Linear Equation
- The Fourth term of an Arithmetic Progression is 37 and the Sixth term is 12 more than the Fourth term. What is the sum of the Second and Eight terms?
-
View Hint View Answer Discuss in Forum
Let us assume the first number is a and common difference is d.
According to question,
4th term of A.P = 37
a + ( n - 1 ) x d = 37
Put the value of a , n and d, we will get,
a + (4 - 1 ) x d = 37
a + 3d = 37..................(1)
sixth term is 12 more than the fourth term,
6th term = 12 + 4th term
a + ( n - 1 ) x d = 12 + 37
a + ( 6- 1 ) x d = 39
a + 5d = 39................(2)
Solve the equation and get the answer.Correct Option: C
Let us assume the first number is a and common difference is d.
According to question,
4th term of A.P = 37
a + ( n - 1 ) x d = 37
Put the value of a , n and d, we will get,
a + (4 - 1 ) x d = 37
a + 3d = 37..................(1)
sixth term is 12 more than the fourth term,
6th term = 12 + 4th term
a + ( n - 1 ) x d = 12 + 37
a + ( 6- 1 ) x d = 39
a + 5d = 39................(2)
subtract the equation (1) from (2)
a + 5d - a - 3d = 39 - 37
5d - 3d= 2
2d = 2
d = 1
Put the value of d in equation (1), we will get
a + 3 x 1 = 37
a = 37 - 3
a = 34
Second term = a + (n - 1) x d = 34 + (2 - 1) x 1 = 34 + 1 = 35
Six term = a + (n - 1) x d = 34 + (6 - 1) x 1 = 34 + 5 = 39
Sum of Second and Six term = 35 + 39 = 74
Sum of Second and Six term = 74
Answer is 74.
- The sum of two numbers is 25 and their difference is 13. Find their product.
-
View Hint View Answer Discuss in Forum
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)
Solve the above equation.Correct Option: B
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
- Ram and Mohan are friends. Each has some money. If Ram gives ₹ 30 to Mohan, Then Mohan will have twice the money left with Ram. But if Mohan gives ₹ 10 to Ram, Then Ram will have thrice as much as is left with Mohan. How much money does each have?
-
View Hint View Answer Discuss in Forum
Let us assume Ram has R rupees and Mohan has M rupees.
According to question,
If Ram gives 30 rupees to Mohan, then
Ram has left money = R - 30 and Mohan has money = M + 30
Again According to question,
if Mohan gives 10 rupees to Ram, then
Mohan has left the money = M - 10 and Ram has the money = R + 10
Solve the above equation.Correct Option: A
Let us assume Ram has R rupees and Mohan has M rupees.
According to question,
If Ram gives 30 rupees to Mohan, then
Ram has left money = R - 30 and Mohan has money = M + 30
Then Mohan will have twice the money left with Ram,
M + 30 = 2(R - 30)
M + 30 = 2R - 60
2R - M = 90................................(1)
Again According to question,
if Mohan gives 10 rupees to Ram, then
Mohan has left the money = M - 10 and Ram has the money = R + 10
Then According to question,
Ram will have thrice as much as is left with Mohan,
R + 10 = 3 (M - 10 )
⇒ R + 10 = 3M - 30
⇒ 3M - R = 10 + 30
⇒ 3M - R = 40..................................(2)
After Multiplying 2 with Equation (2) , add with the equation (1),
6M - 2R + 2R - M = 80 + 90
⇒ 6M - M = 170
⇒ 5M = 170
⇒ M = 170/5
⇒ M = 34
Put the value of M in equation (1), we will get
⇒ 2R - 34 = 90
⇒ 2R = 90 + 34
⇒ 2R = 124
⇒ R = 124/2
⇒ R = 62
- In the certain party, there was a bowl of rice for every two guests , a bowl of juice for every three of them and a bowl of meat for every four of them. If there were all 65 bowls of food , then how many guests were there in the party ?
-
View Hint View Answer Discuss in Forum
Let the number of rice bowls be a, the number of juice bowls be b, and the number of meat bowls be c.
According to question,
a + b + c = 65........................(1)
The total number of guests = 2a
The total number of guests = 3b
The total number of guests = 4c
So The total number of guests will be same.
2a = 3b = 4c..........................(2)
Solve the equation with the help of (1) and (2).Correct Option: C
Let the number of rice bowls be a, the number of juice bowls be b, and the number of meat bowls be c.
According to question,
a + b + c = 65........................(1)
The total number of guests = 2a
The total number of guests = 3b
The total number of guests = 4c
So the total number of guests will be same in the party.
2a = 3b = 4c..........................(2)
As per Equation (2)
b = 2a/3................................(3)
c = 2a/4 = a/2......................(4)
Now put the value of b and c from Equation (3), (4) in Equation (1),
a + 2a/3 + a/2 = 65
(6a + 4a + 3a)/6 = 65
13a = 65 x 6
a = 5 x 6 = 30
Put the value of a in equation (3) and (4) in order to get the value of b and c,
b = 2 x 30/3 = 2 x 10 = 20
c = 30/2 = 15
The Total number of Guests = 2a = 3b = 4c = 60
- A driver's income consists of his salary and tips. during one week his tips were 5/4 of his salary. what fraction of his income came from tips?
-
View Hint View Answer Discuss in Forum
Let us assume the salary of driver be ₹ R
Then his income during one week = Salary + Tips
Then his income during one week = R + (5R/4)Correct Option: B
Let us assume the salary of driver be ₹ R
Then Tips of week = R x 5/4
According to question,
his income during one week = Salary + Tips
⇒ Total Income during one week = R + (5R/4)
⇒ Total Income during one week = (4R + 5R)/4 = 9R/4
Required Fraction = Tips in a week/Total Income in a week
⇒ Required Fraction = 5R/4 / 9R/4
⇒ Required Fraction = 5R/4 x 4/9R
⇒ Required Fraction = 5/9
