Class 8

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 6 Square Roots and Cube Roots Ex 6.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 6th Lesson Square Roots and Cube Roots Exercise 6.1

Question 1.
What will be the units digit of the square of the following numbers?
(i) 39
(ii) 297
(iii) 5125
(iv) 7286
(v) 8742
Solution:

Question 2.
Which of the following numbers are perfect squares?
(i) 121
(ii) 136
(iii) 256
(iv) 321
(v) 600
Solution:

Question 3.
The following numbers are not perfect squares. Give reasons?
(i) 257
(ii) 4592
(iii) 2433
(iv) 5050
(v) 6098
Solution:
i) 257 → The units digit of the number is 7. So it is not a perfect square number.
ii) 4592 → The units digit of the number is 2. So it is not a perfect square number.
iii) 2433 → The units digit of the number is 3. So, it is not a perfect square number.
Iv) 5050 → The last two digits of the number are not two zero’s. So, it is not a perfect square number.
v) 6098 → The units digit of the number is 8. So It is not a perfect square number

Question 4.
Find whether the square of the following numbers are even or odd?
(i) 431
(ii) 2826
(iii) 8204
(iv) 17779
(v) 99998
Solution:

Question 5.
How many numbers lie between the square of the following numbers
(i) 25; 26
(ii) 56; 57
(iii) 107;108
Solution:
The numbers lie between the square of the numbers are:
1) 25,26 → 2 x 25=50
ii) 56, 57 → 2 x 56 = 112
iii) 107, 108 → 2 x 107 = 214

Question 6.
Without adding, find the sum of the following numbers
(i) 1 + 3 + 5 + 7 + 9 =
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 =
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 =
Solution:
(i) 1 + 3 + 5 + 7 + 9 = (5)² = 5 x 5 = 25
[∵ Sum of ‘n’ consecutive odd number = n²]
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 9² = 9 x 9 = 81
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 = 13² = 13 x 13 = 169

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures Ex 8.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 8th Lesson Exploring Geometrical Figures Exercise 8.2

Question 1.
Cut the bold type English alphabets (capital) and paste in your note book. Draw possible number of lines of symmetry for each of the letter.
(i) How many letters have no linear symmetry?
(ii) How many letters have one line of symmetry?
(iii) How many letters have two lines of symmetry?
(iv) How many letters have more than two lines of symmetry?
(v) Which of them have rotational symmetry?
(vi) Which of then have point symmetry?
Solution:

i) letters don’t have line of symmetry
→ F, G, J, L, N, P, Q, R, S, Z.
ii) Liters which are having 1 line of symmetry
→ B, C, D, E, H, I, K, M, 0, T, U, V,
iii) Letirs which are having 2 lines of symmetry
→ H, I, O, X.
iv) Letters which are having more than two lines of symmetry O, X.
v) Letters which are having rotational symmetry
→ B, D, E, H, I, M, 0, S, T, W, X, Z.
vi) Letters which are having point symmetry
→ 0, X, M, W, H, 1, E, D.

Question 2.
Draw lines of symmetry for the following figures. Identify which of them have point symmetry. Is there any implication between lines of symmetry and point symmetry?

Solution:

The given figures has point of symmetry. Which are having line of symmetry are also having point of symmetry.

Question 3.
Name some natural objects with faces which have atleast one line of symmetry.
Solution:
1) Moon (Full)
2) Face of a human being
3) Orange fruit
4) Sunflower
5) Butterfly

Question 4.
Draw three tessellations and name the basic shapes used on your tessellation.
Solution:
I may notice that the basic shapes used to draw tessellations are pentagon, rectangle, squares, equilateral triangles, hexagons etc. Most tessellations can’t be formed with these shapes.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures Ex 8.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 8th Lesson Exploring Geometrical Figures Exercise 8.1

Question 1.
Name five pairs of congruent objects, you use daily.
Solution:
Solution:
i) Pair of ear rings.
ii) Wheels of a cycle.
iii) Equal lengths of scales.
iv) Pair of bangles.
v) Mathematics textbooks of same class.

