The Maths Tailor

A mean, a percentage or a proportion hides the quantity you can actually add up. Convert it back: a mean of $\overset{x}{ˉ}$ over $k$ items means the total is $\overset{x}{ˉ} \times k$ , and a percentage means a fixed amount of the whole. Once everything is an absolute total (or a whole-number count), you can add, subtract and conserve it directly, which is far safer than juggling fractions.

Trigger: the data is described only by means, ratios or percentages, and you need to combine, exchange or recombine the groups.

Instances:
(i) for a weighted mixture, write each component's contribution as volume times fraction and sum them, then divide by the total volume;
(ii) when several proportions act on one population, choose a convenient total (a common multiple of the denominators) so every proportion becomes a whole-number count;
(iii) when items move between groups, use that the grand total is conserved: set the total before equal to the total after to pin down the unknown size or mean.

P, Q, and R are each mixtures of red and white paint. The percentage by volume of red paint in P is 30

(A)

25%

(B)

23%

(C)

13 \frac{1}{3} %

(D)

19 \frac{1}{2} %

(E)

9 \frac{3}{4} %

(F) It is impossible to achieve this result.

Let the fraction of red paint in R be

x

. Total volume is

12 + 5 + 3 = 20

parts. Red paint contribution: from P:

12 \times 0.30 = 3.6

; from Q:

5 \times 0.20 = 1

; from R:

3 x

. The mixture is 25

60

(A)

\frac{1}{5}

(B)

\frac{7}{25}

(C)

\frac{1}{3}

(D)

\frac{11}{25}

(E)

\frac{3}{5}

Take 300 members total. Women: 180, Men: 120. Male cricketers:

\frac{2}{5} \times 120 = 48

. Since

\frac{2}{3}

of cricketers are women, the 48 male cricketers are

\frac{1}{3}

of all cricketers, so total cricketers

= 144

, of which 96 are women. Women tennis players

= 180 - 96 = 84

. Probability

= \frac{84}{300} = \frac{7}{25}

. Hence the answer is B.

There are two sets of data: the mean of the first set is 15, and the mean of the second set is 20. One of the pieces of data from the first set is exchanged with one of the pieces of data from the second set. As a result, the mean of the first set of data increases from 15 to 16, and the mean of the second set of data decreases from 20 to 17. What is the mean of the set made by combining all the data?

(A)

16 \frac{1}{4}

(B)

16 \frac{1}{3}

(C)

16 \frac{1}{2}

(D)

16 \frac{2}{3}

(E)

16 \frac{3}{4}

Let the first set have

n

items and the second set

m

items. Original totals:

15 n

and

20 m

. After the exchange, the totals become

16 n

and

17 m

respectively (since the number of items in each set does not change). The overall total is conserved:

15 n + 20 m = 16 n + 17 m

, giving

n = 3 m

. The combined total is

15 n + 20 m = 45 m + 20 m = 65 m

and the total number of items is

n + m = 4 m

. Combined mean

= \frac{65 m}{4 m} = \frac{65}{4} = 16 \frac{1}{4}

. Hence the answer is A.