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Exercise 10.1

Question 1.
Explain the concept of direct variation.
Solution:
If two quantifies a and b vary with each other in such a way that the ratio ab remains constant and is positive, then we say that a and b vary directly with each other or a and b are in direct variation.

Question 2.
Which of the following quantities vary directly with each other ?
(i) Number of articles (x) and their price (y).
(ii) Weight of articles (x) and their cost (y).
(iii) Distance x and time y, speed remaining the same.
(iv) Wages (y) and number of hours (x) of work.
(v) Speed (x) and time (y) (distance covered remaining the same).
(vi) Area of a land (x) and its cost (y).
Solution:
(i) It is direct variation because more articles more price and less articles, less price.
(ii) It is direct variation because, more weight more price, less weight, less price.
(iii) It is not direct variation. The distance and time vqry indirectly or inversely.
(iv) It is direct variation as more hours, more wages, less hours, less wages.
(v) It is not direct variation, as more speed, less time, less speed, more time.
(vi) It is direct variation, as more area more cost, less area, less cost.
Hence (i), (ii), (iv) and (vi) are in direct variation.

Question 3.
In which of the following tables x and y vary directly ?

Solution:

All are different.
It is not in direct variation.
Hence (i) and (ii) are in direct variation.

Question 4.
Fill in the blanks in each of the following so as to make the statement true :
(i) Two quantities are said to vary ……….. with each other if they increase (decrease) together in such a way that the ratio of the corresponding values remains same.
(ii) x and y are said to vary directly with each other if for some positive number k = k.
(iii) If u = 3v, then u and v vary ……….. with each other.
Solution:
(i) Two quantities are said to vary directly with each other if they increase (decrease) together in such a way that the ratio of the corresponding values remains same.
(ii) x and y are said to vary directly with each other if for some positive number k, xy = k.
(iii) If u = 3v, then u and v vary directly with each other.

Question 5.
Complete the following tables given that x varies directly as y.

Solution:

Question 6.
Find the constant of variation from the table given below :

Solution:

Set up a table and solve the following problems. Use unitary method to verify the answer.
Question 7.
Rohit bought 12 registers for Rs. 156, find the cost of 7 such registers.
Solution:
Price of 12 registers = Rs. 156
Let cost of 7 registers = Rs. x. Therefore

Question 8.
Anupama takes 125 minutes in walking a distance of 100 metre. What distance would she cover in 315 minutes.
Solution:
For walking 100 m, time is taken = 125 minutes
Let in 315 minutes, distance covered = m
Therefore,

Question 9.
If the cost of 93 m of a certain kind of plastic sheet is Rs. 1395, then what would it cost to buy 105 m of such plastic sheet.
Solution:
Cost of 93 m of plastic sheet = Rs. 1395
Let cost of 105 m of such sheet = Rs. x
Therefore,

Question 10.
Suneeta types 1080 words in one hour. What is GWAM (gross words a minute rate) ?
Solution:
1080 words were typed in = 1 hour = 60 minutes
Let x words will be typed in 1 minute
Therefore,

Question 11.
A car is travelling at the average speed of 50 km/hr. How much distance would it travel in 12 minutes.
Solution:
Speed of car = 50 km/hr = 50 km in 60 minutes
Let it travel x km in 12 minutes. Therefore

Question 12.
68 boxes of a certain commodity require a shelf length of 13.6 m. How many boxes of the same commodity would occupy a shelf of 20.4 m ?
Solution:
For 68 boxes of certain commodity is required a shelf length of 13.6 m
Let x boxes are require for 20.4 m shelf Then

Question 13.
In a library 136 copies of a certain book require a shelf length of 3.4 metre. How many copies of the same book would occupy a shelf-length of 5.1 metres ?
Solution:
For 136 copies of books require a shelf of length = 3.4 m
For 5.1 m shelf, let books be required = x Therefore :

Question 14.
The second class railway fare for 240 km of journey is Rs. 15.00. What would be the fare for a journey of 139.2 km ?
Solution:
Fare of second class for 240 km = Rs. 15.00
Let fare for 139.2 km journey = Rs. x
Therefore :

Question 15.
If the thickness of a pile of 12 cardboards is 35 mm, find the thickness of a pile of 294 cardboards.
Solution:
Thickness of a pile of 12 cardboards = 35 mm.
Let the thickness of a pile of 294 cardboards = x mm
Therefore :

Question 16.
The cost of 97 metre of cloth is Rs. 242.50. What length of this can be purchased for Rs. 302.50 ?
Solution:
Cost of 97 m of cloth = Rs. 242.50
Let x m can be purchase for Rs. 302.50
Therefore :

Question 17.
men can dig 634 metre long trench in one day. How many men should be employed for digging 27 metre long trench of the same type in one day ?
Solution:
11 men can dig a trench = 634 m long
Let x men will dig a trench 27 m long.
Therefore,

Question 18.
A worker is paid Rs. 210 for 6 days work. If his total income of the month is Rs. 875, for how many days did he work ?
Solution:
Payment for 6 day’s work = Rs. 210
Let payment for x day’s work = Rs. 875
Therefore :

Question 19.
A worker is paid Rs. 200 for 8 days work. If he works for 20 days, how much will he get ?
Solution:
Labour for 8 days work = Rs. 200
Let x be the labour for 20 days work, then

Question 20.
The amount of extension in an elastic string varies directly as the weight hung on it. If a weight of 150 gm produces an extension of 2.9 cm, then what weight would produce an extension of 17.4 cm ?
Solution:
150 gm of weight produces an extension = 2.9 cm
Let x gm of weight will produce an extension of 17.4 cm
Therefore :

Question 21.
The amount of extension in an elastic spring varies directly with the weight hung on it. If a weight of 250 gm produces an extension of 3.5 cm, find the extension produced by the weight of 700 gm.
Solution:
A weight of 250 gm produces an extension of 3.5 cm.
Let a weight of 700 gm will produce an extension of x cm. Therefore :

Question 22.
In 10 days, the earth picks up 2.6 x 108 pounds of dust from the atmosphere. How much dust will it pick up in 45 days.
Solution:
In 10 days dust is picked up = 2.6 x 108 pounds
Let x pounds of dust is picked up in = 45 days
Therefore,

Question 23.
In 15 days, the earth picks up 1.2 x 108 kg of dust from the atmosphere. In how many days it will pick up 4.8 x 10s kg of dust ?
Solution:
Dust of 1.2 x 108 kg is picked up in = 15 days
Let the dust of 4.8 x 108 will be picked up in x days
Therefore,

Exercise 10.2

Question 1.
In which of the following tables x and y vary inversely :



Solution:


We see that it in 15 x 4 and 3 x 25 are not equal to 36 others are 72
In it x and y do not vary.

Question 2.
It x and y vary inversely, fill in the following blanks :

Solution:

Question 3.
Which of the following quantities vary inversely as each other ?
(i) The number of x men hired to construct a wall and the time y taken to finish the job.
(ii) The length x of a journey by bus and price y of the ticket.
(iii) Journey (x km) undertaken by a car and the petrol (y litres) consumed by it.
Solution:
(i) Here x and’y var inversely
More men less time and more time less men.
(ii) More journey more price, less journey less price
x and y do not vary inversely.
(iii) More journey more petrol, less journey, less petrol
x and y do not vary inversely.
In (i) x and y, vary inversely.

Question 4.
It is known that for a given mass of gas, the volume v varies inversely as the pressure p. Fill in the missing entries in the following table :

Solution:

Question 5.
If 36 men can do a piece of work in 25 days, in how many days will 15 men do it ?
Solution:
Here less men, more days.
Let in x days, 15 men can finish the work
Therefore.

Question 6.
A work force of 50 men with a contractor can finish a piece of work in 5 months. In how many months the same work can be completed by 125 men.
Solution:
Let in x months, the work will be completed by 125 men

Question 7.
A work-force of 420 men with contractor can finish a certain piece of work in 9 months. How many extra men must he employ to complete the job in 7 months?
Solution:
Let total x men can finish the work in 7 months.
Therefore,


Total men = 540
Number of men already employed = 420
Extra men required = 540 – 420 = 120

Question 8.
1200 men can finish a stock of food in 35 days. How many more men should join them so that the same stock may last for 25 days ?
Solution:
Let x men can finish the stock, then

Total men required = 1680
Already men working = 1200
More men required = 1680 – 1200 = 480

Question 9.
In a hostel of 50 girls, there are food provisions for 40 days. If 30 more girls join the hostel. How long will these provisions last ?
Solution:
Number of girls in the beginning = 50
More girls joined = 30
Total number of girls = 50 + 30 = 80
Let the provisions last for x days.

Question 10.
A car can finish a certain journey in 10 hours at the speed of 48 km/hr. By how much should its speed be increased so that it may take only 8 hours to cover the same distance ?
Solution:
Let x km/hr be the speed. Then

Speed required = 60 km/hr.
Already speed = 48 km/hr
Speed to be increase = 60 – 48 = 12 km/hr

Question 11.
1200 soldiers in a fort had enough food for 28 days. After 4 days, some soldiers were transferred to another fort and thus the food lasted now for 32 more days. How many soldiers left the fort ?
Solution:
Period = 28 days
After 4 day, the remaining period = 28 – 4 = 24 days
In the beginning number of soldiers in the fort = 1200
Period for which the food lasted = 32 days
Let for x soldier, the food was sufficient, then

Question 12.
Three spraying machines working together can finish painting a house in 60 minutes. How long will it take 5 machines of the same capacity to do the same job ?
Solution:
Let in x minutes, 5 machines can do the work
Now

Question 13.
A group of 3 friends staying together, consume 54 kg of wheat every month. Some more friends join this group and they find that the same amount of wheat lasts for 18 days. How new many numbers are there in this group now ?
Solution:
Let x members can finish the wheat in 18 day.
Therefore :


5 member can consume the wheat
Number of members already = 3
5 – 3 = 2 more member joined them.

Question 14.
55 cows can graze a field in 16 days. How many cows will graze the same field in 10 days ?
Solution:
Let number of cows required = x
Therefore :

Question 15.
18 men can reap a field in 35 days. For reaping the same field in 15 days, how many men are required ?
Solution:
Let x men are required,
Therefore,

Question 16.
A person has money to buy 25 cycles worth Rs. 500 each. How many cycles he will be able to buy if each cycle is costing Rs. 125 more ?
Solution:
Price of one cycle = Rs. 500
Number of cycle purchased = 25
New price of the cycle = Rs. 500 + Rs. 125 = Rs. 625
Let number of cycle will be purchase = x

Question 17.
Raghu has enough money to buy 75 machines worth Rs. 200 each. How many machines can he buy if he gets a discount of Rs. 50 on each machine ?
Solution:
Price of each machine = Rs. 200
Price after given discount of Rs. 50 = Rs. 200 – 50 = Rs. 150
Let machine can be purchase = x

Number of machines can be purchased = 100

Question 18.
If x and y vary inversely as each other and
(i) x = 3 when y = 8, find y when x = 4
(ii) x = 5 when y = 15, find x when y = 12
(iii) x = 30, find y when constant of variation = 900.
(iv) y = 35, find x when constant of variation = 7.
Solution:
x and y vary inversely
x x y is constant of variation
(i) x = 3, y = 8
Constant = xy = 3 x 8 = 24

Exercise 5.1

Question 1.
Without performing actual addition and division, write the quotient when the sum of 69 and 96. is divided by
(i) 11
(ii) 15
Solution:
Two numbers are 69 and 96 whose digits are reversed Here a = 6,= 9
(i) Sum if 69 + 96 is divisible by 11, then quotient = a + 6 = 6 + 9 = 15
(ii) If it is divided by a + b i.e., 6 + 9 = 15, then quotient = 11

Question 2.
Without performing actual computations, find the quotient when 94 – 49 is divided by
(i) 9
(ii) 5
Solution:
Two given numbers are 94 and 49. Whose digits are reversed.
(i) If 94 – 49 is divided by 9, then the quotient = a-b = 9-4 = 5
(ii) and when it is divided by a – b i.e. 9-4 = 5, then quotient will be = 9

Question 3.
If sum of the number 985 and two other numbers obtained by arranging the digits of 985 in cyclic order is divided by 111, 22 and 37 respectively. Find the quotient in each case.
Solution:
The given number is 985
The other two numbers by arranging its digits
in cyclic order, will be 859, 598 of the form
abc¯¯¯¯¯¯¯,bca¯¯¯¯¯¯¯,cba¯¯¯¯¯¯¯
Therefore,
If 985 + 859 + 598 is divided by 111, then quotient will bea + 6 + c = 9 + 8 + 5 = 22
If this sum is divided by 22, then the quotient = 111
and if it is divided by 37, then quotient = 3 (a + b + c) = 3 (22) = 66

Question 4.
Find the quotient when difference of 985 and 958 is divided by 9.
Solution:
The numbers of three digits are
985 and 958 in which tens and ones digits are reversed, then
abc¯¯¯¯¯¯¯−acb¯¯¯¯¯¯¯ = 9 (b – c)
985 – 958 = 9 (8 – 5) = 9 x 3
i. e., it is divisible by 9, then quotient = b-c =8-5=3

Exercise 5.2

Question 1.
Given that the number  35a64¯¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 3, where a is a digit, what are the possible volues of a ?
Solution:
The number 35a64¯¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 3
∵The sum of its digits will also be divisible by 3
∴ 3 + 5 + a + b + 4 is divisible by 3
⇒ 18 + a is divisible by 3
⇒ a is divisible by 3 (∵ 18 is divisible by 3)
∴ Values of a can be, 0, 3, 6, 9

Question 2.
If x is a digit such that the number 18×71¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 3,’ find possible values of x.
Solution:
∵ The number 18×71¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
is divisible by 3
∴ The sum of its digits will also be divisible by 3
⇒ l + 8+ x + 7 + 1 is divisible by 3
⇒ 17 + x is divisible by 3
The sum greater than 17, can be 18, 21, 24, 27…………
∴ x can be 1, 4, 7 which are divisible by 3.

