COMPETING COMPARATIVE PROBLEMS EASILY EASY

COMPETING COMPARATIVE PROBLEMS EASILY EASY

Comparison is the concept of comparing two or more magnitudes to see which one is greater, whichever is smaller, how much difference, how much is it and so on. For example, comparing the number of chickens with the price, many workers with time finished work and others. The stratified comparison is comparing more than two magnitudes, either three or four or more. For junior level, the stratified comparison is only up to level three. Ability of average junior high students still not good in solving the problem of stratified comparison of sola. This is evident from the results of the UN 2017 yesterday which averages put a matter of stratified comparison into the category of difficult questions. In this case required a new breakthrough by us as a teacher or mathematical mathematics UN. With such an easy solution it is expected that the matter of stratified comparisons is no longer a scourge for our students. This opportunity allows admins to share the multilevel comparative learning experiences that our students find far more easily understood

Problem-comparison comparison of multilevel has two categories, then admin will give an example. Of course, about the problem that we discussed become an important part in the Grid of the National Examination from 2016 until 2019 using SKL slices of curriculum 2006 and the curriculum 2013, about the matter of storied comparison will always come out adorn the queried about the National Exam junior high. Immediately, we go to the example of multi-level comparisons. A. Category matter first tiered comparison 1. Comparison of money A and B is 2: 5, while the ratio of money B and C is 3: 4. If the sum of money them three Rp820.000,00, then the difference of money A and C is .... A. Rp100.000 , 00 C. Rp280.000,00 B. Rp180.000,00 D. Rp300.000,00

Solved 

Then A: B: C = 6: 15: 20

(6 + 15 + 20) / (20-6) = (820 000) / x

41/14 = (820 000) / x

X = (820 000 x 14) / 41 = 280 000, where x is the difference A with C

So the difference A with C is Rp 280.000,00

Categories of second-tier comparison

The comparison of Furqon, Aldi and Hidayat marbles is 4: 1: 5. If the difference between many 

marbles Hidayat and aldi 320 grains, the number of marbles they  ...

540 items
640 items
800 items
900 items
 
Answer
Difference between Hidayat and Aldi = 4
Difference of marbles Hidayat with Aldi = 320
Number of Third Comparisons = 10
Number of marbles three = x
4/320 = 10 / x
 
x = (320 x 10) / 4 = 800 items

Let's practice the matter of otherstratified comparisons. 




1. Comparison of money A: B: C is 6: 3: 5. If the amount of money A and C is Rp88.000,00,

 then the amount of money is three ....


A. Rp122.000,00 

B. Rp102.000 , 00 

C. Rp112.000,00 

D. Rp92.000,00 




2. Comparison
of money Beni and Rita money is 2: 3, while the ratio of money Rita and Susi 4 : 5. If ​​their money 

amount Rp140.000,00, then the difference between Rita and Susi money is ....

 A. Rp12.000,00       C. Rp28.000,00 

 B. Rp16.000,00        D. Rp32.000,00


It's easy to do. Hopefully with this simple way learners will be easy in
solving this stratified comparison problem easily

solve the problem of widespread two dimensional wake mathematics

solve the problem of widespread two dimensional wake mathematics

The development of mathematical concepts needs to be emphasized in a simpler form that is more easily understood by learners. As good as any educator who teaches with thousands of scientific treasures will not be enough to make students understand the concept of mathematics. The focal point of educating mathematics lies in the simplicity of the teaching style and the content taught, of course this is only relevant to the level of junior high school education. A higher level of education may require deep deductive axiomatic standards in mathematical study.

Step educator in simplifying the concept of mathematical applications in the start of doing the discovery activities of the concept of concept in question. At this time I will try to share to the father / mother or younger sister about how to simplify the calculation of Reduction Area two-dimentional figure a straight build using smart solution. My experience during teaching with this technique is more understandable to students.

Consider the Case below.

Look at the following garden sketch drawings!

Given a plot of parallelogram parallelogram EFGH. Inside the garden is made garden with long 
ABCD the rest is the road. Define Street area!
 
