The weight will be 250g plus (1.55 - 0.65)/0.10 units of 100g
250 + 900 = 1150
This is the maximum weight that can be sent at that price. But, weights exceeding
250 + 800 will also get charged this amount (that is what the ?part thereof? implies).
Hence a package weighing 1145 will be charged $1145
, which is always divisible by 6 and 12 both, since n(n+1) is always even.
We know that,
A+7B=112
it is clear that A = 14, then it becomes A>=B, but A is smallest angle) given
so, range of A is 0.0001 to 13.9999 ( I am taking upto 4 decimal places)
so, range of B becomes 14 to 16 ( after rounding off to 4 decimal places)
so, range of C becomes 152 to 164 ( after rounding off to 4 decimal places)
By trial and error method, we get
2880/3 = 960 is not a perfect square
2880/4 = 720 is not a perfect square
2880/5 = 576 which is perfect square of 24
Hence, 5 is the least number by which 2880 must be divided in order to make it into a perfect square.
Let the original number is 82 i.e 8 + 2 = 10 and when the digits are reversed i.e 28
The difference is 82 - 28 = 54 i.e, the number is decreased by 54.
The value of 4 in 475 means the place value of 4 in 475 and not its face value. Face value means the digit itself though it is at any place in the given number. But place value means the value of digit in its place in the given number.
Here the place value of 4 in 475 can be determined by as 4 is in 100's place in 475.
Hence, the place value of 4 in 475 is 4x100 = 400.
To solve a pair of simultaneous equations such as those given we can add or subtract them.
Adding we get 4x + 4y = 20
Therefore 2x + 2y = 10
Two digit numbers: The two digits can be 2 and 9: Two possibilities 29 and 92.
Three-digit numbers: The three digits can be 1, 2 and 9 => 3! Or 6 possibilities.
We cannot have three digits as (3, 3, 2) as the digits have to be distinct.
We cannot have numbers with 4 digits or more without repeating the digits.
So, there are totally 8 numbers.
When is divided by 9, we have
, remainder = 7
, remainder = 4
, remainder = 1
, remainder = 7
, remainder = 4
, remainder = 1
So, we have cyclicity of 3 factors i.e 7,4,1.
Hence only 3 remainders are possible.
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