Ram and Mohan are friends. Each has some money. If Ram gives ? 30 to Mohan, Then Mohan will have twice the money left with Ram. But if Mohan gives ? 10 to Ram, Then Ram will have thrice as much as is left with Mohan. How much money does each have?

Let the number of rice bowls be a, the number of juice bowls be b, and the number of meat bowls be c.
According to question,
a + b + c = 65........................(1)
The total number of guests = 2a
The total number of guests = 3b
The total number of guests = 4c
So the total number of guests will be same in the party.
2a = 3b = 4c..........................(2)
As per Equation (2)
b = 2a/3................................(3)
c = 2a/4 = a/2......................(4)
Now put the value of b and c from Equation (3), (4) in Equation (1),
a + 2a/3 + a/2 = 65
(6a + 4a + 3a)/6 = 65
13a = 65 x 6
a = 5 x 6 = 30
Put the value of a in equation (3) and (4) in order to get the value of b and c,
b = 2 x 30/3 = 2 x 10 = 20
c = 30/2 = 15

The Total number of Guests = 2a = 3b = 4c = 60

Let us assume the salary of driver be ? R
Then Tips of week = R x 5/4
According to question,
his income during one week = Salary + Tips
? Total Income during one week = R + (5R/4)
? Total Income during one week = (4R + 5R)/4 = 9R/4

Required Fraction = Tips in a week/Total Income in a week
? Required Fraction = 5R/4 / 9R/4
? Required Fraction = 5R/4 x 4/9R
? Required Fraction = 5/9

Let us assume the number is a.
Given that 7 ¹/3 = 22/3
According to question,
a x 1/2 + a x 1/5 = a x 1/3 + 22/3
? a/2 + a/5 = a/3 + 22/3
? a/2 + a/5 - a/3 = 22/3
? (15a + 6a - 10a)/30 = 22/3
? (15a + 6a - 10a) = 30 x 22/3
? 11a = 10 x 22
? a = 10 x 2
? a = 20

Let us assume the number is n.
According to question,
If 24 is subtracted, it becomes 4/7 of the number.
n - 24 = n x 4/7
? n - 4n/7 = 24
? (7n - 4n)/7 = 24
? 3n/7 = 24
? 3n = 24 x 7
? n = 24 x 7/3
? n = 8 x 7
? n = 56
Sum of the digits of the number = 5 + 6 = 11

Let us assume the first term is a and common difference is d.
Formula for n^th term of Arithmetic Progression = a + ( n - 1 ) x d
So first term t₁ = a + ( n - 1 ) x d
t₁ = a + (1 - 1) x d = a + 0 x d = a
Fifth term t₅ = a + ( n - 1 ) x d = a + (5 - 1) x d = a + 4d
According to question,
Sum of the first term and the fifth term = 10
t₁ + t₅ = 10
? a + a + 4d = 10
? 2a + 4d = 10
? a + 2d = 5.............................(1)

Formula of sum of n terms for Arithmetic Progression,
S_n = n/2[ 2a + ( n - 1 ) x d ]
Put the value of n = 10, Since total number of terms is 10 .
? S₁₀ = 10/2[ 2a + ( 10 - 1 ) x d ]
? S₁₀ = 5[ 2a + 9 x d ] = 10a + 45d
Again According to question,
The sum of all terms of the Arithmetic Progression, except for the 1st term = 99
S₁₀ - first term = 99
? 10a + 45d - a = 99
? 9a +45d = 99
? a + 5d = 11.................................(2)
Subtracts Equation (1) from Equation (2), we will get,
? a + 5d - a - 2d = 11 - 5
? 3d = 6
? d = 2
Put the value of d in equation (1), we will get
? a + 2 x 2 = 5
? a + 4 = 5
? a = 5 - 4
? a = 1
Since t_n = a + ( n - 1) x d
find the third term by putting the value of a , n and d. we will get,
t₃ = 1 + ( 3 - 1) x 2 = 1 + 2 x 2 = 1 + 4 = 5
So 3rd term of Arithmetic Progression is 5.

