There are 200 question in a 3 hour examination. Among 200 question, 50 are Mathematics, 100 are from GK and 50 are from Sciences. Ram spent twice as much time on each Mathematics question for each other question. How many minute did he spend on Mathematics questions?

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 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

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

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