## Linear Equation

#### Linear Equation

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

1. 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

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

1. 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
Solve the equation and find the answer.

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
Solve the equation and find the answer.

##### Correct Option: C

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

1. The present ages of Vikas and Vishal are in the ratio 15:8. After ten years , their ages will be in the ratio 5:3. Find their present ages

1. 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
solve the equation and get the answer.

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
solve the equation and get the answer.

##### Correct Option: A

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

1. The auto-rickshaw fare consists of a fixed charge together with the charge for the distance covered. For a journey of 10 km, the charge paid is 85 and for a journey of 15 km, the charge paid is 120. The fare for a journey of 25 km will be

1. 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

1. The numbers are such that the square of one is 224 less than 8 times the square of the other. If the numbers be in the ratio 3 : 4, the numbers are

1. 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