A number consist of two digit whose sum is 10 If the digit of the number are reversed then the number decreased by 36. Which of the following is correct.

I. The number is divisible by a composite number.
II. The number is a multiple of a prime number.

By trial, we find that the smallest number consisting entirely of fives and exactly divisible by 13 is 555555. On dividing 555555 by 13, we get 42735 as quotient.
? Req. smallest number =42735.

Let the number be x.
44² < x < 45²
? 1936 < x < 2025 ...(i)

From equation (i), the required number will be any number between 1936 and 2025. Since one part of the number is the square of 6 means one factor is 36.

? L, C, M of 36 and 5 = 180
? Number will be multiple of 180 i,e.,
180 x 11 = 1980 the only value which satisfies the equation (i).

Let first number = P
And second number = Q
According to the question
P - Q = 10 ....(i)
And 3P - 4Q = 18...(ii)

on multiply eq (i) by 3 and then subtract eq (ii) from it
3P - 3Q = 54
3P - 4Q = 18
Q = 36

On putting the value of Q in the eq (i)
P = 18 + Q = 18 +36
? P = 54
Required sum = P + Q = 54 +36 =90

Number of trees that can be planted on one side of road = (1760 / 20) + 1
= 88 + 1
=89

? Trees on the both side = 2 x 89=178

Let the third number = N
Then the first number = 3N
And second number = 3N/2
According to the question [N + 3N + 3N/2] / 3=154
? (2N +6N +3N) / 6 = 154
? N = (154 x 6) / 11 = 84
? Required difference = 3N - N = 2N
= 2 x 84 = 168

The only one even prime number is 2 and this is not in the form of 3k + 1.
Thus any prime number of the form 3k + 1 is an odd number.
This means 3k must be even number which implies that k must be even number also.
Let k = 2m.
Then a prime of the form 3k + 1 is of the form 3(2m) + 1 = 6m+1.
Every prime number of form 3k + 1 can be represented in the form 6m + 1 only, when k is even.

Let the two digit number be 10x + y.
Now, according to the question,
x + y = 10............(i)
if digits reversed , number get decreased by 36,
and ( x +10y) = ( 10x + y ) - 36
? - 9y + 9x = 36
? x - y = 4 ..................... (ii)
On adding Eqs. (i) and (ii), we get
2x = 14
? x = 7
? x = 7 and y = 3
? Required number is 73.
So,the number is a multiple of a prime number.

Let the fraction be x / y.
According to the question,
( x+2)/(y+1) = 1/2
? 2x + 4 = y+1
? 2x - y = -3 .........................(i)
Again according to the question,
(x+1)/(y-2) = 3/5
? 5x +5 = 3y - 6
? 5x - 3y = -11..........................(ii)
on multiplying Eq. (i) by 3 and then subtracting from Eq. (ii) we get,
multiplying Eq. (i) by 3
6x - 3y = -9
subtracting from Eq. (ii)
5x- 3y - ( 6x - 3y ) = -11 - (-9)
5x- 3y - 6x + 3y = -11 + 9
- x = - 2
x= 2
On putting the value of x in Eq. (i) we get
2x- y = -3
2 x 2 - y = -3
? 2 x 2 -y = -3
? -y = -3 - 4
? y = 7
? Original fraction = x/y =2/7

Method 1 to solve this question.
Let p be the first odd number.
Then next odd number is p + 2.
According to question,
Sum of five consecutive odd numbers 175
? p + p + 2 + p + 4 + p + 6 + p + 8 = 175
? 5p + 20 = 175
? p = (175 - 20)/5 = 31
According to question,
Sum of the second largest and square of smallest one = (31 + 6) + (31)²
Sum of the second largest and square of smallest one = 37 + 961 = 998

Method 2 to solve this question.
Let p be the even number.
Then next odd number is p + 1. so on p + 3 , p + 5................ P + 9
According to question,
Sum of five consecutive odd numbers 175
? p + 1 + p + 3 + p + 5 + p + 7 + p + 9 = 175
? 5p + 25 = 175
? p = (175 - 25)/5
? p = (150)/5
? p = 30
First odd number = p + 1 = 30 + 1 = 31
According to question,
Sum of the second largest and square of smallest one = (30 + 7) + (30 + 1)²
Sum of the second largest and square of smallest one = 37 + 961 = 998

Let the third numbers = p
Then, the first number = 3p
and second number = 3p/2
According to question,
( p + 3p + 3p/2 )/3 = 154
? ( 2p + 6p + 3p )/6 = 154
? p = 154 x 6/11 = 84
? Required difference = 3p - p = 2p
= 2 x 84 = 168