Question 3.
If is a digit of the number 66784x¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ such that it is divisible by 9, find the possible values of x.
Solution:
∵ The number 66784 x is divisible by 9
∴ The sum of its digits will also be divisible by 9
⇒ 6+6+7+8+4+x is divisible by 9
⇒ 31 + x is divisible by 9
Sum greater than 31, are 36, 45, 54………
which are divisible by 9
∴ Values of x can be 5 on 9
∴ x = 5

Question 4.
Given that the number 67y19¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 9, where y is a digit, what are the possible values of y ?
Solution:
∵ The number 67y19¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 9
∴The sum of its digits will also be divisible by 9
⇒ 6 + 7+ y+ 1+ 9 is divisible by 9
⇒ 23 + y is divisible by 9
∴ The numbers greater than 23 are 27, 36, 45,……..
Which are divisible by 9
∴y = A

Question 5.
If 3×2¯¯¯¯¯¯¯¯ is a multiple of 11, where .v is a digit, what is the value of * ?
Solution:
∵ The number 3×2¯¯¯¯¯¯¯¯ is multiple of 11
∴ It is divisible by 11
∴ Difference of the sum of its alternate digits is zero or multiple of 11
∴ Difference of (2 + 3) and * is zero or multiple of 11
⇒ If x – (2 + 3) = 0 ⇒ x-5 = 0
Then x = 5

Question 6.
If 98125×2¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ is a number with x as its tens digits such that it is divisible by 4. Find all the possible values of x.
Solution:
∵ The number 98125×2¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 4
∴ The number formed by tens digit and units digit will also be divisible by 4
∴ x2¯¯¯¯¯ is divisible by 4
∴ Possible number can be 12, 32, 52, 72, 92
∴ Value of x will be 1,3, 5, 7, 9

Question 7.
If x denotes the digit at hundreds place of the number 67×19¯¯¯¯¯¯¯¯¯¯¯¯¯ such that the
number is divisible by 11. Find all possible values of x.
Solution:
∵ The number 67×19¯¯¯¯¯¯¯¯¯¯¯¯¯ is divisible by 11
∴ The difference of the sums its alternate digits will be 0 or divisible by 11
∴ Difference of (9 + x + 6) and (1 + 7) is zero or divisible by 11
⇒ 15+x-8 = 0, or multiple of 11,
7 + x = 0 ⇒ x = -7, which is not possible
∴ 7 + x = 11, 7 + x = 22 etc.
⇒ x=11-7 = 4, x = 22 – 7
⇒ x = 15 which is not a digit
∴ x = 4

Question 8.
Find the remainder when 981547 is divided by 5. Do this without doing actual division.
Solution:
A number is divisible by 5 if its units digit is 0 or 5
But in number 981547, units digit is 7
∴ Dividing the number by 5,
Then remainder will be 7 – 5 = 2

Question 9.
Find the remainder when 51439786 is divided by 3. Do this without performing actual division.
Solution:
In the number 51439786, sum of digits is 5 + 1+ 4 + 3 + 9 + 7 + 8 + 6 = 43 and the given number is divided by 3.
∴ The sum of digits must by divisible by 3
∴ Dividing 43 by 3, the remainder will be = 1
Hence remainder = 1

Question 10.
Find the remainder, without performing actual division when 798 is divided by 11.
Solution:
Let n = 798 = a multiple of 11 + [7 + 8 – 9] 798 = a multiple of 11 + 6
∴ Remainder = 6

Question 11.
Without performing actual division, find the remainder when 928174653 is divided by 11.
Solution:
Let n = 928174653
= A multiple of 11+(9 + 8 + 7 + 6 + 3)-(2 + 1+4 + 5)
= A multiple of 11 + 33 – 12
= A multiple of 11 + 21
= A multiple of 11 + 11 + 10
= A multiple of 11 + 10
∴ Remainder =10

Question 12.
Given an example of a number which is divisible by :
(i) 2 but not by 4.
(ii) 3 but not by 6.
(iii) 4 but not by 8.
(iv) both 4 and 8 but not 32.
Solution:
(i) 2 but not by 4
A number is divisible by 2 if units do given is even but it is divisible by 4 if the number formed by tens digit and ones digit is divisible by 4.
∴ The number can be 222, 342 etc.
(ii) 3 but not by 6
A number is divisible by 3 if the sum of its digits is divisible by 3
But a number is divisible by 6, if it is divided by 2 and 3 both
∴ The numbers can be 333, 201 etc.
(iii) 4 but not by 8
A number is divisible by 4 if the number formed by the tens digit and ones digit is divisible by 4 but a number is divisible by 8, if the number formed by hundreds digit, tens digit and ones digit is divisible by 8.
∴ The number can be 244, 1356 etc.
(iv) Both 4 and 8 but not by 32
A number in which the number formed by the hundreds, tens and one’s digit, is divisible by 8 is divisible by 8. It will also divisible by 4 also.
But a number when is divisible by, 4 and 8 both is not necessarily divisible by 32 e.g., 328, 5400 etc.

Question 13.
Which of the following statements are true ?
(i) If a number is divisible by 3, it must be divisible by 9.
(ii) If a number is divisible by 9, it must be divisible by 3.
(iii) If a number is divisible by 4, it must be divisible by 8.
(iv) If a number is divisible by 8, it must be divisible by 4.
(v) A number is divisible by 18, if it is divisible by both 3 and 6.
(vi) If a number is divisible by both 9 and 10, it must be divisible by 90.
(vii) If a number exactly divides the sum of two numbers, it must exactly divide the numbers separately.
(viii) If a number divides three numbers exactly, it must divide their sum exactly.
(ix) If two numbers are co-priirie, at least one of them must be a prime number.
(x) The sum of two consecutive odd numbers is always divisible by 4.
Solution:
(i) False, it is not necessarily that it must divide by 9.
(ii) Trae.
(iii) False, it is not necessarily that it must divide by 8.
(iv) True.
(v) False, it must be divisible by 9 and 2 both.
(vi) True.
(vii) False, it is not necessarily.
(viii)True.
(ix) False. It is not necessarily.
(x) True.

Exercise 5.3

Solve each of the following cryptarithms.
Question 1.

Solution:

Values of A and B be from 0 to 9 In ten’s digit 3 + A = 9
∴ A = 6 or less.
∴ 7 + B = A = 6 or less
∴ 7 + 9 or 8 = 16 or 15
∴ But it is two digit number
B = 8
Then A = 5

Question 2.

Solution:
Values of A and B can be between 0 and 9
In tens digit, A + 3 = 9
∴ A = 9 – 3 = 6 or less than 6
In ones unit B + 7 = A = 6or less
∴ 7 + 9 or 8 = 16 or 15
But it is two digit number
∴ B = 8 and
∴ A = 5

Question 3.

Solution:

Value of A and B can be between 0 and 9 In units place.
1+B = 0 ⇒1+B = 10
∴ B = 10 – 1 = 9
and in tens place
1 + A + 1 = B ⇒ A + 2 = 9
⇒ A = 9 – 2 = 7

Question 4.

Solution:
Values of A and.B can be between 0 and 9
In units place, B+1 = 8 ⇒ B = 8-1=7
In tens place A + B= 1 or A + B = 11
⇒ A + 7 = 11 ⇒ A =11-7 = 4

Question 5.

Solution:

Values of A and B can be between 0 and 9
In tens place, 2 + A = 0 or 2 + A=10
A = 10-2 = 8
In units place, A + B = 9
⇒ 8 + B = 9 ⇒ B = 9- 8 = 1

Question 6.

Solution:

Values of A and B can be between 0 and 9
In hundreds place,

Question 7.
Show that cryptarithm 4 x AB¯¯¯¯¯¯¯¯=CAB¯¯¯¯¯¯¯¯¯¯¯ does not have any solution.
Solution:

It means that 4 x B is a numebr whose units digit is B
Clearly, there is no such digit
Hence the given cryptarithm has no solution.

Exercise 4.1

Question 1.
Find the cubes of the following numbers:
(i) 7
(ii) 12
(iii) 16
(iv) 21
(v) 40
(vi) 55
(vii) 100
(viii) 302
(ix) 301
Solution:
(i) (7)³ = 7 x 7 x 7 = 343
(ii) (12)³ = 12 x-12 x 12 = 1728
(iii) (16)³ = 16 x 16 x 16 = 4096
(iv) (21)³ = 21 x 21 x 21 = 441 x 21 =9261
(v) (40)³ = 40 x 40 x 40 = 64000
(vi) (55)³ = 55 x 55 x 55 = 3025 x 55 = 166375
(vii) (100)³ = 100 x 100 x 100 =1000000
(viii)(302)³ = 302 x 302 x 302 = 91204 x 302 = 27543608
(ix) (301)³ = 301 x 301 x 301 = 90601 x 301 =27270901

Question 2.
Write the cubes of all natural numbers between 1 and 10 and verify the following statements :
(i) Cubes of all odd natural numbers are odd.
(ii) Cubes of all even natural numbers are even.
Solution:
Cubes of first 10 natural numbers :
(1)³ = 1 x 1 x 1 = 1
(2)³ = 2 x 2 x 2 = 8
(3)³ = 3 x 3 x 3 = 27
(4)³= 4 x 4 x 4 = 64
(5)³ = 5 x 5 x 5 = 125
(6)³ = 6 x 6 x 6 = 216
(7)³ = 7 x 7 x 7 = 343
(8)³ = 8 x 8 x 8 = 512
(9)³ = 9 x 9 x 9= 729
(10)³ = 10 x 10 x 10= 1000
We see that the cubes of odd numbers is also odd and cubes of even numbers is also even.

Question 3.
Observe the following pattern :

Write the next three rows and calculate the value of 1³ + 2³ + 3³ +…. + 9³ + 10³ by the above pattern.
Solution:
We see the pattern

Question 4.
Write the cubes of 5 natural numbers which are multiples of 3 and verify the followings :
The cube of a natural number which is a multiple of 3 is a multiple of 27′
Solution:
5 natural numbers which are multiples of 3
3,6,9,12,15.
(3)³ = 3 x 3 x 3 = 27
Which is multiple of 27
(6)³ = 6 x 6 x 6 = 216 ÷ 27 = 8
Which is multiple of 27
(9)³ = 9 x 9 x 9 = 729 + 27 = 27
Which is multiple of 27
(12)³= 12 x 12 x 12 = 1728 ÷ 27 = 64
Which is multiple of 27
(15)³ = 15 x 15 x 15 = 3375 ÷ 27 = 125
Which is multiple of 27
Hence, cube of multiple of 3 is a multiple of 27

Question 5.
Write the cubes of 5 natural numbers which are of the form 3n+ 1 (e.g.,4, 7, 10, …………) and verify the following :
‘The cube of a natural number of the form 3n + 1 is a natural number of the same form i.e. when divided by 3 it leaves the remainder 1’.
Solution:
3n + 1
Let n = 1, 2, 3, 4, 5, then
If n = 1, then 3n +1= 3 x 1+1= 3+1= 4
If n = 2, then 3n +1=3 x 2+1=6+1=7
If n = 3, then 3n + 1= 3 x 3 + 1= 9 + 1 = 10
If n = 4, then 3n + 1= 3 x 4+1 = 12 + 1= 13
If n = 5, then 3n +1=3 x 5 + 1 = 15 +1 = 16
Now
(4)³ = 4 x 4 x 4 = 64
Which is 643=21, Remainder = 1
(7)³ = 7 x 7 x 7 = 343
Which is 3433 =114, Remainder = 1
(10)³ = 10 x 10 x 10 = 1000 ÷ 3 = 333, Remainder = 1
(13)³ = 13 x 13 x 13 = 2197 ÷ 3 = 732, Remainder = 1
(16)³ = 16 x 16 x 16 = 4096 ÷ 3 = 1365, Remainder = 1
Hence cube of natural number of the form, 3n + 1, is a natural of the form 3n + 1

Question 6.
Write the cubes of 5 natural numbers of the form 3n + 2 (i.e. 5, 8, 11,……… ) and verify the following :
‘The cube of a natural number of the form 3n + 2 is a natural number of the same form i.e. when it is dividend by 3 the remainder is 2’.
Solution:
Natural numbers of the form 3n + 2, when n
is a natural number i.e. 1, 2, 3, 4, 5,………….
If n = 1, then 3n + 2 = 3 x 1+2 = 3+ 2 = 5
If n = 2, then 3n + 2 = 3 x 2 + 2 = 6 + 2 = 8
If n = 3, then 3n + 2 = 3 x 3 + 2 = 9 + 2 = 11
If n = 4, then 3n + 2 = 3 x 4 + 2 = 12 + 2 = 14
and if n = 5, then 3n + 2 = 3 x 5 + 2 = 15 + 2= 17
Now (5)³ = 5 x 5 x 5 = 125
125 + 3 = 41, Remainder = 2
(8)² = 8 x 8 x 8 = 512 512 -s- 3 = 170, Remainder = 2
(11)³ = 11 x 11 x 11 = 1331
1331 + 3 = 443, Remainder = 2
(14)³ = 14 x 14 x 14 = 2744
2744 + 3 = 914, Remainder = 2
(17)³ = 17 x 17 x 17 = 4913
4913 = 3 = 1637, Remainder = 2
We see the cube of the natural number of the
form 3n + 2 is also a natural number of the
form 3n + 2.

Question 7.
Write the cubes of 5 natural numbers of which are multiples of 7 and verify the following :
‘The cube of a multiple of 7 is a multiple of 7^3′.
Solution:
5 natural numbers which are multiple of 7,are 7, 14, 21, 28, 35
(7)³ = (7)³ which is multiple of 7³
(14)³ = (2 x 7)3 = 2³ x 7³, which is multiple of 7³
(21)³ = (3 x 7)3 = 33 x 7³_, which is multiple of 7³
(28)³ = (4 x 7)³ = 4³ x 7³, which is multiple of 7³ (35)³ = (5 x 7)³ = 5³ x 73 which is multiple of 7³
Hence proved.

Question 8.
Which of the following are perfect cubes?
(i) 64
(ii) 216
(iii) 243
(iv) 1000
(v) 1728
(vi) 3087
(vii) 4608
(viii) 106480
(ix) 166375
(x) 456533
Solution:
(i) 64 = 2 x 2 x 2 x 2 x 2 x 2

Grouping the factors in triplets of equal factors, we see that no factor is left
∴ 64 is a perfect cube
(ii) 216 = 2 x 2 x 2 x 3 x 3 x 3

Grouping the factors in triplets of equal factors, we see that no factor is left
216 is a perfect cube.
(iii) 243 = 3 x 3 x 3 x 3 x 3

Grouping the factors in triplets, we see that two factors 3 x 3 are left
∴ 243 is not a perfect cube.
(iv) 1000 = 2 x 2 x 2 x 5 x 5 x 5

Grouping the factors in triplets of equal factors, we see that no factor is left
∴ 1000 is a perfect cube.
(v) 1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3

Grouping the factors in triplets of the equal factors, we see that no factor is left
∴ 1728 is a perfect cube,
(vi) 3087 = 3 x 3 x 7 x 7 x 7

Grouping the factors in triplets of the equal factors, we see that two factor 3×3 are left
∴ 3087 is not a perfect cube.
(vii) 4608 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3

Grouping the factors in triplets of equal factors, we see that two factors 3, 3 are left
∴ 4609 is not a perfect cube.
(viii) 106480 = 2 x 2 x 2 x 2 x 5 x 11 x 11 x 11

Grouping the factors in triplets of equal factors, we see that factors 2, 5 are left
∴ 106480 is not a perfect cube.
(ix) 166375 = 5 x 5 x 5 x 11 x 11 x 11

Grouping the factors in triplets of equal factors, we see that no factor is left
∴ 166375 is a perfect cube.
(x) 456533 = 7 x 7 x 7 x 11 x 11 x 11

Grouping the factors in triplets of equal factors, we see that no factor is left
∴ 456533 is a perfect cube.

