Category: Class 8

Students can Download Maths Chapter 1 Life Mathematics Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Additional Questions

Additional Questions and Answers

Question 1.
Fill in the blanks

Question (a)
Percent means ……….
Answer:
Per hundred or out of hundred.

Question (b)
Percent is useful in ………
Answer:
Comparing quantities easily

Question (c)
The formula to find the increased quantity ………
Answer:
I = (1 + \(\frac{x}{100}\))

Question (d)
The formula to find the decreased quantity ………
Answer:
D = (1 + \(\frac{x}{100}\))

Question (e)
Gain or profit % ……..
Answer:
(\(\frac{Profit}{C.P}\) x 100)%

Question (f)
Loss % = ………..
Answer:
(\(\frac{Loss}{C.P}\) x 100)%

Question (g)
S.P = ………. (if gain % is given)
Answer:

Question (h)
C.P = ……… (if gain % is given)
Answer:

Question (f)
S.P = ………. (if loss % is given)
Answer:

Question (h)
C.P = ………. (if loss % is given)
Answer:

Question (k)
Selling price = Marked price – …………
Answer:
Discount

Question (l)
Cost price = Cost price + ……….
Answer:
Over head expenses

Question 2.
If y% of ₹ 1000 is 600, find the value ofy.
Solution:
y% of 1000 = 600
\(\frac{y}{100}\) x 1000 = 600
y = \(\frac{600}{10}\)
y = 60

Question 3.
A number when decreased by 10% becomes 900. Then find the number.
Solution:
Let the number be ‘x’
Given x – \(\frac{10}{100}\)x = 900
\(\frac{100x-10x}{100}\) = 900
\(\frac{90x}{100}\) = 900
x = \(\frac{900×100}{90}\) = 1000

Question 4.
If the population in a city has increased from 5,00,000 to 7,00,000 in a year, find the percentage increase in population.
Solution:
Increase in population = 7,00,000 – 5,00,000 = 2,00,000
Percentage increase in population = \(\frac{2,00,000}{5,00,000}\) x 100 = 40%

Question 5.
If the selling price of a refrigerator is equal to \(\frac{10}{8}\) of its cost price, then find the gain/ profit percent.
Solution:
Let the C.P of the refrigerator be x
S.P = \(\frac{10}{8}\)x
Profit = S.P – C.P = \(\frac{10}{8}\)x – x = \(\frac{10x-8x}{8}\) = \(\frac{2x}{8}\)

Question 6.
Karnan bought a dishwasher for ₹ 32,300 and paid ₹ 2700 for its transportation. Then he sold it for X 38,500. Find his gain or loss percent.
Solution:
Total C.P of the dishwasher = C.P + Overhead expenses.
= ₹ 32300 + ₹ 2700 = ₹ 35000
S.P = ₹ 38500
Therefore, we find S.P > C.P

Question 7.
The value of a car 2 years ago was ₹ 1,40,000. It depreciates at the rate of 4% p.a. Find its present value.
Solution:
Depreciated value = P(1 + \(\frac{r}{100}\))ⁿ = 1,40,000(1 – \(\frac{4}{100}\))²
= 1,40,000(\(\frac{96}{100}\)) x (\(\frac{96}{100}\)) = ₹ 1,29,024

Question 8.
Find the difference in C.I and S.I for P = ₹ 10,000, r = 4% p.a, n = 2 years.
Solution:
C.I – S.I = P(\(\frac{r}{100}\))² = 10,000(\(\frac{4}{100}\))²
= 10,000 x \(\frac{4}{100}\) x \(\frac{4}{100}\) = ₹ 16

Question 9.
Find the C.I for the given Principal = ₹ 8,000, r = 5% p.a, n = 2 years
Amount A = P(1 + \(\frac{r}{100}\))ⁿ = 8000(1 + \(\frac{5}{100}\))²
= 8000 x \(\frac{105}{100}\) x \(\frac{105}{100}\)
= 8000 x \(\frac{21}{20}\) x \(\frac{21}{20}\)
A = ₹ 8820
Cl = A – P = 8820 – 8000 = 820

Question 10.
Find the S.I for the principal P = ₹ 16,000, r = 5% p.a, n = 3 years
Solution:
P = ₹ 16,000, n= 3 years, r = 5%
S.I = \(\frac{Pnr}{100}\) = \(\frac{16000x3x5}{100}\) = ₹ 2400

Students can Download Maths Chapter 1 Life Mathematics Ex 1.4 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.4

Miscellaneous Practise Problems

Question 1.
Nanda’s marks in 3 Math tests Tl, T2 and T3 were 38 out of 40,27 out of 30 and 48 out of 50. In which test did he do well? Find his overall percentage in all the 3 tests.
Solution:
Nanda’s marks are as follows
In Test 1 → T₁ \(\frac{38}{40}\)
Test 2 → T₂ \(\frac{27}{30}\)
Test 3 → T₃ \(\frac{48}{50}\)
For percentage, multiply by 100
T₁ = \(\frac{38}{40}\) x 100 = 95%
T₂ = \(\frac{27}{30}\) x 100 = 90%
T₃ = \(\frac{48}{50}\) x 100 = 96%
Hence, he has scored highest percentage in test 3
∴ He is done well in T3
Overall percentage is the average of the 3 percentages.
i.e \(\frac{90+95+96}{3}\) = \(\frac{281}{50}\) = 93\(\frac{2}{3}\)%

Question 2.
Sultana bought the following things from a general store. Calculate the total bill amount to be paid by her.
(i) Medicines costing ₹ 800 with GST @ 5% ………..

(ii) Cosmetics costing ₹ 650 with GST @ 12% ………….

(iii) Cereals costing ₹ 900 with GST @ 0% …………

(iv) Sunglass costing ₹ 1750 with GST @ 18 % ………..

(v) Air Conditioner costing ₹ 28500 with GST @ 28% ………..

Solution:

(i) Medicine : bill amount is 800(1 + \(\frac{5}{100}\)) = 800 x \(\frac{105}{100}\) = 840
(ii) Cosmetics: Bill amount is 650(1 + \(\frac{12}{100}\)) = 650 x \(\frac{112}{100}\) = 728
(iii) Cereals : Bill amount is 900(1 + \(\frac{0}{100}\)) = 900
(iv) Sunglass: bill amount is 1750(1 + \(\frac{18}{100}\)) = 1750 x \(\frac{118}{100}\) = 2065
(v) AC : Bill amount is 28500(1 + \(\frac{28}{100}\)) = 28500 x \(\frac{128}{100}\) = 36480
∴ Total bill amount = 840 + 728 + 900 + 2065 + 36480
= ₹ 41,013 (total bill amount)

Question 3.
P’s income is 25% more than that of Q. By what percentage is Q’s income less than P’s?
Solution:
Let Q’s income be 100.
P’s income is 25% more than that of Q
∴ P’s income = 100 + \(\frac{25}{100}\) x 100 = 125
Q’s income is 25 less than that of P
In percentage terms, Q’s income is less than P’s with respect to P’s income is
\(\frac{P-Q}{P}\) x 100 = \(\frac{125-100}{125}\) x 100 = \(\frac{25}{100}\) x 100 = 20%

Question 4.
Gopi sold a laptop at 12% gain. If it had been sold for ₹ 1200 more, the gain would have been 20%. Find the cost price of the laptop.

