Average: Formulas, Shortcuts and Solved Examples — MCQs with Answers (Set 1)

Let the two numbers be x and y, with reciprocals 1/x and 1/y. Average of the reciprocals = (1/x + 1/y) / 2. Combine the fractions inside: 1/x + 1/y = (y+x) / (xy). So the average = [(x+y)/(xy)] / 2 = (x+y) / (2xy). Check: for x=2 and y=4, the reciprocals are 1/2 and 1/4, averaging to 3/8. The formula gives (2+4) / (2x2x4) = 6/16 = 3/8, which matches.

Let a set have n numbers with average X, so their total sum = n x X. One of the numbers equals X exactly. Removing it leaves n-1 numbers with total sum = nX – X = (n-1)X. Average of the rest = (n-1)X / (n-1) = X. So the average of the remaining numbers is still X, unchanged. Check: this only holds because the removed value exactly equals the original average; removing any other value would shift the average up or down.

Let the 20 numbers have an average of 0, so their total sum = 20 x 0 = 0. Suppose 19 of the numbers are positive, for example each equal to 1, contributing +19 to the total. The 20th number can then be -19, making the total 19 + (-19) = 0. So 19 positive numbers is achievable while keeping the sum at zero. All 20 numbers cannot be positive, because then the total would be positive, not zero. Check: the maximum count of positive numbers is 20 – 1 = 19, leaving exactly one number to balance the sum back to zero.

Average, or arithmetic mean, summarizes a set of numbers with one representative value. It is computed in two steps. First, add up every observation to get the total sum. Second, divide that sum by the number of observations. This is different from the mode, which is the most frequent value, and the median, which is the middle value when sorted.

Statement I says P + Q = 50. Statement II says R = 30. The average of P, Q, and R needs only their total, not each value separately. Total of P, Q, and R = 50 + 30 = 80. Average = total / 3 = 80/3 years. Statement I alone gives no value for R, and Statement II alone gives no value for P+Q, so neither is sufficient by itself. Check: 80/3 is fixed regardless of how P and Q individually split their sum of 50, since only the sum matters for the average.

Statement I says A + B + C = 60. Statement II says the average of A and C is 22. Average of A and C = 22 means A + C = 22 x 2 = 44. Substituting into Statement I: B = 60 – (A+C) = 60 – 44 = 16. Statement I alone does not fix B, since A and C are unknown individually. Statement II alone does not fix B either, since it says nothing about B. Both statements together are needed. Check: A+C=44 and B=16 add up to 44+16=60, matching Statement I.

Let the known values be 12, 12, 16, 40, 27, 24, 38, plus one unknown value x. There are 8 values in total. Arithmetic mean = sum of values / count, so sum of values = mean x count. Sum of all 8 values = 24 x 8 = 192. Sum of the known values = 169. x = total sum – sum of known values = 192 – 169 = 23. Check: (169 + 23) / 8 = 192/8 = 24, matching the given mean.

Let the known values be 9, 21, 38, 22, 17, 6, 25, plus one unknown value x. There are 8 values in total. Arithmetic mean = sum of values / count, so sum of values = mean x count. Sum of all 8 values = 25 x 8 = 200. Sum of the known values = 138. x = total sum – sum of known values = 200 – 138 = 62. Check: (138 + 62) / 8 = 200/8 = 25, matching the given mean.

Let the known values be 9, 22, 25, 29, 29, 7, plus one unknown value x. There are 7 values in total. Arithmetic mean = sum of values / count, so sum of values = mean x count. Sum of all 7 values = 20 x 7 = 140. Sum of the known values = 121. x = total sum – sum of known values = 140 – 121 = 19. Check: (121 + 19) / 7 = 140/7 = 20, matching the given mean.

Let the known values be 12, 16, 17, 31, 38, 12, 26, 23, 20, plus one unknown value x. There are 10 values in total. Arithmetic mean = sum of values / count, so sum of values = mean x count. Sum of all 10 values = 27 x 10 = 270. Sum of the known values = 195. x = total sum – sum of known values = 270 – 195 = 75. Check: (195 + 75) / 10 = 270/10 = 27, matching the given mean.

