primer on quantitative aptitude

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1.Tenyearsago,PwashalfofQinage.Iftheratiooftheirpresentagesis3:4,whatwillbethetotaloftheirpresentages?A.45 B.40C.35 D.30

Hereistheanswerandexplanation

Answer:OptionC

Explanation:

LetthepresentageofPandQbe3xand4xrespectively.

Tenyearsago,PwashalfofQinage

=>(3x10)=(4x10)

=>6x20=4x10

=>2x=10

=>x=5

totaloftheirpresentages=3x+4x=7x=75=35

2.FatherisagedthreetimesmorethanhissonSunil.After8years,hewouldbetwoandahalftimesofSunil'sage.Afterfurther8years,howmanytimeswouldhebeofSunil'sage?A.4times B.4timesC.2times D.3times


Answer:OptionC

Explanation:

AssumethatSunil'spresentage=x.

Thenfather'spresentage=3x+x=4x

After8years,fathersage=2timesofSunils'age

=>(4x+8)=2(x+8)



=>4x+8=(5/2)(x+8)

=>8x+16=5x+40

=>3x=4016=24

=>x=24/3=8

Afterfurther8years,

Sunil'sage=x+8+8=8+8+8=24

Father'sage=4x+8+8=48+8+8=48

Father'sage/Sunil'sage=48/24=2

3.Aman'sageis125%ofwhatitwas10yearsago,but831/3%ofwhatitwillbeafterten10years.Whatishispresentage?A.70 B.60C.50 D.40


Answer:OptionC

Explanation:

Lettheagebefore10years=x

Then125x/100=x+10

=>125x=100x+1000

=>x=1000/25=40

Presentage=x+10=40+10=50

4.Amanis24yearsolderthanhisson.Intwoyears,hisagewillbetwicetheageofhisson.Whatisthepresentageofhisson?A.23years B.22yearsC.21years D.20years



Answer:OptionB

Explanation:

Letthepresentageoftheson=xyears

Thenpresentagetheman=(x+24)years

Given that in 2 years,man's agewill be twice the age of hisson

=>(x+24)+2=2(x+2)

=>x=22

5.PresentagesofKiranandSyamareintheratioof5:4respectively.Threeyearshence,theratiooftheirageswillbecome11:9respectively.WhatisSyam'spresentageinyears?A.28 B.27C.26 D.24


Answer:OptionD

Explanation:

RatioofthepresentageofKiranandSyam=5:4

=>LetthepresentageofKiran=5x

PresentageofSyam=4x

After3years,ratiooftheirages=11:9

=>(5x+3):(4x+3)=11:9

=>(5x+3)/(4x+3)=11/9

=>9(5x+3)=11(4x+3)

=>45x+27=44x+33

=>x=3327=6



Syam'spresentage=4x=46=24

6.Thesumofagesof5childrenbornattheintervalsof3yearseachis50years.Findouttheageoftheyoungestchild?A.6years B.5yearsC.4years D.3years


Answer:OptionC

Explanation:

Lettheageoftheyoungestchild=x

Thentheagesof5childrencanbewrittenasx,(x+3),(x+6),(x+9)and(x+12)

X+(x+3)+(x+6)+(x+9)+(x+12)=50

=>5x+30=50

=>5x=20

=>x=20/5=4

7.AistwoyearsolderthanBwhoistwiceasoldasC.ThetotaloftheagesofA,BandCis27.HowoldisB?A.10 B.9C.8 D.7


Answer:OptionA

Explanation:

LettheageofC=x.Then

AgeofB=2x

AgeofA=2+2x

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ThetotalageofA,BandC=27

=>(2+2x)+2x+x=27

=>5x=25

=>25/5=5

B'sage=2x=25=10

8.TheAverageageofaclassof22studentsis21years.Theaverageincreasedby1whentheteacher'sagealsoincluded.Whatistheageoftheteacher?A.48 B.45C.43 D.44

9. A father said to his son, "I was as old as you are at the present at the time of your birth". If thefather'sageis38yearsnow,whatwastheson'sagefiveyearsback?A.20years B.18yearsC.14years D.22years


Answer:OptionC

Explanation:

Lettheson'spresentagebexyears.

Then,(38x)=x

=>2x=38

=>x=38/2=19

Son'sage5yearsback=195=14

10.Ayisha'sageis1/6thofherfather'sage.Ayisha'sfather'sagewillbetwicetheageofShankar'sageafter 10 years. If Shankar's eight birthdays was celebrated two years before, then what is Ayisha 'spresentage.A.10years B.12years



C.8years D.5years


Answer:OptionD

Explanation:

ConsiderAyisha'spresentage=x

Thenherfather'sage=6x

Given that Ayisha 's father's age will be twice the age ofShankar'sageafter10years

=>Shankar'sageafter10years=(6x+10)=3x+5

Also given that Shankar's eight birthdays was celebrated twoyearsbefore=>

Shankar'sageafter10years=8+12=20

=>3x+5=20

=>x=15/3=5

=>Ayisha'spresentage=5years

11.Thesumofthepresentagesofasonandhisfatheris60years.Sixyearsago,father'sagewasfivetimestheageoftheson.After6years,whatwillbeson'sage?A.23years B.22yearsC.21years D.20years


Answer:OptionD

Explanation:

Letthepresentageoftheson=x,then

Presentageofthefather=60x

Sixyearsagofather'sagewas5timestheageoftheson



=>(60x)6=5(x6)

=>84=6x

=>x=84/6=14

Son'sageafter6years=x+6=14+6=20

12.KiarnisyoungerthanBineeshby7yearsandtheiragesareintherespectiveratioof7:9,howoldisKiran?A.25 B.24.5C.24 D.23.5


Answer:OptionB

Explanation:

LettheagesofKiranandBineeshare7xand9xrespectively

7x=9x7

=>x=7/2=3.5

Kiran'sage=7x=73.5=24.5

13.Sixyearsago,theratiooftheagesofVimalandSarojwas6:5.Fouryearshence,theratiooftheirageswillbe11:10.WhatisSaroj'sageatpresent?A.18 B.17C.16 D.15


Answer:OptionC

Explanation:

Given that ,sixyearsago, the ratioof theagesofVimalandSaroj=6:5


Hencewecanassumethat

TheageofVimalsixyearsago=6x

TheageofSarojsixyearsago=5x

After4years,theratiooftheirages=11:10

=>(6x+10)/(5x+10)=11/10

=>10(6x+10)=11(5x+10)

=>5x=10

=>x=10/5=2

Saroj'spresentage=5x+6=52+6=16

14.Atpresent,theratiobetweentheagesofShekharandShobhais4:3.After6years,Shekhar'sagewillbe26years.FindouttheageofShobhaatpresent?A.15years B.14yearsC.13years D.12years


Answer:OptionA

Explanation:

After6years,Shekhar'sagewillbe26years

=>PresentageofShekhar=266=20

LetpresentageofShobha=x

Then

20/x=4/3

=>x=203/4=15

15.Mybrotheris3yearseldertome.Myfatherwas28yearsofagewhenmysisterwasbornwhilemy



motherwas26 years of agewhen Iwas born. Ifmy sisterwas4 years of agewhenmybrotherwasborn,thenwhatwastheagemyfatherwhenmybrotherwasborn?A.35years B.34yearsC.33years D.32years


Answer:OptionD

Explanation:

Letmyage=x

Then

Mybrother'sage=x+3

Mymother'sage=x+26

Mysister'sage=(x+3)+4=x+7

MyFather'sage=(x+7)+28=x+35

=>agemyfatherwhenmybrotherwasborn=x+35(x+3)=32

16.ThepresentagesofA,BandCare inproportions4:7:9.Eightyearsago, thesumof theirageswas56.Whataretheirpresentages(inyears)?A.Insufficientdata B.16,30,40C.16,2840 D.16,28,36


Answer:OptionD

Explanation:

Let's take the present age of A,B and C as 4x, 7x and 9xrespectively

Then

(4x8)+(7x8)+(9x8)=56

=>20x=80

=>x=4



Hence the present age of A, B and C are 44, 74 and 94respectively

ie,16,28and36respectively.

17.Aperson'spresentageistwofifthoftheageofhismother.After8years,hewillbeonehalfoftheageofhismother.Whatisthepresentageofthemother?A.60 B.50C.40 D.30


Answer:OptionC

Explanation:

Letthepresentageoftheperson=x.

Thenpresentageofthemother=5x/2

Given that , after 8 years, the personwill be onehalf of theageofhismother.

=>(x+8)=(1/2)(5x/2+8)

=>2x+16=5x/2+8

=>x/2=8

=>x=16

Presentageofthemother=5x/2=516/2=40

18.AisasmuchyoungerthanBandheisolderthanC.IfthesumoftheagesofBandCis50years,whatisdefinitelythedifferencebetweenBandA'sage?A.Datainadequate B.3yearsC.2years D.5years


Answer:OptionA

Explanation:


AgeofCSobha's father'sageSobha'smother'sage=(x+38)[(x4)+36]

=x+38x+436

=6

20.Theageof father10yearsagowasthricetheageofhisson.Tenyearshence,father'sagewillbetwicethatofhisson.Whatistheratiooftheirpresentages?A.7:3 B.3:7C.9:4 D.4:9




Answer:OptionA

Explanation:

Let the age of the son before 10 years = x and age of thefatherbefore10years=3x

Nowwecanwriteas

(3x+20)=2(x+20)

=>x=20

=>AgeofFatherthesonatpresent=x+10=20+10=30

Ageofthefatheratpresent=3x+10=320+10=70

Requiredratio=70:30=7:3

21.Theagesof twopersonsdifferby16years.6yearsago, theelderonewas3 timesasoldas theyoungerone.Whataretheirpresentagesoftheelderperson?A.10 B.20C.30 D.40


Answer:OptionC

Explanation:

Let'stakethepresentageoftheelderperson=x

andthepresentageoftheyoungerperson=x16

(x6)=3(x166)

=>x6=3x66

=>2x=60

=>x=60/2=30



22.Thepresentageofafatheris3yearsmorethanthreetimestheageofhisson.Threeyearshence,father'sagewillbe10yearsmorethantwicetheageoftheson.Whatisfather'spresentage?A.30years B.31yearsC.32yeas D.33years


Answer:OptionD

Explanation:

Letthepresentagetheson=x.

Thenpresentageofthefather=3x+3

Given that ,three years hence, father's age will be 10 yearsmorethantwicetheage

oftheson

=>(3x+3+3)=2(x+3)+10

=>x=10

Father'spresentage=3x+3=310+3=33

23.Kamalwas4 timesasoldashisson8yearsago.After8years,Kamalwillbe twiceasoldashisson.FindoutthepresentageofKamal.A.40years B.38yearsC.42years D.36years


Answer:OptionA

Explanation:

Lettheageofthesonbefore8years=x.