Solution:
Let the cost price of the laptop be ‘x’
Gain = 12%

If the selling price was 1200 more
i.e \(\frac{112}{100}\)x +1200, the gain is 20%
i.e new selling price = x(1 + \(\frac{2}{100}\))
= \(\frac{112}{100}\)x + 1200 = x(100 + \(\frac{20}{100}\)) = \(\frac{112}{100}\)x
∴ 1200 = \(\frac{120}{100}\)x = \(\frac{112}{100}\)x = \(\frac{8}{100}\)x
∴ x = \(\frac{120×100}{8}\) = 15000
Cost price of the laptop is ₹ 15000

Question 5.
Vaidegi sold two sarees for ₹ 2200 each. On one she gains 10% and on the other she loses 12%. Calculate her gain or loss percentage in the sales.

Solution:
Saree 1:
The selling price is ₹ 2200, let cost price be CP₁, gain is 10%
Cost price ₹ Using the formula

Saree 2:
The selling price is 2200, let cost price be CP₂, loss is given as 12%. We need to find CP₂ using the formula as before,

Cost price of both together is CP₁ + CP₂
= 2000 + 2500 = 4500 ……(1)
Selling price of both together is 2 x 2200 = 4400 …….(2)
Since net selling price is less than net cost price, there is a loss.
loss% = \(\frac{loss}{cost price}\) x 100
Loss = Net cost price – Net selling price
(1) – (2) = 4500 – 4400 = 100
∴ loss% = \(\frac{100}{4500}\) x 100 = \(\frac{100}{45}\) = \(\frac{20}{9}\) = 2\(\frac{2}{9}\)%
= 2\(\frac{2}{9}\)% loss

Question 6.
A sum of money becomes ₹ 18000 in 2 years and ₹ 40500 in 4 years on compound interest. Find the sum.
Solution:
Let the sum of money be ‘P’
Given that sum ‘P’ becomes 18000 in 2yrs.
i.e Amount (A) = P(1 + \(\frac{r}{100}\))ⁿ
18000 = P(1 + \(\frac{r}{100}\))²
(1 + \(\frac{r}{100}\))² = \(\frac{18000}{P}\) ⇒ (1 + \(\frac{r}{100}\))⁴ = (\(\frac{18000}{P}\))² ……(1)
Also given that sum ‘P’ becomes46500 in 4 yrs.
i.e Amount (A) = P(1 + \(\frac{r}{100}\))ⁿ
40500 = P(1 + \(\frac{r}{100}\))⁴
Substituting (1) in (2), we get

Question 7.
Find the difference in the compound interest on ₹ 62500 for 1\(\frac{1}{2}\) years at 8% p.a compounded annually and when compounded half-yearly.
Solution:
Case 1:
P = ₹ 62500
n = 1\(\frac{1}{2}\)yrs. (a\(\frac{b}{c}\)) formula
r = 8% Compound annully
CI = A – P

CI = A – P = 70200 – 62500 = 7700 ……(1)
CAse 2:
P = ₹ 62500
n = 1\(\frac{1}{2}\)yrs.
r = 8% p.a when compound half yearly
r = 4% compound half yearly

= 70304 – 62500 = ₹ 7804
Difference between case 1 & case 2 is (2) – (1)
∴ (2) – (1) = 7804 – 7700 = ₹ 104

Challenging Problems

Question 8.
If the first number is 20% less than the second number and the second number is 25% more than 100, then find the first number.
Solution:
Second number is 25% more than 100
∴ 2nd number is 100 + \(\frac{25}{100}\) x 100= 125
First number is 20% less than second no.

1st number is 100

Question 9.
A shopkeeper gives two successive discounts on an article whose marked price is ? 180 and selling price is ? 108. Find the first discount percent if the second discount is 25%.
Solution:
Marked price is given as ? 180
Let 1^st discount be d₁% = ? (to find)
2nd discount be d₂% = 25%
Selling price is 108 (given)
Price after 1^st discount = 180(1 – \(\frac { d_{ 1 } }{ 100 } \)) = P₁
Price after 2^nd discount = P₁(1 – \(\frac { d_{ 2 } }{ 100 } \)) = 108
Substituting for P₁ from (1), we get

∴ 1 – \(\frac { d_{ 1 } }{ 100 } \) = \(\frac{4}{5}\)
∴ \(\frac { d_{ 1 } }{ 100 } \) = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
∴ d₁ = \(\frac{1}{5}\) x 100 = 20%

Question 10.
A man bought an article on 30% discount and sold it at 40% more than the marked price. Find the profit made by him.
Solution:
Let marked price be ‘P’
Discounted price = P(1 – \(\frac{d}{100}\)) where d is the discount %
∴ Discounted price = P(1 – \(\frac{30}{100}\)) = \(\frac{70}{100}\)P → this is the cost price.
Selling price = 40% more than marked price

Question 11.
Find the rate of compound interest at which a principal becomes 1.69 times itself in 2 years.
Solution:
Let principal be ‘P’
Amount is given to be 1.69 times principal
i.e 1.69 P
Time period is 2yrs. = (n)
Rate of interest = r = ? (required)
Applying the formula,
Amount = Principal (1 + \(\frac{r}{100}\))ⁿ
Substituting, 1.69 P = P (1 + \(\frac{r}{100}\))²
∴ (1 + \(\frac{r}{100}\))² = \(\frac{1.69P}{P}\)
Taking square root on both sides, we get
\(\sqrt{1.69}\) = 1 + \(\frac{r}{100}\)
∴ 1 + \(\frac{r}{100}\) = 1.3
∴ \(\frac{r}{100}\) = 1.3
r = 30%
∴ rate of compound interst is 30%

Question 12.
The simple interest on a certain principal for 3 years at 10% p.a is ₹ 300. Find the compound interest accrued in 3 years.
Solution:
Let principal be ‘P’
Rate of interest is 10% p.a (r)
Time period (n) = 3 yrs.
Formula for simple interest
∴ P = \(\frac{300×100}{10×3}\) = 1000
Compound Interest = CI = A – P
Amount (A) = P (1 + \(\frac{r}{100}\))ⁿ
= 1000(1 + \(\frac{10}{100}\))³ = 1000 x (\(\frac{10o+10}{100}\))³
= 1000 x \(\frac{110}{100}\) x \(\frac{110}{100}\) x \(\frac{110}{100}\) = 1331
Compound Interest (Cl) = Amount – Principal
= 1331 – 1000 = 331
Compound Interest = ₹ 331

Students can Download Maths Chapter 1 Life Mathematics Ex 1.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.3

Question 1.
Fill in the blanks

Question (i)
The compound interest on ₹ 5000 at 12% p.a for 2 years compounded annually is ………..
Answer:
₹ 1272
Hint:
Compound Interest (Cl) formula is
Cl = Amount – Principal
Amount = A (1 + \(\frac{r}{100}\))ⁿ = 5000 (1 + \(\frac{12}{100}\))²
= 5000 (1 + \(\frac{112}{100}\))² = 6272
∴ Cl = 6272 – 5000 = ₹ 1272

Question (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 (1 + \(\frac{r}{100}\))²ⁿ [2n as it is compounded half yearly]
r = 10% p.a, for half yearly r = 1 + \(\frac{10}{2}\) = 5
A = 8000 (1 + \(\frac{5}{100}\))^2×1 = 8000 x (\(\frac{105}{100}\))² = 8820
CI = Amount – principal
= 8820 – 8000 = ₹ 820

