Real Numbers

$\displaystyle \textbf{Question 1. }\text{Find the LCM and HCF of the following pairs of integers and verify that:}$
$\displaystyle \mathrm{LCM}\times\mathrm{HCF}=\text{Product of the integers.}$
$\displaystyle (i)\ 336\text{ and }54\qquad (ii)\ 404\text{ and }96\qquad [\mathrm{CBSE}\ 2018]$
$\displaystyle (iii)\ 96\text{ and }120\qquad [\mathrm{CBSE}\ 2023]\qquad (iv)\ 72\text{ and }120$
$\displaystyle \text{Answer:}$
$\displaystyle (i)\ 336=2^{4}\times3\times7$
$\displaystyle 54=2\times3^{3}$
$\displaystyle \therefore \mathrm{HCF}=2\times3=6$
$\displaystyle \mathrm{LCM}=2^{4}\times3^{3}\times7=3024$
$\displaystyle \mathrm{LCM}\times\mathrm{HCF}=3024\times6=18144$
$\displaystyle 336\times54=18144$
$\displaystyle \therefore \mathrm{LCM}\times\mathrm{HCF}=336\times54$
$\\$
$\displaystyle (ii)\ 404=2^{2}\times101$
$\displaystyle 96=2^{5}\times3$
$\displaystyle \therefore \mathrm{HCF}=2^{2}=4$
$\displaystyle \mathrm{LCM}=2^{5}\times3\times101=9696$
$\displaystyle \mathrm{LCM}\times\mathrm{HCF}=9696\times4=38784$
$\displaystyle 404\times96=38784$
$\displaystyle \therefore \mathrm{LCM}\times\mathrm{HCF}=404\times96$
$\\$
$\displaystyle (iii)\ 96=2^{5}\times3$
$\displaystyle 120=2^{3}\times3\times5$
$\displaystyle \therefore \mathrm{HCF}=2^{3}\times3=24$
$\displaystyle \mathrm{LCM}=2^{5}\times3\times5=480$
$\displaystyle \mathrm{LCM}\times\mathrm{HCF}=480\times24=11520$
$\displaystyle 96\times120=11520$
$\displaystyle \therefore \mathrm{LCM}\times\mathrm{HCF}=96\times120$
$\\$
$\displaystyle (iv)\ 72=2^{3}\times3^{2}$
$\displaystyle 120=2^{3}\times3\times5$
$\displaystyle \therefore \mathrm{HCF}=2^{3}\times3=24$
$\displaystyle \mathrm{LCM}=2^{3}\times3^{2}\times5=360$
$\displaystyle \mathrm{LCM}\times\mathrm{HCF}=360\times24=8640$
$\displaystyle 72\times120=8640$
$\displaystyle \therefore \mathrm{LCM}\times\mathrm{HCF}=72\times120$
$\\$

$\displaystyle \textbf{Question 2. }\text{Find the LCM and HCF of the following integers by applying the prime }$
$\displaystyle \text{factorisation method: }(i)\ 12,\ 15\text{ and }21\qquad (ii)\ 17,\ 23\text{ and }29$
$\displaystyle (iii)\ 8,\ 9\text{ and }25\qquad (iv)\ 40,\ 36\text{ and }126\qquad (v)\ 26,\ 65\text{ and }117\qquad [\mathrm{CBSE}\ 2023]$
$\displaystyle \text{Answer:}$
$\displaystyle (i)\ 12=2^{2}\times3$
$\displaystyle 15=3\times5$
$\displaystyle 21=3\times7$
$\displaystyle \therefore \mathrm{HCF}=3$
$\displaystyle \mathrm{LCM}=2^{2}\times3\times5\times7=420$
$\\$
$\displaystyle (ii)\ 17=17$
$\displaystyle 23=23$
$\displaystyle 29=29$
$\displaystyle \therefore \mathrm{HCF}=1$
$\displaystyle \mathrm{LCM}=17\times23\times29=11339$
$\\$
$\displaystyle (iii)\ 8=2^{3}$
$\displaystyle 9=3^{2}$
$\displaystyle 25=5^{2}$
$\displaystyle \therefore \mathrm{HCF}=1$
$\displaystyle \mathrm{LCM}=2^{3}\times3^{2}\times5^{2}=1800$
$\\$
$\displaystyle (iv)\ 40=2^{3}\times5$
$\displaystyle 36=2^{2}\times3^{2}$
$\displaystyle 126=2\times3^{2}\times7$
$\displaystyle \therefore \mathrm{HCF}=2$
$\displaystyle \mathrm{LCM}=2^{3}\times3^{2}\times5\times7=2520$
$\\$
$\displaystyle (v)\ 26=2\times13$
$\displaystyle 65=5\times13$
$\displaystyle 117=3^{2}\times13$
$\displaystyle \therefore \mathrm{HCF}=13$
$\displaystyle \mathrm{LCM}=2\times3^{2}\times5\times13=1170$
$\\$

