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

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

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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
 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

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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
 The autorickshaw 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

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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
 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

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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}
16r^{2} + 224 = 8 x 9r^{2}
16r^{2} + 224 = 72r^{2}
72r^{2}  16r^{2} = 224
56r^{2} = 224
r^{2} = 224/56
r^{2} = 4
r = 2
First number = 3r = 3 x 2 = 6
Second number = 4r = 4 x 2 = 8