Case analysis
Because ABCD and EFGH are similar then EFGH is the result of enlargement of ABCD.
The scale of the magnification (s) = EF / AB
 EF = s. AB and KH = s. DJ
 
Then the Area of ​​EFGH = EF. KH
                           = s.AB. s. DJ
                           = s2 AB. DJ
                              = 〖(EF / AB)〗 ^ 2.Large ABCD
 
Area of ​​Road = Area of ​​EFGH - Area ABCD
 
= 〖(EF / AB)〗 ^ 2. Area ABCD - Area ABCD
 
= ((〖EF〗 ^ 2 / 〖AB〗 ^ 2) - 1) Area ABCD
= ((〖EF〗 ^ 2- 〖AB〗 ^ 2) / 〖AB〗 ^ 2) Area of ​​ABCD
By the same analogy
 
Area of ​​road = ((〖EF〗 ^ 2- 〖AB〗 ^ 2) / 〖EF〗 ^ 2) Area of ​​EFGH
 
 
Example Use of the formula to solve mathematical problems
 
(UN Junior High School 2016)


Look at the following garden sketch drawings!


A plot of parallelogram parallelogram. Inside the garden is a garden with length AB = 20 m and 

length DJ = 15 m. Around the park will be made way. If the gardens and parks are similar, then 
the wide road is ....
A. 66 m2 C. 300 m2
B. 132 m2 D. 360 m2
 
Answer
From the image AB = 20 m and EF = 20 + 4 = 24 m
Then AB: EF = 20: 24 = 5: 6
Area of ​​the Road = ((6 ^ 2-5 ^ 2) / 5 ^ 2) 20 x 15
= (36-25) / 25 x 300
= 11/25 x 300 = 132 m2
2. A photo is affixed to a carton as shown


On the top left part of the photo

There is 5 cm cardboard remaining. If the photo
and cartons of widespread carton
which is not closed photo is ...
a. 750 cm2
b. 850 cm2
c. 1,050 cm2
d. 1.350 cm2


Answer
Photo width = 40-5-5 = 30
Photo: carton = 30: 40 = 3: 4
Uncovered cardboard area of ​​photograph = ((4 ^ 2-3 ^ 2) / 4 ^ 2) .40 x 60
= (16-9) / 16 x 240
= 7/16 x 240 = 105 cm2
 
That's the smart solution to solving the widespread problem of 

two-dimentional figure. Hope to benefit.

smart solution solves the problem of AVERAGE math subjects

smart solution solves the problem of AVERAGE math subjects

Good night Mr maternal mathematics mathematician who is surfing looking for a quick way to solve the problem of the average frequency distribution. Mathematics problems concerning the average problem does have its own challenges for students. There are those who think it is also difficult to solve the problem that contains the average problem with the distribution of frequency tables. For students with the ability to calculate the good multiplication and meticulous it is not a problem but for those of you who have the ability to count that is not too good sometimes because the lack of telitian causes an incorrect count until the correct answer was not also obtained. Especially if you're grappling with the national exam that is almost always present every year adorn the test paper.

Consider the tips to be true in working out the matter of the average frequency distribution

1. Always improve student's simple counting skills.

2. Diligently facilitate the child to practice about the average frequency distribution to improve the accuracy of the calculation

3. Help students to have confidence while working on the problem. This is so that children can focus on working on the problem, not depend on other friends.

4. Find a lot of references on how to quickly work out the question of the average form of frequency distribution.

        Well of the four tips above admin will tangent item no 4 that is how to quickly work out the matter of the average form of frequency distribution. The fast way here does not mean there is a special formula inside but only simplifies the calculation into smaller calculations. This will really help you not to get caught in a miscalculation.

Consider the following reviews.

Notice the table below.

Nilai	50	60	70	80	90
Frekuensi	2	4	x	6	3

The table above shows the data of mathematical repetition value of a group of students. If the average 

value of the data is 72 then many students who score below the average value are ....

