KSEEB Solutions for Class 8 Maths Chapter 9 Commercial Arithmetic Additional Questions

Students can Download Maths Chapter 9 Commercial Arithmetic Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 8 Maths Chapter 9 Commercial Arithmetic Additional Questions

Question 1.
Four alternative options are given for each of the following statements. Select the correct option.
(a) Nine per cent of Rs. 700 is
A. Rs. 63
B. Rs. 630
C. Rs. 6.3
D. Rs. 0.63
Solution:
A. Rs. 63

(b) What per cent of 50 meters is 12 meters?
A. 20%
B. 60%
C. 24%
D. 32%
Solution:
C. 24%

(c) The number of whose 8% is 12 is
A. 12
B. 150
C. 130
D. 140
Solution:
B. 150

(d) An article costing Rs. 600 is sold for Rs. 750. The gain percent is
A. 20
B. 25
C. 30
D. 35
Solution:
B. 25

KSEEB Solutions for Class 8 Maths Chapter 9 Commercial Arithmetic Additional Questions

(e) By selling a notebook for Rs. 22 a shopkeeper gains 10%. The cost price of the book is
A. Rs. 18
B. Rs. 30
C. Rs. 20
D. Rs. 22
Solution:
C. Rs. 20

(f) The percentage of loss, when an article worth Rs. 10,000 was sold for Rs. 9000 is
A. 10
B. 20
C. 15
D. 25
Solution:
A. 10

(g) A radio marked Rs. 1000 is given away for Rs. 850. The discount is
A. Rs. 50
B. Rs. 100
C. Rs. 150
D. Rs. 200
Solution:
C. Rs. 150

(h) A bookmarked Rs. 250 was sold for Rs. 200 after discount. The percentage of discount is
A. 10
B. 30
C. 20
D. 25
Solution:
C. 20

(i) The marked price of an article is 2. Rs. 200. If 15% of discount is allowed on it its selling price is
A. Rs. 185
B. Rs. 170
C. Rs. 215
D. Rs. 175
Solution:
B. Rs. 170

(j) One sells his bike through a broker by paying Rs. 200 brokerage. The rate of brokerage is 2%. The selling price of the bike is
A. Rs. 12,000
B. Rs. 10,000
C. Rs. 14,000
D. Rs. 12,500
Solution:
B. Rs. 10,000

(k) The brokerage amount for a deal is Rs. 25,000 at 2% rate of commission is
A. Rs. 500
B. Rs. 250
C. Rs. 5000
D. Rs. 2500
Solution:
A. Rs. 500

(l) If Rs. 1,600 is the commission at 8% for goods sold through a broker the selling price of the goods is
A. Rs. 18,000
B. Rs. 20,000
C. Rs. 22,000
D. Rs. 24,000
Sol:
B. Rs. 20,000

(m) The simple interest on Rs. 5000 at 2% per month for 3 months is
A. Rs. 100
B. Rs. 200
C. Rs. 300
D. Rs. 400
Solution:
C. Rs. 300

KSEEB Solutions for Class 8 Maths Chapter 9 Commercial Arithmetic Additional Questions

(n) The time in which simple interest on a certain sum be 0.15 times the principal at 10% per annum is
A. 1.5 years
B. 1 year
C. 2 years
D. 2.5 years
Solution:
A. 1.5 years

(o) The principal that yields a simple 4. interest of Rs. 1280 at 16% per annum for 8 months is
A. Rs. 10,000
B. Rs. 12,000
C. Rs. 12,800
D. Rs. 14,000
Solution:
B. Rs. 12,000

Question 2.
A time interval of 3 minutes and 20 seconds is wrongly measured as 3 minutes and 25 seconds what is the percentage error.
Solution:
3 min 20 seconds = 180 + 20 = 200 seconds.
Difference in measurement = 205 – 200 = 5 seconds
For 200 seconds the difference in 5 seconds
For 100 seconds the difference in
\frac{5}{200} × 100 = 2.5
∴ The percentage error is 2.5%

KSEEB Solutions for Class 8 Maths Chapter 9 Commercial Arithmetic Additional Questions

Question 3.
Hari reads 22% of the pages of a book on the first day, 53% on the second day and 15% on the third day. If the number of pages remaining to be read is 30 find the total number of pages in the book.
Solution:
The percentage of the pages read = 22 + 53 + 15 = 90%
The percentage of the pages to be read = 100 – 90 = 10%
Let the number of pages in the book be ‘x’
Given 10% of ‘x’ is 30
\frac{10}{100} × x = 30
\frac{10}{100} = 30 x = 30 × 10 = 300
∴ The number of pages in the book is 300

Question 4.
If 55% of students in a school are girls and the number of boys is 270 find the number of girls in the school.
Solution:
Percentage of boys = (100 – 55) = 45%
No. of boys in the school is 270 Let the no. of students in the class be ‘x’
45% of x is 270
\frac{45}{100} × x = 270
x = \frac{270 \times 100}{45} = 600
No. of students = 600
∴ no. of girls = 600 – 270 = 330

KSEEB Solutions for Class 8 Maths Chapter 9 Commercial Arithmetic Additional Questions

Question 5.
By selling an article for Rs. 920 a shopkeeper gains 15%. Find the cost price of the article.
Solution:
S.P. = Rs. 920, gain % = 15%, C.P. = ?

KSEEB Solutions for Class 8 Maths Chapter 9 Commercial Arithmetic Additional Questions 1

Question 6.
Amit sells a watch at a 20% gain. Had he sold it for Rs. 36 more, he would have gained 23%. Find the cost price of the watch.
Solution:
Let the C.P. be Rs. x, gain %
= 20%

KSEEB Solutions for Class 8 Maths Chapter 9 Commercial Arithmetic Additional Questions 2

Had he sold it for Rs. 36 more the new
S.P. would be Rs. \frac{120 x}{100} + 36.
This is equal to the S.P. when the gain percent is 23%.
∴ S.P., when gain percent is 23%, is

KSEEB Solutions for Class 8 Maths Chapter 9 Commercial Arithmetic Additional Questions 3
∴ The C.P. of the watch is Rs. 1200

KSEEB Solutions for Class 8 Maths Chapter 9 Commercial Arithmetic Additional Questions

Question 7.
On selling apples at Rs. 40 per kg, a vendor incurs 10% loss. If he incurs a total loss of Rs. 120 calculate the quantity (in kg) of apples he sold.
Solution:
Loss = Rs 120, Loss % = 10%. S.P. = ?
Let the C.P. be 100 then S.P =, (100 – 10) = Rs. 90
If the loss Rs. 10 then S.P. is Rs. 90
If the loss Rs. 120 then S.P. = \frac{90 \times 120}{10}
= Rs. 1080
The total selling price
Let the quantity (in kg) of apples be ‘x’. The SP of 1 kg = Rs. 40. The S.P. of x kg = 40 x.
∴ 40x = 1080
x = \frac{1080}{40} = 27
He sold 27 kg of apples.

Question 8.
A dealer allows a discount of 20% and still gains 20%. Find the marked price of an article that costs the dealer Rs. 720.
Solution:
C.P. = Rs. 720, gain % = 20% discount % = 20%
Gain on Rs. 720 = \frac{20}{100} × 720 = Rs. 144
∴ S.P of the article = C.R + gain = 720 + 144 = Rs. 864
Discount percent = 20%
Let the M.R be 100 then the S.R is 100 – 20 = 80
If the S.P is Rs. 80 then M.P is Rs. 100
If the S.P is Rs. 864 the M.P is \frac{100 \times 864}{80} = Rs. 1080
The marked price of the article is Rs. 1080

KSEEB Solutions for Class 8 Maths Chapter 9 Commercial Arithmetic Additional Questions

Question 9.
A shopkeeper buys an article for Rs. . 600 and marks 25% above the cost price. Find (i) the selling price if he sells the article at 10% discount (ii) the percentage of discount if it is sold for Rs. 600.
Solution:
C.P = Rs. 600 and Discount rate is 25%
∴ S.P = C.P + Rate of discount × C.P.
= 600 + 25% × 600
= 600 + \frac{25}{100} × 600
= 600 + 150 = Rs. 750
Marked price = Rs. 750

(i) Rate of discount = 10%, M.P. = Rs. 750
S.P = M.P. – discount
= 750 – \frac{10}{100} × 750
= 750- 75
= Rs. 675

(ii) S.P = Rs. 690, M.P = 750
Discount = M.P – S.P.
= 750 – 690 = Rs. 60
If the M.P. is Rs. 750 the discount is Rs. 60. If the M.P. is 100 the discount is
\frac{60 \times 100}{750} = 8%

Question 10.
A retailer purchases goods worth Rs. 33,600 and gets a discount of 14% from a wholesaler. For paying cash, the wholesaler gives an additional discount of 1.5% on the amount to be paid after the first discount. What is the net amount the retailer has to pay?
Solution:
M.P. = Rs. 33,600, Discount rate = 14%
Discount = \frac{14}{100} × 33,600 = 4704
S.P = M.P – discount
= 33600 – 4704
S.P. = 28,896
S.P. after 14% discount is Rs. 28,896. He gets an additional discount of 1.5% on Rs. 28,896
Discount = \frac{1.5}{100} × 28896
= 1.5 × 288.96
= 433.44
∴ S.P = 28896 – 433.44 = 28,462.56
The retailer has to pay Rs. 28,462.56

KSEEB Solutions for Class 8 Maths Chapter 9 Commercial Arithmetic Additional Questions

Question 11.
An old car was disposed of through a broker for Rs. 42,000. If the brokerage is 2\frac{1}{2} % find the amount the owner gets.
Solution:
S.P = Rs. 42,000, commission rate
= 2\frac{1}{2} %
Commission = \frac{2.5}{100} × 42000 – 1050
= Rs. 1050
The owner gets = 42000 – 1050 = Rs. 40,950

Question 12.
A milk-man sells 20 liters of milk every day at Rs. 22. He receives a commission of 4% every litre. Find the total commission he receives in a month of 30 days.
Solution:
The quantity of milk sold in a month of 30 days = 30 × 20 = 600
The cost of 1 liter = Rs. 22
The cost of 600 litres = 22 x 600 = Rs. 13,200
Commission rate = 4%
Commission = \frac{4}{100} × 13200 = 528
The commission received by him in Rs. 528.

Question 13.
A bike was sold for Rs. 48,000 and a commission of Rs. 8640 was received by the dealer. Find the rate of commission.
Solution:
S.P = 48,00, commission = Rs. 8640
If the S.P. is 48,000 the commission is Rs. 8640
If the SP is 100 the commission is \frac{8640 \times 100}{48000} = 1.8%
Rate of commission = 1.8%

Question 14.
In how many years will a sum of money becomes three times at the rate of interest 10% per annum?
Solution:
Let the principal be x. It amounts to 3x.
Simple interest = Amount – principal = 3x – x
Simple interest = 2x

KSEEB Solutions for Class 8 Maths Chapter 9 Commercial Arithmetic Additional Questions 4

T = 20 years It takes 20 years.

KSEEB Solutions for Class 8 Maths Chapter 9 Commercial Arithmetic Additional Questions

Question 15.
In what time will the simple interest on a certain sum be 0.24 times the principal at 12% per annum?
Solution:
Let the principal be ‘x’. If the simple interest is 0.24 times the principal then. simple interest = 0.24 x.

KSEEB Solutions for Class 8 Maths Chapter 9 Commercial Arithmetic Additional Questions 5
100 × 0.24x = 12Tx
24x = 12Tx
∴ T = \frac{24 x}{12 x}
T = 2 years
In 2 years the simple interest will be 0.24 time the principal.

