How to Get the Greatest Common Factor of Numbers the numbers that can divide an integer is called its factor or divisor. For For example, the factors of 4 are are 1, 2, and and 4 becau because se these these are the the numb numbers ers that that divi divide de 4 with withou outt havi having ng a remainder. Another example is 6 which has factors 1, 2, 3, and 6. t is clear that each number has alwa!s 1 and itself as factors. "ote that in this discussion, when sa ! number, number, mean positive integer. integer. f we select more than one number, we can observe that the! have common factors #$ust li%e having common multiples&. multiples &. 'et(s have the following examples. How to Get the Greatest Common Factor of Numbers Example 1: )hat are the common factors of 12 and 1*+ Factors of 12 12 1, 2, 3, 4, 6, 12 Factors of 18: 1, 18: 1, 2, 3, 6, -, 1* f we examine the factors of 12 and 1*, we see that there are 4 common factors 1, 2, 3 and 6. Among the factors, 6 is the largest. herefore, we sa! that 6 is the greatest common factor #/0F& factor #/0F& or greatest common diisor #/0& diisor #/0& of 12 and 1*. Example 2 : Find the /0F of 2, 32, 2*. Factors of 2!: 1, 2!: 1, 2, 3, 4, , 1, 2 Factors of "2: 1, "2: 1, 2, 4, *, 16, 32 Factors of 28: 1, 28: 1, 2, 4, , 14, 2* As we can see, the common factors of 2, 32, and 2* are 1, 2, and 4. he /0 or /0F of the three numbers is 4. Anoth Another er wa! wa! to get get the the great greatest est comm common on facto factorr of numb number ers s is to write write thei theirr prim prime e factori factori5at 5ation ion.. rime rime factori factori5at 5ation ion is the process process of express expressing ing a number number as product product of prim prime e numb numbers ers.. A prim prime e numb number er is a numb number er which which is onl! onl! divi divisi sibl ble e b! 1 and and itsel itself f #read #ntroduction to $rime Numbers if Numbers if !ou don(t %now what is a prime number&. he first 1 prime numbers are 2, 3, , , 11, 13, 1, 1-, 23, and 2-. )e will use the examples above and use prime factori5ation in order to get their greatest common factor. Example ": Find ": Find the /0F of 12 and 1* using prime factori5ation. rime factori5ation of 12 2 7 2 7 3 rime Factori5ation of 1* 2 7 3 7 3 "ow to get the greatest common factor, we multipl! the common factors to both numbers. he common factors factors to both are 2 and 3, therefore, the greatest greatest common factor of 12 and 1* is 2 7 3 8 6. Example %: Find %: Find the /0F of 12 and 1* using prime factori5ation. $rime factori&ation of 2!: 2 2!: 2 7 2 7 $rime factori&ation of "2: 2 "2: 2 7 2 7 2 7 2 $rime factori&ation of 28: 2 28: 2 7 2 7 2 n this example, 2 and 2 are common to all the three numbers, so the /0 or /0 of these three numbers is 2 7 2 which is e9ual to 4. he difference between the two methods is that in the first method, !ou list all the factors and find the largest number. n the second method, !ou list the prime factori5ation and the multipl! the factors that are common to all numbers. 'hat(s the use of greatest common factor) )ell, /0F are used a lot in mathematics, but in the 0ivil :ervice ;xam, !ou will use it when !ou reduce fractions to lowest terms. terms . For example, !our final answer is
1
and is not on the choices. hen, !ou %now that !ou have to get the greatest common factor of 12 and 1* and divide both the numerator and denominator b! it. :o, the answer is
How to Get the *east Common +ultiple of Numbers n mathematics, a multiple is a product of an! number and an integer. he numbers 16, <4* and 2 are multiples of * because * x 2 8 16, * x <3 8 <4* and * x - 8 2. :imilarl!, the first five positive positive multiples of are the following ,- 1%- 21- 28- ".. ". . n this post, we will particularl! tal% about positive integers integers and positive multiples. his is in preparation for the discussions on addition and subtraction of fractions. )e can alwa!s find a common multiple given two or more numbers. For example, if we list all the positive multiples of 2 and 3, we have 2, 4, /, *, 1, 12, 12, 14, 16, 18, 18, 2 and 3, /, -, 12, 12, 1, 18, 18, 21, 24, 2, 3. As we can see, in the list, 6, 12 and 1* are common multiples of 2 and 3. f we continue further, further, there are still other multiples, and in fact, we will never run out of multiples. 0an !ou predict the next five multiples of 2 and 3 without listing+ he he mo most st impo importa rtant nt am amon ong g the the mult multipl iples es is the the least common common multiple multiple.. he he leas leastt common multiple is the smallest among all the multiples. 0learl!, the least common multiple of 2 and 3 is 6. =ere are some examples. Example 1: Find the least common multiple of 3 and >ultiples of 3 3, 6, -. 12, 1.- 1* 1.- 1* >ultiples of , 1, 1., 1., 2, 2,3 As we can see, 1. appeared as the first common multiple, so 1 is the least common multiple of 3 and . Example 2: Find the least common multiple of 3, 4, and 6.
2
and is not on the choices. hen, !ou %now that !ou have to get the greatest common factor of 12 and 1* and divide both the numerator and denominator b! it. :o, the answer is
How to Get the *east Common +ultiple of Numbers n mathematics, a multiple is a product of an! number and an integer. he numbers 16, <4* and 2 are multiples of * because * x 2 8 16, * x <3 8 <4* and * x - 8 2. :imilarl!, the first five positive positive multiples of are the following ,- 1%- 21- 28- ".. ". . n this post, we will particularl! tal% about positive integers integers and positive multiples. his is in preparation for the discussions on addition and subtraction of fractions. )e can alwa!s find a common multiple given two or more numbers. For example, if we list all the positive multiples of 2 and 3, we have 2, 4, /, *, 1, 12, 12, 14, 16, 18, 18, 2 and 3, /, -, 12, 12, 1, 18, 18, 21, 24, 2, 3. As we can see, in the list, 6, 12 and 1* are common multiples of 2 and 3. f we continue further, further, there are still other multiples, and in fact, we will never run out of multiples. 0an !ou predict the next five multiples of 2 and 3 without listing+ he he mo most st impo importa rtant nt am amon ong g the the mult multipl iples es is the the least common common multiple multiple.. he he leas leastt common multiple is the smallest among all the multiples. 0learl!, the least common multiple of 2 and 3 is 6. =ere are some examples. Example 1: Find the least common multiple of 3 and >ultiples of 3 3, 6, -. 12, 1.- 1* 1.- 1* >ultiples of , 1, 1., 1., 2, 2,3 As we can see, 1. appeared as the first common multiple, so 1 is the least common multiple of 3 and . Example 2: Find the least common multiple of 3, 4, and 6.
2
n this example, we find the least multiple that are common to the three numbers. >ultiples of 3 3, 6, -, 12 12,, 1 >ultiples of 4 4, *, 12 12,, 16, 2 >ultiples of 6 6, 12 12,, 1*, 24, 3 :o, the least common multiple of 3, 4, and 6 is 12 12.. Find the least common multiple of 3, * and 12. Example 3: 3: Find >ultiples of 3 3, 6, -, 12, 1, 1*, 21, 2% >ultiples of * *, 16, 2% 2%,, >ulitples of 12 12, 2% 2%,, 36, 4*, 6 :o, the least common multiple of 3, 4 and 6 is 2% 2%..
How to 0dd $ositie and a nd Negatie #ntegers ?ne of the topics in basic mathematics mathematics that will li%el! be included in the the hilippine 0ivil :ervice ;xam both professional and subprofessional are operations on integers. Although a few 0ivil :ervice test items ma! be be given from this this topic, it is important that !ou master it because a lot of calculation in other topics will need %nowledge of integers and its operations #addition, subtraction, multiplication, division&. For example, solving some word problems in mathematics and solving e9uations will need %nowledge on operations of integers. ntegers are whole numbers that are either positive or negative. ;xamples of integers are <, 6, , and 1. f we place this on the number line, negative integers are the integers that are below #left of &, while w hile the positive integers are the integers above #right of &.
Adding ntegers that Are @oth ositive )hen !ou add integers that are both positive, it is $ust li%e adding whole numbers. @elow are the examples. ;xample 1 2 4 8 6 ;xample 2 - 41 - 8 6
3
;xample 3 12 13 12 8 14 Although we have created a small before the number to indicate that it is positive, in realit!, onl! negative numbers have signs. his means that 2 4 8 6 is $ust written as 2 4 8 6. Adding ntegers that Are @oth "egative Adding number that are both negative is $ust the same as adding numbers that are both positive. he onl! difference is that if !ou add two negative numbers, the result is negative. ;xample 1 B B* 8 B13 ;xample 2 B1 B1* B32 8 B6 ;xample 3 B22 B11 B16 8 B24 =ow to Add ositive and "egative ntegers @efore adding, !ou should alwa!s remember that 1 and B1 cancel out each other, other, or 1 B1 is . :o the strateg! is to pair the positive and negative numbers and ta%e out what(s left. Example 1: )hat is 13 B*+ :olution )e pair * positives and * negatives to cancel out. hen what(s left is of 13 is . n e9uation form, we have
B B B 13 * 8 * * 8 #* *& 8 #& 8
B Example 2: )hat 2: )hat is 1 2+
:olution B B )e pair 1 negatives and 1 positives. )hat(s left of 2 is 3. n e9uation form, we have
B B B B 1 B 2 8 1 # 1 B3& 8 # 1 1& 3 8 3 8 B 3
B B Example 3 3 )hat is 16 3 2 3 B - +
n answering 9uestions with multiple addends, combine all the positives and the negatives then add. B B B hat is 16 3 8 3 and 2 3 8 B 32.