A. 11

B. 10

C. 9

D. 6

Method 1

If done the usual way using a table like the following

the results obtained the same that students who get 70 is 5 people. So that got a score 
below the average 72 as many as 11 students
 
 
The calculation of the second way is fewer, fewer steps, fewer numbers. With such an 
assumption it will make the students more easily understand and solve the problem of 
the average form of frequency distribution table.

Smart Solutions Complete the geometry sequence

In solving the problem of geometry sequences is not an easy thing, it takes a smart way so that children can quickly absorb the material. Based on the observation and experience, the national exam always uses a fairly easy ratio for junior level, usually in geometry sequence using ratio 2, 3, 1/2 or 1/3. Rarely use more ratio than that. It can not be separated from the assessment of junior high level with all the limitations of its ability to count. So when we teach our students about the geometry sequence then simply tell them "kids do not have to worry, the geometry sequence is easy as long as you can multiplication 2 or 3, 1/2 and or 1/3 then about the geometry sequence also will be done easily ". No need to use memorized this formula and that too children can still with the child's own natural ability. Remember Math is not just memorizing but also developing intuition you know !. Believe?????

Smart Solution Analysis 1

  ........ ,    6        , ......... ,    24 ,     ,   ........    ........    ........ ,  ........ ,  ........ ,  ........ 

   U1          U2         U3        U4           U5        U6       U7        U8       U9      U10

step work, please  try to use the right ratio

Because the line up then the chances ratio 2 and 3

try using ratio 2

Results

 ........ ,    6        , .12  .. ,    24 ,     ,   ........    ........    ........ ,  ........ ,  ........ ,  ........ 

  U1        U2         U3        U4           U5        U6       U7        U8       U9      U10

Already fulfilled, no need to try ratio 3.

U10 can easily be found

........ ,    6        , .12  .. ,    24 ,     ,   .48...    ..96..    .192... ,.384.. ,  .768.. ,  .1536.. 

  U1        U2         U3        U4           U5        U6       U7        U8        U9         U10

Easy does not need to memorize the formula, no need to search a and r children can still

 finish correctly. If using the formula, the length of the process is also the same. if you can 

say the burden of thought.

Smart Solution Analysis 2

we can also use quantum bridge to solve it.

Notice the schematic below screenshot

Just as easy is not it?

Sample UN question - complete geometry sequence in 2016

    First and fifth tribes of a sequence of geometries 5 and 80 respectively. The 9th row of 

the sequence is ....

A. 90      C. 940

B. 405     D. 1.280

Smart Solution Analysis 1

   ..5... ,                , ......... ,   ..... ,     ,   .80...    ........    ........ ,  ........ ,  ........ ,   

   U1          U2         U3        U4           U5        U6       U7        U8       U9     

Because the line up then the chances ratio 2 and 3

try using ratio 2

..5... ,       10    , ..20... ,   .40 ,     ,   .80...    ........    ........ ,  ........ ,  ........ ,   

   U1          U2         U3        U4           U5        U6       U7        U8       U9     

Already fulfilled, no need to try ratio 3.

U10 can easily be found

  ..5... ,    10    , ..20.. ,   .40. ,     ,   .80...    .160..    .320... ,  ..640.. ,  ..1280.. 

   U1        U2        U3        U4           U5        U6       U7          U8           U9       

Done ....... easy is not it?  Smart Solution Analysis 2 we can also use quantum bridge to solve it. Notice the schematic below screenshot

Learning the material of geometric sequence numbers with intelligent methods utilizing quantum techniques will be 
very useful for students. Completion using the scheme will 

make students easy to remember and remember in their brain

 recording for a long time. 

We as teachers should always innovate to provide a fun 

learning service for students, with the aim of ultimately 

the students are able to independently explore their own 

patterns of mathematics so that our children will always be 

creative in facing the problems of everyday life and could be 

our children someday to be the inventor of the future 

inventor. 

Similarly a short article on How to intelligently solve

 the problem of geometry sequence, 

Hopefully this article is beneficial to all of us.