Question 16.
Find the amount of Rs. 30,000 from 15th January 2010 to 10th August 2010 at 12% per annum.
Solution:
P = Rs. 30,000, R = 12%,
T = 15th Jan 2010 to 10th Aug 2010
= 17 + 28 + 31 + 30 + 31 + 30 + 31 + 10 = 208 days

KSEEB Solutions for Class 8 Maths Chapter 9 Commercial Arithmetic Additional Questions 6
Amount = Principal + interest = 30000 + 2051.50
= Rs. 32051.50

KSEEB Solutions for Class 8 Maths Chapter 9 Commercial Arithmetic Additional Questions

Question 17.
A person purchases electronic items worth Rs. 2,50,000. The shop keeper charges him a sales tax of 21% instead of 12%. The purchaser does not realize that he has overpaid. But after some time he finds that he has paid excess and asks the shopkeeper to return the excess ‘ money. The shopkeeper refuses and the purchaser moves the consumer court. The court with due learning orders the shop-keeper to pay the purchaser the excess money paid by the way of sales tax with an interest of 12% per annum. If the whole deliberation takes 8 months what is the money that the purchaser gets back?
Solution:
S.P = 2,50,000
Tax percent = 21%
Tax at the rate of 21%
= \frac{21}{100} × 2,50,000 = 52,500
Rs. 52500
Tax at the rate of 12%
= \frac{12}{100} × 2,50,000 = 30,000 100
Rs. 30,000
Excess paid = 52,500 – 30,000
= Rs. 22,500
Now P = 22,500, R = 12%,
T = 8 months = \frac{8}{12} years

KSEEB Solutions for Class 8 Maths Chapter 9 Commercial Arithmetic Additional Questions 7

Amount = Principal + Interest = 22,500 + 1800 = 24,300
The purchaser gets back Rs. 24,300.

KSEEB Solutions for Class 8 Maths Chapter 9 Commercial Arithmetic Additional Questions

KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions

Students can Download Maths Chapter 8 Linear Equations in One Variable Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions

Question 1.
Choose the correct answer
a. The value of ‘x’ in the equation 5x – 35 = 0 is
A. 2
B. 7
C. 8
D. 11
Solution:
(B) 7

b. If 14 is taken away from one fifth of a number, the result is 20. The equation expressing this statement is
A. \left(\frac{x}{5}\right)-14=20
B. x-\left(\frac{14}{5}\right)=\frac{20}{5}
C. x-14=\left(\frac{20}{5}\right)
D. x+\left(\frac{14}{5}\right)=20
Solution:
A. \left(\frac{x}{5}\right)-14=20

c. If five times a number is increased by 8 is 53, the number is
A. 12
B. 9
C. 11
D. 2
Solution:
B. 9

d. The value of x in the equation 5 (x – 2) = 3 (x – 3) is
A. 2
B. \frac{1}{2}
C. \frac{3}{4}
D. 0
Solution:
B. \frac{1}{2}

e. If the sum of two numbers is 84 and their difference is 30, the numbers are
A. – 57 and 27
B. 57 and 27
C. 57 and – 27
D. – 57 and – 27
Solution:
(B) 57 and 27

f. If the area of a rectangle whose length is twice its breadth is 800 m2, then the length and breadth of the rect-angle are
A. 60 m and 20 m
B. 40 m and 20m
C. 80 m and 10 m
D. 100 m and 8m
Solution:
(B) 40 m and 20 m

g. If the sum of three consecutive odd numbers is 249, the numbers are
A. 81, 83, 85
B. 79, 81, 83
C. 103, 105, 10
D. 95, 97, 99
Solution:
(A) 81, 83, 85

h. If \frac{(x+0.7 x)}{x} = 0.85 the value of x is
A. 2
B. 1
C. – 1
D. 0
Solution:
B. 1

i. If 2x – (3x – 4) = 3x – 5 then x equals
A. \frac{4}{9}
B. \frac{9}{4}
C. \frac{3}{2}
D. \frac{2}{3}
Solution:
B. \frac{9}{4}

KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions

Question 2.
Solve
(i) \frac{(3 x+24)}{(2 x+7)}=2
Solution
\frac{(3 x+24)}{(2 x+7)}=2
3x + 24 = 2( 2x + 7)
3x + 24 = 4x + 14
24 – 14 = 4x – 3x
10 = x
or x = 10

(ii) \frac{(1-9 y)}{(11-3 y)}=\frac{5}{8}
Solution
\frac{(1-9 y)}{(11-3 y)}=\frac{5}{8}
8 (1 – 9y) = 5 ( 11 – 3y)
8 – = 55 – 15y
8 – 55 = 72 y – 15 y
– 47 = 57 y
\frac{-47}{57} = y
or y = \frac{-47}{57}

Question 3.
The sum of two numbers is 45 and their ratio is 7 : 8. Find the numbers.
Solution:
Ratio of the number = 7 : 8
Let the numbers be 7x and 8x
sum of the numbers = 45
7x + 8x = 45
15x = 45
x = \frac{45}{15}
7x = 7 x 3 = 21, 8x = 8 x 3 = 24
∴ The numbers are 21 and 24

KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions

Question 4.
Shona’s mother is four times as old as shona. After 5 years,her mother will be three times as old as shona (at that time). What are their present age?
Solution:
Let shona’s present age be ‘x’ years.
Her mother’s present age = 4x years
5 years later shona’s age = x + 5 and
shona’s mother’s age = 4x + 5
Given that 4x + 5 = 3 (x + 5)
4x + 5 = 3x + 15
4x – 3x = 15 – 5
x = 10
Shonas present age = x = 10 years
Her mothers present age = 4x = 4 x 10 = 40 years.

Question 5.
The sum of 3 consecutive even numbers is 336. Find them.
Solution:
‘ Let the first even number be ‘x’
The consecutive even numbers are x + 2 and x + 4
Their sum = 336
x + x + 2 + x + 4 = 335
3x + 6 = 336
3x = 336 – 6
3x = 330
x = \frac{330}{3} = 110
x = 110, x + 2 = 110 + 2
= 112, x + 4 = 110 + 4 = 114.
∴ The numbers are 11, 112, 114.

KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions

Question 6.
Two friend’s A and B start a joint business with a capital Rs. 60,000. If As share is twice that of B how much have each invested?
Solution:
Let ‘B’ Invest Rs. x. Then ‘A’s investment is 2x.
Total Investment = 60,000
x + 2x = 60,000
3x = 60,000
x = \frac{60000}{3}
x = 20,000
‘B’s investment = x = Rs. 20,000
As investment = 2x = 2 x 20,000 = Rs. 40,000

Question 7.
Which is the number when 40 is subtracted gives one third of the original number.
Solution:
Let the number be ‘x’. One third the
\frac{1}{3} x number is x = \frac{x}{3}
Given that x – 40 = \frac{x}{3}
3 (x – 40) = x
3x – 120 = x
3x – x = 120
2x = 120
x = \frac{120}{2}
x = 60
∴ The original number is 60

Question 8.
Find the number whose 6th part exceeds its eighth part by 3.
Solution:
Let the number be ‘x’. Its 6th part is \frac{x}{6} and 8th part is \frac{x}{8}
KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions 1
KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions 2
x = 72, The number is 72

KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions

Question 9.
A house and a garden together cost Rs. 8,40,000. The price of the garden is \frac{5}{12} times the total cost. Find the price of the house and the garden.
Solution:
The price of the garden is
\frac{5}{12} × 8,40,000 . 12
= 5 × 70,000
= Rs. 3,50,000
The cost of the house is 8,40,000 – 3,50,000 = 4,90,000
The cost of the house is Rs. 4,90,000

Question 10.
Two farmers A and B together own a stock of grocery. They agree to divide it by value. Farmer A takes 72 bags while B takes 92 bags and gives Rs. 8,000 to A. What is the cost o each bag?
Solution:
Let the cost of each bag be ‘x’.
A takes 72 bags and Receives Rs. 8000
∴ The value of the stock = 72x + 8000
B takes 92 bags and gives Rs. 8000
The value of the stock = 92x – 8000
∴ 72x + 8000 = 92x – 8000
8000 + 8000 = 92x – 7x
16000 = 20x
∴ x = \frac{16000}{20}
x = 800
∴ The cost of each bag is Rs. 800

KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions

Question 11.
A fathers age is four times that of his son. After 5 year s, it will be three times that of his son. How many more years will take if fathers age is to be twice that of his son?
Solution:
Let the sons age be x. Fathers age is 4x .
After 5 years sons age = x + 5 years and fathers age = 4x + 5
Given (4x + 5) = 3 (x + 5)
4x + 5 = 3x + 15 4x – 3x = 15 – 5 x = 10
Sons present age = 10 years
Fathers present age = 4 × x = 4 × 10 = 40 years.
Let in ‘y’ years fathers age will be twice as his sons age.
After ‘y’ years son will be (10 + y) years and father will be (40 + y) years.
40 + y = 2 (10 + y)
40 + y = 20 + 2y
40 – 20 = 2y – y
20 = y
y = 20 ∴ 20 – 5 = 15 years
In 15 years fathers age wi11 be twice his sons age.

Question 12.
Find a number which when multiplied by 7 is as much above 132 as it was origina11y below it.
Solution:
Let the number be x. When its multi, plied by 7 the product is 7x. This number is above 132 by (7x – 12). This is same as it was origina11y below it i.e., (132 – x)
∴ 7x – 132 = 132 – x
7x + x = 132 + 132
8x = 264
x = \frac{264}{8} ; x = 33
The number is 33

KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions

Question 13.
A person buys 25 pens worth Rs. 250, each of equal cost. He wants to keep 5 pens for himself and se11 the remaining to recover his money what should be the price of each pen?
Solution:
Number pens bought = 25
The cost of 25 pens = Rs. 250
∴ The cost of each pen
\frac{250}{25} = Rs. 10
He keeps 5 pens with him and se11s the remaining (25 – 5) 20 pens to recover the same amount. Let the S.R of each pen be x.
20x = 250
x = \frac{250}{20} = Rs. 12.5
The price of each pen should be Rs. 12.50

Question 14.
The sum of the digits of a two digit number is 12. If the new number formed by reversing the digits is greater than the original number by 18. Find the original number. Check your solution.
Solution:
Let the digit in the units place be ‘x’ and digit in the ten’s place be ‘y
Sum of the digits = 12
x + y = 12
y = 12 – x
∴ The original number is 10 y + x
The number got by reversing the digits is 10 x + y
Given that 10 x + y = 10 y + x + 18
10x + (12 – x ) = 10 (12 – x) + x 18
10x + 12 – x = 120 – 10 x + x + 18
9x + 12 = 138 – 9x
9x + 9x = 138 – 12
18x = 126
x = \frac{126}{18} = 7
y = 12 – x
= 12 – y = 5
∴ The original number is (10 x 5) +7 = 50 + 7 = 57

Verification :
Original number = 57
The new number = 75
75 – 57 = 18

KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions

Question 15.
The distance between two stations is 340 km. Two trains start simaltaneously from these stations on para11el tracks and cross each other. The speed of one of them is greater than that of the other by 5km / hr. If the distance between two trains after 2 hrs of their start is 30 km find the speed of each train.
Solution:
Let the speed of one train be x km /hr. The speed of the other train is (x + 5) km / hr
Distance covered by the trains in 2 hrs are 2x and 2 (x + 5) respectively. After crossing each other the distance between the trains is 30 km. Therefore the total distance covered = 340 + 30 = 370 km
2x + 2 (x + 5) = 370
2x + 2x + 10 = 370
4x + 10 = 370
4x = 370 – 10
4x = 360
x = \frac{360}{4} = 90
x = 90 km / hr
The speeds of the trains are 90 km/hr and 90 + 5 = 95 km/ hr.

Question 16.
A steamer goes down stream and coveres the distance between two ports in 4 hrs while it coveres the same distance upstream in 5 hrs. If the speed of the steamer up stream is 2km / hr find the speed of the steamer in sti11 water.
Solution
Speed upstream = 2km / hr
Time taken up stream = 5 hr
∴ Distance between two ports
= 2 x 5 = 10 km

KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions 3

Time taken down stream = 4 hrs Distance between two ports = 10 km
speed down stream

KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions 4

Let the speed of the steamer in sti11 water be ‘x’ km / hr and speed of the stream be y km / hr.

KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions 5
x = 2.25 km / hr
∴ The spped of the steamer in sti11 water is 2.25 km / hr.

KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions

Question 17.
The numerator of rational number is less than its denaminator by 3. If the numerator becomes 3 times and denominator is increased by 20 the new number becomes \frac{1}{8}. Find the original number.
Solution:
Let the denominator be ‘x. The numera-tor is x – 3. The original fraction is \frac{x-3}{x}
If the numerator becomes 3 times it is 3 (x – 3) and denominator is increased by 20 it is x + 20 new number is – \frac{1}{8}

KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions 6
8(3x – 9) = x + 20
24x – 72 = x + 20
24x – x = 20 + 72
23x = 92
x = \frac{92}{23}
x = 4
∴ original fraction = \frac{x-3}{x}=\frac{4-3}{4}=\frac{1}{4}

Question 18.
The digit at the tens place of a two digit number is three times the digit at the units place. If the sum of this number and the number formed by reversing the digits is 88 find the number.
Solution:
Let the digit in the units place be x and tens place be y.
Given y = 3x ∴ The original number is 10 y + x
The number formed by reversing the digits is 10 x + y.
Sum of the numbers = 88
10 y + x + 10 x + y = 88
11y + 11x = 88
but y = 3x
∴ 11 (3x) + 11x = 88
33x + 11x = 88
44x = 88
x = \frac{88}{44}
x = 2
y = 3x = 3 x 2 = 6
The original number is 10 y + x = 10 x 6 + 2 = 62

KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions

Question 19.
The altitude of a triangle is five thirds the length of its corresponding base. If the altitude is increased by 4cm and the base decreased by 2 cm the area of the triangle would remain the same. Find the base and altitude of the triangle.
Solution:
Let the base of the triangle be ‘x’ cm
The altitude is \frac{5x}{3} cm
Area of the traingle = \frac{1}{2} x base x height

KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions 8

If the height is increased by 4cm it is \left(\frac{5 x^{2}}{3}+4\right) and the base is decreased by 2 cm it is (x – 2) the area remains the same.
KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions 7
x(5x + 12)- 2(5x +12) = 6 x \frac{5 x^{2}}{6}
5x2 + 12x – 10x – 24 = 5x2
5x2 + 2x – 24 – 5x2 = 0
2x = 24
x = \frac{24}{2} = 12
x = 12
\frac{5 x}{3}=\frac{5 \times 12}{3}=\frac{60}{3}=20 \mathrm{cm}
The base of the triangle is 12 cm and altitude is 20 cm

Question 20.
One of the angles of a triangle is equal to the sum of the other two angles. If the ratio of the other two angles of the triangle is 4 : 5 find the angles of the triangles.
Solution:
The ratio of the two angles is 4 : 5 Let the angles be 4x and 5x. Sum of the two angles = 4x + 5x = 9x. The third angle is equal to sum of the other , two angles i.e., 9x.
Sum of the angles of a traingle
= 180°
9x + 9x = 180
18x = 180
∴ x = \frac{180}{8}
x = 10°
4x = 4 x 10 = 40°
5x = 5 x 10 = 50°
9x = 9 x 10 = 90° .
∴ The angles of the triangle are 90°, 40° and 50°.

KSEEB Solutions for Class 8 Maths Chapter 8 Linear Equations in One Variable Additional Questions

KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions

Students can Download Maths Chapter 7 Rational Numbers Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 8 Maths Chapter 7 Rational Numbers Additional Questions

I. Fill in the blanks:

Question 1.
(a) The number 0 is not in the set of Natural numbers
(b) The least number in the set of all whole numbers is O
(c) The least number in the set of al even natural numbers is 2
(d) The successor of 8 in the set of all-natural numbers is 9
(e) The sum of two odd integers is even
(f) The product of two odd integers is odd

KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions

Question 2.
State whether the following statements are true or false.
(a) The set of all even natural numbers is a finite set.
Answer:
False.

(b) Every non-empty subset of Z has the smallest element.
Answer:
False

(c) Every integer can be identified with a rational number
Answer:
True

(d) For each rational number, one can find the ne×t rational number
Answer:
False

(e) There is the largest rational number
Answer:
False

(f) Every integer is either even or odd
Answer:
True

(g) Between any two rational numbers, there is an integer.
Answer:
False

KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions

Question 3.
Simplify:
(i) 100(100 – 3) – (100 × 100 – 3)
Solution:
100(97) – (10000 – 3)
= 9700 – (9997) = – 297

(ii) (20 – (2011 – 201) + 2011 – (201 – 20))
= 20 – 2011 + 201 + 2011 – 201 + 20 = 40

Question 4.
Suppose m is an integer such that m * -1 and m ≠ – 2. Which is larger \frac{m}{m+1} \text { or } \frac{m+1}{m+2} ? State your reasons.
Solution:
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 1

KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions

Question 5.
Define as operation * on the set of all rational numbers Q as follows r * s = r + s – (r × s) for any two rational numbers r, s. Answer the following with justification.
(i) Is Q closed under the operatiron ?
(ii) Is * an associative operation on Q?
(iii) Is * commutative operation on Q?
(iv) What is a * 1 for any a in Q?
(v) Find two integers a ≠ 0, b ≠ 0 such that a * b = 0
Solution:
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 2
Q is closed under the operation *. because ∀r, s ∈ Q r * s = r + s – (r × s) is also a rational number.

(ii) Let r, s and p ∈ Q
r * (s * p) = r * (s + p – (s × p))
= r + s + p – (s × p) – (r × (s + p – (s × p)
= r + s + p – sp – (rs + rp – rsp)
= r + s + p – sp – rs – rp + rsp …(i)
(r * s) * p = r + s – (r × s) * p
= r + s – (r × s) + p – (r + s – (r × s) × p)
= r + s – rs + p – (rp – sp – rsp)
= r + s + p – rs – rp – sp – rsp …(ii)
From (i) and (ii) r * (s * p) = (r * s) * p
∴ * is an associative operation on Q.

(iii) Let r, s, ∈ Q
r * s = r + s – (r × s) = r + s – rs …(i)
s * r = s + r – (s × r) = s + r – sr = r + s – rs …(ii)
From (i) and (ii) r * s = s * r
∴ * is commutative operation on Q.

(vi) r * s = r + s – (r × s)
∴ a * 1 = a + 1 – (a × 1)
= a + 1 – a
a * 1 = 1

(v) r * s = r + s – (r × s)
a * b = a + b – (a × b)
if a = 2, and b = 2
2 * 2 = 2 + 2 – (2 × 2) = 4 – 4
2 * 2 = 0
∴ If a = 2 and b = 2 a * b = 0

KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions

Question 6.
Find the multiplicative inverse of the following rational numbers.
\frac{8}{13}
Solution:
\frac{13}{8}

\frac{-13}{11}
Solution:
\frac{-11}{13}

\frac{12}{17}
Solution:
\frac{17}{12}

\frac{-101}{100}
Solution:
\frac{-100}{101}

\frac{26}{23}
Solution:
\frac{23}{26}

KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions

Question 7.
Write the following in the increasing order.
\frac{10}{13}, \frac{20}{23}, \frac{5}{6}, \frac{40}{43}, \frac{25}{28}, \frac{10}{11}
Solution:
Take the LCM of the denominators
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 3
LCM = 2 × 13 × 23 × 3 × 43 × 14 × 11
= 11879868
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 4

Question 8.
Write the following in decreasing order
\frac{21}{17}, \frac{31}{27}, \frac{13}{11}, \frac{41}{37}, \frac{51}{47}, \frac{9}{8}
Solution:
Take the LCM of the denominators
LCM = 17 × 27 × 11 × 37 × 47 × 8 = 70241688
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 5

Question 9.
(a) What is the additive inverse of O?
Solution:
O

(b) What is the multiplicative inverse of 1?
Solution:
1

(c) Which integers have a multiplicative inverse?
Solution:
1 and -1

KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions

Question 10.
In the set of all rational numbers, give 2 examples each illustrating the following properties.
(i) Associativity
(a)
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 6
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 7
Associative property of addition in satisfied.
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 8
Associative property of multiplication is satisfied.

KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 9
Associative property of addition is satisfied.

KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 10

From (i) and (ii)

KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 11

Associative property of multiplication is satisfied.
(ii) Commutativity
(a)
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 12
Comutative property of addition is satis¬fied.

KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 13
Commutative property of multiplication is satisfied

(b)
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 14
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 15
Commutative property of addition is satisfied.

KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 16

Commutative property of multiplication is satisfied.

(iii) Distributivity of multiplication over addition.
(a)
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 17

Multiplication is distributive over addition.

(b)
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 18

∴ Multiplication is distributive over addition.

KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions

Question 11.
Simplify the following using distributive property
(1) \frac{2}{5} \times\left(\frac{1}{9}+\frac{2}{5}\right)
Solution:
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 19

(2) \frac{5}{12} \times\left(\frac{25}{9}+\frac{32}{5}\right)
Solution:
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 20
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 21

(3) \frac{8}{9} \times\left(\frac{11}{2}+\frac{2}{9}\right)
Solution:
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 22

Question 12.
Simplify the following:
(i) \left(\frac{25}{9}+\frac{12}{3}\right)+\frac{3}{5}
Solution:
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 23

(ii) \left(\frac{22}{7}+\frac{36}{5}\right) \times \frac{6}{7}
Solution:
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 24

(iii) \left(\frac{51}{2}+\frac{7}{6}\right) \div \frac{3}{5}
Solution:
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 25

(iv) \left(\frac{16}{7}+\frac{21}{8}\right) \times\left(\frac{15}{3}-\frac{2}{9}\right)
Solution:
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 26

Question 13.
Which is the property that is there in the set of all rational numbers but which is not in the set of all integers?
Solution:
Every non zero rational number is invertible but only + 1 are invertible integers.

Question 14.
What is the value of 1+\frac{1}{1+\frac{1}{1+1}}
Solution:
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 27

Question 15.
Find the value of \left(\frac{1}{3}-\frac{1}{4}\right) /\left(\frac{1}{2}-\frac{1}{3}\right)
Solution:
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 28

KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions

Question 16.
Find all rational numbers each of which is equl to its reciprocal.
Solution:
The rational numbers + 1 and – 1 and their reciprocals are equal.

Question 17.
A bus shuttles between two neighbouring towns every two hours. It starts from 8 AM in the morning and the last trip in at 6 PM. On one day the driver observed that the first trip had 30 passengers and each subsequent trip had one passenger less than the previous trip. How many passengers travelled on that day?
Solution:
The number of trips = 6 The number of passengers in the first rip is 30, second trip in 29, third trip is 28 and so on
∴ No of passenger travelled on that day = 30 + 29 + 28 + 27 + 26 + 25 = 165

Question 18.
How many rational numbers p/q are there between 0 and 1 for which q < P?
Solution:
There are no rational numbers between 0 and 1 such that q < p.

Question 19.
Find all integers such that \frac{3 n+4}{n+2} is also an integer.
Solution:
Let n = 0
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 29
It is an integer if n = 0, – 1, – 2, – 3, – 4.

Question 20.
By inserting parenthesis (that is brackets), you can get several values for 2 × 3 + 4 × 5 (For example ((2 × 3) + 4) × 5 is one way of inserting parenthesis) How many such values are there?
Solution:
(i) (2 × 3) + (4 × 5) = 6 + 20 = 26
(ii) 2 × (3 + 4) × 5 = 2 × 7 × 5 = 70
(iii) 2 × (3 + (4 × 5)) – 2 × (3 + 20) = 2 × 23 = 46
(iv) ((2 × 3) + 4) × 5 = (6 + 4) × 5 = 10 × 5 = 50 There are 4 values

KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions

Question 21.
Suppose \frac{p}{q} is a positive rational in its lowest form prove that \frac{1}{q}+\frac{1}{p+q} is also in its lowest form.
Solution:
\frac{p}{q} is in its lowest form means p and q do not have any common factors other than 1.
Consider \frac{1}{q}+\frac{1}{p+q}=\frac{p+q+q}{q(p+q)}=\frac{p+2 q}{q(p+q)} p + 2q and q (p + q) do not have common factors other than 1 ∴ \frac{p+2 q}{q(p+q)} is in its lowest form, that means \frac{1}{q}+\frac{1}{p+q} is in its lowest form.

