC = BC (common side)
∴ ΔABC ≅ ΔDBC
( ∵ by SSS congruence)

Activity

Question 1.
Construct a right angled triangle with hypotenuse 5 cm. and one side 3 cm. long. How many different triangles can be constructed? Compare your triangle with those of the other members of your class. Are the triangles congruent? Cut them out and place one triangle over the other with equal side placed on each other. Turn the triangle if necessary what do you observe? You will find that two right triangles are congruent, if side and hypotenuse of one triangle are respectively equal to the corresponding side and hypotenous of other triangle.
[Page No. 165]

Activity

Question 1.
Draw a triangle ABC mark a point A’ on CA produced (new position of it)
So A’C > AC (Comparing the lengths) Join A to B and complete the triangle ABC.

What can you say about ∠A’BC and ∠ABC ?
Compare them. What do you observe?

Clearly, ∠A’BC > ∠ABC Continue to mark more points on CA (extended) and draw the triangles with the side BC and the points marked. You will observe that as the length of the side AC is increases (by taking different positions of A), the angle opposite to it, that is ∠B also increases.
[Page No. 169]

Question 2.
Construct a scalene triangle ABC (that is a triangle in which all sides are of different lengths). Measure the lengths of the sides.

Now, measure the angles. What do you observe?
In ΔABC figure, BC is the longest side and AC is the shortest side. ‘
Also, ∠A is the largest and ∠B is the smallest.
Measure angles and sides of each of the above triangles, what is the rela tion between a side and its opposite angle when compared with another pair?

Activity

Question
Draw a line segment AB. With A as centre and some radius, draw an arc and mark different points say P, Q, R, S, T on it.
Solution:

Join each of these points with A as well as with B (see figure). Observe that as we move from P to T, ∠A is becoming larger and larger. What is happening to the length of the side opposite to it? Observe that the length of the side is also increasing; that is ∠TAB > ∠SAB > ∠RAB > ∠QAB > ∠PAB
and TB > SB > RB > QB > PB.
Now, draw any triangle with all angles unequal to each other. Measure the lengths of the sides (see figure).

Observe that the side opposite to the largest angle is the longest. In figure, ZB is the largest angle and AC is the longest side
Repeat this activity for some more triangles and we see that the converse of the above Theorem is also true.

Measure angles and sides of each triangle given below. What relation you can visualize for a side and its opposite angle in each triangle.

In this way, we arrive at the following theorem. [Page No. 170]

Do This

Question
Draw a triangle ABC and measure its sides. Find the sum of the sides AB + BC, BC + AC; and AC + AB, compare it with the length of the third side. What do you observe ?
[Page No. 171]
Solution:

AB + BC = 4 + 3 = 7
= 7 > 4 = AC
BC + CA > AB;
3 + 4 > 4
CA + AB > BC;
4 + 4 > 3

DE + EF > DF
EF + DF > DE
FD + DE > EF
∴ Sum of any two sides of a triangle is greater than the third side.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.4 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 2nd Lesson Linear Equations in One Variable Exercise 2.4

Question 1.
Find the value of ’x’ so that l || m

Solution:
Given l|| m. Then 3x – 10° = 2x + 15°
[Vertically opposite angles and corresponding angles are equal.]
⇒ 3x – 10 = 2x + 15
⇒ 3x – 2x = 15 + 10
∴ x = 25°

Question 2.
Eight times of a number reduced by 10 is equal to the sum of six times the number and 4. Find the number.
Solution:
Let the number be ‘x’ say.
8 times of a number = 8 × x = 8x
¡f10 is reduced from 8x then 8x – O
6 times of a number = 6 × x = 6x
If 4 is added to 6x then 6x + 4
According to the sum,
8x – 10 = 6x + 4
⇒ 8x – 6x = 4 + 10
⇒ 2x = 14
⇒ x = 7
∴ The required number = 7