Question 9.
Which of the following are cubes of even natural numbers ?
216, 512, 729,1000, 3375, 13824
Solution:
We know that the cube of an even natural number is also an even natural number
∴ 216, 512, 1000, 13824 are even natural numbers.
∴ These can be the cubes of even natural number.

Question 10.
Which of the following are cubes of odd natural numbers ?
125, 343, 1728, 4096, 32768, 6859
Solution:
We know that the cube of an odd natural number is also an odd natural number,
∴ 125, 343, 6859 are the odd natural numbers
∴ These can be the cubes of odd natural numbers.

Question 11.
What is the smallest number by which the following numbers must be multiplied, so that the products are perfect cubes ?
(i) 675
(ii) 1323
(iii) 2560
(iv) 7803
(v) 107311
(vi) 35721
Solution:
(i) 675 = 3 x 3 x 3 x 5 X 5

Grouping the factors in triplet of equal factors, 5 x 5 are left without triplet
So, by multiplying by 5, the triplet will be completed.
∴ Least number to be multiplied = 5
(ii) 1323 = 3 x 3 x 3 x 7 x 7

Grouping the factors in triplet of equal factors. We find that 7 x 7 has been left
So, multiplying by 7, we get a triplet
∴ The least number to be multiplied = 7
(iii) 2560 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5

Grouping the factors in triplet of equal factors, 5 is left.
∴ To complete a triplet 5 x 5 is to multiplied
∴ Least number to be multiplied = 5 x 5 = 25
(iv) 7803 = 3 x 3 x 3 x 17 x 17

Grouping the factors in triplet of equal factors, we find the 17 x 17 are left
So, to complete the triplet, we have to multiply by 17
∴ Least number to be multiplied = 17
(v) 107811 = 3 x 3 x 3 x 3 x 11 x 11 x 11

Grouping the factors in triplet of equal factors, factor 3 is left
So, to complete the triplet 3 x 3 is to be multiplied
∴ Least number to be multiplied = 3 x 3 = 9
(vi) 35721 = 3 x 3 x 3 x 3 x 3 x 3 x 7 x 7

Grouping the factors in triplet of equal factors, we find that 7 x 7 is left
So, in order to complete the triplets, we have to multiplied by 7
∴ Least number to be multiplied = 713&ifi=8&uci=a!8&btvi=7&fsb=1&xpc=6d2Qjo9kVP&p=https%3A//www.learninsta.com&dtd=6298

Question 12.
By which smallest number must the following numbers be divided so that the quotient is a perfect cube ?
(i) 675
(ii) 8640
(iii) 1600
(iv) 8788
(v) 7803
(vi) 107811
(vii) 35721
(viii) 243000
Solution:
(i) 675 = 3 x 3 x 3 x 5 x 5

Grouping the factors in triplet of equal factors, 5 x 5 is left
5 x 5 is to be divided so that the quotient will be a perfect cube.
∴ The least number to be divided = 5 x 5 = 25
(ii) 8640 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3

Grouping the factors in triplets of equal factors, 5 is left
∴ In order to get a perfect cube, 5 is to divided
∴ Least number to be divided = 5
(iii) 1600 = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5

Grouping the factors in triplets of equal factors, we find that 5 x 5 is left
∴ In order to get a perfect cube 5 x 5 = 25 is to be divided.
∴ Least number to be divide = 25
(iv) 8788 = 2 x 2 x 13 x 13 x 13

Grouping the factors in triplets of equal factors, we find that 2 x 2 has been left
∴ In order to get a perfect cube, 2 x 2 is to be divided
∴ Least number to be divided = 4
(v) 7803 = 3 x 3 x 3 x 17 x 17

Grouping the factors in triplets of equal factors, we see that 17 x 17 has been left.
So, in order to get a perfect cube, 17 x 17 is be divided
∴ Least number to be divided = 17 x 17 = 289
(vi) 107811 = 3 x 3 x 3 x 3 x 11 x 11 x 11

Grouping the factors in triplets of equal factors, 3 is left
∴ In order to get a perfect cube, 3 is to be divided
∴ Least number to be divided = 3
(vii) 35721 = 3 x 3 x 3 x 3 x 3 x 3 x 7 x 7

Grouping the factors in triplets of equal factors, we see that 7 x 7 is left
So, in order to get a perfect cube, 7 x 7 = 49 is to be divided
∴ Least number to be divided = 49
(viii) 243000 = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 5 x 5 x 5

Grouping the factors in triplets of equal factors, 3 x 3 is left
∴ By dividing 3 x 3, we get a perfect cube
∴ Least number to be divided = 3 x 3=9

Question 13.
Prove that if a number is trebled then its cube is 27 times the cube of the given number.
Solution:
Let x be the number, then trebled number of x = 3x
Cubing, we get:
(3x)³ = (3)³ x³ = 27x³
27x³ is 27 times the cube of x i.e., of x³

Question 14.
What happenes to the cube of a number if the number is multiplied by
(i) 3 ?
(ii) 4 ?
(iii) 5 ?
Solution:
number (x)³ = x³
(i) If x is multiplied by 3, then the cube of
∴ (3x)³ = (3)³ x x³ = 27x³
∴ The cube of the resulting number is 27 times of cube of the given number
(ii) If x is multiplied by 4, then the cube of
(4x)³ = (4)³ x x³ = 64x³
∴ The cube of the resulting number is 64 times of the cube of the given number
(ii) If x is multiplied by 5, then the cube of
(5x)³ = (5)³ x x³ = 125x³
∴ The cube of the resulting number is 125 times of the cube of the given number

Question 15.
Find the volume of a cube, one face of which has an area of 64 m².
Solution:
Area of one face of a cube = 64 m²
∴ Side (edge) of cube = √64
= √64 = 8 m
∴ Volume of the cube = (side)³ = (8 m)³ = 512 m³

Question 16.
Find the volume of a cube whose surface area is 384 m².
Solution:
Surface area of a cube = 384 m² Let side = a
Then 6a² = 384 ⇒ a² = 3846= 64 = (8)²
∴ a = 8 m
Now volume = a³ = (8)³ m³ = 512 m³

Question 17.
Evaluate the following :

Solution:

Question 18.
Write the units digit of the cube of each of the following numbers :
31,109,388,833,4276,5922,77774,44447, 125125125.
Solution:
We know that if unit digit of a number n is
= 1, then units digit of its cube = 1
= 2, then units digit of its cube = 8
= 3, then units digit of its cube = 7
= 4, then units digit of its cube = 4
= 5, then units digit of its cube = 5
= 6, then units digit of its cube = 6
= 7, then the units digit of its cube = 3
= 8, then units digit of its cube = 2
= 9, then units digit of its cube = 9
= 0, then units digit of its cube = 0
Now units digit of the cube of 31 = 1
Units digit of the cube of 109 = 9
Units digits of the cube of 388 = 2
Units digits of the cube of 833 = 7
Units digits of the cube of 4276 = 6
Units digit of the cube of 5922 = 8
Units digit of the cube of 77774 = 4
Units digit of tl. cube of 44447 = 3
Units digit of the cube of 125125125 = 5

Question 19.
Find the cubes of the following numbers by column method :
(i) 35
(ii) 56
(iii) 72
Solution:

Question 20.
Which of the following numbers are not perfect cubes ?
(i) 64
(ii) 216
(iii) 243
(iv) 1728
Solution:
(i) 64 = 2 x 2 x 2 X 2 x 2 x 2

Grouping the factors in triplets, of equal factors, we see that no factor is left
∴ 64 is a perfect cube.
(ii) 216 = 2 x 2 x 2 x 3 x 3 x 3

Grouping the factors in triplets, of equal factors, we see that no factor is left
∴ 216 is a perfect cube.
(iii) 243 = 3 x 3 x 3 x 3 x 3

Grouping the factors in triplets, of equal factors, we see that 3 x 3 are left
∴ 243 is not a perfect cube.
(iv) 1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3

Grouping the factors m triplets, of equal factors, we see that no factor is left.
∴ 1728 is a perfect cube.

Question 21.
For each of the non-perfect cubes, in Q. 20, find the smallest number by which it must be
(a) multiplied so that the product is a perfect cube.
(b) divided so that the quotient is a perfect cube.
Solution:
In qustion 20, 243 is not a perfect cube and 243 = 3 x 3 x 3 x 3 x 3
Grouping the factors in triplets, of equal factors, we see that 3 x 3 is left.
(a) In order to make it a perfect cube, 3 is to be multiplied which makes a triplet.
(b) In order to make it a perfect cube, 3 x 3 or 9 is to be divided.

Question 22.
By taking three different values of n verify the truth of the following statements :
(i) If n is even, then n³ is also even.
(ii) If n is odd, then n³ is also odd.
(iii) If n leaves remainder 1 when divided by 3, then it3 also leaves 1 as remainder when divided by 3.
(iv) If a natural number n is of the form 3p + 2 then n³ also a number of the same type.
Solution:
(i) n is even number.
Let n = 2, 4, 6 then
(a) n³ = (2)³ = 2 x 2 x 2 = 8, which is an even number.
(b) (n)³= (4)³ = 4 x 4 x 4 = 64, which is an even number.
(c) (n)³ = (6)³ = 6 x 6 x 6 = 216, which is an even number.

(ii) n is odd number.
Letx = 3, 5, 7
(a) (n)³ = (3)³ = 3 x 3 x 3 = 27, which is an odd number.
(b) (n)³ = (5)³ = 5 x 5 x 5 = 125, which is an odd number.
(c) (n)³ = (7)³ = 7 x 7 x 7 = 343, which is an odd number.

(iii) If n leaves remainder 1 when divided by 3, then n³ is also leaves 1 as remainder,
Let n = 4, 7, 10 If n = 4,
then «³ = (4)³ = 4 x 4 x 4 = 64
= 64 + 3 = 21, remainder = 1
If n = 7, then
n³ = (7)³ = 7 x 7 x 7 = 343
343 + 3 = 114, remainder = 1
If n – 10, then
(n)³ = (10)³ = 10 x 10 x 10 = 1000
1000 + 3 = 333, remainder = 1

(iv) If the natural number is of the form 3p + 2, then n³ is also of the same type
Let p =’1, 2, 3, then
(a) If p = 1, then
n = 3p + 2 = 3 x 1+2=3+2=5
∴ n³ = (5)³ = 5 x 5 x 5 = 125
125 = 3 x 41 + 2 = 3p +2

(b) If p = 2, then
n = 3p + 2 = 3 x 2 + 2 = 6 + 2 = 8
∴ n³ = (8)³ = 8 x 8 x 8 = 512
∴ 512 = 3 x 170 + 2 = 3p + 2

(c) If p = 3, then
n = 3p + 2 = 3 x 3 + 2 = 9 + 2 = 11
∴ n³ = (11)³ = 11 x 11 x 11 = 1331
and 1331 =3 x 443 + 2 = 3p + 2
Hence proved.

Exercise 4.2

Question 1.
Find the cubes of:
(i) -11
(ii) -12
(iii) – 21
Solution:
(i) (-11)³=(-11)³=(11 x 11 x 11) =-1331
(ii) (-12)³=(-12)³=(12 x 12 x 12) =  -1728
(iii) (-21)³=(-21)³=(21 x 21 x 21) = -9261

Question 2.
Which of the following numbers are cubes of negative integers.
(i) -64
(ii) -1056
(iii) -2197
(iv) -2744
(v)  -42875
Solution:


∴ All factors of 64 can be grouped in triplets of the equal factors completely.
∴ -64 is a perfect cube of negative integer.

All the factors of 1056 can be grouped in triplets of equal factors grouped completely
∴ 1058 is not a perfect cube of negative integer.

All the factors of -2197 can be grouped in triplets of equal factors completely
∴ 2197 is a perfect cube of negative integer,

All the factors of -2744 can be grouped in triplets of equal factors completely
∴ 2744 is a perfect cube of negative integer

All the factors of -42875 can be grouped in triplets of equal factors completely
∴ 42875 is a perfect cube of negative integer.

Question 3.
Show that the following integers are cubes of negative integers. Also, find the integer whose cube is the given integer :
(i) -5832 (ii) -2744000
Solution:


Grouping the factors in triplets of equal factors, we see that no factor is left
∴ -5832 is a perfect cube
Now taking one factor from each triplet we find that
-5832 is a cube of – (2 x 3 x 3) = -18
∴ Cube root of-5832 = -18

Grouping the factors in tuplets of equal factors, we see that no factor is left. Therefore it is a perfect cube.
Now taking one factor from each triplet, we find that.
-2744000 is a cube of – (2 x 2 x 5 x 7) ie. -140
∴ Cube root of -2744000 = -140

Question 4.
Find the cube of :

Solution:

Question 5.
Which of the following numbers are cubes of rational numbers :

Solution:

Exercise 4.3

Question 1.
Find the cube roots of the following numbers by successive subtraction of numbers : 1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397,
(i) 64
(ii) 512
(iii) 1728
Solution:
(i) 64
64 – 1 = 63
63 – 7 = 56
56 – 19 = 37
37 – 37 = 0
∴ 64 = (4)³
∴ Cube root of 64 = 4

(ii) 512
512 -1 =511
511- 7 = 504
504 – 19 = 485
485 – 37 = 448
448 – 61 = 387
387 – 91 =296
296 – 127 = 169
169 – 169 = 0
∴ 512 = (8)³
∴ Cube root of 512 = 8

(iii) 1728
1728 – 1= 1727
1727 -7 = 1720
1720 -19 = 1701
1701 -37= 1664
1664 – 61 = 1603
1603 – 91 = 1512
1512 -127= 1385 .
1385 – 169= 1216
1216 – 217 = 999
999 – 271 =728
728 – 331 = 397
397 – 397=0
∴ 1728 = (12)³
∴ Cube root of 1728 = 1²

Question 2.
Using the method of successive subtraction, examine whether or not the following numbers are perfect cubes :
(i) 130
(ii) 345
(iii) 792
(iv) 1331
Solution:
(i) 130
130 – 1 = 129
129 -7 = 122
122 -19 = 103
103 -37 = 66
66 – 61 = 5
We see that 5 is left
∴ 130 is not a perfect cube.