$\displaystyle \textbf{Question 3. }\text{Find by prime factorisation the LCM of the numbers }18180\text{ and }7575.$
$\displaystyle \text{Also, find the HCF of the two numbers.}\qquad [\mathrm{CBSE}\ 2023]$
$\displaystyle \text{Answer:}$
$\displaystyle 18180=2^{2}\times3^{2}\times5\times101$
$\displaystyle 7575=3\times5^{2}\times101$
$\displaystyle \therefore \mathrm{HCF}=3\times5\times101=1515$
$\displaystyle \mathrm{LCM}=2^{2}\times3^{2}\times5^{2}\times101=90900$
$\\$

$\displaystyle \textbf{Question 4. }(i)\ \text{Given that }\mathrm{HCF}(306,657)=9,\text{ find }\mathrm{LCM}(306,657).$
$\displaystyle (ii)\ \text{Write the smallest number which is divisible by both }306\text{ and }657.\qquad [\mathrm{CBSE}\ 2019]$
$\displaystyle \text{Answer:}$
$\displaystyle (i)\ \mathrm{LCM}\times\mathrm{HCF}=306\times657$
$\displaystyle \therefore \mathrm{LCM}=\frac{306\times657}{9}$
$\displaystyle =22338$
$\displaystyle (ii)\ \text{The smallest number divisible by both }306\text{ and }657$
$\displaystyle =\mathrm{LCM}(306,657)$
$\displaystyle =22338$
$\\$

$\displaystyle \textbf{Question 5. }\text{Can two numbers have }16\text{ as their HCF and }380\text{ as their LCM? Give reason.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{No.}$
$\displaystyle \text{The HCF of two numbers must always divide their LCM.}$
$\displaystyle 380\div16=23.75$
$\displaystyle \text{Since }16\text{ does not divide }380,\text{ such two numbers cannot exist.}$
$\\$

$\displaystyle \textbf{Question 6. }\text{The HCF of two numbers is }145\text{ and their LCM is }2175.\text{ If one number}$
$\displaystyle \text{is }725,\text{ find the other.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Product of two numbers}=\mathrm{HCF}\times\mathrm{LCM}$
$\displaystyle 725\times\text{Other number}=145\times2175$
$\displaystyle \text{Other number}=\frac{145\times2175}{725}$
$\displaystyle =435$
$\\$

$\displaystyle \textbf{Question 7. }\text{The HCF of two numbers is }16\text{ and their product is }3072.\text{ Find their LCM.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Product of two numbers}=\mathrm{HCF}\times\mathrm{LCM}$
$\displaystyle 3072=16\times\mathrm{LCM}$
$\displaystyle \mathrm{LCM}=\frac{3072}{16}=192$
$\\$

$\displaystyle \textbf{Question 8. }\text{The LCM and HCF of two numbers are }180\text{ and }6\text{ respectively. If one of the}$
$\displaystyle \text{numbers is }30,\text{ find the other number.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Product of two numbers}=\mathrm{LCM}\times\mathrm{HCF}$
$\displaystyle 30\times\text{Other number}=180\times6$
$\displaystyle \text{Other number}=\frac{180\times6}{30}$
$\displaystyle =36$
$\\$

