There are two examination halls P and Q. If 10 students shifted P to Q, then the number of students will be equal in both the examination halls. If 20 students shifted from Q to P, then the students of P would be doubled to the students of Q. The number of students would be in P and Q respectively are?

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 first number = p
and second number = q
According to the question,
p - q = 18 ................................(i)
and 3p - 4q = 18.....................(ii)
On multiplying Eq .(i) by 3 and subtracting Eq. (ii) from it,we get
3p - 3q = 54
3p - 4q = 18
q = 36
On putting the value of q in Eq, (i) we get
p = 18 + q
? p = 18 + 36 = 54
? Required sum = p + q = 54 + 36 = 90

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

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 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

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 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 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

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 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