Counting of Figures

Counting of Figures
(1) Counting of Triangles:-

In the counting of triangles, we have a figure. From the given figure we will have to

 identify how many triangles in a given known figure . 

 TYPE 1:-
Example 1. consider the following figure

                     There are 3 triangles in figure.

                     In single horizontal line sum of inside triangles 1+2=3

Example 2. consider the following figure

                    There are 6 triangles in figure.1,2,3,4,1+2,2+3

                    In single horizontal line sum of inside triangles 1+2+3=6

Example 3. consider the following figure

      There are 10 triangles in figure.1,2,3,4,1+2,2+3,3+4,1+2+3,2+3+4,(1+2+3+4) greater triangle

                    In single horizontal line sum of inside triangles 1+2+3+4=10

 TYPE 2:-

Example 4. consider the following figure

There are 6 triangles in figure.1,2,1+2,1+3,2+4,(1+2+3+4) greater triangle

                    In two horizontal lines sum of inside triangles 2(1+2)=6.

Example 5. consider the following figure

There are 20 triangles in figure.2✖{1,2,3,4,1+2,2+3,3+4,1+2+3,2+3+4,(1+2+3+4) greater triangle} 

                    In two horizontal lines sum of inside triangles 2(1+2+3+4)=20.

Example 6. consider the following figure

There are 30 triangles in figure.3✖{1,2,3,4,1+2,2+3,3+4,1+2+3,2+3+4,(1+2+3+4) greater triangle} 

                    In three horizontal lines sum of inside triangles 3(1+2+3+4)=30.

 TYPE 3:-

Example 7. consider the following figure

There are 5 triangles in figure.

Triangle inside triangle 1 hence number of triangles = 4✖1 +1 (greater triangle) = 5 

Example 8. consider the following figure

There are 9 triangles in figure.

Triangle inside triangles 2 hence number of triangles = 4✖2 +1 (greater triangle) = 9 

Example 9. consider the following figure

There are 13 triangles in figure.

Triangle inside triangles 3 hence number of triangles = 4✖3 +1 (greater triangle) = 9 

TYPE 4:-

Example 10. consider the following figure

There are 11 triangles in figure.

Triangle inside triangle 1 hence number of triangles = 4✖1 +1 (greater triangle) = 5 

and triangles 6 are (1,2,3,4,5,6) 

Hence total triangles 5+6 = 11

Example 11. consider the following figure

There are 15 triangles in figure.

Triangle inside triangle 2 hence number of triangles = 4✖2 +1 (greater triangle) = 9 

and triangles 6 are (1,2,3,4,5,6) 

Hence total triangles 9+6 = 15

TYPE 5:-

Example 12. consider the following figure

Triangles 1,2,3,4,1+2,2+3,3+4,4+1

i.e. 4✖2=8

There are 8 triangles

Example 13. consider the following figure

Example 14. consider the following figure

Example 15. consider the following figure

Example 16. consider the following figure

Example 17. consider the following figure

Example 18. consider the following figure

Example 19. consider the following figure

Example 20. consider the following figure

(2) Counting of Square:-

In the counting of squares, we have a figure. From the given figure we will have to

 identify how many squares in a given known figure . 
Type-1
If number of rows = number of columns =n
Then number of squares = ∑n²

Example 1.:-consider the following figure

Type-2
If number of rows = n,number of columns =m

Then number of squares = m✖n +(m-1)✖(n-1) + ............

Example 2.:-consider the following figure

Type-3

Example 3. :-consider the following figure

Alligation and Mixture

 Alligation and Mixture

                            Alligation is the rule that find the ratio in which two or more ingredients 

                            at the given price must be mixed to produce a mixture of a desired price.

Mean Price :-The cost price of unit quantity of the mixture.

Rule of alligation :-If two ingredients are mixed,then

      Cheaper quantity : Dearer quantity = (D-M) : (M-C)                

Replacement of part of mixture :-Suppose a container contains a solution from which
                                                     some quantity of solution is taken out and replaced with
                                                    one of the ingredients.This process is repeated n times then  

Example 1.:-In what ratio must rice at Rs.40 /kg to mix rice at Rs.60 /kg,so that  the average cost of mixture is Rs.55 /kg ?

Explanation :-

Hence the answer is 60-55 : 55-40 = 5:15

                                                             1:3

 Example 2.:-

In what ratio must a grocer mix two varieties of pulses costing Rs. 15 and  

                       Rs. 20 per kg respectively so as to get a mixture worth Rs. 16.50 kg?

Explanation :-

              

Hence the answer is 20-16.50:16.50-15=3.50 :1.50

                                                             7:3

Example 3.:-How many kilogram of sugar costing Rs. 9 per kg must be mixed with 27 kg

                      of sugar costing Rs. 7 per kg so that there may be a gain of 10% by selling

                      the mixture at Rs. 9.24 per kg?

Explanation :-S.P. of 1 kg of mixture = Rs. 9.24, 

                        Gain 10%.

C.P. of 1 kg of mixture = Rs.		100	x 9.24		= Rs. 8.40
		110

By the rule of alligation, we have:

           Ratio of quantities of 1^st and 2^nd kind = 1.40 : 0.60 = 7 : 3.

           Let x kg of sugar of 1^st be mixed with 27 kg of 2^nd kind.

          Then, 7 : 3 = x : 27

               x = (27✖7)/3 = 63

Example 4.:-Two containers A and B contain milk and water in the ratio of 5 : 2 and 

                     7 : 6 respectively. Find the ratio in which these two mixtures can be mixed

                     so that a new mixture formed in the container C is in the ratio of 8 : 5. 