Question 3.
A number consists of two digits whose sum is 9. If 27 is subtracted from the number its digits are reversed. Find the number.
Solution:
Let a digit of two digit number be x.
The sum of two digits = 9
∴ Another digit = 9 – x
The number = 10 (9 – x) + x
= 90 – 10x + x
= 90 – 9x
If 27 is subtraçted from the number its digits are reversed.
∴ (90 – 9x) – 27 = 10x + (9 – x)
63 – 9x = 9x + 9
9x + 9x = 63 – 9
18x = 54
∴ x = \(\frac { 54 }{ 18 }\) = 3
∴ Units digit = 3
Tens digit = 9 – 3 = 6
∴ The number = 63

Question 4.
A number is divided into two parts such that one part is 10 more than the other. If the two parts are in the ratio 5:3, find the number and the two parts.
Solution:
If a number is divided into two parts in he ratio of 5 : 3, let the parts be 5x, 3x say.
According to the sum,
5x = 3x + 10
⇒ 5x – 3x = 10
⇒ 2x = 10
∴ x = \(\frac { 10 }{ 2 }\)
∴ x = 5
∴ The required number be
x + 3x = 8x
= 8 × 5 = 40
And the parts of number are
5 = 5 × 5 = 25
3 = 3 × 5 = 15

Question 5.
When I triple a certain number and add 2, I get the same answer as I do when I subtract the number from 50. Find the number.
Solution:
Let the number be x’ say.
3 times of a number = 3 × x = 3x
If 2 is added to 3x then 3x + 2
If ‘xis subtracted from 50 then it becomes 50 – x.
According to the sum,
3x + 2 = 50 – x
3x + x = 50 – 2
4x = 48 .
x = 12
∴ The required number 12

Question 6.
Mary is twice older than her sister. In 5 years time, she will be 2 years older than her sister. Find how old are they both now.
Solution:
Let the age of Marys sister = x say.
Mary’s age = 2 × x = 2x
After 5 years her sister’s age
= (x + 5) years
After 5 years Mary’s age
= (2x + 5) years
According to the sum,
2x + 5 = (x + 5) + 2
= 2x – x = 5 + 2 – 5
∴ The age of Mary’s sister = x = 2 years
Mary’s age = 2x = 2 x 2 = 4 years

Question 7.
In 5 years time, Reshma will be three times old as she was 9 years ago. How old is she now?
Solution:
Reshma’s present age = ‘x’ years say.
After 5 years Reshmats age
= (x + 5) years
Before 9 years Reshma’s age
=(x – 9) years
According to the sum
= x+ 5 = 3(x – 9) = 3x – 27
x – 3x = -27-5
-2x = -32
x = \(\frac{-32}{-2}\) = 16
∴ x = 16
∴ Reshma’s present age = 16 years.

Question 8.
A town’s population increased by 1200 people, and then this new population decreased 11%. The town now had 32 less people than it did before the 1200 increase. Find the original population.
Solution:
Let th population of a town after the increase of 1200 is x say.
11% of present population

The present population of town
= 11,200 – 1200 = 10,000

Question 9.
A man on his way to dinner shortly after 6.00 p.m. observes that the hands of his watch form an angle of 110°. Returning before 7.00 p.m. he notices that again the hands of his watch form an angle of 1100. Find the number of minutes that he has been away.
Solution:
Let the number be ‘x ray.
\(\frac { 1 }{ 3 }\) rd of a number = \(\frac { 1 }{ 3 }\) x x = \(\frac { x }{ 3 }\)
\(\frac { 1 }{ 5 }\) th of a number = \(\frac { 1 }{ 5 }\) x x = \(\frac { x }{ 5 }\)
According to the sum

∴ x = 30
∴ The required number is 30.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 2nd Lesson Linear Equations in One Variable Exercise 2.3

Question
Solve the following equations:
1. 7x – 5 = 2x
2. 5x – 12 = 2x – 6
3. 7p- 3 = 3p + 8
4. 8m + 9 = 7m + 8
5. 7z + 13 = 2z + 4
6. 9y + 5 = 15y – 1
7. 3x + 4 = 5(x – 2)
8. 3(t – 3) = 5(2t – 1)
9. 5(p – 3) = 3(p – 2)
10. 5(z + 3) = 4(2z + 1)
11. 15(x – 1) + 4(x + 3) = 2 (7 + x)
12. 3 (5z – 7) +2 (9z – 11) = 4 (8z – 7) – 111
13. 8(x – 3) – (6 – 2x) = 2(x + 2) – 5 (5 – x)
14. 3(n – 4)+2(4n – 5) = 5(n + 2) + 16
Solution:
1. 7x – 5 = 2x
⇒ 7x – 2x = 5
⇒ 5x = 5
⇒ x = \(\frac { 5 }{ 5 }\) = 1
∴ x = 1

