Category: Class 8

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 6 Statistics Ex 6.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 6 Statistics Ex 6.1

Question 1.
Fill in the blanks:
(i) Data has already been collected by some other person is ________ data.
Answer:
Secondary

(ii) The upper limit of the class interval (25-35) is ________ .
Answer:
35

(iii) The range of the data 200, 15, 20, 103, 3, 196, is ________ .
Answer:
197

(iv) If a class size is 10 and range is 80 then the number of classes are ________ .
Answer:
8

(v) Pie chart is a ________ graph.
Answer:
circular

Question 2.
Say True or False:
(i) Inclusive series is a continuous series.
Answer:
False

(ii) Comparison of parts of a whole may be done by a pie chart.
Answer:
True

(iii) Media and business people use pie charts.
Answer:
True

(iv) A pie diagram is a circle broken down into component sectors.
Answer:
True

Question 3.
Represent the following data in ungrouped frequency table which gives the number of children in 25 families.
1, 3, 0, 2, 5, 2, 3, 4, 1, 0, 5, 4, 3, 1, 3, 2, 5, 2, 1, 1, 2, 6, 2, 1, 4
Answer:
The data given is raw data.
Ascending order : 0, 1, 2, 3, 4, 5, 6

∴ Tabulating in frequency distribution table we get

Question 4.
Form a continuous frequency distribution table for the marks obtained by 30 students in a X std public examination.
328, 470, 405, 375, 298, 326, 276, 362, 410, 255, 391, 370, 455, 229, 300, 183, 283, 366, 400, 495, 215, 157, 374, 306, 280, 409, 321, 269, 398, 200.
Answer:
Maximum mark obtained = 495
Minimum marks obtained = 157
Range = Maximum value – Minimum value
Range = 495 – 157
= 338

If we take the class size as 50 then the number of class intervals possible

= \(\frac{338}{50}\) = 6.76
≅ 7

The percentage difference calculator is here to help you compare two numbers.

Question 5.
A paint company asked a group of students about their favourite colours and made a pie chart of their findings. Use the information to answer the following questions.
(i) What percentage of the students like red colour?
(ii) How many students liked green colour?
(iii) What fraction of the students liked blue?
(iv) How many students did not like red colour?
(v) How many students liked pink or blue?
(vi) How many students were asked about their favourite colours?

Answer:
Total percentage of students = 100 %
∴ 50students = 100% – (30% + 20% + 25% + 15%)
= 100% – 90%
50 students = 10%
10% of total students = 50
∴ \(\frac { 10 }{ 100 }\) (Total students) = 50
Total students = \(\frac{50 \times 100}{10}\) = 500.
Total students = 500.

(i) 20% of the students like red colour.

(ii) 15% of the students liked green colour.
\(\frac{15}{100}\) × 500 = 75 students liked green colour.

(iii) 25% students liked blue students liked blue.
⇒ \(\frac{25}{100}\) students liked blue.
⇒ \(\frac{1}{4}\) students liked blue.

(iv) Percentage of students liked red colour
= 20%
Percentage of students did not like red colour
= 100% – 20%
= 80%
∴ Number of students did not like red colour
= 80% of 500
= \(\frac{80}{100}\) × 500 = 400
400 students did not like red colour.

(v) Students liked pink or blue = students liked pink + students liked blue.
= 30% of 500 + 25% of 500
= \(\frac{30}{100}\) × 500 + \(\frac{25}{100}\) × 500
= 150 + 125
= 275

(vi) Total number of students = 500
500 students were asked about their favourite colour.

Question 6.
A survey gives the following information of food items preferred by people. Draw a Pie chart.

Answer:
Total number of people = 160 + 90 + 80 + 50 + 30 + 40 = 450
Converting the number of people prefer various food items into components part of 360°

Food items are preferred by people.

Question 7.
Income from various sources for Government of India from a rupee is given below. raw a pie chart.

Answer:

Income from various sources for Government of India in a rupee.

Question 8.
Monthly expenditure of Kumaran’s family is given below. Draw a suitable Pie chart.

Also
1. Find the amount spent for education if Kumaran spends ₹ 6000 for Rent.
2. What is the total salary of Kumaran?
3. How much did he spend more for food than education?
Answer:

Monthly expenditure of kumaran’s family.

1. Given Kumaran spends ₹ 6000 for Rent.
∴ 15% of’ total expenditure = 6000
\(\frac{15}{100}\) (Total Expenditure) = 6000
Total Expenditure = \(\frac{6000 \times 100}{15}\)
Total Expenditure = ₹ 40,000
Amount spend l’or education = 20% of total expenditure.
\(\frac{20}{100}\) × 40,000
= ₹ 8000

2. Total salary of Kumaran = ₹ 40,000

3. Amount spend for food = 50% of (40,000)
Amount spend for the food than education
= 20,000 – 8,000
= ₹ 12,000

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 4 Life Mathematics Ex 4.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 4 Life Mathematics Ex 4.3

This calculator computes any of the values in the half-life formula given the rest values.

Question 1.
Fill in the blanks:
(i) The compound interest on ₹ 5000 at 12% p.a for 2 years, compounded annually is ________ .
Answer:
₹ 1272
Hint:
Compound Interest (CI) formula is
CI = Amount – Principal

∴ 6272 – 5000 = ₹ 1272

(ii) The compound interest on ₹8000 at 10% p.a for 1 year, compounded half yearly is ________ .
Answer:
₹ 820
Hint:
Compound interest (CI) = Amount – Principal
Amount = p \(\left(1+\frac{r}{100}\right)^{2 n}\) [2n as it is compounded half yearly]
r = 10% p.a, for half yearly r = \(\frac{10}{2}\) = 5
∴ A = 8000 \(\left(1+\frac{5}{100}\right)^{2 \times 1}\) = 8000 × \(\left(\frac{105}{100}\right)^{2}\) = 8820
CI = Amount – principal = 8820 – 8000 = ₹ 820

(iii) The annual rate of growth in population of a town is 10%. If its present population is 26620, then the population 3 years ago was ________ .
Answer:
₹ 20,000
Hint:
Rate of growth of population r = 10%
Present population = 26620
Let population 3 years ago be x
∴ Applying the formula for population growth which is similar to compound interest,

The population 3 years ago was ₹ 20,000

(iv) If the compound interest is calculated quarterly, the amount is found using the formula ________ .
Answer:
A = \(P\left(1+\frac{r}{400}\right)^{4 n}\)
Hint:
Quarterly means 4 times in a year.
∴ The formula for compound interest is
A = \(P\left(1+\frac{r}{400}\right)^{4 n}\)

(v) The difference between the C.I and S.I for 2 years for a principal of ₹ 5000 at the rate of interest 8% p.a is ________ .
Answer:
₹ 32
Hint:
Difference between S.I & C.I is given by the formula
CI – SI = \(\left(\frac{r}{100}\right)^{2}\)
Principal (P) = 5000. r = 8% p.a
∴ CI – SI = 5000\(\left(\frac{8}{100}\right)^{2}\) = 5000 × \(\left(\frac{8}{100}\right)^{2}\) × \(\left(\frac{8}{100}\right)^{2}\) = ₹ 32

CBSE Class 10 Maths formulas and equations are available chapter wise.

Question 2.
Say True or False.
(i) Depreciation value is calculated by the formula, \(P\left(1-\frac{r}{100}\right)^{n}\).
Answer:
True
Hint:
Depreciation formula is \(P\left(1-\frac{r}{100}\right)^{n}\)

(ii) If the present population ola city is P and it increases at the rate of r% p.a, then the population n years ago would be \(P\left(1-\frac{r}{100}\right)^{n}\)
Answer:
False
Hint:
Let the population ‘n’ yrs ago be ‘x’
∴ Present popuLation (P) = x × \(\left(1+\frac{r}{100}\right)^{n}\)
∴ x = \(\frac{P}{\left(1+\frac{r}{100}\right)^{n}}\)

(iii) The present value of a machine is ₹ 16800. It depreciates at 25% p.a. Its worth after 2 years is ₹ 9450.
Answer:
True
Hint:
Present value of machine = ₹ 16800
Depreciation rate (n) = 25%

(iv) The time taken for ₹ 1000 to become ₹ 1331 at 20% p.a, compounded annually is 3 years.
Answer:
False
Hint:
Pnncipal money = 1000
rate of interest = 20%
Amount = 1331, applying in formula we get

(v) The compound interest on ₹ 16000 for 9 months at 20% p.a, compounded quarterly is ₹ 2522.
Answer:
True
Hint:
Principal (P) = 16000
n = 9 months = \(\frac{9}{12}\) years
r = 20% p.a
For compounding quarterly, we have to use below formula,
Amount(A) = P × \(\left(1+\frac{r}{100}\right)^{4 n}\)
Since quarterly we have to divide ‘r’ by 4

∴ Interest A – P = 18522 – 16000 = 2522 (True)

Question 3.
Find the compound interest on ₹ 3200 at 2.5 % p.a for 2 years, compounded annually.
Answer:
Principal (P) = ₹ 3200
r = 2.5% p.a
n = 2 years comp. annually
∴ Amount (A) = \(\left(1+\frac{r}{100}\right)^{n}\)
= 3200 \(\left(1+\frac{25}{100}\right)^{2}\)
= 3200 × (1.025)² = 3362
Compound interest (CI) = Amount – Principal
= 3362 – 3200 = 162

Question 4.
Find the compound interest for 2\(\frac { 1 }{ 2 }\) years on ₹ 4000 at 10% p.a, if the interest is compounded yearly.
Answer:
Principal (P) = ₹ 4000
r = 10 %p.a
Compounded yearly
n = 2\(\frac { 1 }{ 2 }\) years. Since it is of the form a \(\frac{b}{c}\) years

∴ CI = Amount – principal = 5082 – 4000 = 1082

Question 5.
A principal becomes ₹ 2028 in 2 years at 4% p.a compound interest. Find the principal.
Answer:
n = 2 years
r = rate of interest = 4% p.a
Amount A = ₹ 2028

Question 6.
In how many years will ₹ 3375 become ₹ 4096 at 13 % p.a if the interest is compounded half-yearly?
Answer:
Principal = ₹ 3375
Amount = ₹ 4096
r = 13\(\frac{1}{3}\)%p.a = \(\frac{40}{3}\)%p.a

Let no. of years be n
for compounding half yearly, formula is

Taking cubic root on both sides,

Question 7.
Find the CI on ₹ 15000 for 3 years if the rates of interest are 15%, 20% and 25% for the I, II and III years respectively.
Answer:
Principal (P) = ₹ 15000
rate of interest 1 (a) = 15% for year I
rate of interest 2 (b) = 20% for year II
rate of interest 3 (c) = 25% for year III
Formula for amount when rate of interest is different for different years is

Substituting in the above formula, we get

∴ Compound Interest(CI) = A – P = 25,875 – 15,000 = ₹ 10,875
CI = ₹ 10.875

Question 8.
Find the difference between C.I and S.I on ₹ 5000 for 1 year at 2% p.a, if the interest is compounded half yearly.
Answer:
Principal (P) = ₹ 5000
time period (n) = 1 yr.
Rate of interest (r) = 2%p.a
for half yearly r = 1%
Difference between CI & SI is given by the formula

Question 9.
Find the rate of interest if the difference between C.I and S.I on ₹ 8000 compounded annually for 2 years is ₹ 20.
Answer:
Principal (P) = ₹ 8000
time period (n) = 2 yrs.
rate of interest (r) = ?
Difference between CI & SI is given by the formula
CI – SI = \(p\left(1+\frac{r}{100}\right)^{n}\)
Difference between CI & SI is given as 20
∴ 20 = 8000 × \(\left(\frac{r}{100}\right)^{2}\)
∴ \(\left(\frac{r}{100}\right)^{2}=\frac{20}{8000}=\frac{1}{400}\)
Taking square root on both sides
\(\frac{r}{100}=\sqrt{\frac{1}{400}}=\frac{1}{20}\)
∴ r = \(\) = 5 %

Question 10.
Find the principal if the difference between C.I and S.l on it at 15% p.a for 3 years is ₹ 1134.
Answer:
Rate of interest (r) = 15% p.a
time period (n) = 3 years
Difference between CI & SI is given as 1134
Principal = ? → required to find
Using formula for difference
Simple Interest SI = \(\frac{P n r}{100}\)
Compound Interest CI = p(1 + i)ⁿ – p

1134 = P[(1.15)³ – 1 – 0.45] = P(1.52 – 1.45) = P (0.07)
∴ p = \(\frac{1134 \times 100}{0.07 \times 100}=\frac{113400}{7}\)
P = ₹ 16200

Objective Type Questions

Question 11.
The number of conversion periods in a year, if the interest on a principal is compounded every two months is _________ .
(A) 2
(B) 4
(C) 6
(D) 12
Answer:
(C) 6
Hint:
Conversion period is the time period after which the interest is added to the principal. If principal is compounded every two months then in a year, there will be 6\(\left(\frac{12}{2}\right)\) conversion periods.

Question 12.
The time taken for ₹ 4400 to become ₹ 4851 at 10%, compounded half yearly is _________ .
(A) 6 months
(B) 1 year
(C) 1\(\frac{1}{2}\) years
(D) 2 years
Answer:
(B) 1 year
Hint:
Principal = ₹ 4400
Amount = ₹ 4851
Rate of interest = 10% p.a
for half yearly, divide by 2,
r = \(\frac{10}{2}\) = 5 %
Compounded half yearly, so the formula is
A = P\(P\left(1+\frac{r}{100}\right)^{2 n}\)
Substituting in the above formula, we get

Taking square root on both sides, we get
\(\left(\frac{21}{20}\right)^{2 n}=\left(\frac{21}{20}\right)^{2}\)
Equating power on both sides
∴ 2n = 2, n = 1

Question 13.
The cost of a machine is ₹ 18000 and it depreciates at 16\(\frac{2}{3}\)% annually. Its value after 2 years will be _________ .
(A) ₹ 2000
(B) ₹ 12500
(C) ₹ 15000
(D) ₹ 16500
Answer:
(B) ₹ 12500
Hint:
Cost of machine = 18000
Depreciation rate = 16\(\frac{2}{3}\)% = \(\frac{50}{3}\)%p.a
time period = 2 years
∴ As per depreciation formula,
Depriciated value = Original value \(\left(1-\frac{r}{100}\right)^{n}\)
Substituting in above formula, we get
Depreciated value after 2 years

Question 14.
The sum which amounts to ₹ 2662 at 10% p.a in 3 years, compounded yearly is _________ .
(A) ₹ 2000
(B) ₹ 1800
(C) ₹ 1500
(D) ₹ 2500
Answer:
(A) ₹ 2000
Hint:
Amount = ₹ 2662
rate of interest = 10 % p.a
Time period = 3 yrs. Compounded yearly
Principal (P) → required to find?