Let the known values be 38, 9, 36, 12, 25, 28, 21, 19, plus one unknown value x. There are 9 values in total. Arithmetic mean = sum of values / count, so sum of values = mean x count. Sum of all 9 values = 29 x 9 = 261. Sum of the known values = 188. x = total sum – sum of known values = 261 – 188 = 73. Check: (188 + 73) / 9 = 261/9 = 29, matching the given mean.

Let the known values be 24, 29, 18, 12, 28, 16, 9, plus one unknown value x. There are 8 values in total. Arithmetic mean = sum of values / count, so sum of values = mean x count. Sum of all 8 values = 18 x 8 = 144. Sum of the known values = 136. x = total sum – sum of known values = 144 – 136 = 8. Check: (136 + 8) / 8 = 144/8 = 18, matching the given mean.

Let the known values be 32, 8, 12, 18, 23, 5, 7, 37, plus one unknown value x. There are 9 values in total. Arithmetic mean = sum of values / count, so sum of values = mean x count. Sum of all 9 values = 22 x 9 = 198. Sum of the known values = 142. x = total sum – sum of known values = 198 – 142 = 56. Check: (142 + 56) / 9 = 198/9 = 22, matching the given mean.

Let the known values be 33, 36, 37, 32, 28, plus one unknown value x. There are 6 values in total. Arithmetic mean = sum of values / count, so sum of values = mean x count. Sum of all 6 values = 33 x 6 = 198. Sum of the known values = 166. x = total sum – sum of known values = 198 – 166 = 32. Check: (166 + 32) / 6 = 198/6 = 33, matching the given mean.

Let the known values be 15, 26, 23, 22, 35, 30, 29, 6, 31, 20, 26, plus one unknown value x. There are 12 values in total. Arithmetic mean = sum of values / count, so sum of values = mean x count. Sum of all 12 values = 25 x 12 = 300. Sum of the known values = 263. x = total sum – sum of known values = 300 – 263 = 37. Check: (263 + 37) / 12 = 300/12 = 25, matching the given mean.

Let the known values be 20, 7, 26, 19, 11, 37, 18, 32, 20, plus one unknown value x. There are 10 values in total. Arithmetic mean = sum of values / count, so sum of values = mean x count. Sum of all 10 values = 25 x 10 = 250. Sum of the known values = 190. x = total sum – sum of known values = 250 – 190 = 60. Check: (190 + 60) / 10 = 250/10 = 25, matching the given mean.

Let the one-way distance be d = 90 x 20 = 1800 km, chosen so both speeds divide it evenly. Time for the outward trip = d / 90 = 1800/90 = 20 hours. Time for the return trip = d / 20 = 1800/20 = 90 hours. Total distance = 2 x 1800 = 3600 km. Total time = 20 + 90 = 110 hours. Average speed = total distance / total time = 3600 / 110 = 360/11 km/h. This matches the harmonic-mean shortcut, 2 x 90 x 20 / (90+20). Check: it is less than the simple average (90+20)/2 = 55, since harmonic mean is always at most the arithmetic mean.

Let the one-way distance be d = 50 x 90 = 4500 km, chosen so both speeds divide it evenly. Time for the outward trip = d / 50 = 4500/50 = 90 hours. Time for the return trip = d / 90 = 4500/90 = 50 hours. Total distance = 2 x 4500 = 9000 km. Total time = 90 + 50 = 140 hours. Average speed = total distance / total time = 9000 / 140 = 450/7 km/h. This matches the harmonic-mean shortcut, 2 x 50 x 90 / (50+90). Check: it is less than the simple average (50+90)/2 = 70, since harmonic mean is always at most the arithmetic mean.

