On dividing we get
75)8485(113
75
------
98
75
------
235
225
------
10
------
Required number = (8485-10)=8475.
Let,
no. of rice bowl = x
no. of dal bowl = y
no. of meat bowl = z
Then, x + y + z = 65 ....(1)
Also given that, 2x = 3y = 4z
substituting value of y & z in terms of x i.e. y = 2x/3 and z = x/2 in eq (1),
We get x = 30.
As 1 rice bowl is shared between 2 guests,
Therefore, there are 60 guests.
Let n=4*q + 3. Then, 2*n = 8*q + 6 = 4(2*q + 1) + 2.
Thus when 2*n is divided by 4, the reminder is 2.
Given that,
Now Power Cycle of
4 is 4, 6
6 is 6
8 is 8, 4, 2, 6
1 is 1
- 198 is even number , So the unit digit value is 6
- It is always 6
- 66/4 the remainder is 2 , So the second value is 4
- 11 is odd number, So the unit digit value is 4
1 - It is always 1
Total = 21
Therefore, The unit digit is 1
The maximum number of bows will be 4 yards (= 4 x 36 inches) divided by 15 inches.
This gives 9.6. But as a fraction of a bow is no use, we can only make 9 bows.
Sum of digits = (5 + 1 + 7 + x + 3 + 2 + 4) = ( 22 + x ), which must be divisible by 3.
When x=2. [(22 + 2) = 24 is divisible by 3]
So, the answer is 2
The pattern is , ,
The next number =
i.e. 17
Since 653xy is divisible by 5 as well as 2, so y = 0.
Now, 653x0 must be divisible by 8.
So, 3x0 must be divisible by 8. This happens when x = 2
x + y = (2 + 0) = 2.
Here a = 3 and r = 6/3 = 2. Let the number of terms be n.
Then, t = 384 => a * r^(n-1) = 384
=> 3 * 2^(n-1) = 384 => 2^(n-1) = 128 = 2^(7)
=> n-1 = 7 => n = 8.
Probability of left handed = 4/5
total student = 40
total left handed = 40 x(4/5)
= 32
(kx22)/100 = 340 - 166.64 = 173.36
k = (173.36 x 100)/22
k = 788
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