4
B :o, the final e9uation is 3 32. )e pair 32 positives and 32 negatives leaving 21 positives.
n e9uation form, we have B B 3 32 8 21 32 32 8 21 # 32 B 32& 8 21 8 21
How to ubtract $ositie and Negatie #ntegers his is the continuation of the series of 0ivil :ervice review in mathematics particularl! on operations of integers. n this post, we are going to discuss the most complicated operation on integers. have taught people of all ages about this topic and it seems that for man!, this is the most difficult among the four operations. n this post, we are going to learn how to subtract positive and negative integers or signed numbers. "ote that in subtracting integers, there are onl! four forms. f a and b are positive, the subtraction are of the following forms. 0ase 0ase 0ase 0ase
1 2 3 4
positive minus positive #a B b& negative minus positive # Ba B b& positive minus negative #a B Bb& negative minus negative # Ba B Bb&
=ow to :ubtract ositive and "egative ntegers )hat most people don(t %now that a B b is the same as a Bb, or subtracting a number is the same as adding its negative. hat means that !ou onl! have to memori5e the steps in addition of integers. /iven a subtraction sentence, !ou then transform it into addition. =ere are a few examples. 0ase 1 ;xampe 1 B * :ubtracting is the same as adding its negative, so B * 8 B*. "ote that B* is alread! addition and B* 8 B3. 0ase 2 ;xample B1 B 4
he expression B1 B 4 is the same as B1 B4 8 B14. Cemember also that if !ou see two consecutive B signs or a minus and a negative sign, !ou can transform it to . hat is, <#Ba& 8 a and <#Ba& a. n most exam, the negative signs are not usuall! superscript, so !ou will li%el! <#
5
0ase 3 ;xample B B6 he above expression might be written in < <6 or <#<6&. n an! case, two negative signs, a minus and a negative sign can be transformed into a plus sign so, B #B6& 8 6 8 11. "otice that the last e9uation is also an addition sentence. 0ase 4 ;xample B* B B6 he expression B* B B6 8 B* 6 8 B2.
?bserve that the four forms are alread! c ompleted in the examles. From the strateg! above, we onl! remember two strategies #1& transform an! subtraction sentence to addition sentence and #2& replace two consecutive negatives or a minus and a negative with sign. How to +ultipl igned Numbers n the two previous post in >athematics, we have discussed how to add and subtract signed numbers. n this post, we are going to learn how to multipl! signed numbers particularl! integers. :igned means positive and negative. $ositie #nteger x $ositie #nteger 0learl!, the product is positive. )e had been multipl!ing positive integers since /rade school and we all %now that the product is positive. $ositie #nteger x Negatie #nteger )hen !ou multipl!, notice that !ou are actuall! adding repeated. )hen we sa! 2 x 3, we are actuall! sa!ing twice three or 2 groups of 3 or 3 3. )hen we sa!, thrice 11, we are sa!ing 11 11 11. )ith this in mind, 3 x B 8 < < <. :ince we are adding integers which are negative, the sum is also negative or <1. his means that 3 x < 8 <1. f we generali5e this, we can sa! that the product of a positive integer and a negative integer is negative.
Negatie #nteger x $ositie #nteger
f !ou can remember, multiplication is commutative. his means that the order of the number !ou multipl! does not matter, their product will alwa!s be the same. For example, 4 x 3 x is e9ual to x 4 x 3 or 3 x 4 x or an! other arrangement using the three numbers.
6
his means that 3 x < 8 < x 3. :o, a negative integer multiplied b! a positive integer is also negative. Negatie #nteger x Negatie #nteger For multiplication of two negative integers, we can use patterns to %now their product and generali5e. <3 x 2 8 <6 <3 x 1 8 <3 B3x8 "ow, what is <3 x <1+ f we loo% at the pattern in the product, we are actuall! adding b! 3 each step so the next number is 3. All other numbers from <1, <2, <3 and so on will be positive #)h!+&. herefore, the product of two negative numbers is positive. ummar: 3ules on how to +ultipl igned Numbers From the above discussion, we summari5e the multiplication of integers. ositive nteger x ositive nteger 8 ositive nteger ositive nteger x "egative nteger 8 "egative nteger "egative nteger x ositive nteger 8 "egative nteger "egative nteger x "egative nteger 8 ositive nteger )e can also sa! that if we multipl! two numbers with the same sign, the answer is positive. f we multipl! two numbers with different signs, the answer is negative. 4iiding $ositie and Negatie #ntegers n the previous post on integers, we have learned the rules in multipling positie integers and negatie integers. n this post, we are going to learn how to divide positive and negative integers. f !ou have observed, in the post on subtracting integers, we have converted the Dminus signE to a Dplus negative sign.E thin% it is safe for us to sa! that subtraction is some sort of Ddisguised addition.E :imilarl!, we can also convert a division expression to multiplication. For example, we can turn
7
to
.
to
.
n general, the division
From the discussion above, we can as% the following 9uestion 0an we use the rules in multipl!ing integers when dividing integers+ he answer is a big ;:. he rules are ver! much related. positive integer G positive integer 8 positive integer positive integer G negative integer 8 negative integer negative integer G positive integer 8 negative integer negative integer G negative integer 8 positive integer "otice that the! are ver! similar to the rules in multipl!ing integers. positive integer x positive integer 8 positive integer positive integer x negative integer 8 negative integer negative integer x positive integer 8 negative integer negative integer x negative integer 8 positive integer =ere are some examples wor%ed examples. 1. 1* G 3 8 6 2.36 G <12 8 B 3 3. <1 G 2 8 B . 4.< * G <4 8 2 From the discussion and the wor%ed examples above, we can therefore conclude that in dividing positive and negative integers, we onl! need to memori5e the rules in multipl!ing integers and appl! them in dividing integers. ##5 F30C6#7N A /entle ntroduction to Fractions
8
Fractions is one of the mathematics topics that man! people have difficult! with. =owever, unfortunatel!, it is also one of the most important topics that must be mastered. his is because examination 9uestions in mathematics alwa!s include fractions. For example, in the 0ivil :erviceCeview "umerical Ceasoning tests, fractions appear in almost ever! test basic arithmetic, number se9uences, e9uations and problem solving. n this post, we are going to discuss the basics about fractions particularl! about the terminologies used. ?f course, !ou don(t reall! have to memori5e them now, but !ou can refer to this post in the following discussions. n the future discussions, !ou will use the vocabular! that !ou have learned here. #ntroduction to Fractions n la!man(s language, a fraction is reall! a part of a whole. n the figure below, the part which is shaded is one out of four, so we sa! that H of the s9uare is shaded. )e can also sa! that three out of four or I of the s9uare is not shaded. )e can also sa! that adding H and I e9uals one whole.
Fractions can also be a subset of a set5 f 3 out of 1 students are girls, then we sa! that 3J1 of the students are girls. A fraction could also mean division. For example, wen we sa! J1, we can also mean, divided b! 1.
A fraction is composed of a numerator, the number above the bar, and a denominator, the number below the bar. . Fractions whose numerator are less than the denominator are called proper fractions. Fractions whose numerator are greater than the numerator are called improper fractions. mproper fractions can be converted to mixed fractions or fractions that contain whole numbers.
9
Kust li%e other numbers, we can perform operations on fractions. n the next four posts, we will be discussing the different operations on fractions. )e will learn how to add, subtract, multipl!, and divide fractions. =ow to /et the 'east 0ommon >ultiple of "umbers n mathematics, a multiple is a product of an! number and an integer. he numbers 16, <4* and 2 are multiples of * because * x 2 8 16, * x <3 8 <4* and * x - 8 2. :imilarl!, the first five positive multiples of are the following ,- 1%- 21- 28- ".. n this post, we will particularl! tal% about positive integers and positive multiples. his is in preparation for the discussions on addition and subtraction of fractions. )e can alwa!s find a common multiple given two or more numbers. For example, if we list all the positive multiples of 2 and 3, we have 2, 4, /, *, 1, 12, 14, 16, 18, 2 and 3, /, -, 12, 1, 18, 21, 24, 2, 3. As we can see, in the list, 6, 12 and 1* are common multiples of 2 and 3. f we continue further, there are still other multiples, and in fact, we will never run out of multiples. 0an !ou predict the next five multiples of 2 and 3 without listing+ he most important among the multiples is the least common multiple. he least common multiple is the smallest among all the multiples. 0learl!, the least common multiple of 2 and 3 is 6. =ere are some examples. Example 1: Find the least common multiple of 3 and >ultiples of 3 3, 6, -. 12, 1.- 1* >ultiples of , 1, 1., 2, 2,3
10
As we can see, 1. appeared as the first common multiple, so 1 is the least common multiple of 3 and . Example 2: Find the least common multiple of 3, 4, and 6. n this example, we find the least multiple that are common to the three numbers. >ultiples of 3 3, 6, -, 12, 1 >ultiples of 4 4, *, 12, 16, 2 >ultiples of 6 6, 12, 1*, 24, 3 :o, the least common multiple of 3, 4, and 6 is 12. Example 3: Find the least common multiple of 3, * and 12. >ultiples of 3 3, 6, -, 12, 1, 1*, 21, 2% >ultiples of * *, 16, 2%, >ulitples of 12 12, 2%, 36, 4*, 6 :o, the least common multiple of 3, 4 and 6 is 2%. n the next part of this series, we will discuss about =ow to Add Fractions.