Question 15.
The difference between compound and simple interest on a certain sum of money for 2 years at2%p.ais U. The sum of money is _________ .
(A) ₹ 2000
(B) ₹ 1500
(C) ₹ 3000
(D) ₹ 2500
Answer:
(D) ₹ 2500
Hint:
Difference between CI and SI is given as Re I
Time period (n) = 2 yrs.
Rate of interest (r) = 2 % p.a
Formula for difference is
CI – SI = \(P \times\left(1+\frac{r}{100}\right)^{n}\)
Substituting the values in above formula, we get
1 = p × \(\left(\frac{2}{100}\right)^{2}\)
∴ p = 1 × \(\left(\frac{100}{2}\right)^{2}\) = 1 × (50)² = ₹ 2500

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 1 Numbers Ex 1.4 Textbook Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 1 Numbers Ex 1.4

Question 1.
Fill in the blanks:
(i) The ones digit in the square of 77 is ________ .
Answer:
9

(ii) The number of non-square numbers between 242 and 252 is ________ .
Answer:
48

(iii) The number of perfect square numbers between 300 and 500 is ________ .
Answer:
5

(iv) If a number has 5 or 6 digits in it, then its square root will have ________ digits.
Answer:
3

(v) The value of Jii lies between integers ______ and ________ .
Answer:
13, 14

Question 2.
Say True or False:
(i) When a square number ends in 6, its square root will have 6 in the unit’s place.
Answer:
True

(ii) A square number will not have odd number of zeros at the end.
Answer:
True

(iii) The number of zeros in the square of 91000 is 9.
Answer:
False

(iv) The square of 75 is 4925.
Answer:
False

(v) The square root of 225 is 15.
Answer:
True

Question 3.
Find the square of the following numbers.
(i) 17
(ii) 203
(iii) 1098
Answer:
(i) 17

(ii) 203

(iii) 1098

Question 4.
Examine if each of the following is a perfect square.
(i) 725
(ii) 190
(iii) 841
(iv) 1089
Answer:
(i) 725
725 = 5 × 5 × 29 = 5² × 29
Here the second prime factor 29 does not have a pair.
Hence 725 is not a perfect square number.

(ii) 190
190 = 2 × 5 × 19
Here the factors 2, 5 and 9 does not have pairs.
Hence 190 is not a perfect square number.

(iii) 841
841 = 29 × 29
Hence 841 is a perfect square

(vi) 1089
1089 = 3 × 3 × 11 × 11 = 33 × 33
Hence 1089 is a perfect square

The factors of 144 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144.

Question 5.
Find the square root by prime factorisation method.
(i) 144
(ii) 256
(iii) 784
(iv) 1156
(v) 4761
(vi) 9025
Answer:
(i) 144
144 = 2 × 2 × 2 × 2 × 3 × 3
√144 = 2 × 2 × 3 = 12

(ii) 256
256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
√256 = 2 × 2 × 2 × 2 = 16

(iii) 784
784 = 2 × 2 × 2 × 2 × 7 × 7
√784 = 2 × 2 × 2 × 2 × 7 × 7 = 28

(iv) 1156
1156 = 2 × 2 × 17 × 17
1156 = 2² × 17²
1156 = (2 × 17)²
∴ \(\sqrt{1156}\) = \(\sqrt{(2 \times 17)^{2}}\) = 2 × 17 = 34
∴ \(\sqrt{1156}\) = 34

(v) 4761
4761 = 3 × 3 × 23 × 23
4761 = 3² × 23²
4761 = (3 × 23)²
√4761 = \(\sqrt{(3 \times 23)^{2}}\)
√4761 = 3 × 23
√4761 = 69

(vi) 9025
9025 = 5 × 5 × 19 × 19
9025 = 5² × 19²
9025 = (5 × 19)²
√925 = \(\sqrt{(5 \times 19)^{2}}\) = 5 × 19 = 95

Question 6.
Find the square root by long division method.
(i) 1764
(ii) 6889
(iii) 11025
(iv) 17956
(v) 418609
Answer:
(i) 1764

√1764 = 42

(ii) 6889

√6889 = 83

(iii) 11025

√11025 = 105

(iv) 17956

√17956 = 134

(v) 418609

√418609 = 647

Roots Calculator is a free online tool that displays the roots of the given quadratic equation.

Question 7.
Estimate the value of the following square roots to the nearest whole number:
(i) √440
(ii) √800
(iii) √1020
Answer:
(i) √440
we have 20² = 400
21²= 441
∴ √440 ≃ 21

(ii) √800
we have 28² = 784
29²= 841
∴ √800 ≃ 28

(iii) √1020
we have 31² = 961
32²= 1024
∴ √1020 ≃ 32

Question 8.
Find the square root of the following decimal numbers and fractions.
(i) 2.89
(ii) 67.24
(iii) 2.0164
(iv) \(\frac{144}{225}\)
(v) \(7 \frac{18}{49}\)
Answer:
(i) 2.89

√2.89 = 1.7

(ii) 67.24

√67.24 = 8.2

(iii) 2.0164

√2.0164 = 1.42

(iv) \(\frac{144}{225}\)

(v) \(7 \frac{18}{49}\)

\(\sqrt{7 \frac{18}{49}}=2 \frac{5}{7}\)

Question 9.
Find the least number that must be subtracted to 6666 so that it becomes a perfect square. Also, find the square root of the perfect square thus obtained.
Let us work out the process of finding the square root of 6666 by long division method.

The remainder in the last step is 105. Is if 105 be subtracted from the given number the
remainder will be zero and the new number will be a perfect square.
∴ The required number is 105. The square number is 6666 – 105 = 6561.

Question 10.
Find the least number by which 1800 should be multiplied so that it becomes a perfect square. Also, find the square root of the perfect square thus obtained.
Answer:
We find 1800 = 2 × 2 × 3 × 3 × 5 × 5 × 2
= 2² × 3² × 5² × 2
Here the last factor 2 has no pair. So if we multiply 1800 by 2, then the number becomes a perfect square.

∴ 1800 × 2 = 3600 is the required perfect square number.
∴ 3600 = 1800 × 2
3600 = 2² × 3² × 5² × 2 × 2
3600 = 2² × 3² × 5² × 2²
= (2 × 3 × 5 × 2)²
\(\sqrt{3600}=\sqrt{(2 \times 3 \times 5 \times 2)^{2}}\)
= 2 × 3 × 5 × 2 = 60
∴ √3600 = 60

Objective Type Questions

Question 11.
The square of 43 ends with the digit .
(A) 9
(B) 6
(C) 4
(D) 3
Answer:
(A) 9
Hint:
Ones digit = 3 × 3 = 9

Question 12.
_______ is added to 24² to get 25².
(A) 4²
(B) 5²
(C) 6²
(D) 7²
Answer:
(D) 7²
Hint:
25² = 25 × 25 = 625
24² = 24 × 24 = 576

Question 13.
√48 is approximately equal to .
(A) 5
(B) 6
(C) 7
(D) 8
Answer:
(C) 7
Hint:
√49 = 7

Question 14.
\(\sqrt{128}-\sqrt{98}+\sqrt{18}\)
(A) √2
(B) √8
(C) √48
(D) √32
Answer:
(D) √32

Question 15.
The number of digits in the square root of 123454321 is ______.
(A) 4
(B) 5
(C) 6
(D) 7
Answer:
(B) 5
Hint:
\(\frac{n+1}{2}=\frac{10}{2}=5\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 3 Algebra Ex 3.6 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra Ex 3.6

To find the value of x, bring the variable to the left side and bring all the remaining values to the right side. Simplify the values to find the result.

Question 1.
Fill in the blanks:
(i) The value of x in the equation x + 5 = 12 is ________ .
Answer:
7
Hint:
Given, x + 5 = 12
x = 12 – 5 = 7 (by transposition method)
Value of x is 7

(ii) The value of y in the equation y – 9 = (-5) + 7 is ________ .
Answer:
11
Hint:
Given, y – 9 = (-5) + 7
y – 9 = 7 – 5 (re-arranging)
y – 9 = 2
∴ y = 2 + 9 = 11 (by transposition method)

(iii) The value of m in the equation 8m = 56 is ________ .
Answer:
7
Hint:
Given, 8m = 56
Divided by 8 on both sides
\(\frac{8 \times m}{8}=\frac{56}{8}\)
∴ m = 7

(iv) The value of p in the equation \(\frac{2 p}{3}\) = 10 is ________ .
Answer:
15
Hint:
Given, \(\frac{2 p}{3}\) = 10
Multiplying by 3 on both sides

Dividing by 2 on both sides

∴ p = 15

(v) The linear equation in one variable has ________ solution.
Answer:
one

Question 2.
Say True or False.
(i) The shifting of a number from one side of an equation to other is called transposition.
Answer:
True

(ii) Linear equation in one variable has only one variable with power 2.
Answer:
False
[Linear equation in one variable has only one variable with power one – correct statement]

Question 3.
Match the following

(A) (i),(ii), (iv) ,(iii),(v)
(B) (iii), (iv), (i) ,(ii), (v)
(C) (iii),(i) ,(iv), (v), (ii)
(D) (iii) , (i) , (v) ,(iv) ,(ii)
Answer:
(C) (iii),(i) ,(iv), (v), (ii)

a. \(\frac{x}{2}\) = 10, multiplying by 2 on both sides, we get
\(\frac{x}{2}\) × 2 = 10 × 2 ⇒ x = 20

b. 20 = 6x – 4 by transposition ⇒ 20 + 4 =6x
6x = 24 dividing by 6 on both sides,
\(\frac{6 x}{6}=\frac{24}{6}\) ⇒ x = 4

c. 2x – 5 = 3 – x
By transposing the variable ‘x’, we get
2x – 5 + x = 3
by transposing – 5 to other side,
2x + x = 3 + 5

∴ x = \(\frac{8}{3}\)

d. 7x – 4 – 8x = 20
by transposing – 4 to other side,
7x – 8x = 20 + 4
– x = 24
∴ x = – 24

e. \(\frac{4}{11}-x=\frac{-7}{11}\)
Transposing \(\frac{4}{11}\) to other side,
– x = \(\frac{-7}{11} \frac{-4}{11}=\frac{-7-4}{11}=\frac{-11}{11}\) = – 1
∴ – x = – 1 ⇒ x = 1

Question 4.
Find x:
(i) \(\frac{2 x}{3}-4=\frac{10}{3}\)
Answer:
Transposing – 4 to other side, it becomes + 4

(ii) \(y+\frac{1}{6}-3 y=\frac{2}{3}\)
Answer:
Transposing \(\) to the other side,

(iii) \(\frac{1}{3}-\frac{x}{3}=\frac{7 x}{12}+\frac{5}{4}\)
Answer:
Transposing \(\frac{-x}{3}\) to the other side, it becomes + \(\frac{x}{3}\)
∴ \(\frac{1}{3}=\frac{7 x}{12}+\frac{5}{4}+\frac{x}{3}\)
Transposing \(\frac{5}{4}\) to the other side, it becomes \(\frac{-5}{4}\)
\(\frac{1}{3}-\frac{5}{4}=\frac{7 x}{12}+\frac{x}{3}\)
Multiply by 12 throughout [we look at the denominators 3, 4, 12, 3 and take the LCM, which is 12]

4 – 15 = 7x + x × 4
– 11 = 7x + 4x
11x = – 11
x = -1

Question 5.
Find x
(i) -3(4x + 9) = 21
Answer:

Expanding the bracket,
-3 × 4 + (-3) × 9 = 21
∴ -12x + (-27) = 21
– 12x – 27 = 21
Transposing – 27 to other side, it becomes +27
– 12 x = 21 + 27 = 48
∴ – 12x = 48 ⇒ 12x = – 48
Dividing by 12 on both sides

(ii) 20 – 2 (5 – p) = 8
Answer:

Expanding the bracket,
20 – 2 × 5 – 2 × (-p) = 8
20 – 10 + 2p = 8
(- 2 × – p = 2p)
10 + 2p = 8 transposing lo to other side
2p = 8 – 10 = – 2
∴ 2p = – 2
∴ p = – 1

(iii) (7x – 5) – 4(2 + 5x) = 10(2 – x)
Answer:

Expanding the brackets,
7x – 5 – 4 × 2 – 4 × 5x = 10 × 2 + 10 × (-x)
7x – 5 – 8 – 20x = 20 – 10x
7x – 13 – 20x = 20 – 10x
Transposing 10x & – 13, we get
7x – 13 – 20x + 10x = 20
7x – 20x + 10x = 20 + 13, Simplifying,
– 3x = 33
3x = – 33
x = \(\frac{-33}{3}\) = – 11
x = – 11

Question 6.
Find x and m:
(i) \(\frac{3 x-2}{4}-\frac{(x-3)}{5}=-1\)
Answer:

(ii) \(\frac{m+9}{3 m+15}=\frac{5}{3}\)
Answer:
Cross multiplying, we get

∴ (m + 9) × 3 = 5 × (3m + 15)
m × 3 + 9 × 3 = 5 × 3m + 5 × 15

27 – 75 = 15m – 3m
– 48 = 12m

⇒ m = – 4

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 7 Information processing Ex 7.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 7 Information processing Ex 7.2

Question 1.
Using repeated division method, find the HCF of the following:
(i) 455 and 26
Answer:

Step 1: The larger number should be dividend 455 & smaller number should be divisor = 26
Step 2: After 1st division, the remainder becomes new divisor & the previous divisor becomes next dividend.
Step 3: This is done till remainder is zero.
Step 4: The last divisor is the HCF L.
∴ Ans: HCF is 13.

(ii) 392 and 256
Answer:
256 is smaller, so it is the 1st divisor

∴ HCF = 8

(iii) 6765 and 610
Answer:

∴ HCF = 5

(iv) 184, 230 and 276
Answer:
First let us take 184 & 230

∴ 46 is the HCF of 184 and 230
Now the HCF of the first two numbers is the dividend for the third number.