Question 22.
Show that for each natural number n the fraction \frac{14 n+3}{21 n+4} is in its lowest form
Solution:
If
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 30
is in the lowest form
If
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 31
is in the lowest form
If
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 32
is in the lowest form
If
KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions 33
is in the lowest form.

Similarly by substituting natural numbers
for n we can show that \frac{14 n+3}{21 n+4} is in its lowest form.

Question 23.
Find all integers n for which the number (n + 3) (n – 1) is also an integer.
Sol:
If n = 0, (n + 3) (n – 1) = (0 + 3) (0 – 1) = 3x-l = -3 is an integer If n = 1, (n + 3) (n – 1) = (1 + 3) (1 – 1) = 4×0 = 0 is an integer.
If n = – 1, (n + 3) (n – 1) = (- 1 + 3) (- 1 – 1) = 3 (- 2) = – 6 is an integer Similarly when n is replaced by any integer we find that (n + 3) (n – 1) is also an integer
∴ n = { ……….. – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4 ……….}

KSEEB Solutions for Class 8 Maths Chapter 7 Rational Numbers Additional Questions

KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions

Students can Download Maths Chapter 6 Theorems on Triangles Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions

Question 1.
Fill up the blanks to make the following statements true
a. Sum of the angles of a triangle in 180°
b. An exterior angle of a triangle is equal to the sum of interior opposite angles.
c. An exteior angle of a triangle is always more than either of the interior opposite angles.
d. A triangle can not have more than one right angles.
e. A triangle can not have more than one obtuse angle.

KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions

Question 2.
Choose the correct answer from the given alternatives.
a. In a triangle ABC, ∠A = 80° and AB = BAC then ∠B is __________
A. 50°
B. 60°
C. 40°
D. 70°
Solution:
A. 50°

b. In right-angled triangle, ∠A is right angle and ∠B = 35° then ∠C is _______
A. 65°
B. 55°
C. 75°
D. 45°
Solution:
B. 55°

c. In ∆ABC, ∠B = ∠C = 45°, then the triangle is ________
A. Right-angled
B. Acute angled
C. Obtuse angled
D. Equilateral triangle
Solution:
A. Right-angled

d. In an equilateral triangle, each exterior angle is ________
A. 60°
B. 90°
C. 120°
D. 150°
Solution:
C. 120°

e. Sum of the three exterior angles of a triangle is __________
A. two right angles
B. three right angles
C. one right angle
D. four right angles
Solution:
D. Four right angles

KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions

Question 3.
In a triangle ABC, ∠B = 70° find ∠A + ∠C
Solution:
∠A + ∠B + ∠C = 180° [Sum of the angles of a triangle is 180°]
∠A + 70 + ∠C = 180°
∠A + ∠C = 180° – 70
∠A + ∠C = 110°

Question 4.
In a triangle ABC, ∠A =110° and AB = AC find ∠B and ∠C
Solution:
AB = AC
∴ ∠C = ∠B [Base angles of an isosceles triangle]
∠A + ∠B + ∠C = 180° [Sum of the angles of a triangle is 180°]
110° + ∠B + ∠B = 180°
2∠B = 180 – 110
2∠B = 70°
∠B = \frac{70}{2}
∠B = 35°
∠B = ∠C = 35°

KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions

Question 5.
If three angles of a triangle are in the ratio 2 : 3 : 5 determine three angles.
Solution:
Let the common ratio be x The three angles are 2x, 3x and 5x 2x + 3x + 5x = 180° [Sum of the angles of triangle is 180°]
10x = 180°
x = \frac{180}{10}
2x = 2 ×18 = 36°
3x = 3 × 18 = 54°
5x = 5 × 18 = 90°

Question 6.
The angles of a triangle are arranged in ascending order of magnitude. If the difference between consecuttive angles is 15° find the three angles.
Solution:
Let the first angle be x then the second angle is x + 15 and third angle is x + 30.
x + x + 15 + x + 30 = 180° [Sum of the angles]
3x + 45 = 180
3x = 180 – 45
3x = 135
x = \frac{135}{3} = 45°
First angle = x = 45°
Second angle = x + 15 = 45 + 15 = 60°
Third angle = x + 30 = 45 + 30 = 75°

Question 7.
The sum of two angles of a triangle is equal to its third angle. Determine the measure of third angle.
Solution:
Let the sum of two angles be x and the third angle be y
x + y = 180° [Sum of the angles of the triangle]
y + y = 180° [∴ sum of two angles = third angle]
2y = 180°
y = \frac{180}{2}
y = 90°
∴ The third angle is 90°

KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions

Question 8.
In a triangle ABC if 2∠A = 3∠B = 6∠C determine ∠A , ∠B and ∠C
Solution:
KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions 1
∠A + ∠B + ∠C = 180° [Sum of the angles of a triangle]
KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions 2
KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions 3

Question 9.
The angles of a triangle are x – 40°, x – 20 and \frac{1}{2} x + 15° find the value of x
Solution:
x – 40 + x – 20 + \frac{1}{2} x + 15 = 180° [Sum of the angles of a triangle]
x + x + \frac{1}{2} x – 60 + 15 = 1 80°
x + x + \frac{1}{2} x – 45 = 180°
2x + \frac{1}{2} x = 180 + 45
\frac{4 x+x}{2} = 225
5x = 225 x 2
5x = 450
x = \frac{450}{5}
x = 90°

KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions

Question 10.
In ∆ABC ∠A – ∠B = 15° and ∠B – ∠C = 30° find ∆ABC ∠A , ∠B and ∠C
Solution:
∠A – ∠B = 15°
∴ ∠A = 15 + ∠B
∠B – ∠C = 30
∴ ∠B = 30 + ∠C
∠A + ∠B + ∠C = 180° [Sum of the angle of a triangle]
15 + ∠B + ∠B + ∠C = 180° (∴ ∠A = 15 + ∠B)
15 + 2∠B + ∠C = 180°
15 + 2[30 + ∠C + ∠C = 180° (∴ ∠B = 30 + ∠C)
15 + 60 + 2∠C] + ∠C = 180°
75 + 3∠C = 180°
3∠C = 180 – 75
3∠C = 105°
∠C = \frac{105^{\circ}}{3}
∠C = 35°
Now ∠B = 30 + ∠C
∠B = 30 + 35
∠B = 65°
∠A = 15 + ∠B
∠A = 15 + 65
∠A = 180°

KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions

Question 11.
The sum of two angles of a triangle 80° and their difference is 20° find the angles of the triangle.
Solution:
Let the angles be x and y
KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions 4
x = \frac{100}{2} ; x = 50°
x + y = 80°
50 + y = 80°
y = 80 – 50
y = 30°
Let the third angle be z
x + y + z = 180° [Sum of the angles]
50 + 30 + z = 180°
80 + z = 180°
z = 180 – 80
z = 100°

Question 12.
In a triangle ABC, ∠B = 60°, ∠C = 80°. Suppose the bisectors of ∠B and ∠C meet at L find ∠BLC.
Solution:
KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions 5
BL bisects ∠B
KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions 6
∠LBC + ∠BLC + ∠LCB = 180° (Sum of the angles of a triangle)
30 + ∠BLC + 40° = 180°
70 + ∠BLC = 180°
∠BLC = 180 – 70
∠BLC = 110°

KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions

Question 13.
In a triangle, each of the smaller angles is half of the largest angle. Find the angles.
Solution:
Let the smallest angle be x then the largest angle is 2x.
x + x + 2x = 180° [Sum of the angles of triangle]
4x = 180°
x = \frac{180}{4}
x = 45°
2x = 2 × 45° = 90°
∴ The angles are 45°, 45° and 90°

Question 14.
In a triangle each of the bigger angle is twice the third angle find the angles.
Solution:
Let the third angle be x. then the bigger angles is 2x 2x + 2x + x = 180° [Sum of the angles of triangle]
5x = 180°
x = \frac{180}{5}
x = 36°
2x = 2 × 36° = 72°
∴ The angles are 72°, 72° and 36°

Question 15.
In a triangle ABC, ∠B = 50° and ∠A = 60°. Suppose BC is extended D. Find ∠ACD
Solution
∠ACD = ∠ ABC + ∠BAC [Exterior angle = sum of interior opposite angles]
KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions 7
∠ACD = 50 + 60
∠ACD = 110°

Question 16.
In an isosceles triangle, the vertex angle is twice the sum of the base angles. Find the angles of the triangle.
Solution:
KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions 8
In ∆ABC, AB = BC
∴ ∠B = ∠C
∠Base angles of an isosceles triangle]
Given ∠A = 2(∠B + ∠C)
∠A +∠B + ∠C = 180°
[Sum of the angles of a triangle]
2(∠B + ∠C) + ∠B + ∠C = 180°
[Substituted for ∠A ]
2∠B + 2∠C + ∠B + ∠C = 180°
2∠B + 2∠B + ∠B + ∠B = 180° [∴ ∠B = ∠C]
6∠B = 180° ∠B = \frac{180}{6} = 30°
∠B = 30°, ∠C = 30° and ∠A = 2(∠B + ∠C)
= 2 (30 + 30) = 2 (60) = 120°

KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions

Question 17.
Find the sum of all the angles at the five vertices of the adjoining star.
KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions 9
Solution:
Since all the sides of the star are equal to each other the Afes AGL, CDB, EFD, GHF and KHL are isosceles triangles. Let each base angle equal to ‘x’.
In the figure, BDFHL is a pentagon & its sides are extended in order. ∴ The sum of exterior angles = 360°
∠DBC + ∠FDE + ∠GFH + ∠KHL + ∠BLA = 360°
x + x + x + x + x = 360°
5x = 360°
x = \frac{360^{\circ}}{5} = 72°
In ∆ABL,
∠A + ∠ABL + ∠ALB = 180°
∠A + x + x = 180°
∠A + 72 + 72 = 180°
∠A + 144 = 180°
∠A = 180° – 144 ∠A = 36°
∠C, ∠E, ∠G and ∠K = 36°
∴ Sum of all the angles at the five vertices = 36° × 5 = 180°

KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Additional Questions

KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions

Students can Download Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions

Question 1.
Match the following numbers ¡n the column A with their squares ¡n the column B.
KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions 1
Solution:
(1) (a) 25
(2) (e) 64
(3) (f) 4
(4) (c) 36
(5) (d) 484
(6) (b) 144

Question 2.
Choose the correct option:
(a) The number of perfect squares from 1 to 500 is
A. 1
B. 16
C. 22
D. 25
Solution:
C. 22

KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions

(b) The last digit of a perfect square can never be
A. 1
B. 3
C. 5
D. 9
Solution:
B. 3

(c) If a number ends in 5 zeros, its square ends in
A. 5 zeros
B. 8 zeros
C. lo zeros
D. 12 zeros
Solution:
C. 10 zeros

(d) Which could be the remainder among the following when a perfect square is devided by 8?
A. 1
B. 3
C. 5
D. 7
Solution:
A. 1

(e) The 6th triangular number is
A. 6
B. 10
C. 21
D. 28
Solution:
C. 21

KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions

Question 3.
Consider all integers froni – 10 to 5 and square each of them. How many distinct numbers you get?
Solution:
(-10)2 = 100, (-9)2 = 81, (-8)2
= 64,(-7)2 = 49
(-6)2 = 36, (5)2 = 25, (-4)2 16, (-3)2 =9,
(2)2 =4, (-1)2 = 1, 02 = 0, 12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25
There are 11 distinct numbers.

Question 4.
Write the digit in units place when the following numbers are squared.
Solution:
KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions 2

Question 5.
Write all the numbers from 400 to 425 which end in 2, 3, 7 or 8. Check if any of these in a perfect square.
Solution:
402, 403, 407, 408, 412, 413, 417, 418, 422, 423
None of these is a perfect square.

