3 Fraction

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3·2 )ua ntitative A ptitu cle ior C om pe titive E xa min ation s

7 7Exampll ': Thus, is greater than

13 19

Rule 3

When [WO or more fractions with different denominators and different numerators are to be compared, then

the following simple technique is to be used:

Step 1 Among all the given fractions,

let the maximum number of digits in the numerator = II

the maximum number of digits in the denominator = d

Step 2 Find (d ~ 1 1 ) .

Step 3 If (d ~ 1/) = 0 or I, multiply each fraction by 10.

If (d ~ 1 1 ) = 2, 3,4 ... multiply each given fraction by 102• 10'. 10~. . respectively.

Step 4 After multiplication, find only the integer value of the resultant fraction.

Step 5 If in step 4. any of the two fractions have the same integer value, then find the next decimal

place and so on.

Step 6 Compare the integer/decimal values obtained in step 4 or step 5. The fraction having the

maximum value is the greatest fraction.

Note: In order to write the given fraction in ascending order the smallest fraction is written first. then the next greater

one and so on. In order to write the given fraction in descending order. the greatest fraction is written first. then the

next smaller and so on.

E 1 Arr7 493 87 123 . d di d

~ x a l l l p e: ange . . . 1 1 1 escen mg or cr

13 971 165' 235

I.~()//Ili(}l/: Here. maximum no. of digits in the numerator 11= 3 (in 493 or in 123) and maximum no. of digits

in the denominator = d = 3 (in 971 or in 235)

Now. d ~ 1 1 = 3 ~ 3 = 0

So, multiply the given fractions by 10.

L x 10. 493 x 10. 87 x 10. 123 x 10.13 971 165 235

5 5 5 5 (integer values)

Since the integer values are same, so, find the next decimal digit for these fractions,

5.3 5.0 5.2 5.2

L_-same--!

Sinee the value of two fractions are same. find the second decimal digit for these two fractions only, by

dividing further

5.3 5.0 5.27 5.23.

Now, 5.3 > 5.27 > 5.23 > 5.0

7 87 123 493(in descending order)::::} > > >

13 165 235 971

3.2 FRACTIONAL PART OF A NUM BER

Fractional part of a number (or quantity) is simply the product of the related fraction and the given number.

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Fraction 3-3

Example: Consider a given number as 60, then

:.Two-thirds of 60 = x 60 = 40

3

~- rd of 60 is 40 (fractional part)3

I Fractional part of any number = number x its related fraction.j (1)

3.2.1 Different Fractional Parts of the Same Number

Consider any number, say, 36

then 1 . th of 36= 27 (fractional part of 36)4

1h of 36 = 4 (fractional part of 36)9

')

-: rd of 36 = 24 (fractional part of 36)3

From this, we find that as the fraction changes. the fractional part of the same number also changes.

In our earlier examples. we find that

27 4 24

2/3. = 36 (Fixed)

iI . .

"-----+ Original number

In such cases, equation (1) can be re-written as

3/4 1/9

Any fractional number - I '

= Original numberits related fraction I ,

~ I~~ it is fixed.)

(2)

1 3["ample: A man travels - th part by scooter. - th bv car and rest 48 km by bus. Find the total distance

4 8 .

covered.

<oluuon: Here, total distance (i.e. original quantity) is to be found out.

Fraction related to rest 48 km = 1 1 - ( ± + ~)

rrelated fraction to total distance

Lsing the relation (2)

Any fractional number= Original number.

its related fraction

Here, we find,

Rest distance

---------------- = total distanceits related fraction

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3-4 Q uantitative A ptitude for C om petitive Exam inations

-1-8= total distance

total distance = 483/8

128 km

':x amp lt': I f l_ th of a number is U :. then find its two-third.1 7

')

."(JIII/loll: Let z: rd of number = .r,

3

3 2Since both the fractions and are to be found out for the same number. the relation (2) can be used

17 3

as'"

Fractional number


Another fractional number

its related fraction= original number (always)

18 x

3/17 2/3

. v = 68

Two-third of the number is 68 .

* Note: This relation can be used to find another fractional part directly without finding the original numher.