2. 5x – 12 = 2x – 6
⇒ 5x – 2x = – 6 + 12
⇒ 3x = 6
⇒ x = \(\frac { 6 }{ 3 }\) = 2
∴ x = 2

3. 7p – 3 = 3p + 8
⇒ 7p – 3p = 8 + 3
⇒ 4p = 11
⇒ p = \(\frac { 11 }{ 4 }\)

4. 8m + 9 = 7m + 8
⇒ 8m – 7m = 8 – 9
∴ m = – 1

5. 7z + 13 = 2z + 4
⇒ 7z – 2z = 4 – 13
⇒ 5z = – 9
∴ z = \(\frac { -9 }{ 5 }\)

6. 9y + 5 = 15y – 1
⇒ 9y – 15y = – 1 – 5
⇒ -6y = -6
⇒ y = \(\frac { -6 }{ -6 }\)
∴ y = 1

7. 3x + 4 = 5(x – 2)
⇒ 3x + 4 =5x – 10
⇒ 3x – 5x= – 10 – 4
⇒ – 2x = – 14
:. x = 71

8. 3(t – 3) = 5(2t – 1)
⇒ 3t – 9 = 10t – 5
⇒ 3t – 10t = – 5+ 9
⇒ – 7t = 4
∴ t = \(\frac { -4 }{ 7 }\)

9. 5 (p – 3) = 3 (p – 2)
⇒ 5p – 15 = 3p – 6
⇒ 5p – 3p = -6 + 15
⇒ 2p = 9
∴ p = \(\frac { 9 }{ 2 }\)

10. 5(z + 3) = 4(2z + 1)
⇒ 5z + 15 = 8z + 4
⇒ 5z – 8z = 4 – 15
⇒ – 3z = – 11
⇒ z = \(\frac { -11 }{ 3 }\)
∴ z = \(\frac { -11 }{ 3 }\)

11. 15(x – 1) + 4(x + 3) = 2(7 + x)
⇒ 15x – 15 + 4x + 12= 14 + 2x
⇒ 19x – 3 = 14 + 2x
⇒ 19x – 2x = 14 + 3
⇒ 17x = 17 ,
x = \(\frac { 17 }{ 17 }\) = 1
∴ x = 1

12. 3(5z – 7)+2(9z – 11) = 4(8z – 7) – 111
⇒ 15z – 21 + 18z – 22 = 32z – 28 – 111
⇒ 33z – 43 = 32z – 139
⇒ 33z – 32z = – 139 + 43
∴ z = – 96

13. 8(x – 3) – (6 – 2x)=2(x+2)-.5(5 – x)
⇒ 8x – 24 – 6 + 2x = 2x + 4 – 25 +5x
⇒ 8x – 30 = 5x – 21
⇒ 8x – 5x= – 21 +30
⇒ 3x = 9
∴ x = 3

14. 3(n – 4) + 2(4n – 5) = 5(n + 2) + 16
⇒ 3n – 12 + 8n – 10 = 5n + 10 + 16
⇒ 11n – 22 = 5n + 26
⇒ 11n – 5n = 26 + 22
⇒ 6n =48
⇒ n = \(\frac { 48 }{ 6 }\) = 8
∴ n = 8

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 6 Linear Equation in Two Variables InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 6th Lesson Linear Equation in Two Variables InText Questions

