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___________
1)Let present age of Aftab be x
And, present age of daughter is represented byy
Then Seven years ago,
Age of Aftab = x -7
Age of daughter = y-7
According to the question,
(x - 7) = 7 (y – 7 )
x – 7 = 7 y – 49
x- 7y = - 49 + 7
x – 7y = - 42 …(i)
x = 7y – 42
Putting y = 5, 6 and 7, we get
x = 7 × 5 - 42 = 35 - 42 = - 7
x = 7 × 6 - 42 = 42 – 42 = 0
x = 7 × 7 – 42 = 49 – 42 = 7
x-707y567
Three years from now ,
Age of Aftab = x +3
Age of daughter = y +3
According to the question,
(x + 3) = 3 (y + 3)
x + 3 = 3y + 9
x -3y = 9-3
x -3y = 6 …(ii)
x = 3y + 6
Putting, y = -2,-1 and 0, we get
x = 3 × - 2 + 6 = -6 + 6 =0
x = 3 × - 1 + 6 = -3 + 6 = 3
x = 3 × 0 + 6 = 0 + 6 = 6
x036y-2-10
Algebraic representation
From equation (i) and (ii)
x – 7y = – 42 …(i)
x - 3y = 6 …(ii)
2)Let the number of boys = x
Number of girls = 4 more than the number of boys
= x + 4
According to the problem given ,
Total students participated in quiz = 10
Number of boys + Number of girls = 10
x + x + 4 = 10
2x + 4 = 10
2x = 10 - 4
2x = 6
x = 6 / 2
6)Let x be the fixed charge and y be the additional charge per km,
Thus, the charge for 10 km,
x + 10 y,
And, the charge for 15 km,
x + 15 y
According to the question,
x + 10 y = 105 -------(1)
x + 15 y = 155 --------(2),
Equation (2) - Equation (1),
5 y = 50
⇒ y = 10,
From Equation (1),
We get, x = 5
Hence, the charges for 25 km = x + 25 y = 5 + 25 × 10 = 5 + 250 = 255 rupees.
x = 3
Therefore
Number of boys = x = 3
Number of girls = x + 4
= 3 + 4
= 7
3)Let the two numbers be a and b
The difference between two numbers is given as 26
x – y = 26 …(1)
Given one number is three times the other, hence
x = 3y …(2)
Substitute (2) in (1), the equation becomes
3y – y = 26
2y = 26
y = 13 …(3)
Substitute (3) in (2)
x = 3(13)
x = 39
The two numbers are 13 and 39.
4)let the two angle be /x and /y
according to given condition-
/x+/y = 180 ----(1)
/x = /y + 18 -----(2)
substituting the value of (2) in (1)
(1) - /x+/y = 180
- (/y+18)+/y = 180
- 2/y+ 18 = 180
-2/y = 162
-/y= 81
substituting the value of y in (2)
/x=/y+18
/x=81+18
/x=99
Click to let oth
5)Let cost of each bat = Rs x
Cost of each ball = Rs y
Given that coach of a cricket team buys 7 bats and 6 balls for Rs 3800.