∴ Ans: HCF of 184, 230 & 276 is 46

Question 2.
Using repeated subtraction method, find the HCF of the following:
(i) 42 and 70
Answer:
Let number be m & n m > n
We do, m – n & the result of subtraction becomes new ‘m’. if m becomes less than n,
we do n – m and then assign the result as n. We should do this till m n. When m = n then ‘m’ is the HCF.
42 and 70
m = 70 n = 42
70 – 42 = 28

now m = 42, n = 28
42 – 28 = 14.

now m = 28, n = 14
28 – 14 = 14.

now m = 14. n = 14;
we stop here as m = n
∴ HCF of 42 & 70 is 14

(ii) 36 and 80
Answer:

28 – 8 = 20
20 – 8 = 12
12 – 8 = 4
8 = 4 = 4
now m = n = 4
∴ HCF is 4

(iii) 280 and 420
Answer:
Let m = 420, n = 280
m – n = 420 – 280 = 140
now m = 280, n = 140
m – n = 280 – 140 = 140
now m = n = 140
∴ HCF is 140

(iv) 1014 and 654
Answer:
Let m = 1014, n = 654
m – n = 1014 – 654 = 360
now m = 654, n = 360
m – n = 654 – 360 = 294
now m = 360, n = 294
m – n = 360 – 294 = 66
now m = 294 n = 66
m – n = 294 – 66 = 228
now m = 66, n = 228
n – m = 228 – 66 = 162
now m = 162, n = 66
= 162 – 66 = 96
n – m = 96 – 66 = 30
Similarly, 66 – 30 = 36
36 – 30 = 6
30 – 6 = 24
24 – 6 = 18
18 – 6 = 12
12 – 6 = 6 now m = n
∴ HCF of 1014 and 654 is 6

Question 3.
Do the given problems by repeated subtraction method and verify the result.
(i) 56 and 12

Answer:
Let n = 56 & n = 12
m – n = 56 – 12 = 44
now m = 44, n = 12
m – n = 44 – 12 = 32
m – n = 32 – 12 = 20
m – n = 20 – 12 = 8
n – m = 12 – 84
m – n = 8 – 4 = 4. now m = n
∴ HCF of 56 & 12 is 4

(ii) 320, 120 and 95

Answer:
Let us take 320 & 120 first m = 320, n = 120
m – n = 320 – 120 = 200
m = 200, n = 120
∴ m – n = 200 – 120 = 80
120 – 80 = 40
80 – 40 = 40
∴ m = n = 40 → HCF of 320, 120
Now let us find HCF of 40 & 95
m = 95, n = 40
∴ m – n = 95 – 40 = 55
55 – 40 = 15
40 – 15 = 25
25 – 15 = 10
15 – 10 = 5
HCF of 40 & 95 is 5
10 – 5 = 5
∴ HCF of 320 120 & 95 is 5

Question 4.
Kalai wants to cut identical squares as big as she can, from a piece of paper measuring 168mm and by 196mm. What is the length of the side of the biggest square? (To find HCF using repeated subtraction method)
Answer:
Sides are 168 & 196
To find HCF of 168 & 196, we are to use repeated subtraction method.
∴ m = 196, n = 168
m – n = 196 – 168 = 28

now n = 28, m = 168
m – n = 168 – 28 = 140

now m = 140, n = 28
m – n = 140 – 28 = 112

now m = 112, n = 28
m – n = 112 – 28 = 84

now m = 84, n = 28
m – n = 84 – 28 = 56

now m = 56, n = 28
m – n = 56 – 28 = 28
∴ HCF is 28
∴ Length of biggest square is 28

Objective Type Questions

Question 5.
What is the eleventh Fibonacci number?
(a) 55
(b) 77
(c) 89
(d) 144
Answer:
(c) 89
Hint:

∴ 11^th Fibonacci number is 89

Question 6.
If F(n) is a Fibonacci number and n = 8, which of the following is true?
(a) F(8) = F(9) + F( 10)
(b) F(8) = F(7) + F(6)
(c) F(8) = F(10) × F(9)
(d) F(8) = F(7) – F(6)
Answer:
(b) F(8) = F(7) + F(6)
Hint:
Given F(n) is a Fibonacci number & n = 8
∴ F(8) = F(7) + F (6) as any term in Fibonacci series is the sum of preceding 2 terms

Question 7.
Every 3^rd number of the Fibonacci sequence is a multiple of_______
(a) 2
(b) 3
(c) 5
(d) 8
Answer:
(a) 2
Hint:
Every 3^rd number in Fibonacci sequence is a multiple of 2

Question 8.
Every _____ number of the Fibonacci sequence is a multiple of 8
(a) 2^th
(b) 4^th
(c) 6^th
(d) 8^th
Answer:
(c) 6^th

Question 9.
The difference between the 18^th and 17^th Fibonacci number is
(a) 233
(b) 377
(c) 610
(d) 987
Answer:
(d) 987
Hint:
F(18) = F(17) + F(16)
F(18) – F(17) = F(16) = F(15) + F(14)
= 610 + 377 = 987

Factors of 70 are the list of integers that we can split evenly into 70.

Question 10.
Common prime factors of 30 and 250 are
(a) 2 × 5
(b) 3 × 5
(c) 2 × 3 × 5
(d) 5 × 5
Answer:
(a) 2 × 5
Prime factors of 30 are 2 × 3 × 5
Prime factors of 250 are 5 × 5 × 5 × 2
∴ Common prime factors are 2 × 5

Question 11.
Common prime factors of 36,60 and 72 are
(a) 2 × 2
(b) 2 × 3
(c) 3 × 3
(d) 3 × 2 × 2
Answer:
(d) 3 × 2 × 2
Hint:
Prime factors of 36 are 2 × 2 × 3 × 3
Prime factors of 60 are 2 × 2 × 3 × 5
Prime factors of 72 are 2 × 2 × 2 × 3 × 3
∴ Common prime factors are 2 × 2 × 3

Question 12.
Two numbers are said to be co-prime numbers if their HCF is
(a) 2
(b) 3
(c) 0
(d) 1
Answer:
(d) 11

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 1 Numbers Ex 1.5 Textbook Questions and Answers, Notes.

Simplifying Rational Expressions Calculator.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 1 Numbers Ex 1.5

Question 1.
Fill in the blanks
(i) The ones digits in the cube of 73 is __________ .
Answer:
7

(ii) The maximum number of digits in the cube of a two digit number is __________ .
Answer:
6

(iii) The smallest number to be added to 3333 to make it a perfect cube is __________ .
Answer:
42

(iv) The cube root of 540×50 is __________ .
Answer:
30

(v) The cube root of 0.000004913 is __________ .
Answer:
0.017

Question 2.
Say True or False.
(i) The cube of 24 ends with the digit 4.
Answer:
True

(ii) Subtracting 103 from 1729 gives 93.
Answer:
True

(iii) The cube of 0.0012 is 0.000001728.
Answer:
False

(iv) 79570 is not a perfect cube.
Answer:
True

(v) The cube root of 250047 is 63.
Answer:
True

Question 3.
Show that 1944 is not a perfect cube.
Answer:

1944 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3

= 2³ × 3³ × 3 × 3
There are two triplets to make further triplets we need one more 3.
∴ 1944 is not a perfect cube.

Question 4.
Find the smallest number by which 10985 should be divided so that the quotient is a perfect cube.
Answer:

We have 10985 = 5 × 13 ×13 × 13
= 5 × 13 ×13 × 13
Here we have a triplet of 13 and we are left over with 5.
If we divide 10985 by 5, the new number will be a perfect cube.
∴ The required number is 5.

Question 5.
Find the smallest number by which 200 should be multiplied to make it a perfect cube.
Answer:


Grouping the prime factors of 200 as triplets, we are left with 5 × 5
We need one more 5 to make it a perfect cube.
So to make 200 a perfect cube multiply both sides by 5.