How to 0dd Fractions Fractions whose denominators are the same are called similar fractions. Fractions that are not similar are called dissimilar fractions. =ence, the fractions , , and fractions, while the fractions learn how to add fractions.
and
are similar
are dissimilar fractions. n this post, we are going to
How to 0dd imilar Fractions Adding similar fractions is ver! eas!. n adding similar fractions, !ou $ust add the numerator and cop! the denominator. =ere are a few examples. Example 1
11
Example 2
Example 3
n most cases, improper fractions or fractions whose denominator is less than its numerator such as the third example is converted to mixed form. he mixed form of discuss how to ma%e such conversion in the near future.
is
. )e will
How to 0dd 4issimilar Fractions Addition of dissimilar fractions is a bit more complicated than adding similar fractions. n adding dissimilar fractions, !ou must determine the least common multiple #'0>& of their denominator which is %nown as the least common denominator. "ext, !ou have to convert all the addends to e9uivalent fractions whose denominator is the '0>. =aving the same denominator means that the fractions are alread! similar. =ere are a few examples. Example 1
Solution a. Get the least common multiple (LCM) of 2 and 3. >ultiples of 2 2, 4, /, *, 1, 12 >ultiples of 3 3, /, -, 12, 1 '0> of 2 and 3 is /. b. Cone!t the f!actions into f!actions "hose denominato! is the LCM "hich is 6.
First Addend . :o, the e9uivalent of
is .
12
:econd Addend
:o, the e9uivalent fraction of
is .
c. #dd the e$uialent f!actions
.
:o,
.
Example 2
Solution a. Get the LCM of 3 and %. >ultiples of 3 3, 6, -, 12, 1., 1* >ultiples of , 1, 1., 2 herefore, the '0> of 3 and is 1.. b. Cone!t the &ien f!actions into e$uialent f!actions "hose denominato! is 15 .
First Addend
:o, the e9uivalent fraction of
is
.
:econd Addend
13
:o, the e9uivalent fraction of
is
.
c. Add the e9uivalent fractions
.
:o, Example 3
Solution a. Get the LCM of 3, ' and . >ultiples of 3 3, 6, -, 12, 1, 1*, 21, 2% >ultiples of 6 6, 12, 1*, 2%, 3 >ultiples of * *, 16, 2%, 32, 4 '0> of 3, 6 and * is 24. b. Cone!t the &ien f!actions into e$uialent f!actions "hose denominato! is 24.
First Addend . herefore, the e9uivalent fraction of
is
:econd Addend
herefore, the e9uivalent fraction of
is
14
hird Addend
herefore, the e9uivalent fraction of
is
.
c. #dd the e$uialent f!actions
n the next post, we will have more examples and e xercises regarding addition of similar and dissimilar fractions. will also give !ou some tips in getting the least common multiple of two or more numbers without listing.
How to +ultipl Fractions Among the four fundamental operations on fractions, multiplication is the easiest. t is $ust simple. >ultipl! the numerator and then the denominator. ?f course, if the given fractions can be conerted to lowest terms, the easier the multiplication will be. n this post, we are going to learn how to multipl! fractions. ou must master this operation, as well as other fundamental operations on fractions because !ou will use them in higher mathematics and solving word problems. @elow are some examples. Example 1
Solution
.
Answer
.
Example 2
15
Solution
)e reduce the fraction to lowest term b! dividing both the numerator and the denominator b! 2. his results to Llatex
which is the final answer.
Answer Example 3
Solution First, we reduce
b! dividing both the numerator and the denominator b! 3. his results
to . )e now multipl!
.
Answer
.
Example 4
Solution n this example, we need to convert the mixed fraction into improper fraction. o do this, we multipl! the denominator of the mixed fraction to the whole number and the product to the denominator. hat is
. "ow, let us multipl! the two fractions.
16
Answer How to 4iide Fractions )e have alread! discussed addition and multiplication of fractions and what we have left are subtraction and division. n this post, we learn how to divide fractions. o divide fractions, we must get the reciprocal of the divisor. his is $ust the same as swapping the numerator and the denominator. For example, the reciprocal of getting the reciprocal, $ust multipl! the fractions.
is . After
Example 1
Solution First, we get the reciprocal of , the divisor. his is . hen, we multipl! the fractions.
Answer Example 2
Solution First, we get the reciprocal of
which is
. >ultipl!ing the fractions, we have
)e reduce the answer to lowest terms b! dividing both the numerator and denominator b! resulting to
.
17
Answer Example 3
Solution n dividing fractions, the dividend and the divisor must not be mixed fractions. herefore, we need to convert the mixed fraction to improper fraction. o do this, we multipl! b! and then add . he result becomes the numerator of the mixed fraction. :o, the the e9uivalent of
is
.
>ultipl!ing the fractions, we have
)e can convert the improper fraction to mixed f orm which is e9ual to
Answer Example 4 . Solution f the divisor is a whole number, the reciprocal will be 1 DoverE that number. n the given, the reciprocal of fractions
is . After getting the reciprocal of the divisor, we multipl! the two
. Answer How to ubtract Fractions
18
)e have alread! learned the three operations on fractions namel! addition, multiplication, and diision. n this post, we are going to learn the last elementar! operation subtraction. f !ou have mastered addition of fractions, this will not be a problem for !ou because the process is $ust the same. 'et(s subtract fractionsM
.
Example 1: Solution
he given is a similar fraction #fraction whose denominators are the same&, so $ust li%e in addition, we $ust perform the operation on the numerators. herefore, we $ust have to subtract the numerator and cop! the denominator. hat is,
. )e reduce to lowest term b! dividing both the numerator and denominator of his results to
Example 2:
b! .
which is the final answer.
.
Solution he two fractions are dissimilar, so we must find their least common denominator. o do this, we find the least common multiple of and . he common multiples of 2 are and so on and the common multiples of
are
and so on. As we can see from the lists above,
is the least common multiple of
)e now change the denominator of both fractions to
and .
.
First, we find the e9uivalent fraction of . hat is,
.
19
o find the value of
, divide
b!
and then multipl! to . he result is
the numerator of the e9uivalent fraction. :o, the e9uivalent fraction of confused with this process, please read How to 0dd Fractions. "ow, we get the e9uivalent fraction of b!
or we find the value of
which becomes is
in
and then multipl! it b! , which gives us . :o, the e9uivalent fraction of
. f !ou are
. )e divide is
.
)e now subtract the fractions.
. he final answer is
.
Example 3: Solution First, we convert
to improper fraction. hat is,
. to get
. he least common multiple of
and
is
#tr! listing as in example 2&.
"ow, to get the e9uivalent fraction, we have means, the e9uivalent fraction
. "ow,
. his
. )e also convert
to
which is e9ual to
.
"ow, we subtract the fractions.
.
20
0onverting the answer which is an improper fraction to mixed number, we have
. How to Conert Fractions to *owest 6erms #n the Ciil erice Examination and in man mathematics examinations- results that are fractions are usuall reuired to be conerted to their lowest terms5 6he numerator and the denominator of a fraction in lowest terms cannot be diided b an similar integer5 9nowledge of diisibilit rules can be helpful in this process5 Example 1: Conert
to lowest terms5
#n the first example- we can see that the numerator and the denominator are both diisible b "5 4iiding both the numerator and the denominator b " gies us 2"5
Note that diiding both the numerator and the denominator b the same integer does not change the alue of the fraction5 Example 2: Conert
to lowest terms5
ometimes- it is hard to find the largest common diisor of fractions with large numerator and denominator so ou need to perform diision repeatedl5 #n this example- the most obious clue is that both numbers are diisible b 25 4iiding both the numerator and denominator b 2 gies us
5 0gain- both the denominator
and the denominator are diisible b "- ma;ing the lowest term 5 7f course- ou could hae diided the numerator and the denominator b / in the first step5 Example ": Conert
to lowest terms5
6here are cases where ou are as;ed to conert an improper fractions
6he alue of
in mixed form is
5
21
#t ta;es time and practice to master conerting fractions to lowest terms5 9nowledge on diisibilit and familiarit of characteristics of numbers will help ou perform the computation faster5 How to Conert #mproper Fractions to +ixed Forms #n #ntroduction to Functions- we hae learned about proper and improper fractions5 0 fraction whose numerator
and
are proper fractions5
7n the other hand- a fraction whose numerator is greater than its denominator is called an improper improper fractions5
fraction5
6herefore
the fractions
-
and
are
#n the Ciil erice Examinations- some fractions need to be conerted from one form to another5 For example- in answering a number series test- ou might need to conert an improper fraction to mixed form in order to compare it to other fractions in mixed form5 #n this post- we learn this method: how to conert an improper fraction to mixed form5 #n conerting improper fractions to mixed form ou will >ust hae to diide the fraction- find its uotient and its remainder5 3emember that the fraction means "% diided b .5
also
'hen we diide "% b .we call . the diisor5 6he uotient to this diision is / with a remainder of %5 From the method- we can obsere the following: 6he uotient / is the whole number on the mixed fraction5 6he diisor . is the denominator of the mixed fraction5
22
6he remainder % goes to the numerator of the mixed fraction5 Now- for the second example- let us conert into mixed fraction5 #f we diide 28 b "- the diisor is "- the uotient is ? and the remainder is 15 6herefore- the euialent of the improper fraction
is
5
#@5 $E3CEN6 #ntroduction to the Concept of $ercentage Now that we hae alread studied fractions and decimals- we discuss percentage5 Aou are li;el to be aware that the concept of percentage is er useful in dail life5 'e alwas go to stores where there are discounts and we do not want loans with high interest5 6hese calculations inole the concept of percentage5 'hat is percentage reall) $ercentage is a number ratio expressed as a fraction of 1!!5 'hen we sa 1! percent- what we reall mean is 1! out of 1!!- or in fraction notation 1!1!!5 6herefore- when we see that a shirt is sold for a .! percent discount- we actuall sa .! out of 1!! or .!1!!5 Notice that .!1!! when reduced to lowest terms is 12 which means that we onl hae to pa half of the price of the shirt5 0s we all ;now- we use the smbol B to denote percent5 Converting Percent to Fractions for Faster Calculations Numbers in their percent form can be conerted to fractions for uic;er calculations5 For example- when we sa that a $hp2%!!5!! wristwatch has a 2.B discount- we can easil calculate b conerting 2.B to fraction5 6he euialent of 2.B discount is 1% in fraction- so- we deduct 1% of 2%!!