Try This

Question 1.
Express the following linear equations in the form of ax + by + c = 0 and indicate the value of a, b, c in each case. [Page No. 128]
i) 3x + 2y = 9
Solution:
3x + 2y = 9
⇒ 3x + 2y – 9 = 0
Here a = 3, b = 2, c = -9

ii) – 2x + 3y = 6
Solution:
– 2x + 3y = 6
⇒ – 2x + 3y – 6 = 0
Here a = -2, b = 3, c = -6

iii) 9x – 5y = 10
Solution:
9x- 5y = 10
⇒ 9x- 5y – 10 = 0
Here a = 9, b = -5, c = -10

iv) \(\frac{x}{2}-\frac{y}{3}-5=0\)
⇒ \(\frac{3 x-2 y-30}{6}=0\)
⇒ 3x – 2y – 30 = 0
Here a = 3, b = – 2 and c = – 30

v) 2x = y
Solution:
2x = y
⇒ 2x – y = 0
Here a = 2, b = – 1 and c = 0

Try These

Take a graph paper, plot the point (2, 4), and draw a line passing through it. Now answer the following questions. [Page No. 135]

Question 1.
Can you draw another line that passes through the point (2, 4) ?
Solution:
Yes, We can draw another line passing through (2, 4).

Question 2.
How many such lines can be drawn ?
Solution:
Infinitely many lines can be drawn.

Question 3.
How many linear equations in two variables exist for which (2, 4) is a solution ? Solution:
Infinitely many linear equations in two variables exist.
Graph is given in next page.

Do This

Question 1.
i) Draw the graph of following equations. [Page No. 145]
a) x = 2 b) x = -2 c) x = 4 d) x = -4

ii) Are the graphs of all these equations parallel to Y-axis? .
Solution:
Yes; all these lines are parallel to Y-axis.

iii) Find the distance between the graph and the Y-axis in each case.
Solution:

Question 2.
i) Draw the graph of the following equations. [Page No. 145]
a) y = 2 b) y = -2 c) y = 3 d) y = – 3
Solution:

ii) Are all these parallel to the X-axis?
Solution:
Yes, these are all parallel to X-axis.

iii) Find the distance between the graph and the X-axis in each case.

Try These

Question
Find the equation of the Y – axis. [Page No. 146]
Solution:
All points on Y – axis have their x – co-ordinates as zero.
∴ Equation of Y- axis is x = O.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 2nd Lesson Linear Equations in One Variable Exercise 2.2

Question 1.
Find ‘x’ in the following figures?


Solution:
1) In a triangle the exterior angle is equal to the sum of its opposite interior angles.
∴ ∠ACD = ∠B + ∠A
⇒ 123°= x + 56°
⇒ x = 123°- 56° = 67°
∴ x = 67°

ii) Sum of three angles of a triangle = 180°
∠P + ∠Q +∠R = 180°
⇒ 45° + 3x + 16°+ 68° = 180°
⇒ 3x + 129° = 180°
3x = 180 – 129 = 51
∴ x = \(\frac{51}{3}\)
∴ x = 17°

iii) ∠A + ∠B + ∠C = 180°
⇒ 25° + x + 30° = 180°
x + 55°= 180°
x = 180 – 55 = 125°
∴ x = 125°

iv) In ΔXYZ, \(\overline{\mathrm{XY}}=\overline{\mathrm{XZ}}\) then ∠Y = ∠Z
∴ 2x + 7° = 45°
⇒ 2x = 45 – 7
⇒ 2x = 38
⇒ x = \(\frac{38}{2}\)
∴ x = 19°

v) From ΔBOA
\(\overline{\mathrm{AB}}=\overline{\mathrm{OA}} \) ⇒ ∠B = ∠O = 3x + 10° ………(1)
From ΔCOD
\(\overline{\mathrm{OC}}=\overline{\mathrm{CD}} \) ⇒ ∠O = ∠D …………………(2)
[∵ The angles which are opposite to the equal sides are equal].
from (1) & (2)
∠BOA = ∠COD
[∵ Vertically opposite angles are equal.]
But ∠COD = 90 – x
(∵ 2x + ∠O + ∠D = 180
⇒ 2x + ∠O + ∠O = 180 (∵∠O = ∠D)
⇒ 2∠O = 180 – 2x
∠O = \(\frac{180-2 x}{2}\) = 90 – x]
∴ From ∠BOA = ∠COD
⇒ 3x + 10 = 90 – x
⇒ 3x + x = 90 – 10
⇒ 4x = 80
∴ x = 20°

