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The additional ₹50 is for 5 km. Therefore, charge per kilometre = 50/5 = ₹10.
For 10 km, total charge is ₹105. Fixed charge = 105 - 10(10) = ₹5.
Total parts = 9. First income = 5/9 × 36000 = ₹20,000.
Let width = x and length = x + 5. Then 2(2x + 5) = 50, giving x = 10 and length = 15 m.
Let the fraction be x/y. Solving (x + 1)/(y + 1) = 2/3 and (x - 1)/(y - 1) = 1/2 gives x = 5 and y = 7.
x + y = 10 and (10x + y) - (10y + x) = 18, giving x - y = 2. Hence x = 6 and y = 4, so the number is 64.
Let the numbers of ₹50 and ₹100 notes be x and y. Then x + y = 25 and 50x + 100y = 2000. Solving gives x = 10.
Extra days = 7 - 3 = 4. Extra charge = 4 × 3 = ₹12. Fixed charge = ₹27 - ₹12 = ₹15.
The second equation must be twice the first: 4x + 6y = 10. Hence k = 6.
The second equation must be twice the first, so k = 2 × 12 = 24.
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