Factors and Multiples

Factors

        Factors can be defined as the numbers which we can multiply together to get another number.

Example

2 * 3 = 6 , here 2 and 3 are the factors 6

2  *4 = 8 , 2 and 4 are the factors of 8.

Note : A number can have many factors

Example

1 * 12 = 12

2 * 6 = 12

3 * 4 = 12

 so the factors of 12 are 1 , 2 , 3 , 4 , 6 and 12.

Multiples

        A multiple is the result of multiplying a number by an integer.

Example

Multiples of 2

2 * 1 = 2

2 * 2 = 4

2 * 3 = 6 ,...

so the multiples of 2 are 2 , 4 , 6 , 8 , ...

Note : Factors and multiples are different but they both involve multiplication.

Class 6 : Excersice 10.3

10. By splitting the following figures into rectangles, find their areas ( The measures are given in cms )

a.

Solution: 

First we have to broken the given figure into rectangles as shown above (in blue ).

now lets find each rectangles area

Area of 1st Rectangle :
length = 4cm , width = 2cm
Area = 4 * 2 = 8 cm square

Area of 2nd rectangle :
length = 6cm , width = 1cm
Area = 6 * 1 = 6 cm square

Area of 3rd rectangle :
length = 3cm , width = 2cm
Area = 3 * 2 = 6 cm square

Area of 4th rectangle :
length = 4cm , width = 2cm
Area = 4 * 2 = 8 cm square

Area of the given figure = area of 1st rectangle + area of 2nd rectangle
                                         + area of 3rd rectangle + area of 4th rectangle  
                                       = 8 + 6 + 6 + 8 cm square
                                       = 28 cm square

Class 6 : Excersice 10.3

12. How many tiles whose length and breath are 12cm and 5cm respectively will be needed to fit in a rectangular region whose length and breath are respectively :

a. 100cm and 144cm.

Solution

first let we find
 the total area of the region = 100 * 144
                                            = 14400 cm square
area of one tile = 5 * 12
                         = 60 cm square
Number of tiles required = 14400 / 60
                                        = 240

b. 70 cm and 36 cm

Solution

first let we find
 the total area of the region = 70 * 36
                                            = 2520 cm square
area of one tile = 5 * 12
                         = 60 cm square
Number of tiles required = 2520 / 60
                                        = 42

Class 6 CBSE : Chapter 10

11. Split the following shapes into rectangles and find their areas. ( The measurement given in cm).

a.

in the above picture,

first let we name the corners as shown in the figure
here, we have to shapes into rectangle
so, to split the given shape into rectangles lets draw a line (in orange) 
and name that as CX.
given, AB = 2cm
hence CX also 2cm 
since ABCX forms rectangle (opposite sides are equal in length)
BC = 10cm which means AX = 10cm
1st rectangle ABXC
area of reactangle ABXC = 10 * 2
                             = 20 cm square 
2nd rectangle XDFE
given, AD = 12, we knew AX = 10
so, XD = AD - AX = 12 - 10
       XD = 2cm
XD = EF = 2cm
given, DF = 10cm
XE = XC + CE = 2 + 8 =10
so, XE = 10cm
now we know all the sides measurement
lets find its area now
Area of rectangle XDFE = 10 * 2
                                        = 20 cm square
we have to find  
Area of given shape = area of rectangle ABXC + area of rectangle XDFE
                                 =              20                       +               20
                                 =   40 cm square.

b. 

in the above figure 
first let we name all the corners as ABCDEFGHIJKL
lets split the above shape into square ABLC, rectangle KDEJ and another square IFGH
 Area of square ABLC  = 7 * 7 = 49 cm square
Area of square IFGH = 7 * 7 = 49 cm square
Area of rectangle KDEJ 
here, KD = KL + LC + CD = 7 + 7 + 7 = 21
Area of rectangle KDEJ = 21 * 7 = 147 cm square   
Area of  ABCDEFGHIJKL = Area of square ABLC  +  Area of square IFGH + 
                                                                                                      Area of rectangle KDEJ  
                                                        =  49 + 49 + 147
                                                        = 245 cm square

c.  

 













In the above figure,
lets first name the corners ABCDEFGH
we split the above given shape into two rectangles
ABCD and EFGH
Area of rectangle ABCD :
CD = CE + EF + FD = 2 + 1 + 2 = 5 ( GH = 1,so EF =1 )
AB = CD = 5 , AC = BD = 1
Area of rectangle ABCD = 5 * 1 = 5 cm square

Area of rectangle EFGH :
EF = GH = 1
EG = FG = 4
Area of rectangle EFGH = 1 * 4 = 4 cm square

Area of the given shape = Area of rectangle ABCD + Area of rectangle EFGH
                                       =      5   +   4
                                       = 9 cm square

Grade 1 : Numbers

Numbers :

Numbers are the basic thing of mathematics. Numbers are used to count and to measure.