Question (iii)
The annual rate of growth in population of a town is 10%. If its present population is 26620, 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,
26620 = x (1 + \(\frac{r}{100}\))³
∴ 26620 = x (1 + \(\frac{10}{100}\))³ = x (\(\frac{110}{100}\))³
∴ x = 26620 x (\(\frac{110}{100}\))³
= ₹ 20,000
The population 3 years ago was ₹ 20,000

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

Question (v)
The difference between the S.I and C.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 = P (\(\frac{r}{100}\))²
Principal (P) = 5000, r = 8% p.a
∴ CI – SI = 5000 (\(\frac{8}{100}\))² = 5000 x \(\frac{8}{100}\) x \(\frac{8}{100}\) = ₹ 32

Question 2.
Say True or False

Question (i)
Depreciation value is calculated by the formula P (1 – \(\frac{r}{100}\))ⁿ
Answer:
True
Hint:
Depreciation formula is P (1 – \(\frac{r}{100}\))ⁿ

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

Question (iii)
The present value of a machine is ₹ 16800. It depreciates @25% p.a. Its worth after 2 years is ₹ 9450.
Answer:
True
Hint:
Present value of machine = ₹ 16800
Depreciation rate (n) = 25%
Value after 2 years = P (1 – \(\frac{r}{100}\))ⁿ = 16800 (1 – \(\frac{25}{100}\))²
= 16800 x (1 – \(\frac{1}{4}\))² = 16800 x \(\frac{3}{4}\) x \(\frac{3}{4}\) = 9450

Question (iv)
The time taken for ₹ 1000 to become ₹ 1331 @20% p.a compounded annually is 3 years.
Answer:
False
Principal money = 1000
rate of interest Amount = 20%
Amount = 1331, applying in formula we get
A = (1 + \(\frac{r}{100}\))ⁿ
1331 = 1000(1 + \(\frac{r}{100}\))ⁿ
∴ \(\frac{1331}{1000}\) = (1 – \(\frac{1}{5}\))ⁿ
\(\frac{1331}{1000}\) = (\(\frac{6}{5}\))ⁿ
∴ n ≠ 3 (False)

Question (v)
The compound interest on ₹ 16000 for 9 months @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 x (1 + \(\frac{r}{100}\))⁴ⁿ
Since quarterly we have to divide r by 4
r = \(\frac{20}{4}\)

∴ 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.
Solution:
Principal (P) = ₹ 3200
r = 2.5% p.a
n = 2 years comp, annually
∴ Amount (A) = (1 + \(\frac{r}{100}\))ⁿ = (1 + \(\frac{2.5}{100}\))²
= 3200 x (1.025)² = 3362
Compound interest (Cl) = 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.
Solution:
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

= 4000x 1.1 x 1.1 x 1.05 = 5082
∴ Cl = Amount – Principal = 5082 – 4000 = 1082

Question 5.
Magesh invested ₹ 5000 at 12% p.a for one year. If the interest is compounded half yearly, find the amount he gets at the end of the year.
Solution:
Principal (P) = ₹ 5000
Interest compounded half yearly
r = 12% p.a = \(\frac{12}{2}\) = 6% for half yearly
t = 1 yr.
Since compounded half yearly, the formula to be used is
Amount A = P (1 + \(\frac{r}{100}\))²ⁿ
A = 5000 (1 + \(\frac{6}{100}\))^2×1 = 5000 x (\(\frac{106}{100}\))² = ₹ 5618

Question 6.
At what time will a sum of ₹ 3000 will amount to ₹ 3993 at 10% p.a compounded annually?
Solution:
Amount A = ₹ 3993
Principal = ₹ 3000
r = 10% p.a
n = ?

Question 7.
A principal becomes ₹ 2028 in 2 years at 4% p.a compound interest. Find the Principal.
Solution:
n = 2 years
r = rate of interest = 4% p.a
Amount A = ₹ 2028
Amount (A) = P (1 + \(\frac{r}{100}\))ⁿ
2028 = P (1 + \(\frac{4}{100}\))ⁿ
2028 = P (\(\frac{r}{100}\))²
∴ P = \(\frac{2028x100x100}{104×104}\) = ₹ 1875

Question 8.
At what rate percentage p.a will ₹ 5625 amount to ₹ 6084 in 2 years at compound interest?
Solution:
Principal (P) = ₹ 5625
Amount (A) = ₹ 6084
n = 2 years
r = ?
Amount (A) = P (1 + \(\frac{r}{100}\))ⁿ [Applying in formula]
6084 = 5625 (1 + \(\frac{r}{100}\))²
(1 + \(\frac{r}{100}\))² = \(\frac{6084}{5625}\)
Taking square root on both sides, we get
1 + \(\frac{r}{100}\) = \(\frac{78}{75}\)
\(\frac{r}{100}\) = \(\frac{78}{75}\) – 1 = \(\frac{3}{75}\) = \(\frac{1}{25}\)
∴ r = \(\frac{1}{25}\) x 100 = 4%

Question 9.
In how many years will ₹ 3375 amount to ₹ 4096 at 13\(\frac{1}{3}\)% p.a where interest is compounded half-yearly?
Solution:
Principal = ₹ 3375
Amount = ₹ 4096
r = 13\(\frac{1}{3}\)% p.a = \(\frac{40}{3}\)% p.a
Compounded half yearly r = \(\frac { \frac { 40 }{ 3 } }{ 2 } \) = \(\frac{2}{3}\)
Let no. of years be n
for compounding half yearly, formula is

Question 10.
Find the C.I on ₹ 15000 for 3 years if the rates of interest is 15%, 20% and 25% for I, II and III years respectively.
Solution:
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

Compound Interest (Cl) = A – P = 25,875 – 15,000 = 10,875
Cl = ₹ 10,875

Question 11.
The present height of a tree is 847 cm. Find its height two years ago, if it increases at 10 % p.a.

Solution:
Present height of tree = 847 cm
Present height = ‘h’
n = 2 yrs
rate of growth = 10% p.a
Applying in formula, we get

∴ Original height of tree = 70 cm

Question 12.
Find the difference between the C.I and the S.I on ₹ 5000 for 1 year at 2% p.a, if the interest is compounded half yearly.
Solution:
Principal (P) = ₹ 5000
time period (n) = 1 yr.
Rate of interest (r) = 2% p.a
for half yearly r = 1%
Difference between Cl & SI is given by the formula
CI – SI = P (\(\frac{r}{100}\))²ⁿ [for half yearly compounding]
CI – SI = P (\(\frac{1}{100}\))^2×1
= 5000 x \(\frac{1}{100}\) x \(\frac{1}{100}\) = ₹ 0.50

Question 13.
What is the difference in simple interest and compound interest on 115000 for 2 years at 6% p.a compounded annually.
Solution:
Principal (P) = ₹ 15,000
Time period (n) = 2 yrs.
Rate of interest (r) = 6% p.a compounded annually
Difference between CI and SI given by
CI – SI = P (\(\frac{r}{100}\))ⁿ = 15000 (\(\frac{6}{100}\))²
= 15000 x \(\frac{6}{100}\) x \(\frac{6}{100}\)
= ₹ 54

Question 14.
Find the rate of interest if the difference between the C.I and S.I on ₹ 8000 compounded annually for 2 years is ₹ 20.
Solution:
Principal (P) = ₹ 8000
time period (n) = 2 yrs.
rate of interest (r) = ?
Difference between Cl & SI is given by the formula
CI – SI = P (1 + \(\frac{r}{100}\))ⁿ
Difference between Cl & SI is given as 20
∴ 20 = 8000 x (\(\frac{r}{100}\))²
∴ (\(\frac{r}{100}\))² = \(\frac{20}{8000}\) = \(\frac{1}{400}\)
Taking square root on both sides
\(\frac{r}{100}\) = \(\sqrt { \frac { 1 }{ 400 } } \) = \(\frac{1}{20}\)
∴ r = \(\frac{100}{20}\)

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

Objective Type Questions

Question 16.
The number of conversion periods, 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\(\frac{12}{2}\) conversation periods.