(ii) 345
345 – 1 = 344
344 – 7 = 337
337 – 19 = 318
318 – 37 = 281
81 – 61 =220
220 – 91 = 129
129 – 127 = 2
We see that 2 is left
∴ 345 is not a perfect cube.

(iii) 792
792 – 1 = 791
791 – 7 = 784
784 – 19 = 765
765 – 37 = 728
728 – 61 = 667
667 – 91 = 576
576 – 127 = 449
449 – 169 = 280
∴ We see 280 is left as 280 <217
∴ 792 is not a perfect cube.

(iv) 1331
1331 – 1 = 1330
1330 -7 = 1323
1323 – 19 = 1304
1304 – 37 = 1267
1267 – 61 = 1206
1206 – 91 = 1115
1115 – 127 = 988
988 – 169 = 819
819 – 217 = 602
602 – 271 = 331
331 – 331 =0
∴ 1331 is a perfect cube

Question 3.
Find the smallest number that must be subtracted from those of the numbers in question 2, which are not perfect cubes, to make them perfect cubes. What are the corresponding cube roots ?
Solution:
We have examined in Question 2, the numbers 130, 345 and 792 are not perfect cubes. Therefore
(i) 130
130 – 1 = 129
129 -7= 122
122 -19 = 103
103 – 37 = 66
66 – 61 = 5
Here 5 is left
∴ 5 < 91 5 is to be subtracted to get a perfect cube.
Cube root of 130 – 5 = 125 is 5

(ii) 345
345 – 1 = 344
344 -7 = 337
337 – 19 = 318
318 – 37 = 281
281 – 61 =220
220 – 91 = 129
129 – 127 = 2
Here 2 is left ∵ 2 < 169
∴ Cube root of 345 – 2 = 343 is 7
∴ 2 is to be subtracted to get a perfect cube.

(iii) 792
792 – 1 = 791
791 – 7 = 784
784 – 19 = 765
765 – 37 = 728
728 – 61 =667
667 – 91 = 576 5
76 – 127 = 449
449 – 169 = 280
280 – 217 = 63
∴ 63 <217
∴ 63 is to be subtracted
∴ Cube root of 792 – 63 = 729 is 9

Question 4.
Find the cube root of each of the following natural numbers :
(i) 343
(ii) 2744
(iii) 4913
(iv) 1728
(v) 35937
(vi) 17576
(vii) 134217728
(viii) 48228544
(ix) 74088000
(x) 157464
(xi) 1157625
(xii) 33698267
Solution:

Question 5.
Find the smallest number which when multiplied with 3600 will make the product a perfect cube. Further, find the cube root of the product.
Solution:

Grouping the factors in triplets of equal factors, we see that 2, 3 x 3 and 5 x 5 are left
∴ In order to complete the triplets, we have to multiply it by 2, 3 and 5.
∴ The smallest number to be multiplied = 2×2 x 3 x 5 = 60
Now product = 3600 x 60 = 216000 and cube root of 216000

Question 6.
Multiply 210125 by the smallest number so that the product is a perfect cube. Also, find out the cube root of the product.
Solution:

Grouping the factors in triplets of equal factors, we see that 41 x 41 is left
∴ In order to complete the triplet, we have to multiply it by 41
∴ Smallest number to be multiplied = 41
∴ Product = 210125 x 41 = 8615125
∴ Cube root of 8615125

Question 7.
What is the smallest number by which 8192 must be divided so that quotient is a perfect cube ? Also, find the cube root of the quotient so obtained.
Solution:

Grouping the factors in triplets of equal factors, we see that 2 is left
∴ Dividing by 2, we get the quotient a perfect cube
∴ Perfect cube = 8192 + 2 = 4096

Question 8.
Three numbers are in the ratio 1:2:3. The sum of their cubes is 98784. Find the numbers.
Solution:
Ratio in numbers =1:2:3
Let first number = x
Then second number = 2x
and third number = 3x
∴ Sum of cubes of there numbers = (x)³ + (2x)³+(3x)³

Question 9.
The volume of a cube is 9261000 m³. Find the side of the cube.
Solution:

Exercise 4.4

Question 1.
Find the cube roots of each of the following integers :
(i) -125
(ii) -5832
(iii) -2744000
(iv) -753571
Solution:

Question 2.
Show that :

Solution:

Question 3.
Find the cube root of each of the following numbers :
(i) 8 x 125
(ii) -1728 x 216
(iii) -27 x 2744
(iv) -729 x -15625
Solution:

Question 4.
Evaluate :

Solution:

Question 5.
Find the cube root of each of the following rational numbers.

Solution:

Question 6.
Find the cube root of each of the following rational numbers :
(i) 0.001728
(ii) 0.003375
(iii) 0.001
(iv) 1.331
Solution:

Question 7.
Evaluate each of the following :


Solution:

Question 8.
Show that :

Solution:

Question 9.
Fill in the Blanks :

Solution:

Question 10.
The volume of a cubical box is 474.552 cubic metres. Find the length of each side of the box.
Solution:

Question 11.
Three numbers are to one another 2:3: 4. The sum of their cubes is 0.334125. Find the numbers.
Solution:

Question 12.
Find side of a cube whose volume is

Solution:

Question 13.
Evaluate :

Solution:

Question 14.
Find the cube root of the numbers : 2460375,20346417,210644875,57066625 using the fact that
(i) 2460375 = 3375 x 729
(i) 20346417 = 9261 x 2197
(iii) 210644875 = 42875 x 4913
(iv) 57066625 = 166375 x 343
Solution:

Question 15.
Find the units digit of the cube root of the following numbers ?
(i) 226981
(ii) 13824
(iii) 571787
(iv) 175616
Solution:
(i) 226981
In it unit digit is 1
∴The units digit of its cube root will be = 1
(∵ 1 x 1 x 1 = 1)
∴Tens digit of the cube root will be = 6
(ii) 13824
∵ The units digit of 13824 = 4
(∵ 4 X 4 X 4 = 64)
∴Units digit of the cube root of it = 4
(iii) 571787
∵ The units digit of 571787 is 7
∴The units digit of its cube root = 3
(∵ 3 x 3 x 3 = 27)
(iv) 175616
∵ The units digit of 175616 is 6
∴The units digit of its cube root = 6
(∵ 6 x 6 x 6=216)

Question 16.
Find the tens digit of the cube root of each of the numbers in Question No. 15.
Solution:
(i) In 226981
∵ Units digit is 1
∴Units digit of its cube root = 1
We have 226
(Leaving three digits number 981)
6³ = 216 and 7³ = 343
∴6³ ∠226 ∠ T
∴The ten’s digit of cube root will be 6
(ii) In 13824
Leaving three digits number 824, we have 13
∵ (2)³ = 8, (3)³ = 27
∴2³ ∠13 ∠3′
∴Tens digit of cube root will be 2
(iii) In 571787
Leaving three digits number 787, we have 571
8³ = 512, 9³ = 729
∴ 8³ ∠571 ∠9³
Tens digit of the cube root will be = 8
(iv) In 175616
Leaving three digit number 616, we have 175
∵ 5³ = 125, 6³ = 216
∴5³ ∠175 ∠6³
∴Tens digit of the cube root will be = 5

Exercise 4.5

Making use of the cube root table, find the cubes root of the following (correct to three decimal places) 
Question 1.
7
Solution:
7–√3 =1.913 (From the table)

Question 2.
70
Solution:
70−−√3 =4.121 (From the table)

Question 3.
700
Solution:
700−−−√3=7×100−−−−−−√3= 8.879 (from 10x−−−√3)

Question 4.
7000
Solution:
7000−−−−√3=70×100−−−−−−−√3 = 19.13 (from 100x−−−−√3)

Question 5.
1100
Solution:
1100−−−−√3=11×100−−−−−−−√3 = 10.32 (from 100x−−−−√3)

Question 6.
780
Solution:
780−−−√3=78×100−−−−−−−√3 = 9.205 (from 10x−−−√3)

Question 7.
7800
Solution:
7800−−−−√3=78×100−−−−−−−√3 = 19.83 (from 100x−−−−√3)

Question 8.
1346
Solution:

Question 9.
250
Solution:

Question 10.
5112
Solution:

Question 11.
9800
Solution:

Question 12.
732
Solution:

Question 13.
7342
Solution:

Question 14.
133100
Solution:

Question 15.
37800
Solution:

Question 16.
0.27
Solution:

Question 17.
8.6
Solution:

Question 18.
0.86
Solution:

Question 19.
8.65
Solution:

Question 20.
7532
Solution:

Question 21.
833
Solution:

Question 22.
34.2
Solution:

Question 23.
What is the length of the side of a cube whose volume is 275 cm³. Make use of the table for the cube root.
Solution:

Exercise 3.1

Question 1.
Which of the following numbers are perfect squares ?
(i)484
(ii) 625
(iii) 576
(iv) 941
(v) 961
(vi) 2500
Solution:

Grouping the factors in pairs, we have left no factor unpaired
∴ 484 is a perfect square of 22

∴ Grouping the factors in pairs, we have left no factor unpaired
∴ 625 is a perfect square of 25.

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 576 is a perfect square of 24
(iv) 941 has no prime factors

∴ 941 is not a perfect square.
(v) 961 =31 x 31
Grouping the factors in pairs, we see that no factor is left unpaired
∴ 961 is a perfect square of 31

Grouping the factors in pairs, we see that no factor is left impaired
∴ 2500 is a perfect square of 50 .

Question 2.
Show that each of the following* numbers is a perfect square. Also find the number whose square is the given number in each case :
(i) 1156
(ii) 2025
(iii) 14641
(iv) 4761
Solution:

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 1156 is a perfect square of 2 x 17 = 34

Grouping the factors in pairs, we see that no factor is left unpaired
2025 is a perfect square of 3 x 3 x 5 =45

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 14641 is a perfect square of 11×11 = 121

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 4761 is a perfect square of 3 x 23 = 69

Question 3.
Find the smallest number by which the given number must be multiplied so that the product is a perfect square.
(i) 23805
(ii) 12150
(iii) 7688
Solution:

Grouping the factors in pairs of equal factors, we see that 5 is left unpaird
∴ In order to complete the pairs, we have to multiply 23805 by 5, then the product will be the perfect square.
Requid smallest number = 5
(ii) 12150 = 2 x 3 x 3×3 x 3×3 x 5×5

Grouping the factors in pairs of equal factors, we see that factors 2 and 3 are left unpaired
∴ In order to complete the pairs, we have to multiply 12150 by 2 x 3 =6 i.e., then the product will be the complete square.
∴ Required smallest number = 6

Grouping the factors in pairs of equal factors, we see that factor 2 is left unpaired
∴ In order to complete the pairs we have to multiply 7688 by 2, then the product will be the complete square
∴ Required smallest number = 2

Question 4.
Find the smallest number by which the given number must be divided so that the resulting number is a perfect square.
(i) 14283
(ii) 1800
(iii) 2904
Solution:

Grouping the factors in pairs of equal factors, we see that factors we see that 3 is left unpaired
Deviding by 3, the quotient will the perfect square.

Grouping the factors in pair of equal factors, we see that 2 is left unpaired.
∴ Dividing by 2, the quotient will be the perfect square.

Grouping the factors in pairs of equal factors, we see that 2 x 3 we left unpaired
∴ Dividing by 2 x 3 = 6, the quotient will be the perfect square.

Question 5.
Which of the following numbers are perfect squares ?
11, 12, 16, 32, 36, 50, 64, 79, 81, 111, 121
Solution:
11 is not a perfect square as 11 = 1 x 11
12 is not a perfect square as 12 = 2×2 x 3
16 is a perfect square as 16 = 2×2 x 2×2
32 is not a perfect square as 32 = 2×2 x 2×2 x 2
36 is a perfect square as 36 = 2×2 x 3×3
50 is not a perfect square as 50 = 2 x 5×5
64 is a perfect square as 64 = 2×2 x 2×2 x 2×2
79 is not a perfect square as 79 = 1 x 79
81 is a perfect square as 81 = 3×3 x 3×3
111 is not a perfect square as 111 = 3 x 37
121 is a perfect square as 121 = 11 x 11
Hence 16, 36, 64, 81 and 121 are perfect squares.

Question 6.
Using prime factorization method, find which of the following numbers are perfect squares ?
∴ 189,225,2048,343,441,2916,11025,3549
Solution:

Grouping the factors in pairs, we see that are 3 and 7 are left unpaired
∴ 189 is not a perfect square

Grouping the factors in pairs, we see no factor left unpaired
∴ 225 is a perfect square

Grouping the factors in pairs, we see no factor left unpaired
∴ 2048 is a perfect square

Grouping the factors in pairs, we see that one 7 is left unpaired
∴ 343 is not a perfect square.

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 441 is a perfect square.

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 2916 is a perfect square.

Grouping the factors in pairs, we see that no factor is left unpaired
∴ 11025 is a perfect square.

Grouping the factors in pairs, we see that 3, no factor 7 are left unpaired
∴ 3549 is a perfect square.

Question 7.
By what number should each of the following numbers be multiplied to get a perfect square in each case ? Also, find the number whose square is the new number.
(i) 8820
(ii) 3675
(iii) 605
(iv) 2880
(v) 4056
(vi) 3468
Solution:

Grouping the factors in pairs, we see that 5 is left unpaired
∴ By multiplying 8820 by 5, we get the perfect square and square root of product will be
= 2 x 3 x 5 x 7 = 210

Grouping the factors in pairs, we see that 3 is left unpaired
∴ Multiplying 3675 by 3, we get a perfect square and square of the product will be
= 3 x 5 x 7 = 105

Grouping the factors in pairs, we see that 5 is left unpaired
∴ Multiplying 605 by 5, we get a perfect square and square root of the product will be
= 5 x 11 =55

Grouping the factors in pairs, we see that 5 is left unpaired
∴ Multiplying 2880 by 5, we get the perfect square.
Square rooi of product will be = 2 x 2 * 2 – 3 x 5 = 120

Grouping the factors in pairs, we see that 2 and 3 are left unpaired
∴ Multiplying 4056 by 2 x 3 i.e., 6, we get the perfect square.
and square root of the product will be
= 2 x 2 x 3 x 13 = 156

Grouping the factors in pairs, we see that 3 is left unpaired
∴ Multiplying 3468 by 3 we get a perfect square, and square root of the product will be 2 x 3 x 17 = 102

Grouping the factors in pairs, we see that 2 and 3 are left unpaired
∴ Multiplying 7776 by 2 x 3 or 6 We get a perfect square and square root of the product will be
= 2 x 2 x 2 x 3 x 3 x 3 = 216

Question 8.
By what numbers should each of the following be .divided to get a perfect square in each case ? Also find the number whose square is the new number.
(i) 16562
(ii) 3698
(iii) 5103
(iv) 3174
(v) 1575
Solution:

Grouping the factors in pairs, we see that 2 is left unpaired
∴ Dividing by 2, we get the perfect square and square root of the quotient will be 7 x 13 = 91

Grouping the factors in pairs, we see that 2 is left unpaired,
∴ Dividing 3698 by 2, the quotient is a perfect square
and square of quotient will be = 43

Grouping the factors in pairs, we see that 7 is left unpaired
∴ Dividing 5103 by 7, we get the quotient a perfect square.
and square root of the quotient will be 3 x 3 x 3 = 27

Grouping the factors iq pairs, we see that 2 and 3 are left unpaired
∴ Dividing 3174 by 2 x 3 i.e. 6, the quotient will be a perfect square and square root of the quotient will be = 23

Grouping the factors in pairs, we find that 7 is left unpaired i
∴ Dividing 1575 by 7, the quotient is a perfect square
and square root of the quotient will be = 3 x 5 = 15

Question 9.
Find the greatest number of two digits which is a perfect square.
Solution:
The greatest two digit number = 99 We know, 92 = 81 and 102 = 100 But 99 is in between 81 and 100
∴ 81 is the greatest two digit number which is a perfect square.