1000 = 2 × 2 × 2 × 5 × 5 × 5 × 5 × 5
Now 1000 is a perfect cube.
∴ The required number is 5.

Question 6.
Find the cube root 24 × 36 × 80 × 25.
Answer:

Question 7.
Find the cube root of 729 and 6859 prime factorisation.
Answer:
(i)


= 3 × 3
\(\sqrt[3]{729}\) = 9

(ii) \(\sqrt[3]{6859}\) = \(\sqrt[3]{19 \times 19 \times 19}\)
\(\sqrt[3]{6859}\) = 19

Question 8.
What is the square root of cube root of 46656?
Answer:

We have to find out \(\sqrt{(\sqrt[3]{46656})}\)
First we will find \(\sqrt[3]{46656}\)

∴ The required number is 6.

Question 9.
If the cube of a squared number is 729, find the square root of that number.
Answer:

(729)^1/3 = 3 × 3 = 9
∴ The cube of 9 is 729.
9 = 3 × 3 [ie 3 is squared to get 9]

We have to find out √3,
√3 = 1.732

Question 10.
Find two smallest perfect square numbers which when multiplied together gives a perfect cube number.
Answer:
Consider the numbers 2² and 4²
The numbers are 4 and 16.
Their procluct 4 × 16 = 64
64 = 4 × 4 × 4
∴ The required square numbers are 4 and 16

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th English Guide Pdf Supplementary Chapter 1 The Woman on Platform 8 Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 8th English Solutions Supplementary Chapter 1 The Woman on Platform 8

8th English Guide The Woman on Platform 8 Text book Back Questions and Answers

8th English Guide The Woman on Platform 8 Textual Exercise Questions and Answers

A. Choose the best answer. (Text Book Page No. 31)

1. Satish’s mother handed to her son …………….
a) bag of pencil
b) bag of vegetables
c) big box of chocolates
d) cricket ball
Answer:
c) big box of chocolates

2. The train would come at …………….
a) one o’clock
b) twelve o’clock
c) two o’clock
d) eleven o’clock
Answer:
b) twelve O’clock

Age calculator is online tool to calculate age from date of birth to current date. It can also be used to calculate time difference between two dates.

3. The strange lady gave ……………. to Arun.
a) coffee and vada
b) tea and bajji
c) samosas and jalebis
d) black tea and cake
Answer:
c) samosas and jalebis

4. Arun was sitting on platform ……………. .
a) no. 7
b) no. 8
c) no. 4
d) no. 3
Answer:
b) no. 8

5. Satish and Arun were ……………. years old boys.
a) 12
b) 11
c) 13
d) 10
Answer:
a) 12

B. Match the following. (Text Book Page No. 32)

1. woman in white	mother of Satish
2. train	boy of same age
3. bowler	dressed simply
4. Satish	waves of stream
5. spectacles	Arun

Answer:

1. woman in white	dressed simply
2. train	waves of stream
3. bowler	Arun
4. Satish	boy of same age
5. spectacles	mother of Satish

C. Identify the character.

1. I am glad to know that.
Answer:
Mother of Satish

2. Are you all alone, my son?
Answer:
strange lady

3. Yes, I am going to school.
Answer:
Arun

4. He is one of my friends.
Answer:
Satish

5. Goodbye mother.
Answer:
Arun

D. Answer the following questions.

1. Where was arun sitting?
Answer:
Arun was sitting on platform no. 8 at Ambala station.

2. What was the expected time of train’s arrival?
Answer:
The train would arrive at the midnight.

3. What was the sight Arun had seen on the platform?
Answer:
Arun saw a tide of people, the cries of various vendors and the newspaper boy.

4. What did the vendors sell?
Answer:
The vendors were selling curds and lemon, the sweet meet, the newspaper.

5. How did the woman appear?
Answer:
The woman had a pale face and dark kind eyes. She wore no jewels and was dressed very simply in a white saree.

6. Where was Arun travelling to?
Answer:
Arun was traveling to his boarding school.

7. What did the woman buy for him?
Answer:
The woman bought samosas and jalebis. She also ordered tea for Arun.

8. What was the advice of Sathish’s mother?
Answer:
The mother of Sathish advised him not to talk to the strangers.

9. What were Arun’s last words?
Answer:
Aruns last words were “Goodbye – mother”.

10. What was the reaction of the woman at the end?
Answer:
The strange woman held the hand of Arun and she smiled in a gentle understanding way.

Step to Success (Text Book Page No. 33)

Sentence Rearrangement Common Example:
A. Miss Sullivan arrived at the Keller home when Helen was seven.
B. The deaf and blind Helen learned to communicate verbally.
C. But, eventually, Miss Sullivan’s effort was rewarded.
D. Before Helen Keller was two years old, she lost her sight and her hearing.
E. Miss Sullivan worked closely with Helen, her new student.
F. At times the teacher became frustrated.
a) DEFACB
b) DAEFCB
c) ACFDEB
d) CFDABE
e) FDACEB
Answer:
b) DAEFCB

Connecting to Self (Text Book Page No. 33)

Think you are in the following situation and write what would you do and why?

Will you ignore/take and give to its owner/ take and keep it yourself.
Answer:
I will take the purse and give it to the owner of the purse. It is the duty of an honest citizen.

Will you ignore/go and help him/ stand watching him feeling shy to help
Answer:
I will go and help him. I will try to free him of heavy luggage. I will also help him in carrying his heavy luggage.

Will you ignore/try to stop them fighting/be afraid and go away from there.
Answer:
I will try to stop them from fighting. If I am not able to stop them, I will inform the police.

The Woman on Platform 8 Summary in English

A schoolboy waits on platform no. 8 at Ambala station for the train. He meets a stranger. A dignified but simply dressed woman. She is kind to the boy and gets him tiffin. She gives him company till he gets into the train. Though she is a stranger she introduces herself to be the mother of Arun, to the mother of Sathish. Sathish is the classmate of Arun. Aran, the schoolboy bids farewell to the stranger with a loving kiss. He watches her till she is lost in the crowd.

The Woman on Platform 8 Summary in Tamil

பள்ளியில் படிக்கும் பையன் அருண், அம்பாலா ரயில் நிலையத்தில் நடைமேடை 8இல் ரயிலுக்கு காத்து இருக்கிறான். ஒரு புதிய பெண்ணைச் சந்திக்கிறான். அன்பான அந்தப் பெண், பையனை அழைத்துச் சென்று உணவளிக்கிறாள். ரயில் வரும் வரை அவன் பக்கத்தில் அமர்ந்து பேச்சுத் துணையாய் இருக்கிறாள். சதீஷ், அருணின் வகுப்புத் தோழன் தன் அம்மாவுடன் வரும்போது அந்தப் பெண் தன்னை அருணின் அம்மா என்று அறிமுகப்படுத்திக் கொள்கிறாள். அருண் மனம் இளகுகிறது. அந்தப் புதிய பெண்ணை முத்தமிட்டு தன் அன்பை வெளிப்படுத்துகிறான். அவள் கூட்டத்தில் மறையும்வரை அந்தப் பெண்ணையே உற்றுநோக்குகிறான்.

