4 simplification

8/2/2019 4 Simplification

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4 S IM P L I F IC A T IO N A N DA P P R O X I M A T I O N",--~~~------------------- ----4.1 OPERATION ORDER SEQUENCEFor simplifying an expression containing various types of fractions. the order of various operation- involvedshould be strictly maintained. A simple technique for arranging the expression in the proper sequence, isby placing them in the order of the first letter appearing in VBODMAS where.

1. V Stands for vinculum or bar as ( - )2. B stands for bracket and operation of brackets in the order (), {} and then [J3. 0 stands for 'of'4. D stands for division (-:-)S . M stands for multiplication ( x )6. A stands for addition (+)7. S stands for subtraction (-)

4.2 APPLICATION FOR ALGEBRAIC FORMULASome algebraic formulae are used in solving the problems on simplification. Following important formulaare to be memorised:

1. (a + b)2 = a2 + b2 + 2ab.2. (a - b)2 = a2 + b2 - lob.3. (a + b)2 + (a - b)2 = 2(a2 + b2).4. (a + b)2 - (a - b)2 = 4ab.5. (a + b) x (a - b) = a2 - b2 6. (0 + b)3 = 03 + 3a2b + 3ab2 + b3

= a3 + b3 + 3ab (a + b).7. (a - b)3 = a3 - 3a2b + 3ab2 _ b3

= a3 - b3 - 3ab (a - b).8. a3 + b3 = (a + b) (a2 - ab + b2 )9. a3 - b3 = (a - b) (a2 + ab + b2)10. am x a ll = am + II1 1 . a m -+ a " :::::a m - II12. a3 + b3 + e3 - 3abe = (a + b + c) (a2 + b2 + e2 - ab - be - cal

if a + b + e = 0, then the above identity reduces to a3 + b3 + e3 = 3abe


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4-2 Quantitative A ptitude for C om petitive E xam inations- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -4.2.1 Square Root and SquareWhen a number is multiplied by itself, the product obtained is called the square of the number since6 x 6 = 36.. . 36 is the square of 6 or 62 = 36Also. - 3 x - 3 = 9 =} (- 3 )2 = 9. . 9 is the square of - 3The square root of a given number is equal to the value whose square is the given numl--r and the signfor square root is'F'Since 62 = 36 then

rrr-:. .J 36 = 6. i.e. square root of 36 is 6r:":- - J 25 = 5, i.e. square root of 25 is 5

) - 25 = an imaginary quantitySquare root of a negative number is an imaginary quantity1. J? = y

Then the required number = yl2. ) a2 x b2 = ab3. ) a3 x b3 = ab.r;;;;

4.2.2 Division Method for Finding the Square Root.j64009 =?

6 40 09Step 1 Pairing the digits from right to left, we get

Step 2 Then take the first pair, here it is only '6' and find the largest whole number whose square isequal to 6 or less than 6. Such a whole number is 2.Step 3 Hence, write 2 in the quotient and also in the divisor. (see next page)Step 4 Subtract 2 x 2 = 4 from 6. The remainder is then 2.Step 5 Bring down the second pair of digits (i.e. 40) double the quotient (i.e. 2 x 2 = 4) and write theresult on the left of 240 and then repeat Step-2 till the remainder is zero. The whole process can beenumerated step-by-step as shown in the following table.


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Simplification and Approximation 4-3

2 6 40 09+ 2 4

4S ") 40-+ 5 [ 2 2S503 15 09

IS 090

2S3 = Quatient

(since 4S x 5 = 225)i (since 503 x 3 = 1509)

,,64009 = 2S3 (i.e. Quotient!4.2.3 Properties of a Perfect Square Number!,. number whose exact square root (which must be a whole number ca n be obtained. is called a perfect

'a,1 A number ending with 2, 3. 7 or 8 cannot be a perfect square.I'bl The last digit of a perfect square must be 0, 1. 4, 5. 6 or 9.'c) A number ending with odd number of zeroes cannot be a perfect square, e.g. 9000. 25000, 16000,

etc. are not perfect squares.,d) A perfect square number is either exactly divisible by 3 or leaves a remainder of 1, when divided

by 3.e.g. 64 if divided by 3, will leave a remainder of 136 is exactly divisible by 3.

I,e) A perfect square number is either exactly divisible by 4 or leaves a remainder of 1. when dividedby 4e.g. 81 if divided by 4, will leave a remainder of 1.

100 is exactly divisible by 4.';,' te: Above properties are very useful to check if a given number is a perfect square or not.4.2.4 Square Root of Vulgar Fraction

!31- =?V 7Step 1 Multiply the numerator and the denominator by the donominator.Step 2 Find the square root of the new numerator and divide it by the new denominator.

v ! . ~ . = )~, x ~ = ~ = 4.~82 = 0.6544.2.5 Square Root of a Rational Decimal Fraction,,387.09126 =?Step 1 Pair the integer part first

387


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4-4 Quantitative Aptitude (or Competitive F.xamlnation,Step 2 Check the number of decimal places.If it is odd. then affix a zero on the extreme right of decimal part to make the even number of decimalplaces.Here. no. of decimal places = : ' 1 . \0. after placing a zero. it becomes .09126()Step 3 Pair the decimal part accordingly

.ell) 12 60Step 4 Start finding the square root h) the division method as explained in +.2.2 a.id .t the decimalpoint in the square J"()O[ as ,0011 as tilt: integer part is exhausted.

+3 ~7 0912601

19.67-+

29 287+ 9 261 Iintcuer part---_----+-------=-.-~-_-386 2609___6_L 23163927

7[ 29312

27489 I since 3927 x 7 = 2748939344 I 182360 1--

____ --ic-- .1_5_7_3_76 - - 4 I _ _ s_il1_ce39344 x 4 ::= 157376

i~ oyer) since 29 x 9 = 261since 386 x 6 = 2316----i----------- ---

The square root of 387.09126 = -1387.0912619.674

4.3 SIMPLIFICATION OF DECIMAL FRACTIONThe number of digits which are present on the RIGHT OF A DECIMAL POINT is called the number ofdecimal places.

That is, 32.cl075 has four digits on the right of the decimal point. Therefore, the number is expressedto four decimal places.

A WHOLE NUMBER can also be written as a decimal fraction by putting a decimal after its LASTDIGIT and adding as many zeros as are required.e.g. 12 = 12.0 ::= 12.000 and so 011.

4.3.1 AdditionFor addition of a decimal number with another decimal number or with another whole number write thegiven number in such a way that the number of decimal places are equal for all the numbers.

e.g. 1 + 0.59 + 0.008Here maximum number of decimal places = 3 (three) in 0.008.Convert all the number so that they have 3 decimal places.. . I + 0.59 + 0.008 = 1.000 + 0.590 + 0.008 = 1.5984.3.2 SubtractionIn subtraction also, the given numbers are to be wirtten in such a way that the number of decimal placesbecome equal for all the numbers (empty places are filled up with zeroes).

. . . .


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>!lI7J)/;Ticdti(J!1 mci Approximation 4-5e.g. 2 - 0.283I n 0.283. number of decimal places = 3In 2. number of decimal places = 0So. make 2 as having 3 decimal places. i.e. 2000.. 2 - 0.283 = 2.000 - 0.283 = 1.717.

