The sum of three consecutive odd numbers is 20 more than the first of these numbers.What is the middle number?

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 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 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 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 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

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

Method 1
Let us assume the number be 3p , 3(p+1) and 3(p+2)
According to question,
3p + 3(p+1) + 3(p+2) = 72
? 3p + 3p + 3 +3p + 6 = 72
? 9p +9 = 72
? 9p = 72 - 9
? 9p = 63
? p = 63/9 = 7
? Largest number = 3(p + 2)
Put the value of p in above equation.
? Largest number = 3 x ( 7 + 2 )
? Largest number = 3 x 9
? Largest number = 27

Method 2
Let us assume the numbers P , P + 3 ,P+ 6
According to Question,
sum of three numbers = 72
P + P + 3 + P + 6 = 72
? 3P + 9 = 72
? 3P = 72 - 9
? 3P = 72 - 9
? P = 63/3
? P = 21
So largest Number = P + 6 = 21 + 6 = 27

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.