3.3 TO FIND THE FRACTION RELATED TO BAlANCE (REST) AMOUNT

Conventionally. we have learnt that

Fraction related to balance (rest) part = 1 - (sum of all other fractions)

It is used when all fractions are independent. Following example will Illustrate the fact.

Example: A person spends ~ th pan of his salary on food, /2 th pan of his salary on education, ~ th part

of his salary on clothing. He is now left with Rs. 550. Find his total salary .

.in/III/01l. Here. the spending on each item is independent. because each fraction has been indicated as outof total salary (original number).

fraction related to rest part I- ( 2 + _ _ ! _ . + l _ )8 12 4

5

24

Rest amounttotal salary

fraction related to rest pan(Refer equation 2 ojsec. 3.2.11

550 = Rs 26-1-0 .5/24

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Fraction 3-5

Hence, for' independent fractions.

Fraction for balance (rest) part = 1 - (sum of all independent fractions) (3)

,: .< . con sider another example.

h.:.mple: A person spends i s t h part of his salary on food. _ _ ! _th of the rest part on education and 1 th of1 2 4

= : : remainder on clothing. He is now left with Rs 550. Find his total salary.

... ';,,11: Here, spending on the second item (i.e. education) depends on the amount left afte SPl nding on

:= : : ~lrst item (i.e. food). Similarly. spending on the third item (i.e. clothing) depends on the amount left

:"';":TlJ.ining) after spending on the first item and the second item.

Here. spending on each item (except the first item) depends on the amount remaining. after ,j nding on

- : = previous item.

l . : : ,uch cases, all fractions (except the first one) are dependent on the previous fractions.

fur dependent fractions, ')

Fraction forbalance (rest) part = (1 - first fraction) x (1 - second fraction) j

S'.:.,in our example, using the relation 4

( l _ i ) ( I - _ _ ! _ V I - _ l _ )S 12J~ 4

5 I I 3= -x-x-S 12 4

~"'.::.~tionfor balance (rest) part =

55

128

Rest amounttotal salary =

Fraction related to rest part

550 = Rs 1280551128

Sore: Observe the difference in the language of the two examples under 3.3

3.4 TO INSERT ANY NUMBER OF FRACTIONS IN BElWEEN lWO GIVEN FRACTIONS

Let two given fractions be - hQ

and ~. To insert a fraction lvinz between _,(/and 2.. the following steps arev - - ) ,. -

taken. .

Step 1 The numerators of two given fractions are added to get the numerator of the result fraction, i.e.

numerator of the result fraction = a + x

Step 2 The denominators are also added to get denominator of the result fraction. That is, denominator

of the result fraction = h + y

Step 3a+x

Result fraction =b+y

(4)

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3-6 Quantitative Aptitude to r Competitive Examinations

Hence, the result fraction so obtained has its magnitude (value) lying between the two given fractions.

By this method. any number of fractions can be inserted between two given fractions.

Solved Example.

E-l Arrange the following fractions in decreasing (desceuding j order:

(i)S 3 S 6

( i i )1 3 . 3 2 1

6' 4' 8' 7") . 5' 1 0 . 50

(iii)7 5 17 I

(iv)3 5 13 97

1 2 . 16' 36 .--

3 S · 7' 16 . 104

2 5 22 13(v) 91 ' 1 77 ' 1091' 558

S-1 Using the method 3.1.1

(i)Here 1 1 = 1 d = 1 :. d - 1 1 = 0

So. multiply the given fractions by 10.

5 10 8 3 5 6 0 .. . '6 x "". 4 x 10 '" 7. 8 x 10 '" 6. 7 x I. '" 8 (integer value)

Since two fractions have the same integer value (= 8), find the next decimal digit for these two

fractions only namely

8.3 7 6 8.5

6 S 3 S-->- > ..>7 6 4 8

in descending order.ow, 8.5 > 8.3 > 7 > 6 =;>

(ii) Here n = 2 d = 2 .. d - n = O .

max" no. of digits in numerator = / ) = 2 (in ~~ )

max III no. of digits in denominator = d = 2 ( . 3 2 1 )11 - or-10 50

So, multiply the given fraction by 10.

. . 1 _ x 10 '" 5. ,} x 10 '" 6. 3 x 10 '" 3.2 5 10

Now. 6> 5> 4> 3 => 3 > _ 1 > 21 > 3S 2 50 10

(iii) Here 1 1 = 2 d = 2 . . d - n = O.

So. multiply the given fraction by 10.

7 5 17 I. . T 2 x 10 '" 5. 16 x 10 '" 3, 36 x 10 '" 4, 3 x 10 '" 3 (integer values)

Since the two fractions have the same integer value (=3), find the next decimal place for these

two fractions only.

r.e. 5 3.1 4 3.3

2 1 x 10 '" 4 (integer value)50

Now,S> 4 > 3.3 > 3.1 =;>7 1 7 1 5

12> 36> 3 > 1 6

(iv) Using the method 3.1 1. here, maximum number of digits in numerator = 2

(.

' 13)n- .

1 6

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Fraction 3-7

Maximum number of digits in denominator = 3 (in 97 'J .IO-i

So, /1 = 2 d = 3 . . d - 11 = I

So, multiply the numerator of the given fraction by 1 0

:. ~ x 10 '" 6. ~ x 10 '" 7 . i ~ x 10 '" 8. l~: x 10 "" 9 (integer value)

Now, 9 > 8 > 7 > 6 ~97 13 5 3

--- > . > > ...104 16 7 5

(v) Hcren = 2 (in 2 2 or in 1 3) d = 4 (in 1 (91 ) :. d -/1 = 2

So, multiply the numerator of given fractions by I ()C

:. il x 1 0 0 '" 2, I~j x 1 0 0 "" 2. ' f~~l x 1 0 0 '" 2. S i S ' S x ]00 ec 2 t integer va. e)

All the fractions have the same integer value, so, find the next decimal place. i.c.

2 .] 2 .8 2.0 2 .3

Now. .., 8 ') 3 .., 1 .., 0 5 > 13 > 2 > 2 1 _. ~. > _. > ~. >... ~ 1 7 7 558 91 1091

E-2 (i part of what amount will be equal to 3 1 part of Rs 100.

S-2 Let the amount be Rs x

l_x = 31 I 00 ~12 4

x = 1£x)5 x 1005 4

5.r

12='SxlOO

4

~ x = Rs 900

Required amount is Rs 900.

E-3 What fraction is 6 bananas in 5 dozens')

S-3 Required fraction = 6 out of 5 dozen

6 1

Sxl2 10'

7There are 40 students in a class. One day only T O students were present. Find the number of

absentees on that day.

In solving the problem on fraction. the whole quantity is alwavs considered as I.

. . Number of absentees = Fraction of absentees x Total number

= ( 1 - 170) x 40 = 12 students.

E -5 A man spent ~ of his savings and still has Rs 1.000 left with him. What were his savings')

S-5 In this type of problem, if balance amount is given. then this amount is to be related to the balance

part (fraction). Using relation 2. for savings.

E-4

S-4

Savings =fraction related to balance part

balance amount

savmgs =1000

= Rs 1400.

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Using the relation 4. the fraction related to balance part =

3-8 Quantitative Aptitude tor Competitive Examinations

E-6 A man reads ~ of a book on a day and ~ of the remainder. on the second day. If the number of pages

still unread arc 40. how many pages did the book contain')

4S-6 It is a dependent activit», because on the second day he reads of the remaining pages.

5

using relation 2.

Total pages =fraction related to pages unread

Pages unread

= 40 = 3201 /8

Total pages = 320

E7 2 fh . 3 f h i I I f I I h- A man spends '5 0 is salary on food, 1 0 0 hIS sa ary on house rent and 8 0 the sa ary on c ot es.

He still has Rs 1.400 left with him. Find his salary.

S-7 The expenditure incurred on each item is expressed as part of the total amount (salary), so it consists

of independent fractions.

Using the relation 3, we get

Fraction related to balance part = 1 - (sum of independent fractions)

= 1- ( ~ + l~ +

n= 1- !~:0

Using relation 2, we get.