4.3.3 Multiplication0.005 x 0.08 x 0.4= ?Step 1 Multiply the number only. i.c. 5 .:< .8 x _j . = 1 6UStep 2 Add the total number of decimal place, in the gi\ en I1UmbLT. I.e. 3 - :: -< - I hStep 3 Write the result of Step 1 and convert it to a number with decimal places a o obtained inStep 2 by shifting the dcci mal point to the left.

i.e. by six decimal places. we then get0.005 x O .Ws x 0.-1 = 0.000160 = 0 .00016

Similarly 0.03 x 0.7 x : 2 = 0.0-12Total of 3 decimal places

4.3.4 Division of Decimals(a) When the Divisor (or Denominator) is a Whole Number

3.0056e.g. -7-Step 1 Simply divide the number without considering the decimal points gi\ en i.e. 7) 30056 (4 293.7Step 2 Count the no. of decimal places in the given number. Here it has - I decimal places in 3.0056.Step 3 Shift the decimal point in the quotient obtined to the same no. of decimal places as in Step-2

ShiftHence the result becomes.' -12937 = 0..+2937(b) When the Divisor (Denominator) is also a Decimal Number

12.598e.g. 27.08 x 1.417Step 1 Shift the decimal point to the right of the numerator and of the denominator such that- total decimal point shift in numerator = total decimal point shift in denominator.- there is no decimal place left after the shift.

Here. no. of decimal place in numerator (in 12.598) = 3no. of decimal place in denominator (in n.08 and 1...+17) = 2 + 3 = 5since 5 > 3. so. total shift in decimal point to be made Iin numerator and denominator) = 5

Now, 5 decimal point shifts are made,5 -hitt,.------,

1259~00 = 12598002708 x 141708 x 1417LJ L___Zsluft .hhiti


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4-6 Qua ntita tive A ptitu de fo r C om p etitive E xa m in atio nsStep 2 Division process is continued with the resulting fraction obtained in step 1.

4.4 SIMPLIFICATION OF A MIXED FRACTIONA mixed fraction consists of two parts, the integer part and the fractional part.e.g.

[n fact) ___- 18

) .2. 7has 2 as an integer and-18 18 as a fraction.72 +

4.4.1 Addition18

Step 1Step 2Step 3

Add the integer part only i.e. 12 + 13 = 25Add the fractional part only i.e. . 2 . + 2 . = I 238 II 88Add the results obtained in Step 1 and Step 2I '" ) 5.:.- +8 13 7_ = 25 + I2l11 88

4.4.2 Subtration

Step 1Step 2Step 3Hence,

Similarly

Subtract the integer part only, i.e. 10- 18 = - 8Subtract the fraction part only, i.e. 1_'! ' = . ! . . ! .7 6 42Add the result obtained in Step 1 and Step 2

101 _ 18.!. = - 8 + l _ l = -8 + I _ I + I_} = -(8- 1) _ (42 -11) = -7 _ 317 6 42 42 42 42; ; ; ; ( 7 + !~) [Please refer to Section 4.4.3]=-7l.! .42

12~-15~=(12-15)+ (~l-~)17 . 88 - 17 * 2 _71= -3 + - = _ (3 - 1)88 88 88

"Explanation of in between steps of adding 1 & subtracting 1 have been explained in the previous problem, however,this step can be directly obtained after little practice.


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Step 2 Subtract the numerator from denominator and write in the numerator i.e 314848 - 31

1 81748

5 im p liiica tio n a ncl A pp ro xim a tion 4-7

(iii) 122_102 =02-10)+ ( 2 _ _ ~ ) =2+ _l 2 J7 3 \7 3 21 21(iv) 5~ - 21..+ 6l = (5 - 2 + 6) + ( l _ _ !_ + l)3 7 8 3 7 8

(112 - 24 -+ - 63 \ 151= 9 + l 168 ) = 91684.4.3 Subtraction of a Whole Number and Fraction5_3/ =?48Step 1 Subtract 1 from the whold number i.e. + (5 - 1) = (4)

Step 3 Add the results obtained in Step I and Step 2. _ 5 _ 31 = 4 1 1 .

48 481 1 52 -II 11.l2 - - = (12 - I) -- =52 52 5223 36 - 23 _ 5 13- ~

-6+ = - (6- 1)"--'-' - ~-- =36 36 364.4.4 Easy Method For Simplification5542 + ? + 1369 = 4200Step 1 First always put 'x' for (?)Step 2 Proceed and follow the rules to find the value for 'x' (or finding the value of")4.4.5 Multiplication of a Whole Number and a Fraction4 x /6 2 =?3Step 1 Multiply the integer part by the whole number

4 x 16 = 642 ")Step 2 Multiply the fraction part by the whole number i.e. 4 x 3 2 I

Step 3 Add the results obtained in Step 1 and Step 264 + 2l = 66?3 34 x 16~ = 66}3 3 .


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4-8 Quantitative Aptitude tor Cumpetitive Examinations4.4.6 Division of Mixed Fraction by a Whole Number2/6--+4=?3Step 1 Divide the integer part by the whole number,

i.e. _ l _ ( ) _ = 44Step 2 Divide the fractional part by the whole number.

~ + 4 = I3 6Add the results obtained Step I and Step 2.

r.e.

Step 3

6 416I.e. 4+

")

16~ + 4 = 4~

4.5 CONTINUED FRACTIONS AND ITS SIMPLIFICATIONFractions of the form(a) 7 + _1_ or

4+ _I_5+1 3 8+_1_

4- 5arc called continued fractionsSimplification RuleTo simplify a continued fraction begin at the bottom and work upwards.Exampl t> : Simplify _1--:-__

3 + _1_-:---_5 +

1+ 1-63+

5 +76

3 + 417

3+ _765+

3+741


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=

=13041

41131

4.6 RECURRING DECIMAlSA decimal fraction in which a digit or set of digits is repeated continually is called a Recurring or Periodicdecimal.~ , ~ _ ! _ -~!;. 3 - 0.333 ...Here. on performing the division, it is found that the remainder is always Iand in the quotient. the digit~ rs continually repeated. Hence it is written as 0.3. where the dot over 3 indicates that the 3 has to be~ontinually repeated.- 'I I 1.~uru ar y, - IS7

7) 1.000000 (0.142857 142857 ...5,0. If we continue the division, we shall get the same set of figures 142857 again and again and in the same.rder.

17Tae repeated digits or repeated set of digits is called the period of the recurring decimal.

Taere are two types of Recurring Decimals,.::.'Pure recurring decimal: Such a decimal in which all the decimal digits recur. e.g. 0.142857:" Mixed recurring decimal: Such a decimal in which all the decimal digits do not recur, e.g. u . 7 1 6 7

Taerefore O_i42857 or 0.142857

"6.1 Conversion of a Pure Recurring Decimal to the form pqSteps (a) Write the decimal part without the decimal point as the numerator.

(b) Write as many 9s as there are different repeating digits for the denominator.587 - 17e.g. 0.587 = -, 3.17 = 3 + -999 J . . 99

(remains unchanged as it is integer part)

u.2 Conversion of a Mixed Recurring Decimal to the form pqStep (1) First, write the decimal part without the decimal point and subtract the non-repeating part

from it and write the result in the numerator


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Example: Express 7.0081 as vulgar fraction jI \-- form jq /

4-10 Quantitative Aptitude tot Competitive Examination,Step ( 2) Write a number in the denominator with as man) 9 s as there are repeating digits in the'

decimal part and followed by as many zeroes as there are non-i epeating digits in the decimalpart.

. ' ; ( J / 1 I 1 i , , " 7.0481481 - 047---9 9 0 0

7 477_9 9 0 0

4.7 IMPORTANT DERIVATIONS

(i) ( ~ r m l n = ( ; r nlill(ii) ( _ ( 1 _ ) - m l l l = (_ b ) + m l l l

b \. a'(iii) If a -7- b. then a = c x b and a -7- c = b

4.8 APPROXIMATE VALUEIn this type of questions, candidates do not have 10 find out the exact value. but all they have to do isStep 1 To round off the numbers given in the questionStep 2 To simplifyStep 3 To round off the result obtained in Step-ZVery 11l1{JOr!Ulll: In some of the questions, the choices given arc very close 10 each other. In such case,

Step-I is to be avoided, and we should go directly to Step-Z,4.8.1 Rounding Off NumbersOn some occasions for case in simplification, we require only a rough estimation and not the exact value.