So that 7x + 6y = 3800
6y = 3800 – 7x
Divide by 6 we get
y = (3800 – 7x) /6 … (1)
Given that she buys 3 bats and 5 balls for Rs 1750.so that
3x + 5y = 1750
Plug the value of y
3x + 5 ((3800 – 7x) /6) = 1750
Multiply by 6 we get
18 x + 19000 – 35 x = 10500
-17x =10500 - 19000
-17x = -8500
x = - 8500 / - 17
x = 500
Plug this value in equation first we get
y = ( 3800 – 7 * 500) / 6
y = 300/6
y = 50
Hence cost of each bat = Rs 500 and cost of each balls is Rs 50
sorry mate
i can provide only these answers
there are 23 questions AND all answer cannot fit
answer next time ok
Answer:
plz do mark my answer as brainliest
i have written all verified answers to your questions
___________
1)Let present age of Aftab be x
And, present age of daughter is represented byy
Then Seven years ago,
Age of Aftab = x -7
Age of daughter = y-7
According to the question,
(x - 7) = 7 (y – 7 )
x – 7 = 7 y – 49
x- 7y = - 49 + 7
x – 7y = - 42 …(i)
x = 7y – 42
Putting y = 5, 6 and 7, we get
x = 7 × 5 - 42 = 35 - 42 = - 7
x = 7 × 6 - 42 = 42 – 42 = 0
x = 7 × 7 – 42 = 49 – 42 = 7
x-707y567
Three years from now ,
Age of Aftab = x +3
Age of daughter = y +3
According to the question,
(x + 3) = 3 (y + 3)
x + 3 = 3y + 9
x -3y = 9-3
x -3y = 6 …(ii)
x = 3y + 6
Putting, y = -2,-1 and 0, we get
x = 3 × - 2 + 6 = -6 + 6 =0
x = 3 × - 1 + 6 = -3 + 6 = 3
x = 3 × 0 + 6 = 0 + 6 = 6
x036y-2-10
Algebraic representation
From equation (i) and (ii)
x – 7y = – 42 …(i)
x - 3y = 6 …(ii)
2)Let the number of boys = x
Number of girls = 4 more than the number of boys
= x + 4
According to the problem given ,
Total students participated in quiz = 10
Number of boys + Number of girls = 10
x + x + 4 = 10
2x + 4 = 10
2x = 10 - 4
2x = 6
x = 6 / 2
6)Let x be the fixed charge and y be the additional charge per km,
Thus, the charge for 10 km,
x + 10 y,
And, the charge for 15 km,
x + 15 y
According to the question,
x + 10 y = 105 -------(1)
x + 15 y = 155 --------(2),
Equation (2) - Equation (1),
5 y = 50
⇒ y = 10,
From Equation (1),
We get, x = 5
Hence, the charges for 25 km = x + 25 y = 5 + 25 × 10 = 5 + 250 = 255 rupees.
x = 3
Therefore
Number of boys = x = 3
Number of girls = x + 4
= 3 + 4
= 7
3)Let the two numbers be a and b
The difference between two numbers is given as 26
x – y = 26 …(1)
Given one number is three times the other, hence
x = 3y …(2)
Substitute (2) in (1), the equation becomes
3y – y = 26
2y = 26
y = 13 …(3)
Substitute (3) in (2)
x = 3(13)
x = 39
The two numbers are 13 and 39.
4)let the two angle be /x and /y
according to given condition-
/x+/y = 180 ----(1)
/x = /y + 18 -----(2)
substituting the value of (2) in (1)
(1) - /x+/y = 180
- (/y+18)+/y = 180
- 2/y+ 18 = 180
-2/y = 162
-/y= 81
substituting the value of y in (2)
/x=/y+18
/x=81+18
/x=99
Click to let oth
5)Let cost of each bat = Rs x
Cost of each ball = Rs y
Given that coach of a cricket team buys 7 bats and 6 balls for Rs 3800.
So that 7x + 6y = 3800
6y = 3800 – 7x
Divide by 6 we get
y = (3800 – 7x) /6 … (1)
Given that she buys 3 bats and 5 balls for Rs 1750.so that
3x + 5y = 1750
Plug the value of y
3x + 5 ((3800 – 7x) /6) = 1750
Multiply by 6 we get
18 x + 19000 – 35 x = 10500
-17x =10500 - 19000
-17x = -8500
x = - 8500 / - 17
x = 500
Plug this value in equation first we get
y = ( 3800 – 7 * 500) / 6
y = 300/6
y = 50
Hence cost of each bat = Rs 500 and cost of each balls is Rs 50
sorry mate
i can provide only these answers
there are 23 questions AND all answer cannot fit
answer next time ok
Step-by-step explanation: