Question
Answer the following
a. A firm produces a good in two qualities, A and B. For the comingyear, the estimated output of A is 50% higher than that of B. Theprofit per unit sold of the two qualities is $300 for Aand $200 forB. If the profit target is $13,000 over the next year, how much ofeach of the two qualities must be produceds.
b. At the beginning of the year a person had a total of $10,000 intwo accounts. The interest rates were 5% and 7%, respectively. Ifthe person has made no transfers during the year, and has earned atotal of $676 in interest, what was the initial balance in each ofthe two accounts.
c. Five tables and 20 chairs cost $1800, whereas 2 tables and 3chairs cost $420. what is the price of each table and eachchair?
d. Find two number whose sum is 52 and whose difference is 26.
Explanation / Answer
part a: output: if output of B=1 unit output of A=1.5 unit(50% greater) it can also be the case A=B+0.5B=1.5B profit: due to A=$300*A(units) due to B=$200*B(units) profit target=$13000 so $13000=$300*A+$200*B $13000=$300*(1.5B)+$200*B 13000=450B+200B 13000=(450+200)B=650B B=13000/650=20 and A=1.5*20=30 ans: 30 units of A and 20 units of B. PART B: let amount in one account =x in the other one is=y so at start of the year: x+y=$10000.....eq1 now interest rate of 1st account(x)=5% so increment after completion of year=5% of x=(5*x)/100=0.05x and in second account interest rate=7% of y=(7*y)/100=0.07y total interest=0.05x+0.07y $676=0.05x+0.07y..............eq2 now solve eq1 and eq2: put x=10000-y in eq2 676=0.05(10000-y)+0.07y 676=500-0.05y+0.07y 676-500=0.02y 176=0.02y y=176/0.02 y=8800 put in x: x=10000-8800=1200 ans: he has $1200 in the account having 5% interest rate and$8800 in the one having 7% interest rate. part c: let no. of tables=x and chairs=y as per the 1st condition, 5 tables and 20 chairs cost$1800: 5x+20y=1800.........eq1 2nd condition: 2 tables and three chairs cost $420: 2x+3y=420.........eq2 Now lets solve the two equations:(by elimination) by 1st equation: 5x=1800-20y x=(1800-20y)/5 x=360-4y putting in eq2: 2(360-4y)+3y=420 720-8y+3y=420 -5y=420-720 -5y=-300 y=(-300)/(-5) y=60 put in x: x=360-4y=360-4*60 x=360-240 x=120 ans: each table costs $120 and chair costs $60 part d: let the two numbers be "x" and "y": then according to given conditions: x+y=52.......eq1 and x-y=26...........eq2 now solving these two equations through eliminationmethod: by eq2: x=26+y putting this in equation1 (26+y)+y=52 26+2y=52 2y=52-26 2y=26 y=26/2 y=13 and x=26+y=26+13 x=39 ans: the two numbers are 39 and 13.