In such cases we round off the values to the nearest tens. or hundreds or thousands,Rounding off a number to the nearest ten. hundred or thousand means finding the multiple of 10. 100

or 1000 which is closest to (or approximate) the original number. It can be done by the following procedure.(a) rounding off to the nearest 10: Replace the digit at unit's place hy 0 If the replaced digit is .5 or

more. then add 1 to the digit at lens place. otherwise digit at tens place remains unchanged.


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S irnp iific a tion and Approxima t ion 4-11roundedoff47 50.g .

J _>5, so I is added to digit at tens place i.e. 1 + - + = 5 (digit at tens place after rounding oft)

3Ul. 92 ~ 901less than 5. so. 9 (digit at tens place I remains unchanged.75 ~ 801equal to 5, so J is added to 7 i.e. J + 7 = 8 becomes the ten's place dig.:295 ~ 300

\. rounding off to the nearest 100: Replace the digit at unit and tens places by 00. If the replacvd digit.:: tens place is 5 or more, then add J to the digit at hundreds place, otherwise the digit at hundreds:-:.lce remains unchanged.. + I ~'= 264 ~ 3001

>5, so I is added to digit at hundreds place i.e. + 2 = 3,~) ----t 5700. because 660 is rounded off as 700.~l ----t 800

c rounding off to the nearest 1000: Replace the ones, tens and hundreds digits by 000. If the replaced.:.git at hundreds place is 5 or more. then add I to the digit at thousands place. otherwise not.197 3 ~ 2000.

~. ;~.,,[hera given number is to be rounded off to the nearest 10. or 100 or 1000.11 depend, OIl the other numbersF': :': in the simplification. It will be explained in the examples provided subsequently..A1 Rounding off a Number to a Decimal PlaceT: ~._:1J off a number to the rth decimal place: following steps are to be checked.Szep 1 Check the digit immediately. next right to the nh place.Step 2 If the next right digit is 5 or more. then add I to the digit in the rth place, otherwise the digit~~., unchanged.Step 3 Delete all the digits in places to the right of the rth place.

rounded off 5.79to 2nd place~-95 ----t 5.8

umph:: What approximate value should come in place of e n in the following equation0~76 -i- 24.96 + 215.005 - ') = 309.85

... 9876 + 24.96 + 215.005 - 'J = 309.85-~ .n place of ? then approximating the terms to the nearest values.

~X) -i- 25 + 215 - x = 3109900 + 215 - 3 10.1'= 25


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4-12 Quantita tive Aptitude for Com petitive Exam ina tions=> x = 396 + 215 - 3JO

= 301 '" 300Therefore the approximate value in place of (? is 300.Example: What approximate value should come in place of (") in the following equation,

96895 + 589 + 22497 = 0(a) 120000 (b) 125000 (C) 122000 (d) 99000 (e) 130000 .

. . . . .0 1 1 1 / / 0 , , Here. out of the 5 choices. the values in three choices are very close 10 e ch other.Now, put (x) in place of ('J) and after approximating. we get

x = 96895 + 589 + 22497'" 96900 + 600 + 22500= 120000

Hence answer is (a)Solved Example.

E-1 Simplify1 0 + - [S1+{6-(7-6-4)}]s .r IOl Isl+{6-(7-2)}]2 L : .

1 [S1+{6-5}]0- -2IO l [81+ 1 ]2

6 - 4 = 2 (VI

(7 - 2) = 5 (8)

{6 - 5} = 1 (B)= IO l - 9 1 = 1.:. 2

E -2 0,75 x 0.75 + 0.25 x 0.75 x 2 + 0.25 x 0.25S-2 Let 0.75 = a and 0.25 = b

By 4 .2 (1). we have (/2 + Lab + 1 / = (0 + /))2=


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Simpliticatiun and Approximation 4-13

- - 4 . . Required Number = (9840)2410= (24)2 = 576.

Simplify (402 - 302) = 10 x 'J-5 Let 40 = a, 30 = b, required numher = x

a2 - b" = (a + b) (a - b)

(Refer 4 .2 .11

40= - 30 210

( 40 + 30) ( 40 - 30 ) = 70.10Required Numher (x) =.6 3" x 36 + 39 = '?-6 here a = 3 and the base is the same. Now, on multiplication. the exponents are added. and Oil d: ision

the difference of the exponents are taken. Now applying 4.2 (10) and (11). we getthe required number (x) = 3 f.) ... n - 9) = 312+_1_+_1_ = ')

2+12 )2-2

-7 c + J 2 { ~ ~ ) ~ % ~ Ji 2 + (AGE '93)

[Since a = 1 2 b = 2212 2n= 2 + 1 2 +--- => 2 + 1 2 + --- = 2.2-4 ")

g If x * y = (x + 2)2 (y - 2) then 7 * 5 = '?-8 Substituting x = 7 and v = 2 , we get.7 * 5 = (7 + 2)2 (5 - 2 )

= (9): x 3 = 243.9 If In and n are whole numbers sueh that Ill" = 1 2 1 . then ( III _ I)" + I = 'J-9 Given that m " = 12 1 => m ' = (1 1 )2

Hence m = 1 1 and II = 2, and substituting these values.(m - 1)" + J = (11 - 1)2 + J = 10' = 1000.

3 6 (y-x)10 If ~ = - . then - + - ')v 4 7 (Y+1)lO Substruting .r = 3 k

\' = 4k we get.6 (Y - x)-+7 (y+x)

6 4k - 3k+7 4k + 3k6 1= - +- = 1.7 7


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4-14 Q ua ntita tive A ptitude for C om p etitive E xa m tna tionsE-ll If Fx -.r; = 1 and Fx +h = 17. then ,-/'-;; = ~

(i)

and, , l-; -.r; = (ii)Adding equations (i) and ( i i )Subtracting equation (ii) from (i}, '-/.' = 8Substituting these values,

- V X I '? + 12

= Fx xh9 x 8 = 7 2 ,E-12 = 2

0,2 x 3,6S-12 Putting .r in place of ? we getx +12

0,2 x 3.6 =2

=? x -i- 12 = 2 x 0.2 x 3,6 =? _2_ = 2 x 0.2 x 3.612=? x = 12 x 2 x 0.2 x 3.6 = 17.28.

E-13 f ? X 7 x 18 = 84S-13 Subsituting x for ? we get.FX7 x 18 = 84

8418

(squaring both sides)

x x 7 = (~~}84x 84

=? x = = 3.11.18x18x7E-14 ( 2 ~ ) x ( y ) = 7 ~S-14 Taking the quotients 2. y and 7. we get

(MBA, '82)

2 y = 7, which gives the quotient as 3 ( Sincev = 3. Substituting the value of v, we get,zlx3.l 73x 2 4


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Simplification anciApproximdtJon 4-15-_--------734 ,1. , ,Now. ,' ,,= > -3 I r 14 .\2.\"= 14 and \ = 3

E-15 22' := : 256S-15 Putting x fo r 'J. we get

)2 = 256 := : 28(lA, '87)

.r = 3.E-16 2' + 13 = 4' + 2 then r = 'JS-16 2' + 1< = (22), - .'

~ 2 . \ " + L~ = 2 2t - -+~ x + 1 3 = 2x + 4 ~ .r = 9 .

E-17 3m -,,)75 +012 -.,S-17 35--175 + f0.

3 11 , , 1, ''j - " ! . . , , 0= . \j. x., X _ -"j _X :l X _ + \j_x _ x .'

r==(9-5+2),;3 = 6,/3E-18 If 53 x v I = 19. then the value of rv . v) = 'J.\' " .S-18 Equating the quotients o f bo th sides.

