Real Numbers

$\displaystyle \textbf{Question 1. }\text{Define HCF of two positive integers and find the HCF of the following}$
$\displaystyle \text{pairs of numbers: }(i)\ 475\text{ and }495\qquad (ii)\ 75\text{ and }243\qquad (iii)\ 240\text{ and }6552$
$\displaystyle (iv)\ 155\text{ and }1385\qquad (v)\ 100\text{ and }190\qquad (vi)\ 105\text{ and }120\qquad [\mathrm{CBSE}\ 2009]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{HCF of two positive integers is the largest positive integer which divides both the}$
$\displaystyle \text{given integers exactly.}$
$\displaystyle (i)\ 495=475\times1+20$
$\displaystyle 475=20\times23+15$
$\displaystyle 20=15\times1+5$
$\displaystyle 15=5\times3+0$
$\displaystyle \therefore \mathrm{HCF}(475,495)=5$
$\displaystyle (ii)\ 243=75\times3+18$
$\displaystyle 75=18\times4+3$
$\displaystyle 18=3\times6+0$
$\displaystyle \therefore \mathrm{HCF}(75,243)=3$
$\displaystyle (iii)\ 6552=240\times27+72$
$\displaystyle 240=72\times3+24$
$\displaystyle 72=24\times3+0$
$\displaystyle \therefore \mathrm{HCF}(240,6552)=24$
$\displaystyle (iv)\ 1385=155\times8+145$
$\displaystyle 155=145\times1+10$
$\displaystyle 145=10\times14+5$
$\displaystyle 10=5\times2+0$
$\displaystyle \therefore \mathrm{HCF}(155,1385)=5$
$\displaystyle (v)\ 190=100\times1+90$
$\displaystyle 100=90\times1+10$
$\displaystyle 90=10\times9+0$
$\displaystyle \therefore \mathrm{HCF}(100,190)=10$
$\displaystyle (vi)\ 120=105\times1+15$
$\displaystyle 105=15\times7+0$
$\displaystyle \therefore \mathrm{HCF}(105,120)=15$
$\\$

$\displaystyle \textbf{Question 2. }\text{Use Euclid's division algorithm to find the HCF of}$
$\displaystyle (i)\ 135\text{ and }225\qquad (ii)\ 867\text{ and }255\qquad (iii)\ 1260\text{ and }7344\qquad [\mathrm{CBSE}\ 2019]$
$\displaystyle (iv)\ 2048\text{ and }960\qquad [\mathrm{CBSE}\ 2019]$
$\displaystyle \text{Answer:}$
$\displaystyle (i)\ 225=135\times1+90$
$\displaystyle 135=90\times1+45$
$\displaystyle 90=45\times2+0$
$\displaystyle \therefore \mathrm{HCF}(135,225)=45$
$\displaystyle (ii)\ 867=255\times3+102$
$\displaystyle 255=102\times2+51$
$\displaystyle 102=51\times2+0$
$\displaystyle \therefore \mathrm{HCF}(867,255)=51$
$\displaystyle (iii)\ 7344=1260\times5+1044$
$\displaystyle 1260=1044\times1+216$
$\displaystyle 1044=216\times4+180$
$\displaystyle 216=180\times1+36$
$\displaystyle 180=36\times5+0$
$\displaystyle \therefore \mathrm{HCF}(1260,7344)=36$
$\displaystyle (iv)\ 2048=960\times2+128$
$\displaystyle 960=128\times7+64$
$\displaystyle 128=64\times2+0$
$\displaystyle \therefore \mathrm{HCF}(2048,960)=64$
$\\$

$\displaystyle \textbf{Question 3. }\text{If the HCF of }408\text{ and }1032\text{ is expressible in the form }$
$\displaystyle 1032m-408\times5, \text{ find }m.$
$\displaystyle \text{Answer:}$
$\displaystyle 1032=408\times2+216$
$\displaystyle 408=216\times1+192$
$\displaystyle 216=192\times1+24$
$\displaystyle 192=24\times8+0$
$\displaystyle \therefore \mathrm{HCF}(408,1032)=24$
$\displaystyle 1032m-408\times5=24$
$\displaystyle 1032m-2040=24$
$\displaystyle 1032m=2064$
$\displaystyle m=2$
$\\$

$\displaystyle \textbf{Question 4. }\text{If the HCF of }657\text{ and }963\text{ is expressible in the form }$
$\displaystyle 657x+963\times(-15), \text{ find }x.$
$\displaystyle \text{Answer:}$
$\displaystyle 963=657\times1+306$
$\displaystyle 657=306\times2+45$
$\displaystyle 306=45\times6+36$
$\displaystyle 45=36\times1+9$
$\displaystyle 36=9\times4+0$
$\displaystyle \therefore \mathrm{HCF}(657,963)=9$
$\displaystyle 657x+963\times(-15)=9$
$\displaystyle 657x-14445=9$
$\displaystyle 657x=14454$
$\displaystyle x=22$
$\\$

$\displaystyle \textbf{Question 5. }\text{Find the largest number which divides }615\text{ and }963\text{ leaving remainder }6$
$\displaystyle \text{in each case.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Required number divides }615-6\text{ and }963-6.$
$\displaystyle 615-6=609$
$\displaystyle 963-6=957$
$\displaystyle \therefore \text{Required number}=\mathrm{HCF}(609,957)$
$\displaystyle 957=609\times1+348$
$\displaystyle 609=348\times1+261$
$\displaystyle 348=261\times1+87$
$\displaystyle 261=87\times3+0$
$\displaystyle \therefore \mathrm{HCF}(609,957)=87$
$\displaystyle \therefore \text{The largest number is }87.$
$\\$

$\displaystyle \textbf{Question 6. }\text{During a sale, colour pencils were being sold in packs of }24\text{ each and crayons}$
$\displaystyle \text{in packs of }32\text{ each. If you want full packs of both and the same number of pencils and crayons,}$
$\displaystyle \text{how many of each would you need to buy?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Required number of pencils and crayons}=\mathrm{LCM}(24,32)$
$\displaystyle 24=2^{3}\times3$
$\displaystyle 32=2^{5}$
$\displaystyle \therefore \mathrm{LCM}(24,32)=2^{5}\times3=96$
$\displaystyle \therefore \text{Number of pencils}=96$
$\displaystyle \therefore \text{Number of crayons}=96$
$\displaystyle \text{Number of pencil packs}=\frac{96}{24}=4$
$\displaystyle \text{Number of crayon packs}=\frac{96}{32}=3$
$\displaystyle \therefore \text{You need to buy }4\text{ packs of colour pencils and }3\text{ packs of crayons.}$
$\\$

$\displaystyle \textbf{Question 7. }\text{144 cartons of Coke Cans and }90\text{ cartons of Pepsi Cans are to be stacked in a Canteen.}$
$\displaystyle \text{If each stack is of the same height and is to contain cartons of the same drink, what would be}$
$\displaystyle \text{the greatest number of cartons each stack would have?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Greatest number of cartons in each stack}=\mathrm{HCF}(144,90)$
$\displaystyle 144=90\times1+54$
$\displaystyle 90=54\times1+36$
$\displaystyle 54=36\times1+18$
$\displaystyle 36=18\times2+0$
$\displaystyle \therefore \mathrm{HCF}(144,90)=18$
$\displaystyle \therefore \text{Each stack would have }18\text{ cartons.}$
$\\$

$\displaystyle \textbf{Question 8. }\text{Two brands of chocolates are available in packs of }24\text{ and }15\text{ respectively.}$
$\displaystyle \text{If I need to buy an equal number of chocolates of both kinds, what is the least number of boxes}$
$\displaystyle \text{of each kind I would need to buy?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Least equal number of chocolates}=\mathrm{LCM}(24,15)$
$\displaystyle 24=2^{3}\times3$
$\displaystyle 15=3\times5$
$\displaystyle \therefore \mathrm{LCM}(24,15)=2^{3}\times3\times5=120$
$\displaystyle \text{Number of boxes of first brand}=\frac{120}{24}=5$
$\displaystyle \text{Number of boxes of second brand}=\frac{120}{15}=8$
$\displaystyle \therefore \text{The least number of boxes required are }5\text{ and }8\text{ respectively.}$
$\\$

