Let use assume the fixed charge = ₹ a
and charge for 1 km = ₹ 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)
Solve the equation (1) and (2) to get the answer.

Correct Option: B

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
Solve the equation and get the answer.

Correct Option: B

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
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)
Solve the equation and get the answer.

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
Solve the equation and get the answer.

Correct Option: B

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 - 12D = 360D
⇒ - 3D + 1440 - 12D = 360D - 360 x D
⇒ - 3D + 1440 - 12D = 0
⇒ 3D - 1440 + 12D = 0
⇒ D - 480 + 4D = 0
⇒ D + 4D - 480 = 0
⇒ D + 24D - 20D - 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(d² + 4d)
⇒ 1440 = 3d² + 12d
⇒ 3d² + 12d – 1440 = 0
⇒ d² + 4d – 480 = 0
⇒ d² + 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
Solve the equation and find the answer.

Correct Option: C

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
According to question,
Two-fifths of one -fourth of three -seventh of a number = 15
2/5 x 1/4 x 3/7 x P = 15
Solve the above equation.

Correct Option: D

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