23
such as shown in the table below5 Aou can memori&e them if ou want- but the conersion method is fairl eas that ou can do them mentall5
How to Conert $ercent to Fraction #n Ciil erice Examinations- as well as other examinations in basic mathematics;nowing how to conert percent- fractions- and decimals to each other is er adantageous especiall if ou can do it mentall5 *et us tr with the following example5 A P64 s!irt is mar"e# 25$ #iscount% &o' muc! 'ill (ou !ave to pa( for it) #t seems that ou need a pencil for this problem- but ou can actuall do it in our head5 3ead it to beliee it5
6he euialent of 2.B in fraction is 1%- therefore- ou hae to ta;e awa the fourth of the price5 Now- 1% of /%! seems difficult but what if we tr to split it to /!! %!) Now- 1% of /!! is 1.!- which means that from the /!!- ou hae %.! left5 Now- 1% of %! is 1!- which means that ou hae "! left5 o- %.! "! is %8! and that is the discounted price of the tDshirt5 Now- with a little bit of practice- ou would be able to do this on our own and ou won(t hae to use a pen to perform calculations for problems such as this5 How to Conert $ercent to Fraction 6here is one important concept to remember when conerting percent to fraction5 6hat is- when ou sa percent- it means per hundred5 6he word cent comes from
24
the *atin word centum which means hundred5 #n effect- when ou sa- /!B- it means /! per hundred- !5%B means !5% per hundred- 12.B means 12. per hundred5 'hen ou sa x per hundred- ou can also represent it b the fraction x1!!5 6his means that the percentages aboe can be represented as
respectiel5 Now- all we hae left to do is to conert these fractions to lowest terms5 Example 1: 3ecall that to conert a fraction to lowest terms- we find the greatest common factor
6herefore- the euialent of /!B in fraction is 5 Example 2: #n this example- we hae a decimal point at the numerator and a whole number at the denominator5 'e hae to get rid of the decimal point5 6o do this- we can multipl both the numerator and the denominator b 1!
5 Now- the greatest common factor of % and 1!!! is %- so we diide both the numerator and the denominator b %5 6he final result is 6herefore- the euialent fraction of !5%B is
5
5
Example ": 6he greatest common factor of 12. and 1!! is 2.- so we diide both the numerator and the denominator b 2.5 #n doing this- we get 5 6herefore- the euialent fraction of 12.B is ummar 6here are three steps to remember in conerting percent to fractions5
25
15 +a;e a fraction from the gien percent with the gien as numerator and 1!! as denominator5 25 Eliminate the decimal points ust memori&e=5 6he first method can be used for fractions whose denominators can be easil related to 1!! b multiplication or diision5 3ecall that from Conerting $ercent to Fraction- # hae mentioned that when we sa percent it means per hundred5 #n effect- nB can be represented b n1!!5 6herefore- if ou hae a fraction and ou can turn it into n1!!
Now- since we hae 1!! as denominator- the answer in percent is therefore the numerator5 6herefore- the euialent of 1. in percent is 2!B5 Example 2: 'hat is "2. in percent) 0gain- how do ou related 2. to 1!!) multipling it b %5 6herefore-
6herefore- the euialent of "2. in percent is 12B5 Example ": 'hat is 2"2!! in percent) #n this example- we can relate 2!! to 1!! b diiding it b 25 o- we also diide the numerator b 25 6hat is
26
6herefore- the answer is 115.B 6here are two important things to remember in using the method aboe5 <1= in changing the form the fractions to n1!!- the onl operations that ou can use are multiplication and diision and <2= whateer ou do to the numerator- ou also do to the denominator5 Note that multipling the denominator
5 6he euation will result to which is eual to 2!5 Now- this is >ust the same as multipling both the numerator and the denominator b 2!5 Note that the method of relating to 1!! b multiplication or diision can onl wor; easil for denominators that diides 1!! or can be diided b 1!!5 7ther fractions
27
Now- notice how it is related to the new method5 #n this method- we related 1. to n1!!5 6hat is- what is the alue of in
5 6o simplif the euation- we multipl both sides of the euation b 1!!- and we get
implifing and switching the position of the expressions- we get the means that
5 6his
5
7f course- $art 1 seems to be easier- but the good thing about putting it into euation is that it applies to all fractions5 For instance- it is uite hard to conert ,12 using the method in part 15 Example 2: 'hat is the euialent of 'e set up the euation with
in percent)
on the left5
6o eliminate the fraction- multipl both sides b denominator5 6his results to
or about .85""B5 6he curl eual sign means approximatel eual to since " is a nonDterminating decimal5 Now- tr to examine the expression
because this is where the deried the rule5 3ecall the rule in conerting fraction to percent: 4iide the fraction and then multipl the result to 1!!5 6hat is exactl it5 o- when ou hae the fractionresult to 1!!5 6hat is-
>ust diide it manuall- and then multipl the
5
28
4o not forget though that the diisor during diision is the denominator <. in 2.=5 as shown below5
6hat(s it5 # thin; we don(t hae to hae the third part- since we a lread deried the rule here5
V. FRACTION-DECIMAL-PERCENT CONVERSION How to Conert 4ecimals to Fractions $art 1 'e hae learned how to conert fractions to decimals and in this post- we are going to learn how to conert decimals to fractions5 efore doing this- we need to reiew the meaning of place alue5 #n the decimal number !5."2- . is the tent!s place- " is the !un#re#t!s place- 2 is the t!ousan#t!s place% 6he number . tenths is the same as
- " hundredths is the same as
and 2 thousandths is the same as 5 #n conerting decimals to fractions- we hae to see the place alue of the last digit of the decimal place5 Example 1 Conert
to fraction5
Solution !5, is , tenths or
6herefore- the euialent of
5
in fraction is the same as
Example 2 Conert
to fraction5
29
!5/ is 'e reduce the fraction to lowest terms b diiding both the numerator and the denominator b the greatest common factor of / and 1! which is 25
6herefore- the euialent fraction of
is 5
Example " Conert
to fraction
6he last digit of the decimal is in the hundredths place- so we can read this as 12 hundredths5
6wele hundredths is
5
'e conert this fraction to lowest terms b diiding both the numerator and denominator b the greatest common factor of 12 and 1!! which is eual to %5 o-
5
6herefore- the euialent of !512 in fraction is Example % Conert
to fraction5
Solution 6he last digit of the decimal number aboe is in the thousandths place5 o- we can read it as ",. thousandths5
Now- ",. thousandths is the same as
5
'e conert ",. thousandths to lowest terms b diiding both its numerator and denominator b the greatest common factor of ",. and 1!!! which is eual to 12.5 6hat is-
5
30
6herefore- the euialent fraction of
is 5
How to Conert Fraction to $ercent $art 1 #n the preious post- we hae learned how to conert percent to fraction 5 #n these series of posts- we learn the opposite: how to conert fraction to percent5 # am going to teach ou three methods- the last one would be used if ou forgot the other two methods- or if the first two methods would not wor;5 $lease be reminded though to understand the concept ust memori&e=5 6he first method can be used for fractions whose denominators can be easil related to 1!! b multiplication or diision5 3ecall that from Conerting $ercent to Fraction- # hae mentioned that when we sa percent it means per hundred5 #n effect- nB can be represented b n1!!5 6herefore- if ou hae a fraction and ou can turn it into n1!!
