1/8 part of a pencil is black and 1/2 part of the remaining is white. If the remaining part is blue and length of this blue part is 3¹/₂ cm, then find the length of the pencil.

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
a^m = 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 p^X = p^Y 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