Ramesh and Ganesh had some oranges initially. If Ramesh gave 5 oranges to Ganesh, then Ganesh will have thrice as many oranges as Ramesh. Instead of that, if Ganesh was to give 5 oranges to Ramesh, then they both will have the same number of oranges. Find the ratio of oranges that were distributed between Ramesh and Ganesh?

Let the distance covered by walking be x km.
Speed while walking = 2 km/h
Time spent on walking = x/2 h
Now, distance covered by runnig = (18 - x) km
Speed while running = 6 km/h
Time spent on running = (18 - x)/6 h
?x/2 + (18 -x)/6 = 5
? (3x + 18x - x)/6 = 5
? 2x + 18 = 30
? 2x = 12
? x = 6
? The distance covered by running = 18 - 6 = 12 km

Let the number be x. Then,
5x/2 - 2x/5 = 5208
? (25x - 4x)/10 = 5208
? 21x/10 = 5208
? 21x = 5208 x 10
? x = (5208 x 10)/21
? x = 2480
So, the original numbers was 2480.

Since, p(x) = ax³ + 4x² + 3x - 4 and q(x) = x³ -4x + a divided by x - 3, hence p(3) = q(3)
? a x 3³ + 4 x 3² + 3 x 3 - 4 = 3³ - 4 x 3 + a
? a = -1

When f(x) = x⁴ - 6x³ + 2x² + 3x + 1 is divided by (x - 2), then remainder can be calculated by putting x - 2 = 0
i.e. x = 2

Now, put x = 2 in f(x)
f(2) = (2)⁴ - 6(2)³ + 2(2)² + 3 x 2 + 1
= 16 - 48 + 8 + 6 + 1 = -17
? The remainder will be - 17.

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.

Let the digit in the unit's place be y and digit in the ten's place be x.
? Number = 10x + y
Then, x = 2y + 2
? x - 2y - 2 = 0 ...(i)

Number obtained by reversing the digits = 10y + x
Then, 10y + x = 5 + 3(x + y)
? 2x - 7y + 5 = 0 ...(ii)

On multiplying Eq.(i) by 2 and then subtracting it from Eq. (ii), we get y = 3
On substituting the value of y in Eq.(i), we get x = 8
So, the two-digit number is 83.

Let the number of correct answers marked be c and the number of wrong answers marked be w.
According to the question,
c + w = 60 ...(i)
Also, 3c - w = 120 ...(ii)
On adding Eqs.(i) and (ii), we get c = 45
So, the questions marked correct are 45.

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.