Let the amount of water to be mixed = x ml new mixture = 237+x The new mixture is diluted to 75% strength i.e. 75/100 75/100 = 3/4 Therefore, 237/(237+x) = 3/4 4 *237 =3 (237+x) 948=711+3x 3x=948-711 3x=237 x=237/3 x=79 The amount of water to be mixed with ammonia = 79ml
Let the x ml of water be mixed with 237 ml of ammonia whose strength is 100%, in order to create a mixture of 75% strength. The equation required to solve this will be: 237/(237+x)=75/100 solving for x we get: 237/(237+x)=3/4 237×4=3(237+x) 948=711+3x 3x+711=948 3x=948-711 3x=237 x=237/3 x=79 ml therefore the amount of water to be added will be 79 ml
1. D 2. B 3. either c or d, cant tell without graphic. 4 C 5. also cant tell without graphic. is FA congruent to AC or twice as much or something else? its impossible to answer without this knowledge.