Question 2.
(a) Draw two congruent figures. Are they similar? Explain
(b) Take two similar shapes. If you slide rotate or flip one of them, does the similarity remain?
Solution:
a) ΔABC ≅ ΔPQR

From ΔABC & ΔPQR
AB = PQ
AC = PR
BC = QR
∠A = ∠P
∠B = ∠Q
∠C = ∠R
∴ Congruent triangles are similar to each other.
But similar triangles are not congruent.

b)

From fig. ΔXYZ and ΔSTU are similar triangles. If these triangles are rotated even they are similar to each other.

Question 3.
If Δ ABC ≅ Δ NMO, name the congruent sides and angles.
Solution:

From the congruent triangles
ΔABC ≅ ΔNMO
AB = NM
BC = MO
AC = NO
∠A = ∠N
∠B =∠M
∠C = ∠O

Question 4.
State whether the following statements are true. Explain with reason.
(i) Two squares of side 3 cm each and one of them rotated through 45° are congruent.
(ii) Any two right triangles with hypotenuse 5 cm, are congruent.
(iii) Any two circles of radii 4 cm each are congruent.
(iv) Two equilateral triangles of side 4 cm each but labeled as A ABC and A LHN arc not congruent.
(v) Mirror image of a polygon is congruent to the original.
Solution:
i) True. If the square is rotated 45° then it is remaining congruent.
ii) True. If the hypotenuse of two right triangles are equal, then the two triangles are congruent to each other, [∵ Their corresponding sides and angles will also be equal]

(iii) True
Since they are having equal radii (r₁ = r₂ = 4 cm )
= 4 cm).
∴ The given circles are congruent.

iv) False.
If two sides of one triangle are equal to the corresponding two sides of another triangle, then the third sides are in the proportion.
∴ ΔABC ≅ ΔLHN

But given tXABC,’L\LHN. Then it is false.

v) True.
Always the mirror images of a polygon is congruent to the original.

Question 5.
Draw a polygon on a square dot sheet. Also draw congruent figures in different directions and mirror image of it.
Solution:

Question 6.
Using a square dot sheet or a graph sheet draw a rectangle and construct a similar figure.
Find the perimeter and areas of both and compare their ratios with the ratio of their
corresponding sides.
Solution:

Question 7.
7 pillars are used to hold a slant iron gudder as shown in the figure. If the distance between every two pillars is 1 m and height of the last piller is 10.5 m. Find the height of pillar.

Solution:

The height of the 1st pillar
h₁ = \(\frac{1}{7}\) x 10.5 = 1.5m.

The height of 2nd pillar
h₂ = \(\frac{2}{7}\) x 10.5 = 3m.

The height of 3rd pillar
h₃ = \(\frac{3}{7}\) x 10.5 = 4.5 m.

The height of 4th pillar
h₄ = \(\frac{4}{7}\) x 10.5 = 6 m.

The height of 5th pillar
h₅ = \(\frac{5}{7}\) x 10.5 = 7.5 m.

The height of 6th pillar
h₆ =\(\frac{6}{7}\) x 10.5 = 9 m.

Question 8.
Standing at 5 m apart from a vertical pole of height 3 m, Sudha observed a building at the back of the piller that tip of the pillar is in line with the top of the building. If the distance
between pillar and building is 10 m, estimate the height of the building. [Here height of
Sudha is neglected]
Solution:
ΔAOAB ~ ΔOCD
The corresponding sides of similar triangles are in a same proportion^?

⇒ h = 3 x 3 = 9
∴ h = height of a building = 9 mts.

Question 9.
Draw a quadrilateral of any measurements. Construct a dilation of scale factor 3. Measure their corresponding sides and verify whether they are similar.
Solution:

Observe the following dilation ABCD, it is a square drawn on a graph sheet.
All vertices A, B, C, D are joined from the sign ’O’ and produced to 3 times the length upto A’, B’, C’ and D’ respec¬tively. Then A’, B’, C’, D’ are joined to form a rectangle which 3 times has enlarged sides of ABCD. Here, 0 is called centre of dilation and \(\frac{\mathrm{OA}^{\prime}}{\mathrm{OA}}=\frac{3}{1}\) = 3 is called scale factor.
∴ □ ABCD ~ □ A’B’C’D’
[ ∵ their corresponding sides are equal]

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 1 Rational Numbers Ex 1.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 1st Lesson Rational Numbers Exercise 1.3

