If the area of three Adjacent faces of a cuboidal box are 120 cm², 72 cm², and 60 cm² respectively, then the volume of the box is

Let the weight of vesel = x kg
Then, 1/4 of volume of water = (5.25 - x)kg
Full volume of water = (16.5 - x)
? 4(5.25 - x) = (16.5 - x)
? 21 - 4x = 16.5 - x
? 21 - 16.5 = 3x
? x = 4.5/3 = 1.5 kg

Let A has ? 200 at the starting of game, then B has ? 100 at the starting of game.
After the game, money left with A = 200 - 200/3 = ? 400/3
? Total money with B = 100 + 200/3 = ? 500/3
Let the fraction of money lost by B = N
Then, 500/3 - N x 500/3 = 400/3 + N x 500/3
? 500/3 - 400/3 = 500N/3 + 500N/3
? 100/3 = 1000N/3
? N = (100/3) x (3/1000) = 1/10
so, the fraction of money lost is 1/10.

Let the fraction be x, y and z respectively in decreasing order.
Then, according to the question,
x/z = 7/6
? x = 7z/6

And y = 7/6 - 1/3 = 7-2/6 = 5/6

Now, x + y + z = 2¹¹/₂₄
? 7z/6 + 5/6 + z = 59/24
? (13z + 5)/ 6 = 59/24
? 13z = 59/4 - 5 =39/4
? z = 3/4
? x =7z/6 = 7/8
Hence, the fraction are 7/8, 5/6, 3/4.

Let the original fraction be p/q.
Numerator is increased by 200%.
? Numerator = p + 200% of p
= p + 200p/100
= p + 2p
= 3p
and denominator of the fraction is increased by 150%.
Denominator = q + 150q/100
= (100q + 150q)/100
= 250q/100
= 5q/2
Then, according to the question,
3p/5q/2 = 9/35
3p x 2/5q = 9/35
6p/5q = 9/35
? p/q = 9 x 5/35 x 6
p/q = 3/14

Let A, B, C and D pay ?x, ?y, ?z and ?a
According to the question,
x = 1/2(y + z + a)
2x = (y + z + a) ..........(i)
y = 1/3(x + z + a)
3y = (x + z + a) ...........(ii)
z = 1/4(x + y + a)
4z = (x + y + a) ..........(iii)
Also, x + y + z + a = 60 ...............(iv)
Now, put the value of x + y + a = 4z from (iii) in (iv)
Then, 4z + z = 60 ? 5z = 60
? z = 12
Similarly, on putting the value of x + z + a = 3y from (ii) in (iv), we get
3y + y = 60 = ? 4y = 60
? y = 15
Again, on putting the value of (y + z + a) = 2x from (i) in (iv), we get
2x + x = 60 ? 3x = 60
&ther4; x = 20
Now, x + y + z + a = 60
On putting the value of x, y and z, we get
12 + 15 + 20 + a = 60
? a = 60 - 47 = ? 13

x³ + 1 + 2x =6x + 1/x
x³ + 1 + 2x = (6x² + 1)/x
(x³ + 1 + 2x)x = 6x² + 1
x⁴ - 4x² - 6x² -1 =0
x⁴ - 4x² + x - 1 =0
Degree of polynomial is highest exponent degree term i.e.,4.

x = 1 + ?2
? x⁴ - 4x³ + 4x² = x²(x² - 4x + 4)
= x²(x - 2)²
= (1 + ?2)²(1 + ?2 - 2)²
=(?2 + 1)² (?2 - 1)²
=[(?2)² - (1)²]²
=(2 - 1)² =1

x - y = 0.9 ...(i)
and 11(x + y)^-1=2
? 11/ (x + y) = 2
? 2(x + y) =11
? x + y = 11/2 ...(ii)
On solving Eqs.(i) and (ii),we get
x = 3.2
and y = 2.3

Let two alternate odd integers odd integers be (2x+1) and (2x+5).
Then according to the question,
(2x + 1) (2x + 5) = 3(2x + 1) + 12
? (2x + 1) (2x + 5 - 3) = 12
? 2x² + 3x - 5 = 0
On solving this quadratic equation,we get
x = 1 and x = -5/2
x = -5/2 is not a integer ? x = 1
Then, larger integer = 2x + 5 = 2 x 1 + 5 = 7

Let the number of 25 paise coins and 50 paise cons be x and y,respectively.
According to the question,
x + y = 38
Also, 25x + 50y = 1350

On multiplying Eq.(i) by 25 and substracting from Eq.(ii), we get y =16
On substituting the value of y in Eq.(i),we get y=22
So,the number of 25 paise coins in 22.