The Woman on Platform 8 About the Author in English

Ruskin Bond, short story writer, novelist, and poet, the favourite writer of Indian children. His first novel, Room on the Roof, was published when he was still in his teens. This novel won him the John Rhys Memorial Award in 1957. He also writes about children and the simple hill folk of Uttarakhand. Simplicity and fluency of language and an insight into human nature are hallmarks of his style. His major writings include An Island of Trees, A Bond with the Mountains and The India I Love. He has also been honoured with the Sahitya Akademi Award for his contribution to Indian literature.

The Woman on Platform 8 About the Author in Tamil

ரஸ்கின் பான்ட், சிறுகதை எழுத்தாளர், நாவலாசிரியர், கவிஞர், இந்திய குழந்தைகளின் அபிமான எழுத்தாளர். கூரையின் மேல் உள்ள அறை அவரது முதல் புத்தகம். இப்புத்தகம் அவர் சிறுவனாய் இருக்கும்போதே எழுதப்பட்டது. இப்புத்தகம் 1957இல் ஜான்ரிஸ் மெமோரியல் பரிசு பெற்றது. உத்தரகாண்ட் மலைமீது உள்ள எளிய மக்களின் குழந்தைகள் பற்றி எழுதுபவர். எளிமை, வளமை, ஆத்மாவை ஊடுறுவும் பார்வை இவையே இவரது முத்திரைகள். மரங்களின் தீவு, மலைகளோடு பிணைப்பு, நான் விரும்பும் இந்தியா (An Island of Trees, A Bond with the Mountains, The India I Love) இவையே அவரது முதன்மையான புத்தகங்கள். இந்திய இலக்கியத்திற்கான சாகித்ய அகாடமி பரிசு அவரது படைப்புகளுக்கு அளிக்கப்பட்டுள்ளது.

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 5 Geometry Ex 5.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 5 Geometry Ex 5.1

Question 1.
Fill in the blanks with the correct term from the given list.
(in proportion, similar, corresponding, congruent, shape, area, equal)
(i) Corresponding sides of similar triangles are _________ .
Answer:
in proportion

(ii) Similar triangles have the same ________ but not necessarily the same size.
Answer:
Shape

(iii) In any triangle _______ sides are opposite to equal angles.
Answer:
equal

(iv) The symbol is used to represent _________ triangles.
Answer:
congruent

(v) The symbol ~ is used to represent _________ triangles.
Answer:
similar

Question 2.
In the figure, ∠CIP ≡ ∠COP and ∠HIP ≡ ∠HOP. Prove that IP ≡ OP.

Answer:

Statements	Reasons
1. CI = CO	∵ CIP ≡ COP, by CPCTC
2. IP =  OP	By CPCTC
3. CP = CP	By CPCTC
4. Also HI = HO	CPCTC ∆HIP ≡ HOP given
5. IP = OP	By CPCTC and (4)
6. ∴ IP ≡ OP	By (2) and (4)

Question 3.
In the given figure, AC ≡ AD and ∠CBD ≡ ∠DEC. Prove that ∆ BCF ≡ ∆ EDF.

Answer:

Statements	Reasons
1. ∠BCF = ∠EFD	Vertically opposite angles
2. ∠CBD = ∠DEC	Angles on the same base given
3. ∠BCF = ∠EDF	Remaining angles of ∆BCF and ∆EDF
4. ∆BCF ≡ ∆EDF	By (1) and (2) AAA criteria

Question 4.
In the given figure, △ BCD is isosceles with base BD and ∠BAE ≡∠DEA. Prove that AB ≡ ED.

Answer:

Statements	Reasons
1. ∠BAE ≡ ∠DEA	Given
2. AC = EC	By (1) sides opposite to equal angles are equal
3. BC = DC	Given BCD is isosceles with base BD
4. AC – BC = EC – DC	2 – 3
5. AB ≡ ED	By 4

Midpoint Calculator is a free online tool that displays the midpoint of the line segment.

Question 5.
In the given figure, D is the midpoint of OE and ∠CDE = 90°. Prove that △ODC ≡ △EDC

Answer:

Statements	Reasons
1. OD = ED	D is the midpoint OE (given)
2. DC = DC	Common side
3. ∠CDE = ∠CDO = 90°	Linear pair and given ∠CDE = 90°
4. △ODC ≡ △EDC	By RHS criteria

Question 6.
Is △PRQ ≡ △QSP? Why?

Answer:
In △PRQ and △PSQ
∠PRQ = ∠PSQ = 90° given
PR = QS = 3 cm given
PQ = PQ = 5 cm common
It satisfies RHS criteria
∴ △PRQ congruent to △QSP.

Question 7.
From the given figure, prove that △ABC ~ △EDF

Answer:
From the △ABC, AB = AC
It is an isosceles triangle
Angles opposite to equal sides are equal
∴ ∠B = ∠C = 65°
∴ ∠B + ∠C = 65° + 65°
= 130°
We know that sum of three angles is a triangle = 180°
∠A + ∠B + ∠C = 180°
∠A + 130° = 180°
∠A = 180° – 130°
∠A = 50°
From △EDF, ∠E = 50°
∴ Sum of Remaining angles = 180° – 50° = 130°
DE = FD
∴ ∠D = ∠F
From △ABC and △EDF ∴ △D = \(\frac{130}{2}\) = 65°
∠A = ∠E = 50°
∠B = ∠D = 65°
∠C = ∠F = 65°
∴ By AAA criteria △EDF ~ △ABC

Question 8.
In the given figure YH || TE. Prove that △WHY ~ △WET and also find HE and TE.

Answer:

Statements	Reasons
1. ∠EWT = ∠HWY	Common angle
2.  ∠ETW = ∠HYW	Since YH || TE, corresponding angles
3. ∠WET = ∠WHY	Since YH || TE corresponding angles
4. △WHY ~ △WET	By AAA criteria

Also △ WHY ~ △WET
∴ Corresponding sides are proportionated

⇒ 6+HE = \(\frac{6}{4}\) × 16
⇒ 6 + HE = 24
∴ HE = 24 – 6
HE = 18
Again \(\frac{4}{\mathrm{ET}}=\frac{4}{16}\)
ET = \(\frac{4}{4}\)
ET = 16

Question 9.
In the given figure, if △EAT ~ △BUN, find the measure of all angles.

Answer:
Given △EAT ≡ △BUN
∴ Corresponding angles are equal
∴ ∠E = ∠B …… (1)
∠A = ∠U …… (2)
∠T = ∠N ……. (3)
∠E = x°
∠A = 2x°
Sum of three angles of a triangle = 180°
In △ EAT, x + 2x + ∠T = 180°
∠T = 180° – (x° + 2x°)
∠T = 180° – 3x°
Also in △BUN
(x + 40)° + x° + ∠U = 180°
x + 40° + x + ∠U = 180°
2x° + 40° + ∠U = 180°
∠U = 180° – 2x – 40° = 140° – 2x°
Now by (2)
∠A = ∠U
2x = 140° – 2x°
2x + 2x = 140°
4x = 140°
x = \(\frac{140}{4}\) = 35°
∠A = 2x° = 2 × 35° = 70°
∠N = x + 40° = 35° + 40° = 75°
∴ ∠T = ∠N = 75°
∠E = ∠8 = 35°
∠A = ∠U = 70°

Question 10.
In the given figure, UB || AT and CU ≡ CB Prove that △CUB ~ △CAT and hence △CAT is isosceles.