Question 2.
The difference between two numbers is 8. if 2 is added to the bigger number the result will be three times the smaller number. Find the numbers.
Solution:
Let the bigger number be x.
The difference between two numbers 8
∴ Smaller number = x – 8
If 2 is added to the bigger number the result will be three times the smaller
number.
So x + 2 = 3(x – 8)
x + 2 = 3x – 24
x – 3x = -24 – 2
– 2x = -26
∴ x = \(\frac{26}{2}\) = 13
∴ Bigger number = 13
Smaller number = 13 – 8 = 5

Question 3.
What are those two numbers whose sum is 58 and difference is 28’?
Solution:
Let the bigger number be ‘x’.
The sum of two numbers = 58
∴ Smaller number = 58 – x
The difference of two numbers = 28
∴ x – (58 – x) = 28
x – 58 + x = 28
2x = 28 + 58 = 86
∴ x = \(\frac{86}{2}\) = 43
∴ Bigger number or one number = 43
Smaller number or second number = 58 – 43 = 15

Question 4.
The sum of two consecutive odd numbers is 56. Find the numbers.
Solution:
Let the two consecutive odd numbers
be 2x 1, 2x + 3 say.
Sum of the odd numbers
=(2x + 1) + (2x + 3) = 56
= 4x + 4 = 56
⇒ 4x = 56 – 4 = 52
x = \(\frac{52}{4}\) = 13
∴ x = 13
∴ 2x + 1 = 2 × 13 + 1
= 26 + 1 = 27
2x + 3 = 2 × 13 + 3
= 26 + 3 = 29
∴ The required two consecutive odd numbers be 27, 29.

Question 5.
The sum of three consecutive multiples of 7 is 777. Find these multiples.
(Hint: Three consecutive multiples of 7 are ‘x’, ‘x+ 7’, ‘x+ 14’)
Solution:
Let the three consecutive multiples of 7 be x, x + 7, x + 14 say.
According to the sum,
The sum of three consecutive multiples of 7 is 777.
⇒ x + (x + 7) + (x + 14)= 777
⇒ 3x + 21 = 777
⇒ 3x = 777 – 21 = 756
x = \(\frac { 756 }{ 3 }\) = 252
x+ 7 = 252 + 7 = 259
x + 14 252 + 14 = 266
∴ The required three consecutive multiples of 7 are 252, 259, 266

Question 6.
A man walks 10 km, then travels a certain distance by train and then by bus as far as twice by the train. 1f the whole journey is of 70km, how far did he travel by train?
Solution:
The distance travelled by walk = 10 km
Let the distance travelled by train = x km say.
The distance travelled by bus
= 2 × x = 2x km
∴ 10 + x + 2x = 70
⇒ 3x = 70 – 10
⇒ 3x = 60
⇒ x = \(\frac { 60 }{ 3 }\) = 20
⇒ x = 20
∴ The distance travelled by train = 20 km.

Question 7.
Vinay bought a pizza and cut it into three pieces. When the weighed the first piece he found that it was 7g lighter than the second piece and 4g.heavier than the third piece. If the whole pizza weighed 300g. How much did each of the three pieces weigh?

(Hint: weight of normal piece be ‘x’ then weight of largest piece is ‘x+ 7’, weight of the smallest piece is ‘x-4’)
Solution:
If pizza is cut into three pieces.
Let the weight of first piece he ‘x’ gm say.
Weight of the second piece = (x + 7) gm
Weight of the third piece = (x – 4) gm
According to the sum
∴ x + (x + 7) + (x – 4) = 300
⇒ 3x + 3 = 300
⇒ 3x = 300 – 3 = 297
⇒ x = \(\frac { 297 }{ 3 }\) = 99
∴ x= 99
x + 7= 99 + 7 =106
x – 4 = 99 – 4 = 95
∴ The required 3 pieces of pizza weighs 95 gm. 99 gm, 106 gm.

