Given exp. = 1/(log2 ?) + 1/(log6 ?)
= log? 2 + log? 6
= log? (2 x 6)
= log?12
Since 12 > ? so the value of given expression is more than 1.
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.
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
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
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
Average will increase by 8 times.
So average of new members = (21 x 8 ) = 168
Sum of 50 member = (50 x 38 )
= 1900
Sum of 48 member = [1900- (45 + 55 ) ]
= 1800
? Required Average = 1800/48
= 225/6
= 37.5
Average height of the whole class = sum of height of 40 girls / 40
= (30 x 160 + 10 x 156) / 40
= 159 cms.
Since a, b, c, d, e are 5 consecutive numbers so
b = a + 2.
c = a + 4.
d = a + 6.
e = a + 8.
Average = (a + b + c + d + e)/ 5
= [a + (a + 2 ) + (a +4 ) + (a+6 )+ (a+8)] / 5
= (5a+20) / 5
= a + 4
Let second number = x
So, first number = 2x and third number= 4x
? ( x + 2x + 4x ) / 3 = 56
? 7x = 168
? x = 24
So the numbers are 48,24, 96,
weight of (A + B ) = (2 X 40 ) Kg = 80 kg
weight of (B + C ) = (2 X 43) kg = 86 kg
weight of (A + 2B +C ) = (80 + 86) kg = 166 kg
weight of (A + B + C ) = (3 X 45) kg = 135 kg
weight of B = (166-135) kg = 31 kg
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