Question 17.
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) 2years
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 (1 + \(\frac{r}{100}\))²ⁿ
Substuting in the above formula, we get

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

Question 18.
The cost of a machine is ₹ 18000 and it depreciates at 16\(\frac{2}{3}\)% annually. Its value after 2 years will be ………..
(a) ₹ 12000
(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 (1 – \(\frac{r}{100}\))ⁿ
Substituting in above formula, we get

Question 19.
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?
Applying formula A = P (1 + \(\frac{r}{100}\))ⁿ

Question 20.
The difference between simple and compound interest on a certain sum of money for 2 years at 2% p.a is ₹ 1 the sum of money is ……….
(a) ₹ 2000
(b) ₹ 1500
(c) ₹ 3000
(d) ₹ 2500
Answer:
(d) ₹ 2500
Difference between Cl and SI is given as Re 1
Time period (n) = 2 yrs.
Rate of interest (r) = 2% p.a
Formula for difference is
CI – SI = P x (1 + \(\frac{r}{100}\))ⁿ
Substituting the values in above formula, we get
1 = P x (\(\frac{2}{100}\))²
∴ P = 1 x (\(\frac{100}{2}\))²
= 1 x (50)² = ₹ 2500

Students can Download Maths Chapter 1 Life Mathematics Ex 1.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.2

Question 1.
Fill in the blanks:

Question (i)
Loss or gain percent is always calculated on the ………
Answer:
Cost price

Question (ii)
A mobile phone is sold for ? 8400 at a gain of 20%. The cost price of the mobile phone is …………
Answer:
₹ 7000
Hint:
Let cost price of mobile be ₹ x. Given that selling price is ? 8400 and gain is 20%

Question (iii)
An article is sold for ₹ 555 at a loss of 7\(\frac{1}{2}\)% the cost price qfthe article is ……….
Answer:
₹ 600
Hint:
Given selling price is ₹ 555 & loss is 7\(\frac{1}{2}\)%
as per formula,

Question (iv)
The marked price of a mixer grinder is ₹ 4500 is sold for ₹ 4140 after discount. The rate of discount is ……..
Answer:
8%
Hint:
Marked price is X 4500. Discounted price in ₹ 4140
∴ Discount = Marked price – Discounted price = 4500 – 4140 – 360

Question (v)
The total bill amount of a shirt costing ₹ 575 and a T-shirt costing ₹ 325 with GST of 5% is ………..
Answer:
₹ 945
Hint:
Cost price of shirt = ₹ 575 (CP)
GST = 5%

= 575 x (\(\frac { (100+5) }{ 100 } \)) = 575 x \(\frac { 105 }{ 100 } \)
= ₹ 603.75
Cost price of T-shirt = ₹ 325 (CP)
GST = 5%

= 325 x (\(\frac { (100+5) }{ 100 } \))
Total bill amount = ₹ 603.75 + ₹ 341.25 = ₹ 945

Question 2.
If selling an article for ₹ 820 causes 10% loss on the selling price, find its cost price.
Solution:
Given that selling price (SP) = ₹ 820
Loss % = 10 %

Question 3.
If the profit earned on selling an article for ₹ 810 is the same as loss on selling it for ₹ 530, then find the cost price of the article.
Solution:
Case 1: Profit = Selling price (SP) – Cost price (CP)
Case 2: Loss = Cost price (CP) – Selling price (SP)
Given that profit of case 1 = loss of case 2
∴ P = 810 – CP
L = CP – 530
Since profit (P) = loss (L)
810 – CP = CP – 530
∴ 2CP = 810 + 530 = 1340 ⇒ C.P = \(\frac{1340}{2}\)
∴ CP = 670

Question 4.
Some articles are bought at 2 for ₹ 15 and sold at 3 for ₹ 25. Find the gain percentage.
Solution:
Let cost price of one article be C.P
Given that 2 are bought for ₹ 15
∴ 2 x CP = 15 ⇒ CP = \(\frac{15}{2}\)
Let selling price of one article be SP
Given that 3 are sold for ₹ 25
∴ 3 x CP = 25 ⇒ SP = \(\frac{25}{3}\)
∴ Gain = SP – CP = \(\frac{25}{3}\) – \(\frac{15}{2}\) = \(\frac{50-45}{6}\) = \(\frac{5}{6}\)

Question 5.
If the selling price of 10 rulers is the same as the cost price of 15 rulers, then find the gain percentage.
Solution:
Let cost price of one ruler be x
Given that selling price (SP) of 10 rulers
i.e., same as cost price (CP) of 15 rulers
∴ SP of 10 rulers = 15 × x = 15x
SP of 1 ruler = \(\frac{15x}{10}\) = 1.5x
∴ Gain = SP of 1 ruler – CP of ruler = 1.5x -x = 0.5x

Question 6.
By selling a speaker for ₹ 768, a man loses 20%. In order to gain 20% how much should he sell the speaker?

Solution:
Selling price (SP) of speaker = ₹ 768
Loss % = 20%
as per formula

For gain of 20%, we should now calculate the selling price

= 960 (\(\frac{100+20}{100}\)) = 960 x \(\frac{120}{100}\)
= 96 x 12 = ₹ 1152

Question 7.
A man sold two gas stoves for ₹ 8400 each. He sold one at a gain of 20% and the other at a loss of 20%. Find his gain or loss % in the whole transaction.

Solution:
Let the CP of gas stove 1 be x and gas stove 2 be y
Given that selling price (SP) for both is the same = ₹ 8400
Case i:
First gastove: Cost price (CP) = x
Selling Price (SP) = ₹ 8400
Gain % = 20%

Cost price of first gas stove = ₹ 7000

Case 2:
Second gastove: Cost price (CP) = y
Selling price (SP) = ₹ 8400
loss % = 20%

∴ Cost price of second stove = ₹ 10,500
From (1) and (2),
Total cost price = Cost of stove 1 + Cost of stove 2
= 7000+ 10500 = 17,500
Total selling price = SP of stove 1 + SP of stove 2
= 8400 + 8400 = 16,800
Now, we find that total selling price is less than total cost price, therefore it is a loss
∴ Loss = CP – SP= 17,500 – 16,800 = 700
Loss % = \(\frac{loss}{CP}\) x 100 = \(\frac{700}{17500}\) x 100 = 4%
Loss % = 4 %

Question 8.
Find the unknowns x, y and z

Solution:
(i) Book marked price = ₹ 225 discount = 8%

∴ 225 x (\(\frac { (100-8) }{ 100 } \))
= 225 x (\(\frac { 92 }{ 100 } \)) = ₹ 207

(ii) LED TV selling price = 11970 discount = 5%,

∴ 11970 = y x \(\frac { (100-d%) }{ 100 } \)
∴ y = \(\frac { 11970×100 }{ 95 } \) = 126 x 100 = ₹ 12,600

(iii) Digital clock marked price (MP) = ₹ 750, MP = ₹ 12,600
Selling price (SP) = ₹ 615, Discount = z

∴ 615 = 750 x \(\frac { (100-z) }{ 100 } \)
∴ (100 – z) = \(\frac { 615×100 }{ 700 } \)
100 – z = 82
∴ z = 100 – 82
Discouont = 18%

Question 9.
Find the total bill amount for the data below.