Balance amountTotal salary =

fraction related to balance part

= 1400 = 800{)7/40

Total salary = Rs 8000

E -8A man spends

*of his income on food, of the rest

!on house rent and :) on cloth. He still has

Rs 1760 left with him. Find his income.

S-8 Here. of the rest amount (after spending on food), · l is spent on house rent and'~ is spent on clothes.

So, spending on these last two items are independent of each other, but dependent on the expenditure

incurred on the first item. It is a problem both on dependent and independent activities.

Using relation 3 and 4 together. we get,

Fraction related to balance part = ( 1 - ~ ) x [ I - ( ~ + ; ) ]~

I~independent

2 II=-x-3 20

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Fraction 3-9

Using relation 2, we find, total income = 1760 =480011130

total income = Rs 4800

E -9 % of a pole is in the mud. When 3 of it is pulled out. 250 cm. of the pole is still in the mud. Find

the full length of the pole.

S-9 Using the relation 2 , we get.

Length in mudtotal length of pole =

Part in mud

250

4 1

7 3

1050

Length of pole = 1,050 cm.

E-10 After covering five-eighth of my journey. I find that J have travelled 60 km. How much journey is

left?

S-10 Using the relation 2. we get*

Journey covered Journey left

=Fraction related to journey covered

60

5/8


Journey left

1- 58

Journey left = 36 km

*Note: This relation is equal to total journey.

E-11 How much is to be added with 0.685 of 325 to get 300?

S-l1 Let 'x' is to be added, then

x + (0.685 x 325) = 300 : : : : : > x = 300 - 222.625 = 77.375.

E-12 A man distributes 0.375 of his money to his wife and 0.4 to his son. He has still Rs 3,375 left with

him. How much initial money did the man have') How much did his wife get?

S-12 Using the relation 3 and 2, we can write directly

[1 - (0.375 + 0.4)] x total money = balance money

: : : : : > [1 - 0.775J x Total money = 3375 : : : : : > Total money = 3375 = Rs 150001- 0.775

" Wife's share = 0.375 x total money = 0.375 x 15000 = Rs 5625.

E-13 Insert one fraction between

(i)1 4

(ii)I

-::;-and -- 2 and 32, 5

S-13 (i)4

3' 5

1 1+4 4 .'3 ' 3 + 5 ' 5 (using the method 3.4)

154---

3' 8' 5

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3-10 Q uantitative Aptitude tor Competitive Exarninations

5 4Thus, the resulting fraction 8 is more than 3 and less than 5 in magnitude (value).

(ii) 2, 3 ~

2 7

1 . 2

2 2 + 7 7= T' 1 + 2 '" 2 (using the method 3.4)

7= 2,3, 2

Hence. the resulting fraction is 3.

1 4E-14 Insert three fraction between 3 and s

4S-14 :1 ' 5

_ 1 1+4 4

-3'3+5'5( inserting one fraction between t and ~.)

I 1+5 55+4 4

3' 3+8'8' 8+5'5

6 5 9 43'11'8 '13'5 '

( inserting one fraction between * and

one fraction between ~ and ~ )

( three fractions inserted between 1 and 4 Il 3 5 )

58 " ' and

E-15 Which one of the following fractions is less than t ' J (MBA '82)

2 2(a) .

63

4(b) --II

. 15(c)-

46

33(d)-"-

98

S-15 Step 1 Reverse the test fraction, i.e. } becomes ~ = 3.

3Step 2 Reverse each alternative and find which alternative is greater than l '

63 3 I 1 3 46 3 98 3---< ... - <-- ..> ... -<-2 2 I' 4 I ' 15 I' 33 1

R. d f . . 15eqUlre racuon IS -

46

Note: Reversing method is used because division process becomes easier when the numerator is greater than

the denominator.