ThenageofKamalbefore8yearsago=4x



After8years,Kamalwillbetwiceasoldashisson

=>4x+16=2(x+16)

=>x=8

PresentageofKamal=4x+8=48+8=40

24.If6yearsaresubtractedfromthepresentageofAjayandtheremainderisdividedby18,thenthepresentageofRahulisobtained.IfRahulis2yearsyoungertoDeniswhoseageis5years,thenwhatisAjay'spresentage?A.50years B.60yearsC.55years D.62years


Answer:OptionB

Explanation:

PresentageofDenis=5years

PresentageofRahul=52=3

LetthepresentageofAjay=x

Then(x6)/18=presentageofRahul=3

=>x6=318=54

=>x=54+6=60

25.Theratiooftheageofamanandhiswifeis4:3.Atthetimeofmarriagetheratiowas5:3andAfter4yearsthisratiowillbecome9:7.Howmanyyearsagoweretheymarried?A.8years B.10yearsC.11years D.12years


Answer:OptionD

Explanation:

Let the present age of the man and his wife be 4x and 3xrespectively.


After4yearsthisratiowillbecome9:7

=>(4x+4)/(3x+4)=9/7

=>28x+28=27x+36

=>x=8

=>Presentageoftheman=4x=48=32

Presentageofhiswife=3x=38=24

Assumethattheygotmarriedbeforetyears.Then

(32t)/(24t)=5/3

=>963t=1205t

=>2t=24

=>t=24/2=12

26.TheproductoftheagesofSyamandSunilis240.IftwicetheageofSunilismorethanSyam'sageby4years,whatisSunil'sage?A.16 B.14C.12 D.10


Answer:OptionC

Explanation:

LettheageofSunil=xandageofSyam=y.

Then

xy=240(1)

2x=y+4

=>y=2x4

=>y=2(x2)(2)

Substitutingequation(2)inequation(1).Weget

x2(x2)=240

=>x(x2)=240/2

=>x(x2)=120(3)

Wegotaquadraticequationtosolve.

Alwaystimeispreciousandobjectivetestsmeasuresnotonlyhowaccurateyouarebutalsohowfastyouare.Wecaneithersolve this quadratic equation in the traditionalway.Butmoreeasyway is just substitute the values given in the choices inthe quadratic equation (equation 3 ) and see which choicesatisfytheequation.

Here the option A is 10. If we substitute that value in thequadraticequation,x(x2)=108whichisnotequalto120

NowtryoptionBwhichis12.Ifwesubstitutethatvalueinthequadraticequation,x(x2)=1210=120.See,wegotthatx=12.

HenceSunil'sage=12

(Or else,we cal solve the quadratic equation by factorizationas,

x(x2)=120

=>x22x120=0

=>(x12)(x+10)=0=>x=12or10.Sincexisageandcannotbenegative,x=12

Or by using quadratic formula as

Sinceageispositive,x=12)

x = =b 4acb2

2a

2 (2 4 1 (120))2

2 1

= = = = 12or 102 4 + 480

22 484

22 22

2



27. One year ago, the ratio of Sooraj's and Vimal's agewas 6: 7 respectively. Four years hence, thisratiowouldbecome7:8.HowoldisVimal?A.32 B.34C.36 D.38


Answer:OptionC

Explanation:

LettaketheageofSoorajandVimal,1yearagoas6xand7xrespectively.

Giventhat,fouryearshence,thisratiowouldbecome7:8.

=>(6x+5)/(7x+5)=7/8

=>48x+40=49x+35

=>x=5

Vimal'spresentage=7x+1=75+1=36

28.ThetotalageofAandBis12yearsmorethanthetotalageofBandC.CishowmanyyearyoungerthanA?A.10 B.11C.12 D.13


Answer:OptionC

Explanation:

GiventhatA+B=12+B+C

=>AC=12+BB=12

=>CisyoungerthanAby12years



29.Sachin'sageafter15yearswillbe5timeshisage5yearsback.FindoutthepresentageofSachin?A.10years B.11yearsC.12years D.13years


Answer:OptionA

Explanation:

LetthepresentageofSachin=x

Then(x+15)=5(x5)

=>4x=40

=>x=10

30.Sandeep's age after six yearswill be threeseventh of his father's age. Ten years ago the ratio oftheirageswas1:5.WhatisSandeep'sfather'sageatpresent?A.30years B.40yearsC.50years D.60years


Answer:OptionC

Explanation:

LettheageofSandeepandhisfatherbefore10yearsbexand5xrespectively.

GiventhatSandeep'sageaftersixyearswillbethreeseventhofhisfather'sage

=>x+16=(3/7)(5x+16)

=>7x+112=15x+48

=>8x=64

=>x=8

Sandeep'sfather'spresentage=5x+10=58+10=50

Comments(244) Newest

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danielbishop 02Feb201512:37AM

MyBrotherisfouryearsolderthanIam.Thesumofouragesis34.Howoldarewe?

Jay 11Feb20159:46PM

taketheagesasxand(x+4)sum=2x+4=34=>x=15agesare15and19

amritava 29Jan201512:48PM

themeanoftheagesoffatherandhissonis27years.after18years,fatherwillbetwiceasoldashisson.whatwouldbetheirpresentages?

Raj 09Feb201511:51PM

letageoffatherbexandhissonbey(x+y)/2=27x+y=54(eq1)

After18years,fatherwillbetwiceasoldashisson(x+18)=2(y+18)x+18=2y+36x2y=18(eq2)

Subtractingeq2fromeq13y=36y=12

x=54y=42

Presentageofthefather=42Presentageoftheson=12

ImportantFormulas-Area

1.

PythagoreanTheorem(Pythagoras'theorem)

Inarightangledtriangle,thesquareofthehypotenuseisequaltothesumofthesquaresoftheothertwosides

c2=a2+b2

wherecisthelengthofthehypotenuseandaandbarethelengthsoftheothertwosides

2. Piisamathematicalconstantwhichistheratioofacircle'scircumferencetoitsdiameter.Itisdenotedby

3. GeometricShapesandsolidsandImportantFormulas

GeometricShapes Description Formulas

Rectangle

l=Length

b=Breadth

d=Lengthofdiagonal

Area=lb

Perimeter=2(l+b)

d=

Square

a=Lengthofaside

d=Lengthofdiagonal

Perimeter=4a

d=

Parallelogram

bandcaresides

b=base

h=height

Area=bh

Perimeter=2(b+c)

Rhombus

a=lengthofeachside

b=base

Area=bh(Formula1forarea)

Area=

3.14 227

+l2 b2

Area= =a212

d 2

a2

12

d1d2

h=height

d1,d2arethediagonal

(Formula2forarea)

Perimeter=4a

Tr iangle

a,bandcaresides

b=base

h=height

(Formula1forarea)

(Formula2forarea-Heron'sformula)

Perimeter=a+b+c

RadiusofincircleofatriangleofareaA=

Equilateral

Tr iangle

a=side

Perimeter=3a

Radiusofincircleofanequilateraltriangleofsidea=

Radiusofcircumcircleofanequilateraltriangleofsidea=

Baseaisparalleletobaseb

Trapezium

(TrapezoidinAmerican

English)

h=height

Circle

r=radius

d=diameter

d=2r

Area= bh12

Area= S(S a)(S b)(S c)

whereSisthesemiperimeter=a + b + c

2

whereSisthesemiperimeter=AS

a + b + c2

Area=3

4a2

a

2 3

a

3

Area= (a + b)h12

Area= = r 214

d 2

Circumference=2r = d

= Circumference

d

Area,A =

..

(ifanglemeasureisindegrees) 2

Sector ofCircle

r=radius

=centralangle

Ellipse

Majoraxislength=2a

Minoraxislength=2b

Area=ab

Rectangular

Solid

l=length

w=width

TotalSurfaceArea=2lw+2wh+2hl=2(lw+wh+hl)

Volume=lwh

Area,A =

(ifanglemeasureisindegrees)

360r 2

(ifanglemeasureisinradians)12

r 2

ArcLength,s=

r(ifanglemeasureisindegrees)

180

r(ifanglemeasureisinradians)

Plesenotethatintheradiansystemforangularmeasurement,

2radians=360

1radian=180

1= radians

180

Hence,

AngleinDegrees=AngleinRadians 180

AngleinRadians=AngleinDegrees

180

Perimeter 2+a2 b2

2

h=height

Cube

s=edge

TotalSurfaceArea=6s2

Volume=s3

Right

Circular

Cylinder

h=height

r=radiusofbase

LateralSurfaceArea=(2r)h

TotalSurfaceArea=(2r)h+2(r2)

Voulme=(r2)h

Pyramid

h=height

B=areaofthebase

TotalSurfaceArea=B+Sumoftheareasofthetrianguarsides

Right

Circular

Cone

h=height

r=radiusofbase

Sphere

r=radius

d=diameter

d=2r

4. ImportantpropertiesofGeometricShapes

I. P ropertiesofTriangle

i. Sumoftheanglesofatriangle=180

Volume= Bh13

LateralSurfaceArea=r = rs+r 2 h 2

wheresistheslantheight= +r 2 h 2

TotalSurfaceArea=r + = rs + +r 2 h 2 r 2 r 2

SurfaceArea=4 = r 2 d 2

Volume= = 43

r 316

d 3

ii. Sumofanytwosidesofatriangleisgreaterthanthethirdside.

iii. Thelinejoiningthemidpointofasideofatriangletothepositivevertexiscalledthemedian

iv. Themedianofatriangledividesthetriangleintotwotriangleswithequalareas

v. Centroidisthepointwherethethreemediansofatrianglemeet.

vi. Centroiddivideseachmedianintosegmentswitha2:1ratio

vii. Areaofatriangleformedbyjoiningthemidpointsofthesidesofagiventriangleisone-fourthoftheareaofthegiventriangle.

viii. Anequilateraltriangleisatriangleinwhichallthreesidesareequal

ix. Inanequilateraltriangle,allthreeinternalanglesarecongruenttoeachother

x. Inanequilateraltriangle,allthreeinternalanglesareeach60

xi. Anisoscelestriangleisatrianglewith(atleast)twoequalsides

xii. Inisoscelestriangle,altitudefromvertexbisectsthebase.

II. P ropertiesofQuadrilaterals

A.Rectangle

i. Thediagonalsofarectangleareequalandbisecteachother

ii. oppositesidesofarectangleareparallel

iii. oppositesidesofarectanglearecongruent

iv. oppositeanglesofarectanglearecongruent

v. Allfouranglesofarectanglearerightangles

vi. Thediagonalsofarectanglearecongruent

B.Square

i. Allfoursidesofasquarearecongruent

ii. Oppositesidesofasquareareparallel

iii. Thediagonalsofasquareareequal

iv. Thediagonalsofasquarebisecteachotheratrightangles

v. Allanglesofasquareare90degrees.