Solution:

For bill amount, we should apply GST on the discounted value of the items.

Total bill amount = Bill amount of School bag + Stationary + Cosmetics + Hair drier
= 532 + 252 + 1357 + 2304
= ₹ 4,445

Question 10.
A shopkeeper buys goods at \(\frac { 4 }{ 5 } \) of its marked price and sells them at \(\frac { 4 }{ 5 } \) of the marked price find his profit percentage.
Solution:
Let marked price be MP
Given that he buys good at \(\frac { 4 }{ 5 } \) of marked price
∴ CP (cost price) = \(\frac { 4 }{ 5 } \) MP
Given that selling price (SP) = \(\frac { 7 }{ 5 } \) x MP
∴ Profit = Selling price – Cost price = \(\frac { 7 }{ 5 } \) MP – \(\frac { 4 }{ 5 } \) MP = \(\frac { 3 }{ 5 } \)  MP

Question 11.
A branded AC has a marked price of ₹ 37250. There are 2 options given for the customer.

1. Selling Price is ₹ 37250 along with attractive gifts worth ₹ 3000 (or)
2. Discount of 8% but no free gifts. Which offer is better?

Marked price of AC = ₹ 37,250
Option 1:
Selling price = ₹ 37250 & gifts worth ₹ 3000
∴ Net gain for customer = ₹ 3000 as there is no discount on AC

Option 2:
Discount of 8%, but no gift

= 37250 x \(\frac { (100-8) }{ 100 } \) = 37250 x 0.92 = 34270
∴ Savings for customer = 37250 – 34270 = ₹ 2980
Therefore, the customer gets 3000 gift in option 1 where as he is able to save only ₹ 2980 in option 2. Therefore, option 1 is better.

Question 12.
If a mattress is marked for ₹ 7500 and is available at two successive discount of 10% and 20%, find the amount to be paid by the customer.
Solution:
Marked price of mattress = ₹ 7500
Discount d₁ = 10%
Discount d₂ = 20%

= 7500 x \(\frac { (100-10) }{ 100 } \) = 7500 x \(\frac { 90 }{ 100 } \) = 6750

= 6750 x \(\frac { (100-20) }{ 100 } \) = ₹ 5400

Objective Type Questions

Question 13.
A fruit vendor sells fruits for ₹ 200 gaining ₹ 40. His gain percentage is –
(a) 20%
(b) 22%
(c) 25%
(d) 16\(\frac { 2 }{ 3 } \)
Answer:
(c) 25%
Hint:
Selling price = ₹ 200
Gain = 40
∴ CP = Selling price – gain = 200 – 40 = 160
Gain % = \(\frac { Gain }{ CP } \) x 100 = \(\frac { 40 }{ 160 } \) x 100 = 25%

Question 14.
By selling a flower pot for ₹ 528, a woman gains 20%. At what price should she sell it to gain 25%?
(a) ₹ 500
(b) ₹ 550
(c) ₹ 553
(d) ₹ 573
Answer:
(b) ₹ 550
Hint:
If selling price (SP) = ₹ 528
Gain % = 20%
∴ CP = ?

∴ 528 = CP x \(\frac { 100+20 }{ 100 } \)
∴ CP = \(\frac { 528×100 }{ 120 } \)
If gain % = 25 %, Selling ?

= 440 x \(\frac { (100+gain%) }{ 100 } \) = 440 x \(\frac { 125 }{ 100 } \)
= ₹ 550

Question 15.
A man buys an article for ₹ 150 and makes overhead expenses which are 12% of the cost price. At what price must he sell it to gain 5%?
(a) ₹ 180
(b) ₹ 168
(c) ₹ 176.40
(d) ₹ 85
Answer:
(c) ₹ 176.40
Hint:
Cost price of article = ₹ 150
Over head expenses = 12% of cost price = \(\frac { 12 }{ 100 } \) x 150 = ₹ 85
∴ Effective cost of article = 150 + 18 = ₹ 168
Now, to gain 5%, he has to sell at

Question 16.
The price of a hat is ₹ 210. What was the marked price of the hat if it is bought at 16% discount?
(a) ₹ 243
(b) ₹ 176
(c) ₹ 230
(d) ₹ 250
Answer:
(d) ₹ 250
Hint:
Let marked price be MP
Discounted price = ₹ 210
Rate of discount = 16%
As per formula:

Question 17.
The single discount which is equivalent to two successive discount of 20% and 25% is –
(a) 40%
(b) 45%
(c) 5%
(d) 22.5%
Answer:
(a) 40%
Let marked price be MP, after discount 1 of 20%,

After discount 2 of 25%,

Comparing with formula, we get
∴ This is equivalent to a single discount of 40%

Students can Download Maths Chapter 1 Rational Numbers Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 1 Rational Numbers Additional Questions

Additional Questions And Answers

Exercise 1.1

Very Short Answers [2 Marks]
Question 1.
Add \(\frac{3}{5}\) and \(\frac{13}{5}\)
Solution:

Question 2.
Add \(\frac{7}{9}\) and \(\frac{-12}{9}\)
Solution:

Question 3.
Add \(\frac{-3}{7}\) and \(\frac{-17}{7}\)
Solution:

Question 4.
Add \(\frac{4}{-13}\) and \(\frac{7}{13}\)
Solution:

Question 5.
Subtract \(\frac{3}{4}\) and \(\frac{7}{4}\)
Solution:
\(\frac{7}{4}-\frac{3}{4}=\frac{7-3}{4}=\frac{4}{4}\) = 1

Short Answers [3 Marks]

Question 1.
Add \(\frac{4}{-3}\) and \(\frac{8}{15}\)
Solution:

Question 2.
Simplify \(\frac{9}{-27}+\frac{18}{39}\)
Solution:

Long Answers [5 Marks]

Question 1.
By what number should we multiply \(\frac{3}{-14}\), so that the product may be \(\frac{5}{12}\)
Solution:
Let the number to be multiplied by x

∴ The number to be multiplied = \(\frac{-35}{18}\)

Question 2.
Simplify

Solution:

Exercise 1.2

Very Short Answers [2 Marks]
Question 1.
Verify addition of rational number is closed using \(\frac{1}{4}\) and \(\frac{2}{3}\)
Solution:

∴ Addition of rational numbers is closed

Question 2.
Is subtraction is commutative for rational numbers. Given an example.
Solution:
No, subtraction is not commutative for rational numbers.
Example: Let a = \(\frac{1}{2}\) and b = \(\frac{5}{6}\)

From (1) and (2)
a – b ≠ b – a for rational numbers

Very Short Answers [5 Marks]

Question 1.
Verify associative property for addition of rational numbers for a = \(\frac{5}{6}\), b = \(\frac{-3}{4}\), c = \(\frac{4}{7}\)
Solution:
Given a = \(\frac{5}{6}\), b = \(\frac{-3}{4}\), c = \(\frac{4}{7}\)

From (1) and (2) we have (a + b) + c = a + (b + c)
∴ Associative property is true for addition of rational numbers.