5\ = 19, we get the quotient as 3._:. v = 3Hence. 51. x 3 _ I_ 19x ")

s l 19 , 3- . ' 'i~ -- ~ )-.r 7 .r 72

.v = 7 so .r = 7 and v = 3.E-19 g _ l _ + 8J- + '.' = 2014 . 2 8S-19 Substituting .r for ", we get,

g J + 8 I4 2 +.\"= 2() I8

I ,


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4 -1 6 Qua ntita tive Aptitude tor Com petitive Exa m ina tions

::::::. = 201- 8!- - 818 4 2= (20 - 8 - 8) + (1_1 _1)\8 4 2= 4 + U = ~ - 4 ) [R efer 4 .4 .2 J=4+ -58 = (4 - I)

8-58 [Refer 4.4.3}

= 3~.8E-20 -J1296 = _?_

? 2.25S-20 Putting x for (?), we get.fl296 x 2.25 = x2

[R efer 4 .4 .4 ]

36 x 2.25 = X C : : : : : : . x = ,j 36 x 2.25 ::::::. x = 6 x 1.5:. x = 9.

E-21 65% of? = 124.90 - 63.15S-21 Putting x for (?),65 of x = 61.75100::::::.= 61.75 x 100 ::::::.x = 95.65

(Since - J 1296 = 36)

17 0 +hE-22 If _o_ = -. what is -_ equal to?a + h 23 a - bS-22 Given that ..: = 120+ b 2 3

i.e. if 0 = 17. then 0 + h =23or b = 6:. a - b = 17 - 6 = 11

(lA, '79)

a+h 23hence a - 11J 2 4 + - J 2 1 6E-23 = ?J%

S-23 Putting x for (") and solving for .r, we getx= 8 J 6:::::: .x = -------4 J 6

.. x = 2.


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Simplification and Approximation 4-17E-24 Simplify (x + y - ;:)" - (x _ y + ;:)2S-24 Using the formula

a2 - b 2 = (a + b) (a - b)=> (x + y - ::} - (x _ y + ;:)2= (x + y - ; : + x - y + z) (r + y - ; : - x + y - ::) = 2x (2 y - 2 z)= 4xy - 4xz.

E-25 If x = 5 and v = - 2, then what is the value of (x - 2):)1/,,)S-25 (x - 2 y) lIY . .= (5 + 4)1/-2 :::: (9 )1/-2 ::::

(9)1/2

1 1J9 32 4 1E-26 2 - -i- - of?:::: -3 5 6

S-26 Putting x for e n and solving for .r, we get,8 4=> --;--x::::3 5 683

1 4 8x6x5- x x => x::::6 5 3 x 4=> x = 20.

E-27 Find the value of(i) 5 of of 24 rupees.6 20

S-27 (i) 5 of of 24 rupees.6 205 x _ ! _ x 24 rupees ::::1 rupee.6 20

E-28 (11 -;-2 t) - ;-~l of 2 1 - 2S-28 Applying VBODMAS Rules. we get= [ 1 1 X 2 IJ -i- 1 ~ of 2 ~ - 2

[Refer 4.2 -t

(ii) 4 of 14 times of 1 kg.2 --7 4(ii) 4 of 14 times of 2 1 - kg.7 4

4 x 14 x 9= - kg. ::::1.8 kg.7 4

[Refer 4.1}

[B)

[0]


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4-18 Quantitative Aptitude for Competitive Examinations

=5X1.x 11= 315.22

l +I _ _LE-29 3 5 lO~ x l + l582

20+ 6 - 330S-29 = ---- =

_! +I10 223 10 5= - x-= 1---30 6 18

~ _ 2 = 125 - 22 22 (D)[M]( S ]

23301+5lO

E-30 C ? 8 ) x C ~ 2 ) = IS-30 Putting x for (?) and solving it for .r, we get

x x1 8 x 162 = 1~ X l = 18 x 162~ ; . = 18 x 18 x 9 ~ x = 18 x 3.. x = 54.(0.55)2 + (0.07)2 + (0.027 )2E-31

(BSRB, '92)

[Refer 4.4.2.J

= ?(0.055)2 + (0.007)2 + (0.0027)2S-31 Let 0.55 = a, 0.07 = b and 0.027 :: c

Then, the given expression becomesa2 + b2 + c2 [a 2 + b2 + c2 ]-.----,-.--~--

0.01 (a2 + b2 + c2 ]0.1 x a)2 + (0.1 x b )2 + (0.1 X c)2~1_ = 100.0.01

E-32 137 x 137 x 137 + 133 x 133 x 133 =?137 x 137 -137 x 133 + 133 x 133

S-32 Let 137 = a and 133 = bThen, the given expression becomes

(a + b)( a 2 - ab + b2 )~ . .- - - - - - - - = a+ b(al +ab +b2) [Refer 4.2 (8)J


19/39

'\

Simplification and Approximation 4-19Putting the value of a and b, we get= 137 + 133 = 270.20.25 x 2.80

E-33 ---- 28.352012.804S-33 ---- 28.35 (Since 20.. 5 = 20)

E-37 ~ 6~.6 = 0.26S-37 Putting x for (?) we get

(SBI, '80)

_ ! x 2.84 81xO.728.358.35

= 56.70 = 2.28.35E-3454 .;- 66 .;- 33 = ?S-34 54 .;- 3366

9 .;-33 = 1_ x _ l_ 311 11 33 121E-35 2? 9 = 916

4 16S-35 Putting x for (?) we get

2x9 916 2x9 229--=- ~4 16 4 4x=2

E-36 f l i = 0.011S-36 Putting x for (?) we get

r r ; = 0.011 (Since x is under square root)~ = (0.011)210x = 10 x 0.000121x = 0.00121.



20/39

4-20 Qua ntita tive A ptitud e fo r C om p etitive E xa m ina tion s

~ =0.26~".~-.


x = = 67.6 ~ x=1000.0.0676.E.38 ~ 0.324 =?10S-38 Putting x for 0), we get

~0.~~4 = xHere x is not under square root, hence squaring is not done .~ . J0 . 0 324 = x

~18 x 18~ x= ~ x=000 100 x 100

18 x = = 0.18.x= ~IO( 63 + 36)2 + ( 63 - 36 )2E-39 632 + 362 = ?

S39 Putting 63 = a and 36 = b in the given expression, we get(a +b v : + (a _b)2

(ITI, '83)

.1:=

x= - -._-- (Since (a + h)2 + (a - b)2 = 2(a2 + b2))

x = 2.E40 44.60 x 2.50 = . )S40 Putting x for 0), we get

x = = 44.60 x 2.50 ~ x = = 44.60 x 104

(Refer 4.2 (iii)j

(Since 2.50 = = l~ )~ x = = 11.15 x 10~ x = = 111.5.14 x 14 - 46E-41 = = ?llx6-(4)2


21/39

Simplification and Approximation 4-21S-41 Putting x for (7) and applying VBODMAS Rule,

We get,x = 196 - 46 => x = _ ! 2 Q _ = 3.

66 -16 50E-42 2002 - 2002 -i- 10.10 = ?S-42 Putting x for e n and applying VBODMAS Rule. we get

x = 2002 _ .~Q0210.10=> x = 2002 - 200 => x = 1802.

(lA, '80)

18-3x4+2E-43 - ~6x5-3x8S-43 Putting x for (7) and applying VBODMAS Rule. we get

18 -12 + 2-------30- 2420-1230 - 24

18+2-12..'__-_ ...30 - 24 [M][A]> x= => x = => x =8=> x= 64.. x= 3

E-44 Express 5005 into its prime factors.S-44 5005 = 5 x 10015 x 7 x 143 = 5 x 7 x 11 x 13.

( sse , '86)

E-45 (43)4 +(42)3 x(45)0 = 7S-45 Put x for 0) , Since all base are equal to 4, hence. put a = 4.

