Problems on H.C.F and L.C.M 252 can be expressed as a product of primes as: 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 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 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 Reduce 391/667 to lowest terms . 11/13 17/29 23/37 31/45 11/13 17/29 23/37 31/45 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP H.C.F. of 391 and 667 is 23.On dividing the numerator and denominator by 23, we get :391 = 391 ¸23 = 17667 667¸ 23 29
Problems on H.C.F and L.C.M The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is: 322 276 345 299 322 276 345 299 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Clearly, the numbers are (23 x 13) and (23 x 14). Larger number = (23 x 14) = 322.
Problems on H.C.F and L.C.M Three numbers are in the ratio 3:4:5 and their L.C.M is 2400. Their H.C.F is: 200 120 80 40 200 120 80 40 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M The greatest number of four digits which is divisible by 15, 24, 40 and 75 is : 9400 9600 9000 9800 9400 9600 9000 9800 ANSWER DOWNLOAD EXAMIANS APP
Problems on H.C.F and L.C.M An electronic device makes a beep after every 60 sec. Another device makes a beep after every 62 sec. They beeped together at 10 a.m. The time when they will next make a beep together at the earliest, 10 : 31 am 9 : 31 am 12 : 31 am 11 : 31 am 10 : 31 am 9 : 31 am 12 : 31 am 11 : 31 am ANSWER EXPLANATION DOWNLOAD EXAMIANS APP L.C.M. of 60 and 62 seconds is 1860 seconds1860/60 = 31 minutesThey will beep together at 10:31 a.m.Sometimes questions on red lights blinking comes in exam, which can be solved in the same way
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