Let the number of sides be n.
The number of diagonals is given by nC2 - n
Therefore, nC2 - n = 44, n>0
n(n - 1) / 2 - n = 44
n2 - 3n - 88 = 0
n2 -11n + 8n - 88 = 0
n(n - 11) + 8(n - 11) = 0
n = -8 or n = 11.
Parallelograms are formed when any two pairs of parallel lines (where each pair is not parallel to the other pair) intersect.
Hence, the given problem can be considered as selecting pairs of lines from the given 2 sets of parallel lines.
Therefore, the total number of parallelograms formed = 7C2 x 6C2 = 315
Let Kumar’s present age be x years and Sailesh’s present age be y years.
Then, according to the first condition,
x - 10 = 3(y - 10) or, x - 3y = - 20 ..(1)
Now. Kumar’s age after 10 years = (x + 10) years
Sailesh’s age after 10 years = (y + 10)
(x + 10) = 2 (y + 10) or, x - 2y = 10 ..(2)
Solving (1) and (2), we get x = 70 and y = 30
Kumar’s age = 70 years and Sailesh’s age = 30 years
Let son’s age = x. Then father’s age = 2x
12 (x - 20) = (2x - 20)
10x = 220
x = 22
Father’s present age = 44 years.
logb(c+a) + logb(c-a) =logb(c^2-a^2) = logb b^2 = 2.
X = (875)^10 = (87.5 x 10)^10
Therefore, log10(X) = 10(log10(87.5 + 1))
= 10(1.9421 + 1)
= 10(2.9421) = 29.421
X = antilog(29.421)
Therefore, number of digits in X = 30.
Cost price of one kg of orange in which the three varieties of oranges are mixed in the ratio 2 : 3 : 5 is equal to S where
S = 0.2 x 20 + 0.3 x 40 + 0.5 × 50
= 4 + 12 + 25
= Rs 41
Selling price per kg of oranges to ensure there is a net profit of 20%
= 1.2 x 41
= Rs 49.2
Assume A be the cost price.
Therefore,
[{(3/5) x A x (10/100)} – {(2/5) x A x 5/100}] = 1500
Or A = Rs 37,500
We have ,
12% of X = 6% of Y
=> 2% of X = 1% of Y
=>(2 x 9)% of X = ( 1 x 9)% of Y
Thus, 18% of X = 9% of Y.
Take a number 100,
Then the other number is 120
% the smaller number is less than the first = [(20/(120)) x 100]% = 16(2/3)%.