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.
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.
The pattern is , ,
The next number =
i.e. 17
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.
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
Let the three consecutive odd numbers be x, x+2, x+4
Then,
x + x + 2 + x + 4 = 93
=> 3x + 6 = 93
=> 3x = 87
=> x = 29 => 29, 31, 33 are three consecutive odd numbers.
Therefore, the middle number is 31.
Given the angles of a quadrilateral are in the ratio of 2:4:7:5
Let the angles of a quadrilateral are 2x, 4x, 7x, 5x
But we know that sum of the angles = 360 degrees.
=> 2x + 4x + 7x + 5x = 360
=> x = 20
Therfore, the smallest angle of the quadrilateral = 2x = 2x20 = 40 degrees.
One of the angle of the triangle = 2 x 40 = 80 degrees
The other angle is 180 - (40 + 80) = 60 degrees.
Hence the second largest angle of the triangle is 60 degrees.
We have to rearrange the equation to make R the subject.
Start by cross multiplying by (r + R); V (r + R) = 12R
Multiply out the bracket Vr + VR = 12R
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