Question 10.
Find the least number of three digits which is perfect square.
Solution:
The smallest three digit number =100
We know that 92 = 81, 102 = 100, ll2 = 121
We see that 100 is the least three digit number which is a perfect square.

Question 11.
Find the smallest number by which 4851 must be multiplied so that the product becomes a perfect square.
Solution:
By factorization:

Grouping the factors in pairs, we see that 11 is left unpaired
∴ The least number is 11 by which multiplying 4851, we get a perfect square.

Question 12.
Find the smallest number by which 28812 must be divided so that the quotient becomes a perfect square.
Solution:
By factorization,

Grouping the factors in pairs, we see that 13 is left unpaired
∴ Dividing 28812 by 3, the quotient will be a perfect square.

Question 13.
Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also find the number whose square is the resulting number.
Solution:
By factorization,

Grouping the factors in pairs, we see that one 2 is left unpaired.
∴ Dividing 1152 by 2, we get the perfect square and square root of the resulting number 576, will be 2 x 2 x 2 x 3 = 24

Exercise 3.2

Question 1.
The following numbers are not perfect squares. Give reason :
(i) 1547
(ii) 45743
(iii) 8948
(iv) 333333

Solution:
We know that if the units digit is 2, 3, 7 or 8 of a number, then the number is not a perfect square.

(i) ∴ 1547 has 7 as units digit.
∴ It is not a perfect square.

(ii) 45743 has 3 as units digit
∴ It is not a perfect square.

(iii)  ∴ 8948 has 8 as units digit
∴ It is not a perfect square.

(iv)  ∴ 333333 has 3 as units digits
∴ It is not a perfect square.

Question 2.
Show that the following numbers are not perfect squares :
(i) 9327
(ii) 4058
(iii) 22453
(iv) 743522

Solution:
(i) 9327
∴ The units digit of 9327 is 7
∴ This number can’t be a perfect square.

(ii) 4058
∴ The units digit of 4058 is 8
∴ This number can’t be a perfect square.

(iii) 22453
∴ The units digit of 22453 is 3
.∴ This number can’t be a perfect square.

(iv) 743522
∴ The units digit of 743522 is 2
∴ This number can’t be a perfect square.

Question 3.
The square of which of the following numbers would be an odd number ?
(i) 731
(ii) 3456
(iii) 5559
(iv) 42008
Solution:
We know that the square of an odd number is odd and of even number is even. Therefore
(i) Square of 731 would be odd as it is an odd number.
(ii) Square of 3456 should be even as it is an even number.
(iii) Square of 5559 would be odd as it is an odd number.
(iv) The square of 42008 would be an even number as it is an even number.
Therefore suqares of (i) 731 and (ii) 5559 will be odd numbers.

Question 4.
What will be the units digit of the squares of the following numbers ?
(i) 52
(ii) 977
(iii) 4583
(iv) 78367
(v) 52698
(vi) 99880
(vii) 12796
(viii) 55555
(ix) 53924

Solution:

(i) Square of 52 will be 2704 or (2)² = 4
∴ Its units digit is 4.

(ii) Square of 977 will be 954529 or (7)² = 49 .
∴ Its units digit is 9

(iii) Square of 4583 will be 21003889 or (3)² = 9
∴ Its units digit is 9

(iv) IS 78367, square of 7 = 7² = 49
∴ Its units digit is 9

(v) In 52698, square of 8 = (8)² = 64
∴ Its units digit is 4

(vi) In 99880, square of 0 = 0² = 0
∴ Its units digit is 0

(vii) In 12796, square of 6 = 6² = 36
∴ Its units digit is 6

(viii) In In 55555, square of 5 = 5² = 25
∴ Its units digit is 5

(ix) In 53924, square pf 4 = 4² = 16
∴ Its units digit is 6

Question 5.
Observe the following pattern
1 + 3 = 2²
1 + 3 + 5 = 3²
1+34-5 + 7 = 4²
and write the value of 1 + 3 + 5 + 7 + 9 +…………upto n terms.
Solution:
The given pattern is
1 + 3 = 2²
1 + 3 + 5 = 3²
1+3 + 5 + 7 = 4²
1+3 + 5 + 7 + 9 +……………… upto n terms (number of terms)² = n²

Question 6.
Observe the following pattern :
2² – 1² = 2 + 1
3² – 2² = 3 + 2
4² – 3² = 4 + 3
5² – 4² = 5 + 4
Find the value of
(i) 100² – 99²
(ii) 111² – 109²
(iii) 99² – 96²
Solution:
From the given pattern,
2² – 1² = 2 + 1
3² – 2² = 3 + 2
4² – 3² = 4 + 3
₅² – 4² = 5 + 4
Therefore
(i) 100²-99° = 100 + 99

(ii) 111² – 109² = 111² – 110²– 109²
= (111² – 110²) + (110² – 109²)
= (111 + 110) + (110+ 109)
= 221 + 219 = 440

(iii) 99² – 96² = 99² – 98² + 98² – 97² + 97² – 96²
= (99² – 98²) + (98² – 97²) + (97² – 96²)
= (99 + 98) + (98 + 97) + (97 + 96)
= 197 + 195 + 193 = 585

Question 7.
Which of the following triplets are Pythagorean ?
(i) (8, 15, 17)
(ii) (18, 80, 82)
(iii) (14, 48, 51)
(iv) (10, 24, 26)
(vi) (16, 63, 65)
(vii) (12, 35, 38)
Solution:
A pythagorean triplet is possible if (greatest number)² = (sum of the two smaller numbers)

(i) 8, 15, 17
Here, greatest number =17
∴ (17)² = 289
and (8)² + (15)² = 64 + 225 = 289
∴ 8² + 152 = 172
∴ 8, 15, 17 is a pythagorean triplet

(ii) 18, 80, 82
Greatest number = 82
∴ (82)² = 6724
and 18² + 80² = 324 + 6400 = 6724
∴ 18² + 80² = 82²
∴ 18, 80, 82 is a pythagorean

(iii) 14, 48, 51
Greatest number = 51
∴ (51)² = 2601
and 14² + 48² = 196 + 2304 = 25 00
∴ 51²≠ 14² + 48²
∴ 14, 48, 51 is not a pythagorean triplet

(iv) 10, 24, 26
Greatest number is 26
∴ 26² = 676
and 102 + 242 = 100 + 576 = 676
∴ 26² = 10² + 24²
∴ 10, 24, 26 is a pythagorean triplet

(vi) 16, 63, 65
Greatest number = 65
∴ 65² = 4225
and 16² + 63² = 256 + 3969 = 4225
∴ 65² = 16² + 63²
∴ 16, 63, 65 is a pythagorean triplet

(vii) 12, 35, 38
Greatest number = 38
∴ 38² = 1444
and 12² + 35² = 144 + 1225 = 1369
∴ 38² ≠12² + 35²
∴ 12, 35, 38 is not a pythagorean triplet.

Question 8.
Observe the following pattern

Solution:
From the given pattern

Question 9.
Observe the following pattern

and find the values of each of the following :
(i) 1 + 2 + 3 + 4 + 5 +….. + 50
(ii) 31 + 32 +… + 50
Solution:
From the given pattern,

Question 10.
Observe the following pattern

and find the values of each of the following :
(i) 1² + 2² + 3² + 4² +…………… + 10²
(ii) 5² + 6² + 7² + 8² + 9² + 10² + 12²
Solution:
From the given pattern,

Question 11.
Which of the following numbers are squares of even numbers ?
121,225,256,324,1296,6561,5476,4489, 373758
Solution:
We know that squares of even numbers is also are even number. Therefore numbers 256, 324,1296, 5476 and 373758 have their units digit an even number.
∴ These are the squares of even numbers.

Question 12.
By just examining the units digits, can you tell which of the following cannot be whole squares ?

1. 1026
2. 1028
3. 1024 
4. 1022
5. 1023
6. 1027

Solution:
We know that a perfect square cam at ends with the digit 2, 3, 7, or 8
∴ By examining the given number, we can say that 1028, 1022, 1023, 1027 can not be perfect squares.

Question 13.
Write five numbers for which you cannot decide whether they are squares.
Solution:
A number which ends with 1,4, 5, 6, 9 or 0
can’t be a perfect square
2036, 4225, 4881, 5764, 3349, 6400

Question 14.
Write five numbers which you cannot decide whether they are square just by looking at the unit’s digit.
Solution:
A number which does not end with 2, 3, 7 or 8 can be a perfect square
∴ The five numbers can be 2024, 3036, 4069, 3021, 4900

Question 15.
Write true (T) or false (F) for the following statements.
(i) The number of digits in a square number is even.
(ii) The square of a prime number is prime.
(iii) The sum of two square numbers is a square number.
(iv) The difference of two square numbers is a square number.
(vi) The product of two square numbers is a square number.
(vii) No square number is negative.
(viii) There is not square number between 50 and 60.
(ix) There are fourteen square number upto 200. 

Solution:
(i) False : In a square number, there is no condition of even or odd digits.
(ii) False : A square of a prime is not a prime.
(iii) False : It is not necessarily.
(iv) False : It is not necessarily.
(vi) True.
(vii) True : A square is always positive.
(viii) True : As 7² = 49, and 8² = 64.
(ix) True : As squares upto 200 are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196 which are fourteen in numbers..

Exercise 3.3

Question 1.
Find the squares of the following numbers using column method. Verify the result by finding the square using the usual multiplication :
(i) 25
(ii) 37
(iii) 54
(iv) 71
(v) 96
Solution:
(i) (25)²

Question 2.
Find the squares of the following numbers using diagonal method :
(i) 98
(ii) 273
(iii) 348
(iv) 295
(v) 171
Solution:

Question 3.
Find the squares of the following numbers :
(i) 127
(ii) 503
(iii) 451
(iv) 862
(v) 265
Solution:
(i) (127)² = (120 + 7)²
{(a + b)² = a² + lab + b²}
= (120)² + 2 x 120 x 7 + (7)²
= 14400+ 1680 + 49 = 16129

(ii) (503)² = (500 + 3)²
{(a + b)² = a² + lab + b¹}
= (500)² + 2 x 500 x 3 + (3)²
= 250000 + 3000 + 9 = 353009

(iii) (451)² = (400 + 51)²
{(a + b)² = a² + lab + b²}
= (400)² + 2 x 400 x 51 + (5l)²
= 160000 + 40800 + 2601 = 203401

(iv) (451)² = (800 + 62)²
{(a + b)² = a² + lab + b²}
= (800)² + 2 x 800 x 62 + (62)²
= 640000 + 99200 + 3844 = 743044

(v) (265)²
{(a + b)² = a² + 2ab + b²}
(200 + 65)² = (200)² + 2 x 200 x 65 + (65)²
= 40000 + 26000 + 4225 = 70225

Question 4.
Find the squares of the following numbers
(i) 425
(ii) 575
(iii) 405
(iv) 205
(v) 95
(vi) 745
(vii) 512
(viii) 995
Solution:
(i) (425)²
Here n = 42
∴ n (n + 1) = 42 (42 + 1) = 42 x 43 = 1806
∴ (425)² = 180625

(ii) (575)²
Here n = 57
∴ n (n + 1) = 57 (57 + 1) = 57 x 58 = 3306
∴ (575)² = 330625

(iii) (405)²
Here n = 40
∴ n (n + 1) = 40 (40 + 1) -40 x 41 = 1640
∴ (405)² = 164025

(iv) (205)²
Here n = 20
∴ n (n + 1) = 20 (20 + 1) = 20 x 21 = 420
∴ (205)² = 42025

(v) (95)²
Here n = 9
∴ n (n + 1) = 9 (9 + 1) = 9 x 10 = 90
∴ (95)² = 9025

(vi) (745)²
Here n = 74
∴ n (n + 1) = 74 (74 + 1) = 74 x 75 = 5550
∴ (745)² = 555025

(vii) (512)²
Here a = 1, b = 2
∴ (5ab)² = (250 + ab) x 1000 + (ab)²
∴ (512)² = (250 + 12) x 1000 + (12)²
= 262 x 1000 + 144
= 262000 + 144 = 262144

(viii) (995)²
Here n = 99
∴ n (n + 1) = 99 (99 + 1) = 99 x 100 = 9900
∴ (995)² = 990025

Question 5.
Find the squares of the following numbers using the identity (a + b)1 = a2 + lab + b1
(i) 405
(ii) 510
(iii) 1001
(iv) 209
(v) 605
Solution:
a + b)² = a² + lab + b²

(i) (405)² = (400 + 5)²
= (400)² + 2 x 400 x 5 + (5)²
= 160000 + 4000 + 25 = 164025

(ii) (510)² = (500 + 10)²
= (500)² + 2 x 500 x 10 x (10)²
= 250000 + 10000 + 100
= 260100

(iii) (1001)² = (1000+1)²
= (1000)² + 2 X 1000 x 1 + (1)
= 1000000 + 2000 + 1
=1002001

(iv) (209)² = (200 + 9)²
= (200)² + 2 x 200 x 9 x (9)²
= 40000 + 3600 +81
= 43681

(v) (605)² = (600 + 5)²
= (600)² + 2 x 600 x 5 +(5)²
= 360000 + 6000       25
=366025