Question 8.
The distance around a rectangular field is 400 meters.The length of the field is 26 meters more than the breadth. Calculate the length and breadth of the field’?
Solution:
Let the breadth of a rectangular field = x m
Length = (x + 26) m.
Perimeter of a rectangular field
= 2(l + b) = 400
l + b = 200
x + 26 + x = 200
2x = 200 – 26 = 174
x = \(\frac { 174 }{ 2 }\)
∴ x = 87
∴ The length = x +26
= 87 + 26
= 113 m
Breadth = x = 87 m.

Question 9.
The length of a rectangular field is 8 meters less than twice its breadth. If the perimeter of the rectangular field is 56 meters, find its length and breadth’?
Solution:
Let the breadth of a rectangular field = xm.
Length = 2 × x – 8 = (2x – 8) m.
Perimeter of a field = 56 m.
∴ 2(l + b) = 56
l + b = 28
2x – 8 + x = 28
3x = 28 + 8 = 36
x = \(\frac { 36 }{ 3 }\)
∴ x= 12
∴ Breadth = 12 m
Length = 2x – 8
= 2 × 12 – 8
= 24 – 8 = 16m.

Question 10.
Two equal sides of a triangle are each 5 meters less than twice the third side. if the perimeter of the triangle is 55 meters, find the length of its sides’?
Solution:
A triangle in which the length of the third side = x m. say.
The length of remaining two equal sides = 2 × x – 5 = (2x – 5) m.
Perimeter of a triangle = 55 m.
∴ (2x – 5) + (2x -5) + x = 55
⇒ 5x – 10 = 55
⇒ 5x = 65
⇒ x = \(\frac { 65 }{ 5 }\)
∴ x = 13m
2x – 5 = 2 × 13 – 5 = 26 – 5 = 21m.
∴ The lengths of three sides of a triangle are 13, 21, 21. (in m.)

Question 11.
Two complementary angles differ by 12°, find the angles’?
Solution:
Let one angle in two complementary angles be x.
Sum of the two complementary angles = 90°
∴ Second angle = 90° – x
Here two complementary angles differ by 12°.
∴ x – (90°- x) = 12°
x – 90° + x = 12°
2x = 12° + 90° = 102°
∴ \(\frac{102^{\circ}}{2}\) = 51°
one angle = 51°
Second angle 90° – 51° = 39°

Question 12.
The ages of Rahul and Laxmi arc in the ratio 5:7. Four years later, the sum of their ages will
be 56 years. What are their present ages’?
Solution:
The ratio of ages of Rahul and Lakshmi = 5:7
Let their ages be 5x, 7x say.
After 4 years Rahuls agt. = 5x + 4
After 4 years Lakshmis age = 7x + 4
According to the sum,
After 4 years the sum of their ages = 56
⇒ (5x + 4) + (7x + 4) = 56
⇒ 12x + 8 = 56
⇒ 12x = 56 – 8 = 48
⇒ x= \(\frac { 48 }{ 12 }\) = 4
∴ x = 4
∴ Rahuls present age
= 5x = 5 × 4 = 20 years
∴ Lakshmi’s present age
= 7x = 7 × 4 = 28 years

Question 13.
There are 180 multiple choice questions in a test. A candidate gets 4 marks for every correct answer, and for every un-attempted or wrongly answered questions one mark is deducted from the total score of correct answers. If a candidate scored 450 marks in the
test how many questions did he answer correctly ?
Solution:
Number of questions attempted for correct answers = x say
umber of questions attempted for wrong answers = 180 – x
4 marks are awarded for every correct answer.
Then numl)er of marks ohtained for correct answers 4 × x = 4x
1 mark is deducted for every wrong answer.
∴ Number of marks deducted for wrong answers
=(180 – x) × 1 = 180 – x
According to the sum.
4x – (180 – x) = 450
⇒ 4x – 180 + x = 450
⇒ 5x = 450 + 180
⇒ 5x = 630
x = \(\frac { 630 }{ 5 }\)
∴ x = 126
∴ Number of questions attempted for correct answers = 126