=> x = (a3)4 -;- (a2)3 x ( (5)0=> x = a12 -i- a6 xl=> x = a12 - 6 => X = a6

(ITI, '84)

(Since am -;- a" = am - ") [R efer 4.2 (9)]x = 46.

E-46 Find the value of x in the equation. ( sse , '86)R5- . x-- +-44 12S-46 144 + 25 1+~144 12

r--~169 1+~ 13 1+ -~-> = => --_144 12 12 12=> x = 1.

[ S ] i'


22/39


b 1 a-b 2E-47 If - = - , what is the value of the expression -- + -a 2 a+b 3S-47 Given that ba 2

i.e. if b = I, then a = 2 (assume it, for convenience)a=b 2 2-1 2So, _- + - = -- + - = 1.a+b 3 2+1 3

E-48 If J18 x 14 x x = 84, then x is equal to ?S-48 J18x14xx = 84

Since x is under square root, so, squaring both sides we get84 x 84

(Auditors, '8

18 x 14 x x = 84 x 84 ~ x = - - -18 x 14.. x = 28 .

E-49 (243)8 + (243)4 = ?S-49 Putting x for (?) and 243 = a, we get,

x = (a)08 -i- (a)04 = (a)0.8 - 0.4~ X = aOA

x = (243).4X = (243)4 /10 = = = > X = (35) 2/ 5 =::} X = 32

=::} X = 9.E-SO 1.2 x 1.2 + 0.8 x 0.8 + 2.4 x 0.8 = ?S-SO Putting x for 0) and 1.2 = a, 0.8 = b,

we getx = a2 + b2 + 2ab = = = > x = (a + b)2

= = = > X = (l.2 + 0.8)2 =::} X = 4.

(BSRB, '8[Refer 4.2 (i.

64 9E-S l 121 64 =?8 3- +-11 8

(SBI, 'I

(642-9xI21) 8xl1S-SI x = X -----121x64 (8x8+3xl1)( 642 - 3 x 3 x 11 x 11) 8 x 11= = = > x = x -- -._-(11 x II x 8 x 8) (64 + 33)


23/39

Simplification and Approximation 423

~ x= (64+33)(64-33) 1------- x ---llx8 (64+33)

~ x=3188

0.1 + 0.75 ( 1 )E52 + 0.125 + -.-2 .S + O.OS 4 .8S-52 Putting x for (7)

~ x = 0.85 + ( 1 . + . ! Q . )2.55 8 48~ x= 1.

= 7

( 2 1 6 ) - 2 / 3 . ( 2 7 ) - 4 1 3E53 - 'T"-1 1 = 7 (Bank PO, '79)S-53 Putting x for (7), we get

~ e ~ 6 r 2 / 3 + e 1 7 r 4 / 3 = x [ . ( a ) - m i n ( b ) m l n JSince -) =-b. a~ x= 49

E5485/ 3 + ( l2S t2 /3 = 7S54 Putting x for (7)(2 3 )5/3 x (125)2/3 = X

~ X = 25 X (53 )2 /3 ~ X = 32 x 25(Since am + b- n = am X bn)

x = 800 .E55 ( _ I _ J X (81)-3/4 = 76-2S55 Putting x for (7) and solving6 ; X ( 8 \ f 4 = X [ . ( a ) - m l n _ ( b ) m l n JSince - - -b a

62 6x6~ x= ~ x=(3 4) 3 1 4 3x3x34

X= -3


24/39

4-24 Qua ntita tive Aptitude for C om p etitive Exa m ina tionsE-56 ~ 147 + J27 = = ~ x . . / 38-56 Putting x for (0) and solving it for x,

)3X7X7 +J3x3X3 = = xX f3=? X x f3 = = 7J3 + 3J3 =? x.j3 = 1O~ 3

x = 10.E-57 J 9 8 - 50 = ? x 1 28-57 Putting x for ('1) and solving it for .r,

)7x7x2-)5x5x2 =xX 1 2=? 712-SJ2=xxJ2.. .r = 2 .

(Bank PO, '80)

E-58 [9+ ) 3 1 =?{25 168-58 l+ {49 = x =?5 {16 3 7x= -+-5 4 =? x= ,.J_~ 20E-59 )0.01 +.JQ.0064 = ?8-59 Putting x for (") and solving it for x,

~0.09 = x0.01 + 0.08 = = x ~:. x = = 0.3

E_60)2S.6..;- 12.1 = = ?36.1 81xO.l8-60 Putting x for ("), we get~ . . ; - P : l 1 = = X ~

16x9 144 . x = = - - =11 x 19 209E-61 Express the number 51 as the difference of squares of two numbers.

8-61 Using the formula

X = = . ! 2 . + D .19 9

(Bank PO, '82)

[ N 1 ] 2 [ N 1 ] 2N = +-T ,where N = = original numberPutN=SI~ 51 = [ ~ 12~~r - [ 512- Ir=? 51 = (26)2 - (25)2.


25/39

Simplification and Approximation 4-25Find the number whose seventh part multiplied by its eleventh part gives 1,232Let x be the number such that~ x ~ = = 12327 11==} . r : : 7 x 11 x 1232 ==} x2 = = 7 x 11 x 7 x 11 x 4 x 4==} x = = 7 x 11 x 4 ==} 308

the number is 308.

Find the square root of ( 3 l r - ( 4 ~ r( 3 ~ y - ( 4 ~ rLet (3~J = = a and ( 4 ~ - r = = b. thenGiven expression = = - - - a-b = = a + b

169 + 169 ::: 169 ( j _ _ + 1 ) = = 169 x 2516 9 16 9 144

Required square root = = 169x~ 1 4 465 5= = = = 5---12 12

4 The highest score in an innings was ~ of the total score and the next highest was ~ of the remainder.9 9These scores differed by 8 runs. What was the total score in the innings') (NDA, '88)64 Let the total score be x runs, such that

2 2 7-x--X-x = = 89 9 92 2- x x - = = 8 ==} x:::: 1629 9

. , The total score in the innings was 16265 Simplify

( 6 1 4 r + (64r1 l2 + (_ 32)4/5 ( sse , '94)


26/39


S-65 ( _ ! _ ) o +(64)-112+(_32)41564 .= = 1 + (82r112 + (-1 x 32)415= = 1 + 8-1 + [(- 1)415 X (32)415= = 1 + 1 . + [(_12)215 X (25)415] = = 1 + 1 + [24] = = 17!.8 8 8

E-66 Simplify(243)12 x (243)008

S-66 (243f12 x (243)008= = (243)0.12 + 0.08 = = (243)02= (35 )115 = = 3.

E-67 If En = 64, then find the value of nS-67 En = = 64

n = = 62n = = 12.

(1 ) - 2 1 3E-68 2~6 = = ?

S-68 Putting x for (?), we getx = = h ~ ~ r 2 / 3

=> x = = 36.E-69 (-2)- (2)' -0} = = ?S-69 Putting x for (?)

X= = (_2)-(2)'-"

x = [ ~ r 2=> x=(4)2=>x=16.

( sse , '94)

(AGE, '90)

(Since bases are same)

( sse , '94)

( sse , '94)

[Since (-; r mll1 =( -~ r i l l 1[Since (~l = = ~ J


27/39

S-72 l +l a+b=a b ab~ a2 + b2 + 2ab= ab~45 + 2 x 18 9 !18 18 2

(Since a+ b= ~(a+b)2)


E - 7 0 1 1 . ! . x 4 _ _ _ - + - ? =22l3 10 3S - 7 0 Putting x for (?), we get

11. ! .x4~-+-x = 22l3 1 0 3

(NDA, '83)

1 1 . ! . x 4 _ _ _ = 22l x x3 10 3 (If a -i- b = e then 1= b xc)1 8x = - x4-2 10 r Sincel 1 1

1-~ 2 i ; ~~ x = 2.4.

