SOLUTION: The peanut pub has pecans that sell for $7.50 per kilogram and almonds that sell for $8.75. The owner wants to make a 10-kg mixture of these two varieties, which will sell for $8.0

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: The peanut pub has pecans that sell for $7.50 per kilogram and almonds that sell for $8.75. The owner wants to make a 10-kg mixture of these two varieties, which will sell for $8.0      Log On


   



Question 688262: The peanut pub has pecans that sell for $7.50 per kilogram and almonds that sell for $8.75. The owner wants to make a 10-kg mixture of these two varieties, which will sell for $8.00 per kg. How many kg of each variety will he need to use to produce the desired mixture. Thank you for the help in advance.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +a+ = kg of pecans needed
Let +b+ = kg of almonds needed
given:
(1) +a+%2B+b+=+10+
(2) +%28+7.5a+%2B+8.75b+%29+%2F+10+=+8+
----------------------------
(2) +7.5a+%2B+8.75b+=+80+
(2) +750a+%2B+875b+=+8000+
Multiply both sides of (1) by +750+
and subtract (1) from (2)
(2) +750a+%2B+875b+=+8000+
(1) +-750a+-+750b+=+-7500+
+125b+=+500+
+b+=+4+
and, since
(1) +a+%2B+b+=+10+
(1) +a+%2B+4+=+10+
(1) +a+=+6+
6 kg of pecans are needed
4 kg of almonds are needed
check:
(2) +%28+7.5a+%2B+8.75b+%29+%2F+10+=+8+
(2) +%28+7.5%2A6+%2B+8.75%2A4+%29+%2F+10+=+8+
(2) ++45+%2B+35+=+80+
(2) +80+=+80+
OK