Average Questions and Answers

Answer: C

The value of m is 7.

Total production over m days is 100m. Adding today's 180 units gives m + 1 days averaging 110:

\[ \begin{aligned} 100m + 180 &= 110(m + 1) \\ 180 - 110 &= 110m - 100m \\ m &= 7 \end{aligned} \]

Check: 7 days at 100 is 700, plus 180 is 880, and 880/8 = 110. Writing 110m on the right instead of 110(m + 1) gives m = 18, the first option, and the other options do not satisfy the equation.

Answer: B

The average is \(\frac{a+b}{2ab}\).

The reciprocals are \(\frac{1}{a}\) and \(\frac{1}{b}\). Their average is half their sum:

\[ \begin{aligned} \tfrac{1}{2}\left(\tfrac{1}{a} + \tfrac{1}{b}\right) &= \tfrac{1}{2} \cdot \tfrac{a+b}{ab} \\ &= \tfrac{a+b}{2ab} \end{aligned} \]

\(\frac{2(a+b)}{ab}\) multiplies by 2 instead of dividing. \(\frac{2ab}{a+b}\) is the reciprocal of the answer, the harmonic mean of a and b. \(\frac{a+b}{a-b}\) has no connection with the average.

Answer: A

The total spent was ₹ 1,640.

Let the average spending of all 20 be x, so the total is 20x. Fifteen spent ₹ 80 each, and the other five spent \(x + 6\) each:

\[ \begin{aligned} 20x &= 15 \times 80 + 5(x + 6) \\ 15x &= 1230 \\ x &= 82 \end{aligned} \]

The total is \(20 \times 82 = 1640\). Check: \(1200 + 5 \times 88 = 1640\). ₹ 1,600 assumes everyone spent the average of 80.

Answer: D

The average is 25. The six numbers add up to \[(2x+3)+(3x+2)+25+(x+6)+26+(x+5) = 7x + 67\] Their average is 17, so the sum is 6 x 17 = 102. Then 7x = 35 and x = 5. Hence 4x + 3 = 23 and 7x - 8 = 27, and the average of 23 and 27 is 25. The other options come from slips such as wrong x or adding instead of averaging.

Answer: A

The average height of the boys is 160.5 cm. Take 4 boys and 5 girls (girls are 25 % more). Let the boys' average be b, so the class average is b - 2.5. Total height: \[5(156) + 4b = 9(b - 2.5)\] \[780 + 4b = 9b - 22.5\] \[5b = 802.5\] so b = 160.5. Check: the class average is 158, and (780 + 642)/9 = 158. The other options do not satisfy the condition that the boys' average is 2.5 cm above the class average.

Answer: D

The total of all 21 results is the average times the number of results, and the 11th result is what is left after taking out the first ten and the last ten.

\[ 21 \times 15 = 315,\quad 10 \times 12 = 120 \] \[ 10 \times 14 = 140 \]

\[ \text{11th result} = 315 - 120 - 140 = 55 \]

The values 50, 52 and 54 would come from arithmetic slips in the sums, for example forgetting that the middle result belongs to neither group. The answer is 55, option D.

Answer: D

The first six multiples of 6 are 6, 12, 18, 24, 30 and 36. Their sum is 126, and the average is the sum divided by the number of terms.

\[ \frac{6 + 12 + 18 + 24 + 30 + 36}{6} = \frac{126}{6} = 21 \]

It can also be found as \(6 \times \frac{1 + 2 + \cdots + 6}{6} = 6 \times 3.5 = 21\). The values 6 and 12 are the first and second multiples, and 19 is not related to the series. The answer is 21, option D.

Answer: A

The average is \(3\frac{1}{2}\). Add the fractions. The whole parts give 3 + 5 + 2 + 2 = 12 and the fractional parts give \[\frac{1}{2} + \frac{1}{4} + \frac{3}{4} + \frac{1}{2} = 2\] so the sum is 14. Dividing by the four numbers gives 14/4 = 3.5, which is \(3\frac{1}{2}\). The options \(2\frac{1}{2}\), \(3\frac{3}{4}\) and \(2\frac{3}{4}\) come from dropping a term or adding the fractional parts wrongly.

Answer: A

When every number of a set is multiplied by 14, the average is multiplied by the same 14. The average of the ten numbers is 8, so the new average is 8 x 14 = 112.

The figure 84 comes from multiplying by 10.5, and 19 and 14 come from adding or from forgetting to multiply the original average. The sum of the new numbers is 1120, and divided by 10 it gives 112.

Answer: A

The average weight is 58 kg. The average of three values is their sum divided by three.

\[ 54 + 56 + 64 = 174 \]

\[ \frac{174}{3} = 58 \text{ kg} \]

The option 56 kg is the weight of B alone and 54 kg is the weight of A alone, so neither is the mean of the three. The option 60 kg would need a total of 180 kg, which is 6 kg more than the actual total. Only 58 kg matches the sum of 174 kg shared among the three people.