E-71 (1.06 + 0.04)2 - ? = 4 x 1.06 x 0.04S - 7 1 Putting x for (?) and solving for it

(1.06 + 0.04)2 - X = 4 x 1.06 x 0.04Assume, 1.06 = a and 0.04 = b. . (a + b)2 - X = 4ab.. x = (a b)2 = 0.06 - 0.04)2 = 1.0404.

E- 72 If a2 + b2 = 45 and ab = 18, find l+la b

(CDS, '80)

[(a + b)2 - (a - b)2 = 4ab](MBA, '87)

E-73 If ab a +b= - . then find the value of in terms of e and d onlycd a+b (MBA, '87)

ahcda2 + b2 + 2abe2 + d2 + 2ed

a2 + b2 - 2abe2 + d2 - 2ed (By componendo and dividendo)

a e b c+ d-- =--a -b c-d


28/39

4-2 8 Q ua ntita tive Aptitude for Com petitive Exa m ina tionsE-74 Simplify

a 1/2 + a -1/2 1 _ a -112-----+---I-a 1+.[;; (MBA, '87)

S-74 al/2+a-1/21_a-1I2-----+---I-a 1+.[;;al/2 + a-l1 2 1- aln= +---0+aI/2)(1-a1l2) l+a1 /2

( 1 + a In ) ( 1 - a 1/2 )

= (1- a)2

I-ar-E-75 Solve 5,{; + 12' \ = 13"-' (MBA, '87)

t; I t:S-75 5v\ +12"\ = 13"-'The given equation is of the form52+122=132

Comparing the two equations, we find; - ; = 2

~ x= 4.E-76 Directions (i)-(iv): What approximate value should come in place of the question mark (?) in the

following questions:

(By the Pythagoras theorem of numbers)

(i) 139% of 459 + 51.. of 384 = ?2(ii) - J 2000 x 0.7 = e)2(iii) ?% of 8999 + 599 = 263003

(BSRB Bombay PO, '97)

(BSRB Bangalore PO, '97)(BSRB Bangalore PO, '97)

(iv) 3.9% 99 + iof 700 = 400/( of ?9 (BSRB Bangalore PO, '97)S-76 (i) Assuming x for') and approximating the terms to closest values, we get

1 1x = 140% of 460 +2 x 384


29/39


X=140 x 460

100II+ x 3842

:=} x = 644 + 2112 = = 2756 '" 2800. . the required value is 2800.

(ii) Assuming x for'? and approximating the terms to its closest values. we get(x)2 = J2000 x 0.7

= 20J"S x 0.7= 20 x 2.23 x 0.7= 31.22 "" 31

x = J3 l = 5.56 '" 5.6.(iii) Assuming x for') and approximating the terms to its nearest values. we get

x% of 9000 = 26300 _ 600326300 - 200~ x= xl009000

= 290.(iv) Assuming x for '? and approximating the terms to its nearest values, we get

40% of x = 4 0 / c of 100 + 4 x 7009

2+(2x2)(1) = '?(2+2)x2

(UTI, '90)

2 xx=4+3115315 x 5 1575 = 7872 2 790.=

I REG ULAR PROBLEM S

(a) 2 (b) 1 (e) 4 (d) 4 (e) None of these(2) Find the missing number:

~x1_+') 47 12 9

(RRB Mumbai, '98)

(a) 23 (b) 322(e) 9 Cd) 92 (e) I3


30/39

4 3 0 Q ua ntita tive Aptitude for C om petitive Exa m ina tions

(a) 1 (b) 318 (d) 118 (e J 214(4) If ,J13 .69 = = 3.7 then

,J1369 + . J0 .1369 + .J0.001369 =?(a) 37.407 (b) 34.307 (c) 37.470

(5) If 462 - 442 = = 45p, then the value of 'p' will be:(a) 10 (b) 15 (c) 4196 ?(6)? 36

\.SM Exam, '01)

(d) 34.707 (e) 37.737Cd) 12 (e)

(lA, '82)(a) 28 (b) 84 (c) 56 (d) 16.3 (e) 24

(7) If a b = = a + b + , r ; ; b , then 6 * 24 is equal to(a) 41 (b) 42 (c) 43

(8) Find the value of 453 - 653 + 203(a) -175500 (b) 165500 (c) 0

(d) 44(RRB Bhopal, '98)

(e) 45(RRB Ajmer, '98)(e) -140055d) -174500

Hint: Here, a + b + c = 45 - 65 + 20 = = O. Then use the identity no. 12 of 4.20.538 x 0.538 - 0.462 x 0.462- - - - - - - - - = = ?1- 0.924 .

(a) 2 (b) 1.08 (c) 0.076(10) What should come in place of the ('l) mark?

( 9) (Bank PO, '83)(d) 0.987 (e) 1 (SBI PO, '99)

? 7 224 -j?(a) 12 (b) 16 (c) 114 (d) 144 (e) None of these

(11) Find the value of (512r2/ 9(a) 4 (b) 1 (c) 3 (d) 5 (e) 7- - - -4 4 4 4

(12) 1 28 x 12- - 75 = 1- x ?2 3(a) 15 (b) 16 (c) 25 Cd) 3 (e) 21.5 3Hint: 8 x 121 = 1002

(13) )121+131.+? =64 4(a) 10 (b) 16 (e) 12 (d) 36 (e) 18

(14) 7 - [? - {4 - 7}. - {5 - (4 - 5) + 2}] = = 16(a) 2 (b) 6 (e) 4 Cd) 3 (e) -4


31/39

Simplification and Approximation 431

(15) If F x - J Y = 1 and F x + fY = 17 then . . J x Y = ?(a) 51 (b) 16 (e) f72 (d) 72f32+f48(16) If In In = 2, then (") is,,8+,,?

(BSRB, '93)(e) JU

(a) 2J3 (b) 24 (e) 12 (d) 144 (e) 3" 2(17) A man eats 'x' bananas in a week. In how many days will he eat 84 bananas?

84 588(a) - (b) 12x (e) ---x x(18) 10 raised to the fifth power may be expressed as:

(d) 12x (e) 84x

(a) 10 x 5(d) 10 x 10 x 10 x 10 x 10

(b) 5 1 0(e) (5)10/5

(c) v r w(19) Which of the following is the same as 50 + 12 ?

(a) 10 (5 + 3) (b) (50 + 6) + (50 + 6)(d) 50 + 4 x 3 (e) 50 + 4 + 3

(20) If ( - J 3 f = 1, then the value of x is(c) 25 + 12 x 2

(a) 1 (b) _1- J 3 (c) J3 (d) 0 (e) 2(21) (1616 x 236) is equal to

(a) 64 (b) 16 (c) 2 1(d) 2 (e) 1(22) If 0.5 x A = 0.0003, then the value of A will be

(a) 0.6 (b) 0.06 (c) 0.0006 (d) 0.006 (e) 0.175Hint : Put 0.5 = J , for easy calculation2

(23) Reciprocal of ~ is equal to5(a) -3 (b) 15 2(c) -5 (d) 14 I(e) -3

(24) A decimal fraction is multiplied by itself. If the product is 477 .4225, then the fraction is(a) 19.325 (b) 23.715 (c) 22.75 (d) 21.85 (e) 18.65

(25) Simplify: 2.5 - ------3.25 2_.5_0.75 + 0.50

(Goods Guard Ex, '99)

(a) 0.50 (b) 1.70 (c) 1.25 (d) 0.80 (el 1.18


32/39

4-3 2 Qua ntita tive Aptitude tor Com petitive Exa m ina tions0.3 x 1.06(26) Simplify: 0.5 x 0.4

(a) 1.4318 (b) 0.28 (c) 1.032 (d) 1.64 (e) 1.88(27) If ~ V 0.000001 x x ::::0.4, then the value of x is

(a) 21/6 (b) 4096 (c) 41/6 (d) 64 (e) _ /3(28) 31x 31x .4 121, then the value of 'x' is2

(a) I (b) I3 (c) 2 (d) 3 (el(29) If 3.2 -i- 64 * 10:::: 2.45 - 1.95. which of the following should replace the asterisk (* l?