Question 1.
Express each of the following decimal in the \(\frac{p}{q}\) form.
(i) 0.57 (ii) 0.176 (iii) 1.00001 (iv) 25.125
Solution:
(i) 0.57 = \(\frac{57}{100}\) (∵ two digits are there after the decimal poing)
(ii) 0.176 = \(\frac{176}{1000}\)
(iii) 1.00001 = \(\frac{100001}{100000}\)
(iv) 25.125 = \(\frac{25125}{1000}\)

Question 2.
Express each of the following decimals in the rational form \(\frac{p}{q}\)
(1) \(0 . \overline{9}\)
(ii) \(0 . \overline{57}\)
(iii) \(0 .7 \overline{29}\)
(iv) \(12.2 \overline{8}\)
Solution:
(i) \(0 . \overline{9}\)
Let x = \(0 . \overline{9}\)
⇒ x = 0.999 ………………. (1)
Here periodicity is 1. So, equation (1) should be multiplied both sides with
= 10 × x = 10 × 0.999
10 x = 9.999 ………….. (2)

\(0 . \overline{9}\) = 1

Second Method:
\(0 . \overline{9}=0+\overline{9}=0+\frac{9}{9}\)
= 0 + 1 = 1

(ii) \(0 . \overline{57}\)
x = \(0 . \overline{57}\) ⇒ x = 0.5757…………(1)
Here periodicity is 2. So, we should multiply with 100
⇒ 100 × x = 100 x 0.5757 …………..
100 × =57.57 ……………………. (2)

(iii) \(0 .7 \overline{29}\)
x = \(0 .7 \overline{29}\)
x = \(0 .7 \overline{29}\) ⇒ x = 0.7979…………(1)
Here periodicity is 2. So, equation (1) should multiply with 100
⇒ 100 × x = 100 × 0.72929 …………..
100 × = 72.929 …………………… (2)

(iv) \(12.2 \overline{8}\)
x = (iv) \(12.2 \overline{8}\)
⇒ x = 12.288 ………..(1)
Here periodicity is 1. So, equation (1) should multiply with 10
⇒ 100 × x = 100 × 12.288 …………..
10 x = 122.888 …………………… (2)

Question 3.
Find(x + y) ÷ (x – y) if
(i) x = \(\frac{5}{2}\), y = \(-\frac{3}{4}\)
(ii) x = \(\frac{1}{4}\), y = \(\frac{3}{2}\)
Solution:
If x = \(\frac{5}{2}\), y = \(-\frac{3}{4}\) then

ii) x = \(\frac{1}{4}\), y = \(\frac{3}{2}\)

Question 4.
Divide the sum of \(-\frac{13}{5}\) and \(\frac{12}{7}\) by the product of \(-\frac{13}{7}\) and \(-\frac{1}{2}\)
Solution:
Sum of \(-\frac{13}{5}\) and \(\frac{12}{7}\)

the product of \(-\frac{13}{7}\) and \(-\frac{1}{2}\)

Question 5.
If \(\frac{2}{5}\) of a number exceeds \(\frac{1}{7}\) of the same number by 36. Find the number.
Solution:
Let the number be ‘x’ say.
\(\frac{2}{5}\) part of x = \(\frac{2}{5}\) × x = \(\frac{2x}{5}\)
\(\frac{1}{7}\) part of x = \(\frac{1}{7}\) × x = \(\frac{x}{7}\)
∴ According to the sum,

Question 6.
Two pieces of lengths 2\(\frac{2}{5}\) m and 3\(\frac{3}{10}\) mare cut off from a rope 11 m long. What is the length of the remaining rope?
Soltuion:
The length of the remaining rope

∴ The length of remaining rope
= 5\(\frac{1}{10}\) mts.

Question 7.
The cost of 7\(\frac{2}{3}\) meters of cloth is ₹12\(\frac{3}{4}\) . Find the cost per metre.
Solution:
The cost of 7\(\frac{2}{3}\) mts (\(\frac{23}{3}\) mts ) of cloth
= ₹ \(12 \frac{3}{4}\) = ₹ \(\frac{51}{4}\)
∴ The cost of 1m cloth
= \(\frac{51}{4} \div \frac{23}{3}=\frac{51}{4} \times \frac{3}{23}=\frac{153}{92}\) = ₹ 1.66

Question 8.
Find the area of a rectangular park which is 18\(\frac{3}{5}\)m long and 8\(\frac{2}{3}\) in broad.
Solution:
The length of the rectangular park
= 18\(\frac{3}{5}\)m = \(\frac{93}{5}\)
Its width / breath = 8\(\frac{2}{3}\) m = \(\frac{26}{3}\) m
∴ Area of the rectangular park
(A) = l × b

Question 9.
What number should \(-\frac{33}{16}\) be divided by to get \(-\frac{11}{4}\)
Solution:
Let the dividing number be ‘x’ say.

