Exercise1.2
Q1.Expresseachnumberasproductofitsprimefactors:
i.
140
ii.
156
iii.
3825
iv.
5005
v.
7429
Ans.
I.
140=2x2x5x7
II.
156=2x2x3x13
III.
3825=3x3x5x5x17
IV.
5005=5x7x11x13
V.
7429=17x19x23
Q2. FindtheLCMandHCFofthefollowingpairsofintegersandverifythat
LCM×HCF=productofthetwonumbers.
I.
26and91
II.
510and92
III.
336and54
Ans. I.
26and91
26=2x13
91=7x13
HCF=13
LCM=2x7x13=182
Productofthetwonumbers=26x91=2366
HCFxLCM=13x182=2366
Hence,productoftwonumbers=HCFxLCM
II.
510and92
510=23x5x17
92=2x2x23
HCF=2
LCM=2x2x3x5x17x23=23460
Productofthetwonumbers=510x92=46920
HCFxLCM=2x23460
=46920
Hence,productoftwonumbers=HCFxLCM
III.
336and54
336=2x2x2x23x7
54=2x3x3x3
LCM=24x33x7=3024
Productoftwonumbers=336x54=18144
HCFxLCM=6x3024=18144
Hence,productoftwonumbers=HCFxLCM
Q3.FindtheLCMandHCFofthefollowingintegersbyapplyingtheprime
factorisationmethod
i.
12,15and21
ii.
17,23and29
iii.
8,9and25
Ans.
Q4. GiventhatHCF(306,657)=9,findLCM(306,657).
Ans.
Q.5 Checkwhether6ncanendwiththedigit0foranynaturalnumbern.
Ans. Ifanynumberendswiththedigit0,itshouldbedivisibleby10orinother
words,itwillalsobedivisibleby2and5as10=2×5
Primefactorisationof6n=(2×3)n
Itcanbeobservedthat5isnotintheprimefactorisationof6n.
Hence,foranyvalueofn,6nwillnotbedivisibleby5.
Therefore,6ncannotendwiththedigit0foranynaturalnumbern.
Q6. Explainwhy7×11×13+13and7×6×5×4×3×2×1+5are
compositenumbers.
Ans. Numbersareoftwotypes-primeandcomposite.Primenumberscanbe
dividedby1andonlyitself,whereascompositenumbershavefactors
otherthan1anditself.
Itcanbeobservedthat
7×11×13+13=13×(7×11+1)=13×(77+1)
=13×78
=13×13×6
Thegivenexpressionhas6and13asitsfactors.Therefore,itisa
compositenumber.
7×6×5×4×3×2×1+5=5×(7×6×4×3×2×1+1)
=5×(1008+1)
=5×1009
1009cannotbefactorisedfurther.Therefore,thegivenexpressionhas5
and1009asitsfactors.Hence,itisacompositenumber.
Q7. Thereisacircularpatharoundasportsfield.Soniatakes18minutesto
driveoneroundofthefield,whileRavitakes12minutesforthesame.
Supposetheybothstartatthesamepointandatthesametime,andgoin
thesamedirection.Afterhowmanyminuteswilltheymeetagainatthe
startingpoint?
Ans. ItcanbeobservedthatRavitakeslessertimethanSoniaforcompleting1
roundofthecircularpath.Astheyaregoinginthesamedirection,they
willmeetagainatthesametimewhenRaviwillhavecompleted1round
ofthatcircularpathwithrespecttoSonia.Andthetotaltimetakenfor
completingthis1roundofcircularpathwillbetheLCMoftimetakenby
SoniaandRaviforcompleting1roundofcircularpathrespectivelyi.e.,
LCMof18minutesand12minutes.
18=2×3×3
And,12=2×2×3
LCMof12and18=2×2×3×3=36
Therefore,RaviandSoniawillmeettogetheratthestartingpointafter36
minutes.