The sum of three consecutive multiples of 3 is 72. What is the largest number?

Method 1
Let us assume the ratio factor is x.
Therefor the present ages of Vikas and Vishal be 15x years and 8x years.
After 10 years
vikas's age = 15x + 10 and Vishal' age = 8x + 10
According to question,
(15x+10)/(8x+10) = 5/3
? 3(15x + 10) = 5(8x + 10)
? 45x + 30 = 40x + 50
? 5x =20
? x = 20/5
? x = 4
Therefore Present age of Vikas = 15x = 15 x 4 = 60 years
and Present age of Vishal = 8x = 8 x 4 = 32 years

Method 2
Let us assume the present age of Vikas = x years and Vishal's present year = y years
According to question,
Present age of Vikas/ Present age of Vishal = 15/8
? x/y = 15/8
? 8x = 15y
? 8x - 15y = 0
? x = 15y/8 .................................(1)
After 10 years Vikash age = x + 10 and vishal age = y + 10
Ratio of age after 10 years = 5/3
(x + 10)/(y + 10) = 5/3
? 3(x + 10) = 5(y + 10)
? 3x + 30 = 5y + 50
? 3x - 5y = 50 - 30
? 3x - 5y = 20 ................................(2)
Put the value of x from equation (1) in above equation (2).
? 45y/8 - 5y = 20
? 45y - 40y = 20 x 8
? 5y = 20 x 8
? y = 4 x 8 = 32
Put the value of Y in equation (1)
x = 15 x 32/8 = 15 x 4 = 60

Therefore Present age of Vikas = x = 60 years
and Present age of Vishal = y = 32 years

Let use assume the fixed charge = ? a
and charge for 1 km is ? = b
According to question,
for 10 KM journey charge paid = 85
a + 10 x b = 85
a + 10b = 85 .........................(1)
for 15 KM journey charge paid = 120
a + 15b = 120.........................(2)
Subtract the equation (1) from equation (2). we will get,
a + 15b - a - 10b = 120 - 85
5b = 35
b = 7
Put the value of b in equation (1). we will get
a + 10 x 7 = 85
a = 85 - 70
a = 15
Charges for 25 km = a + 25 x b
Put the value of a and b in above equation.
Charges for 25 km =15 + 25 x 7 = 15 + 175 = 190
Charges for 25 km =?190

Let us assume the first Odd number is a.
then 2nd odd number = a + 2
and 3rd odd number = a + 4
According to question,
Sum of three consecutive odd numbers is 20 more than the first of these numbers;
a + (a + 2) + (a + 4) = a + 20
? a + a + 2 + a + 4 - a = 20
? 2a + 6 = 20
? 2a + 6 = 20
? 2a = 20 - 6
? 2a = 14
? a = 7
The First odd number a = 7
The second odd number = a + 2 = 7 + 2 = 9
Second odd number is middle number = 9

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.

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

Let us assume father age is F and his son age is S.
According to question,
The sum of the ages of father and his son is 4 times the age of the son,
F + S = 4S
F = 3S.............. (1)
Average age of father and son is 28.
(F + S)/2 = 28
F + S = 56
Put the value of F from equation from (1),
3S + S = 56
4S = 56
S = 14 years
Age of Son = 14 years.

Let the two numbers be x and y .
According to question,
? xy = 192...................... (1)
x + y = 28........................(2)
As we know that,
(x - y)² = (x + y)² - 4xy
Put the value of xy and x + y from equation (1) and (2), we will get
(x - y)² = (28)² - 4 x 192
? (x - y)² = 784 - 786
? (x - y)² = 16
? x - y = 4......................(3)
Add the equation (2) and (3), we will get
x + y + x - y = 28 + 4
2x = 32
x = 16
Put the value of x in equation (2), we will get
16 + y = 26
y = 28 - 16
y = 12

So numbers are x = 16 and y = 12.
Smaller number is 12.

Let the middle number = x , then first number = x - 2 and last number = x + 2
(x - 2) + x + (x + 2) = 176/4 - 14
x - 2 + x + x + 2 = 44 - 14
3x = 30
x = 10
So middle number = x = 10

Let us assume the ratio factor is x.
Then the three numbers be 5x , 9x and 11x respectively.
Then according to question,
5x + 9x + 11x = 300
? 25x = 300
? x = 12
So, the second number is 9x = 9 x 12 =108

Let the number be x.
According to question,
x - 20 = 7x/12
? x - 7x/12 = 20
? (12x - 7x)/12 = 20
? 5x/12 = 20
? x = 20 x 12/5
? x = 4 x 12
? x = 48
Sum of the digits of 48 = 4 + 8 = 12