Question 6.
Find the squares of the following numbers using the identity (a – b)² = a² – 2ab + b² :
(i) 395
(ii) 995
(iii) 495
(iv) 498
(v) 99
(vi) 999
(vii) 599
Solution:
a – b)² = a² – lab + b²

(i) (395)² = (400 – 5)²
= (400)² – 2 x 400 x 5 + (5)²
= 160000-4000 + 25
= 160025-4000
= 156025

(ii) (995)² = (1000 – 5)²
= (1000)² – 2 x 1000 x 5 + (5)²
= 1000000- 10000 + 25
= 1000025- 10000
= 990025

(iii) (495)² = (500 – 5)²
= (500)² – 2 x 500 x 5 + (5)²
= 250000 – 5000 + 25
= 250025 – 5000
= 245025

(iv) (498)² = (500 – 2)²
= (500)² – 2 x 500 x 2 + (2)²
= 250000 – 2000 + 4
= 250004 – 2000
= 248004

(v) (99)² = (100 – l)²
= (100)² – 2 x 100 x 1 + (1)²
= 10000 – 200 + 1
= 10001 – 200
= 9801

(vi) (999)² = (1000- l)²
= (1000)² – 2 x 1000 x 1+ (1)2
= 1000000-2000+1
= 10000001-2000=998001

(vii) (599)² = (600 – 1)²
= (600)² -2 x 600 X 1+ (1)2
= 360000 -1200+1
= 360001 – 1200 = 358801

Question 7.
Find the squares of the following numbers by visual method :
(i) 52
(ii) 95
(iii) 505
(iv) 702
(v) 99
Solution:
(a + b)² = a² – ab + ab + b²
(i) (52)² = (50 + 2)²
= 2500 + 100 + 100 + 4
= 2704

(ii) (95)² = (90 + 5)²
= 8100 + 450 + 450 + 25
= 9025

(iii) (505)² = (500 + 5)²
= 250000 + 2500 + 2500 + 25
= 255025

(iv) (702)² = (700 + 2)²
= 490000 + 1400+ 1400 + 4
= 492804

(v) (99)² = (90 + 9)²
= 8100 + 810 + 810 + 81
= 9801

Exercise 3.4

Question 1.
Write the possible unit’s digits of the square root of the following numbers. Which of these numbers are odd square roots ?
(i) 9801
(ii) 99856
(iii) 998001
(iv) 657666025
Solution:
(i)  In 9801−−−−√ ∴ the units digits is 1, therefore, the units digit of the square root can be 1 or 9
(ii) In 799356−−−−−−√ ∴ the units digit is 6
∴ The units digit of the square root can be 4 or 6
(iii) In 7998001−−−−−−−√ ∴ the units digit is 1
∴ The units digit of the square root can be 1 or 9
(iv) In 657666025
∴ The unit digit is 5
∴ The units digit of the square root can be 5

Question 2.
Find the square root of each of the following by prime factorization.
(i) 441
(ii) 196
(iii) 529
(iv) 1764
(v) 1156
(vi) 4096
(vii) 7056
(viii) 8281
(ix) 11664
(x) 47089
(xi) 24336
(xii) 190969
(xiii) 586756
(xiv) 27225
(xv) 3013696
Solution:

Question 3.
Find the smallest number by which 180 must be multiplied so that it becomes a perfect square. Also, find the square root of the perfect square so obtained.
Solution:
Factorising 180,

180 = 2 x 2 x 3 x 3 x 5
Grouping the factors in pairs we see that factor 5 is left unpaired.
∴ Multiply 180 by 5, we get the product 180 x 5 = 900
Which is a perfect square
and square root of 900 = 2 x 3 x 5 = 30

Question 4.
Find the smallest number by which 147 must be multiplied so that it becomes a perfect square. Also, find the square root of the number so obtained.
Solution:
Factorising 147,

147 = 3 x 7×7
Grouping the factors in pairs of the equal factors, we see that one factor 3 is left unpaired
∴ Multiplying 147 by 3, we get the product 147 x 3 = 441
Which is a perfect square
and its square root = 3×7 = 21

Question 5.
Find the smallest number by which 3645 must be divided so that it becomes a perfect square. Also, find the square root of the resulting number.
Solution:
Factorising 3645

3645 = 3 x 3 3 x 3 x 3 x 3 x 5
Grouping the factors in pair of the equal factors, we see t at one factor 5 is left unpaired
∴ Dividing 3645 by 5, the quotient 729 will be the perfect square and square root of 729 = 27

Question 6.
Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also, find the square root of the number so obtained.
Solution:
Factorsing 1152,

1152 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3
Grouping the factors in pairs of the equal factors, we see that factor 2 is left unpaired.
∴ Dividing by 2, the quotient 576 is a perfect square .
∴ Square root of 576, it is 24

Question 7.
The product of two numbers is 1296. If one number is 16 times the others find the numbers.
Solution:
Product of two numbers = 1296
Let one number = x
Second number = 16x

∴ First number = 9
and second number = 16 x 9 = 144

Question 8.
A welfare association collected Rs. 202500 as donation from the residents. If each paid as many rupees as there were residents find the number of residents.
Solution:
Total donation collected = Rs. 202500
Let number of residents = x
Then donation given by each resident = Rs. x
∴ Total collection = Rs. x x x

Question 9.
A society collected Rs. 92.16. Each member collected as many paise as there were members. How many members were there and how much did each contribute?
Solution:
Total amount collected = Rs. 92.16 = 9216 paise
Let the number of members = x
Then amount collected by each member = x
paise

∴ Number of members = 96
and each member collected = 96 paise

Question 10.
A school collected Rs. 2304 as fees from its students. If each student paid as many paise as there were students in the school, how many students were there in the school ?
Solution:
Total fee collected = Rs. 2304
Let number of students = x
Then fee paid by each student = Rs. x
∴ x x x = 2304 => x² = 2304
∴ x = 2304−−−−√

Question 11.
The area of a square field is 5184 m². A rectangular field, whose length is twice its breadth has its perimeter equal to the perimeter of the square field. Find the area of the rectangular field.
Solution:
The area of a square field = 5184 m²
Let side of the square = x

∴ side of square= 72 m
∴ Perimeter, of square field = 72 x 4 m = 288 m
Perimeter of rectangle = 288 m
Let breadth of rectangular field (b) = x
Then length (l) = 2x
∴ Perimeter = 2 (l + b)
= 2 (2x + x) = 2 x 3x = 6x
= 2 (2x + x) = 2 x 3x = 6x

∴ Length of rectangular field = 2x = 2 x 48 = 96 m
and breadth = 48 m
and area = l x b = 96 x 48 m²
= 4608 m²

Question 12.
Find the least square number, exactly divisible by each one of the numbers :
(i) 6, 9,15 and 20
(ii) 8,12,15 and 20
Solution:

LCM of 6, 9, 15, 20 = 2 x 3 x 5 x 3 x 2 = 180
=2 x 2 x 3 x 3 x 5
We see that after grouping the factors in pairs, 5 is left unpaired
∴ Least perfect square = 180 x 5 = 900

We see that after grouping the factors,
factors 2, 3, 5 are left unpaired
∴ Perfect square =120 x 2 x 3 x 5 = 120 x 30 = 3600

Question 13.
Find the square roots of 121 and 169 by the method of repeated subtraction.
Solution:

Question 14.
Write the prime factorization of the following numbers and hence find their square roots. ^
(i) 7744
(ii) 9604
(iii) 5929
(iv) 7056
Solution:
Factorization, we get:
(i) 7744 = 2 x 2 x 2 x 2 x 2 x 2 x 11 x 11

Grouping the factors in pairs of equal factors,

Question 15.
The students of class VIII of a school donated Rs. 2401 for PM’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.
Solution:
Total amount of donation = 2401
Let number of students in VIII = x
∴ Amount donoted by each student = Rs. x

Question 16.
A PT teacher wants to arrange maximum possible number of 6000 students in a Held such that the number of rows is equal to the number of columns. Find the number of rows if 71 were left out after arrangement.
Solution:
Number of students = 6000
Students left out = 71
∴ Students arranged in a field = 6000 – 71=5929

Exercise 3.5

Question 1.
Find the square root of each of the long division method.
(I) 12544
(ii) 97344
(iii) 286225
(iv) 390625
(v) 363609
(vi) 974169
(vii) 120409
(viii) 1471369
(ix) 291600
(x) 9653449
(xi) 1745041
(xii) 4008004
(xiii) 20657025
(xiv) 152547201
(jcv) 20421361
(xvi) 62504836
(xvii) 82264900
(xviii) 3226694416
(xix)6407522209
(xx) 3915380329
Solution:
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Question 2.
Find the least number which must be subtracted from the following numbers to make them a perfect square :
(i) 2361
(ii) 194491
(iii) 26535
(iv) 16160
(v) 4401624
Solution:
(i) 2361
Finding the square root of 2361

We get 48 as quotient and remainder = 57
∴ To make it a perfect square, we have to subtract 57 from 2361
∴ Least number to be subtracted = 57
(ii) 194491
Finding the square root of 194491

We get 441 as quotient and remainder = 10
∴ To make it a perfect square, we have to subtract 10 from 194491
∴ Least number to be subtracted = 10
(iii) 26535
Finding the square root of 26535

We get 162 as quotient and 291 as remainder
∴ To make it a perfect square, we have to subtract 291 from 26535
∴ Least number to be subtracted = 291
(iv)16160
Finding the square root of 16160

We get 127 as quotient and 31 as remainder
∴ To make it a perfect square, we have to subtract 31 from 16160
∴ Least number to be subtracted = 31
(v) 4401624
Find the square root of 4401624

We get 2098 as quotient and 20 as remainder
∴ To make it a perfect square, we have to subtract 20 from 4401624
∴ Least number to be subtracted = 20

Question 3.
Find the least number which must be added to the following numbers to make them a perfect square :
(i) 5607
(ii) 4931
(iii) 4515600
(iv) 37460
(v) 506900
Solution:
(i) 5607

Finding the square root of 5607, we see that 74² = 5607- 131 =5476 and 75² = 5625
∴ 5476 < 5607 < 5625
∴ 5625 – 5607 = 18 is to be added to get a perfect square
∴ Least number to be added = 18
(ii) 4931

Finding the square root of 4931, we see that 70²= 4900
∴ 71² = 5041 4900 <4931 <5041
∴ 5041 – 4931 = 110 is to be added to get a perfect square.
∴ Least number to be added =110
(iii) 4515600

Finding the square root of 4515600, we see
that 2124² = 4511376
and 2 1 25² = 45 1 56 25
∴ 4511376 <4515600 <4515625
∴ 4515625 – 4515600 = 25 is to be added to get a perfect square.
∴ Least number to be added = 25
(iv) 37460

Finding the square root of 37460
that 193² = 37249, 194² = =37636
∴ 37249 < 37460 < 37636
∴ 37636 – 37460 = 176 is to be added to get a perfect square.
∴ Least number to be added =176
(v) 506900

Finding the square root of 506900, we see that
711² = 505521, 712² = 506944
∴ 505521 < 506900 < 506944
∴ 506944 – 506900 = 44 is to be added to get a perfect square.
∴ Least number to be added = 44

Question 4.
Find the greatest number of 5 digits which is a perfect square.
Solution:
Greatest number of 5-digits = 99999 Finding square root, we see that 143 is left as remainder

∴ Perfect square = 99999 – 143 = 99856 If we add 1 to 99999, it will because a number of 6 digits
∴ Greatest square 5-digits perfect square = 99856

Question 5.
Find the least number of four digits which is a perfect square.
Solution:
Least number of 4-digits = 10000
Finding square root of 1000
We see that if we subtract 39

From 1000, we get three digit number
∴ We shall add 124 – 100 = 24 to 1000 to get a
perfect square of 4-digit number
∴ 1000 + 24 = 1024
∴ Least number of 4-digits which is a perfect square = 1024

Question 6.
Find the least number of six-digits which is a perfect square.
Solution:
Least number of 6-digits = 100000

Finding the square root of 100000, we see that if we subtract 544, we get a perfect square of 5-digits.
So we shall add
4389 – 3900 = 489
to 100000 to get a perfect square
Past perfect square of six digits= 100000 + 489 =100489

Question 7.
Find the greatest number of 4-digits which is a perfect square.
Solution:
Greatest number of 4-digits = 9999

Finding the square root, we see that 198 has been left as remainder
∴ 4-digit greatest perfect square = 9999 – 198 = 9801

Question 8.
A General arranges his soldiers in rows to form a perfect square. He finds that in doing so, 60 soldiers are left out. If the total number of soldiers be 8160, find the number of soldiers in each row.
Solution:
Total number of soldiers = 8160 Soldiers left after arranging them in a square = 60
∴ Number of soldiers which are standing in a square = 8160 – 60 = 8100

Question 9.
The area of a square field is 60025 m². A man cycle along its boundry at 18 km/hr. In how much time will be return at the starting point.
Solution:
Area of a square field = 60025 m²

Question 10.
The cost of levelling and turfing a square lawn at Rs. 250 per m² is Rs. 13322.50. Find the cost of fencing it at Rs. 5 per metre ?
Solution:
Cost of levelling a square field = Rs. 13322.50
Rate of levelling = Rs. 2.50 per m²

and perimeter = 4a = 4 x 73 = 292 m Rate of fencing the field = Rs. 5 per m
∴ Total cost of fencing = Rs. 5 x 292 = Rs. 1460

Question 11.
Find the greatest number of three digits which is a perfect square.
Solution:
3-digits greatest number = 999

Finding the square root, we see that 38 has been left
∴ Perfect square = 999 – 38 = 961
∴ Greatest 3-digit perfect square = 961

Question 12.
Find the smallest number which must be added to 2300 so that it becomes a perfect square.
Solution:
Finding the square root of 2300

We see that we have to add 704 – 700 = 4 to 2300 in order to get a perfect square
∴ Smallest number to be added = 4

Exercise 3.6

Question 1.
Write the possible unit’s digits of the square root of the following numbers. Which of these numbers are odd square roots ?
(i) 9801
(ii) 99856
(iii) 998001
(iv) 657666025
Solution:
(i)  In 9801−−−−√ ∴ the units digits is 1, therefore, the units digit of the square root can be 1 or 9
(ii) In 799356−−−−−−√ ∴ the units digit is 6
∴ The units digit of the square root can be 4 or 6
(iii) In 7998001−−−−−−−√ ∴ the units digit is 1
∴ The units digit of the square root can be 1 or 9
(iv) In 657666025
∴ The unit digit is 5
∴ The units digit of the square root can be 5

