A two-digit number is such that the ten's digit exceeds twice the unit's digit by 2 and the number obtained by interchanging the digits is 5 more than three times the sum of the digits. Find the two-digits number?

Let the number of oranges Ramesh has be x and the number of oranges Ganesh has be y.

According to the first condition,
y + 5 = 3(x - 5)
? 3x - y + 10 = 0 ...(i)

According to the second condition,
x + 5 = y - 5
? x - y + 10 = 0 ...(ii)

On subtracting Eq. (ii) from Eq. (i), we get x = 15
On substituting the value of x in Eq.(ii), we get y = 25
? The ratio of oranges to be distributed Ramesh and Ganesh = 15:25 = 3:5

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 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.

Let the two numbers be x and y.
Then, x² + y² = 234 ...(i)
and (x - y)² = 144 ...(ii)
Now, (x - y)² = 144
? x² + y² - 2xy = 144
? 234 - 2xy = 144 [using Eq.(i)]
On substituting the value of x² + y² from Eq.(i) in Eq.(ii),
we get
2xy = 90
? xy = 45
So, the product of the two numbers is 45.