$\displaystyle \textbf{Question 9. }\text{Use Euclid's division algorithm to find the HCF of}$
$\displaystyle (i)\ 184,\ 230\text{ and }276\qquad (ii)\ 136,\ 170\text{ and }255$
$\displaystyle \text{Answer:}$
$\displaystyle (i)\ \text{First find the HCF of }184\text{ and }230.$
$\displaystyle 230=184\times1+46$
$\displaystyle 184=46\times4+0$
$\displaystyle \therefore \mathrm{HCF}(184,230)=46$
$\displaystyle \text{Now find the HCF of }46\text{ and }276.$
$\displaystyle 276=46\times6+0$
$\displaystyle \therefore \mathrm{HCF}(184,230,276)=46$
$\displaystyle (ii)\ \text{First find the HCF of }136\text{ and }170.$
$\displaystyle 170=136\times1+34$
$\displaystyle 136=34\times4+0$
$\displaystyle \therefore \mathrm{HCF}(136,170)=34$
$\displaystyle \text{Now find the HCF of }34\text{ and }255.$
$\displaystyle 255=34\times7+17$
$\displaystyle 34=17\times2+0$
$\displaystyle \therefore \mathrm{HCF}(136,170,255)=17$
$\\$

$\displaystyle \textbf{Question 10. }\text{What is the largest number that divides }626,\ 3127\text{ and }15628\text{ and leaves}$
$\displaystyle \text{remainders of }1,\ 2\text{ and }3\text{ respectively?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Required number divides }626-1,\ 3127-2\text{ and }15628-3.$
$\displaystyle 626-1=625$
$\displaystyle 3127-2=3125$
$\displaystyle 15628-3=15625$
$\displaystyle \therefore \text{Required number}=\mathrm{HCF}(625,3125,15625)$
$\displaystyle 3125=625\times5+0$
$\displaystyle 15625=625\times25+0$
$\displaystyle \therefore \mathrm{HCF}(625,3125,15625)=625$
$\displaystyle \therefore \text{The largest number is }625.$
$\\$

$\displaystyle \textbf{Question 11. }\text{Find the greatest number that will divide }445,\ 572\text{ and }699\text{ leaving remainders}$
$\displaystyle 4,\ 5\text{ and }6\text{ respectively.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Required number divides }445-4,\ 572-5\text{ and }699-6.$
$\displaystyle 445-4=441$
$\displaystyle 572-5=567$
$\displaystyle 699-6=693$
$\displaystyle \therefore \text{Required number}=\mathrm{HCF}(441,567,693)$
$\displaystyle 567=441\times1+126$
$\displaystyle 441=126\times3+63$
$\displaystyle 126=63\times2+0$
$\displaystyle \therefore \mathrm{HCF}(441,567)=63$
$\displaystyle 693=63\times11+0$
$\displaystyle \therefore \mathrm{HCF}(441,567,693)=63$
$\displaystyle \therefore \text{The greatest number is }63.$
$\\$

$\displaystyle \textbf{Question 12. }\text{Using Euclid's division algorithm, find the largest number that divides }$
$\displaystyle 1251,\ 9377 \text{ and }15628\text{ leaving remainders }1,\ 2\text{ and }3\text{ respectively.}\qquad [\mathrm{CBSE}\ 2019]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Required number divides }1251-1,\ 9377-2\text{ and }15628-3.$
$\displaystyle 1251-1=1250$
$\displaystyle 9377-2=9375$
$\displaystyle 15628-3=15625$
$\displaystyle \therefore \text{Required number}=\mathrm{HCF}(1250,9375,15625)$
$\displaystyle 9375=1250\times7+625$
$\displaystyle 1250=625\times2+0$
$\displaystyle \therefore \mathrm{HCF}(1250,9375)=625$
$\displaystyle 15625=625\times25+0$
$\displaystyle \therefore \mathrm{HCF}(1250,9375,15625)=625$
$\displaystyle \therefore \text{The largest number is }625.$
$\\$