C.Parallelogram

i. Theoppositesidesofaparallelogramareequalinlength.

ii. Theoppositeanglesofaparallelogramarecongruent(equalmeasure).

iii. Thediagonalsofaparallelogrambisecteachother.

iv. Eachdiagonalofaparallelogramdividesitintotwotrianglesofthesamearea

D.Rhombus

i. Allthesidesofarhombusarecongruent

ii. Oppositesidesofarhombusareparallel.

iii. Thediagonalsofarhombusareunequalandbisecteachotheratrightangles

iv. Oppositeinternalanglesofarhombusarecongruent(equalinsize)

v. Anytwoconsecutiveinternalanglesofarhombusaresupplementaryi.e.thesumoftheirangles=180(equalinsize)

Otherpropertiesofquadrilaterals

i. Thesumoftheinterioranglesofaquadrilateralis360degrees

ii. Asquareandarhombusonthesamebasewillhaveequalareas.

iii. Aparallelogramandarectangleonthesamebaseandbetweenthesameparallelsareequalinarea.

iv. Ofalltheparallelogramofgivensides,theparallelogramwhichisarectanglehasthegreatestarea.

v. Eachdiagonalofaparallelogramdividesitintotwotrianglesofthesamearea

III. Sum ofInteriorAnglesofapolygon

i. Thesumoftheinterioranglesofapolygon=180(n-2)degreeswheren=numberofsides

Example1:Numberofsidesofatriangle=3.Hence,sumofthe interioranglesofatriangle=180(3-2)=1801=180

Example2 :Numberofsidesofaquadrilateral=4.Hence,sumof the interioranglesofanyquadrilateral=180(4-2)=1802=360



1.Anerror2%inexcessismadewhilemeasuringthesideofasquare.Whatisthepercentageoferrorinthecalculatedareaofthesquare?A.4.04% B.2.02%C.4% D.2%


Answer:OptionA

Explanation:

Error=2%whilemeasuringthesideofasquare.

Letthecorrectvalueofthesideofthesquare=100

CorrectValueoftheareaofthesquare=100100=10000

CalculatedValueoftheareaofthesquare=102102=10404

Error=10404-10000=404

2.Arectangularpark60mlongand40mwidehastwoconcretecrossroadsrunninginthemiddleoftheparkandrestoftheparkhasbeenusedasalawn.Theareaofthelawnis2109sq.m.whatisthewidthoftheroad?A.5m B.4mC.2m D.3m


Answer:OptionD

Explanation:

Thenthemeasuredvalue=100 = 102(error2%inexcess)(100 + 2)

100

PercentageError= 100 = 100 = 4.04%Error

ActualValue404

10000

Pleasereferthediagramgivenabove.

Areaofthepark=6040=2400m2

Giventhatareaofthelawn=2109m2

Areaofthecrossroads=2400-2109=291m2

Assumethatthewidthofthecrossroads=x

Thentotalareaofthecrossroads= Area of road 1 + area of road 2 - (Common Area of thecrossroads)

=60x+40x-x2

(Let's look in detail how we got the total area of the cross

roadsas60x+40x-x2

Asshown in thediagram,areaof the road1=60x.Thishastheareasoftheparts1,2and3giveninthediagram

Areaoftheroad2=40x.Thishastheparts4,5and6

You can see that there is an area which is intersecting (i.e.part2andpart5)


andtheintersectionarea=x2.

Since60x+40xcoversthe intersectingarea(x2)twotimes(part2andpart5)

,weneedtosubtracttheintersectingareaof(x2)oncetimetogetthetotalarea.

.Hencetotalareaofthecrossroads=60x+40x-x2)

Now,wehave

Totalareasofcrossroads=60x+40x-x2

Butareaofthecrossroads=291m2

Hence60x+40x-x2=291

=>100x-x2=291

=>x2-100x+291=0=>(x-97)(x-3)=0=>x=3(xcannotbe97asthepark isonly60m longand40mwide)

3.A towel,when bleached, lost 20% of its length and 10% of its breadth.What is the percentage ofdecreaseinarea?A.30% B.28%C.32% D.26%


Answer:OptionB

Explanation:

---------------------------------------------------------Solution1---------------------------------------------------------Letoriginallength=100andoriginalbreadth=100Thenoriginalarea=100100=10000

Newarea=8090=7200

Lost20%oflength

Newlength=Originallength = 100 = 80(100 20)

10080100

Lost10%ofbreadth

Newbreadth=Originalbreadth = 100 = 90(100 10)

10090100

Decreaseinarea=OriginalArea-NewArea=10000-7200=2800

---------------------------------------------------------Solution2---------------------------------------------------------

Letoriginallength=landoriginalbreadth=b

Thenoriginalarea=lb

Percentageofdecreaseinarea= 100 = 100 = 28%DecreaseinArea

OriginalArea280010000

Lost20%oflength

Newlength=Originallength = l =(100 20)

10080100

80l100

Lost10%ofbreadth

Newbreadth=Originalbreadth = b =(100 10)

10090100

90b100

Newarea= = =80l100

90b100

7200lb10000

72lb100

Decreaseinarea=OriginalArea-NewArea=lb =72lb100

28lb100

Percentageofdecreaseinarea= 100DecreaseinArea

OriginalArea

= 100 = = 28%

( )28lb100

lb28lb 100

100lb


4. If the lengthofa rectangle ishalvedand itsbreadth is tripled,what is thepercentagechange in itsarea?

A.25%Increase B.25%DecreaseC.50%Decrease D.50%Increase


Answer:OptionD

Explanation:

---------------------------------------------------------Solution1---------------------------------------------------------Letoriginallength=100andoriginalbreadth=100Thenoriginalarea=100100=10000

Newarea=50300=15000

Increaseinarea=NewArea-OriginalArea=15000-10000=5000

---------------------------------------------------------Solution2---------------------------------------------------------


Thenoriginalarea=lb

Lengthoftherectangleishalved

Newlength= = = 50Originallength

21002

breadthistripled Newbreadth=Originalbreadth 3 = 100 3 = 300

PercentageofIncreaseinarea= 100 = 100 = 50%IncreaseinArea




5.Apersonwalkeddiagonallyacrossasquareplot.Approximately,whatwas thepercentsavedbynotwalkingalongtheedges?A.35% B.30%C.20% D.25%


Answer:OptionB

Explanation:

---------------------------------------------------------Solution1---------------------------------------------------------

Consider a square plot as shown above and let the length of


Newlength= =Originallength

2l2

breadthistripled Newbreadth=Originalbreadth 3 = 3b

Newarea= 3b =l2

3lb2

Increaseinarea=NewArea-OriginalArea= lb =3lb2

lb2

PercentageofIncreaseinarea= 100IncreaseinArea

OriginalArea

= 100 = = 50%

( )lb2

lblb 100

2lb

eachside=1

Distancetravelledifwalkedalongtheedges=BC+CD=1+1=2

DistanceSaved=2-1.41=.59

---------------------------------------------------------Solution2---------------------------------------------------------

Considerasquareplotasshownaboveandletthelengthofeachside=x

Distancetravelledifwalkedalongtheedges=BC+CD=x+x=2x

Thenlengthofthediagonal= =(1 + 1) 2

Distancetravelledifwalkeddiagonally=BD= = 1.412

Percentdistancesaved= 100 = .59 50 30%.592

Thenlengthofthediagonal= =(x + x) 2x

Distancetravelledifwalkeddiagonally=BD= = 1.41x2x



DistanceSaved=2x-1.41x=.59x

6.Arectangularfieldhastobefencedonthreesides leavingasideof20feetuncovered.Iftheareaofthefieldis680sq.feet,howmanyfeetoffencingwillberequired?A.95 B.92C.88 D.82


Answer:OptionC

Explanation:

Giventhatareaofthefield=680sq.feet=>lb=680sq.feet

Length(l)=20feet

=>20b=680

Requiredlengthofthefencing=l+2b=20+(234)=88feet

7.Arectangularparkingspaceismarkedoutbypaintingthreeofitssides.Ifthelengthoftheunpaintedsideis9feet,andthesumofthelengthsofthepaintedsidesis37feet,findouttheareaoftheparkingspaceinsquarefeet?A.126sq.ft. B.64sq.ft.C.100sq.ft. D.102sq.ft.


Answer:OptionA

Explanation:

Letl=9ft.

Thenl+2b=37=>2b=37-l=37-9=28

=>b=28/2=14ft.

Percentdistancesaved= 100 = .59 50 30%.59x2x

b = = 34feet68020



Area=lb=914=126sq.ft.

8.Theareaofarectangleplotis460squaremetres.Ifthelengthis15%morethanthebreadth,whatisthebreadthoftheplot?A.14metres B.20metresC.18metres D.12metres


Answer:OptionB

Explanation:

lb=460m2------(Equation1)

Letthebreadth=b

FromEquation1andEquation2,

9.A largefieldof700hectares isdivided intotwoparts.Thedifferenceoftheareasofthetwoparts isone-fifthoftheaverageofthetwoareas.Whatistheareaofthesmallerpartinhectares?A.400 B.365C.385 D.315


Answer:OptionD

Explanation:

Lettheareasofthepartsbexhectaresand(700-x)hectares.

Thenlength,l=b = ------(Equation2)(100 + 15)

100115b100

b = 460115b100

= = 400b246000115

b = = 20m400

Differenceoftheareasofthetwoparts=x-(700-x)=2x-700



Giventhatdifferenceoftheareasofthetwoparts=one-fifthoftheAverageofthe

twoareas

=>2x-700=70

=>2x=770

Hence,Areaofsmallerpart=(700-x)=(700385)=315hectares.

10.The lengthofaroom is5.5mandwidth is3.75m.What isthecostofpayingthe floorbyslabsattherateofRs.800persq.metre.A.Rs.12000 B.Rs.19500C.Rs.18000 D.Rs.16500.


Answer:OptionD

Explanation:

Area=5.53.75sq.metre.Costfor1sq.metre.=Rs.800

Hencetotalcost=5.53.75800=5.53000=Rs.16500

11. The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth isincreased by 5 cm, the area of the rectangle is increased by 75 sq.cm. What is the length of therectangle?

A.18cm B.16cmC.40cm D.20cm


Differenceoftheareasofthetwoparts=x-(700-x)=2x-700

one-fifthoftheAverageofthetwoareas=15

[x + (700 x)]

2

= = = 7015

7002

3505

x = = 3857702


Answer:OptionC

Explanation:

Letbreadth=xcmThenlength=2xcm

Area=lb=x2x=2x2

Newlength=(2x-5)Newbreadth=(x+5)NewArea=lb=(2x-5)(x+5)

Butgiventhatnewarea=initialarea+75sq.cm.