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Fraction 3-11

REG ULAR PROBLEM S

0) A badminton player, won 6 games and lost 4. The fraction of the games he won is:

3 7 3 2(a) (b) - (c) (d) (e)-

2 3 5 2 5

(2) What fraction of 2 hours is 12 seconds ')

(a)I

(b)I I

(d)1 3

- -~ Ic) Ie) -

600 1 2 60 5 50

(3) A rope is 25lm long. How many pieces each of I I m long can be cut from it?2 7

(a) 1 6 (h) 21 (e) 13 (d ) II (e) 1 7

I(a) 4- m. 6

(b) 10 m3

2(e) 16 m

3(d) 4 m (e) 20 m

(4) A lamp post has half of its length in mud. I of its length in water and 31m ahove the water. The3 3

total length of the post is:

2(5) A man pays off of his debt and still has to pay Rs 240 to payoff the debt completely. The total

5

amount of debt is:

(a) Rs 600 (b) Rs 400 (e) Rs 960 (d) Rs 480 (e) Rs 1200

(6) A drum of water is 3 full. When 38 litres are drawn from it. it is just 1 full. The half capacity of5 8

drum in litres is:

(a) 40 (b) 80 (c) 152 (d) 21.7 (e) 76

(7) The monthly salary of a man is Rs 480 and he spends 7 _ _ of it. His income increases by l of the8 6

present salary and his spending also increases by 2 of the present expenditure. His savings will now7

(a) increase by Rs 45 (b) decrease by Rs 40 (c) increase by Rs 40

(d) decrease by Rs 80 (e) decrease by Rs 60

(8) Which of the following fraction is the smallest?

(a)7

(b)1 4

(c)II

(dl8

(e)9

- -

13 33 25 15 11

Hint: See 3.1.1

( 9 ) Which of the following fraction is the greatest?

(a)1 6

(b)II

(c)1 6

(d)1 6

(e)11

- - -21 14 1 9 23 17

__------~-

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(a 3A) --

20

o » 9A

10

A(c)

10

(d) 3A

10

3A(e)

2

3-12 Quantitative Aptitude for Competitive Examinations

(10) A man pays off l._ of his debt cverv month. At the end of 6 months. his remaining debt is Rs A.• 20· ~

How much amount has he cleared off in every month (in Rs)?

(11) } part of a kerosene tin is filled. If 6 bottles are taken out of it and 3 bottles arc j led again, then5

half the tin is full. What is the capacity of the tin') (in bottles)

(a) 20 (b) 30 (e) 45 (d) 50

(RRB (-uwabati, '97)

(e) 4()

(12) Reciprocal of sum of the reciprocals of 3 and 7 IS:5 :,

1Cal ~4

2 1(b) --

44

4(c) -

5(d).t

4

- 21(e) 15

44

Hint : Start solving from backwards. First make reciprocals of 3 and 75 3

5 and 3.e,

3 7

Then sum it, as 5 + 3, 8 then find reciprocal of the sum3 7

numbers')

1 . , then what will be the sum of the2

(RRB Trivendrum (Tech), '97)

(13) If the product of two numbers is 5 and one of the number is

1(a) 4 ~

2(b) 6 ! _

24

5(c) ~

6(d) 9 (e) 4 1

3

(14) In an examination, a student was asked to find l_ of a certain number. By misake. he found 3 of14 4

it. His answer was 150 more than the correct answer. The given number is:

(a) 450 (b) 300 (c) 270 (d) 180 (e) 280

2 3(15) One of the rational numbers between ~ and is:

7 14

(a) 5

14

Hint : Refer

(16) Which of the following fractions is the greatest?

3

(b) 49(c)

4(d)

2(e) None

(RBI, '98)

(a) 219

337

(b) 221

335

(c) 217

339(d) .?!J

341(e) 222. 339

Hint : Do not try to calculate. The greatest fraction can be found out by eliminating first the fractions

ith l d d . . 215217 d 219 Th 221 andWIt ower numerator an greater enommator, I.e, - - an --- en compare341' 339' 337 . 335

222

339

~,-.,-., ..--,.--,-- _ ----------J

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3-14 Quantitative Aptitude for Competitive Examinatiom

(8) A man left t of his property to his daughter and the remaining to hi' sons to be equally divided

among them. if the share of the each son be double of that of the daughter. find the number of sons.