$\displaystyle \textbf{Question 13. }\text{A merchant has }120\text{ litres of oil of one kind, }180\text{ litres of another kind}$
$\displaystyle \text{and }240\text{ litres of the third kind. He wants to sell the oil by filling the three kinds}$
$\displaystyle \text{of oil in tins of equal capacity. What should be the greatest capacity of such a tin?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Required capacity of the tin}=\mathrm{HCF}(120,180,240)$
$\displaystyle 180=120\times1+60$
$\displaystyle 120=60\times2+0$
$\displaystyle \therefore \mathrm{HCF}(120,180)=60$
$\displaystyle 240=60\times4+0$
$\displaystyle \therefore \mathrm{HCF}(120,180,240)=60$
$\displaystyle \therefore \text{The greatest capacity of the tin is }60\text{ litres.}$
$\\$

$\displaystyle \textbf{Question 14. }\text{Find the HCF of the following pairs of integers and express it as a linear}$
$\displaystyle \text{combination of them. }(i)\ 963\text{ and }657\qquad (ii)\ 592\text{ and }252\qquad (iii)\ 506\text{ and }1155$
$\displaystyle \text{Answer:}$
$\displaystyle (i)\ 963=657\times1+306$
$\displaystyle 657=306\times2+45$
$\displaystyle 306=45\times6+36$
$\displaystyle 45=36\times1+9$
$\displaystyle 36=9\times4+0$
$\displaystyle \therefore \mathrm{HCF}(963,657)=9$
$\displaystyle 9=45-36$
$\displaystyle =45-(306-45\times6)$
$\displaystyle =7\times45-306$
$\displaystyle =7(657-306\times2)-306$
$\displaystyle =7\times657-15\times306$
$\displaystyle =7\times657-15(963-657)$
$\displaystyle =22\times657-15\times963$
$\displaystyle \therefore 9=(-15)\times963+22\times657$
$\\$

$\displaystyle (ii)\ 592=252\times2+88$
$\displaystyle 252=88\times2+76$
$\displaystyle 88=76\times1+12$
$\displaystyle 76=12\times6+4$
$\displaystyle 12=4\times3+0$
$\displaystyle \therefore \mathrm{HCF}(592,252)=4$
$\displaystyle 4=76-12\times6$
$\displaystyle =76-(88-76)\times6$
$\displaystyle =7\times76-6\times88$
$\displaystyle =7(252-88\times2)-6\times88$
$\displaystyle =7\times252-20\times88$
$\displaystyle =7\times252-20(592-252\times2)$
$\displaystyle =47\times252-20\times592$
$\displaystyle \therefore 4=(-20)\times592+47\times252$
$\\$

$\displaystyle (iii)\ 1155=506\times2+143$
$\displaystyle 506=143\times3+77$
$\displaystyle 143=77\times1+66$
$\displaystyle 77=66\times1+11$
$\displaystyle 66=11\times6+0$
$\displaystyle \therefore \mathrm{HCF}(506,1155)=11$
$\displaystyle 11=77-66$
$\displaystyle =77-(143-77)$
$\displaystyle =2\times77-143$
$\displaystyle =2(506-143\times3)-143$
$\displaystyle =2\times506-7\times143$
$\displaystyle =2\times506-7(1155-506\times2)$
$\displaystyle =16\times506-7\times1155$
$\displaystyle \therefore 11=16\times506+(-7)\times1155$
$\\$

$\displaystyle \textbf{Question 15. }\text{Express the HCF of }468\text{ and }222\text{ as }468x+222y\text{ where }x,y$
$\displaystyle \text{are integers in two different ways.}$
$\displaystyle \text{Answer:}$
$\displaystyle 468=222\times2+24$
$\displaystyle 222=24\times9+6$
$\displaystyle 24=6\times4+0$
$\displaystyle \therefore \mathrm{HCF}(468,222)=6$
$\displaystyle 6=222-24\times9$
$\displaystyle =222-(468-222\times2)\times9$
$\displaystyle =222-468\times9+222\times18$
$\displaystyle =(-9)\times468+19\times222$
$\displaystyle \therefore 6=468(-9)+222(19)$
$\displaystyle \text{One way is }x=-9,\ y=19.$
$\displaystyle \text{Also, }6=468(28)+222(-59)$
$\displaystyle \text{Another way is }x=28,\ y=-59.$
$\\$