Question 14.
A sum of ₹ 500 is in the form of denominations of ₹ 5 and ₹ 10. If the total number of notes is 90 find the number of notes of each denomination.
(Hint: let the number of 5 rupee notes be ‘x’, then number of 10 rupee notes = 90 – x)
Solution:
Number of ₹ 5 notes = x say.
Number of ₹ 10 notes = 90 – x
5x + 10(90 – x) = 500
5x + 900 – 10x = 500
– 5x = 500 – 900 = – 400
x = \(\frac { -400 }{ -5 }\)
∴ x = 80
∴ Number of ₹ 5 notes = 80
Number of ₹ 10 notes
= 90 – x = 90 – 80 = 10

Question 15.
A person spent ₹ 564 in buying geese and ducks,if each goose cost ₹ 7 and each duck ₹ 3 and if the total number of birds bought was 108, how many of each type did he buy?

Solution:
Let the number of pens be x.
The total number of things = 108
∴ The number of pencils = 108 – x
Thecostofpens of x = ₹7 × x = ₹ 7x
The cost of pencils of (108- x)
= ₹3(108 – x) = ₹ (324 – 3x)
Total amount to t)uy eflS and Pencils = ₹564
∴ 7x + (324 – 3x) 5M
7x + 324 – 3x = 564
4x = 564 – 324 = 240
∴ x = \(\frac { 240 }{ 4 }\) = 60
The number of pens = 60
The number of pencils = 108 – 60 = 48

Question 16.
The perimeter ofa school volleyball court is 177 ft and the length is twice the width. What are the dimensions of the volleyball court’?

Solution:
Breadth of a volleyball court = x feet say.
∴ Its length = 2 × x = 2 x feet.
The perimeter of a court = 177 feet.
⇒ 2(l + b) = 177
⇒ 2(2x + x) = 177
⇒ 2 x 3x = 177
⇒ 6x = 177
⇒ x =\(\frac { 177 }{ 6 }\)
∴ x = 29.5
∴ The breadth of a volleyball court = 29.5 ft
The length of a volleyball court = 2x = 2 × 29.5 = 59ft

Question 17.
The sum of the page numbers on the facing pages of a book is 373. What are the page numbers?
(Hint :Let the page numbers of open pages are x and (x + 1)

Solution:
Number of first page of a opened book x
Number ol second page = x + 1
∴ The surï of the numbers of two pages = 373
⇒ x + (x + 1) = 373
⇒ 2x + 1 = 373
2x = 372
⇒ x = 186
∴x + 1 = 186 + 1 = 187
∴ Numbers of two c(nlsecutive pages = 186, 187

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 15th Lesson Proofs in Mathematics InText Questions

Do This

Question
Make 5 more sentences and check whether they are statements or not ? * Give reasons. [Page No. 311]
Solution:
1) 9 is a prime number – False
This is a statement because we can judge the truthness of this sentence. Clearly it is a false statement as 9 has factors other than 1 and 9, hence it is a composite number.
2) x is less than 5 – can’t say True or False
This is not a statement. The truthness can t be verified unless the value of x is known. Hence it is a sentence only.
3) 3 + 5 = 8 – True
The above sentence is a statement. It is a true statement as 5 + 3 = 8.
4) Sum of two odd numbers is even – True
The above sentence can be verified as a true sentence by taking ex¬amples like 3 + 5 = 8, 5 + 7 = 12 etc. Hence it is a true statement.
5) \(\frac{\mathrm{X}}{2}\) +3 = 9- can’t say True or False.
The above sentence is not a state¬ment. Its truthness can’t be
verified without the value of x.

Try This

1. 3 is a prime number.
2. Product of two odd integers is even.
3. For any real number x; 4x T x = 5x
4. The earth has one moon.
5. Ramu is a good driver.
6. Bhaskara has written a book “Leelavathi ”.
7. All even numbers are composite.
8. A rhombus is a square.
9. x > 7.
10. 4 and 5 are relative primes.
11. Silver fish is made of silver.
12. Humans are meant to rule the earth.
13. For any real number .v. 2x > x.
14. Havana is the capital of Cuba.