Let the one-way distance be d = 90 x 45 = 4050 km, chosen so both speeds divide it evenly. Time for the outward trip = d / 90 = 4050/90 = 45 hours. Time for the return trip = d / 45 = 4050/45 = 90 hours. Total distance = 2 x 4050 = 8100 km. Total time = 45 + 90 = 135 hours. Average speed = total distance / total time = 8100 / 135 = 60 km/h. This matches the harmonic-mean shortcut, 2 x 90 x 45 / (90+45). Check: it is less than the simple average (90+45)/2 = 135/2, since harmonic mean is always at most the arithmetic mean.

Let the one-way distance be d = 50 x 40 = 2000 km, chosen so both speeds divide it evenly. Time for the outward trip = d / 50 = 2000/50 = 40 hours. Time for the return trip = d / 40 = 2000/40 = 50 hours. Total distance = 2 x 2000 = 4000 km. Total time = 40 + 50 = 90 hours. Average speed = total distance / total time = 4000 / 90 = 400/9 km/h. This matches the harmonic-mean shortcut, 2 x 50 x 40 / (50+40). Check: it is less than the simple average (50+40)/2 = 45, since harmonic mean is always at most the arithmetic mean.

Let the one-way distance be d = 80 x 20 = 1600 km, chosen so both speeds divide it evenly. Time for the outward trip = d / 80 = 1600/80 = 20 hours. Time for the return trip = d / 20 = 1600/20 = 80 hours. Total distance = 2 x 1600 = 3200 km. Total time = 20 + 80 = 100 hours. Average speed = total distance / total time = 3200 / 100 = 32 km/h. This matches the harmonic-mean shortcut, 2 x 80 x 20 / (80+20). Check: it is less than the simple average (80+20)/2 = 50, since harmonic mean is always at most the arithmetic mean.

Let the one-way distance be d = 80 x 48 = 3840 km, chosen so both speeds divide it evenly. Time for the outward trip = d / 80 = 3840/80 = 48 hours. Time for the return trip = d / 48 = 3840/48 = 80 hours. Total distance = 2 x 3840 = 7680 km. Total time = 48 + 80 = 128 hours. Average speed = total distance / total time = 7680 / 128 = 60 km/h. This matches the harmonic-mean shortcut, 2 x 80 x 48 / (80+48). Check: it is less than the simple average (80+48)/2 = 64, since harmonic mean is always at most the arithmetic mean.

Let the one-way distance be d = 72 x 40 = 2880 km, chosen so both speeds divide it evenly. Time for the outward trip = d / 72 = 2880/72 = 40 hours. Time for the return trip = d / 40 = 2880/40 = 72 hours. Total distance = 2 x 2880 = 5760 km. Total time = 40 + 72 = 112 hours. Average speed = total distance / total time = 5760 / 112 = 360/7 km/h. This matches the harmonic-mean shortcut, 2 x 72 x 40 / (72+40). Check: it is less than the simple average (72+40)/2 = 56, since harmonic mean is always at most the arithmetic mean.

Let the one-way distance be d = 30 x 20 = 600 km, chosen so both speeds divide it evenly. Time for the outward trip = d / 30 = 600/30 = 20 hours. Time for the return trip = d / 20 = 600/20 = 30 hours. Total distance = 2 x 600 = 1200 km. Total time = 20 + 30 = 50 hours. Average speed = total distance / total time = 1200 / 50 = 24 km/h. This matches the harmonic-mean shortcut, 2 x 30 x 20 / (30+20). Check: it is less than the simple average (30+20)/2 = 25, since harmonic mean is always at most the arithmetic mean.

Let the one-way distance be d = 24 x 50 = 1200 km, chosen so both speeds divide it evenly. Time for the outward trip = d / 24 = 1200/24 = 50 hours. Time for the return trip = d / 50 = 1200/50 = 24 hours. Total distance = 2 x 1200 = 2400 km. Total time = 50 + 24 = 74 hours. Average speed = total distance / total time = 2400 / 74 = 1200/37 km/h. This matches the harmonic-mean shortcut, 2 x 24 x 50 / (24+50). Check: it is less than the simple average (24+50)/2 = 37, since harmonic mean is always at most the arithmetic mean.