Chapter 2
Introduction to Linear Polynomials
CBSE Class 9
Mathematics Solutions
Educart Mathematics class Class 9 NCERT Exemplar cover
Question:

For how many km will the fare be ₹ 130 (Refer to Example 8 on NCERT page no. 23)?                       [Page No. 23]

Answer: Verified

For n≥2n \geq 2 km the fare is 15n−515n - 5. Set it equal to 130:

• 15n−5=130→15n=135→n=915n - 5 = 130 \rightarrow 15n = 135 \rightarrow n = 9.

So, the fare is ₹ 130 for a journey of 9 km.

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(All Q's of this Chapter solved)
More NCERT Questions
What is the cost of travelling 15 km? For how many km will the cost be ₹ 700 (Refer to Example 9 on NCERT page no. 24)? [Page No. 24] Differentiate between the graphs of the equations y = 3x + 1, and y = -3x + 1. [Page No. 33] Identify other points on the line by completing the following table. [Page No. 28] x 1 2 5 7 9 12 20 y 3 15 Can you guess what the numbers 20 and 150 in the equation y = 20x + 150 represent (Refer to Example 11 on NCERT page no. 26)? [Page No. 27]