=>(2x-5)(x+5)=2x2+75

=>2x2+10x-5x-25=2x2+75=>5x-25=75=>5x=75+25=100

=>x=100/5=20cm

Length=2x=220=40cm

12.Ifasquareandarhombusstandonthesamebase,thenwhatistheratiooftheareasofthesquareandtherhombus?A.equalto B.equaltoC.greaterthan1 D.equalto1


Answer:OptionD

Explanation:

A square and a rhombus on the same basewill have equalareas.

Hence ratioof theareasof the squareand the rhombusw illbeequalto1sincetheystandonthesamebase

================================================================Note:Please find theproofof the formulagivenbelowwhichyoumayliketogothrough


LetABCDbethesquareandABEFbetherhombus

Considertheright-angledtrianglesADFandBCE

WeknowthatAD=BC(sidesofasquare)

AF=BE(sidesofarhombus)

DF=CE[DF2=AF2-AD2andCE2=BE2-BC2]

HenceADF=BCE

=>ADF+TrapeziumABCF=BCE+TrapeziumABCF

=>AreaofsquareABCD=AreaofrhombusABEF

13.Thebreadthofarectangularfieldis60%ofitslength.Iftheperimeterofthefieldis800m,findouttheareaofthefield.

A.37500m2 B.30500m2

C.32500m2 D.40000m2


Answer:OptionA

Explanation:

Giventhatbreadthofarectangularfieldis60%ofitslength

perimeterofthefield=800m

=>2(l+b)=800

b = =60l100

3l5


14.A room5m44cm longand3m74cmbroadneeds tobepavedwith square tiles.Whatwillbe theleastnumberofsquaretilesrequiredtocoverthefloor?A.176 B.124C.224 D.186


Answer:OptionA

Explanation:

l=5m44cm=544cmb=3m74cm=374cm

Area=544374cm2

Nowweneed to findoutHCF(HighestCommonFactor)of544and374.Let's find out theHCF using long divisionmethod for quickerresults)

374)544(1374

170)374(2340

2 (l + ) = 8003l5

l + = 4003l5

= 4008l5

= 50l5

l = 5 50 = 250m

b= = = 2 50 = 150m3l5

3 2505

Area=lb=250 150 = 37500m2


34)170(5170

0

Hence,HCFof544and374=34

Hence,sidelengthoflargestsquaretilewecantake=34cm

Areaofeachsquaretile=3434cm2

15.Thelengthofarectangularplotis20metresmorethanitsbreadth.Ifthecostoffencingtheplot@Rs.26.50permetreisRs.5300,whatisthelengthoftheplotinmetres?A.60m B.100mC.75m D.50m


Answer:OptionA

Explanation:

Lengthoftheplotis20metresmorethanitsbreadth.Hence,let'stakethelengthaslmetresandbreadthas(l-20)metres

Lengthofthefence=perimeter=2(length+breadth)=2[l+(l-20)]=2(2l-20)metresCostpermeter=Rs.26.50Totalcost=2(2l-20)26.50

TotalcostisgivenasRs.5300=>2(2l-20)26.50=5300=>(2l-20)26.50=2650=>(l-10)26.50=1325

=>(l-10)=1325/26.50=50=>l=50+10=60metres

Numberoftilesrequired= = 16 11 = 176544 37434 34



16.Theratiobetween the lengthand thebreadthofarectangularpark is3:2.Ifamancyclingalongtheboundaryoftheparkatthespeedof12km/hrcompletesoneround in8minutes,thenwhat istheareaofthepark(insq.m)?A.142000 B.112800C.142500 D.153600


Answer:OptionD

Explanation:

l : b = 3 : 2 ------------------------------------------(Equation1)

Perimeteroftherectangularpark=Distancetravelledbythemanatthespeedof12km/hrin8minutes

=speedtime=128/60(8minute=8/60hour)

=8/5km=8/51000m=1600m

Perimeter=2(l+b)

=>2(l+b)=1600

=>l+b=1600/2=800m---------------------------(Equation2)

From(Equation1)and(Equation2)

l=8003/5=480m

b=8002/5=320m(Orb=800-480=320m)

Area=lb=480320=153600m2

17.Whatisthepercentageincreaseintheareaofarectangle,ifeachofitssidesisincreasedby20%?A.45% B.44%C.40% D.42%


Answer:OptionB

Explanation:

---------------------------------------------------------Solution1---------------------------------------------------------

Letoriginallength=100andoriginalbreadth=100Thenoriginalarea=100100=10000

Newarea=120120=14400

Increaseinarea=NewArea-OriginalArea=14400-10000=4400

---------------------------------------------------------Solution2---------------------------------------------------------


Thenoriginalarea=lb

Increasein20%oflength

Newlength=Originallength = 100 = 120(100 + 20)

100120100

Increasein20%ofbreadth

Newbreadth=Originalbreadth = 100 = 120(100 + 20)

100120100

Percentageincreaseinarea= 100 = 100 = 44%IncreaseinArea




18.Ifthedifferencebetweenthe lengthandbreadthofarectangle is23mand itsperimeter is206m,whatisitsarea?

A.2800m2 B.2740m2

C.2520m2 D.2200m2


Answer:OptionC

Explanation:

l-b=23...................(Equation1)

perimeter=2(l+b)=206=>l+b=103.............(Equation2)

(Equation1)+(Equation2)=>2l=23+103=126

=>l=126/2=63metre


Newlength=Originallength = l =(100 + 20)

100120100

120l100

Increasein20%ofbreadth

Newbreadth=Originalbreadth = b =(100 + 20)

100120100

120b100

Newarea= = =120l100

120b100

14400lb10000

144lb100

Increaseinarea=NewArea-OriginalArea= lb =144lb100

44lb100

Percentageofincreaseinarea= 100IncreaseinArea

OriginalArea

= 100 = = 44%

( )44lb100

lb44lb 100

100lb


Substitutingthisvalueoflin(Equation1),weget63-b=23=>b=63-23=40metre

Area=lb=6340=2520m2

19.Theratiobetweentheperimeterandthebreadthofarectangle is5:1.Iftheareaoftherectangleis216sq.cm,whatisthelengthoftherectangle?A.16cm B.18cmC.14cm D.20cm


Answer:OptionB

Explanation:

=>2l+2b=5b

=>2l=3b

Alsogiventhatarea=216cm2

=>lb=216cm2

20.Whatistheleastnumberofsquarestilesrequiredtopavethefloorofaroom15m17cmlongand9m2cmbroad?A.814 B.802

Giventhat = 52(l + b)

b

=> b =2l3

Substitutingthevalueofb,weget,l = 2162l3

= = 3 108 = (3 3) 36l23 216

2

l = 3 6 = 18cm


C.836 D.900


Answer:OptionA

Explanation:

l=15m17cm=1517cmb=9m2cm=902cm

Area=1517902cm2

NowweneedtofindoutHCF(HighestCommonFactor)of1517and902.Let's find out theHCF using long divisionmethod for quickerresults)

902)1517(1902

615)902(1615

287)615(2574

41)287(7287

0

Hence,HCFof1517and902=41

Hence,sidelengthoflargestsquaretilewecantake=41cm

Areaofeachsquaretile=4141cm2

21.Thediagonalofthefloorofarectangularroomis feet.Theshortersideoftheroomis4 feet.

Numberoftilesrequired= = 37 22 = 407 2 = 8141517 902

41 41

712

12


Whatistheareaoftheroom?A.27squarefeet B.22squarefeetC.24squarefeet D.20squarefeet


Answer:OptionA

Explanation:

22. The diagonal of a rectangle is cm and its area is 20 sq. cm. What is the perimeter of therectangle?

Diagonal,d=7 feet = feet12

152

Breadth,b=4 feet = feet12

92

Intheright-angledtrianglePQR,

= l2 ( )152

2

( )92

2

= =2254

814

1444

l = = feet=6feet1444

122

Area=lb=6 = 2792

feet 2

41


A.16cm B.10cmC.12cm D.18cm


Answer:OptionD

Explanation:

Forarectangle, = +d 2 l2 b2

wherel=length,b=breadthandd=diagonaloftheoftherectangle

d = cm41

= +d 2 l2 b2

+ = = 41........(Equation1)l2 b2 ( )41 2

............(Equation2)Area=lb=20cm 2

Solving(Equation1)and(Equation2)

(a + b = + 2ab +)2 a2 b2

usingtheaboveformula,wehave

(l + b = + 2lb + = ( + ) + 2lb = 41 + (2 20) = 81)2 l2 b2 l2 b2

(l + b) = = 9cm81

perimeter=2(l + b) = 2 9 = 18cm


23.Atankis25mlong,12mwideand6mdeep.Whatisthecostofplasteringofitswallsandbottomattherateof75paisepersq.m?A.Rs.558 B.Rs.502C.Rs.516 D.Rs.612


Answer:OptionA

Explanation:

Consider a rectangular solid of length l, w idth w andheighth.Then

1.TotalSurfaceareaofarectangularsolid,S=2lw +2lh+2w h=2(lw +lh+w h)

2.Volumeofarectangularsolid,V=lw h

Inthiscase,l=25m,w=12m,h=6mandallsurfaceneedstobeplasteredexceptthetop

Hencetotalareaneedstobeplastered=TotalSurfaceArea-AreaoftheTopface=(2lw+2lh+2wh)-lw


=lw+2lh+2wh=(2512)+(2256)+(2126)=300+300+144

=744m2

Costofplastering=74475=55800paise=Rs.558

24. It isdecided to constructa2metrebroadpathwayarounda rectangularploton the inside. If theareaof theplots is96sq.m.and therateofconstruction isRs.50persquaremetre.,whatwillbe thetotalcostoftheconstruction?A.Rs.3500 B.Rs.4200C.InsufficientData D.Rs.4400


Answer:OptionC

Explanation:

Let length and width of the rectangular plot be l and brespectivelyTotalAreaoftherectangularplot=96sq.m.

Widthofthepathway=2m

Lengthoftheremainingareaintheplot=(l-4)breadthoftheremainingareaintheplot=(b-4)Areaoftheremainingareaintheplot=(l-4)(b-4)

Areaofthepathway= Total Area of the rectangular plot - remaining area in theplot


=96-[(l-4)(b-4)]=96-[lb-4l-4b+16]=96-[96-4l-4b+16]=96-96+4l+4b-16]=4l+4b-16=4(l+b)-16

Wedonotknow thevaluesof landbandhence totalareaoftherectangularplotcannotbe foundout.Sowecannot findout totalcostof theconstruction.