Question 2.
Find the square root of each of the following by prime factorization.
(i) 441
(ii) 196
(iii) 529
(iv) 1764
(v) 1156
(vi) 4096
(vii) 7056
(viii) 8281
(ix) 11664
(x) 47089
(xi) 24336
(xii) 190969
(xiii) 586756
(xiv) 27225
(xv) 3013696
Solution:

Question 3.
Find the smallest number by which 180 must be multiplied so that it becomes a perfect square. Also, find the square root of the perfect square so obtained.
Solution:
Factorising 180,

180 = 2 x 2 x 3 x 3 x 5
Grouping the factors in pairs we see that factor 5 is left unpaired.
∴ Multiply 180 by 5, we get the product 180 x 5 = 900
Which is a perfect square
and square root of 900 = 2 x 3 x 5 = 30

Question 4.
Find the smallest number by which 147 must be multiplied so that it becomes a perfect square. Also, find the square root of the number so obtained.
Solution:
Factorising 147,

147 = 3 x 7×7
Grouping the factors in pairs of the equal factors, we see that one factor 3 is left unpaired
∴ Multiplying 147 by 3, we get the product 147 x 3 = 441
Which is a perfect square
and its square root = 3×7 = 21

Question 5.
Find the smallest number by which 3645 must be divided so that it becomes a perfect square. Also, find the square root of the resulting number.
Solution:
Factorising 3645

3645 = 3 x 3 3 x 3 x 3 x 3 x 5
Grouping the factors in pair of the equal factors, we see t at one factor 5 is left unpaired
∴ Dividing 3645 by 5, the quotient 729 will be the perfect square and square root of 729 = 27

Question 6.
Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also, find the square root of the number so obtained.
Solution:
Factorsing 1152,

1152 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3
Grouping the factors in pairs of the equal factors, we see that factor 2 is left unpaired.
∴ Dividing by 2, the quotient 576 is a perfect square .
∴ Square root of 576, it is 24

Question 7.
The product of two numbers is 1296. If one number is 16 times the others find the numbers.
Solution:
Product of two numbers = 1296
Let one number = x
Second number = 16x

∴ First number = 9
and second number = 16 x 9 = 144

Question 8.
A welfare association collected Rs. 202500 as donation from the residents. If each paid as many rupees as there were residents find the number of residents.
Solution:
Total donation collected = Rs. 202500
Let number of residents = x
Then donation given by each resident = Rs. x
∴ Total collection = Rs. x x x

Question 9.
A society collected Rs. 92.16. Each member collected as many paise as there were members. How many members were there and how much did each contribute?
Solution:
Total amount collected = Rs. 92.16 = 9216 paise
Let the number of members = x
Then amount collected by each member = x
paise

∴ Number of members = 96
and each member collected = 96 paise

Question 10.
A school collected Rs. 2304 as fees from its students. If each student paid as many paise as there were students in the school, how many students were there in the school ?
Solution:
Total fee collected = Rs. 2304
Let number of students = x
Then fee paid by each student = Rs. x
∴ x x x = 2304 => x² = 2304
∴ x = 2304−−−−√

Question 11.
The area of a square field is 5184 m². A rectangular field, whose length is twice its breadth has its perimeter equal to the perimeter of the square field. Find the area of the rectangular field.
Solution:
The area of a square field = 5184 m²
Let side of the square = x

∴ side of square= 72 m
∴ Perimeter, of square field = 72 x 4 m = 288 m
Perimeter of rectangle = 288 m
Let breadth of rectangular field (b) = x
Then length (l) = 2x
∴ Perimeter = 2 (l + b)
= 2 (2x + x) = 2 x 3x = 6x
= 2 (2x + x) = 2 x 3x = 6x

∴ Length of rectangular field = 2x = 2 x 48 = 96 m
and breadth = 48 m
and area = l x b = 96 x 48 m²
= 4608 m²

Question 12.
Find the least square number, exactly divisible by each one of the numbers :
(i) 6, 9,15 and 20
(ii) 8,12,15 and 20
Solution:

LCM of 6, 9, 15, 20 = 2 x 3 x 5 x 3 x 2 = 180
=2 x 2 x 3 x 3 x 5
We see that after grouping the factors in pairs, 5 is left unpaired
∴ Least perfect square = 180 x 5 = 900

We see that after grouping the factors,
factors 2, 3, 5 are left unpaired
∴ Perfect square =120 x 2 x 3 x 5 = 120 x 30 = 3600

Question 13.
Find the square roots of 121 and 169 by the method of repeated subtraction.
Solution:

Question 14.
Write the prime factorization of the following numbers and hence find their square roots. ^
(i) 7744
(ii) 9604
(iii) 5929
(iv) 7056
Solution:
Factorization, we get:
(i) 7744 = 2 x 2 x 2 x 2 x 2 x 2 x 11 x 11

Grouping the factors in pairs of equal factors,

Question 15.
The students of class VIII of a school donated Rs. 2401 for PM’s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.
Solution:
Total amount of donation = 2401
Let number of students in VIII = x
∴ Amount donoted by each student = Rs. x

Question 16.
A PT teacher wants to arrange maximum possible number of 6000 students in a Held such that the number of rows is equal to the number of columns. Find the number of rows if 71 were left out after arrangement.
Solution:
Number of students = 6000
Students left out = 71
∴ Students arranged in a field = 6000 – 71=5929

Exercise 3.7

Find the square root of the following numbers in decimal form :

Question 1.
84.8241
Solution:

Question 2.
0.7225
Solution:

Question 3.
0.813604
Solution:

Question 4.
0.00002025
Solution:

Question 5.
150.0625
Solution:

Question 6.
225.6004
Solution:

Question 7.
3600.720036
Solution:

Question 8.
236.144689
Solution:

Question 9.
0.00059049
Solution:

Question 10.
176.252176
Solution:

Question 11.
9998.0001
Solution:

Question 12.
0.00038809
Solution:

Question 13.
What is that fraction which when multiplied by itself gives 227.798649 ?
Solution:

Question 14.
square playground is 256.6404 square metres. Find the length of one side of the playground.
Solution:
Area of square playground = 256.6404 sq. m

Question 15.
What is the fraction which when multiplied by itself gives 0.00053361 ?
Solution:

Question 16.
Simplify :


Solution:

Question 17.
Evaluate 50625−−−−−√ and hence find the value of 506.25−−−−−√+5.0625−−−−−√.
Solution:

Question 18.
Find the value of 103.0225−−−−−−−√ and hence And the value of

Solution:

Exercise 3.8

Question 1.
Find the square root of each of the following correct to three places of decimal :
(i) 5
(ii) 7
(iii) 17
(iv) 20
(v) 66
(vi) 427
(vii) 1.7
(vii’) 23.1
(ix) 2.5
(x) 237.615
(xi) 15.3215
(xii) 0.9
(xiii) 0.1
(xiv) 0.016
(xv) 0.00064
(xvi) 0.019
(xvii) 78
(xviii) 512
(xix) 2 12
(xx) 287 88
Solution:

Question 2.
Find the square root of 12.0068 correct to four decimal places.
Solution:

Question 3.
Find the square root of 11 correct to five decimal places.
Solution:
https://googleads.g.doubleclick.net/pagead/ads?client=ca-pub-7601472013083661&outp/www.learninsta.com&dtd=1834

Question 4.
Given that √2, = 1-414, √3 = 1.732, √5 = 2.236 and √7 = 2.646. Evaluate each of the following :

Solution:

Question 5.
Given that √2 = 1-414, √3 = 1-732, √5= 2.236 and √7= 2.646, find the square roots of the following :

Solution:

Exercise 3.9

Using square root table, find the square roots of the following :

Question 1.
7
Solution:

Question 2.
15
Solution:

Question 3.
74
Solution:

Question 4.
82
Solution:

Question 5.
198
Solution:

Question 6.
540
Solution:

Question 7.
8700
Solution:

Question 8.
3509
Solution:

Question 9.
6929
Solution:

Question 10.
25725
Solution:

Question 11.
1312
Solution:

Question 12.
4192
Solution:

Question 13.
4955
Solution:

Question 14.
99144
Solution:

Question 15.
57169
Solution:

Question 16.
101169
Solution:

Question 17.
13.21
Solution:

Question 18.
21.97
Solution:

Question 19.
110
Solution:

Question 20.
1110
Solution:

Question 21.
11.11
Solution:

Question 22.
The area of a square field is 325 m². Find the approximate length of one side of the field.
Solution:

Question 23.
Find the length of a side of a square, whose area is equal to the area bf the rectangle with sides 240 m and 70 m.
Solution:

Exercise 2.1

Question 1.
Express the following numbers in standard form :
(i) 6020000000000000
(ii) 0.00000000000942
(iii) 0.00000000085
(iv) 846 X 10⁷
(v) 3759 x 10^-4
(vi) 0.00072984
(vii) 0.000437 x 10⁴ 
(Viii) 4 + 100000
Solution:

Question 2.
Write the following numbers in the usual form :
(i) 4.83 x 10⁷
(ii) 3.02 x 10^-6
(iii) 4.5 x 10⁴
(iv) 3 x 10^-8
(v) 1.0001 x 10⁹
(vi) 5.8 x 10²
(vii) 3.61492 x 10⁶
(viii) 3.25 x 10^-7
Solution:

Exercise 2.2

Question 1.
Write each of the following in exponential form :

Solution:

Question 2.
Evaluate :

Solution:

Question 3.
Express each of the following as a rational number in the form pq:

Solution:

Question 4.
Simplify :

Solution:

Question 5.
Express each of the following rational numbers with a negative exponent :

Solution:

Question 6.
Express each of the following rational numbers with a positive exponent :

Solution:

Question 7.
Simplify :

Solution:

Question 8.
By what number should 5^-1 be multiplied so that the product may be equal to (-7)^-1 ?
Solution:

Question 9.
By what number should (12)−1 be multiplied so that the product may be equal to (−47)−1 ?
Solution:

Question 10.
By what number should (-15)^-1 be divided so that the quotient may be equal to (-5)^-1 ?
Solution:

Question 11.
By what number should (53)−2 be multiplied so that the product may be (73)−1 ?
Solution:

Question 12.
Find x, if

Solution:

Question 13.

Solution:

Question 14.
Find the value of x for which 5^2x + 5^-3 = 5⁵.
Solution:

Exercise 2.3

Question 1.
Express the following numbers in standard form :
(i) 6020000000000000
(ii) 0.00000000000942
(iii) 0.00000000085
(iv) 846 X 10⁷
(v) 3759 x 10^-4
(vi) 0.00072984
(vii) 0.000437 x 10⁴ 
(Viii) 4 + 100000
Solution:

Question 2.
Write the following numbers in the usual form :
(i) 4.83 x 10⁷
(ii) 3.02 x 10^-6
(iii) 4.5 x 10⁴
(iv) 3 x 10^-8
(v) 1.0001 x 10⁹
(vi) 5.8 x 10²
(vii) 3.61492 x 10⁶
(viii) 3.25 x 10^-7
Solution:

Exercise 1.1

Question 1.
Add the following rational numbers:

Solution:

Question 2.
Add the following rational numbers:

Solution:

Question 3.
Simplify:

Solution:

Question 4.
Add and Express the sum as mixed fraction:

Solution:

Exercise 1.2

Question 1.
Verify commutativity of addition of rational numbers for each of the following pairs of rational numbers :

Solution:

Question 2.
Verify associativity of addition of rational numbers i.e., (A: + y) + z = x + (y + z), when :

Solution:

Question 3.
Write the additive inverse of each of the following rational numbers :

Solution:

Question 4.
Write the negative (additive inverse) of each of the following :

Solution:

Question 5.
Using commutativity and associativity of addition of rational numbers, express ‘iach of the following as a rational number :

Solution:

Question 6.
Re-arrange suitably and find the sum in each of the following :


Solution:

Exercise 1.3

Question 1.
Subtract the first rational number from the second in each of the following :



Solution:

Question 2.
Evaluate each of the following :

Solution:

Question 3.
The sum of two numbers is 59. If one of the numbers is 13, find the other.
Solution:

Question 4.
The sum of two numbers is −13. If one of the numbers is −123, find the other.
Solution:

Question 5.
The sum of two numbers is −43. If one of the number is -5, find the
Solution:

Question 6.
The sum of two rational numbers is -8. If one of the numbers is −157 find the other.
Solution:

Question 7.
What should be added to so as to −78 get 59 ?
Solution:

Question 8.
What number should be added to −511 so as to get 263 ?
Solution:

Question 9.
What number should be added to −57 to get −23 ?
Solution:

Question 10.
What number should be subtracted from −53 to get 56 ?
Solution:

Question 11.
What number should be subtracted from 37 to get 54 ?
Solution:

Question 12.
What should be added to (23+35) to get −1215 ?
Solution:

Question 13.
What should be added to (12+13+15) to get 3 ?
Solution:

Question 14.
What should be subtracted from (34−23) to get −16 ?
Solution:

Question 15.
Fill in the blanks :

Solution:

Exercise 1.4

Question 1.
Subtract the first rational number from the second in each of the following :



Solution:

Question 2.
Evaluate each of the following :

Solution:

Question 3.
The sum of two numbers is 59. If one of the numbers is 13, find the other.
Solution:

Question 4.
The sum of two numbers is −13. If one of the numbers is −123, find the other.
Solution:

Question 5.
The sum of two numbers is −43. If one of the number is -5, find the
Solution:

Question 6.
The sum of two rational numbers is -8. If one of the numbers is −157 find the other.
Solution:

Question 7.
What should be added to so as to −78 get 59 ?
Solution:

Question 8.
What number should be added to −511 so as to get 263 ?
Solution:

Question 9.
What number should be added to −57 to get −23 ?
Solution:

Question 10.
What number should be subtracted from −53 to get 56 ?
Solution:

Question 11.
What number should be subtracted from 37 to get 54 ?
Solution:

Question 12.
What should be added to (23+35) to get −1215 ?
Solution:

Question 13.
What should be added to (12+13+15) to get 3 ?
Solution:

Question 14.
What should be subtracted from (34−23) to get −16 ?
Solution:

Question 15.
Fill in the blanks :

Solution:

Exercise 1.5

Question 1.
Multiply:

Solution:

Question 2.
Multiply:

Solution:

Question 3.
Simplify each of the following and express the result as a rational number in standard form :

Solution:

Question 4.
Simplify :


Solution:

Question 5.
Simplify :

Solution:

Exercise 1.6

Question 1.
Verify the property : x x y = y x x by taking :

Solution:

Question 2.
Verify the property : x x (y x z) = (x x y) x z by taking :

Solution:

Question 3.
Verify the property :xx(y + 2) = xxy + x x z by taking :

Solution:

Question 4.
Use the distributivity of multiplication of rational numbers over their addition to simplify :

Solution:

