Explanation
Solution 2: Coordinate Geometry
Problem: Triangle formed by X-axis, Y-axis, and 3x+4y=60. Find points P(a,b) inside where b is a multiple of a.
1. Conditions for Point P(a,b) strictly inside:
-
a > 0 aur b > 0 (Kyunki strictly inside hai).
-
3a + 4b < 60 (Line ke niche).
-
b=ka, jahan k ek positive integer hai (Kyunki b, a ka multiple hai).
2. Values check karein (b=ka ko equation mein rakhne par):
3a + 4(ka) < 60 \implies a(3 + 4k) < 60 \implies a < \frac{60}{3 + 4k}
Hume har k (1,2,3...) ke liye a ki possible integer values nikalni hain:
| k (Multiple) |
a<3+4k60 |
Possible a values |
Count |
| k=1 |
a < \frac{60}{7} \approx 8.57 |
1,2,3,4,5,6,7,8 |
8 |
| k=2 |
a < \frac{60}{11} \approx 5.45 |
1,2,3,4,5 |
5 |
| k=3 |
a < \frac{60}{15} = 4 |
1,2,3 (Strictly less than 4) |
3 |
| k=4 |
a < \frac{60}{19} \approx 3.15 |
1,2,3 |
3 |
| k=5 |
a < \frac{60}{23} \approx 2.6 |
1,2 |
2 |
| k=6 |
a < \frac{60}{27} \approx 2.22 |
1,2 |
2 |
| k=7 |
a < \frac{60}{31} \approx 1.93 |
1 |
1 |
| ... |
... |
... |
... |
| k=14 |
a < \frac{60}{59} \approx 1.01 |
1 |
1 |
| k=15 |
a < \frac{60}{63} < 1 |
Koi nahi |
0 |
3. Total Points:
-
For k=7 to k=14, a hamesha 1 hoga (Total 8 points).
-
Ab sabko jodte hain: 8(k=1)+5(k=2)+3(k=3)+3(k=4)+2(k=5)+2(k=6)+8(k=7 to 14).
-
Total =8+5+3+3+2+2+8= 31.
Answer: Aise kul 31 points hain.
