Akram Miya has two types of grapes. One is the fresh grapes containing 80% water and dry grapes containing 25% water. He sells 20 kg dry grapes, by adding water to the dry grapes, at cost price. What is the total profit percentage when after adding water the weight of 20 kg dry grapes increase in the proportion of water in fresh grapes?

Total number of coins = 50
Let ? 1 coins = x and ? 2 coins = y
Now,according to the question,
x + y = 50 ... (i)
and x + 2y = 75 ...(ii)
On solving Eqs.(i) and (ii),we get
x = 25 and y = 25

Let the number of gold coins initially be x and the number of non-gold coins be y.
According to the question, 3x = y .....(i)
When 10 more gold coins are added, then total number of gold coins becomes x + 10 and the number of non-gold coins remain the same as y.
Now,we have 2(10 + x) = y ....(ii)
On solving these two equations, we get x = 20 and y = 60

? Total number of coins in the collection at the end is equal to x + 10 + y = 20 + 10 + 60 = 90.

Let the original price of the book be Rs. x.
? Number of books bought for Rs.200 = 200/x
and reduced price of the book=Rs.(x - 5)
? Number of books bought for Rs.200 = 200/(x - 5)
Now, according to the condition,
200/(x - 5) - 200/x = 2
? (200x - 200x + 1000)/(x - 5)x = 2
? (200x - 200x + 1000)/x(x - 5) = 2
? 1000/x(x - 5) = 2
? 1000/x(x - 5) = 1
? 1000/x(x - 5) = 2
? 500/x(x - 5) = 1
? 500 = x(x-5)
? x² -5x - 50 = 0
? x²- 25 + 20x - 500 = 0
? x(x - 25) + 20(x - 25) = 0
? (x - 25)(x + 20) = 0
? x - 25 = 0 or x + 20 = 0
? x = 25 0r -20
? x = 25 (since,x cannot be negatives)
So, the original price of the bok is Rs.25.

x³ + 1 + 2x = 6x + 1/x
x³ + 1 + 2x = (6x² + 1) / x
(x³ + 1 + 2x)x = 6x²2 + 1
x⁴ + x + 2x² - 6x² - 1 = 0
x⁴ - 4x² + x - 1 = 0
Degree of polynomial is highest exponent degree term i. e., 4.

Let the length of the piece be L m.
Then,rate per metre = ? 120/L
New length = (L + 2) m
Since, the cost remains same.
? New rate per metre =Rs.120/(L + 2)
According to the given condition, 120/L - 120/(L + 2) = 2
? [120(L + 2) - 120L]/L(L + 2) = 2
? [120L + 240 - 120L] / L(L + 2) = 2
? [120L + 240 -120L] / L(L + 2) = 2
? 240/L(L + 2) = 2
? 240 = 2L(L + 2)
? 120 = L(L + 2) = L² + 2L
? L² + 2L - 120 = 0
? L² + 12L - 10L - 120 = 0
? L(L + 12) - 10 x (L + 12) = 0
? (L + 12)(L - 10) = 0
? L + 12 = 0 or L - 10 = 0
? L = -12 or 10
? L = 10(since, L cannot be negative)
? Original rate per metre of cloth = 120/L = 120/10 = ? 12

Let Arjun had x arrows.
? x/2 + 6 + 3 + 4?x + 1 = x
? x = 20 + 8?x
? x - 20 =8?x
On squaring both sides and by solving the formed quadratics equation, we get x = 100

Let the cost of each apple,orange nd pear be Rs.x,y and z, respectively.Then,
2x + 3y + z = 62 ...(i)
5x + 6y + 4z = 20 ...(ii)
On subtracting Eq.(i)from Eq.(ii), we get
3x + 3y + 3z = 58
So, the cost of 3 apple,3 oranges and 3 pears is Rs. 58.

When p is wrong i.e., -b/a = (? + ?) is wrong but c/a =(??) is correct.
Then ?? = c/a = 2 x 6 = 12 ....(i)

Again, when q is wrong i.e.,c/a = (?.? ) is wrong but -b/a = ?+? is correct.
Then, -b/a = ?+?= 2 + (-9) = -7

So, the required correct quadratic equations is
x² -(? + ?)x + ?.?=0
? x² - (-7)x + 12 = 0
? x² + 7x + 12 = 0
and correct roots this equations are -3, -4.

The required greatest number is the HCF of 263-7, 935-7, 1383-7
i.e. 256, 928 and 1376 HCF = 32

If (x+2) is the HCF of x² + ax + b and x² + cx + d.
Then, (-2)² -2a +b = (-2)² -2c + d
b + 2c = 2a + d