Positive Number :

A positive number is a number which is above 0. It can be written as +1, +2, +3,... or just number itself like 1, 2, 3,...

Negative Number :

A negative number is a number which is below 0. It can be written as -1, -2, -3,...

Example :

Positive and negative numbers are used  in everyday life is in measuring temperature.

If the temperature falls below 0 degree C, we can use negative numbers.

Natural Number :

The numbers which we use to count is called as Natural Number.

1, 2, 3, 4,.... are some of natural numbers

Whole Number :

Whole numbers are positive numbers including 0 without any decimal or fraction.

0, 1, 2, 3, 4,... are some of whole numbers.

NOTE : All natural number are Whole number and not vice versa because 0 is not a natural number.

Integers :

An integer is a whole number that can be positive , negative or zero.

that is, it includes

                    * counting numbers ( Natural number )

                    * Zero

                    * negative of the counting numbers (-1, -2, -3,...)

So, ...,-3, -2, -1, 0, 1, 2, 3, ...

NOTE : We can say all natural and whole numbers are integers. 

Even Number : 

Any integer that can be divided by 2 without leaving remainder. 

2, 4, 6, 8,... . Even number can also be in negative.

In simple we can say, even number is an number which we can share equally into two portions.

Odd Number :

 Any integer that cannot be divided by 2 without leaving remainder.

1, 3, 5, 7,...
Odd number can also be in negative.

In simple we can say, odd number is an number which we cannot share equally into two portions.

Math : Word Problem

Two tankers contain 680 liters and 850 liters of diesel respectively. Find the maximum capacity of a container which can measure the diesel of both tankers in exact number of times.

Given,
Total number of tankers = 2
first tanker contains 680 liters
second tanker contains 850 liters
we have to find,
the maximum capacity of a container which can measure the diesel of both tankers
which means we have to find the HCF (Highest common factor)
lets find the factors for 680 and 850
680 =  2*2*2*5*17
850 = 2*5*5*17
HCF = 2*5*17 = 170
So, 170liters container is common for 680liters tanker and 850 liters tanker to fill the tankers in exact number of times
170 *4 = 680
hence to fill 680 liters tankers, 4 times of 170liters container be used
170 * 5 = 850
hence to fill 850 liters tankers, 5 times of 170 liters container be used.

Grade 6 : Math : Exercise 1.2

1. A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final day was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days?

Given,
Exhibition was held for four days
Number of tickets sold on first day = 1094
Number of tickets sold on second day = 1812
Number of tickets sold on Third day = 2050
Number of tickets sold on fourth day = 2751
now, we have to find the total number of tickets sold on all four days
which implies
we have to add all the four days tickets 
that is,
= 1094 + 1812 + 2050 + 2751
= 7,707 tickets

2. Shekar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?

Given,
Total number of scores Shekar scored = 6980 runs
Number of scores he wants to complete = 10,000 runs
we have to find how many more runs he need
that is, he wants to reach 10,000 runs from 6,980 runs
which implies we have to subtract
= 10,000 - 6,980
=3,020 runs

3. In an election, the successful candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. By what margin did the successful candidate win the election?

Given,
Number of votes successful candidate has secured = 5,77,500 votes
Rival has secured = 3,48,700 votes
we have to find at what margin did the successful candidate win the election
which means in what difference did the successful canditate win
that is, = 5,77,500 - 3,48,700
            = 2,28,800 votes

4. Kirti bookstore solds books worth Rs 2,85,891 in the first week of June and books worth Rs 4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much?

Given,
First week, bookstore solds books worth = Rs 2,85,891
Second week, bookstore solds books worth = Rs 4,00,768
The sale for the two weeks together (means add two weeks worth ) 
                                                              = 2,85,891 + 4,00,768
                                                              = Rs 6,86,659
Also we have to find in which week the sale greater
4,00,768 > 2,85,891
so, second week sale is greater
if we find their difference = 4,00,768 - 2,85,891
                                           = Rs 1,14,877 worth greater than first week sale

5. Find the difference between the greatest and the least number that can be written using the digits 6,2,7,4,3 each only once.