Now- since we hae 1!! as denominator- the answer in percent is therefore the numerator5 6herefore- the euialent of 1. in percent is 2!B5 Example 2: 'hat is "2. in percent) 0gain- how do ou related 2. to 1!!) multipling it b %5 6herefore-
6herefore- the euialent of "2. in percent is 12B5 Example ": 'hat is 2"2!! in percent) #n this example- we can relate 2!! to 1!! b diiding it b 25 o- we also diide the numerator b 25 6hat is
6herefore- the answer is 115.B 6here are two important things to remember in using the method aboe5
31
<1= in changing the form the fractions to n1!!- the onl operations that ou can use are multiplication and diision and <2= whateer ou do to the numerator- ou also do to the denominator5 Note that multipling the denominator
5 6he euation will result to which is eual to 2!5 Now- this is >ust the same as multipling both the numerator and the denominator b 2!5 Note that the method of relating to 1!! b multiplication or diision can onl wor; easil for denominators that diides 1!! or can be diided b 1!!5 7ther fractions
Now- notice how it is related to the new method5 #n this method- we related 1. to n1!!5 6hat is- what is the alue of in
5 6o simplif the euation- we multipl both sides of the euation b 1!!- and we get
32
implifing and switching the position of the expressions- we get the means that
5 6his
5
7f course- $art 1 seems to be easier- but the good thing about putting it into euation is that it applies to all fractions5 For instance- it is uite hard to conert ,12 using the method in part 15 Example 2: 'hat is the euialent of 'e set up the euation with
in percent)
on the left5
6o eliminate the fraction- multipl both sides b denominator5 6his results to
or about .85""B5 6he curl eual sign means approximatel eual to since " is a nonDterminating decimal5 Now- tr to examine the expression
because this is where the deried the rule5 3ecall the rule in conerting fraction to percent: 4iide the fraction and then multipl the result to 1!!5 6hat is exactl it5 o- when ou hae the fractionresult to 1!!5 6hat is-
>ust diide it manuall- and then multipl the
5 4o not forget though that the diisor during diision is the denominator <. in 2.=5 as shown below5
33
6hat(s it5 # thin; we don(t hae to hae the third part- since we a lread deried the rule here5
How to Conert $ercent to 4ecimals #n the preious post- we hae learned how to conert decimals to percent 5 #n this post- we learn the opposite of this procedure5 'e learn how to conert percent to decimals5 #f ou can remember from the preious post- we conert decimals to percent b multipling the decimal b 1!!5 o- in this case- we diide percent b 1!! in order to get the decimal alue5 3emember: diision is the inerse operation of multiplication5 Example 1 Conert 8.B to decimals5 Solution 'e diide 8.B b 1!! which means that we will moe two decimal places to the let5 Note that the decimal point is on the immediate right of the ones place
34
0gain- it is a whole number- so the decimal point is at the right of !5 +oing the decimal point two places to the left- we hae 5%! or !5%! 0nswer: !5%! or !5%
35
6he euialent of 1.B to decimal is !51. Now- /!! J !51. ?! 6herefore- the discount is $hp?!5 How to Conert 4ecimal Numbers to $ercent Conersions of decimals- fractions- and percent is a er important basic s;ill in mathematics and man problems in the Ciil Exams reuire this s;ill5 eing able to conert from one form to another will help ou speed up in calculations5 For example- instead of multipling a number b 2.B- ou >ust hae to get its 1% or simpl diide it b %5 $ercent usuall appears in discount and interest problems while fractions and decimals appear in arious tpes of problems5 How to Conert 4ecimals to $ercent 6o conert decimal percent- ou >ust hae to multipl the decimal b 1!!5 Example 1 'hat is !52. in percent) Solution !52. J 1!! 2. o- the answer is 2.B5 Example 2 'hat is !5!8 in percent) !5!8 J 1!! 8 6herefore- the answer is 8B5 7f course- there are cases that the gien is more than one such as the next example Example 3 'hat is 158 in percent) Solution 158 J 1!! 18! 6herefore- the answer is 18!B5 Example 4
36
'hat is !5!!? in percent) Solution !5!!?J 1!! !5?B Notice that some percent can also hae decimal point such as shown in Example %5 #n dealing with man decimals- if we multipl them with 1!!- we >ust moe two decimal places to the right5 @#5 NK+E3 E3#E 0 6easer on 0nswering Number eries Luestions First of all- # would li;e to point out the term series in the Number eries uestions in the Ciil erice Examinations is a bit incorrect5 6echnicall- the list of numbers in the examinations is actuall called a seuence5 0 series is a seuence of sums M well- # will not go into details since it is not included in the examinations5 Aou can clic; the lin;though if ou want to ;now about it5 econd- this is uite a premature discussion since # hae onl written a few posts about integers5 # planned to write about this later- but # thought that a teaser would be nice5 #n this post- # will show ou that it is a must to master all the topics in mathematics because the are all connected5 'e will not discuss the strategies on how to answer the seuence problems here # will hae a separate post about them later5 4on(t stop reading though because ou are going to miss half of our life if ou do <;idding=5 0 seuence or a progression is an ordered list of ob>ects which can be numbersletters- or smbols5 6he list "- ,- 11- 1.- 1? is a seuence where " is the first term and 1? is the fifth term5 7f course- it is eas to see the sixth term is 2" since each term is the sum of % and term before it5 6here are also seuences that are in decreasing order such as 12- .- D2- D?- D1/ and so on5 0s ou can obsere- to get the next term- , is subtracted from the term before it5 Notice also that this seuence needs ;nowledge on subtraction of negatie integers5 6he list
is also an example of a seuence5 6his seuence inoles addition of fractions5 6he next term can be easil soled b conerting the gien into similar fractions which when done will result to
5
37
Clearl- we onl need to add euals
to the last term to get the next term which
5
#n the seuences aboe- we hae onl used two number representations
5
Solution an# Explanation Numbers 1 and 2 are the easiest tpe of seuence to sole5 6his is because the are integers and ou >ust add acent terms <1" P 1! "- 1! P , "- ,D % "= to see if the difference is constant5 #f it is- then ou will ;now that ou will >ust hae to add the same number to get the next term5 6herefore- the next term to the first seuence is 1" " 1/5
38
7f course- seuences can also be decreasing5 #n the second example- the difference is / or it means that / is subtracted from a number to get the next term
6he third example is composed of letters but the principle is the same: constant difference or constant s;ips5 C and F- for instance has two letters in between5 6his is also true between F and # and # and *5 6herefore- the next letter in the seuence is 7 <*- +- N- 7=5
Fractions and decimals are also included in the seuence problems- so it is important that ou master them5 #n the following example- one half is added each time5 0s ou can see- it is not eas to find the next term of this seuence without manuall soling it5 6he first strateg in soling fraction seuences problems is to subtract the ad>acent terms such as which is eual to
#
which is eual to which is eual to
since
5
6here is howeer a better strateg than subtracting the ad>acent terms when it comes to seuences on fractions5 ometimes- it is easier to see the pattern if ou conert them to similar fractions
5
39
From here- it is clear that the next term in the seuence is 5 Note howeer that this strateg is best onl for seuences with constant difference and ma be difficult to use in other tpes of seuences5 #n the next post- we are going to discuss about another tpe of seuence5 How to ole Ciil erice Exam Number eries $roblems 2 #n the preious post- we hae learned how to sole number seuence
First euence: "- /- 12- 2%- %8- OOO #n the first seuence- the first that ou will notice is that the second term is twice the first term5 o- the next thing that ou should as; is- #s the third term twice the second term) Aes- 12 is twice /5 'hat about the next term) Aes5 o- each term in the seuence is multiplied to 2 to get the next term5 6herefore- the missing term is ?/ which is %8 multiplied b 25
#f we loo; at the difference of the numbers in the seuence aboe- we can see that the number we add is also increasing twice5 6o get /- we added "5 6o get 12- we added /5 6o get 2%- we add 12 and so on5 0s we can see- the seuence of the numbers we add
40
econd euence: 18- /- 2- !5//Q- OOO #n the second seuence- the number is reduced each time5 ince the are integersit can either be subtraction or diision5 0s we can see- / is a third of 185 6his means that to get /- 18 is diided b "5 Now- loo; at the next term5 #t(s 25 o it is also a third of /5 Can ou see the pattern now) Each term is diided b " to get the next term5 o- we must diide !5//Q b "5 therefore- the next term is !522Q 6he three dots means that the 2(s are infinitel man5 6hird euence: 1- %- ?- 1/- 2.- OOO 'hat is familiar with this seuence) 6he are all suare numbersR 6hat is-
-
-
and
5
o the next term is
which is "/5
Fourth euence: "- 12- 2,- %8- ,.- OOO 6he fourth seuence seems difficult- but # hae >ust multiplied each number in the third seuence b "5 o- if the seuence is not familiar- tr to see if ou can diide it b an number5 0s ou can see"- 12- 2,- %8- ,. " <1- %- ?- 1/- 2.= or the product of " and the suare numbers5 #n the next post- we are going to discuss alternating seuences5 How to ole Ciil erice Exam Number eries $roblems " 6his is the third part of the soling number series problems5 6he first part includes dealing with patterns that contains addition and subtraction and the second part discusses patterns that contains multiplication or diision5 #n this post- we are going to learn some alternating seuences5 # put a uote in alternating seuence because in mathematics- it has a slightl different meaning5 Note that it is li;el that these tpe of seuence will appear in examinations such as the Ciil erice Exam5 efore we continue with the discussion- tr to see if ou can answer the following uestions5
41
15 2- D.- %- D8- /- D11- 8- D1%- OOO- OOO 25 %- ,- 12- 1.- 2!- 2"- 28- OOOO "5 0- "- 4- 8- G- 1"- OOO- OOO %5 - - - - -
- OOOO- OOOOO
olutions and Explanations First Se+uence: 2, -5, 4, -*, 6, -11, *, -14, ..., ... 6he first seuence seems hard- but it is actuall eas5 #f ou perform addition and subtraction among consecutie terms- ou will surel see a pattern ust put different colors on them- so it is eas to see the pattern5 2- D.- %- D8- /- D11- 8- D1%- OOO- OOO 4o ou see now) Can ou answer the problem) 0s ou can see- the red numbers are >ust increasing b 2 and the blue numbers are decreasing b "5 6herefore- the next numbers are 1! and D1,5 Secon# Se+uence:% 4, /, 12, 15, 2, 23, 2*, .... #n the second seuence- % is increased b " to become ,5 6hen- , is increased b .5 6he increase in the numbers are also in alternating pattern5 o the correct answer is "1 which is eual to 28 "5
Further- what is interesting is that the coloring strateg that we used in the first seuence can be also used in this seuence5 0s ou can see in the colored numbers below- it becomes two seuences as well5 6he seuence composed of blue numbers and the other red5 #n both seuences- the numbers is increased b 85 ince the next number is blue- then it is eual to 2" 8 "15 4, /, 12, 15, 2, 23, 2*, .... 0!ir# Se+uence: A, 3, , *, , 13, ..., ... #n the third seuence- the answers are alread obious after learning the strateg aboe5 6here are two letters in between the letter terms in the seuence < A, ,
42
C, , E, F, , &, , =5 Further- each number term is . greater greater than the preious number term5 o- the correct answer answers are - 1*5 A, 3, , *, , 13, ..., ...
Fourth euence: - - -
- -
- OOOO- OOOOO
euence % is alternating addition5 6he red numbers as shown in the next figure are added b 12 to get the next term while the blue numbers are added b %5
'e hae done seeral examples and it is impossible for us to exhaust all patternsso it is up to ou to be able to spot them5 6he patterns could be different- but the principle of soling them is the same5 How to ole Ciil erice Exam Number eries $roblems %
This is the fourth part of the solving number series problems. The frst part discussed patterns that contains addition and subtraction and the second part discusses patterns that contains multiplication or division. The third part was about alternating patterns. In this post, we are going to discuss some special number patterns. Although there is a small probability that these types of patterns will appear in the Civil Service Examination I didn!t see any when I too" the exams, both professional and subprofessional#, it is better that you "now that such patterns exist. Triangular Numbers
$, %, &, $', $(, )$, )*, %&, +(,
43
Triangular numbers are numbers that are formed by arranging dots in triangular patterns. Therefore, the first term is $, the second term is $ - ), the third term is $ - ) - % and so on. Square Numbers
$, +, , $&, )(, %&, The s/uare numbers is a se/uence of perfect s/uares0 , , , , and so on.