Question 6.
Find the sum of the digits of (1111111111)2.
Solution:
Observe the pattern
KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions 3

KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions

Question 7.
Suppose x2 + y2 = Z2
(i) If x = 4 and y = 3 find z
(ii) If x = 5 and z = 13 find y
(iii) If y = 15 and z = 17 find x.
Solution:
(i) x = 4, y = 3
x2 + y2 = z2
42 + 32 = z2
16 + 9 = z2
25 = z2
∴ z = √25 = 5

(ii) x = 5, z = 13
x2 + y2 = z2
52 + y2 = 132
25+ y2 =169
y2 =169 – 25
y2 = 144
y = √144 = 12

(iii) y = 15, z = 17
x2+ y2 = z2
x2 + 152 = 172
x2 + 225 = 289
x2 = 289 – 225
x2 = 64
x = √ 64 = 8

KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions

Question 8.
A sum of Rs. 2304 is equally distributed among several people. Each gets as many rupees as the number of persons. How much does each one get?
Solution:
KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions 4
Let the number of persons be x. Each person get rupees x.
∴ x × x = 2304
x2 = 2304
x = √2304
2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
= 16 × 3 × 16 × 3
2304 = 48 × 48
√2304 = 48 Each person gets Rs. 48

Question 9.
Define a new addition e on (be set of all natural numbers by m * n = m2 + n2
(i) Is N closed under e?
(ii) Is * commutative on N?
(iii) Is * associative on N?
(iv) Is there an identity element in N with respect * ?
Solution:
(i) Let m 2 and n 5
m * n = m2 + n2
2 * 5 = 22 + 52 = 4 + 25 = 29 ∈ N
∀ m, n ∈ N, m2 + n2 ∈ N
∴ * is closed under N.

(ii) Let m = 4 and n = 7
m * n = m2 + n2 = 42 + 72 = 16 + 49 = 65
n * m = n2 + m2 = 72 + 42 = 49 + 16 = 65
∴ m * n = n * m ∀ m, n ∈ N
Hence * is commutative on N.

KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions

(iii) Lei m = 2, n = 3 and p = 4.
m * (n * P) = 2 * (3*4)
= 2*(32 + 42) = 2*(9 + 16)
= 2*25 = 22 + 252 = 4 + 625
= 629 ….(i)
(m * n) * p = (2 * 3)*4
=(22 + 32) * 4 = (4 * 9) * 4
= 13 * 4 = 132 + 42.
= 169 + 16 = 185 …(ii)
From (i) and (ii) m*(n*p) ≠ (m*n)*p
∴ * is not associative on N.

(iv) Let K be the identity element then m * k = m, m2 + k2 = m2 which means
k2 = m2 – m2 = O
K = 0, O does not belong to N.
There is no identity element in N with respect to *.

KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions

Question 10.
(Exploration) Find all the perfect squares from 1 to 500 each of which is a sum of two perfect squares.
Solution:
Perfect squares from 1 to 500 are 1, 4, 9, 1, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289. 324, 361, 400, 441, 484.
25 = 9 + 16
100 =36 + 64
169 = 25 + 144
289 = 225 + 64

Question 11.
Supposes the area of a square field is 7396 find its perimeter.
Solution:
Let each side of the square be
l2 = 7396 = 2 × 2 × 43 × 43 = 2 × 43 × 2 × 43
l2 =7396 = 86 × 86.

KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions 5
∴ √7396 = 86 = l
Perimeter = 4l = 4 × 86 = 344 m.

Question 12.
Can 1010 be written as a difference of two perfect squares?
Solution:
If a2 – b2 = 1010 for any two integers a and b then either both ‘a’ and ‘b’ are odd or both even. Hence a2 – b2 is divisible by 4. But 1010 is not divisible by 4. Hence 1010 is not the difference of two perfect squares.

KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions

Question 13.
What are the remainders when a perfect cube is divided by 7?
Solution:
When the perfect cubes 8, 27, 64, 125, 216, 343, 512, 729, 1000 are divided by 7 the remainders respectively are 1, 6, 1, 6, 6, 0, 1, 1, 6. Hence the remainders are 0. 1 or 6.

Question 14.
What is the least perfect square which leaves the remainder 1 when divided by 7 as welt as by 11?
Solution:
The least number divisible by both 7 and 11 is 7 × 11 = 77. When 1 is added to 77 we get 78. But 78 is not a perfect square. (77 × 2) + 1 = 154 + 1 = 155 is not a perfect square. In the same way, continue we find that
(77 × 15) + 1 = 1155 + 1 = 1156 in a perfect square.
∴ The number required is 1156 = 342

Question 15.
Find two smallest perfect squares whose product is a perfect cube.
Solution:
4 and 16 are perfect squares. 4 × 16 = 64 is a perfect cube.

Question 16.
Find a proper positive factor of 48 and a proper positive multiple of 48 which add up to a perfect square. Can you prove that there are infinitely many such pairs? By considering 16 as a factor of 48 and 48 as a factor of 240, we can write 16 + 240 = 162.
Solution:
Multiple of 48 is 48 and considering it as 48l where l = m (3m + 2), m – 1, 2, 3 …. we can get infinite numbers.

KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions

Students can Download Maths Chapter 4 Factorisation Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 8 Maths Chapter 4 Factorisation Additional Questions

Question 1.
Choose the correct answer
(a) 4a + 12b is equal to
A. 4a
B. 12b
C. 4(a + 3b)
D. 3a
Answer:
(C) 4 (a + 3b)

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions

(b) The product of two numbers is positive and their sum is negative only when
A. both are positive
B. both are negative
C. one positive other negative
D. one of them is equal to zero
Answer:
(B) both are negative

(c) Factorising x2 + 6x + 8 we get
A. (x + 1) (x + 8)
B. (x + 6) (x + 2)
C. (x + 10) (x – 2)
D. (x + 4) (x + 2)
Answer:
(D) (x + 4) (x + 2)

(d) The denominator of an algebraic fraction should not be
A. 1
B. 0
C. 4
D. 7
Answer:
(B) 0

(e) If the sum of two integers is – 2 and their proiduct is – 24, the numbers are
A. 6 and 4
B. – 6 and 4
C. – 6 and -4
D. 6 and – 4
Answer:
(B) – 6 and 4

(f) The difference (0.7)2 – (0.3)2 simplifies to
A. 0.4
B. 0.04
C. 0.49
D. 0.56
Answer:
(A) 0.4

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions

Question 2.
Factorise the following
(i) x2 + 6x + 9
Answer:
x2 + 6x + 9
= x2 + 2.x.3 + 32
= (x + 3)2

(ii) 1 – 8x + 16x2
Answer:
1 – 8x + 16x2
= 12 – 2.1.4x + (4x)2
= (1 – 4x)2

(iii) 4x2 – 81y2
Answer:
4x2 – 81 y2
= (2x)2 – ( 9y)2
= (2x + 9y) (2x – 9y)

(iv) 4a2 + 4ab + b2
Answer:
4a2 + 4ab + b2
= (2a)2 +2.2a.b + b2
= (2a + b)2

(v) a2b2 + c2d2 – a2c2 – b2d2
Answer:
a2b2 + c2d2 – a2c2 – b2d2
= a2b2 – a2c2 + c2d2 – b2d2
= a2 (b2 – c2) – d2 (b2 – c2)
= (b2 – c2) (a2 – d2)

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 1

Question 3.
Factorise the following
(i) x2 + 7x + 12
Answer:
x2 + 7x + 12 12xz
x2 + 4x + 3x + 12

x (x + 4) + 3 (x + 4)
(x + 4) (x + 3)

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 2

(ii) x2 + x – 12
Answer:
x2 + 4x – 3x – 12
x(x + 4) -3 (x + 4)
(x + 4) (x – 3)

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 3

(iii) x2 – 3x – 18
Answer:
x2 – 6x + 3x – 18
x(x – 6) +3 (x -6)
(x – 6) (x + 3)

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 4

(iv) x2 + 4x – 21
Answer:
x2 + 7x – 3x – 12
x(x + 7) -3 (x + 7)
(x + 7) (x – 3)

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 5

(v) x2 – 4x – 192
Answer:
x2 – 16x + 12x – 192
x(x – 16) +12 (x – 16)
(x -16) (x +12)

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 6

(vi) x4 – 5x2 + 4
Answer:
x4 – 4x2 – x2 + 4
x2(x2 – 4) -1 (x2 – 4)
(x2 – 4) (x2 – 1) = (x + 2) (x – 2)
(x + 1) (x – 1)

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 7

(vii) x4 – 13x2y2 + 36y4
Answer:
x4 -9x2y2 – 4x2y2 +36y4
x2(x2 – 9y2) -4y2
(x2 – 9y2)
(x2 – 9y2) (x2 – 4y2)
= [x2 – (3y)2] [x2 – (2y)2]
= (x + 3y) (x – 3y) (x + 2y) (x – 2y)

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 8

Question 4.
Factorise the following
(i) 2x2 + 7x + 6
Answer:
2x2 + 4x + 3x + 6
2x (x +2) + 3 (x + 2)
(x + 2) (2x + 3)

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 9

(ii) 3x2 – 17x + 20
Answer:
3x2 – 12x – 5x + 20
3x (x – 4) – 5 (x – 4)
(x – 4) (3x – 5)

(iii) 6x2 – 5x – 14
Answer:
6x2 – 12x + 7x – 14
6x (x – 2) + 7 (x – 2)
(x – 2) (6x +7)
KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 10

(iv) 4x2 + 12xy – 5y2
Answer:
4x2 – lOxy + 2xy + 5y2 – 20x2y2
2x (2x + 5y) + y
(2x + 5y)
(2x + 5y) (2x + y)
KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 11

(v) 4x4 – 5x2 + 1
Answer:
4x2 – 4x2 – x2 + 1 4×4
4x2 (x2 – 1) – 1
(x2 – 1)
(x2 – 1) (4x2 – 1)
(x2 – 12) – ((2x)2 – 12)
(x2 – l2) – ((2x)2 – l2)
(x + 1) (x – 1) (2x + 1) (2x – 1)
KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions 12

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions

Question 5.
Factorise the following
(i) x8 – y8
Answer:
(x4)2 – (y4)2
= (x4 + y4) (x4 – y4)
= (x4 + y4) [(x2)2 – (y2)2]
= (x4 + y4) (x2 + y2) (x2 – y2)
= (x4 + y4) (x2 + y2) (x + y)
(x – y)

(ii) a12x4 – a4x12
Answer:
[(a6)2(x2)2 – (a2)2(x6)2]
= [(a6x2)2 – (a2x6)2]
= (a6x2 + a2x6) (a6x2 – a2x6)
= a2x2 (a4 + x4)
[(a3)2 x2 – a2 (x3)2]
= a2x2 (a4 + x4) [(a3x)2 – (ax3)2] = a2x2 (a4 + x4) [(a3x + ax3)
(a3x – ax3)]
= a2x2 (a4 + x4) [ax(a2 + x2).ax (a2 – x2)]
= a2x2 (a4 + x4) [a2x2 (a2 + x2) (a2 – x2)]
= a4x4 (a4 + x4) (a2 + x2) (a + x) (a – x)

(iii) x4 + x2 + 1
Answer:
X4 + X2 + 1 + X2 – X2
= x4 + 2x2 + 1 – x2
= (x2)2 + 2.x2.1 + l2 – x2
= (x2 + l)2 – X2
= (x2 + 1 + x) (x2 + 1 – x)
= (x2 + x + 1) (x2 – X + 1)

(iv) x4 + 5x2 + 9
Answer:
x4 + 5x2 + 9 + x2 – x2
= x4 + 6x2 + 9 – x2
= (x2)2 + 2.x2.3 + 32 – x2 = (x2 + 3)2 – x2
= (x2 + 3 + x) (x2 + 3 – x)
= (x2 + x + 3) (x2 – x + 3)

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions

Question 6.
Factorise x4 + 4y4 use this to prove that 2011 4 + 64 is a composite number Answer:
x4 + 4y4
= (x2)2 + (2y2)2
= (x2 + 2y2)2 – 2x2.2y2
[v a2 + b2 = (a + b)2 – 2ab]
= (x2 + 2y2) – (4x2y2)
= (x2 + 2y2) – (2xy)2
(x2 + 2y2 + 2xy) (x2 + 2y2 – 2xy) [a2 – b2 = (a + b) (a – b)]
∴ x4 + 4y4 = (x2 + 2y2 + 2xy)
(x2 + 2y2 – 2xy)
Similarly 20114 + 64 can be written as
20114 + 4 x 16 = (2011)4 + 4.24
= (20112 + 2.22 + 2.2011.2) (20112 + 2.22 – 2.2011.2)
∴ 20114 + 64 is a composite number.

