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dida
nodriza

treballadors de la restauraciÓ la dida , sccl
treballadors de la restauraciÓ la dida , sccl

then, multiplying and solving: (890 + 510) / 8 = 175 cm you can note that this calculation is exactly like supposing that our class is composed only by 5 men and 3 women, the simplest case according to the question's statement, so we have multiplied by the number of candidates of each size (five and three) and divided by the total number of the group (eight). it may be illustrative too, if you look again at our result for the other class (officers), observing that there our calculation of a direct average was like supposing that the class was composed only by 1 men and 1 women in fact, following its respective proportion ("as many women as men"), so when we did: (180 + 170) / 2 = 175 cm we were actually using the relative weight of that supposition (1 and 1): (180 * 1 + 170 * 1) / (1 + 1) in any case, no matter the process chosen for the calculations, we have proved that both classes, nco and officers, have the same average size, 175 cm, so option c) is right. my advice as we have explained, the easiest way to solve this type of exercises about averages is to suppose that you have a small group, meeting the rule of the question (1 man, 1 woman; 5 men, 3 women), and calculate the joint averages for them: 109
to express this, we only need to multiply both average sizes by their relative weights, and divide it by the total "overweight" applied (five times + three times), in the following way: (178 * 5 + 170 * 3) / (5 + 3)