Answer:

Statements	Reasons
1. ∠CUB = ∠CBU	∵ In △CUB, CU = CB
2. ∠CUB = ∠CAB	∵ UB || AT, Corresponding angle if CA is the transversal.
3. ∠CBU = ∠CTA	CT is transversal UB || AT, Corresponding angle commom angle.
4. ∠UCB = ∠ACT	Common angle
5. △CUB ~ △CAT	By AAA criteria
6. CA = CT	∵ ∠CAT = ∠CTA
7. Also △CAT  is isoceles	By 1, 2 and 3 and sides opposite to equal angles are equal.

Objective Type Questions

Question 11.
Two similar triangles will always have _______ angles
(A) acute
(B) obtuse
(C) right
(D) matching
Answer:
(D) matching

Question 12.
If in triangles PQR and XYZ, \(\frac{\mathrm{PQ}}{\mathrm{XY}}=\frac{\mathrm{QR}}{\mathrm{ZX}}\) then they will be similar if
(A) ∠Q = ∠Y
(B) ∠P = ∠X
(C) ∠Q = ∠X
(D) ∠P ≡∠Z
Ans:
(C) ∠Q = ∠X

Question 13.
A flag pole 15 m high casts a shadow of 3m at 10 a.m. The shadow cast by a building at the same time is 18.6m. The height of the building is
(A) 90 m
(B) 91 m
(C) 92 m
(D) 93 m
Answer:
(D) 93 m

Question 14.
If ∆ABC – ∆PQR in which ∠A = 53° and ∠Q = 77°, then ∠R is
(A) 50°
(B) 60°
(C) 70°
(D) 80°
Answer:
(A) 50°

Question 15.
In the figure, which of the following statements is true?

(A) AB = BD
(B) BD < CD
(C) AC = CD
(D) BC = CD
Answer:
(C) AC = CD

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Tamilnadu State Board New Syllabus Samacheer Kalvi 8th English Guide Pdf Poem 6 Lessons in Life Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 8th English Solutions Poem 6 Lessons in Life

8th English Guide Lessons in Life Text Book Back Questions and Answers

Warm-Up (Text Book Page No. 171)

1. What do you feel when you meet your friend after a long time?
Answer:
I felt very happy and shed tears when I met my friend after a long time.

2. Building a friendship with someone is easy or difficult? Why?
Answer:
Building a friendship with others is somehow difficult. Because, we have to build a friendship with qualities like loyalty, similarity, and selflessness.

8th English Guide Lessons in Life Textual Questions and Answers

A. Comprehension Questions:

1. What is planting a flower is compared to?
Answer:
Planting a flower is compared to the blossoming of a friendship.

2. What does the tiniest creature need?
Answer:
The tiniest creature needs room.

3. What do the smallest gifts deserve?
Answer:
Even the smallest gifts deserve a warm thank you.

4. What will happen if you fail to give importance to others?
Answer:
Everything is really important on earth. If we fail to give importance to others, they will forget us. It will make us feel bad in the future.

5. What do you learn from your lessons in life?
Answer:
I express my friendship with others by respecting them.

6. Pick and write the rhyming words from the third stanza.
Answer:
Sad – bad.

7. “Having a friend is like planting a flower.” Explain.
What is the figure of speech used in the line?
Answer:
The figure of speech used in this line is “simile”.

Lessons in Life Answer the following: (Text Book Page No. 173)

Exercise:

1. Write a sentence using ‘as fast as the wind’.
Answer:
He always runs “as fast as the wind”

2. Write a simile using the word ‘like’.
Answer:
He fought ‘like’ a lion in the battle.

3. Create a simile using the word ‘as’.
Answer:
Her hair is ‘as black as coal’.

4. What does ‘smart as a fox’ mean?
Answer:
It means that ‘looking as foolish outwardly but really very clever’.

Exercise:

1. Which of the given options is a Metaphor?
a) Life is like a chocolate box.
b) Raj is like his twin brother
c) His words are pearls of wisdom.
d) The bus is slow as a snail
Answer:
c) His words are pearls of wisdom

2. What does “The world is a stage” mean?
Answer:
The world is a stage is a Metaphor.

3. Identify the metaphor in the sentence:
Answer:
Her hai is always a rat’s nest in the morning, rat’s nest – is the metaphor.

4. Write a sentence on your own that includes a Metaphor:
Answer:
My mother is the queen of my family.

8th English Guide Lessons in Life Additional Appreciation Questions  and Answers

Read the passage and answer the following questions:
1. Let’s be aware as we walk on this planet Even the tiniest creature needs room.
a) What does ‘this planet’ refer to?
Answer:
‘This planet’ refers to our earth.

b) Who is the speaker?
Answer:
The poet is the speaker.

2. Nothing you’re given will ever come free Even the smallest of gifs deserves “thankyou ”
a) Is anything given free to you?
Answer:
No, nothing is given free to me.

b) Which one deserves ‘thank you’?
Answer:
Even the smallest gifts deserve thank you.

Lessons in Life Summary in English

This poem tells the nobility of friendship. If you show love and kindness, respect, and remembrance to others, you will be repaid, Respect others. Think of others. Remember everyone is important. Nothing is free. Say ‘thank you’ even for the smallest things. The lessons we leam in life are not simple. Even the tiniest creatures need room on the planet. In the same way even the simple things of kindness matter and make a difference.

Lessons in Life Summary in Tamil

இந்த கவிதை நாம் நட்புடன் பிறருடன் பழக வேண்டும் என்பது பற்றியது ஆகும். நாம் பிறரிடம் அன்பு, கருணை, மரியாதை அவர்களது அடையாளத்தை நினைவில் வைத்து பழகுதல் போன்ற பண்புகளோடு பழகினால், நாமும் அன்பையும் மதிப்பையும் பெறுவோம். மற்றவரை மதிக்க வேண்டும். அடுத்தவர் பற்றி நினைக்க வேண்டும். வாழ்வில் எதுவும் இலவசம் இல்லை. சிறிய காரியங்களுக்கு கூட நன்றி கூற வேண்டும். வாழ்வில் நாம் கற்றுக்கொள்ளும் பாடங்கள் எளியவை அல்ல. சிறிய உயிருக்கும் அதேபோல் சிறிய அன்பான காரியங்களுக்கும் முக்கியமானவை ஆகும்.

Lessons in Life About the Author in English

Bridgette Bryant had been in the VP developing and Alumni Relations, Arcadia University U.S. She had her bachelor’s degree in Derklee college of Music.
Daniel Ho is an American musical composer and producer. Some of his albums have been nominated for the Grammy award. He has specialised in innovative approaches to music.

Lessons in Life About the Author in Tamil

பிரிட்ஜெட் ப்ரையண்ட் என்பவர் அமெரிக்காவில் உள்ள ஆர்கேடியா பல்கலைகழகத்தின் பழைய மாணவர் கழகத்தின் விரிவாக்கப் பொறுப்பில் இருந்தவர். மேலும் அவர் பெர்க்லீ இசைக் கல்லூரியில் இளநிலை பட்டதாரி பட்டம் பெற்றவர் ஆவார்.
டேனியல் ஹோ இவர் ஒரு அமெரிக்க இசை இயலாளர் ஆவார். புதிய முறை இசை உருவாக்க வல்லவர். கிராமி அவார்டுக்கு பரிந்துரைக்கப்பட்ட பல இசை ஆல்பங்களை இவர் வெளியிட்டுள்ளார்.