(NABARD, '97)

(a) 2 (b) 3 (c) 6 (d) 4 (e) 7

1(9) A vessel, full of water, weighs 16.5 kg. When the vessel is full, it weighs 5.2':: <g. The weight

4

of the empty vessel (in kg) is:

(a) l.125 (b) 4.5 (c) 1 .5 (d) 3 (e) 2."

')

(10) A scooter before overhauling requires ~ hour service time every 45 days, while, er overhauling

it requires l hour service time every 60 days. What fraction of pre-overhauling service time is saved

3

in the latter case? (MBA, '81)

4(a)-

3(b)

3

(c) 34

C d )4

(el 4

9

(11) Sundari, Kusu and Jyoti took two tests each. Sundari secured 24 marks in the first test and 3260 40

35 54marks in the second test. Kusu secured marks in the first test and marks in the second test.

70 60

1· d 27 ks i h ~ d 4'i . h Wh h diOh secure - mar s III t e first test an - marks III t e second test. 0 among t em id90 50

register maximum progress?

(a) Only Sundari

(d) Both Sundari and Kusu

(b) Only Kusu

(e) Both Kusu and Jyoti

(BSRB Bangalore, 2000)

(c) Only Jyoti

Hint: Tabulate the score in each test with common denominator so that the progress for each person

in second test over the first test can be found & compared.

I II

4 8Sundari -1 -1 -1

1 0 1 0

5 9-1 -1 -1

1 0 1 0

-1

( 1~-1 [) -1

2 times 8 = 2

Kusu Less than 2 times

4

9 < 2

5

Maximum progress, as in II test, score is 3

times the I test in terms of fraction

(12) A boy on being asked 1 3 of a certain fraction had made the mistake of dividing the fraction by 1314 14

lyoti

and so got an answer that exceeded the correct answer by l.The correct is:65

14(a) --

45(b) 12

651 3(e)

45(d) 2

7

196(e)

5 85

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Fraction 3-15

i 1 31 A has twice as much money as B. They play together. and at the end of the first game, B wins one

third of A's money from A: what fraction of the sum that B now has. must A win back in the second

game so that they may have exactly equal money';

la) 13

I(b) ..

S

(c) 14

1

10Ie)

IS

Hint: Assume, before the start of first game. B has Rs 1 and Aha" Rs :: !

,U I bought a number of mangoes at 3 S for : : ! . I divided the whole into two equal parts, one pa of which

1 sold at 1 7 , and the other at 1 8 mangoes per Rs I. Ispent and received an integral nUI11')er f rupees,

but bought the least possible number of mangoes. How many did I buy?

la) 21420 (b) 24120 (C) 22014 (dl 1 : : ! 2 S r e . AI2

Hint: Assume that I buy' 35 mangoes. Then on selling. I get (35 x .L + 3 S x .L ' I = Rs~- ~ ~ 2 17 2 1 8 )

2 2 5

612

~ . . . ~;\0. uf Lilit Lni!

mangoes price price

But the number of rupees is an integer.

1225so, I must receive 612 times

612

Hence no. of mangoes I buy is 612 times 35

! 15 I Find out that minimum fraction which when added to 29 + IS will give a complete number.12 16

(b) l!_

38

(c) 31

48

21lal -

38

1 7(d)- -

48

23(e) .

38t 1 6 1 In a class, 18 boys are there whose height is more than 160 cm. If they are three-fourth of the total

number of boys and the total number of boys is two-third of the total number of students. then how

many girls are there in the class')

(a) 18 (b) 6 (c) 12 (d) 24 (e) 8

I 1 7 1 The fuel indicator in a car shows 1h of the fuel tank as full. When 22 more Iitres of fuel are poured5

into the tank, the indicator rests at the three-fourth of the full mark. The capacity of the fuel tank (in

Iitres) is:

(a) 30 (b) 40 (c) 36 (d) 28 (e) 45

Answers

1. (a)

10. (d)

2 . (e)

11. (c)

3. (c)

12. (c)

4. (d)

13. (d)

5. (b)

14. (a)

6. (C)

15. (C )

7. (a)

16. (c)

8. (b)

17. (b)

9. (c)

3 Fraction

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