25.Theareaofaparallelogram is72cm2and itsaltitude is twice thecorrespondingbase.What is thelengthofthebase?A.6cm B.7cmC.8cm D.12cm


Answer:OptionA

Explanation:

Areaofaparallelogram ,A=bhw here b is the base and h is the height of theparallelogram

Letthebase=xcm.Thentheheight=2xcm(altitudeistwicethebase)

Area=x2x=2x2

Buttheareaisgivenas72cm2

=>2x2=72


=>x2=36=>x=6cm

26.Twodiagonalsofarhombusare72cmand30cmrespectively.Whatisitsperimeter?A.136cm B.156cmC.144cm D.121cm


Answer:OptionB

Explanation:

Remember the following two properties of a rhombus whichwillbeusefulinsolvingthisquestion1.Thesidesofarhombusarecongruent.2. The diagonals of a rhombus are unequal and bisect eachotheratrightangles.

LetthediagonalsbePRandSQsuchthatPR=72cmandSQ=30cm

27.Thebaseofaparallelogramis(p+4),altitudetothebaseis(p-3)andtheareais(p2-4),findoutitsactualarea.A.40sq.units B.54sq.units

PO=OR= = 36cm722

SO=OQ= = 15cm302

PQ=QR=RS=SP= = = =39cm+362 152 1296 + 225 1521

perimeter=439=156cm



C.36sq.units D.60sq.units


Answer:OptionD

Explanation:

Areaofaparallelogram ,A=bhw here b is the base and h is the height of theparallelogram

Hence,wehave

p2-4=(p+4)(p-3)

=>p2-4=p2-3p+4p-12=>-4=p-12=>p=12-4=8

Hence,actualarea=(p2-4)=82-4=64-4=60sq.units

28.Acircle is inscribed inanequilateral triangleofside24cm, touching itssides.What is theareaoftheremainingportionofthetriangle?

A. cm2 B. cm2

C. cm2 D. cm2


Answer:OptionA

Explanation:

144 483 121 363

144 363 121 483

Letr=radiusoftheinscribedcircle.Then

AreaofABC

=AreaofOBC+AreaofOCA+areaofOAB

=(rBC)+(rCA)+(rAB)

=r(BC+CA+AB)

=xrx(24+24+24)

=xrx72=36rcm2------------------------------------------(2)

From(1)and(2),

Areaofanequilateraltriangle=3

4a2

whereaislengthofonesideoftheequilateraltriangle

AreaoftheequilateralABC= = = 144 .............(1)3

4a2

34

242 3 cm2

Areaofatriangle= bh12

wherebisthebaseandhistheheightofthetriangle


29.Arectangularplotmeasuring90metresby50metresneedstobeenclosedbywirefencingsuchthatpolesofthefencewillbekept5metresapart.Howmanypoleswillbeneeded?A.30 B.44C.56 D.60


Answer:OptionC

Explanation:

Perimeterofarectangle=2(l+b)w herelisthelengthandbisthebreadthoftherectangle

Length of thewire fencing= perimeter= 2(90+ 50)= 280metres

Twopoleswillbekept5metresapart.Alsorememberthatthepoleswillbeplacedalong the perimeter of the rectangular plot, not in a singlestraightlinewhichis

144 = 36r3

r = = 4 (3)14436

3 3

Areaofacircle=r 2where=radiusofthecircle

From(3),theareaoftheinscribedcircle= = = 48 (4)r 2 (4 )3 2

Hence,Areaoftheremainingportionofthetriangle

AreaofABCAreaofinscribedcircle

144 483 cm2

ViewAnswer | Notebook |Discuss


veryimportant.

Hencenumberofpolesrequired=2805=56

30.Ifthediagonalsofarhombusare24cmand10cm,whatwillbeitsperimeterA.42cm B.64cmC.56cm D.52cm

31.Whatwill be the length of the longest rodwhich can be placed in a box of 80 cm length, 40 cmbreadthand60cmheight?A. cm B. cmC. cm D. cm


Answer:OptionA

Explanation:

ThelongestroadwhichcanfitintotheboxwillhaveoneendatAandotherendatG(oranyothersimilardiagonal)

Hencethelengthofthelongestrod=AG

Initiallylet'sfindoutAC.ConsidertherightangledtriangleABC

11600 14400

10000 12040

AC2=AB2+BC2=402+802=1600+6400=8000

ConsidertherightangledtriangleACG

AG2=AC2+CG2

AC= cm8000

= + = 8000 + 3600 = 11600( )8000 2 602

AG= cm11600

Thelengthofthelongestrod= cm11600

Hide Answer | Notebook | Discuss


1. An error 2% in excess is made while measuring the side of a square. What is the percentage of error in the calculated area of the square?

A. 4.04 % B. 2.02 %

C. 4 % D. 2 %

Here is the answer and explanation

Answer : Option A

Explanation :

Error = 2% while measuring the side of a square.

Let the correct value of the side of the square = 100

Correct Value of the area of the square = 100 100 = 10000Calculated Value of the area of the square = 102 102 = 10404Error = 10404 - 10000 = 404

2. A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. The area of thelawn is 2109 sq. m. what is the width of the road?

A. 5 m B. 4 m

C. 2 m D. 3 m


Answer : Option D

Explanation :

Please refer the diagram given above.

Area of the park = 60 40 = 2400 m2

Given that area of the lawn = 2109 m2

Area of the cross roads = 2400 - 2109 = 291 m2

Assume that the width of the cross roads = x

Then total area of the cross roads = Area of road 1 + area of road 2 - (Common Area of the crossroads)

= 60x + 40x - x2

(Let's look in detail how we got the total area of the cross

roads as 60x + 40x - x2

As shown in the diagram, area of the road 1 = 60x. This has theareas of the

Then the measured value = 100 = 102 ( error 2% in excess)(100 + 2)

100

Percentage Error = 100 = 100 = 4.04%Error

Actual Value

40410000


parts 1,2 and 3 given in the diagram

Area of the road 2 = 40x. This has the parts 4, 5 and 6

You can see that there is an area which is intersecting (i.e.part 2 and part 5)

and the intersection area = x2.

Since 60x + 40x covers the intersecting area (x2) two times (part 2 and part 5)

,we need to subtract the intersecting area of (x2) once time toget the total area.

. Hence total area of the cross roads = 60x + 40x - x2)

Now, we have

Total areas of cross roads = 60x + 40x - x2

But area of the cross roads = 291 m2

Hence 60x + 40x - x2 = 291

=> 100x - x2 = 291

=> x2 - 100x + 291 = 0=> (x - 97)(x - 3) = 0=> x = 3 (x can not be 97 as the park is only 60 m long and 40m wide)

3. A towel, when bleached, lost 20% of its length and 10% of its breadth. What is the percentage of decrease in area?

A. 30 % B. 28 %

C. 32 % D. 26 %


Answer : Option B

Explanation :

---------------------------------------------------------Solution 1---------------------------------------------------------Let original length = 100 and original breadth = 100Then original area = 100 100 = 10000

New area = 80 90 = 7200

Decrease in area = Original Area - New Area = 10000 - 7200 = 2800

---------------------------------------------------------Solution 2---------------------------------------------------------Let original length = l and original breadth = bThen original area = lb

Lost 20% of length

New length = Original length = 100 = 80(100 20)

10080100

Lost 10% of breadth

New breadth= Original breadth = 100 = 90(100 10)

10090100

Percentage of decrease in area = 100 = 100 = 28%Decrease in Area

Original Area

280010000


4. If the length of a rectangle is halved and its breadth is tripled, what is the percentage change in its area?

A. 25 % Increase B. 25 % Decrease

C. 50 % Decrease D. 50 % Increase


Answer : Option D

Explanation :


New area = 50 300 = 15000

Increase in area = New Area - Original Area = 15000 - 10000= 5000


Lost 20% of length

New length = Original length = l =(100 20)

10080100

80l100

Lost 10% of breadth

New breadth= Original breadth = b =(100 10)

10090100

90b100

New area = = =80l100

90b100

7200lb10000

72lb100

Decrease in area = Original Area - New Area = lb =72lb100

28lb100

Percentage of decrease in area = 100Decrease in Area

Original Area

= 100 = = 28%

( )28lb100

lb28lb 100

100lb

Length of the rectangle is halved

New length = = = 50Original length

21002

breadth is tripled New breadth= Original breadth 3 = 100 3 = 300

Percentage of Increase in area = 100 = 100 = 50%Increase in Area

Original Area

500010000


5. A person walked diagonally across a square plot. Approximately, what was the percent saved by not walking along the edges?

A. 35% B. 30 %

C. 20 % D. 25%


Answer : Option B

Explanation :

---------------------------------------------------------Solution 1---------------------------------------------------------

Consider a square plot as shown above and let the length ofeach side = 1

Distance travelled if walked along the edges = BC + CD = 1 + 1 = 2

Distance Saved = 2 - 1.41 = .59

---------------------------------------------------------Solution 2---------------------------------------------------------

Consider a square plot as shown above and let the length of each side = x

Length of the rectangle is halved

New length = =Original length

2l2

breadth is tripled New breadth = Original breadth 3 = 3b

New area = 3b =l2

3lb2

Increase in area = New Area - Original Area = lb =3lb2

lb2

Percentage of Increase in area = 100Increase in Area

Original Area

= 100 = = 50%

( )lb2

lblb 100

2lb

Then length of the diagonal = =(1 + 1)

2

Distance travelled if walked diagonally = BD = = 1.412

Percent distance saved = 100 = .59 50 30%.592

Then length of the diagonal = =(x + x)

2x




Distance travelled if walked along the edges = BC + CD = x + x = 2x

Distance Saved = 2x - 1.41x = .59x

6. A rectangular field has to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?

A. 95 B. 92

C. 88 D. 82


Answer : Option C

Explanation :

Given that area of the field = 680 sq. feet=> lb = 680 sq. feet

Length(l) = 20 feet

=> 20 b = 680

Required length of the fencing = l + 2b = 20 + (2 34) = 88 feet

7. A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37feet, find out the area of the parking space in square feet?

A. 126 sq. ft. B. 64 sq. ft.

C. 100 sq. ft. D. 102 sq. ft.


Answer : Option A

Explanation :

Let l = 9 ft.

Then l + 2b = 37=> 2b = 37 - l = 37 - 9 = 28

=> b = 28/2 = 14 ft.

Area = lb = 9 14 = 126 sq. ft.