Given,
digits given = 6,2,7,4,3
Greatest number using the given digits only once = 76432 (go from highest to lowest)
Least number using the given digits only once = 23467 (from lowest to greatest)
Their difference = 76432 - 23467
                           = 52,965

6. A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?

Given,
A machine manufactures 2,825 screws per day
we have to find how many it produce in the month of january
there are 31 days in January
so, total number of screws it produced in january = 31*2,825
                                                                                = 87,575 Screws

7. A merchant had Rs 78,592 with her. Sha placed an order for purchasing 40 radio sets at Rs 1200 each. How much money will remain with her after purchase?

Given,
Amount merchant had = Rs 78,592
She wanted to purchase 40 radio sets
The cost of one radio set = Rs 1200
so, the cost of 40 radio sets = 40 * 1200
                                           = Rs 48,000
the remaining amount she had after purchasing = 78,592 - 48,000
                                                                            = Rs 30,592

Problems based on the divisibility test

1. Replace the sign ___ by the smallest number to make them divisible by the given number.

a. 7158__ by 6

By the divisibility test,
if the number is divisible by 2 and 3 then that number is divisible by 6.
if the given number is divisible by 2 the units place must be 0 , 2, 4 , 6 , or 8.
if the given number is divisible by 3 then first we have to add the digits 7+1+5+8=21
21 is divisible by 3
so we can replace the sign by 0 since 71580(unit place 0) which is divisible by 2 and 71580(add will get 21) which is divisible by 3.
hence 71580 is divisible by 6

b. 8152__ by 4

By the divisibility test,
if the last two digits is divisible by 4 then that number is divisible by 4.
if we replace the sign by 0 then the last two digits be 20 which is divisible by 4.
hence 81520 is divisible by 4

c. 86__72 by 11

By the divisibility test,
the sum of the numbers in odd place = 8 + __+2  = 10+__
sum of the numbers in even place = 6 + 7  = 13
10+__  - 13 = 0 (since the difference must be either 0 or divisible by 11)
[we have to write the smallest possible number hence we choose 0]
so __ be 3
13-13=0
hence 86372 is divisible by 11

d. 631__24 by 8

By the divisibility test,
if the last three digit is divisible by 8 then that number is divisible by 8.
__24 , 24 itself is divisible by 8 hence we can replace the sign by 0
hence 631024 is divisible by 8.

2. State whether the following sentences are true or false.

a. 231648 is divisible by 4 and 8

Divisible by 4
lets consider last two digits 48 which is divisible by 4.
Divisible by 8
lets consider last three digits 648 which is divisible by 8
hence 231648 is divisible by 4 and 8 TRUE

b. 976575 is divisible by 3 and 15

Divisible by 3
lets add all the digits 9+7+6+5+7+5=39 which is divisible by 3
Divisible by 15
if the number is divisible by 3 and 5 then that number is divisible by 15
we already found that number number is divisible by 3
the unit place is 5 so the given number is divisible by 5 
hence it is divisible by 15
976575 is divisible by 3 and 15 TRUE


c. 567835 is divisible by 5 and 25  

Divisible by 5
the unit place in the given number 5, so the given number is divisible by 5
Divisible by 25
if the last two digits are 00,50 or 75 then that number is divisible by 25
here the last two digits is 35, so 567835 is not divisible by 25
567835 is divisible by 5 and 25 FALSE  

3. What digit you add each time to make the numbers on the Violet divisible by the numbers given in the RED

a. 67984311 by 2,3,4,5,6,8,9,10

Divisible by 2
if we add last digit as 0 then according to the divisibility test,
679843110 is divisible by 2

Divisible by 3
lets add all the digits 6+7+9+8+4+3+1+1+__ = 39+__
39 itself is divisible by 3
hence we can replace the sign as 0
679843110 is divisible by 3

Divisible by 4
the last two digits 1__, if we replace sign by 2 then 12 which is divisible by 4
hence 679843112 is divisible by 4

Divisible by 5
the last digit must be either 0 or 5
hence it can be 679843110 or 679843115 which are divisible by 5

Divisible by 6
According to the divisibility test,
the number must be divisible by both 2 and 3
if we add 0 that is 679843110 is divisible by both 2 and 3.
hence 679843110 is divisible by 6

Divisible by 8
lets consider the last three digits 11__ ,
110,111 are not divisible by 8
112 is divisible by 8
hence 679843112 is divisible by 8

Divisible by 9
Sum of the digits = 39 + __ 
39 is not divisible by 9
if we add 6 to 39 which gives 45 and 45 be divisible by 9
hence 679843116 is divisible by 9

Divisible by 10
the last digit must be 0
hence 679843110 is divisible by 10