Question 10.
If 36 trousers of equal sizes can be stitched with 64 meters of cloth. What is the length of the cloth required for each trouser?
Solution:
36 trousers of equal sizes can he stitched with 64 mts of cloth, then the length of the cloth ¡s required for each trouser
= 64 ÷ 36
= \(\frac{64}{36}=\frac{16}{9}\) = 1 \(\frac{7}{9}\)

Question 11.
When the repeating decimal 0.363636 …. is written in simplest fractional form\(\frac{p}{q}\) , find the sum p+ q.
Solution:
x = 0.363636………………………….. (1)
Here periodicity is ‘2’. So, equation (1) should be multiplied both sides with 100.
⇒ 100 × x = 100 × 0.363636 …………..
100 x = 36.3636 ………..(2)

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 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 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 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 2nd Lesson Linear Equations in One Variable Exercise 2.1

Question
Solve the following Simple Equations:

(i) 6m = 12
(ii) 14p -42
(iii) -5y = 30
(iv) – 2x = – 12
(v) 34x = – 51
(vi) \(\frac{n}{7}\) = -3
(vn) \(\frac{2x}{3}\) = 8
(vui) 3x+1 = 16
(ix) 3p – 7 = 0
(x) 13 – 6n = 7
(xi) 200y – 51 = 49
(xii) 11n + 1 = 1
(xiii) 7x – 9 = 16
(xiv) 8x + \(\frac{5}{2}\) =13
(xv) 4x – \(\frac{5}{3}\) = 9
(xvi) x – \(\frac{4}{3}\) = 3\(\frac{1}{2}\)
Solution:
i) 6m = 12 ⇒ m = \(\frac{12}{6}\) ⇒ m = 2

ii) 14p = – 42p ⇒ P = \(\frac{-42}{14}\)
∴ p = -3

iii) -5y = 30 ⇒ y = \(\frac{30}{-5}\) = -6
∴ y = -6

iv) -2x = -12
⇒ 2x = 12
x = \(\frac{30}{-5}\)
= 6
∴ x = 6

v) 34x = -51
⇒ \(\frac{-3}{2}\) = \(\frac{-3}{2}\)
∴ x = \(\frac{-3}{2}\)

vi) \(\frac{n}{7}\) = -3
⇒ n = -3 x 7 = -21
∴ n = -21

vii) \(\frac{2x}{3}\) = 18 ⇒ 18 x \(\frac{3}{2}\) = 27
∴ x = 27

viii) 3x + 1 = 16
3x = 16 – 1 = 15
3x = 15
x = \(\frac{15}{3}\)
∴ x = 5

ix) 3p – 7 = 0
⇒ 3p = 7
∴ p = \(\frac{7}{3}\)

x) 13 – 6n = 7 ⇒ -6n = 7 – 13
⇒ -6n = -6 ⇒ n= \(\frac{-6}{-6}\)
∴ n = 1

xi) 200y – 51 = 49
⇒ 200y = 49 + 51
⇒ 200y = 100
⇒ y = \(\frac{100}{200}\)
∴ y = \(\frac{1}{2}\)

xii) 11n + 1 = 1
⇒ 11n = 1 – 1
= 11n = 0
⇒ n = \(\frac{0}{11}\) = 0
∴ n = 0

xiii) 7x – 9 = 16
⇒ 7x = 16 + 9
⇒ 7x = 25
∴ x = \(\frac{25}{7}\)

xiv) 8x + \(\frac{5}{2}\) = 13

xv) 4x – \(\frac{5}{3}\)

xvi) x + \(\frac{4}{3}\) = 3\(\frac{1}{2}\)
⇒ \(x+\frac{4}{3}=\frac{7}{2}\)
⇒\(\frac{7}{2}-\frac{4}{3}=\frac{21-8}{6}\)
∴ x = \(\frac{13}{6}\)

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 9 Area of Plane Figures Ex 9.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 9th Lesson Area of Plane Figures Exercise 9.2

