# Solution

I think there are seven possible answers.

Let the number of ballpoint pens, pencils, and erasers I buy from the dealer be B, P, and E. In all, there are 100 items. Therefore B + P + E = 100. Let this be equation 1.

Each eraser costs 1/4 of a rupee, each pencil costs Re. 1, and each ballpoint pen costs Rs. 10. The total money spent is Rs. 100. Therefore, (1/4)E + P + 10B = 100. Simplifying, we have E + 4P + 40B = 400. Let this be equation 2.

Subtract 1 from 2, we get: 3P + 39B = 300.

I have to buy at least one ballpoint pen, one pencil, and one eraser. So, B, P, and E cannot be equal to zero.

So, let's start with B = 1. If B = 1, then 3P + 39 = 300. So, 3P = 261; therefore P = 87. Substituting these values in equation 1, we get E = 12.

If B = 2, 3P + 78 = 300. So, 3P = 222. Therefore, P = 74. Substituting these values in equation 1, we get E = 24.

You can continue till B = 7, but if you put B = 8, then 39B = 312. If you subtract this from 300, you get -12. But a negative answer is not valid in this case.

1. B = 1, P = 87, E = 12

2. B = 2, P = 74, E = 24

3. B = 3, P = 61, E = 36

4. B = 4, P = 48, E = 48

5. B = 5, P = 35, E = 60

6. B = 6, P = 22, E = 72

7. B = 7, P = 9, E = 84.

So, I have seven options from which to choose. I hope I got it right.

• Union Public Service Commission - www.upsc.gov.in
• IIT-Kharagpur - www.iitkgp.ac.in
• Indian Statistical Institute - www.isical.ac.in
• Indian Institute of Technology Madras - www.iitm.ac.in
• Indian Institute of Management, Ahmedabad - www.iimahd.ernet.in
• Indian Institute of Mass Commission - www.iimc.nic.in
• IIT Bombay - www.iitb.ac.in