8. The area of a rectangle plot is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the plot?

A. 14 metres B. 20 metres

C. 18 metres D. 12 metres


Answer : Option B

Explanation :

lb = 460 m2 ------(Equation 1)

Let the breadth = b

From Equation 1 and Equation 2,

Distance travelled if walked diagonally = BD = = 1.41x2x

Percent distance saved = 100 = .59 50 30%.59x2x

b = = 34 feet68020

Then length, l = b = ------(Equation 2)(100 + 15)

100115b100

b = 460115b100

= = 400b246000115

b = = 20 m400




9. A large field of 700 hectares is divided into two parts. The difference of the areas of the two parts is one-fifth of the average of the two areas. What is the area of thesmaller part in hectares?

A. 400 B. 365

C. 385 D. 315


Answer : Option D

Explanation :

Let the areas of the parts be x hectares and (700 - x)hectares.

Given that difference of the areas of the two parts = one-fifth of the Average of the two areas=> 2x - 700 = 70=> 2x = 770

Hence, Area of smaller part = (700 - x) = (700 385) = 315hectares.

10. The length of a room is 5.5 m and width is 3.75 m. What is the cost of paying the floor by slabs at the rate of Rs. 800 per sq. metre.

A. Rs.12000 B. Rs.19500

C. Rs.18000 D. Rs.16500.


Answer : Option D

Explanation :

Area = 5.5 3.75 sq. metre.Cost for 1 sq. metre. = Rs. 800

Hence total cost = 5.5 3.75 800 = 5.5 3000 = Rs. 16500

11. The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq.cm.What is the length of the rectangle?

A. 18 cm B. 16 cm

C. 40 cm D. 20 cm


Answer : Option C

Explanation :

Let breadth = x cmThen length = 2x cm

Area = lb = x 2x = 2x2

New length = (2x - 5)New breadth = (x + 5)New Area = lb = (2x - 5)(x + 5)

But given that new area = initial area + 75 sq.cm.

=> (2x - 5)(x + 5) = 2x2 + 75

=> 2x2 + 10x - 5x - 25 = 2x2 + 75=> 5x - 25 = 75=> 5x = 75 + 25 = 100

=> x = 100/5 = 20 cm

Length = 2x = 2 20 = 40cm

Difference of the areas of the two parts = x - (700 - x) = 2x - 700

one-fifth of the Average of the two areas = 15

[x + (700 x)]

2

= = = 7015

7002

3505

x = = 3857702



12. If a square and a rhombus stand on the same base, then what is the ratio of the areas of the square and the rhombus?

A. equal to B. equal to

C. greater than 1 D. equal to 1


Answer : Option D

Explanation :

A square and a rhombus on the same base will have equalareas.

Hence ratio of the areas of the square and the rhombuswill be equal to 1 sincethey stand on the same base

================================================================Note : Please find the proof of the formula given below whichyou may like to go through

Let ABCD be the square and ABEF be the rhombus

Consider the right-angled triangles ADF and BCE

We know that AD = BC ( sides of a square)

AF = BE ( sides of a rhombus)

DF = CE [ DF2 = AF2 - AD2 and CE2 = BE2 - BC2]

Hence ADF = BCE

=> ADF + Trapezium ABCF= BCE + Trapezium ABCF

=> Area of square ABCD = Area of rhombus ABEF

13. The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m, find out the area of the field.

A. 37500 m2 B. 30500 m2

C. 32500 m2 D. 40000 m2


Answer : Option A

Explanation :

Given that breadth of a rectangular field is 60% of its length

perimeter of the field = 800 m=> 2 (l + b) = 800

b = =60l100

3l5



14. A room 5m 44cm long and 3m 74cm broad needs to be paved with square tiles. What will be the least number of square tiles required to cover the floor?

A. 176 B. 124

C. 224 D. 186


Answer : Option A

Explanation :

l = 5 m 44 cm = 544 cmb = 3 m 74 cm = 374 cm

Area = 544 374 cm2

Now we need to find out HCF(Highest Common Factor) of 544and 374.Let's find out the HCF using long division method for quickerresults)

374) 544 (1 374 170) 374 (2 340 34) 170 (5 170 0

Hence, HCF of 544 and 374 = 34

Hence, side length of largest square tile we can take = 34 cm

Area of each square tile = 34 34 cm2

15. The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ Rs. 26.50 per metre is Rs. 5300, what is the length of the plot inmetres?

A. 60 m B. 100 m

C. 75 m D. 50 m


Answer : Option A

Explanation :

Length of the plot is 20 metres more than its breadth.Hence, let's take the length as l metres and breadth as (l - 20)metres

Length of the fence = perimeter = 2(length + breadth)= 2[ l + (l- 20) ] = 2(2l - 20) metresCost per meter = Rs. 26.50Total cost = 2(2l - 20) 26.50

Total cost is given as Rs. 5300=> 2(2l - 20) 26.50 = 5300=> (2l - 20) 26.50 = 2650=> (l - 10) 26.50 = 1325

=> (l - 10) = 1325/26.50 = 50

2 (l + ) = 8003l5

l + = 4003l5

= 4008l5

= 50l5

l = 5 50 = 250 m

b = = = 2 50 = 150 m3l5

3 2505

Area = lb = 250 150 = 37500 m2

Number of tiles required = = 16 11 = 176544 37434 34



=> l = 50 + 10 = 60 metres

16. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one roundin 8 minutes, then what is the area of the park (in sq. m)?

A. 142000 B. 112800

C. 142500 D. 153600


Answer : Option D

Explanation :

l : b = 3 : 2 ------------------------------------------(Equation 1)

Perimeter of the rectangular park = Distance travelled by the man at the speed of 12 km/hr in 8minutes

= speed time = 12 8/60 ( 8 minute = 8/60 hour)

= 8/5 km = 8/5 1000 m = 1600 m

Perimeter = 2(l + b)

=> 2(l + b) = 1600

=> l + b = 1600/2 = 800 m ---------------------------

(Equation 2)

From (Equation 1) and (Equation 2)

l = 800 3/5 = 480 m

b = 800 2/5 = 320 m (Or b = 800 - 480 = 320m)

Area = lb = 480 320 = 153600 m2

17. What is the percentage increase in the area of a rectangle, if each of its sides is increased by 20%?

A. 45% B. 44%

C. 40% D. 42%


Answer : Option B

Explanation :


New area = 120 120 = 14400

Increase in area = New Area - Original Area = 14400 - 10000 = 4400


Increase in 20% of length

New length = Original length = 100 = 120(100 + 20)

100120100

Increase in 20% of breadth

New breadth= Original breadth = 100 = 120(100 + 20)

100120100

Percentage increase in area = 100 = 100 = 44%Increase in Area

Original Area

440010000



18. If the difference between the length and breadth of a rectangle is 23 m and its perimeter is 206 m, what is its area?

A. 2800 m2 B. 2740 m2

C. 2520 m2 D. 2200 m2


Answer : Option C

Explanation :

l - b = 23 ...................(Equation 1)

perimeter = 2(l + b) = 206=> l + b = 103.............(Equation 2)

(Equation 1) + (Equation 2) => 2l = 23 + 103 = 126

=> l = 126/2 = 63 metre

Substituting this value of l in (Equation 1), we get63 - b = 23=> b = 63 - 23 = 40 metre

Area = lb = 63 40 = 2520 m2

19. The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?

A. 16 cm B. 18 cm

C. 14 cm D. 20 cm


Answer : Option B

Explanation :

=> 2l + 2b = 5b=> 2l = 3b

Also given that area = 216 cm2

=> lb = 216 cm2

Increase in 20% of length

New length = Original length = l =(100 + 20)

100120100

120l100

Increase in 20% of breadth

New breadth= Original breadth = b =(100 + 20)

100120100

120b100

New area = = =120l100

120b100

14400lb10000

144lb100

Increase in area = New Area - Original Area = lb =144lb100

44lb100

Percentage of increase in area = 100Increase in Area

Original Area

= 100 = = 44%

( )44lb100

lb44lb 100

100lb

Given that = 52(l + b)

b

=> b =2l3

Substituting the value of b, we get, l = 2162l



20. What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?

A. 814 B. 802

C. 836 D. 900


Answer : Option A

Explanation :

l = 15 m 17 cm = 1517 cmb = 9 m 2 cm = 902 cm

Area = 1517 902 cm2

Now we need to find out HCF(Highest Common Factor) of 1517and 902.Let's find out the HCF using long division method for quickerresults)

902) 1517 (1 902 615) 902 (1 615 287) 615 (2 574 41) 287 (7 287 0

Hence, HCF of 1517 and 902 = 41

Hence, side length of largest square tile we can take = 41 cm

Area of each square tile = 41 41 cm2

21. The diagonal of the floor of a rectangular room is feet. The shorter side of the room is 4 feet. What is the area of the room?

A. 27 square feet B. 22 square feet

C. 24 square feet D. 20 square feet


Answer : Option A

Explanation :

Substituting the value of b, we get, l = 2162l3

= = 3 108 = (3 3) 36l23 216

2

l = 3 6 = 18 cm

Number of tiles required = = 37 22 = 407 2 = 8141517 902

41 41

712

12


22. The diagonal of a rectangle is cm and its area is 20 sq. cm. What is the perimeter of the rectangle?

A. 16 cm B. 10 cm

C. 12 cm D. 18 cm


Answer : Option D

Explanation :

23. A tank is 25 m long, 12 m wide and 6 m deep. What is the cost of plastering of its walls and bottom at the rate of 75 paise per sq. m?

A. Rs. 558 B. Rs. 502

C. Rs. 516 D. Rs. 612

Diagonal, d = 7 feet = feet12

152

Breadth, b = 4 feet = feet12

92

In the right-angled triangle PQR,

= l2 ( )152

2

( )92

2

= =2254

814

1444

l = = feet = 6 feet1444

122

Area = lb = 6 = 2792

feet 2

41

For a rectangle, = +d 2 l2 b2

where l = length , b = breadth and d = diagonal of the of the rectangle

d = cm41

= +d 2 l2 b2

+ = = 41........(Equation 1)l2 b2 ( )41 2

............(Equation 2)Area = lb = 20 cm 2

Solving (Equation 1) and (Equation 2)

(a + b = + 2ab +)2 a2 b2

using the above formula, we have

(l + b = + 2lb + = ( + ) + 2lb = 41 + (2 20) = 81)2 l2 b2 l2 b2

(l + b) = = 9 cm81

perimeter = 2(l + b) = 2 9 = 18 cm




Answer : Option A

Explanation :

Consider a rectangular solid of length l, width w andheight h. Then

1. Total Surface area of a rectangular solid, S = 2lw + 2lh+ 2wh = 2(lw + lh + wh)

2. Volume of a rectangular solid, V = lwh

In this case, l = 25 m, w = 12 m, h = 6 m and all surface needs to be plastered except the top

Hence total area needs to be plastered = Total Surface Area - Area of the Top face= (2lw + 2lh + 2wh) - lw= lw + 2lh + 2wh= (25 12) + (2 25 6) + (2 12 6)= 300 + 300 + 144

= 744 m2

Cost of plastering = 744 75 = 55800 paise = Rs.558

24. It is decided to construct a 2 metre broad pathway around a rectangular plot on the inside. If the area of the plots is 96 sq.m. and the rate of construction is Rs. 50 persquare metre., what will be the total cost of the construction?