, ,
Cube Numbers
$, *, )1, &+, $)(, 2ell, from s/uare numbers, you surely have guessed what are cube numbers. They are a se/uence of cube of integers. ,
, ,
,
Fibonacci Sequence
$, $, ), %, (, *, $%, )$, Technically, a 3ibonacci se/uence is a se/uence that starts with ', $#, or $, $#, and each term is the sum of the previous two. 3or example, 44
in the se/uence above, ( is the sum of ) and %, while )$ is the sum of * and $%. In the actual examination, they may give 3ibonacci4li"e se/uences technically called 5ucas Se/uence# where they start with two different numbers. 3or example, a 5ucas se/uence that starts with $ and % will generate $, %, +, 1, $$, $*, 6f course, they can also combine positive and negative numbers to create such se/uences. 3or example, a 5ucas se/uence that starts with 4* and % will generate the se/uence 4*, %, 4(, 4), 41, 4, 2ell, this loo"s li"e a difficult se/uence, but remember that if you can see the pattern, it is easy to loo" for the next terms. VII. ALGEBRA
A Tutorial on Solving E/uations 7art $ Solving equation is one of the most fundamental concepts that you
should learn to be able to solve a lot of mathematical problems such as those in the Civil Service Examinations. 3or example, for you to be able to solve a word problem, you need to translate words into expressions, set up the e/uation, and solve it. Therefore, you should learn this post and its continuation by heart. In this series of posts, we are going to learn how to solve e/uations and then learn how to solve different types of problems number, age, coin, 8eometry, motion, etc#. These types of problems usually appear in the Civil Service Examinations. So! "#at i an equation $eall%&
45
An e/uation are two expressions sometimes more# with the e/ual sign in between. The e/uation
means that the algebraic expression on the left hand side which is has the same value as the numerical expression on the right hand side which is . 9ow, you can thin" of the e/ual sign as a 'alan(e. If you put two different ob:ects and they balance, it means if you ta"e away half of the ob:ect on the left, you also have to ta"e half of the ob:ect on the left. 6r, if you double the amount or weight# of the ob:ect on the left, you also double what!s on the right to "eep the balance.
The fancy name of this ;principle! in mathematics is P$o)e$tie o* Equalit%. It basically means that whatever you do on the left hand side, you also do on the right hand side of the e/uation.
There is really nothing to solve in this example. 2hat will you add to to get . 6f course .
numbers on the other side. Since, is on the left hand side, we we want want to get rid of . So, since was added to , we have to subtract from both sides to get rid of it. So, . This gives us
.
E,a)le / 0
This example can be again solved mentally mentally.. 2hat will you multiply with to get , of course, it!s . =ut, solving it as above, to get rid of % in , since it is multiplication multiplication,, we divide it by . . 6f course, ifif you divide divide the left left hand side side by , you also also divide the right hand side of the e/uation by . This gives us
.
E,a)le 0 0
In this example, is a fraction fraction which which mean that we have to get rid of (. To do this, we multiply both sides by . That is,
Therefore,
.
5i"e Examples $ and ), this can be solved mentally. mentally. E,a)le 12
47
In this example, we have ) times and then added to . 2ell, intuitively,, we can eliminate first by subtracting it from intuitively from both both sides. sides. That is
which results to . 9ow, it!s multiplication, so we eliminate by on the left left hand hand side by dividing both sides by . That is . This results to E,a)le 3 0
.
2e first need to eliminate on the left hand side of the e/uation. Since it is subtraction, to eliminate it, we have to perform addition because # on both sides of the e/uation. >oing this, we have
9ow, we solve for by dividing both sides by . That is . That is,
or
in mixed fraction or
in decimals.
In the next part of part of this series, we are going to learn how to solve more complicated e/uations.
48
A Tu Tutorial torial on Solving Solving E/uations E/uations 7art ) Solvin ving g Equ Equati ations ons art 1 . As I ha This is a co This cont ntin inua uati tion on of Sol have mentioned in that post, being able to solve e/uations is very important since it is used for solving more complicated problems e.g. word problems#.
In th this is po post st,, we ar are e go goin ing g to so solv lve e a sl slig ight htly ly mo more re co comp mplilica cate ted d e/uations. 2e already discussed 5 e!a"ples in the first post, so we start with our sixth example. Example 60
As I have mentioned mentioned in the previous previous examples, we need to isolate on one side of the e/uation and all the numbers on the other side.
6f course,
, so, simplifying, we have
Then, we want want to eliminate eliminate on the left hand side. Since it is multiplication,, we therefore multiplication therefore divide both sides of the e/uation e/uation by .
Therefore,
.
Example 7:
In this this example, example, we want want to avoid a negativ negative e , so it is better better to put all ;s on the right hand side of the the e/uation. e/uation. This means means that we have have 49
to elim eliminat inate e from the left hand side side.. So, So, we subt subtract ract from the left hand side, and of course, the right hand side as well.
9ext, since we want to eliminate all the numbers on the right, the easiest to eliminate first is . To do this, we :ust add on both sides of the e/uation.
. 9ext, we only have one number on the right hand side which is . To To eliminate it, we divide by . 6f course, we also need to divide the other side by .
Therefore, the answer is
.
9otice also that we can add and subtract immediately resulting to ma"ing the process faster. ?ou will be able to discover such strategy on your own if you solve more e/uations. Example 80
In this examp example, le, we have the form in the left hand side of the e/uation. To simplify this, we simply distribute the multiplication of over . That is 50
. This is called the distributive property of multiplication over addition. So, solving the problem above, we have
Adding to both sides of the e/uation, we hhave
>ividing both sides of the e/uation by we have . Example :
In e/uations with fractions, the basic strategy is to eliminate the denominator. In this example, the denominator is . Since means divided by , we cancel out by multiplying the e/uation by (. 9otice how ( is distributed over the left hand side. which is the same as . Simplifying, we have Subtracting
.
from both sides, we have
>ividing both sides by %, we have
. 51
Example !":
.
2e eliminate fraction by multiplying both sides of the e/uation by ). That is
In the left hand side, cancels out , so only hand side, we use distributive property.
Subtracting
is left. 6n the right
from both sides, we have
Subtracting from both sides, we have
A Tutorial on Solving E/uations 7art % This is the third part of the series of tutorials on solving e/uations. In this part, we will solve more complicated e/uations especially those that contain fractions. The frst part and the second part of this series discuss $' sample e/uations. 2e start with the $$th example. Example !!0 4( x @ % 4+ x - $)
This example deals with the /uestion of what if is negativeB 5et us solve the e/uation. 2e want on the left and all the numbers on the right. So, we add 4! to both sides. 4( x @ % - 4 x 4+ x - 4 x - $) 52
@ x @ % $) 9ext, we add % to both sides to eliminate 4% from the left hand side of the e/uation. @ x @ % - % $) - % @ x $( ?ou cannot have a final e/uation li"e this where there is a negative sign on x . To eliminate the negative sign on x , multiply both sides by 4$. That is 4$#4 x # 4$#$(# So, x # $!% is the final answer. Example !&0
This example highlights the distributive property. 9otice that distributive property is also needed on e/uations with fractions. The idea is that if you have an expression that loo"s li"e that is, a multiplied by the /uantity , you must Ddistribute over them. That is, and
.
Solving the e/uation above, we have
9otice on the right hand side that is not distributed to the second because the second is outside the parenthesis. 2e now simplify. . 9ext, we simplify the expression
on the right hand side. 53
9ow, we want to put on the left and all the numbers on the right. 2e do this simultaneously. 2e subtract from the right hand side and add & on the left hand side, so we add to both sides of the e/uation. ?ou can do this separately if you are confused.
6n the left hand side0 side, and
and
This gives us . Fultiplying both sides by Example $$, we have
. 6n the right hand
, as we have done in
as the final answer. Example !':
.
This type of e/uation usually appears in wor" and motion problems which we will discuss later. Gust li"e in solving fractions, all you have to do is get the least co""on deno"inator . 9ow, the least common denominator of ), %, and + is $). So, all we have to do is to multiply everything with $). That is
>ividing both sides by
, we have 54
. Example !(0
This is almost the same the above example. 2e get the least common denominator of and which is e/ual to . Then, we multiply everything with . That is
9ow, on the left hand side, This gives us
and on the right hand side
.
. Simplifying the left hand side, we have
9ow, gives us . Fultiplying both sides by ma"e positive gives us the final answer
to
. Example !%0
This example discusses the /uestion Dwhat if is in the denominatorB If is :ust in the denominator :ust li"e this example, the solution is /uite similar to Example $%.
55
The strategy here is to get the least common denominator of the numbers and then include during the multiplication. In this example, we want to get the least common denominator of and which is . 9ow, we include and the least common denominator of the e/uation above is . 9ow, we multiply everything with . That is,
. Therefore, the answer is
.
This ends the third part of this series, in the next part of this series I am not sure if I will discuss this soon#, we will discuss about dealing e/uations with radicals s/uare root and cube root#.