Question 1.
A rectangular acrylic sheet is 36 cm by 25 cm. From it, 56 circular buttons, each of diameter 3.5 cm have been cut out. Find the area of the remaining sheet.
Solution:
The dimensions of a rectangular acrylic sheet = 36 cm × 25 cm.
∴ Its area = l × b
= 36 × 25 = 900 sq.cm
The diameter of a circular button = 3.5 cm
radius = r = \(\frac{d}{2}=\frac{3.5}{2}\) = 1.75 cm.
∴ Area of each button = πr²
= \(\frac{22}{7}\) × (1.75)²
= \(\frac{22}{7}\) × 1.75 × 1.75
= 9.625 sq.cm.
The area of 56 such buttons = 56 × 9.625 = 539 sq.cm.
∴ The area of remaining sheet
= Area of a rectangular sheet – Area of 56 buttons
= 900 – 539
= 361 sq.cm

Question 2.
Find the area of a circle inscribed in a square of side 28 cm.
[Hint. Diameter of the circle is equal to the side of the square]

Solution:
Diameter of a circle = d = side of a square = 28 cm.
∴ d = 28 cm
⇒ r = \(\frac{d}{2}=\frac{28}{2}\) = 14 cm
∴ Area of a circle = \(\frac{22}{7}\) × 14 × 14
= 22 × 28
= 616 sq.cm.

Question 3.
Find the area of the shaded region in each of the following figures.
[Hint:d + \(\frac{d}{2}\) + d/2 = 42]
d = 21
∴ side of the square 21 cm
i)

Solution:

∴ Side of a square, d = 21 cm.
Radius of a semicircle = r = \(\frac{\mathrm{d}}{2}=\frac{21}{2}\) = 10.5 cm
∴ Area of a semi-circle = \(\frac{\pi r^{2}}{2}\)

= 11 × 1.5 × 10.5
= 173.250 sq.cm.

∴ Area of 4 shaded semi-circles
= 4 × Area of each semi-circle
= 4 × 173.25 = 693 sq.cm.

Diameter of a big circle = 21 m

Diameter of a semicircle = 10.5 m.
Radius = \(\frac{10.5}{2}\) = 5.25 m.
∴ Area of semi-circle = \(\frac{\pi r^{2}}{2}\)

= 11 × 0.75 × 5.25
= 43.3125 sq.m.
Area of two semicircles
= 2 × 43.3125 = 86.6250 sq.m.
∴ Area of shaded region
= Area of big circle – Area of 2 semicircles
= 346.5 – 86.6250
= 259.8750 sq.m.

Question 4.
The adjacent figure consists of four small semi-circles of equal radii and two big semi-circles of equal radii (each 42 cm). Find the area of the shaded region
Solution:
Diameter of a bigger semi circle d = 42 cm
Radius, r = \(\frac { 42 }{ 2 }\) = 21 cm
Area of a bigger semi circle = \(\frac{\pi r^{2}}{2}\)

= 11 × 3 × 21
= 693 sq.cm

∴ Area of the shaded region
= Area of two bigger semi circles – Area of two smaller semi circles + Area of two smaller semi circles
= Area of two bigger semi circles
= 2 × Area of bigger semi circles
= 2 × 693 = 1386 sq.cm.

Question 5.
The adjacent figure consists of four half circles and two quarter circles.
If OA=OB=OC=OD= 14cm. Find the area of the shaded region.

Solution:

OA = OB = OC = OD = 14CM
Radius of \(\frac { 1 }{ 4 }\) of circle BXD, r = 14cm

= 154 sq.cm.
∴ Area of the quarter circle AYC
= 154 sq.cm
.’. Area of the shaded region = Area of quarter circle BXD – Area of semi circle OPB + Area of semi circle OQD + Area of quarter circle AYC – Area of semi circle ARO + Area of semi circle OSC
= Area of quarter circle BXD + Area of quarter circle AYC
= 154 + 154
= 308 sq.cm.

Question 6.
In adjacent figure A, B, C and D are centres of equal circles which touch externally is pairs and ABCD is a square of side 7 cm. Find the area of the shaded region.

