MATHEMATIC
54
(ivThe taxi charges in a city consist of a fixed charge together with the charge for the
distance covered. For a distance of 10 km, the charge paid is 105 and for a
journey of 15 km, the charge paid is * 155. What are the fixed charges and the
charge per km? How much does a person have to pay for travelling a distance of
25 km?
9
A fraction becomes if 2 is added to both the numerator and the denominator
11
5
If, 3 is added to both the numerator and the denominator it becomes
6
fraction.
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Answers
The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 105 and for a journey of 15 km, the charge paid is Rs. 155. What are the fixed charges and the charge per kilometer? How much does a person have to pay for traveling a distance of 25 km?
A Fixed charge is Rs.5 and the charge per kilometer is Rs.10 For 25 km person have to pay Rs.255.
B Fixed charge is Rs.10
. and the charge per kilometer is Rs.5. For25 km person have to pay Rs135.
C Fixed charge is Rs.15
and the charge per kilometer is Rs. 5. For25 km person have to pay Rs.140.
D Fixed charge is
Rs.50. and the charge per kilometer is Rs20. For 25 km person have to pay Rs
50
Fixed charge is Rs.5 and the charge per kilometer is Rs.10For25 km person have to pay Rs.
255
.
Let fixed charge be Rs x and charge per km be Rs y.
⇒
x+
10y=
105...(1)
⇒
x+
15y=
155..(2)
.
Now From eq 2
⇒
15y=
155−
x
⇒
y=
15
155−x
...(3)
Substituting y from eq3 in eq1
⇒
x+
15
10(155−x)
=
105
⇒
15x+
1550−
10x=
1575
⇒
5x=
1575−
1550
⇒
5x=
25
⇒
x=
5
Substituting x in eq2
⇒
5+
15y=
155
⇒
15y=
155−
5
⇒
15y=
150
⇒
y=
10
Hence
x=
5
and
y=
10
Fixed charge is Rs.55 and the charge per kilometer is Rs.1010
For 2525 km person have to pay =
5+
10×
25=
255Rs.