KSEEB Solutions for Class 8 Maths Chapter 4 Factorisation Additional Questions

KSEEB Solutions for Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions

Students can Download Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions

Question 1.
Choose the correct option
(i) If a = 6 and b = a, then b = 60 by _______
A. Axiom 1
B. Axiom 2
C. Axiom 3
D. Axiom 4
Solution:
A. Axiom 1

(ii) Given a point on the plane, one can draw ________
A. unique
B. two
C. finite numberd
D. infinitely many
Solution:
(D) Infinetly many lines through that point.

(iii) Given two points in a plane, the number of lines which can be drawn to pass through these two points is ________
A. Zero
B. exactly one
C. at most one
D. more than one
Solution:
(B) Exactly one

(iv) If two angles are supplementary, then their sum is
A. 90°
B. 180°
C. 270°
D. 360°
Solution:
(B) 180°

(v) The measure of an angle which is 5 times its supplement is _________
A. 30°
B. 60°
C. 120°
D. 150°
Solution:
(D) 150°

KSEEB Solutions for Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions

Question 2.
What is the difference between a pair of supplementary angles and a pair of complementary angles?
Solution:
If the sum of two angles is 180° then they are supplementary angles. If the sum of two angles is 90° then they are complementary angles.

Question 3.
What is the least number of non – collinear points required to determine a plane?
Solution:
Three (3)

Question 4.
When do you say two angles are adjacent?
Solution:
Two angles are said to be adjacent angles if they have a common vertex and a common side.

Question 5.
Let \overline{\mathbf{A B}} be a segment with C and D between them such that the order of points on the segment is A, C, D, B suppose AD = BC prove that AC = DB
Solution:
KSEEB Solutions for Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions 1
AD = BC [data]
AD – CD = BC – CD [Axiom 3]
AC = DB [In the fig AC – CD = AC , BC – CD = DB]

KSEEB Solutions for Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions

Question 6.
Let \overrightarrow{\mathbf{A B}} and \overrightarrow{\mathbf{A B}} be two straight lines intersecting at O. Let \overrightarrow{\mathbf{O X}} be the bisector of ∠BOD. Draw \overrightarrow{\mathbf{O Y}} between \overrightarrow{\mathbf{O D}} and \overrightarrow{\mathbf{O A}} such that \overrightarrow{\mathbf{O Y}}\overrightarrow{\mathbf{O B}} prove that \overrightarrow{\mathbf{O Y}} bisects ∠DOA .
Solution:
KSEEB Solutions for Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions 2
Construction: Produce [/latex] such that \overrightarrow{\mathbf{X O}} to A point Z such that [/latex] such that \overrightarrow{\mathbf{O Z}}.is between [/latex] such that \overrightarrow{\mathbf{O A}} and [/latex] such that \overrightarrow{\mathbf{O C}}
Proof : ∠AOZ = ∠BOX [vertically opposite angles]
∠BOX = ∠POX [\overrightarrow{\mathbf{OX}} bisects ∠DOB]
∴ ∠AOZ = ∠DOX  [Axiom 1] … (i)
[ZOY = [YOX r=90°] … (ii)
(ii) – (i) ∠ZOY – ∠AOZ = ∠YOX – ∠DOX ∠AOY = ∠YOD [Axiom 3]
∴ [/latex] such that \overrightarrow{\mathbf{O Y}} bisects ∠AOD

Question 7.
Let \overrightarrow{\mathbf{A B}} and \overrightarrow{\mathbf{C D}} be two parallel lines and \overrightarrow{\mathbf{P Q}} be a transversal. Let \overrightarrow{\mathbf{P Q}} intersect \overrightarrow{\mathbf{A B}} in L. Suppose the bisector of ∠ALP intersect CD in R and the bisector of ∠PLB intersect CDin S prove that ∠LRS + ∠RSL = 90°
KSEEB Solutions for Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions 3
Solution:
∠ALP + ∠BLP = 180° ∠Linear pair]
\frac{1}{2} ∠ALP + \frac{1}{2} ∠BLP = \frac{1}{2}180°
[Multiplying by \frac{1}{2} ]
∠ELP + ∠FLP = 90° [∴ \overrightarrow{\mathbf{E L}} and \overrightarrow{\mathbf{F L}} are bisectors of ∠ALP and ∠BLP ]
∠ELF = 90°
[SLR = ∠ELF = 90°
[vertically opposite angles]
In ∆ SLR, ∠SLR + ∠LRS + ∠RSL = 180° [sum of the angles of the triangle is 180° ]
90 + ∠LRS + ∠RSL = 180°
∠LRS + ∠SL = 180 – 90
∠LRS + [RSL = 90°

KSEEB Solutions for Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions

Question 8.
In the adjoining figure \overrightarrow{\mathrm{AB}} and \overrightarrow{\mathrm{CD}} are parallel lines. The transversals \overrightarrow{\mathrm{PQ}} and \overrightarrow{\mathrm{RS}} intersect at U on the line \overrightarrow{\mathrm{AB}}. Given ∠DWU = 110° and ∠CVP = 70° find the measure of ∠QUS.
Solution:
KSEEB Solutions for Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions 4

∠ AUV – ∠CVP = 70°
[corresponding angles]
∠QUB – ∠AUV = 70° [vertically opposite angles]
∠BUW + ∠DWU = 180° [Interior angles on the same side of transversal]
∠BUW + 110°= 180°
∠BUW = 180-110° = 70°
∠BUW = ∠AUS= 70° [vertically opposite angles]
∠AUS + ∠SUQ + ∠QUB = 180° ∠AUB is a straight angle]
70 + ∠SUQ + 70 = 180°
∠SUQ + 140° = 180°
∠SUQ = 180° – 140°
∠SUQ = 40°

KSEEB Solutions for Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions

Question 9.
What is the angle between the hours hand and minutes hand of a clock at (i) 1.40 hours (ii) 2.15 hours [use 1° = 60 minutes]
Solution:
i. 1.40 hours – 190°
ii. 2.15 hours – 22° 30′

Question 10.
How much would hour’s hand have moved from its position at 12 noon when the time is 4.24 p.m?
Solution:
144°

Question 11.
Let \overrightarrow{\mathrm{AB}} be a line segment and let C be the midpoint of \overrightarrow{\mathrm{AB}} Extend AB to D such that B lies between A and D. Prove that AD + BD = 2CD
Solution:
KSEEB Solutions for Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions 5
AD + BD = 2CD L.H.S – AD + BD
= AC + CB + BD + BD [∵ AD = AC + CB + BD]
= CB + CB + BD + BD
[∴ AC = CB]
= CB + BD + CB + BD
= CD + CD
= 2CD = RHS
∴ AD + B D = 2CD

KSEEB Solutions for Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions

Question 12.
Let \overrightarrow{\mathbf{A B}} and \overrightarrow{\mathbf{C D}} be two lines . intersecting at a point O. Let \overrightarrow{\mathbf{O X}} be a ray bisecting [BOD. Prove that the extension of \overrightarrow{\mathbf{O X}} to the left of O bisects ∠AOC.
Solution:
KSEEB Solutions for Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions 6

∠DOX = ∠BOX [ ∵\overrightarrow{\mathbf{O X}} bisects ∠BOD]
∠DOX =∠COY [vertically opposite angles] ….(i)
∠BOX = ∠AOY
[vertically opposite anglesj … (ii)
From (i) and (ii)
[vertically opposite] …(ii)
∠AQY – ∠COY [Axiom 1]
\overrightarrow{\mathbf{O X}} bisects ∠AOC

Question 13.
Let \overrightarrow{\mathbf{O X}} be a ray and let \overrightarrow{\mathbf{O A}} and \overrightarrow{\mathbf{O B}} be two rays on the same side of \overrightarrow{\mathbf{O X}} with \overrightarrow{\mathbf{O A}} between \overrightarrow{\mathbf{O X}} and \overrightarrow{\mathbf{O B}} . Let \overrightarrow{\mathbf{O C}} be the bisector of ∠AOB prove that ∠XOA + ∠XOB = 2∠XOC
Solution:
∠XOA + ∠XOB = 2∠XOC
KSEEB Solutions for Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions 7

KSEEB Solutions for Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions

Question 14.
Let \overrightarrow{\mathbf{O A}} and \overrightarrow{\mathbf{O B}} be two rays and let \overrightarrow{\mathbf{O X}} be a ray between \overrightarrow{\mathbf{O A}} and \overrightarrow{\mathbf{O B}} such ∠AOX > ∠XOB . Let \overrightarrow{\mathbf{O C}} be the bisector of ∠AOB prove that ∠AOX – ∠XQB = 2∠COX
Solution:
KSEEB Solutions for Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions 8
KSEEB Solutions for Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions 9

Question 15.
Let \overrightarrow{\mathbf{OA}}, \overrightarrow{\mathbf{O B}}, \overrightarrow{\mathbf{OC}} be three rays such that \overrightarrow{\mathbf{OC}} lies between \overrightarrow{\mathbf{OA}} and \overrightarrow{\mathbf{O A}}. Suppose the bisectors of ∠AOC and ∠COB are perpendicular to each other. Prove that B, O. A are . collinear.
Solution:
∠XOC + ∠YOC – 90° [data]
KSEEB Solutions for Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions 10
∠AOB = 180°
A, O and B are collinear.

KSEEB Solutions for Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions

Question 16.
In the adjoining figure \overrightarrow{\mathrm{AB}} || \overrightarrow{\mathrm{DE}} . Prove that ∠ABC – [DCB + [CDE = 180°
Solution:
Construction: Through C draw \overrightarrow{\mathrm{XY}} || \overrightarrow{\mathrm{AB}}
Proof : ∠ABC + ∠BCX = 180° [sum of interior angles on the same side of trnasversal]
∴ ∠BCX = 180 – ∠ABC
∠CDE + ∠DCY – 180° [sum of interior angles on the same side of transversal]
KSEEB Solutions for Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions 11

∴ ∠DCY = 180° – ∠CDE
∠BCX + ∠BCD + ∠DCY = 180° [∵ XCY is straight angle]
180 – ∠ABC + ∠BCD + 180 – ∠CDE = 180° [By substituting]
360 – ∠ABC + ∠BCD – ∠CDE – 180°
360 – 180 – ∠ABC – ∠BCD + CDE
180° = ∠ABC – ∠BCD + ∠CDE
or
∠ABC – ∠BCD + ∠CDE = 180°

Question 17.
Consider two parallel lines and a transversal among the measures of 8 angles formed how many distinct numbers are there?
Solution:
There will be two distinct numbers.

