Question

there are two families M and N . there are 2 men , 2 women, and 4 childern in family N . and 4 men , 6 women and 2 children in family M . there commended daily allowance for calories is child : 1800, women : 1900 and man : 2400 and for protiens is man :55gm, woman : 45gm and child :33gm. using matrices algebra, calculate the total requirement of protiens and calories for each of the famlies ?

Answer 1

calories
N
men-  2*2400 = 4800
women-  2*1900 = 3800
children- 4*1800 = 7200
total = 19400

M
men- 4*2400 = 9600
women-  6*1900= 11400
children-  2*1800= 3600
total = 24600

proteins
N
men- 2*55 = 110
women 2*45 = 90
children- 4*33= 132
total = 332

M
men-  4*55 = 220
women-  6*45 = 270
children-  2*33 = 66
total = 556

Answer 2

calories for family N
in men=  2X2400 = 4800
in women-  2X1900 = 3800
in children- 4X1800 = 7200
total =4800+3800+7200= 19400

calories for family M
in men= 4X2400 = 9600
in women= 6X1900= 11400
in children=  2X1800= 3600
total = 9600+11400+3600= 24600

proteins for family N
in men= 2X55 = 110
in women = 2X45 = 90
in children= 4X33= 132
total proteins= 110+90+132=332

porteins for family M
in men= 4X55 = 220
in women= 6X45 = 270
 in children=  2X33 = 66
total = 220+270+66=556

therefore, total requirement of calories for N is 19400, for M is24600 and proteins for N is 332, for M is 556