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Answers
Step-by-step explanation:
Situation -1:-
1) The fixed charge for 10 km = Rs. 75
The total charge for 15 km = Rs. 110
Difference in the distance = 15-10 = 5 km
Difference in the charge = 110-75 = Rs. 35
Charge for 5 km = Rs. 35
Charge for 1 km = 35/5 = Rs.7
Charge for each running km after 10 km = Rs.7
It can be written as
=> 110 = 75+35
=> 110 = 75+(15-10)×7
=> 110 = 7.5×10 +(15-10)×7
Refer situation -1
i)
The fixed charge of an auto rickshaw for 10 km= Rs. x
The charge per running km = Rs. y
Let the total distance be n
It can be written as
=> x + (n-10)y
Total Cost be R then
R = x+(n-10)y ------(1)
Where
Where n= total distance
ii) Total distance = 25 km
=>25 km = 10 km + 15 km
The cost for 25 km
=> 75+(25-10)(7)
=> 75+15(7)
=> 75+105
=>Rs. 180
Total charge for 25 km = Rs. 180
iii) Total distance = 50 km
=> 75+(50-10)(7)
=> 75+40(7)
=> 75+280
=>Rs. 355
Refer situation-2:-
Charge of first 8 km = Rs. 91
So let it be fixed
Charge of 14 km = Rs. 145
=> 14 km = 8 km + 6 km
The charge of 14 km
=> 91+charge of 6 km = 145
=> charge of 6 km = 145-91
=> Charge of 6 km = 54
Charge of 1 km = 54/6 = Rs. 9
It can be written as
145 = 91+(14-8) (9)
Let the fixed price for 8 km be Rs. x
Let the running price for each km be Rs. y
Let the total distance be n
Let the total cost be R
R = x + ( n -8) y ------(2)
I) Given total distance = 30 km
We have x = 91
n = 30 km
y = 9 km
=> Total cost = 91+(30-8)(9)
=> 91+(22)(9)
=> 91+198
=>Rs.289
c)
Total distance = 80 m
PY = 80 m
Distance covered by Pushpendra = 1/4th of PY
=> (1/4)×80
=> 80/4
=> 20 m
Distance covered by Pankaj = 1/5rd of PY
=> (1/5)×80
=> 80/5
=> 16 km
I) Coordinates of yellow flag = (3,20)
Coordinates of blue flag = (7,16)
ii)Distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=> √[(7-3)²+(20-16)²]
=> √(4²+4²)
=>√(16+16)
=>√32m
=> 4√2 m
iii)If Raman post the green flag in between the yellow and blue flag then it is at the mid point of the line segment joining the points
Mid point of the line segment joining the points (x1, y1) and (x2, y2) is
({x1+x2}/2 , {y1+y2}/2)
Let (x1, y1) = (3,20) => x1 = 3 and y1 = 20
Let (x2, y2)= (7,16)=> x2 = 7 and y2 = 16
On Substituting these values in the above formula then
=> ({3+7}/2 , {20+16}/2)
=> (10/2 , 36/2)
=> (5,18)
iv)Ratio = 1:2
m1:m2 = 1:2 => m1 = 1 and m2 = 2
Let (x1, y1) = (3,20) => x1 = 3 and y1 = 20
Let (x2, y2)= (7,16)=> x2 = 7 and y2 = 16
If Raman post the flag in the linesegment joining the points in the ratio is 1:2 then
By Section formula ,
({m1x2+m2x1}/(m1+m2),{m1y2+m2y1}/(m1+m2))
=({(1)(7)+(2)(3)}/(1+2), {(1)(16)+(2)(20)}/(1+2))
=> ({7+6}/3 , {16+40}/3)
=> (13/3,56/3)
Used formulae:-
- Mid point of the line segment joining the points (x1, y1) and (x2, y2) is
- ({x1+x2}/2 , {y1+y2}/2)
- Section formula =
({m1x2+m2x1}/(m1+m2),{m1y2+m2y1}/(m1+m2))