$\displaystyle \textbf{Question 9. }\text{Three bells ring at intervals of }6,\ 12\text{ and }18\text{ minutes. If all the three bells rang}$
$\displaystyle \text{at }6\text{ AM, when will they ring together again?}\qquad [\mathrm{CBSE}\ 2023,\ 2024]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{They will ring together again after }\mathrm{LCM}(6,12,18)\text{ minutes.}$
$\displaystyle 6=2\times3$
$\displaystyle 12=2^{2}\times3$
$\displaystyle 18=2\times3^{2}$
$\displaystyle \therefore \mathrm{LCM}=2^{2}\times3^{2}=36$
$\displaystyle \therefore \text{They will ring together again after }36\text{ minutes.}$
$\displaystyle \therefore \text{Required time}=6:36\text{ AM}$
$\\$

$\displaystyle \textbf{Question 10. }\text{Find the smallest number which when increased by }17\text{ is exactly divisible by both } \\ 520\text{ and }468.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Required number}+17=\mathrm{LCM}(520,468)$
$\displaystyle 520=2^{3}\times5\times13$
$\displaystyle 468=2^{2}\times3^{2}\times13$
$\displaystyle \therefore \mathrm{LCM}=2^{3}\times3^{2}\times5\times13=4680$
$\displaystyle \therefore \text{Required number}=4680-17=4663$
$\\$

$\displaystyle \textbf{Question 11. }\text{What is the smallest number that, when divided by }35,\ 56\text{ and }91\text{ leaves}$
$\displaystyle \text{remainders of }7\text{ in each case?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Required number}-7=\mathrm{LCM}(35,56,91)$
$\displaystyle 35=5\times7$
$\displaystyle 56=2^{3}\times7$
$\displaystyle 91=7\times13$
$\displaystyle \therefore \mathrm{LCM}=2^{3}\times5\times7\times13=3640$
$\displaystyle \therefore \text{Required number}=3640+7=3647$
$\\$

$\displaystyle \textbf{Question 12. }\text{A rectangular courtyard is }18\text{ m }72\text{ cm long and }13\text{ m }20\text{ cm broad.}$
$\displaystyle \text{It is to be paved with square tiles of the same size. Find the least possible number of such tiles.}$
$\displaystyle \text{Answer:}$
$\displaystyle 18\text{ m }72\text{ cm}=1872\text{ cm}$
$\displaystyle 13\text{ m }20\text{ cm}=1320\text{ cm}$
$\displaystyle \text{Side of largest square tile}=\mathrm{HCF}(1872,1320)$
$\displaystyle 1872=1320\times1+552$
$\displaystyle 1320=552\times2+216$
$\displaystyle 552=216\times2+120$
$\displaystyle 216=120\times1+96$
$\displaystyle 120=96\times1+24$
$\displaystyle 96=24\times4+0$
$\displaystyle \therefore \mathrm{HCF}(1872,1320)=24$
$\displaystyle \therefore \text{Side of square tile}=24\text{ cm}$
$\displaystyle \text{Number of tiles}=\frac{1872}{24}\times\frac{1320}{24}$
$\displaystyle =78\times55$
$\displaystyle =4290$
$\displaystyle \therefore \text{Least possible number of tiles}=4290$
$\\$

$\displaystyle \textbf{Question 13. }\text{Find the largest number which on dividing }1251,\ 9377\text{ and }15628\text{ leaves}$
$\displaystyle \text{remainders }1,\ 2\text{ and }3\text{ respectively.}\qquad [\mathrm{CBSE}\ 2019]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Required number divides }1251-1,\ 9377-2\text{ and }15628-3.$
$\displaystyle 1251-1=1250$
$\displaystyle 9377-2=9375$
$\displaystyle 15628-3=15625$
$\displaystyle \therefore \text{Required number}=\mathrm{HCF}(1250,9375,15625)$
$\displaystyle 9375=1250\times7+625$
$\displaystyle 1250=625\times2+0$
$\displaystyle \therefore \mathrm{HCF}(1250,9375)=625$
$\displaystyle 15625=625\times25+0$
$\displaystyle \therefore \mathrm{HCF}(1250,9375,15625)=625$
$\displaystyle \therefore \text{The largest number is }625.$
$\\$

