Monday, 22 January 2018

how to solve integer math problem

how to solve integer math problem

integer math problem


Matter integers for students is one of the hard work material especially if the students 
are asked to solve the story problem. The ability to complete the story as a whole for 
junior high school children in Indonesia is still relatively low. This is probably because 
the question of the story requires the ability to understand well every given problem. 
Matter of integers which is the basis of mathematics also contains many stories. And 
strangely when the question of mathematics chapter number given to students the results 
remain disappointing


The ability of students in understanding and analyzing every mathematical problem is 
necessary in the habit. With the aim that students are accustomed to solve problems 
related to the story of the problem of day to day. If the ability to understand and solve 
the problem of the story is always in the habit then the child's ability to understand every
 problem of the story

In the present time the most advanced mathematical ability of students is the ability to solve stories that are closely related to the problems of day-to-day students. . It is intended that students have high-level thinking skills. High-level thinking skills will make students ready to face global competition later.

There are many types of mathematical problems related to the matter of integers.Today we will try some math stories about integers and how to solve them easily.

The provincial OSN contest set rules for students who answered true score 4, wrongly scored (-2) and not answered (-1). From 50 questions given P participants answered 43 questions and 12 of them wrong. The total score obtained by participant P is.....

a.     93                    c. 117
b.    100                  d 124


How to solve the above problems are as follows
 
Number of items about = number of items answered + number of items not answered
50 = 43 + number of unanswered items, this means unanswered item 7
Number of items answered = 
total number of answered items correct + items answered wrong
 
If the number of items answered is correct = B and
the number of items answered is wrong = S then
43 = B + 12 then the number of items answered correctly is 43-12 = 31
Total score = B x 4 + S x (-2) + T x (-1) 
                   = 31 x 4 + 12 x (-2) + 7 x (-1) 
                   = 124 - 24 - 7 
                   = 93
 
Example of integer story 2
 
The temperature in the refrigerator is -13ºC, while the temperature in the room is 32ºC. 
The temperature difference in both places is ....
A. 45ºC
B. 18 ºC
C. -18ºC
D. -45 ºC

completion
Temperature difference = highest temperature - lowest temperature                            
                                       = 32ºC - (-13ºC)                           
                                       = 45 ºC

An example of integer story 3
 
In the Mathematics competition, each correct answer is given a score of 3, the wrong 
answer is given a score of -1, and if not answered a score of 0. Of the 40 questions 
tested, Dedi answered 31 questions, of which 28 questions were answered correctly. 
Dedi's score is ....
A. 81 C. 87
B. 84 D. 93
 
Fri = answered + unanswered
40 = 31 + not answered, means not answered 9
Answered = B + S
31 = 28 + S then the problem is answered incorrectly there are 31-28 = 3
Total score = B x 3 + S x (-1) + T x (0)
                = 28 x 3 + 3 x (-1) + 9 x (0) = 84 - 3 - 0 = 81

Example of integer story 4
 
If the "#" sign means the number of the first five numbers and three times the second 
number, then (-6) # 10 is ....
  A.-60 C. 0
  B.-30 D. 30
 
How to solve
 
-6 # 10 = 5 (-6) + 3 (10) = -30 + 30 = 0
 
An example of integer story 5
If a = -4, b = 3, c = -2 and d = 1, the value of
ab + bc - cd - abcd is ....
A. -44 C. -36
B. -40 D. -24

completion
ab + bc - cd - abcd = (-4) (3) + (3) (- 2) - (-2) (1) - (-4) (3) (- 2) (1)
                                  = -12 + -6 + 2 - 24
                                  = -40
 
 
 
Example of integer story 6
Zika is 3 years older than Yeli. While Yeli 5 years older than Fari. If their age is 
49 years, then the age of Zika is ....
A. 12 years C. 20 years
B. 17 years old D. 22 years old
 completion
 
Complete completion visit here
 
 
So the article about the story of integers and how to solve them. hopefully through 
a series of learning this time we can better understand about the problem of integer 
story.
 
Keep Spirit




























Thursday, 21 December 2017

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 


COMPETING COMPARATIVE PROBLEMS EASILY EASY



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!
solve the problem of widespread two dimensional wake mathematics











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!
solve the problem of widespread two dimensional wake mathematics
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
solve the problem of widespread two dimensional wake mathematics
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.