There are two numbers such that the sum of twice the first and thrice the second is 18, while the sum of thrice the first and twice the second is 17. The larger of the two is?

Let the number be a and b .
Then a² + b² = 68 ....(i)
and (a - b)² = 36
Now, (a - b)² = 36 ....(ii)
? a² + b² - 2ab = 36
? 68 - 2ab = 36
? 2ab = 32
? ab = 16 .

Let the required fraction be p/q.
Then, (p +1) / (q +1) = 4
? p - 4q= 3 ....(i)

And, (p - 1) / (q -1) = 7
? p - 7q = -6 ....(ii)

Solving these equation we get p = 15, q=3

Let unit digit = x and ten's digit = y
3(x + y)= 10y + x .....(i)
10y + x + 45 = 10x + y ....(ii)

2x - 7y = 0.9x -9y = 45
or x - y = 5 .... (iii)

Solving these equations, we get
x=7, y=2
? Required number = 27

Let second number be 3N, then, first one is 6N and the third one is 2N.
From question 3N + 6N + 2N = 132
? 11N = 132
? N = 12

? Second number = 3N = 3 x 12 = 36

Let the value for ? be N
So N x 9 x 7 x 3 = a number with unit digit 1.
Clearly, the minimum value of N is 9.

Let the number be x, y, z
Then,
x + y = 45,
y + z = 55,
z + 3x = 30

Now, y = (45 - x) and
z = 55 - y
? z = 55 - (45 - x ) = 10 + x

? 10 + x + 3x = 90
? x = 20
So, third number = z
= 10 + x = 30

The number is of the form (5x + 3 ) where x is an integer

? [(5x + 3)²] / 5 = ( 25x² + 30x + 9) / 5
= 25x²/ 5 + 30x/5 + (5 +4)/5
? The remainder is 4.

According to the question
Divisor = 2/3 x dividend
and Divisor = 2 x remainder
or 2/3 x dividend = 2 x 5
? Dividend = 2 x 5 x 3 / 2 = 15

Number of one digit pages from
1 to 9 = 9
Number of two digit pages from
10 to 99 = 90
Number of three digit pages from
100 to 200 = 101
? Total number of required figures
= 9 x 1 + 90 x 2 + 101 x 3 = 492

Let 1/2 of the no. = 10x + y
and the no. = 10v + w
From the given conditions,
w= x and v = y-1
Thus the no. = 10 (y-1) + x
? 2(10x + y ) = 10 (y-1) + x
? 8y - 19x = 10 ...(i)
v + w = 7
? y-1 + x = 7
? x + y = 8
Solving equations (i) and (ii) , we get
x = 2 and y = 6

? From equations (A)
Number = 10 (y - 1) + x = 52.