$\displaystyle \textbf{Question 14. }\text{In a morning walk three persons step off together, their steps measure }$
$\displaystyle 80\text{ cm, }85\text{ cm }\text{and }90\text{ cm respectively. What is the minimum distance each }$
$\displaystyle \text{should walk so that he can cover the distance in complete steps?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Minimum distance}=\mathrm{LCM}(80,85,90)$
$\displaystyle 80=2^{4}\times5$
$\displaystyle 85=5\times17$
$\displaystyle 90=2\times3^{2}\times5$
$\displaystyle \therefore \mathrm{LCM}=2^{4}\times3^{2}\times5\times17$
$\displaystyle =12240$
$\displaystyle \therefore \text{Minimum distance}=12240\text{ cm}=122\text{ m }40\text{ cm}$
$\\$

$\displaystyle \textbf{Question 15. }\text{On a morning walk, three persons step out together and their steps measure }$
$\displaystyle 30\text{ cm, }36\text{ cm and }40\text{ cm respectively. What is the minimum distance each }$
$\displaystyle \text{should walk so that each can cover the same distance in complete steps?}\qquad [\mathrm{CBSE}\ 2019]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Minimum distance}=\mathrm{LCM}(30,36,40)$
$\displaystyle 30=2\times3\times5$
$\displaystyle 36=2^{2}\times3^{2}$
$\displaystyle 40=2^{3}\times5$
$\displaystyle \therefore \mathrm{LCM}=2^{3}\times3^{2}\times5$
$\displaystyle =360$
$\displaystyle \therefore \text{Minimum distance}=360\text{ cm}=3\text{ m }60\text{ cm}$
$\\$

$\displaystyle \textbf{Question 16. }\text{The traffic lights at three different road crossings change after every }$
$\displaystyle 48\text{ seconds, }72\text{ seconds and }108\text{ seconds. If they change simultaneously at }$
$\displaystyle 7\text{ a.m., at what time will } \text{they change together next?}\qquad [\mathrm{CBSE}\ 2023]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{They will change together again after }\mathrm{LCM}(48,72,108)\text{ seconds.}$
$\displaystyle 48=2^{4}\times3$
$\displaystyle 72=2^{3}\times3^{2}$
$\displaystyle 108=2^{2}\times3^{3}$
$\displaystyle \therefore \mathrm{LCM}=2^{4}\times3^{3}=432$
$\displaystyle 432\text{ seconds}=7\text{ minutes }12\text{ seconds}$
$\displaystyle \therefore \text{Required time}=7:07:12\text{ a.m.}$
$\\$

$\displaystyle \textbf{Question 17. }\text{Find the smallest number which leaves remainders }8\text{ and }$
$\displaystyle 12\text{ when divided by }28\text{ and }32\text{ respectively.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required number be }N.$
$\displaystyle N\equiv8\pmod{28}$
$\displaystyle N\equiv12\pmod{32}$
$\displaystyle \text{Now, }28-8=20\text{ and }32-12=20.$
$\displaystyle \therefore N+20\text{ is divisible by both }28\text{ and }32.$
$\displaystyle \therefore N+20=\mathrm{LCM}(28,32)$
$\displaystyle 28=2^{2}\times7$
$\displaystyle 32=2^{5}$
$\displaystyle \therefore \mathrm{LCM}=2^{5}\times7=224$
$\displaystyle \therefore N=224-20=204$
$\displaystyle \therefore \text{The smallest number is }204.$
$\\$

$\displaystyle \textbf{Question 18. }\text{Find the greatest number of }6\text{ digits exactly divisible by }24,\ 15\text{ and }36.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Required number must be divisible by }\mathrm{LCM}(24,15,36).$
$\displaystyle 24=2^{3}\times3$
$\displaystyle 15=3\times5$
$\displaystyle 36=2^{2}\times3^{2}$
$\displaystyle \therefore \mathrm{LCM}=2^{3}\times3^{2}\times5=360$
$\displaystyle \text{Greatest }6\text{-digit number}=999999$
$\displaystyle 999999=360\times2777+279$
$\displaystyle \therefore \text{Required number}=999999-279$
$\displaystyle =999720$
$\\$