4ua5$ati( Equation
Introduction to Huadratic E/uation The length of a rectangle is % cm more than its width. Its area is e/ual to (+ s/uare centimeters. 2hat is its length and widthB Solution 5et
56
x width of rectangle x - % length of rectangle The area of a rectangle is the product of the length and width, so we have Area xx - %# which is e/ual to (+. Therefore, we can form the following e/uation0 xx - %# (+. =y the distributive property, we have
Finding t)e value o* x
In the e/uation, we want to find the value of x that ma"es the e/uation true. 2ithout algebraic manipulation, we can find the value of x by assigning various values to x. The e/uation indicates that one number is greater than the other by % and their product is (+. Examining the numbers with product as (+, we have, $ and (+ ) and )1 % and $* & and . 9ote0 2e have excluded the negative e.g. 4$#4(+# (+# numbers since a side length cannot be negative. 9ow, 4& % which means that the side lengths of the rectangle are & and . ?es, their product is (+ and one is % greater than the other. 57
In the e/uation above, subtracting both sides by (+, we have
. The e/uation that we formed above is an example of a /uadratic e/uation. A quadratic equation is of the form , where a, b, and c are real numbers and a not e/ual to '. In the example above, a $, b %, and c 4(+. In the problem above, we got the value of x by testing several values, however, there are more systematic methods. In the next post, we will be discussing one of these methods. These methods are factoring, completing the s/uare, and /uadratic formula. Solving Huadratic E/uations by Extracting the S/uare oot In the previous post, we have learned about /uadratic e/uations or e/uations of the form , where a is not e/ual to '. In this e/uation, we want to find the value of x which we call the root or the solution to the e/uation. There are three strategies in finding the root of the e/uation0 by extracting the roots, by completing the s/uare, and by the /uadratic formula. In this example, we will discuss, how to find the root of the /uadratic e/uation by extracting the root. Gust li"e in solving e/uations, if we want to find the value of x, we put all the numbers on one side, and all the x!s on one side. Since /uadratic e/uations contain the term , we can find the value of x by extracting the s/uare roots. =elow are five examples on how to do this. 58
Example !: Solution
>ividing both sides by ), we have . This gives us . Extracting the s/uare root of both sides, we have . Therefore, the root
.
Example &: Solution
. Adding %& to both side, we have . . Extracting the s/uare root of both sides, we have . . In this example, x has two roots0 x & and x 4&. 59
Example ': Solution
Subtracting *$ from both sides, we have
. In this case, there is no number that when multiplied by itself is negative. 3or example, negative times negative is e/ual to positive, and positive times positive is e/ual to positive. Therefore, there is no real root. There is, however, what we call a complex root as shown in the video below. Example (: Solution
>ividing both sides by (, we have . Extracting the s/uare root of both sides, we have
. 60
Example (0
>ividing both sides by %, we have . Extracting the s/uare root of both sides, we have . Solving Huadratic E/uations by 3actoring In the previous post, we have learned how to solve quadratic equations by extracting t)e roots . In this post, we are going to learn how to solve /uadratic e/uations by factoring. To solve /uadratic e/uations by factoring, we need to use the Jero property of real numbers. It states that the product of two real numbers is Jero if at least one of the two real numbers is Jero. In effect, we need to transfer all the terms to the lefth hand side, let the right hand side be ', e/uate factored form Jero and find the value of x. Example !: Solution
. This means the solutions of
are or
.
Example &:
61
Solution
Subtracting from both sides, we have . 3actoring, we have
. This means the solutions of
are or
.
Example ': Solution
Subtracting .
from both sides,
3actoring out , we have . E/uating both expression to ', we have . This means the solutions of Example (: Solve
are or
.
.
Solution
. This means the solutions of
are
or . 62
Example %: Solve
.
Solution
. This means the solutions of
are K"atex 4$K or
.
Example 6: Solve Solution
. This means the solutions of
are or .
Example 7: Solution
Cross multiplying, we have . Expanding, we have . Transposing all the terms to the left hand side, we have
63
. This means the solutions are or . Solving Huadratic E/uations by Huadratic 3ormula In the previous post, we have learned how to solve /uadratic e/uations by factoring. In this post, we are going to learn how to solve /uadratic e/uations using the /uadratic formula. In doing this, we must identify the values of , , and , in and substitute their values to the /uadratic formula . 9ote that the value of is the number in the term containing , is the number in the term containing , and is the value of the constant without or #. The results in this calculation which are the values of are the roots of the /uadratic e/uation. =efore you calculate using this formula, it is important that you master properties of radical numbers and how to calculate using them. Example $0 3ind the roots of Solution 3rom the e/uation, we can identify
,
, and
.
Substituting these values in the /uadratic formula, we have
64
. 2e "now, that
. So, we have
Therefore, we have two roots or Example )0 3ind the roots of Solution ecall, that it easier to identify the values of , , and if the /uadratic e/uation is in the general form which is . In order to ma"e the right hand side of the e/uation above e/ual to ', subtract $( from both sides of the e/uation by $(. This results to . As we can see,
,
and
.
Substituting these values to the /uadratic formula, we have
65
. =ut
.
Therefore, . 3actoring out ), we have
Therefore, we have two roots0 or
VIII. AREA A$ea o* a Re(tangle Se$ie Calculating 0reas of Geometric Figures 0rea of geometric figures are er common in Ciil erice Exams and also other tpes of examinations5 0rea is basicall the number of suare units that can fit inside a closed region5 #n a closed region- if all the unit suares fit exactl- ou can >ust count them and the number of suares is the area5 For example- the areas of the figures below are %- 1!- 8 and 2! suare units5 6he figures blow are rectangles <es- a suare is a rectangleR=5 Counting the figures and obsering the relationship between their side lengths and their areasit is eas to see that the area is eual to the product of the length and the width <'h)=5
66
6he blue rectangle has length . and width 2- and counting the number of suareswe hae 1!5 7f course- it is eas to see that we can group the suares into two groups of .- or fie groups of 25 From this grouping- we can >ustif wh the formula for the area of a rectangle is described b the formula
where is the area of the rectangle- is the length and suare has the same side length- we can sa that
where
is its area and
is the width5 ince the
is its side length5
6here are also certain figures whose areas are difficult to calculate intuitiel such as the area of a circle- but mathematicians hae alread found was to calculate the areas for these figures5 Challenge: Find the area of the green and blue figure below and estimate the area of the circle5
67
elow are some formulas for the most common shapes used in examinations5 4on(t worr because we will discuss them one b one5 6riangle:
-
$arallelogram:
is base-
is base-
6rape&oid: Circle:
is height5
-
and
is height are the base-
is the height
r is radius
#n this series- we are going to discuss the areas of the most commonl used figure in examinations and we will discuss arious problems in calculating areas of geometric figures5 'e are also going to discuss word problems about them5 Luestions li;e the number of tiles that can be used to tile a room is actuall an area problem5 How to ole 3ectangle 0rea $roblems $art 1 6he area of a rectangle including suare are the simplest to calculate5 0s we hae discussed in the preious post- the can be calculated b multipling their length and the width5 6hat is if a rectangle has area - length and width - thenor simpl
5
#n this post- we are going to sole arious problem inoling area of rectangles5 $roblem 1 6he length of a rectangle is 12 centimeters and its width is . centimeters5 'hat is its area)
68
Solution Ksing the representation aboe-
and
5 Calculating the area- we hae
5 6he area is /! suare centimeters5 $roblem 2 6he area of a rectangular garden is 2! suare meters5 #ts width is 25. meters5 'hat is its length) Solution #n this problem- the missing is the length and the gien are the area and the width5 oand 5 Ksing the formula- we hae 5 ubstituting the alues of
and
- we hae
5 ince we are loo;ing for - we diide both sides of euation b 25.5 6hat is 5 implifing- we hae
5
6herefore- the length of the rectangular garden is eual to 8 meters5 $roblem " 6he floor of a room 8 meters b / meters is to be coered with suare tiles5 6he tiles dimensions is 2. centimeters b 2. centimeters5 How man tiles are needed to coered the room) Note: 1 meter 1!! centimeters Solution 6his problem has at least two solutions5 # will show one solution and leae ou to loo; for another solution5 Ksing the area formula- we can calculate the area of the room in suare meters5 6hat is5 o- the area of the room is %8 suare meters5 Howeer- we are loo;ing for the number of tiles that can coer the room and not the area in suare meters5 Nowthe easiest solution is to find the number of tiles that can fit inside 1 suare
69
meter5 ince the side of a suare is 1 meter which is eual to 1!! centimetersit can fit % tiles as shown below5
1 s+uare meter contains 16 tiles Now- four tiles at the side means 1 suare meter contains ince there are %8 suare meters- the number of tiles needed is
suare tiles5
5 6herefore- we need at least ,/8 suare tiles to coer the entire floor5 How to ole 3ectangle 0rea $roblems $art 2 'e hae alread learned the concept of area of a rectangle and soled sample problems about it5 #n this post- we continue the rectangle area problems series5 'e discuss three more problems about rectangle area5 6he fourth problem below inoles area preseration- the fifth is calculating the area gien its perimeter- and the sixth reuiring the use of uadratic euations5 *et(s begin5 $roblem % 'hat is the area of the figure below)
70
Solution 6he figure aboe can be diided into " rectangles5 7ne wa to do this is to draw the dashed line below
Now- the area of the figure is the sum of the areas of the three rectangles5 0rea of a / cm b % cm is / cm J % cm 0rea of a / cm b % cm is / cm J " cm 0rea of a / cm b % cm is / cm J 1/ cm ?/ suare cm
2% 18
suare suare
cm5 cm5
o- the area of the figure is 2% 18 ?/ 1"8 s5 cm5 $roblem . 6he perimeter of a rectangle .% cm5 #ts length is twice than its width5 'hat is its area)
71