Question 2.
Verify distributive property of multiplication over addition for the rational numbers a = \(\frac{3}{4}\), b = \(\frac{-2}{3}\), c = \(\frac{3}{7}\)
Solution:
Given a = \(\frac{3}{4}\), b = \(\frac{-2}{3}\), c = \(\frac{3}{7}\)
To verify a × (b + c) = (a × b) + (a × c)


From (1) and (2)
a × (b + c) = (a × b) + (a × c)
∴ Distributive property of multiplication over addition is true for the given rational numbers.

Students can Download Maths Chapter 4 Geometry Ex 4.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.1

Question 1.
From the figure given, prove that ∆ABC ~ ∆DEF.

Solution:
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 ∆DEF, ∠D = 50°
∴ Sum of Remaining angles = 180° – 50° = 130°
DE = FD
∴ ∠D = ∠F
From ∆ABC and ∆DEF

∠A = ∠D = 50°
∠B = ∠E = 65°
∠C = ∠F = 65°
∴ By AAA criteria ∆DEF ~ ∆ABC

Question 2.
Prove that ∆GUM ~ ∆ BOX from the given figure.

Solution:


That is their corresponding sides are proportional.
∴ By SSS similarity ∆GUM ~ ∆BOX.

Question 3.
In the given figure YH ||TE Prove that ∆WHY ~ ∆WET and also find HE and TE.

Solution:

Question 4.
In the given figure, if ∆EAT ~ ∆BUN find the measure of all angles.

Solution:
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° …(4)
Also in ∆BUN
(x + 40)° + + ∠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°

∠A = 2x° = 2 × 35° = 70°
∠N = x + 40°
= 35° + 40° = 75°
∴ ∠T = ∠N = 75°
∠E = ∠B = 35°
∠A = ∠U = 70°

Question 5.
From the given figure, UB || AT and CU ≡ CB Prove that ∆CUB ~ ∆CAT and hence ∆CAT is isosceles.

Solution:

Question 6.
In the figure, ∠CIP ≡ ∠COP and ∠HIP ≡ ∠HOP. Prove that IP ≡ OP.

Solution:

Question 7.
In the given triangle, AC ≡ AD and ∠CBD ≡ ∠DEC. Prove that ∆BCF ≡ ∆EDF.

Solution:

Question 8.
In the given figure, ∆ BCD is isosceles with base BD and ∠BAE ≡ ∠DEA. Prove that AB ≡ ED .

Solution:

Question 9.
In the given figure, D is the midpoint of OE and ∠CDE = 90°. Prove that ∆ODC ≡ ∆EDC.

Solution:

Question 10.
In the figure, if SW ≡ SE and ∠NWO ≡ ∠NEO. then, prove that NS bisects WE and ∠NOW = 90°

Proof:

Question 11.
Is ∆PRQ ≡ ∆QSP ? Why ?

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

Question 12.
Fill in the blanks with the most correct term from the given list.
(in proportion, similar, corresponding, congruent shape, area, equal)
Statements Reasons

Question 1.
Corresponding sides of similar triangles are ___.
Solution:
in proportion

Question 2.
Similar triangles have the same ___ but not necessarily the same size.
Solution:
shape

Question 3.
In similar triangles, ___ sides are opposite to equal angles.
Solution:
equal

Question 4.
The symbol ~ is used to represent ___ triangles.
Solution:
congruent

Question 5.
The symbol ~ is used to represent ____ triangles.
Solution:
similar

Objective Type Questions

Question 13.
Two similar triangles will always have ___ angles
(A) acute
(B) obtuse
(C) right
(D) matching
Solution:
(D) matching

Question 14.
If in triangles PQR and XYZ, \(\frac{P Q}{X Y}=\frac{Q R}{Z X}\) then they will be similar if
Solution:
(C) Q = ∠X

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

Question 16.
If ∆ABC ~ ∆PQR in which ∠A = 53° and ∠Q = 77°, then ∠R is
(A) 50°
(B) 60°
(C) 70°
(D) 80°
Solution:
(A) 50°

Question 17.
In the figure, which of the following statements is true?
(A) AB = BD
(B) BD < CD
(C) AC = CD
(D) BC = CD

Solution:
(C) AC = CD

Students can Download Maths Chapter 1 Rational Numbers Ex 1.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 1 Rational Numbers Ex 1.3

Question 1.
Match the following appropriately.

Solution:
(i) 5
(ii) 4
(iii) 2
(iv) 3
(v) 1

Question 2.
Which of the following properties hold for subtraction of rational numbers? Why?
(a) closure
(b) commutative
(c) associative
(d) identity
(e) inverse
Solution:
(i) For subtraction of rational numbers closure property is true.
Because for any two rational number a and b, a + b is in Q.
Eg. \(-\frac{1}{4}+\frac{3}{2}=\frac{-1+6}{4}=\frac{5}{4}\) is rational.
(ii) Commutative fails as \(\frac{1}{3}-\frac{2}{4} \neq \frac{2}{4}-\frac{1}{3}\)
(iii) Associative fails as \(\frac{1}{2}-\left(\frac{1}{3}-\frac{1}{4}\right) \neq\left(\frac{1}{2}-\frac{1}{3}\right)-\frac{1}{4}\)
(iv) Identity fails as 5 – 0 ≠ 0 – 5
(v) Inverse also fails.

Question 3.
Subbu spends \(\frac{1}{3}\) of his monthly earnings on rent, \(\frac{2}{5}\) on food and \(\frac{1}{10}\) on monthly usuals. What fractional part of his earnings is left with him for other expenses?
Solution:

Question 4.
In a constituency, \(\frac{19}{25}\) of the voters had voted for candidate A whereas \(\frac{7}{50}\) had voted for candidate B. Find the fraction of the voters who had voted for other.
Solution:

Question 5.
If \(\frac{3}{4}\) of a box of apples weighs 3 kg and 225 gm, how much does a full box of apples weigh?

Solution:
Let the total weight of a box of apple = x kg.
Weight of \(\frac{3}{4}\) of a box apples = 3 kg 225 gm. = 3.225 kg
\(\frac{3}{4}\) × x = 3225
x = \(\frac{3.225 \times 4}{3}\) kg
= 1.075 × 4 kg = 4.3 kg = 4 kg 300 gm
Weight of the box of apples = 4 kg 300 gm.

Question 6.
Mangalam buys a water jug of capacity 3\(\frac{4}{5}\) litres. If she buys another jug which is 2\(\frac{2}{3}\) times as large as the smaller jug, how many litres can the larger one hold?

Solution:

Question 7.
In a recipe making, every \(1 \frac{1}{2}\) cup of rice requires \(2 \frac{3}{4}\) cups of water. Express this in the ratio of rice to water.
Solution:
For the recipe rice required = \(1 \frac{1}{2}\) cup; water required = \(2 \frac{3}{4}\) cups

∴ rice : water = 6 : 11

Question 8.
Ravi multiplied \(\frac{25}{8}\) and \(\frac{16}{15}\) to obtain \(\frac{400}{120}\). He says that the simplest form of this product is \(\frac{10}{3}\) and Chandru says the answer in the simplest form is \(3 \frac{1}{3}\). Who is correct? or Are they both correct? Explain.
Solution:

Question 9.
A piece of wire is \(\frac{4}{5}\) m long. If it is cut into 8 pieces of equal length, how long will each piece be?
Solution:
Length of the wire = \(\frac{4}{5}\) m = \(\frac{4 \times 100}{5}\) cm = 80 cm
Number of equal pieces made from it = 8 Length of a single piece = 80 ÷ 8 = 10 cm
Length of each small pieces = 10 cm.