KSEEB Solutions for Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Additional Questions

KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions

Students can Download Maths Chapter 2 Algebraic Expressions Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions

Question 1.
Choose the correct answer
(a) Terms having the same literal factor with the same exponents are called
(A) exponents
(B) like terms
(C) factors
(D) unlike term
Answer:
(B) like terms

(b) The coefficient of ab in 2ab is
(A) ab
(B) 2
(C) 2a
(D) 2b
Answer:
(B) 2

(c) The exponential form of a x a x a is
(A) 3a
(B) 3 + a
(C) a3
(D) 3 – a
Answer:
(C) a3

(d) Sum of two negative integers is
(A) negative
(B) positive
(C) zero
(D) infinite
Answer:
(A) negative

(e) What should be added to a2 + 2ab to make it a complete square?
(A) b2
(B) 2ab
(C) ab
(D) 2a
Answer:
(A) b2

KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions

(f) What is the product of (x + 2) (x – 3)?
(A) 2x – 6
(B) 3x – 2
(C) x2 – x – 6
(D) x2 – 6x
Answer:
(C) x2 – x – 6

(g) The value of (7.2)2 is (use an iden-tity to expand)
(A) 49.4
(B) 14.4
(C) 51.84
(D) 49.04
Answer:
(C) 51.84

(h) The expansion of (2x – 3y)2is
(A) 2x2 + 3y2+ 6xy
(B) 4x2 + 9y2 – 12xy
(C) 2x2 +3y2 – 6xy
(D) 4x2 +9y2 + 12xy
Answer:
(B) 4x2 + 9y2 – 12xy

(i) The product of 58 × 62 is (use an identity)
(A) 4596
(B) 2596
(C) 3596
(D) 6596
Answer:
(C) 3596

Question 2.
Take away 8x – 7y – 8p + 10q from 10x + 10y – 7p + 9q
Answer:
KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions 1

KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions

Question 3.
Expand
(i) (4x + 3)2
Answer:
(a + b)2 = a2 + 2ab + b2
Here a = 4x, b = 3
(4x + 3)2 = (4x)2 + 2.4×3 .+ 32
= 16x2 + 24x + 9

(ii) (x + 2y)2
Answer:
(a + b)2 = a2 + 2ab + b2
Here a = x, b = 2y
(x + 2y)2 = x2 + 2.x.2y + (2y)2
= x2 + 4xy + 4y2

(iii) \left(x+\frac{1}{x}\right)^{2}
Answer:
(a + b)2 = a2 + 2ab + b2
Here a = x, b = 1/x
KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions 2

Question 4.
Expand
(i) (2t + 5) (2t – 5).
Answer:
(a + b) (a – b) = a2 – b2
Here = a = 2t, b = 5 ]
(2t + 5) (2t – 5) = (2t)2 – 52
= 4t2 – 25

(ii) (xy + 8) (xy – 8)
Answer:
(a + b) (a -b) = a2 – b2
Here a = xy, b = 8
(xy + 8) (xy – 8) = (xy)2 – 82
= x2y2 – 64

(ii) (2x + 3y) (2x – 3y)
Answer:
(a + b) (a -b) = a2 – b2
Here = a = 2x, b = 3y
(2x + 3y) (2x – 3y) = (2X)2 – (3y)2 = 4x2 – 9y2

KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions

Question 5.
Expand
(i) (n – 1) (n + 1) (n2 + 1)
Answer:
(n – 1) (n + 1) (n2 + 1)
= (n2 – l2) (n2 + 1)
= (n2 – 1) (n2 + 1)
= (n2)2 – l2
= n4 – 1

(ii)\left(n-\frac{1}{n}\right)\left(n+\frac{1}{n}\right)\left(n^{2}+\frac{1}{n^{2}}\right)
Answer:
KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions 3

(iii) (x – 1) (x + 1) (x2 + 1) (x4 + 1)
Answer:
(x – 1) (x + 1) (x2 + 1) (x4 + 1)
= (x2 – l2) (x2 + 1) (x4 + 1)
= (x2 – 1) (x2 + 1) (x4 + 1)
= [(x2)2 – l2)] (x4 + 1)
= (x4 – 1) (x4 + 1)
= (x4)2 – l2
= X8 – 1

(iv) (2x – y) (2x + y) (4x2 + y2)
Answer:
(2x – y) (2x + y) (4x2 + y2)
= [(2x)2 – y2) (4x2 + y2)
= (4x2 – y2) (4x2 + y2)
= [(4x2)2 – (y2)2)]
=16x4 – y4

KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions

Question 6.
Use appropriate formulae and compute
(i) (103)2
Answer:
1032 = (100 + 3)2
(a + b)2 = a2 + 2ab + b2
Here a = 100, b = 3
(100 + 3)2 = 1002 + 2.100.3 + 32
= 10000 + 600 + 9
1032 = 10609

(ii) (96)3
Answer:
962 = (100 – 4)2
(a – b)2 = a2 – 2ab + b2
Here a = 100, b = 4
(100 + 3)2 = 1002 + 2.100.4 + 42
= 10000 – 800 + 16
1032 = 9216

(iii) 107 x 93
Answer:
107 x 93 = (100 + 7)(100 – 7)
(a + b) (a – b) = a2 – b2
Here a = 100, b = 7
(100 + 7) (100 – 7) = 1002 – 72
= 10000 – 49 = 9951

(iv) 1008 x 992
Answer:
1008 x 992 = (1000 + 8)(100 – 8)
(a + b) (a – b) = a2 – b2
Here a = 1000, b = 8
(1000 + 8) (100 – 8) = (1000)2 – 82
= 1000000 – 64
1008 x 992 = 999936

(v) 1852 – 1152
Answer:
1852 – 1152
(a – b)2 = (a + b) (a – b)
Here a = 185, b = 115
1852 – 1152 = (185 + 115) (185 – 115)
= (300) (70)
1852 – 1152 = 21000

Question 7.
If x + y = 7 and xy = 12 find x2 + y2
Answer:
(x + y)2 = x2 + 2xy + y2
(x + y)2 = x2 + y2 + 2xy
72 = x2 + y2 + 2(12)
[Substituting for x +y and xy]
49 = x2 + y2 +24
x2 + y2 = 49 24
x2 + y2 = 25

Question 8.
If x + y = 12 and xy = 32 find x2 + y2
Answer:
(x + y)2 = x2 + y2 + 2xy
122 = x2 + y2 + 2(32)
[Substituting for x +y and xy]
144 = x2 + y2 + 64
x2 + y2 = 144 – 64
x2 + y2 = 80

Question 9.
If 4x2 + y2 = 40 and xy = 6 find 2x + y
Answer:
4x2 + y2 = 40
(2x)2 + y2 = 40
(2x + y)2 = (2x)2 + y2 + 2.2x.y
(2x + y)2 = 4x2 + y2 + 4xy
= 40 + 4 x 6
= 40 + 24
(2x + y)2 = 64
2x + y = ±√64
2x + y = ±√8

KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions

Question 10.
If x – y = 3 and xy = 10 find x2 + y2
Answer:
(x – y)2 = x2 + y2 – 2xy
32 = x2 + y2 + 2(10)
[Substituting for x – y and xy]
9 = x2 + y2 – 20
9 + 20 = x2 + y2
∴ 29 = x2 + y2
x2 + y2 = 29

Question 11.
If x+\frac{1}{x} = 3 find x^{2}+\frac{1}{x^{2}} and x^{3}+\frac{1}{x^{3}}
Answer:
KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions 4
KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions 5

KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions

Question 12.
x+\left(\frac{1}{x}\right)=6 find x^{2}+\frac{1}{x^{2}} and x^{4}+\frac{1}{x^{4}}
Answer:
KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions 6

Question 13.
Simplify (i) (x + y)2 + (x – y)2
Answer:
(x + y)2 + (x – y)2
= x2 + y2 + 2xy + x2 + y2 – 2xy
= x2 + y2 + x2 + y2 = 2x2 + 2y2 ]
= 2(x2 + y2)

(ii) (x + y)2 x (x – y)2
Answer:
(x + y)2 x (x – y)2
= (x2 + y2 + 2xy) (x2 + y2 – 2xy)
= x2 (x2 + y2 – 2xy) + y2 (x2 + y2 – 2xy) + 2xy (x2 + y2 – 2xy)
= x4 + x2y2 – 2x3y + x2y2 + y4 – 2xy3 + 2x3y + 2xy3 – 4x2y2
= X4 + 2x2y2 – 4x2y2 + y4
= x4 – 2x2y2 + y4

KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions

Question 14.
Express the following as difference of two squares
(i) (x + 2z) (2x + z)
Answer:
x(2x + z) + 2z (2x + z)
= 2x2 + xz + 4xz + 2z2
= 2x2 + 2z2 + 4xz + xz
= 2 (x2 + z2 + 2xz) + xz
= 2(x + z)2 + xz
KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions 7
KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions 8

(iii) 4(x + 2y) (2x + y)
Answer:
4(x + 2y) (2x + y)
= 4(2x2 + xy + 4xy + 2y2)
= 4(2x2 + 2y2 + 4xy + xy)
=4 (2 (x2 + y2 + 2xy) + xy)
= 4 (2 (x + y)2 + xy)
= 8 (x + y)2 + 4xy
= 8 (x + y)2 + (x + y)2 – (x – y)2
[∵ 4xy = (x + y)2 – (x – y)2] = 9 (x + y)x^{4}+\frac{1}{x^{4}} – (x – y)2 = [3 (x + y)]2 – (x – y)2

(iii) (x + 98)( (x+ 102)
Answer:
(x + 98)( (x + 102)
= (x + 100 – 2) (x + 100 + 2)
= (x + 100)2 – 22
[∵(a + b) (a – b) = a2 – b2]

(iv) 505 x 495
Answer:
505 x 495 = (500 + 5) (500 – 5)
= (500)2 – 52

Question 15.
If a = 3x – 5y, b = 6x + 3y and c = 2y – 4x find
(i) a + b – c
Answer:
a + b – c = 3x – 5y + 6x + 3y – (2y – 4x)
= 9x – 2y – 2y + 4x
= 13x – 4y

(ii) 2a – 3b + 4c = 2(3x – 5y) – 3 (6x + 3y) + 4 (2y – 4x)
= 6x – 10y – 18x – 9y + 8y – 16 x
= 6x – 18x – 16 x 10y – 9y + 8y
= 6x – 34x – 19y + 8y = – 28x – 11y
= – (28x + 11y)

KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions

Question 16.
The perimeter of a triangle is 15×2 – 23x + 9 and two of its sides are 5×2 + 8x – 1 and 6×2 – 9x + 4. Find the third side
Answer:
Let the sides of the triangle be a, b and c
Let a = 5x2 + 8x – 1,
b = 6x2 – 9x + 4
Perimeter = 15x2 – 23x + 9
a + b + c = 15x2 – -23x + 9 5x2 + 8x – 1 + 6x2 – 9x + 4 + c
= 15x2 – 23x + 9
llx2 – x + 3 + c <= 15x2 – 23x + 9
c = 15x2 – 23x + 9 – (llx2 – x + 3)
= 15x2 – 23x + 9 – llx2 + x – 3
c = 4x2 22x + 6
∴ The third side is 4x2 – 22x + 6

Question 17.
Two adjacent sides of a rectangle are 2x2 – 5xy + 3z2 and 4xy – x2 – z2. Find its perimeter.
Answer:
Let l = 2x2 – 5xy + 3z2 and b = 4xy – x2 – z2
Perimeter of a rectangel = 2(l + b)
= 2 [2x2 – 5xy + 3z2 + 4xy – x2 – z2]
= 4x2 – lOxy + 6z2 + 8xy – 2x2– 2z2]
= 2x2 + 4z2 – 2xy

Question 18.
The base and altitutde of a triangle are (3x – 4y) and (6x + 5y) respectively. Find its area.
Answer:
Area of a triangle = \frac{1}{2} x base x height
= \frac{1}{2} x(3x-4y)(5x + 5y)
= \frac{1}{2}[(3x(6x + 5y) – 4y(6x + 5y) ]
= \frac{1}{2}[18x2 +15y)-24y-20y2]
= \frac{1}{2}[18x2 – 9xy – 20y2]

Question 19.
The sides of a rectangle are 2x + 3y and 3x + 2y. From this a square of side length x + y is removed. What is the area of the remaining region.
Answer:
Area of the remaining region
= Area of rectangle – Area of sqûare
= length x breadth – (side)2
= (2x + 3y) (3x + 2y) – (x + y)2
=2x(3x + 2y) + 3y(3x + 2y) – [x2 + y2 + 2xy]
= 6x2 + 4xy + 9xy + 6y2 – [x2 + y2 + 2xy]
= 6x2 + 13xy + 6y2 – x2 – y2 – 2xy = 5x2 + 11xy + 5y2

Question 20.
If a, b, c are rotational numbers such that a2 + b2 + c2 – ab – be – ca = 0 prove that a = b = c
Answer:
a2 +b2 + c2 – ab – bc – ca = 0 Multiplying both the sides by 2
2(a2 + b2 + c2 – ab – be – ca) = 2 x 0
2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = 0
a2 + a2 + b2 + b2 + c2 + c2 – 2ab – 2bc – 2ca = 0
a2 – 2ab + b2 + b2 – 2bc + c2 + c2 – 2ca + a2 = 0
(a – b)2 + (b – c)2 + (c – a)2 =0
(a – b)2 = 0, (b – c)2 = 0, (c – a)2 = 0 [v If the sum of positive quantities is equal to zero then each quantity is equal to zero]
∴ (a-b)2 = 0 ⇒ a – b = 0 ⇒ a = b
(b-c)2 = 0 ⇒ b – c = 0 ⇒ b = c
(c-a)2 = 0 ⇒ c – a = 0 ⇒ c = a
∴ a = b = c.