Solution:

Side of a square,
AB = BC = CD = DA = 7 cm
Area of a square = s × s
= 7 × 7 = 49 sq.cm.
From the above figure,
Radius of a circle, r = \(\frac{\text { Side of a square }}{2}\)
= \(\frac{7}{2}\) = 3.5
Areas of 4 equal sectors are equal.
Angle of a sector APQ, x = 90°
Radius, r = 3.5 cm
Area of sector APQ

= 5.5 × 0.5 × 3.5
= 9.625 sq.cm.
Total area of 4 equal sectors = 4 × 9.625 = 38.5 sq.cm.
∴ Area of shaded region = Area of square ABCD – Area of 4 equal sectors
= 49 – 38.5
= 10.5 sq.cm.

Question 7.
The area of an equilateral triangle is 49√ cm². Taking each angular point as centre, a circle is described with radius equal to half the length of the side of the triangle as shown in the figure. Find the area of the portion in the triangle not included in the circles.

Solution:
From the given figure,
Δ ABC is an equilateral triangle.
∴ Area of an equilateral triangle
= 49√3 sq.em (given)
3 equal circles (whose radii are equal) are touch externally.
Each angle of a sector in the circle = 60°,
Radius r = 7 cm.
∴ The areas of 3 sectors are equal.
∴ Area of a sector APQ

∴ The area of 3 sectors
= 3 × \(\frac { 77 }{ 3 }\) = 77 sq. cm
∴ The area of the portion in the triangle not included in the circles
= area of ΔABC — area of 3 equal sectors
=49√3 – 77
=49 × 1.7321 – 77 (∵√3 = 1.7321)
= 84.8729 – 77
= 7.8729 sq.cm.

Question 8.
(i) Four equal circles, each of radius ‘a’ touch one another. Find the area between them.
(ii) Four equal circles are described about the four corners of a square so that each circle
touches two of the others. Find the area of the space enclosed between the circumferences
of the circles, each side of the square measuring 14 cm.
Solution:

The side of a square ABCD
AB = BC = CD = DA = a + a = 2a units.
∴ Area of a square = s × s
= 2a × 2a = 4a2 sq.units.
4 equal sectors are there.
In each sector,
The angle of a sector APQ, x = 90°
Radius, r = a units
Area of each sector APQ

∴ Area of 4 equal sectors
= 4 × \(\frac{\pi \mathrm{a}^{2}}{4}\) = πa² sq. units
= 2a × 2a = 4a2 sq.units. 4 equal sectors are there.
∴ The area between the circles
= Area of square sectors
= 4a² – πa²
= a² (4 – π) sq. units
= a² (4 – \(\frac { 22 }{ 7 }\) ) = \(\frac{6 a^{2}}{7}\) sq. units

Question 9.
From a piece of cardbord, in the shape ofa trapezium ABCD, and AB||CD and ∠BCD = 90°,= quarter circle is removed. Given AB = BC = 3.5 cm and DE =2 cm. Calcualte the area of the remaining piece of the cardboard. (Take π to be \(\frac{22}{7}\) )

Solution:
ABCD is a trapezium
AB || CD and ZBCD = 90°
AB = BC = 3.5 cm; DE = 2 cm.
Area of a trapezium ABCD :
Length of parallel sides, AB = 3.5 cm
CD = DE + EC
= 2 + 3.5 = 5.5 cm.

Distance between parallel sides,
BC = 3.5 cm.
∴ Area of a trapezium
= \(\frac { 1 }{ 2 }\) h (a + b)
= \(\frac { 1 }{ 2 }\) × BC (AB + CD)
= \(\frac { 1 }{ 2 }\) × 3.5 (3.5 + 5.5)
= \(\frac { 1 }{ 2 }\) × 3.5 × 9
= 3.5 × 4.5
= 15.75 sq.cm.

Area of quarter circle EBC :
Radius, r = 3.5 cm
∴ Area of quarter circle

= 5.5 × 0.5 × 3.5
= 9.625 sq.cm.

Area of the remaining piece of card board = Area of a trapezium – \(\frac { 1 }{ 2 }\) th of area of a circle
= 15.75 – 9.625
= 6.125 sq.cm.

Question 10.
A horse is placed for grazing inside a rectangular field 70 m by 52 m and is tethered to one corner by a rope 21 m long. How much area can it graze?

Solution:
From the above figure,
The sector represents the grazing area of a horse.
The angle of a sector OPQ, x = 90°
Radius of a sector OPQ, r = 21 m .
∴ Area of a sector


= 346.5 sq.m.
∴ The area of grazing = 346.5 sq.m.