A. Rs.3500 B. Rs. 4200

C. Insufficient Data D. Rs. 4400


Answer : Option C

Explanation :


Let length and width of the rectangular plot be l and brespectivelyTotal Area of the rectangular plot = 96 sq.m.

Width of the pathway = 2 m

Length of the remaining area in the plot = (l - 4)breadth of the remaining area in the plot = (b - 4)Area of the remaining area in the plot = (l - 4)(b - 4)

Area of the pathway = Total Area of the rectangular plot - remaining area in the plot= 96 - [(l - 4)(b - 4)] = 96 - [lb - 4l - 4b + 16]= 96 - [96 - 4l - 4b + 16]= 96 - 96 + 4l + 4b - 16]= 4l + 4b - 16= 4(l + b) - 16

We do not know the values of l and b and hence total area ofthe rectangular plot can not be found out. So we can not find out total cost of theconstruction.

25. The area of a parallelogram is 72 cm2 and its altitude is twice the corresponding base. What is the length of the base?

A. 6 cm B. 7 cm

C. 8 cm D. 12 cm


Answer : Option A

Explanation :

Area of a parallelogram , A = bhwhere b is the base and h is the height of theparallelogram

Let the base = x cm. Then the height = 2x cm ( altitude is twice the base)

Area = x 2x = 2x2

But the area is given as 72 cm2

=> 2x2 = 72

=> x2 = 36=> x = 6 cm



26. Two diagonals of a rhombus are 72 cm and 30 cm respectively. What is its perimeter?

A. 136 cm B. 156 cm

C. 144 cm D. 121 cm


Answer : Option B

Explanation :

Remember the following two properties of a rhombus which willbe useful in solving this question 1. The sides of a rhombus are congruent. 2. The diagonals of a rhombus are unequal and bisect eachother at right angles.

Let the diagonals be PR and SQ such that PR = 72 cm and SQ =30 cm

27. The base of a parallelogram is (p + 4), altitude to the base is (p - 3) and the area is (p2 - 4), find out its actual area.

A. 40 sq. units B. 54 sq. units

C. 36 sq. units D. 60 sq. units


Answer : Option D

Explanation :

Area of a parallelogram , A = bhwhere b is the base and h is the height of theparallelogram

Hence, we have

p2 - 4 = (p + 4)(p - 3)

=> p2 - 4 = p2 - 3p + 4p - 12=> -4 = p - 12=> p = 12 - 4 = 8

Hence, actual area = (p2 - 4) = 82 - 4 = 64 - 4 = 60 sq. units

PO = OR = = 36 cm 722

SO = OQ = = 15 cm 302

PQ = QR = RS = SP = = = = 39 cm+362 152

1296 + 225 1521

perimeter = 4 39 =156 cm



28. A circle is inscribed in an equilateral triangle of side 24 cm, touching its sides. What is the area of the remaining portion of the triangle?

A. cm2 B. cm2

C. cm2 D. cm2


Answer : Option A

Explanation :

Let r = radius of the inscribed circle. ThenArea of ABC = Area of OBC + Area of OCA + area of OAB= ( r BC) + ( r CA) + ( r AB)= r (BC + CA + AB)= x r x (24 + 24 + 24)= x r x 72 = 36r cm2 ------------------------------------------ (2)

From (1) and (2),

29. A rectangular plot measuring 90 metres by 50 metres needs to be enclosed by wire fencing such that poles of the fence will be kept 5 metres apart. How many poles willbe needed?

A. 30 B. 44

C. 56 D. 60


144 483 121 363

144 363 121 483

Area of an equilateral triangle = 3

4a2

where a is length of one side of the equilateral triangle

Area of the equilateral ABC = = = 144 .............(1)3

4a2

34

242 3 cm2

Area of a triangle = bh12

where b is the base and h is the height of the triangle

144 = 36r3

r = = 4 (3)14436

3 3

Area of a circle = r 2

where = radius of the circle

From (3), the area of the inscribed circle = = = 48 (4)r 2 (4 )32

Hence , Area of the remaining portion of the triangle

Area of ABC Area of inscribed circle

144 483 cm2



Answer : Option C

Explanation :

Perimeter of a rectangle = 2(l + b) where l is the length and b is the breadth of the rectangle

Length of the wire fencing = perimeter = 2(90 + 50) = 280metres

Two poles will be kept 5 metres apart. Also remember that thepoles will be placed along the perimeter of the rectangular plot, not in a singlestraight line which isvery important.

Hence number of poles required = 2805 = 56

30. If the diagonals of a rhombus are 24 cm and 10 cm, what will be its perimeter

A. 42 cm B. 64 cm

C. 56 cm D. 52 cm


Answer : Option D

Explanation :

Let the diagonals be PR and SQ such that PR = 24 cm and SQ =10 cm

31. What will be the length of the longest rod which can be placed in a box of 80 cm length, 40 cm breadth and 60 cm height?

A. cm B. cm

C. cm D. cm


Answer : Option A

Explanation :

PO = OR = = 12 cm 242

SO = OQ = = 5 cm 102

PQ = QR = RS = SP = = = = 13 cm+122 52

144 + 25 169

perimeter = 4 13 = 52 cm

11600 14400

10000 12040

The longest road which can fit into the box will have one end atA and other end at G (or any other similar diagonal)

Hence the length of the longest rod = AG

Initially let's find out AC. Consider the right angled triangle ABC

AC2 = AB2 + BC2 = 402 + 802 = 1600 + 6400 = 8000

Consider the right angled triangle ACG

AG2 = AC2 + CG2

AC = cm8000

= + = 8000 + 3600 = 11600( )8000 2

602

AG = cm11600

The length of the longest rod = cm11600

ImportantFormulas-Average

1. Average

Average=

2. AverageSpeed

Ifacarcoversacertaindistanceatxkmphandanequaldistanceatykmph.Then,theaveragespeedofthewholejourney= kmph

Comments(26) Newest

SumofobservationsNumberofobservations

2xyx + y


teja 18Nov20148:49PM

therewere35studentsinahostel.duetotheadmissionof7newstudentstheexpensesforthemesswereincreasedby42/-perdaywhiletheaverageexpenditureperheaddiminishedbyRe1.whatwastheoriginalexpenditureofthemess.

Raj 21Nov20141:00AM

LettheinitialaverageexpenditureperheadbexThentheinitialtotalexpenditure=35x

Aftertheadmissionof7newstudents,numberofstudentswillbecome42averageexpenditureperhead=(x-1)Newtotalexpenditure=42(x-1)

GiventhatExpensesforthemesswereincreasedby4242(x-1)-35x=42

7x=84x=12

Originalexpenditure=35x=Rs.420

HideAnswer | Notebook | Notebook |Discuss


1.Inthefirst10oversofacricketgame,therunratewasonly3.2.Whatshouldbetherunrate intheremaining40overstoreachthetargetof282runs?A.6.25 B.5.5C.7.4 D.5


Answer:OptionA

Explanation:

Totalruns=282

remainingrunstobescored=282-32=250

remainingovers=40

2.AgrocerhasasaleofRs.6435,Rs.6927,Rs.6855,Rs.7230andRs.6562for5consecutivemonths.HowmuchsalemusthehaveinthesixthmonthsothathegetsanaveragesaleofRs.6500?A.4800 B.4991C.5004 D.5000


Answer:OptionB

Explanation:

Letthesaleinthesixthmonth=x

=> 6435+ 6927+ 6855+ 7230+ 6562+ x= 6 6500=39000

=>34009+x=39000

=>x=39000-34009=4991

3.Theaverageof20numbers iszero.Of them,Howmanyof themmaybegreater thanzero ,at themost?

A.1 B.20

Runsscoredinthefirst10overs=10 3.2 = 32

Runrateneeded= = 6.2525040

Then = 65006435 + 6927 + 6855 + 7230 + 6562 + x

6



C.0 D.19


Answer:OptionD

Explanation:

Averageof20numbers=0

=>Sumof20numbers=0

Henceatthemost,therecanbe19positivenumbers.

(Such that if thesumof these19positivenumbers isx,20thnumberwillbe-x)

4.Thecaptainofacricketteamof11membersis26yearsoldandthewicketkeeperis3yearsolder.Iftheagesofthesetwoareexcluded,theaverageageoftheremainingplayers isoneyear lessthantheaverageageofthewholeteam.Findouttheaverageageoftheteam.A.23years B.20yearsC.24years D.21years


Answer:OptionA

Explanation:

Numberofmembersintheteam=11

Lettheaverageageofoftheteam=x

=>Sumoftheagesofallthe11membersoftheteam=11x

Ageofthecaptain=26

Ageofthewicketkeeper=26+3=29

Sumof theagesof9membersof the teamexcludingcaptainandwicketkeeper

=> = 0Sumof20numbers

20

=> = xSumoftheagesofallthe11membersoftheteam

11


=11x-26-29=11x-55

Averageageof9membersof the teamexcludingcaptainandwicketkeeper

=>11x-55=9(x-1)

=>11x-55=9x-9

=>2x=46

5.Theaveragemonthly incomeofAandB isRs.5050.Theaveragemonthly incomeofBandC isRs.6250andtheaveragemonthlyincomeofAandCisRs.5200.WhatisthemonthlyincomeofA?A.2000 B.3000C.4000 D.5000


Answer:OptionC

Explanation:

LetthemonthlyincomeofA=a

monthlyincomeofB=b

monthlyincomeofC=a

(Equation1)+(Equation3)-(Equation2)

=>2a=2(5050+5200-6250)

=11x 55

9

Giventhat = (x 1)11x 55

9

=> x = = 23years462

a + b = 2 5050 (Equation1)

b + c = 2 6250 (Equation2)

a + c = 2 5200 (Equation3)

=> a + b + a + c (b + c) = (2 5050) + (2 5200) (2 6250)


=>a=4000

=>MonthlyincomeofA=4000

6. A car owner buys diesel atRs.7.50,Rs. 8 andRs. 8.50 per litre for three successive years.WhatapproximatelyistheaveragecostperlitreofdieselifhespendsRs.4000eachyear?A.Rs.8 B.Rs.7.98C.Rs.6.2 D.Rs.8.1


Answer:OptionB

Explanation:

TotalCost=4000 3

Totaldieselused= + +40007.5

40008

40008.5

averagecostperlitreofdiesel= =4000 3

( + + )40007.5

40008

40008.5

3

( + + )17.5

18

18.5

Itisimportanthowyouproceedfromthisstage.Remembertimeisveryimportanthere

andifwesolvethiscompletelyinthetraditionalway,itmaytakelotoftime.