KSEEB Solutions for Class 8 Maths Chapter 2 Algebraic Expressions Additional Questions

KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Additional Questions

Students can Download Maths Chapter 1 Playing with Numbers Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 8 Maths Chapter 1 Playing with Numbers Additional Questions

Question 1.
Choose the correct option.
(a) The general form of 456 is
A. (4 × 100) + (5 × 10) + (6 × l)
B. (4 × 100) + (6 × 10) + (5 × 1)
C. (5 × 100) + (4 × 10) + (6 × 1)
D. (6 × 100) + (5 × 10) + (4 × 1)
Answer:
A. (4 × 100) + (5 × 10) + (6 × 1)

(b) Computers use
A. decimal system
B. binary system
C. base 5 system
D. base 6 system
Answer:
B. binary system

(c) If \overline{a b c} is a 3 digit number, then the number.
n=\overline{a b c}+\overline{a c b}+\overline{b a c}+\overline{b c a}+\overline{c a b}+\overline{c b a} is always divisible by
A. 8
B. 7
C. 6
D. 5
Answer:
C. 6

(d) If \overline{a b c} is a 3-digit number then
n=\overline{a b c}-\overline{a c b}+\overline{b a c}-\overline{b c a}+\overline{c a b}-\overline{c b a} is always divisible by
A. 12
B. 15
C. 18
D. 21
Answer:
C. 18

(e) If IK × K1 = K2K, then the letter K stands for the digit.
A. 1
B. 2
C. 3
D. 4
Answer:
A. 1

(f) The number 345111 is divisible by
A. 15
B. 12
C. 9
D. 3
Answer:
D. 3

(g) The number of integers of the form 3AB4, where A, B denote some digits, which are divisible by 11 is
A. 0
B. 4
C. 7
D. 9
Answer:
D. 9

KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Additional Questions

Question 2.
What is the smallest 5 digit number divisible by 11 and containing each of the digits 2, 3, 4, 5, 6?
Answer:
The number is 24365
(∵ (2 + 3 + 5) – (4 + 6) = 0)

Question 3.
How many 5 – digit numbers divisible by 11 are there containing each of the digits 2, 3, 4, 5, 6?
Answer:
The number having the digits 2, 3, 4, 5, 6 is divisible by 11 if 2, 3 and 5 are in the odd places and 4 and 6 are in the even places as (2 + 3 + 5) – (4 + 6) = 0 which is divisible by 11.

If 2 in the first place the numbers can be 24365, 26345, 24563, 26543 (4 numbers)

Similarly if 3 is in the first place we get 4 numbers and 5 is in the first place we get 4 numbers.
Totally there are 12 numbers.

Question 4.
If 49A and A49 where A > 0, have a common factor, find all possible values of A.
Answer:
If A = 1 the numbers are 491 and 149 which do not have a common Tactor. If A = 2, the numbers are 492 and 249 which have 3 as a common factor.
If A = 3 the numbers are 493 and 349 which do not have a common factor.
If A = 4, the numbers are 494 and 449 which do not have a common factor.
If A = 5, the numbers are 495 and 549 which have a common factor 3.
If A = 6, the numbers are 496 and 649 8 which do not have a comon factor.
If A = 7 the numbers are 497 and 749 which have a common factor 7.
If A = 8, the numbers are 498 and 849 which have a common factor 3.
If A = 9, the numbers are 499 and 949 which do not have a. common factor,
∴ the values of K are 2, 5, 7 or 8.

KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Additional Questions

Question 5.
Write 1 to 10 using 3 and 5, each atleast once and using addition and subtraction
(For example 7 = 5 + 5 – 3)
Answer:
1 = 3 + 3 – 5
2 = 5 – 3
3 = 5 + 3 – 5
4 = 3 + 3 + 3 – 5
5 = 3 + 5 – 3
6 = 5 + 5 + 5 – 3 – 3 – 3
7 = 5 + 5 – 3
8 = 5 + 3
9 = 5 + 5 + 5 – 3 – 3
10 = 5 + 5 + 3 – 3

Question 6.
Find all 2-digit numbers each of which is divisible by the sum of its digits.
Answer:
The numbers are 10, 12, 18, 21,
24, 27, 36, 42, 45, 54, 63, 72 and 81.

Question 7.
The page numbers of a book written in a row gives a 216 digit number.
How many pages are there in the book?
Answer:
Page numbers 1 to 9 written in a row form a 9 digit numbers. The page numbers 10 to 99 (= 90 pages) form 180 digit number.
The remaining number digits i.e., (216 – (9 + 180)) = 27
27 is formed by pages 100 to 108 (9 pages)
∴ The total number of pages = 9 + ’90 + 9 = 108

KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Additional Questions

Question 8.
Look at the following pattern.
KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Additional Questions 1

This is called Pascal’s triangle. What is the middle number in the 9th row?
Answer:
Complete the pascal’s triangle
KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Additional Questions 2
The middle number in the 9th row is 70.

Question 9.
Complete the adjoining magic square (Hint. In a 3 × 3 magic square the magic sum is 3 . times the central number)
KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Additional Questions 3
Answer:
Central number is 7. The magic sum is 3 × 7 = 21
KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Additional Questions 4

Question 10.
Find all 3 digit natural numbers which are 12 times as large as the sum of their digits.
Answer:
The number 108. Because 1 + 0 + 8 = 9 and 12 × 9 = 108

KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Additional Questions

Question 11.
Find all digits x, y such that \overline{34 \times 5 y} is divisible by 36.
Answer:
A number is divisible by 36 if it is divisible by both 4 and 9 as 4 × 9 = 36.
If the last two digits number is divisible by 4 then the number is divisible by 4. So in the given number y can be 2 or 6 as 52 and 56 are divisible by 4.

If the sum of all the digits of a number is divisible by 9 then the number is di-visible by 9. In the given number 34x5y if y = 2 and x = 4 the sum of the digits . 3 + 4 + 4 + 5 + 2 = 18 which is
divisible by 9

∴ 34452 is divisible by 4, 9 and 36
Similarly if y = 6 and x = 0 and y = 6 and x = 9 the number is divisible by 36

Question 12.
Can you divide the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 into two groups
such that the product of numbers in one group divides the product of numbers in the other group and the quotient is minimum.
Answer:
Group 1 = {1, 2, 3, 4, 5, 6, 7} the product is 1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040
Group 2 = {8,9,10} the product is 8 × 9 × 10 = 720
\frac{\text { Group } 1}{\text { Group } 2}=\frac{5040}{720}=7

KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Additional Questions

Question 13.
Find all 8 digit numbers 273A49B5
which are divisible by 11 as well as 25.
Answer:
The number is divisible by 25 if the number formed by last 2 digits is divisible by 5. In the number 273A49B5, B is either 2 or 7.
Let B is 2 then the number is divisible by 11 if (sum of oddly placed digits) (sum of evenly place ! digits) = multiple of 11
∴ (2 + 3 + 4 + 2) – (7 + A + 9 + 5) => 11 – (21 + A) = multiple of 11. It is possible when A = 1
∴ The number is 27314925.
Similarly if B = 7 then A must be 6
∴ The number is 27364975

Question 14.
Suppose a, b are integers such that 2 + a and 35 – b are divisible by 11 prove that a + b is divisible by 11.
Answer:
2 + a and 35 – b
are divisible by 11
∴ 2 + a – (35 – b)
is also divisible by 11
2 + a – 35 + b = a + b – 33 is divisible by 11
∴ – 33 is divisible by 11 Hence a + b is divisible by 11

KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Additional Questions

Question 15.
In the multiplication table A8 x 3B = 2730, A and B represnet distinct digits different from 0. Find A + B.
Answer:
A8 × 3B = 2730 The units place of the product is 0 B must be 5.
The product is A8 × 35. 35 × 8 = 280.
We write 0 in the units place and 28 in carried
∴ 35 x A + 28 = 273 ⇒ 35 x A = 273 – 28
= 245 => A = \frac{245}{35} ⇒ A = 7
∴ A = 7 and B = 5
A + B = 7 + 5 = 12

Question 16.
Find the least natural number which leaves the remainder 6 and 8 when divided by 7 and 9 respectively.
Answer:
Since 7 – 6 = 1, 9 – 8 = 1
The remainder in each case in less than the devisor by 1.
Hence if 1 is added to the required number it becomes exactly divisible by 7 and 9.

The least number divisible by 7 and 9 is_ in LCM of 7 and 9 which is equal to 63
The required number is 63 – 1 = 62

KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Additional Questions

Question 17.
Prove that the sum of the cubes of three consecutive natural numbers is always divisible by 3.
Answer:
Let the 3 consecutive natural num-bers be x, x + 1 and x + 2.
Sum of their cubes
= x + (x + l)3 + (x + 2) 3= x3 + x3 + 13 + 3x(x +1) + x3 + 23 + 6x(x + 2)= x3 + x3 +1 + 3x2 + 3x + x3 + 8 + 6x2 + 12x= 3x3 + 9x2 +15x + 9 .= 3(x3 + 3x2 +5x + 3)
Sum of their cubes is a multiple of 3 which means that it is always divisible by 3.

Question 18.
What is the smallest number you have to add to 100000 to get a multiple of 1234?
Answer:
KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Additional Questions 6
∴ 1234 – 46 = 1188 should be added to 100000 to get a multiple of 1234.

Question 19.
Using the digits 4,5,6,7,8, each once, construct a 5-digit number which is divisible by 264.
Answer:
Using divisibility by 4 and 11, we get – 4 numbers.
58476, 48576, 57684, 67584.

KSEEB Solutions for Class 8 Maths Chapter 1 Playing with Numbers Additional Questions

KSEEB Solutions for Class 8 Karnataka State Syllabus

Expert Teachers at KSEEBSolutions.com has created KSEEB Solutions for Class 8 all subjects Pdf Free Download in English Medium and Kannada Medium of 8th Standard Karnataka Textbook Solutions, Textbook Questions and Answers, Guide, Notes Pdf, Model Question Papers with Answers, Study Material, are part of KSEEB Solutions. Here we have given KTBS Karnataka State Board Syllabus for Class 8 Textbook Solutions based on NCERT Syllabus.

Karnataka State Board Syllabus for Class 8 Textbooks Solutions

KSEEB Solutions for Class 8 Karnataka State Syllabus

KSEEB Solutions for Class 8 Karnataka State Syllabus
KSEEB Solutions for Class 8 Karnataka State Syllabus

We hope the given KSEEB Solutions for Class 8 all subjects Pdf Free Download in English Medium and Kannada Medium of 8th Std Karnataka Textbook Solutions, Textbook Questions and Answers, Guide, Notes Pdf, Model Question Papers with Answers, Study Material will help you. If you have any queries regarding KTBS Karnataka State Board Syllabus for Class 8 Textbooks Solutions based on NCERT Syllabus, drop a comment below and we will get back to you at the earliest.