Explanation :-Let the cost price of milk be Rs. 1 per liter.

              Therefore, cost of milk in 1 liter of mixture in

Container A (Milk : Water = 5 : 2) =5/7 ✖Rs.1

=5/7

Container B (Milk : Water = 7 : 6) =

7/13✖Rs.1 = 7/13

Container C (Milk : Water = 8 : 5) =

8/13✖Rs.1 = 8/13

             Now by alligation  rule , 

The required ratio of milk and water = 1/13 : 9/91
                                                               7 : 9

Example 5.:-A container contains 40 litres of milk. From this container 4 litres of milk

                      was taken out and replaced by water. This process was repeated further

                      two times. How much milk is now contained by the container?

Explanation :-

Example 6.:-8 liters are drawn from a  container full of milk. then filled with water

                        This process was repeated three more times. The ratio of the quantity of

                        milk now left in container to that of water is 16:65.How much milk did

                        the container hold originally?

Explanation :-Let the quantity of milk in the container is x liters.

                         Then  the quantity of milk in the container after 4 operations 
                                                                           

                                                                    x = 24 liters

               

Problems On Number

Problems On Number
Example 1.
                A number is as much less than 54 is more than 24.Find the number?
Explanation :-Let the number be x.
                        Given 54-x = x-24
                                        2x = 54 + 24= 78
                                         x  = 78/2= 39
Example 2.

                 If the digits of a two-digit number are interchanged, the number so formed is 18

                 more than the original number. The sum of digits is 8. What is the thrice value of

                 original number?

Explanation :-(1) Let the unit digit of number is y and tens place digit of number is x

                          Then original number is 10x + y.

                          Interchanging digit number is 10y + x.

                           10y + x = 10x + y +18.

                            -9x + 9y = 18

                            -x + y = 2     ............(1)

                             x + y = 8      ...........(2)

                             2y = 10

                               y = 5

                               x = 3

                     Hence the original number is 10x + y = 10✖ 3 + 5 = 35.
                   (2) Let the unit digit of number is x
                         then tens digit f number is 8-x
                        Then original number is 10(8-x) + x = 80-9x.
                         Interchanging digit number is 10x + (8-x) = 8 + 9x
                         8 + 9x = 80 - 9x + 18
                         18x = 90
                             x = 5
                      Hence the original number is 80 - 9x = 80 -9✖ 5 = 35
Example 3.

                 Sum of two numbers is 24 and their product is 140, then find the numbers? 

Explanation :- Let one number be x then another number be 24-x.
                              x✖(24-x) = 140
                              24x-x²= 140                               

                            x²-24x+140=0
                             x²-14x-10x+140=0
                             x(x-14)-10(x- 14)=0
                             (x-14)(x-10) =
                             x=14,     x = 10
Example 4.

                The  ten’s place digit of a two digit number is twice  the unit’s place digit.
                If the sum of this number and the number formed by interchanging  the digits is 66. 
                Find the number?

Explanation :- Let unit’s place digit of number be x 
                        Ten’s place digit = 2x .
                        Then original number is 10(2x) + x = 21x.
                         Interchanging digit number is 10x + 2x = 12x
                        21x + 12x = 66
                        33x = 66
                           x = 2
                     Hence original number is = 21x= 42
Example 5.

                The sum of the digits of a two digit number is 10. 
                The tens place digit is 4 times the unit place digit. Find the number?

Explanation :- Let unit’s place digit of number be x 
                          and tens place digit of number is 10-x

                          Then original number is 10(10-x)x + x = 100-9x.

                          10-x = 4x
                           5x = 10
                             x = 2
                         Then original number is 10(10-x)x + x = 100-9x
                                                                                        = 100 -18
                                                                                        = 82
Example 6.

                The sum of the digits of a two digit number is 9.
                 If the digits are reversed, the new number is 9 less than 3 times
                 the original number. Find the original number.

Explanation :- Let unit’s place digit of number be x 
                          and tens place digit of number is 9-x

                          Then original number is 10(9-x)x + x = 90-9x.
                           Interchanging digit number is 10x +( 9-x )= 9x + 9
                           9x+9 = 3(90-9x) - 9
                             36x = 252
                                x = 7
                        Hence original number is = 90-9x = 90-63= 27.
Example 7.             

              A two-digit number is such that the product of the digit 8. 

              When 18 is added to the number the digit are reversed. What is the number?

Explanation :- Let unit’s place digit of number be x 

                          then  ten’s place digit of number be 8/x
                          Then original number is 10✖8/x + x = 80/x+x.
                           Interchanging digit number is 10x +8/x
                           80/x + x +18 = 10x +8/x
                           80 +x²+18x = 10x²+8
                           9x²-18x-72 = 0
                            x²-2x-8 = 0
                             x²-4x+2x-8 = 0
                             x(x-4)+2(x-4)=0
                             x=4,x=-2
                        Hence original number is = 80/4 +4 .
Example 8.

                 The digits of a two-digit number differ by 5. what is the difference between

                  the number and the number formed by reversing its digits?

                

Explanation :- Let unit’s place digit of number be x 
                          and tens place digit of number is 5+x

                          Then original number is 10(5+x)x + x = 50+11x.
                          Interchanging digit number is 10x +5+x=5+11x
                          (50+11x)-(5+11x)=45
                          Hence the number and interchanging number differ by 45.