X, Y and Z had have dinner together. The cost of the meal of Z was 20% more then that of Y and the cost of the meal of X was 5/6 as much as the cost of the meal of Z. If Y paid ? 100, then what was the total amount that all the three of them had paid?

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 number of students in examination halls P and Q is x and y, respectively

Then as per the first condition,
x - 10 = y + 10
? x - y = 20 ..............(i)
As per the second condition,
? x + 20 = 2( y - 20)
? x - 2y = - 60 ...............(ii)
On subtraction Eq. (ii) from Eq. (i), we get
- y + 2y = 20 + 60
? y = 80
Putting the value of y Eq. (i) we get,
x - 80= 20 ? x = 100
Hence, number of students in examination halls P and Q is 100 and 80, respectively.

Let total number of sweets distributed on children's day = x
According to the question,
x/250 - x/300 = 1
? ( 6x - 5x )/1500 = 1
? x = 1500
? Required number of sweets distributed on children's day = 1500

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

Let hundred's place digit = p
Then according to question,
unit's digit = 4 x hundred's place digit = 4p
and Ten's place digit = 3 x hundred's place digit = 3p
So Number = 100 x p + 10 x 3p + 1 x 4p= 134p

If the digit in the unit's place and the ten's places are interchanged according to question,
unit's digit = 3p
and Ten's place digit = 4p
Now Number = 100 x p + 10 x 4p + 1 x 3p = 143p
According to the question, after interchanging,
143p - 134p = 18
? 9p = 18
? p = 2
Original number = 134p = 134 x 2 = 268
25% of original number = 268 x 25/100 = 67

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