(TC Clerk, '97)(a) + (b)

(30) The value of (9.9)3 is(a) 970.299 (b) 981.009 IC ) 981.999 (d) 998.99 (e) 990.989Hint: Write 9.9 :::: 10 - 0.1 and the use the identity (a - b)3 :::: a3 - b3 - 3ab (a - b)

(31) Square of 21 of 64 is equal to(a) square of 168(d) square of 167

(c) x (d) (e) insufficient data

(b) square of 165(e) square of 1344

(e) square of 179

(32) What is the value of x in 2 5:::: 1(x i : : - O ) " ?x 3x 3(a) cannot be determined (b) -2

(d) -I (e)(33) The value of 06016 x 230) is equal to

(a) 0 (b) I (C) : 2

(c) 3

(d) -1 (e) 4(34) If '+' means 'x'. '-' means +, -i- means + and .x'means '-'. then what will be the value of 20 -i- 40

-4x5+6::::?(a) 0 (b) 13.5 IC ) 60 (d) -15 (e) None of these

(Elec, Appr. Exam, '01)(35) If the given interchanges are made in signs and numbers, which one of the four equations would be

correct? (RRB Secunderabad, '01)Given interchanges: signs x and +. numbers 4 and 9(a) 94 x 7 -i- 47 ::::329 (b) 47 x 9 -i- 94 ::::18 (c) 49 x 7 + 49 ::::7(d) 94 x 7 + 97 ::::324 (e) None

(36) Which of the following equations are equivalent')(1) ( ~ M + ~ N J (2) 4 N 2 1M2 2 M N+- +-9 4 3(3) ( ~ + I N ) ( - ' - M - I N ) (4) l ( M + i N Y. ~ 3 2 3 4 3(a) only (2) and (3) (b) only (I) and (4)(d) only (lLand (3) (e) only (l). (2) and (4)

(Mumbai PO, '99)

(c) only (1) and (2)


33/39

Simplification and Approximation 4-33Directions (37-41): What approximate value will come in place of the question mark (7) in the followingecuation?;37) -./625.04 x 16.96 + 136.001 + 17 = = 7

(a) 4.18 (b) 4.41 (c) 425138) (115 .L + 234.92) x 5 _l_ = = 724 37

(a) 1400 (b) 1750

(RBI, '02)(d) 433 (e) None

(Bank PO, '97)(c) 1350 (d) 1200 (e) 165C

(39) ? % of 6147 = = 21 x 245.762Cal 16 (b) 10 (c) 18

140) 5.6 x 2569 + 2058 = 157% x 6529 + 7(a) 5800 (b) 6300 (c) 6200

1~1) 0.91 + . J9 99 +Jlll + o.li = ?(a) 43 (b) 40

(BSRB Mumbai Hindi Officer's, '97)(d) 20 (e) 15(d) 6500 (e) 6000

r=(a) 2 - . . . J 2 (b) J2 - 1 (c) (dl3-2~2 .e) 2 + J2

(c) 39 (d) 42 Ie) 4.23Answers:1. (d) 2. ( b ) 3. (d) 4. (a) 5. (c) 6. ( b ) 7. ( b ) 8. (a) 9. (e)10. (d) 1 1 . ( b ) 12. (a) 13. (a) 14. (e) 15. C d ) 16. (e) 17. (cl 18. (c)19. (e) 20. (d) 21. (e) 22. (c) 23. ( b ) 24. Cd) 25. ( b ) 26. (a ) 27. (b l28. (d) 29. (c) 30. (a) 31. (a) 32. (e) 33. (e) 34. (a) 35. (d) 36. (e)37. (d) 38. ( b ) 39. ( b ) 40. (c) 41. (a),

REAL PROBLEMS

(1) If x = 0.5 and y = 0.2. then 56 x (3y)' is equal to:(a) 1.0 ( b ) 0.5 (e) 0.6 (d) 1.1 Ie) K8

(2) ~ 1+ J2 . ~ 3 - 2 . . j 2 is equal to:

(3) The value of the expressionx-I XII2 +xl/~

____ ----xqwhen x = = 16. is :(a) 4 (b) 3

(4 ) 999 998 x 13 = = ' )999(al 13888_i_ ( b) 12999 986999 999

(c) :2 (d) 1 (e) 16

(c) 11988~~' (d) 12990-1-999 999 (el 12907904999

998 998 1Hint: 999 - = = 999 + = = 1000 -999 999 999-~---. . .-----------.---.--. -


34/39

4-34 Quantita tiv e Aptitu de for Compeutive har7l l l lation,(5) Arrange the following surds in ascending order of magnitude:

~,~ .. 61 )'(i) ,,3. (ll) ,,7. ( i i i : ,,'-fIl.( a ) (i), (ii), (iii) (b) r iii. (iii r , (i ) (C) (il. r iii i, Iii, (d) (iii). (iii. ( i) (c ) (iii). (i). (ii)Hint: The LC\l of the orders of g iven surds. i.e. -f. 6. 12 are 12. C onvert each one of the given surdsinto a surd of order 12

(6) Simplify:

(a) 256 (b) 1024 (C) 6 -+ (d)

(7 ) 38l divided by 2 can be expressed as:3 8') 1(a) 38-= x-3 4 (b) lx~8 3 / 1 ) (I 1)(c) l-x38 +i-X-='4 ~4 3 116Id) x - t3

(e) (3 8 x 4) +

(a) 11 (b) 121 (e) 169(9) If 4" - 4"-1 = 24. then ( 2h ) " equals: (d) 7 (e) 8 1

r-(a) 25 (b ) 25--./5 ic ) 1 2 5

(10) I f 0' + h~ = O . then: (d) 5 v 5 (e) - . . ; 5(a)a+b=.J20b(d) a+ b= ,jab

(11) Replace the ( " ' ) mark in(e) a + b = v 3 ab

1+_1 _1+_1 __

I + (., I

85

I(a) -:; 3(b) 5 (el 1 (d) _ !_3 1Ie ) 4(a) J 876 (b) 2304 (e) 16 (d) 256 (e) 4096


35/39


(13) Evaluate: I + _1_ + + + up to four places of decimals.lx2 Ix2x4 Ix2x-+x8 lx2x4x8xl6(a) 1.6096 (b) 1.7062 IC ) l.3214 Idl 1.6416 t c ) None of theseHint: Do not waste your time by calculating individual term. Adopt any other convenient method

(14) Find the value of 'b, if b is a natural number and

(a) -+ (b) 16 Ie) 256 Id) 6 - + 11 ' \ 128US)" +(0.28)"(15) Evaluate:(0.45)2 + (0.84 )2

(0.28)' + (0.47)" - (0.75 )'3 (0.28) (0.47) (0.75)

(a) I (b) 0.8 (c) Il9

(16) If ) ~O .OOOOOlx r:= 0.2, then equals to:0.1(a) 14.14 (b) 640 (c) 80

Id) 2.5 Ie)9

(a) 1 2 (b) 10 _ l _ ! _13 (C ) 17.. ,(d) 7-9 (e) -201 7

IJ(d) 20 Ie) None of theseHint: Remove the radicals one by one

Hint: Do not try to find the value of 5x + 7 directly from the given relation(18) What should come in place of the question mark ('J) in the following equation'}

IBSRB Patna PO, '01)

(a) 6 (b) 513 (c) 13 (d) 3 (e) None of these(19) What will replace the question mark (?) in the following equation'?