Solution 'e hae alread discussed how to calculate the perimeter of a rectangle and we hae learned its formula5 0 rectangle with perimeter P - length l and width ' has perimeter P 2l 2' 5 Now- we let the width be eual to x 5 ince the length is twice- it is 2x 5 ubstituting them to the formula aboe- we hae .% 2<2 x = 2 x% implifing- we hae .% / x resulting to x ?5 6herefore- the width is ? and the length which is twice the width is 185 o- the area is ?<18= 1/2 s5 cm5 $roblem / 6he length of a rectangle is . more than its width5 #ts area is 8% suare centimeters5 'hat are its dimensions) Solution uess an# C!ec" 6his problem can be soled using guess and chec; but # wouldn(t recommend it5 For example- ou can choose two numbers where one is . greater than the other and find their product5 Choosing % and ? results to the product "/5 #t is uite small- so ou might want to tr 1! and 1. but the product is 1.!- uite large- soou can go down- and ou will eentuall find , and 12 which is the correct answer5 0nother guess and chec; strateg in this problem is to find the factors of 8%
be the width of the triangle- then it(s length is . greater than the width-
so it is thereforeand width- so-
5 ince the area of a rectangle is the product of its length
5 6his results to the uadratic euation
which is euialent to
5
72
#f ou still remember factoring- then this is an eas problem to factor5 6his gies us
which gies us
which is its width5 6his also gies us the length x . 125
6his solution which uses uadratic euation is a bit adanced- but there is no wa that ou can sole problems li;e the one aboe if ou don(t ;now it5 # am afraid that ou hae to learn it again if ou hae forgotten it5 Aou must practice factoring and memori&e the uadratic formula <# will discuss this after this series=5 6hen and onl then- that ou would be able to sole such problems with better speed and accurac5 Exam 0ip #f ou encounter problems such as this and ou don(t ;now what to do- it is important that ou do not spend too much time on them5 Sust guess the answer first- mar; them- and come bac; to them when ou still hae time at the end of the exam5 Howeer- be sure not to s;ip too man items5
Ci$(le #ntroduction to the asic Concepts of Circles 6he Ciil erice Exams also contain geometr problems- and so farour discussions are mostl algebra problems5 #n this new series of posts- we will discuss how to sole geometr and measurement problems particularl about circles5 Howeer- before we start soling problems- let us first discuss the basic terminologies about circles5 0 circle is a set of points euidistant to a point called the center of the circle5 0s # go around to gie trainings and lectures - # usuall hear the wrong definition below5 # am not sure where this definition originated- but this is wrong5 'rong 4efinition: 0 circle is a polgon with infinite number of sides5 #f ou loo; at the definition of a polgon
73
0 segment from the center of a circle to a point on the circle such as is called radius oining two points on the circle such as
is called chord5
6he longest chord that can be made in a circle passes through the center5 6his chord is called diameter5 #n the figure aboe-
is a diameter of circle 5
ome asic Facts 0bout Circles Notice that diameter is composed of two radiiand 5 6herefore- the diameter of a circle is twice its radius5 o- if we let the diameter - and radius r we can sa that 5 #f we measure the length of the circle- that is if we start from - go aroun# along t!e circle until we reach again- the distance we would hae traeled is called its circumference5 6he formula circumference C is
where
is approximatel "51%1/5
ince and - we can also sa that 5 Note that the circle itself
74
6he circumference of a circle is basicall the distance around the circle itself5 #f ou want to find the circumference of a can- for example- ou can get a measuring tape and wrap around it5 6he animation below shows- the meaning of circumference5 0s we can see- the circle with diameter 1 has circumference or approximatel 5
Note: #f ou want to ;now where
came from- read Calculating the @alue of $i5
Example 1 'hat is the perimeter of a circle with diameter 1 unit) Solution 6he formula of finding the circumference of a circle is with circumference diameter is 5 o-
and
5 Example 2 Find the circumference of a circle with radius 25. cm5 Solution 6he circumference
of a circle with radius
is
o6herefore- the circumference of a circle with radius 25. cm is 1.5, cm5 Example 3 Find the radius of a circle with a circumference 1858% cm5 Kse
5
Solution
75
4iiding both sides b /528- we hae 5 6herefore- the radius of a circle with circumference 1858% cm is " cm5 Example 4 +i;e was >ogging in circular par;5 Halfwa completing the circle- he went bac; to where he started through a straight path5 #f he traeled a total distance of .1% meters- what is the total distance if he >ogged around the par; once)
5
ubstituting- we hae
5
Factoring out - we hae
5 4iiding both sides b .51%- we get 5 Now- we are loo;ing for the distance around the par;
meters5 How to Calculate the 0rea of a Circle *ast wee;- we hae discussed how to calculate the circumference of a circle 5 #n this post- we learn how to calculate the area of a circle5 6he area of a circle which we will denote b is eual to the product of and the suare of its radius 5 $utting it in euation- we hae
76
5 #n the examinations- the alue of
is specified5 6he usuall use
-
or
5
#f ou can recall- the radius is the segment from the center to the point on the circle as shown below5 6he radius is half the diameter5 6he diameter is the longest segment that ou can draw from one point on the circle to another5 #t alwas passes through the center5 Note: 'e also use the term radius to refer ra#ius and diameter as the lengt! of t!e #iameter 5
to
the lengt!
Now that we hae reiewed the basic terminologiessome examples on how to calculate the area of a circle5
let
of
us
t!e
hae
Example 1 'hat is the area of a circle with radius 8 centimeters5 Kse
5
Solution
o- the area of the circle is 2!!5?/ suare centimeters
77
Find the area of a circle with diameter 1% centimeters5 Kse
5
Solution Notice that the gien is the diameter- so we find the radius5 ince the diameter is twice its radius- we diide 1% centimeters b 2 giing us , centimeters as the radius5 Now- let(s calculate the area5
suare centimeters5 Example 3 Find the radius of a circle with area /528 suare meters5 Kse
5
Solution #n this problem- area is gien5 'e are loo;ing for the radius5 'e still use the original formula and ma;e algebraic manipulations later- so we don(t hae to memori&e a lot of formulas5
'e substitute the alue of area and
5
'e are loo;ing for - so we isolate euations=5
to the right side
ince- we hae a suare- we get the suare root of both sides5 6hat is
o- radius is suare root of 2 meters or about 15%1 meters5 #n this calculation- 2 is not a perfect suare5 ince ou are not allowed to use calculator- the probabl won(t let ou calculate for the suare root of number5 o-
78
in this case- the final answer is that the radius of the circle is suare root of 2 meters
T)e +rea o* a Trape,oid Series How to Find the 0rea of a 6rape&oid 'e hae learned how to calculate the areas of a suare- rectangle- parallelogramand circle5 #n this post- we are going to learn how to find the area of a trape&oid5 6his is the first post of Finding the 0rea of a 6rape&oid eries5 0 trape&oid is a polgon whose exactl one pair of sides are parallelT5 6he figure below is a trape&oid where sides a and are parallel5
Notice that if we ma;e another trape&oid which has the same si&e and shape as aboe- flip one trape&oid- and ma;e one pair of the nonDparallel sides meet- we can form the figure below5 6hat figure is a parallelogram5 Can ou see wh)
Now- obsere that the ase of the parallelogram from the figure is a 5 height is !%
#ts
'e hae learned that the area of a parallelogram is the product of its base and height5 o- the expression that describes its area is
79
5 Now- when we calculated for the area of the parallelogram aboe- we actuall calculated the area of two trape&oids5 6herefore- to get the area of a trape&oid- the hae diide the formula aboe b 2 or multipl it b 5 6hat is- if we let area of a trape&oid is
be the
where a and are the base
-
and
)
6he formula for area is
o- substituting we hae
o- the area of the trape&oid is 22. suare units5 How to Find the 0rea of a 6rape&oid $art 2 #n the preious post- we hae learned the formula for finding the area of a trape&oid5 'e deried that the formula for the area of a trape&oid with base and
is
#n this post- which is the second part of Finding the 0rea of a 6rape&oid eries- we are going to continue with some examples5 'e will not onl find the area of a trape&oid- but other missing dimensions such as base and height5 Now- get our paper and pencils and tr to sole the problems on our own before reading the solution5
80
'e hae alread discussed one example in the preious post- so we start with the second example5 Example 2 'hat is the area of a trape&oid whose parallel sides measure / cm and 8 cm and whose altitude is 25. cm) Solution #n this example- the parallel sides are the base- so we can substitute them to and 5 ince we are loo;ing for the sum of and - we can substitute them interchangeabl5 6he term altitu#e is also another term for height5 oand 5 'e now substitute5
o the area is
suare centimeters5
e CarefulR: 0gain- remember that if we tal; about area- we are tal;ing about suare units- and in this case suare centimeters5 #f ou choose an option which is 1,5. centimeters- then it is '37NG5 #t should be ,5. suare centimetersR Example " Find the height of a trape&oid whose base lengths are . and 8 units and whose area is 18 suare units5
81
Solution #n this problem- we loo; for the height5 ut don(t worr- we will still use the same formula- and manipulate the euation later to find 5 o here- we hae - and 5
+ultipling 12 and 12- we hae 5 'e are loo;ing for
- so to eliminate /- we diide both euations b /5 6hat is-
5 o- the height of the trape&oid is " units
of a series on finding the area of a trape&oid here in $H Ciil the first post- we discussed the deriation of the area of a a wor;ed example5 #n the second post- we discussed how to the baseand the height as well as to find the height gien the
82
#n this post- we are going to find the base- gien the height and the area5 'e continue with the fourth example5 Example % 0 trape&oid has area /. suare centimeters- height 1" cm- and base of % cm5 Find the other base5 Solution #n this example- we hae
-
and
5 'e are loo;ing for 5
#n euations with fractions- we alwas want to eliminate the fractions5 #n the euation aboe- we can do this b multipling both sides of the euation b 25 6hat is5 6he product of 2 and 12 is 1- so5 Next- we use distributie propert on the right hand side5 3ecall: 5
5 'e want to find - so we subtract .2 from both sides giing us 5 Next- we diide both sides b 1" 5 o- the other base is / centimeters which is our answer to the problem5 Example . 6he figure below is a trape&oid5 Find the alue of
5
83
