Problems on H.C.F and L.C.M 252 can be expressed as a product of primes as: 2 x 2 x 3 x 3 x 7 2 x 3 x 3 x 3 x 7 2 x 2 x 2 x 3 x 7 3 x 3 x 3 x 3 x 7 2 x 2 x 3 x 3 x 7 2 x 3 x 3 x 3 x 7 2 x 2 x 2 x 3 x 7 3 x 3 x 3 x 3 x 7 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Clearly, 252 = 2 x 2 x 3 x 3 x 7.
Problems on H.C.F and L.C.M Find the least number which when divided by 6,7,8,9, and 12 leave the same remainder 1 each case 505 606 707 404 505 606 707 404 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP 3 | 6 - 7 - 8 - 9 - 12----------------------------- 4 | 2 - 7 - 8 - 3 - 4 --------------------------- 2 | 2 - 7 - 2 - 3 - 1 -------------------------- | 1 - 7 - 1 - 3 - 1 L.C.M = 3 X 2 X 2 X 7 X 2 X 3 = 504.Hence required number = (504 +1) = 505
Problems on H.C.F and L.C.M Three friends Raju , Ramesh and Sunil start running around a circular stadium and complete a single round in 24 s, 36 s and 40 s, respectively. After how many minutes will they meet against at the sta 5 minutes 6 minutes 7 minutes 8 minutes 5 minutes 6 minutes 7 minutes 8 minutes ANSWER EXPLANATION DOWNLOAD EXAMIANS APP 24 = 3 × 2 × 2 × 2 = 3 × 2³ 36 = 3 × 3 × 2 × 2 = 3² × 2²and 40 = 2 × 2 × 2 × 5 = 5¹ × 23 LCM of 24, 36 and 40 = 3² × 2³ × 5 = 9 × 8 × 5 = 360Hence, they will meet again at the starting point after 360 s, i.e., 6 min
Problems on H.C.F and L.C.M Find the greatest number of four digits which when divided by 10, 15, 21 and 28 leaves 4, 9, 15 and 22 as remainders respectively? 9654 9664 9864 9666 9654 9664 9864 9666 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M The maximum number of student amoung them 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is : 1911 910 91 1001 1911 910 91 1001 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M If LCM of two number is 693, HCF of two numbers is 11 and one number is 99, then find other 77 66 55 88 77 66 55 88 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Product of two numbers = Product of their HCF and LCMSo Other number =(693×11/99) = 77
EXAMIANS