In a three-digit number, the digit in the unit's place is four times the digit in the hundred's place. If the digit in the unit's place and the ten's places are interchanged, the new number so formed is 18 more than the original number. If the digit in the hundred's place is one third of the digit in the ten's place, then what is 25% of the original number?

Let total number of candidates in Exam = a
Number of candidates answered 5 questions = a x 5% = a x 5/100 = 5a/100 = a/20
Number of candidates answered not any questions = a x 5% = a x 5/100 = 5a/100 = a/20
? Remaining students = a - ( a/20 + a/20) = a - ( 2a/20 ) = a - ( a/10 ) = (10a - a)/10 = 9a/10
Number of candidates answered only 1 question = ( 9a/10 ) x 25% = ( 9a/10 ) x 25/100 = 9a/40
Number of candidates answered 4 questions = ( 9a/10 ) x 20% = ( 9a/10 ) x 20/100 = 9a/50
Given number of candidates awarded either 2 questions or 3 questions = 396
? a - ( a/20+ a/20 + 9a/40 + 9a/50 ) = 396
? a - ( a/10 + 9a/40 + 9a/50 ) = 396
? a - ( ( a x 20 + 9a x 5 + 9a x 4 )/200) = 396
? a - ( ( 20a + 45a + 36a )/200) = 396
? a - ( 101a/200) = 396
? ( 200a - 101a)/200 = 396
? ( 99a)/200 = 396
? a = 396 x 200/99
? a = 4 x 200 = 800
? a = 800
Hence, number of candidates = 800

Given that,
The cost of meal of Y = ? 100
Now, according to the question,
The cost of the meal of Z = 20% more than that of Y
The cost of the meal of Z = (100 + 100 x 20 %) = (100 + 100 x 20/100 ) = (100 + 20) = ? 120
and the cost of the meal of X = 5/6 as much as the cost of the meal of Z = 5/6 x 120 = ? 100
? Total amount that all the three has to be paid = 100 + 120 + 100
Total amount that all the three has to be paid = ? 320

let the fixed charges = ? p for first 5 km
and the additional charges = ? q per km
according to the question
p + 5q = 350........... (1)
p + 20q = 800...........(2)
on subtracting Eq. 1 from Eq. 2 we get
15q = 450
q = 30
On putting the value of q in Eq. (1) we get
p = 200
? charge for a distance of 30 km = p + 25q
= 200 + 30 x 25 = ? 950

Let Ram spends p minute on each Mathematics question.
According to the question.
50 x p + 100 x p/2 + 50 x p/2 = 3 x 60
p(50+ 50 + 25) = 180
? p = 180/125
? Required time = 50 x 180/125
? Required time = 2 x 180/5
? Required time = 2 x 36 = 72 min

Method 2
Let Ram spends a minute on each sciences and Gk questions.
According to the question,
50 x 2a + 150 x a = 3 x 60
100a + 150a = 180
250a = 180
a = 180/250
So total time spend on Mathematics question = 2a x 50 question
So total time spend on Mathematics question = 2 x 50 x 180/250 = 72 minutes

Let the two number are a and b
According to the question
sum of squares of two numbers = 97
i.e.., a ² + b ² = 97 .....................(i)
and square of their difference = 25
i.e., (a - b) ² = 25 .................(ii)
? a - b = 5...............(iii)
from Eq. (ii)
a² + b² - 2ab = 25
(a² + b²) - 2ab = 25
Now put the value of a ² + b ² from Eq. (i)
? 97 - 2ab = 25
? 2ab = 72
? ab = 36.................(iv)
Now, we have
(a + b)² = ( a² + b²) + 2ab
Now put the value of a ² + b ² from Eq. (i) and ab from Eq. (iv)
(a + b)² = 97 + 72 = 169
a + b = 13.............(v)
Now add the Eqs. (iii) and (v), we get
a - b + a + b = 5 +13
2a = 18
? a = 9
now put the value of a in Eq. (v)
a + b = 13
9 + b = 13
b = 13 - 9
b = 4
? Product of both the numbers = ab = 9 x 4 = 36

Let ten's place digit be a and unit's place digit be a²
Original number = 10 x a + 1 x a² = 10a + a²
The number formed by interchanging the digits,
New number = 10 x a ² + 1 x a = 10a ² + a
According to the question
(10a² + a) - ( 10a + a²) = 54
? 10a² + a - 10a - a ² = 54
? 9a² - 9a = 54
? 9( a² - a) = 54
? ( a² - a) = 54/9
? ( a² - a) = 6
? a² - a - 6 = 0
? a² - 3a + 2a - 6 = 0
? a (a - 3) + 2 (a - 3) = 0
? (a - 3) (a + 2) = 0
? a = 3, - 2
? Ten,s digit = a = 3
Unit's digit = a² = 3² = 9
Original number = 39
? Required number = 39 x 40/100 = 15.6

Number of trees that can be planted on one side of road = (Length of road/gap between two trees ) + 1
Number of trees that can be planted on one side of road = ( 1760/20 )+ 1
Number of trees that can be planted on one side of road = 88 + 1 = 89
? Threes on the both sides
= 2 x 89 = 178

Let the number be P.
According to the question,
P - P/7 = 180
? 6P/7 = 180
? P = 180 x 7/6 = 210

Divisor = 5 x 46 = 230
Also, 10 x Quotient = 230
? Quotient = 23
We know that
Dividend = (Divisor x Quotient ) + Remainder
? Dividend = (230 x 23) + 46
? Dividend = 5290 + 46 = 5336

Let two consecutive odd number be (A + 1) and (A + 3).
According to the question.
(A + 1) (A + 3) = 6723
? A² + 3A + A + 3 = 6723
? A² + 4A + 3 - 6723 = 0
? A² + 4A - 6720 = 0
? A² + 84A - 80A - 6720 = 0
? A(A + 84) - 80 (A + 84) = 0
? (A - 80) (A + 84) = 0
? A = 80, (A = ? - 84)
Hence, the greater number = 80 + 3 = 83