Jack baked some biscuits, He kept 3/7 of the biscuits in container A, 5/8 of the remainder in container B. There are 21 more biscuits in container A than in container C, How many biscuits did Jake bake?
Let T be the total number of biscuits
A = (3/7)*T (He kept 3/7 of the biscuits in container A)
B = (5/8)*(T - A) (5/8 of the remainder in container B)
A = C + 21 (There are 21 more biscuits in container A than in container C)
T = A + B + C (The three tins, A, B, and C hold all of the biscuits)
Substituting for A = (3/7)*T,
B = (5/8)*(T - (3/7)*T)
(3/7)*T = C + 21
T = (3/7)*T + B + C
Simplifying the above 3 eqns by muultiplying out
56B = 20T
3T = 7C + 147
4T = 7B + 7C (or, 32T = 56B + 56C)
Substituting for 56B = 20T,
3T = 7C + 147
32T = 20T + 56C
Simplifying,
3T = 7C + 147
12T = 56C (or, 3T = 14C)
Substituting for 3T = 14C,
14C = 7C + 147
7C = 147
C = 21
Hence T = 98
Hence, B = 35
Hence, A = 42