Question 10.
Find the length of a room whose area is \(\frac{153}{10}\) sq.m and whose breadth is \(2 \frac{11}{20}\) m.
Solution:
Breadth of the room = \( 2\frac{11}{20}\) m; Area of the room = \(\frac{153}{10}\) sq.m
Length of the room × Breadth = Area of the room

Length of the room = 6 m

Challenging Problems

Question 1.
Show that

Solution:

Question 2.
If A walks \(\frac{7}{4}\) km and then jogs \(\frac{3}{5}\) km, find the total distance covered by A. How much did A walk rather than jog?
Solution:
Distance walked by A = \(\frac{7}{4}\) km; Distance jogged by A = \(\frac{3}{5}\) km

Question 3.
In a map, if 1 inch refers to 120km, then find the distance between two cities B and C which are \(4 \frac{1}{6}\) inches and \(3 \frac{1}{3}\) inches from the city A which is in between the cities B and C.

Solution:
1 inch = 120 km

Distance between B and C = 900 km

Question 4.
Give an example for each of the following statements.
(i) The collection of all non-zero rational numbers is closed under division.
(ii) Subtraction is not commutative for rational numbers.
(iii) Division is not associative for rational numbers.
(iv) Distributive of multiplication over subtraction is true for rational numbers, that is a (b – c) = ab – ac.
(v) The mean of two rational numbers is rational and lies between them.
Solution:
(i) Let a = \(\frac{5}{4}\) and b = \(\frac{-4}{3}\) be two non zero rational numbers.
a ÷ b = \(\frac{5}{6} \div \frac{-4}{3}=\frac{5}{6} \times \frac{3}{-4}=\frac{5}{-8}\) is in Q
∴ Collection of non-zero rational numbers are closed under division.

∴ Division is not associative for rational numbers

∴ From (1) and (2)
a × (b – c) = ab – bc
∴ Distributivity of multiplication over subtraction is true for rational numbers.

Question 5.
If \(\frac{1}{4}\) of a ragi adai weighs 120 grams, 4what will be the weight of \(\frac{2}{3}\) of the same ragi adai?
Solution:
Let the weight of 1 ragi adai = x grams given \(\frac{1}{4}\) of x = 120 gm
\(\frac{1}{4}\) × x = 120
x = 120 × 4
x = 480 gm
∴ \(\frac{2}{3}\) of the adai
= \(\frac{2}{3}\) × 480 gm = 2 × 160 gm = 320 gm
\(\frac{2}{3}\) of the weight of adai = 320 gm

Question 6.
Find the difference between the greatest and the smallest of the following rational numbers.
\(\frac{-7}{12}, \frac{2}{-9}, \frac{-11}{36}, \frac{-5}{-6}\)
Solution:
Here \(\frac{-5}{-6}=\frac{5}{6}\) and is a positive rational number.
All other numbers are negative numbers
∴ \(\frac{-5}{-6}\) is the greatest number
LCMof 12, 9, 36 = 3 × 4 × 3 = 36

Question 7.
If p + 2q = 18 and pq = 4o, find \(\frac{2}{p}+\frac{1}{q}\)
Solution:
Given p + 2q = 18 …………… (1)
pq = 40 ………… (2)

Question 8.
Find ‘x’ \(5 \frac{x}{5} \times 3 \frac{3}{4}\) = 21.
Solution:

Question 9.
The difference between a number and its two third is 30 more than one -fifth of the number. Find the numbers.
Solution:
Let the number to be find out = x
Its two third = \(\frac{2 x}{3}\)
Given x – \(\frac{2}{3}\) x = \(\frac{1}{5}\) x + 30

Question 10.
By how much does \(\frac{1}{\frac{10}{11}}\) exceed \(\frac{1}{\frac{10}{11}}\)?
Solution:

Students can Download Maths Chapter 4 Geometry Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 4 Geometry Intext Questions

Answer the following questions:

Question 1.
The sum of the three angles of a triangle is ______
Solution:
1800

Question 2.
The exterior angle of a triangle is equal to the sum of the _______ angles opposite to it.
Solution:
interior

Question 3.
In a triangle, the sum of any two sides is ____ than the third side.
Solution:
greater

Question 4.
The difference between any two sides of a triangle is _______ than the third side.
Solution:
Smaller

Question 5.
Angles opposite to equal sides are ______ and vice-versa.
Solution:
Equal

Question 6.
The angles of a triangle are in the ratio 4 : 5 : 6
(i) Is it an acute, right or obtuse triangle?
(ii) Is it scalene, isosceles or equilateral?
Solution:
(i) Given the angles of a triangle are in the ratio 4 : 5 : 6 Sum of three angles of a
triangle = 180°.
Let the three angles 4x, 5x and 6x
4x + 5x + 6x = 180°
15x = 180° [∵ Vertically opposite angles are equal]

∴ x = 12°
∴ The angles are 4x ⇒ 4 × 12 = 48°
5x ⇒ 5 × 12 = 60°
6x ⇒ 6 × 12 = 72°
∴ The angle of the triangle are 48°, 60°, 72°
∴ It is an acute angles triangle.

(ii) We know that the sides opposite to equal angles are equal.
Here all the three angles are different.
∴ The sides also different.
∴ The triangle is a scalene triangle.

Question 7.
What is ∠A in the triangle ABC?

Solution:
The exterior angle = sum of interior opposite angles.
∴ ∠A + ∠C = 150° in ∆ABC
But ∠C = 40° [∵ Vertically opposite angles are equal]

Question 8.
Can a triangle have two supplementary angles? Why?
Solution:
Sum of three angles of a triangle is 180°.
∴ Sum of any two angles in a triangle will be less than 180°.
∴ A triangle cannot have two supplimentary angles.

Question 9.
________ shapes have the same shapes but different sizes.
Solution:
Similar

Question 10.
shapes are exactly the same in shape and size.
Solution:
Congruent

Exercise 4.1

Try these Page No. 99

Identify the pairs of shapes which are similar and congruent.

Similar shapes:
(i) W and L
(ii) B and J
(iii) A and G
(iv) B and J
(v) B and Y
Congruent shapes:
(i) Z and I
(ii) J and Y
(iii) C and P You can find more.
(iv) B and K
(v) R and S
(vi) I and Z

Try these Page No. 108

Question 1.
Match the following by their congruence

Solution:
1 – (iv)
2 – (iii)
3 – (i)
4 – (ii)

Try this Page No. 108

Question 1.
In the figure, DA = DC and BA = BC. Are the triangles DBA and DBC congruent? Why?
Solution:

Here AD = CD
AB = CB
DB = DB (common)
∆DBA ≅ ∆DBC [∵ By SSS Congruency]
Also RHS rule also bind here to say their congruency.

Exercise 4.3

Try this Page No. 114

Question 1.
Is it possible to construct a quadrilateral PQRS with PQ = 5 cm, QR = 3 cm, RS = 6 cm, PS = 7 cm and PR = 10 cm. If not, why?
Solution:
The lower triangle cannot be constructed as the sum of two sides 5 + 3 = 8 < 10 cm. So this quadrilateral cannot be constructed.

