Given numbers are 585 546 514 404 206 369
Then interchanging the 1st and 3rd digits of the given numbers, we get
585 645 415 404 602 963
Now, arranging them in ascending order
963 645 602 585 415 404
Then the last second number is 415.
Given equation is below
(p/q)n - 1 = (q/p)n - 3
Apply the law of Fractional Exponents and Laws of Exponents
a - m = 1/a m
or
am = 1/a - m
(p/q)n - 1 = (q) - (n - 3) x 1/ (p) - (n - 3)
(p/q)n - 1 = (p) 3 - n x 1/ (q) 3 - n
(p/q)n - 1 = (p/q) 3 - n
if pX = pY then X will be equal to Y. means X = Y;
? n - 1 = 3 - n
? 2n = 4
? n = 2
NA
Let us assume the original fraction be x/y.
According to question,
200x/100 / 200y/100 = 14/5
2x/2y = 14/5
?x/y = 14/5
(a) A = { x : x ? Z and x2 ? 2x ? 3 = 0} = {3, -1}.
? A is a finite set.
(b) B = The set of natural numbers divisible by 2 = { 2, 4, 6, 8, 10,?}.
? B is an infinite set.
(c) Since infinite number of lines pass through a point, C is an infinite set.
(d) D = { -4, -3, -2,?}. Clearly D is an infinite set.
Here, r = 10% and R = 20%
? Required per cent = (r + R)/(100 - r) x 100 %
= [(10 + 20)/(100 - 10)] x 100% = (30 x 100)/90 %
This shows that the marked price of the item is 100/3 % more
than its cost price.
? Marked price of the article = (450 x 400)/300
= ? 600
NA
? Listed price of an article = ? 1500
? Price after first discount
= 1500 x (1 -20/100) = 1500 x 4/5 = ? 1200
Now, second discount = 1200 - 1104 = ? 96
Hence, required percentage = (96/1200) x 100 % = 8 %
Let the maximum aggregate marks be N
According to the question,
10% of N = 296 - 222
? N/10 = 74
? N = (74 x 10) = 740
? = 840.003 ÷ 23.999
Here one number is increased and other is decreased to their nearest whole number
= 840 ÷ 24 = 35
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