Question
Which of the above statements can be tested by giving counter example ?
[Page No. 312]
Solution:
Statements 2, 7, 8, 13 can be tested by giving counter examples 2) Product of two odd integers is even. Counter example.
2) Product of two odd integers 3 and 5 is 3×5 = 15 is not an even number.
7) All even numbers are composite. Counter example : 2 is an even prime.
8) A rhombus is a square.

Counter example: (40°, 140°, 40°. 140°) is a rhombus.
13) For anyx; 2x > x
Counter example : for x = -3:
2x = 2(-3) = – 6
here -6 < – 3

b

Try This

Envied by the popularity of Pythagoras, his younger brother claimed a different relation between the sides of a right angled triangles. [Page No. 319]

Liethagoras Theorem: In any right angled triangle the square of the smallest side equals the sum of the other sides. Check this conjecture, whether It is right or wrong.
Solution:
This conjecture is true for the above
triangles.
i) 3² 5 + 4 ⇒ 9 = 5 + 4
ii) 52 = 25 = 12 + 13
iii) 7² = 49 = 24 + 25
But, when the smallest side happens to be an even integer the conectiire may not hods good.
Eg: 1) 6² = 36 ≠ 10 + 8

ii) 12² = 144 ≠ 20 + 16

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 5 Co-Ordinate Geometry InText Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 5th Lesson Co-Ordinate Geometry InText Questions

Do This

Question
Describe the seating position of any five students in your classroom. [Page No. 108]
Solution:
Oral answer – depend on the classroom.

Activity (Ring Game)

Question
Have you seen ‘Ring game’ in exhibitions ? We throw rings on the objects arranged in rows and columns. Observe the following picture. [Page No. 108]
Complete the following table.
Solution:

Question
Is the object 3rd column and 4th row is same as 4th column 3rd row ?
Solution:
No

Do This

Question
Among the points given below, some of the points lie on X – axis. Identify them.
i) (0, 5) ii) (0, 0) iii) (3, 0) iv) (- 5, 0) v) (- 2, – 3) vi) (- 6,0) vii)(0,6) viii)(0, a) ix) (b, 0) [Page No. 114]
Solution:
The points (ii) (0, 0), (iii) (3, 0) (iv) (- 5, 0) (vi) (- 6, 0) and (ix) (b, 0) lie on X – axis.

Try These

Question 1.
Which axis the points such as (0, x), (0, y), (0, 2) and (0, – 5) lie on ? Why ? [Page No. 114]
Solution:
All the above points lie on Y – axis, since the X – coordinate of all these points is zero.

Question 2.
What is the general form of the points which lie on X – axis. [Page No. 114]
Solution:
The general form of points lying on X – axis is (x, 0).

Do This

Question
Plot the following points on a Cartesian plane.
1. B (- 2, 3) 2. L (5, – 8) 3. U (6, 4) 4. E (-3, – 3) [Page No. 120]
Solution:

Do This

Question
i) Write the co-ordinates of the points A, B, C, D and E. [Page No. 121]
ii) Write the co-ordinates of F, G, H, I and J.

Solution”
i) A(2, 9) ; B(5, 9); C(2, 6) ; D(5, 3) ; E(2, 3)
ii) F(- 6, – 2) ; G(- 4, – 5) ; H(- 3, – 7) ; I(- 9, – 7) ; J(- 8, – 5)

Activity

Question
Study the positions of different cities like Hyderabad, New Delhi, Chennai and Vishakapatnam with respect to longitudes and latitudes on a globe. [Page No. 123]
Solution:
Student Activity

Creative Activity

Question
Take a graph sheet and plot the following pairs of points on the axes and join them with line segments. [Page No. 123]
(1, 0) (0, 9) ; (2, 0) (0, 8) ; (3, 0) (0, 7) ; (4, 0) (0, 6) ;
(5, 0) (0, 5) ; (6, 0) (0, 4) ; (7, 0) (0, 3) ; (8, 0) (0, 2) ; (9, 0) (0, 1).
Try to complete the picture by using above points. What did you observe ?
Solution:
Student Activity

AP Board 9th Class Maths Solutions