Question 5.
Find the multiplicative inverse (reciprocal) of each of the following rational numbers :


Solution:

Question 6.
Name the property of multiplication of rational numbers illustrated by the following statements :

Solution:

Exercise 1.7

Question 1.
Divide :

Solution:

Question 2.
Find the value and express as a rational number in standard form :

Solution:

Question 3.
The product of two rational numbers is15. If one of the numbers is -10, find the other.
Solution:

Question 4.
The product of two rational numbers is−98 if one of the numbers is −415, find other.
Solution:

Question 5.
By what number should we multiply −16 so that the product may be −239 ?
Solution:

Question 6.
By what number should we multiply −1528 so that the product may be −57 ?
Solution:

Question 7.
By what number should we multiply −813 so that the product may be 24 ?
Solution:

Question 8.
Bv what number should −34 multiplied in order to produce 23 ?
Solution:

Question 9.
Find (x +y) + (x – y), if

Solution:

Question 10.
The cost of 7 23 metres of rope is Rs 12 34.Find its cost per metre.
Solution:

Question 11.
The cost of 2 13 metres of cloth is Rs. 75 14Find the cost of cloth per metre.
Solution:

Question 12.
By what number should −3316 be divided to get −114 ?
Solution:

Question 13.
Divide the sum of −135 and 127 by the product of −317 and −12.
Solution:

Question 14.
Divide the sum of 6512 and 127 bv their difference.
Solution:

Question 15.
If 24 trousers of equal size can be prepared in 54 metres of cloth, what length of cloth is required for each trouser ?
Solution:

Exercise 1.8

Question 1.
Find a rational number between -3 and 1.
Solution:

Question 2.
Find any five rational number less than 1.
Solution:

Question 3.
Find four rational numbers between −29 and 59 .
Solution:

Question 4.
Find two rational numbers between 15 and 12 .
Solution:

Question 5.
Find ten rational numbers between 14 and 12 .
Solution:

Question 6.
Find ten rational numbers between −25 and 12 .
Solution:

Question 7.
Find ten rational numbers between 35 and 34 .
Solution:

Exercise 23 A

Page No 238:

Question 1:

The number of students who absented from the class during a week are given below:

Day	Monday	Tuesday	Wednesday	Thursday	Friday
No. of absentees	6	2	4	2	8

Take the scale Figure = 2 absentees.
Draw the pictograph.

ANSWER:

Day	Number of absentees
Monday
Tuesday
Wednesday
Thursday
Friday

Page No 238:

Question 2:

The number of stools in five rooms of a school are given below:

Room number	I	II	III	IV	V
Number of stools	30	40	60	50	20

Taking the scale Figure = 10 stools, draw the pictograph.

ANSWER:

Room Number	Number of stools
I
II
III
IV
V

Page No 238:

Question 3:

In a class test, the number of students passed in various subjects are given below.

Subject	English	Mathematics	Hindi	Drawing
Number of students passed	15	25	10	20

Taking the scale Figure = 5 successful students, draw the pictograph.

ANSWER:

Subject	Number of students passed
English
Mathematics
Hindi
Drawing

Page No 238:

Question 4:

The number of fans sold by a shopkeeper during 6 months are given below:

Month	March	April	May	June	July	August
Number of fans sold	30	40	60	50	20	30

Taking the scale Figure = 10 fans sold, draw the pictograph.

ANSWER:

Month	Number of fans sold
March
April
May
June
July
August

Page No 239:

Question 5:

The following pictograph shows different kinds of trees planted in a park. Each symbol represents 8 trees. Look at the pictograph adn answer the questions given below.

Banyan tree	Figure
Neem tree	Figure
Mango tree	Figure

(i) How many mango trees are there?
(ii) How many banyan trees are there?
(iii) How many neem trees are there?
(iv) How many  trees are there in all?

ANSWER:

(i) Number of mango trees = 3××8 = 24

(ii) Number of banyan trees = 4××8 = 32

(iii) Number of neem trees = 5××8 = 40

(iv) Total number of trees = Number of mango trees + Number of banyan trees + Number of neem trees
                                        = 24+32+40
                                        = 96

Page No 239:

Question 6:

The following pictograph shows the number of scooters sold by a company during a week.

Scale used is Figure = 6 scooters sold.

Day	Figure
Monday	Figure
Tuesday	Figure
Wednesday	Figure
Thursday	Figure
Friday	Figure
Saturday	Figure

Study the pictograph carefully and answer the questions given below.
(i) How many scooters were sold on Monday?
(ii) How many scooters were sold on Tuesday?
(iii) On what day of the week was the sale of the scooters maximum? How many scooters were sold on that day?
(iv) On what day of the week was the sale of the scooters minimum? How many scooters were sold on that day?

ANSWER:

(i) Number of scooters sold on Monday = 5××6 = 30

(ii) Number of scooters sold on Tuesday = 4××6 = 24

(iii) The sale was maximum on Friday.

      Number of scooters sold on Friday = 7××6 = 42

(iv) The sale was minimum on Saturday.

      Number of scooters sold on Saturday = 2××6 = 12
     

Exercise 24A

Page No 243:

Question 1:

Look at the bar graph given below.
Figure
Read it carefully and answer the questions given below:
(i) What information does the bar graph give?
(ii) In which subject is the student poorest?
(iii) In which subject is the student best?
(iv) In which subjects did he get more than 40 marks?

ANSWER:

(i) The given bar graph provides information about the marks obtained by a student in an examinations in four different subjects.

(ii) The student is poorest in science because the height of the bar representing the marks obtained in science is the lowest.

(iii) The student is best in mathematics because the height of the bar representing the marks obtained in mathematics is the highest.

(iv) In Hindi and mathematics he got more than 40 marks. He scored 55 marks in Hindi and 70 marks in mathematics.

Page No 243:

Question 2:

In a survey of 60 families of a colony, the number of members in each family was recorded and the data has been represented by the bar graph given below:
Figure
Read the bar graph carefully and answer the following questions:
(i) What information does the bar graph give?
(ii) How many families have 3 members?
(iii) How many couples have no child?
(iv) Which type of family is the most common?

ANSWER:

(i) The given bar graph provides information about the number of members in 60 families of a colony.

(ii) There are 10 families that have 3 members.

(iii) There are 5 couples that do not have children.

(iv) The families having two children is most common because the bar of the families having four members has the highest height.

Page No 244:

Question 3:

Look at the bar graph given below:
Figure
Study the bar graph carefully and answer the questions given below:
(i) In which week was the production maximum?
(ii) In which week was the production minimum?
(iii) What is the average production during these five weeks?
(iv) How many cycles were produced in the first 3 weeks?

ANSWER:

(i) Production was maximum in the 2nd week.

(ii) Production was minimum in the 4th week.

(iii) Average production of these five weeks =Total production of all weeksNumber of weeksTotal production of all weeksNumber of weeks
     
                                                                            =1000 + 600 + 800 + 500 + 7005=36005=720=1000 + 600 + 800 + 500 + 7005=36005=720

(iv) Number of cycles produced in the first 3 week = Production in the 1st week + Production in the 2nd week + Production in the 3rd week

                                                                                 = 600 + 1000 + 800
                                                                                 = 2400 cycles

Page No 244:

Question 4:

51 students from a locality use different modes of transport to school, as shown by the bar graph given below:
Figure
Look at the bar graph given above and answer the questions given below:
(i) What does the above bar graph show?
(ii) Which mode of transport is used by maximum number of students?
(iii) How many students use bus for going to school?
(iv) How many students of the locality do not use bus for going to school?

ANSWER:

(i) The given bar graph gives information about the different modes of transport used by the students for going to school.

(ii) Since the height of the bar representing the bicycles is the highest, bicycles are used by maximum number of students. 16 students use bicycles for going to school.

(iii) 14 students use bus for going to school.

(iv) Number of students who do not use bus = Total number of students − Number of students who use bus
                                                                      = 51 − 14
                                                                      = 37 students

Page No 213:

Exercise 19A

Question 1:

Mark (✓) against the correct answer in each of Q.1 to Q.6.
A cuboid has
(a) length
(b) length and breadth only
(c) length, breadth and height
(d) thickness only

ANSWER:

A cuboid is a 3-dimensional figure. So, a cuboid has length, breadth and height. 

Hence, the correct answer is option (c).

Page No 213:

Question 2:

Mark (✓) against the correct answer in each of Q.1 to Q.6.
A dice is an example of a
(a) cuboid
(b) cube
(c) cone
(d) cylinder

ANSWER:

(b) cube

Page No 213:

Question 3:

Mark (✓) against the correct answer in each of Q.1 to Q.6.
A gas pipe is an example of a
(a) cuboid
(b) cube
(c) cone
(d) cylinder

ANSWER:

(d) cylinder

Page No 213:

Question 4:

Mark (✓) against the correct answer in each of Q.1 to Q.6.
A football is an example of a
(a) cylinder
(b) cone
(c) sphere
(d) none of these

ANSWER:

(c) sphere

Page No 213:

Question 5:

Mark (✓) against the correct answer in each of Q.1 to Q.6.
A brick is an example of a
(a) cube
(b) cuboid
(c) prism
(d) cylinder

ANSWER:

(b) cuboid

Page No 213:

Question 6:

Mark (✓) against the correct answer in each of Q.1 to Q.6.
An ice-cream cone is an example of a
(a) cuboid
(b) cube
(c) pyramid
(d) none of these

ANSWER:

(d) None of these
Its a cone.

Page No 213:

Question 7:

Fill in the blanks:
(i) An object that occupies space is called a …… .
(ii) A cuboid has …… faces, …… edges and …… vertices.
(iii) The …… faces of a cuboid are identical.
(iv) A …… has no vertex and no edge.
(v) All the faces of a …… are identical.
(vi) A square pyramid has …… lateral triangular faces and …… edges.
(vii) A triangular pyramid has …… triangular lateral faces and …… edges.
(viii) A triangular prism has …… vertices, …… rectangular lateral faces, …… triangular bases and …… edges.

ANSWER:

(i) An object that occupies space is called a solid.
(ii) A cuboid has 6 rectangular faces, 12 edges and 8 vertices.
(iii) The opposite faces of a cuboid are identical.
(iv) A sphere has no vertex and no edge.
(v) All the faces of a cube are identical.
(vi) A square pyramid has 4 lateral triangular faces and 8 edges.
(vii) A triangular pyramid has 3 triangular lateral faces and 6 edges.
(viii) A triangular prism has 6 vertices, 3 rectangular lateral faces, 2 triangular bases and 9 edges.

Page No 213:

Question 8:

Give examples of four objects which are in the shape of:
(a) a cone
(b) a cuboid
(c) a cylinder

ANSWER:

(a) An ice cream cone, a conical tent, a clown’s cap and a conical vessel are in the shape of a cone.

(b) A wooden box, a match box, a brick and an almirah are in the shape of a cuboid.

(c) A measuring jar, a gas cylinder, a test tube and a circular pillar are in the shape of a cylinder.

Page No 218:

Exercise 20A

Question 1:

Mark (✓) against the correct answer in each of Q.1 to Q.8.
A square has
(a) one line of symmetry
(b) two lines of symmetry
(c) three lines of symmetry
(d) four lines of symmetry

ANSWER:

A square has four lines of symmetry. Hence, the correct answer is option (d).

Page No 218:

Question 2:

Mark (✓) against the correct answer in each of Q.1 to Q.8.
A rectangle is symmetrical about
(a) each one of its sides
(b) each one of its diagonals
(c) a line joining the midpoints of its opposite sides
(d) none of these

ANSWER:

(c) a line joining the midpoints of its opposite sides

Page No 218:

Question 3:

Mark (✓) against the correct answer in each of Q.1 to Q.8.
A rhombus is symmetrical about
(a) the line joining the midpoints of its opposite sides
(b) each of its diagonals
(c) perpendicular bisector of each of its sides
(d) none of these

ANSWER:

(b) each of its diagonals

Page No 218:

Question 4:

Mark (✓) against the correct answer in each of Q.1 to Q.8.
A circle has
(a) no line of symmetry
(b) one line of symmetry
(c) two lines of symmetry
(d) an unlimited number of lines of symmetry

ANSWER:

(d) an unlimited number of lines of symmetry

This is because a circle has infinite number of diameters. Also, a circle is symmetrical about each of its diameter.

Page No 218:

Question 5:

Mark (✓) against the correct answer in each of Q.1 to Q.8.
A scalene triangle has
(a) no line of symmetry
(b) one line of symmetry
(c) two lines of symmetry
(d) three lines of symmetry

ANSWER:

(a) no line of symmetry

Page No 219:

Question 6:

Mark (✓) against the correct answer in each of Q.1 to Q.8.
ABCD is a kite in which AB = AD and BC = DC.
The kite is symmetrical about
(a) the diagonal AC
(b) the diagonal BD
(c) none of these
Figure

ANSWER:

(a) the diagonal AC

Page No 219:

Question 7:

Mark (✓) against the correct answer in each of Q.1 to Q.8.
The letter O of the English alphabet has
(a) no line of symmetry
(b) one line of symmetry
(c) two lines of symmetry
(d) none of these

ANSWER:

(c) two lines of symmetry

Page No 219:

Question 8:

Mark (✓) against the correct answer in each of Q.1 to Q.8.
The letter Z of the English alphabet has
(a) no line of symmetry
(b) one line of symmetry
(c) two lines of symmetry
(d) none of these

ANSWER:

(a) no line of symmetry

Page No 219:

Question 9:

Mark (✓) against the correct answer in each of Q.1 to Q.8.
Draw the line (or lines) of symmetry of each of the following figures.
Figure

ANSWER:

(i)


(ii)


(iii)

(iv)

Page No 219:

Question 10:

Mark (✓) against the correct answer in each of Q.1 to Q.8.
Which of the following statements are true and which are false?
(i) A parallelogram has no line of symmetry.
(ii) An angle with equal arms has its bisector as the line of symmetry.
(iii) An equilateral triangle has three lines of symmetry.
(iv) A rhombus has four lines of symmetry.
(v) A square has four lines of symmetry.
(vi) A rectanle has two lines of symmetry.
(vii) Each one of the letters H, I, O, X of the English alphabet has two lines of symmetry.

ANSWER:

(i)   True

(ii)  True

(iii) True
An equilateral triangle is symmetrical about each one of the bisectors of its interior angle. Also, it has three bisectors.

(iv) False
A rhombus has two lines of symmetry. It is symmetrical about each one of its diagonals.

(v)  True
A square is symmetrical about each one of its diagonals and the lines joining the midpoints of the opposite sides.

(vi) True
A rectangle is symmetrical about the lines joining the midpoints of the opposite sides.

(vii) True