SOLVED EXAMPLES (H.C.F. AND L.C.M. OF NUMBERS.) ( in 11 to 18)

Ex. 11. Find the greatest possible length which can be used to measure exactly the lengths 4 m 95 cm, 9 m and 16 m 65 cm.

Sol. 

      required length = H.C.F. of 495 cm, 900 cm and 1665 cm.

     495 = 3² x 5 x 11, 900 = 2² x 3² x 5²x, 1665 = 3² x 5 x 37.

     H.C.F.  = 3 x 5 = 45

     Hence, required length = 45 cm.

Ex. 12. Find the largest number which divides 62, 132 and 237 and 237 to leave the same remainder in each case.

Sol.

             required number = H.C.F. of (132 – 62), (237 – 132) and (237 – 62)

                                        = H.C.F. of 70, 105 and 175 =35.

Ex. 13. Find the largest of four digits exactly divisible by 12, 15. 18 and 27.

Sol. 

     the largest number of four digits is 9999.

     Required number must be divisible by L.C.M. of 12, 15, 18, 27, i.e., 540.

     On dividing 9999 by 540, we get 2790 as remainder.

     Required number (9999 – 279) = 9720.

Ex. 14. Find the largest number of four digits exactly divisible by 16, 24, 36 and 54.

Sol. 

      smallest number of five digits is 10000.

      Required number must be divisible by L.C.M. of 16, 24, 36, 54, ie., 432.

      On dividing 10000 by 432, we get 64 as remainder.

      Required number = 10000 + (432 – 64) = 10368.

Ex. 15. Find the least number which when divided by 20, 25, 35, and 40 leaves remainders 14, 19, 29 and 34 respectively.

Sol.

       here, (20 – 14) 6, (25 -19) = (35 – 29) = 6 and (40 – 34) = 6.

      Required number = (L.C.M. of 20, 25, 35, 40) – 6= 1394.

Ex. 16. Find the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder.

Sol.

       L.C.M. of 5, 6, 7, =840.

       Required number is of the form 840 k + 3.

       Lest value of k for which (840 x 2 + 3) = 1683.

Ex. 17. The traffic lights at three different road crossings after every 48 sec., 72 sec. and 108 sec. respectively. If they all change simultaneously at 8: 20: 00 hours, then at what time will they again change simultaneously?

Sol. 

      interval of change = (C.C.M. of 48, 72, 108) sec. = 432 sec.

      So, the lights will again change simultaneously after every 432 seconds i.er. 7 min. 12 sec.

      Hence, next simultaneous change will take place at 8:27:12 hrs.

Ex. 18. Arrange the fractions 17/18, 31/36, 43/45, 59/60 in the ascending order.

Sol.

      L.C.M. of 18, 36, 45 and 60 = 180.

      Now, 17/18 = 17 x 10/ 18 x 10 = 170/180; 31/36 = 31 x 5/36 x 5 = 155/180;

           43/45 = 43 x 4 /45 x 4 = 172 x 180;   59 x 60 = 59 x 3/60/3 = 177/180

     Since, 155 < 170 < 172 < 177, so, 155 / 180 < 170 / 180 < 172 / 180 < 177? 180.

      Hence, 31/36 < 17/18 < 43/45 <59/60.