Students can Download Maths Chapter 1 Rational Numbers Ex 1.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 1 Rational Numbers Ex 1.2

Question 1.
Fill in the blanks:
(i) The multiplicative inverse of \(2 \frac{3}{5}\) is _____.
(ii) If -3 × \(\frac{6}{-11}=\frac{6}{-11}\) × x, then x is _______.
(iii) If distributive property is true for \(\left(\frac{3}{5} \times \frac{-4}{9}\right)+\left(x \times \frac{15}{17}\right)=\frac{3}{5} \times(y+z)\), then x, y, z are _____, _____ and ____.
(iv) If x × \(\frac{-55}{63}=\frac{-55}{63}\) × x = 1, then x is called the _____ of \(\frac{55}{63}\).
(v) The multiplicative inverse of -1 is ______.
Solution:
(i) \(\frac{5}{13}\)
(ii) -3
(iii) \(\frac{3}{5}, \frac{-4}{9}\) and \(\frac{15}{13}\)
(iv) Mulitplicative inverse
(v) -1

Question 2.
Say True or False.
(i) \(\frac{-7}{8} \times \frac{-23}{27}=\frac{-23}{27} \times \frac{-7}{8}\) illustrates the closure property of rational number.
(ii) Associative property is not true for subtraction of rational numbers.
(iii) The additive inverse of \(\frac{-11}{-17}\) is \(\frac{11}{17}\).
(iv) The product of two negative rational numbers is a positive rational number.
(v) The multiplicative inverse exists for all rational numbers.
Solution:
(i) False
(ii) True
(iii) False
(iv) True
(v) False

Question 3.
Verify the closure property for addition and multiplication of the rational numbers \(\frac{-5}{7}\) and \(\frac{8}{9}\)
Solution:
Closure property for addition.
Let a = \(\frac{-5}{7}\) and b = \(\frac{8}{9}\) be the given rational numbers.

∴ Closure property is true for addition of rational numbers.
Closure property for multiplication

∴ Closure property is true for multiplication of rational numbers.

Question 4.
Verify the associative property for addition and multiplication of the rational numbers \(\frac{-10}{11}, \frac{5}{6}, \frac{-4}{3}\).
Solution:


a × (b × c) = \(\frac{100}{99}\)
From (1) and (2) a × (b × c) = (a × b) × c is true for rational numbers.
Thus associative property is true for addition and multiplication of rational numbers.

Question 5.
Check the commutative property for addition and multiplication of the rational numbers \(\frac{-10}{11}\) and \(\frac{-8}{33}\).
Solution:
Let a = \(\frac{-10}{11}\) and b = \(\frac{-8}{33}\) be the given rational numbers.

From (1) and (2)
a + b = b + a and hence addition is commutative for rational numbers.

From (3) and (4) a × b = b × a
Hence multiplication is commutative for rational numbers.

Question 6.
Verify the distributive property a × (b + c) = (a × b) + (a × c) for the rational numbers a = \(\frac{-1}{2}\) ,b = \(\frac{2}{3}\) and c = \(\frac{-5}{6}\).
Solution:
Given the rational number a = \(\frac{-1}{2}\) ,b = \(\frac{2}{3}\) and c = \(\frac{-5}{6}\).

From (1) and (2) we have a × (b + c) = (a × b) + (a × c) is true.
Hence multiplication is distributive over addition for rational numbers Q.

Question 7.
Evaluate:

Solution:

Question 8.
Evaluate using appropriate properties.

Solution:

Question 9.
Use commutative and distributive properties to simplify \(\frac{4}{5} \times \frac{-3}{8}-\frac{3}{8} \times \frac{1}{4}+\frac{19}{20}\)
Solution:
Since multiplication is commutative

Objective Type Questions

Question 10.
Mulitplicative inverse of 0 (is)
(A) 0
(B) 1
(C) -1
(D) does not exist
Solution:
(D) does not exist

Question 11.
Which of the following illustrates the inverse property for addition?

Solution:
(A) \(\frac{1}{8}-\frac{1}{8}\) = 0

Question 12.
Closure property is not true for division of rational numbers because of the number
(A) 1
(B) -1
(C) 0
(D) \(\frac{1}{2}\)
Solution:
(C) 0

Question 13.
\(\frac{1}{2}-\left(\frac{3}{4}-\frac{5}{6}\right) \neq\left(\frac{1}{2}-\frac{3}{4}\right)-\frac{5}{6}\) illustrates that subtraction does not satisfy the ____ law of rational numbers.
(A) commutative
(B) closure
(C) distributive
(D) associative
Solution:
(D) associative

Question 14.
\(\left(1-\frac{1}{2}\right) \times\left(\frac{1}{2}-\frac{1}{4}\right) \div\left(\frac{3}{4}-\frac{1}{2}\right)\) = ______________

Solution:
(A) \(\frac{1}{2}\)

Students can Download Maths Chapter 5 Information Processing Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 5 Information Processing Additional Questions

Additional Questions And Answers

Question 1.
A fast food restaurant has a meal special ?50 for a drink, sandwich, side item and dessert. The choices are Sandwich : Grilled chicken, All beef patty, Vegeburger and Fill filet.
Side : Regular fries, cheese fries, potato fries
Dessert: Chocolate chip cookie or Apple pie.
Drink: Fanta, Dr. Pepper, Coke, Diet coke and sprite.
How may meal combos are possible?
Solution:
There are 4 stages
1. Choosing a Sandwich
2. Choosing a side
3. Choosing a dessert
4. Choosing a drink
There are 4 different types of sandwich, 3 different types of side two different type of desserts and five different types of drink.
∴ The number of meal combos possible is = 4 × 3 × 2 × 5 = 120

Question 2.
A company puts a code on each different product they sell. The code is made up of 3 numbers and 2 letters. How many different codes are possible?
Solution:
There are 5 stages, Number – 1
Number – 2
Number – 3
Letter – 1
Letter – 2
There are 10 possible numbers 0 to 9
There are 26 possible letters A to Z.
We have 10 × 10 × 10 × 26 × 26 = 6,76, 000 possible codes.

Question 3.
Rani take a survey with five ‘yes’ or ‘No’ answers. How many different ways could she complete the survey?
Solution:
There are 5 stages
Question – 1
Question – 2
Question – 3
Question – 4
Question – 5
There are 2 choices for each question (Yes/No)
∴ Total number of possible ways to answer
= 2 × 2 × 2 × 2 × 2 = 32 ways.

Question 4.
There are 2 vegetarian entry options and 5 meat entry options on a dinner menu. What number of ways one can opt a dinner for any one of it?
Solution:
Number of veg options = 2
Number of meat option = 5
One can opt for any one dinner
∴ Total number of ways = 2 + 5 = 7 ways

Additional Questions And Answers

Question 1.
Colour the graph with minimum number of colours and no two adjacent vertices should have the same colour.
Solution:

Test Yourself

Question 1.
You have three dice, How many possible out comes are there on a toss?
Solution:
8

Question 2.
Your school offers tow English classes three maths classes and 3 history classes, you want to take one of each class. How many different ways are there to organize your schedule?
Solution:
18

Question 3.
A wedding caterer gives 3 choices for main dish, sin starters, five dessert. How many different meals (made up of starter, dinner and dessert and are there?
Solution:
90

Question 4.
In a company ID cards have 5 digit numbers.
(a) How many ID cards can he formed if repetition of the digits allowed?
(b) How many ID cards can be formed if repetition of digits is not allowed?
Solution:
(i) 10,000
(ii) 30,240

Question 5.
A student is shopping for a new computer. He is deciding among 3 desktop and 4 laptop computer. How many ways she can buy a computer?
Solution:
7

Question 6.
Colour the vertices bear the same colour using minimum number of colours.