Instead,wecanfindouttheapproximatevalueeasilyandselecttherightanswerfromthe

givenchoices

Inthiscaseanswer=3

( + + )17.5

18

18.5

83

( + + )18

18

18

3

( )38

Meanswegotthatanswerisapproximatelyequalto8.Fromthegivenchoices,theanswer

canbe8or7.98or8.1.Butwhichonefromthese?

canbe8or7.98or8.1.Butwhichonefromthese?

Itwillbeeasytofigureout.Justseeherethedenominatorwas + +17.5

18

18.5

andweapproximateditas .However38

+ = + =17.5

18.5

18 .5

18 + .5

8 + .5 + 8 .5

(8 .5) (8 + .5)

= [because = (a b) (a + b)]16

( )82 .52a2 b2

=16

(64 .25)

ie, + =17.5

18.5

16

(64 .25)

Weknowthat + = =18

18

14

1664

=> + > +17.5

18.5

18

18

Earlywehadapproximatedthedenominatoras38

Howeverfromtheabovementionedequations,nowyouknowthatactuallydenominatoris

slightlygreaterthan38

Itmeansthatanswerisslightlylowerthat8.Hencewecanpickthechoice7.98as

theanswer

Trytoremembertherelationsbetweennumbersandwhichcanhelpyoutosavealotoftime



7. InKiran'sopinion,hisweight isgreater than65kgbut less than72kg.HisbrotherdoesnotagreewithKiranandhethinksthatKiran'sweightisgreaterthan60kgbutlessthan70kg.Hismother'sviewisthathisweightcannotbegreaterthan68kg.Ifallarethemarecorrectintheirestimation,whatistheaverageofdifferentprobableweightsofKiran?A.70kg B.69kgC.61kg D.67kg


Answer:OptionD

Explanation:

LetKiran'sweight=x.Then

AccordingtoKiran,65


9. A library has an average of 510 visitors on Sundays and 240 on other days.What is the averagenumberofvisitorsperdayinamonthof30daysbeginningwithaSunday?A.290 B.304C.285 D.270


Answer:OptionC

Explanation:

inamonthof30daysbeginningwithaSunday,therewillbe4completeweeksand

anothertwodayswhichwillbeSundayandMonday

Hencetherewillbe5Sundaysand25otherdaysinamonthof30daysbeginningwith

Averageweightof16boys=50.25

TotalWeightof16boys=50.25 16

Averageweightofremaining8boys=45.15

TotalWeightofremaining8boys=45.15 8

Totalweightofallboysintheclass= (50.25 16) + (45.15 8)

Totalboys=16 + 8 = 24

Averageweightofalltheboys=(50.25 16) + (45.15 8)

24

= = (16.75 2) + 15.05 = 33.5 + 15.05(50.25 2) + (45.15 1)

3

= 48.55


aSunday

10.Astudent'smarkwaswronglyenteredas83 insteadof63.Due to that theaveragemarks for theclassgotincreasedbyhalf1/2.Whatisthenumberofstudentsintheclass?A.45 B.40C.35 D.30


Answer:OptionB

Explanation:

AveragevisitorsonSundays=510

Totalvisitorsof5Sundays=510 5

Averagevisitorsonotherdays=240

Totalvisitorsofother25days=240 25

Totalvisitors= (510 5) + (240 25)

Totaldays=30

Averagenumberofvisitorsperday=(510 5) + (240 25)

30

= = (17 5) + (8 25) = 85 + 200 = 285(51 5) + (24 25)

3

Letthetotalnumberofstudents=x


11.Afamilyconsistsoftwograndparents,twoparentsandthreegrandchildren.Theaverageageofthegrandparents is67years, thatof theparents is35yearsand thatof thegrandchildren is6years.Theaverageageofthefamilyis

A. years B. years

C. years D. years


Answer:OptionB

Explanation:

12.TheaverageweightofA,BandC is45kg.IftheaverageweightofAandBbe40kgandthatofBandCbe43kg,whatistheweightofB?

A. kg B. kg

C. kg D. kg

Letthetotalnumberofstudents=x

Theaveragemarksincreasedby duetoanincreaseof83-63=20marks.12

Buttotalincreaseinthemarks= x =12

x2

Hencewecanwriteas

= 20x2

x = 20 2 = 40

3227

3157

2817

3057

Totalageofthegrandparents=67 2

Totalageoftheparents=35 2

Totalageofthegrandchildren=6 3

Averageageofthefamily=(67 2) + (35 2) + (6 3)

7

= = = 31134 + 70 + 18

72227

57

31 2812

32 3012

HideAnswer | Notebook | Notebook |DiscussHereistheanswerandexplanation

Answer:OptionA

Explanation:

13.Iftheaveragemarksofthreebatchesof55,60and45studentsrespectively is50,55,60,what istheaveragemarksofallthestudents?

LettheweightofA,BandCarea,bandcrespectively.

AverageweightofA,BandC=45

a + b + c = 45 3 = 135---equation(1)

averageweightofAandB=40

a + b = 40 2 = 80---equation(2)

averageweightofBandC=43

b + c = 43 2 = 86---equation(3)

equation(2)+equation(3)-equation(1) => a + b + b + c (a + b + c) = 80 + 86 135

=> b = 80 + 86 135 = 166 135 = 31

=> weightofB=31



A.53.23 B.54.68C.51.33 D.50


Answer:OptionB

Explanation:

Averagemarksofbatch1=50

Studentsinbatch1=55


Studentsinbatch2=60


Studentsinbatch3=45

Totalstudents=55+60+45=160

14.Theaverageageofhusband,wifeandtheirchild3yearsagowas27yearsandthatofwifeandthechild5yearsagowas20years.Whatisthepresentageofthehusband?A.40 B.32C.28 D.30


Answer:OptionA

Explanation:

letthepresentageofthehusband=h

presentageofthewife=w

presentageofthechild=c

Totalmarksofbatch1 = 55 50



Averagemarksofallthestudents=(55 50) + (60 55) + (45 60)

160

= = = 54.68275 + 330 + 270

1687516


3yearsago,averageageofhusband,wifeandtheirchild=27

=>(h-3)+(w-3)+(c-3)=81

=>h+w+c=81+9=90-------------------equation(1)

5yearsago,averageageofwifeandchild=20

=>(w-5)+(c-5)=40

=>w+c=40+10=50-------------------equation(2)

Substitutingequation(2)inequation(1)

=>h+50=90

=>h=90-50=40

=>presentageofthehusband=40

15.Theaverageweightof8person's increasesby2.5kgwhenanewpersoncomes inplaceofoneofthemweighing65kg.Whatistheweightofthenewperson?A.75Kg B.50KgC.85Kg D.80Kg


Answer:OptionC

Explanation:

Ifxistheweightofthenewperson,totalincreaseinweight=x-65

=>20=x-65

=>x=20+65=85

16. There are two divisions A and B of a class, consisting of 36 and 44 students respectively. If theaverageweightofdivisionsAis40kgandthatofdivisionbis35kg.Whatistheaverageweightofthe

=> Sumofageofhusband,wifeandtheirchildbefore3years=3 27 = 81

=> Sumofageofwifeandchildbefore5years=2 20 = 40

Totalincreaseinweight=8 2.5 = 20



wholeclass?A.38.25 B.37.25C.38.5 D.37


Answer:OptionB

Explanation:

17.Abatsmanmakesascoreof87runsinthe17thinningandthusincreaseshisaveragesby3.Whatishisaverageafter17thinning?A.39 B.35C.42 D.40.5


Answer:OptionA

Explanation:

Lettheaverageafter17innings=x

Totalrunsscoredin17innings=17x

thenaverageafter16innings=(x-3)

Totalrunsscoredin16innings=16(x-3)

WeknowthatTotalrunsscoredin16innings+87=Totalrunsscoredin17innings

=>16(x-3)+87=17x

=>16x-48+87=17x

=>x=39

18.Astudentneeded to find thearithmeticmeanof thenumbers3,11,7,9,15,13,8,19,17,21,14

TotalweightofstudentsindivisionA=36 40TotalweightofstudentsindivisionB=44 35Totalstudents=36 + 44 = 80

Averageweightofthewholeclass=(36 40) + (44 35)

80

= = = = = 37.25(9 40) + (11 35)

20

(9 8) + (11 7)

472 + 77

41494




andx.Hefoundthemeantobe12.Whatisthevalueofx?A.12 B.5C.7 D.9


Answer:OptionC

Explanation:

=>137+x=144

=>x=144-137=7

19.Arunobtained76,65,82,67and85marks(out in100) inEnglish,Mathematics,Chemistry,BiologyandPhysics.Whatishisaveragemark?A.53 B.54C.72 D.75


Answer:OptionD

Explanation:

20.DistancebetweentwostationsAandB is778km.Atraincoversthe journey fromAtoBat84kmperhourand returnsback toAwithauniformspeedof56kmperhour.Find theaveragespeedof thetrainduringthewholejourney?A.69.0km/hr B.69.2km/hrC.67.2km/hr D.67.0km/hr


Answer:OptionC

Explanation:

-------------------------------------------

Solution1(Quick)

= 123 + 11 + 7 + 9 + 15 + 13 + 8 + 19 + 17 + 21 + 14 + x

12

=> = 12137 + x

12

Averagemark= = = 7576 + 65 + 82 + 67 + 85

53755

--------------------------------------------

-------------------------------------------

Solution2(Fundamentals)

--------------------------------------------

Though it is a good idea to solve the problems quickly usingformulas,youshould

know the fundamentals too. Let's see howwe can solve thisproblemsusingbasics

TraintravelsfromAtoBat84kmperhour

LetthedistancebetweenAandB=x

TraintravelsfromBtoAat56kmperhour

Ifacarcoversacertaindistanceatxkmphandanequaldistanceatykmph.Then,

theaveragespeedofthewholejourney= kmph.2xyx + y

Byusingthesameformula,wecanfindouttheaveragespeedquickly

averagespeed= = =2 84 5684 + 56

2 84 56140

2 21 5635

= = = 67.22 3 56

53365

TotaltimetakenfortravelingfromAtoB= =distancespeed

x84

TotaltimetakenfortravelingfromBtoA= =distance



21.Theaverageageofboysinaclassis16yearsandthatofthegirlsis15years.Whatistheaverageageforthewholeclass?A.15 B.16C.15.5 D.InsufficientData


primer on quantitative aptitude

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