Minimum number of rows = Maximum number of trees in each row
= HCF of 30,45 and 60 = 15
? Number of rows = (30 + 45 + 60)/15 = 9
HCF of 403,456 and 496 = 31
Now, the number of bottles required to put 403 L of petrol = 403/31 = 13
Similarly,the number of bottles required to put 456 L of diesel = 465/31 = 15
The number of bottles required to put 496 L of mobile oil = 496/31 = 16
? Least number of bottles required having 31 L as capacity = 13+15 + 16 = 44
The time which all the three persons meet will be the LCM of the time by each person individually to complete one round.
252 = 22 x 71 x 91
308 = 22 x 71 x 111
198 = 21 x 91 x 111
? LCM of 252, 308 and 198 = 22 x 71 x 91 x 111 = 2772
So,A,B and C will again meet at the starting point in 2772 i.e.,46 min and 12s.
The pieces cut from four rods is least, so length of the pieces is the HCF of 52, 65, 78 and 91 which is 13.
? Number of pieces cut =(52 + 65 + 78 + 91)/13 = 22
x2 - y2 - z2 - 2yz = x2 - (y + z)2 = (x+ y + z)(x - y - z)
x2 - y2 + z2 + 2yz = (x + z)2 - y2 = (x + y + z) (x + y - z)
x2 + y2 - z2 - 2xy = (x - y)2 - z2 = (x - y + z) (x - y - z)
? LCM = (x + y + z) (x - y - z) (x - y + z)
The time at which all the three persons meet will be the LCM of the time taken by each person individually to complete one round.
252 = 22 x 71 x 91
308 = 22 x 71 x 111
198 = 21 x 91 x 111
? LCM of 252, 308 and 198 = 22 x 71 x 91 x 111 = 2772
So, A, B and C will again meet at the starting point in 2772s i.e., 46 min and 12s
Go through options.
Put m = 1 in two given expression,we have
x3 - 10x2 + 31x - 30 and x2 - 8x + 15
x2 - 8x + 15 = (x - 5) (x - 3)
x3 - 10x2 + 31x - 30 = (x - 2) (x - 5) (x - 3)
? (x - 5) (x - 3) = x2 - 8x + 15 is the HCF of both expressions.
? m = 1
(54 - 35) = 19, (80 - 61) = 19 and(119 - 100) = 19
LCM of 54, 80 and 119 is 257040.
Required least number = 257040 + 19 = 257059
Sum of the digits of 257059 = 2 + 5 + 7 + 0 + 5 + 9 = 28
(52 - 33) = 19, (78 - 59) = 19 and (117 - 98) = 19
LCM of 52, 78 and 117 is 468.
Required least number = 468 + 19 = 487
Sum of digits of 487 = 4 + 8 + 7 = 19
According to the question,
1st number = 3M, 2nd number = 4M
where, M = HCF
But given, M = 4
We know that,
LCM = Product of two numbers / HCF
= (3M x 4M) / M
LCM = 12M = 12 x 4 = 48
Let the numbers be 66a and 66b, where a and b are co-primes.
According to the question,
66a + 66b = 1056
? 66(a + b) = 1056
? (a + b) = 1056/66 = 16
? Possible values of a and b are
(a = 1, b = 15), (a = 3, b = 13).
(a = 5, b = 11 ), (a = 7, b = 9)
? Numbers are
(66 x 1, 66 x 15), (66 x 3, 66 x 13),
(66 x 5, 66 x 11), (66 x 7, 66 x 9).
? Possible number of pairs = 4
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