61 lJ x 1532 7 . . J ' - 7 = 216 x 136

(a) 8 (b) 16 (e) 56 (d) 23 (e) 32(20) If 5' = 6 - ' = 307, then the value of XI' IS:x+v

1( a ) 7 (b) 3 (c) 6 (d) 7 (e) I


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4-36 Qua ntita tive Aptitude for Com petitive Exa m ina tions

(21) If 7x + 10 = x2 _ _ !_ = 13x - 2. what is the value Ii V"N(a) _ _ l _

20x(x+2)(22) If =(x+4)4

(b) 20 (c) 18 (d) I18

(e) 22oj x 1 . 0 6----. then the value of .v is:0 . 5 x 0 . 4

(a) 9 (b) 7 . I(e) -3 C d ) I9 (e) r - . . 'ne of theseHint: Do not waste your time by solving it like a quadratic equation

Directions (23-27): Following (A) to IH) are combinations of an operation and an c)rC" .md:(A) means + 3 !B ) means x 3 (C) means - 3 (DJ means + 3(E ) means + :2 (F) means x :2 (G) means _ 2 (H) means + 2

You have been given one or more of these as answer choices for theappropriate choice to replace the question mark in the equations.(23) 42 x 21 _ 12? = 880

(a) A Ib) F (c) G (d) D

following questions. Select the(SBI PO, '99)

(e) None of theseHint: 42 x 21 = 882 . . 12? = 882 _ 880 = 2 Checking the given combination in place of ? Bytrial, (A) ~ 12 + 3 = 4 * 2, thenfor combination (F) ~ 12 x 2 = 24 * 2,

(G) ~ 12 _ 2 = 10 * :2 and so on(24) 36 + 12 ? = 48

(a) A followed by D(d) F followed by H

(b) B followed by G(e) None of these

(C ) A followed by B

Hint: Here the number on right hand side i.e. 48 is comparatively smaller [han previous one(i.e. 880). So. we can directly test the combination of operation as per choices i.e. (a) ~ 36 + 12+ 3 + 3 = 43 * 48

(25) 48 ') + 12 x 4 = 80(a) E followed by B (b) D followed by A (el B followed b~ F(d) F followed by A (e) None of these

(26) 18 x 3 -i- 2 + 3 < 27 ?(a) D followed by G (b) A followed by G (c) D followed b y H(d) D followed by A (e) None of these

(27) (48 + 9) + 19 x 2 = 12 '7(a) B followed by E (bl A followed by H (c) A followed by D(d) e followed by A (e) N one of these

(28) What would be the maximum value of Q in the following equation')2P4 + 7R9 + 4Q7 = 1380

(a) cannot be determined (b) 9 (c) 6 (d) 7 (e) 8Directions (29-36): What approximate value should replace the question mark (") in the following equa-tions:(29) 208.78 of 7 _ l 9 ( + 423.547 of 2439 '7f = 'J5 50

(a) 120 (b) 117 (c) 123 (d) 114 (e) 130Tips: Both the parts have o / c term. so take the % common outside a bracket. Hence calculation of o / cat each step should be avoided .


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S irnp/iiic a rio n and App ro xim a t io n 4-37- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -(30) 159% of 6531.8 + 5.5 x 1015.2 = ? + 5964.9

(a) 11.000 (b) 11,500 IC) 10.000(31) 152/? + 795 = 8226 - 3486(a) 675 (b) 550

(d) 10,800(Mumbai PO, '98)

(e) 12.000

(c) 860 ( d ) 925 (Baroda PO, '99)(e) 500(32) 5 of 1596 + 3015 = ? - 21507

IC) 2 (d) I (e) II8

(a) 5500 ( b) 6300 (c) 49000 (d) 7400 (e) 680f)0(33) 16J"5i~ + 1492 - 250.0521 = ?

(a) 1500 ( b) 1350 (c) 2200 (d) 1800 (e) 1600(34) 857 of 149 c - 5.6 x 12.128 = ') (Chennai PO, 2000)

(a) 45 (b) 52 (c) 65 (d) 60 (e) 40(35) 6.39 x 15.266 + 115.8 of :: . = ?5

(a) 160 (b) 150 (C) 145 (d) 170 (e) 130(36) 33 I % of 768.9 + 250/,;of 161.2 - 68.12 = ')3

(a) 220 (b) 245 (e) 235 (d) 250 (e) 230Hint : Before simplification. put the fractional equivalent of the percentage values as 331 % = ] and3 3

I25% = 4Directions (37-40): Find out the approximate value that should come in place of the question mark in thefollowing questions. (You are not expected to find the exact value).(37) ~ 45689 = 'J

(a) 170 (b) 280 (c) 320 td) 210 te) 430Hint : Do not try to find the square root by conventional method. but. in revet se way. check the squareof which alternative gives closest value to 45689. This checking can be done mentally as square of2, i.e. (2)2 can give the firs! digit as 4. So either (b) or (d) can be the answer. Next you check(28)2 > 252 (625) it is not the starting digit of the given number. So (d) 210 is the choice.(J008.99): rzr:':(38) ------ x ...;3589 x 0.4987 = '?10009.00](a) 300002 7 17 6(39) - +- x - -i- -5 8 19 5(a) 3.L (b) 112 2

(40) 399.89 + 206 x 11.009 = ')(a) 2700 (b) 3100

(b) 900000 (c) 300000 (d) 3000 (e) 60000- 'J

(c) 6566 (d) 4336 (e) 2400


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4-38 Qua ntita tive Aptitude for C om p etitive E \a m ina tionsDirections (41-44): Four of the five parts numbered as (a), (b), (c), (d) and (e) in the following equationsare exactly equal. You have to find out the part that is not equal to the other parts.

v 32..+ + < : 25 6 of 402(a) (b) (c)

J 4 4 1 + Z ' 2197 :: : 15832 + 127..+..+(d) (e)

(42) 8362.64 + 768.3 - 190.57 :: : 593.38 + 60..+.7 + 77"+2.29(a) (b)= 2235.925 x 4 = 9931 _()4 - 990.67 = 17880.74 -+ - 2

(c) (d) (e)Hint: You have to do actual calculation here. Never try to round off

')(43) 10.36 + 69.802 + 2 ..+.938 : : : 2207 .1 -+ - 21 = 16~ (y of 630.6.'( a )32.84375 x 3. 2 = 1 of 1 of ..+729.-+5 9

(d) I e )(44) .r (x + 5) + 2 (3x + 2 ) = (x + 2 )2 + 'lx ::: (\ + 3 ) r .v - 2 ) + 10 (x + 1 )

( a ) (b) (c )= (x - 2)1 - (x - 3)' - 2 (12 - IJI + 8) ::: (.r + 7 ) (I + 2) + 2 (.r - 5)(d) ( e)

(45) If


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Simplification anciAppro\irnation 4-39If 8700 -i- x = 300 and .+590 -i- \ 170. then (x - v) x (x + y) = .)(a) 29 (h) 56 (c I 112 (d) 27 (c ) 81Hint: Refer to .+.7If 8 3 -10. then a .3 _ . ,-a (I 8 a

95 16 2.+ (/2 _ 24(a) -10 (h) - r c I (d ) (e) -----16 95 ([2 _ 8 8 [ 1

IC) 2 . (c) 3. (b: 4 . (b) 5. (C) 6. (d) 7. r d) 8. (bl 9 . d"Ie) 11. (a ) 12. k) 1 3 . (d) 14. (d) 15. (C) 16. (C I 17. (hi 18. (aiIe) 20. d ) 21. (al 22. (h) 23. (e) 24. (C) ,- (d) 26. (C) 27. (bl:- .Ie) 29. (a) 30. (c) 31. (a) 32. (h) 33. Ie ) 34. (b) 35. (c) 36. (e)Id) 38. (c) 39. (d) 40. (a) 41. I e ) 42. (C) 43. (e ) 44. (d) 45. (c)Ie) 47. (b) 48. (e) 49. (b)

4 simplification

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