Let us assume the numbers are a and b.
According to question,
sum of two numbers = 25
a + b = 25......................(1)
difference of two numbers = 13
a - b = 13........................(2)
add the Equation (1) and (2)
a + b+ a - b = 25 + 13
? 2a = 38
? a = 19
Put the value of a in equation in (1)
19 + b = 25
? b = 25 - 19
? b = 6
Product of the numbers = a x b
put the value of a and b,
? Product of the numbers = 19 x 6
? Product of the numbers = 114

Let us assume the numbers are a and b.
According to question,
a² - b² = 256000 .......................(1)
and a + b = 1000 .........................(2)
On dividing the Equation (1) with Equation (2),
(a² - b²)/(a + b) = 256000/1000
(a² - b²)/(a + b) = 256
(a + b)(a - b)/(a + b) = 256
a - b = 256...............................(3)
Add the equation (2) and (3), we will get
a + b + a - b = 1000 + 256
2a = 1256
a = 628
Put the value of a in equation (2), we will get
658 + b = 1000
b = 1000 - 628
b = 372
So answer is 628 , 372.

Let us assume the number is P.
According the question,
three -seventh of a number = 3/7 x P
one -fourth of three -seventh of a number = 1/4 x 3/7 x P
Two-fifths of one -fourth of three -seventh of a number = 2/5 x 1/4 x 3/7 x P
? 2/5 x 1/4 x 3/7 x P = 15
? 2 x 1 x 3 x P = 15 x 7 x 4 x 5
? P = 15 x 7 x 4 x 5 / 2 x 3
? P = 5 x 7 x 2 x 5
? P = 350
? P /2 = 350/2
? P /2 = 175

Let us assume the ratio term is r.
then the number will be 3r and 4r.
According to question,
(4r)² + 224 = 8 x (3r)²
16r² + 224 = 8 x 9r²
16r² + 224 = 72r²
72r² - 16r² = 224
56r² = 224
r² = 224/56
r² = 4
r = 2

First number = 3r = 3 x 2 = 6
Second number = 4r = 4 x 2 = 8

Method 1 to solve this question.
Let us assume the numbers of days of tour before extending the tour are D and daily expenses is E.
According to question,
Total expenses on tour = per day expenses x total number of days
Total expenses on tour = E x D = ED
360 = ED
? ED = 360
? E = 360/D .................(1)
Again according to question,
After extending the tour by 4 days, Number of days of tour = D + 4
Expenses will be reduced by 3 rupees, then everyday expenses = E - 3
so total expenses = (D + 4) x (E - 3)
360 = (D + 4) x (E - 3)
(D + 4) x (E - 3) = 360 .............................................(2)
Put the value of E in Equation (2). we will get,
(D + 4) x (360/D - 3) = 360
? (D + 4) x ( (360 - 3D)/D ) = 360
? (D + 4) x ( (360 - 3D) ) = 360 x D
? (D + 4) x (360 - 3D) = 360 x D
After multiplication by algebra law,
? 360 x D - 3D x D + 4 x 360 - 4 x 3D = 360D
? 360 x D - 3D + 1440 - 12 D = 360 D
? - 3 D + 1440 - 12 D = 360 D - 360 x D
? - 3 D + 1440 - 12 D = 0
? 3 D - 1440 + 12 D = 0
? D - 480 + 4 D = 0
? D + 4 D - 480 = 0
? D + 24 D - 20 D - 480 = 0
? D( D + 24) - 20( D + 24) = 0
? ( D + 24) ( D - 20) = 0
Either ( D + 24) = 0 or ( D - 20) = 0
So D = - 24 or D = 20
But days cannot be negative so D = 20 days.

Method 2 to solve this question.
Let us assume the original day of tour is d days.
Given that his tour is extended for 4 days
Hence daily expenses per days = 360/(d + 4)
Therefore, According to question,
360/d - 360/(d + 4) = 3
? (360(d + 4) - 360d)/d x (d + 4) = 3
? (360(d + 4) - 360d)/d x (d + 4) = 3
? 360(d + 4) - 360d = 3d x (d + 4)
? 360d + 1440 ? 360d = 3(d ² + 4d)
? 1440 = 3d ² + 12d
? 3d ² + 12d ? 1440 = 0
? d ² + 4d ? 480 = 0
? d ² + 24d ? 20d ? 480 = 0
? d(d + 24) ? 20(d + 24) = 0
? (d + 24)(d ? 20) = 0
? (d + 24) = 0 or (d ? 20) = 0
? d = ?24 or d = 20
Since days cannot be negative, d = 20
Hence his original duration of the tour is 20 days.