Shortcuts & Speed Methods

The Assumed-Average (Deviation) Method

Almost every average question rests on one line: sum = average × count. The methods here get you to the answer without building big totals, which matters when you have about 36 seconds a question in Tier 1. Each one ends with the places it goes wrong, such as dividing by the wrong count or turning a ratio the wrong way round, because SSC builds its wrong options from exactly those slips.

Pick a base and average the differences

When to use it: a long list of numbers that sit close together, such as marks, weights or a week's sales.

Fast way: pick a round number near the middle as a base. Write how far each value is above (+) or below (−) the base, add these differences, divide by the count, and add the result to the base.

$$\text{Average} = \text{base} + \frac{\text{sum of differences}}{\text{count}}$$

For example: 64, 71, 68, 59, 73, 66 and 61. With 65 as the base the differences are −1, +6, +3, −6, +8, +1 and −4. They add up to +7, and 7 ÷ 7 = 1, so the average is 65 + 1 = 66. You never had to add up to 462.

A missing value: the differences must cancel

When to use it: the average is given and one value is missing.

Fast way: use the given average as the base. Around the true average the differences always add up to zero, so the missing value has to cancel whatever the others leave over.

For example: five numbers have an average of 48. Four of them are 44, 53, 46 and 51. Their differences from 48 are −4, +5, −2 and +3, which leave +2. The fifth number must be 2 below 48, so it is 46.

The same change to every value

If every value goes up by 7, the average goes up by 7. If every value is multiplied by 3, so is the average. Follow the steps in the order the question gives them: the average of 12 numbers is 25, each number is multiplied by 3 and then 4 is taken away, so the new average is 3 × 25 − 4 = 71.

Where it fails:

• The +7 is the total of the differences, not the shift. Forgetting to divide by the count gives 72 in the first example, and 72 is often one of the options.
• A base far from the values gives big differences and more chance of a slip. Pick one near the middle.
• "Add 5 to each number" moves the average by 5, not by 5 × count. It is the total that moves by 5 × count.

[SSC Tip] The same trick checks an option in seconds. Put the option in as the base. If the differences add up to zero, the option is the average.

The Middle-Term Rule for Consecutive Numbers and APs

The average is the middle term

When to use it: numbers that go up by the same step. That covers consecutive numbers, consecutive odd or even numbers, consecutive multiples of 7, and any arithmetic progression (AP).

Fast way: the average is the middle term. With an odd count it is the middle number. With an even count it is halfway between the two middle numbers. Either way it equals $\frac{\text{first} + \text{last}}{2}$.

From the average you can step out to the ends. For k terms with a common step d,

$$\text{largest} = \text{average} + \frac{(k - 1)d}{2}, \quad \text{smallest} = \text{average} - \frac{(k - 1)d}{2}$$

For example: seven consecutive numbers have an average of 38, so the largest is 38 + 3 = 41. Six consecutive odd numbers have an average of 52, so the smallest is 52 − $\frac{5 \times 2}{2}$ = 47, and the six numbers are 47, 49, 51, 53, 55 and 57.

Taking in the next few terms

When to use it: "If the next m numbers of the same kind are also included, what is the new average?"

Fast way: the middle of the list moves forward by half of the m new steps, so the new average = old average + $\frac{md}{2}$.

For example: eight consecutive odd numbers have an average of 40. Include the next four odd numbers and the average becomes 40 + $\frac{4 \times 2}{2}$ = 44. Check: the list now runs from 33 to 55, and its middle pair is 43 and 45.

A related fact: a block of k consecutive numbers and the next k numbers after it differ in average by exactly k. If five consecutive numbers average 23, the next five average 28. For consecutive odd or even numbers the jump is 2k.

Where it fails:

• Consecutive odd or even numbers have a step of 2. Using a step of 1 for the six odd numbers above gives a smallest value of 49.5, which can't be right, and that is your signal to recheck.
• With an even count of odd numbers, the average (52 here) is even and isn't in the list at all. Don't look for it there.
• The rule needs equal steps. Primes and squares are not APs, so list them or use a formula from the next part.

The sum of an AP is the middle term × the count, so fifteen consecutive numbers with 40 in the middle add up to 600.

Formula Averages for Number Series

Some lists come with a ready-made average. Learn this table once and these questions take five seconds.

When to use it: the question names one of these lists directly.

For example:

• The first 40 odd numbers average 40, and the first 25 even numbers average 26.
• The first 18 multiples of 6 average $\frac{6 \times 19}{2}$ = 57.
• The squares of 1 to 25 average $\frac{26 \times 51}{6}$ = 221.
• The cubes of 1 to 8 average $\frac{8 \times 81}{4}$ = 162.

Numbers between two limits

Fast way: find the first and the last number of the right kind inside the range and take their average. They form an AP, so the middle-term rule does the rest.

For example: the multiples of 8 between 50 and 200 run from 56 to 192, so their average is $\frac{56 + 192}{2}$ = 124. The odd numbers between 30 and 90 run from 31 to 89, so their average is 60.

Primes have no formula, so list them. The primes between 30 and 50 are 31, 37, 41, 43 and 47. They add up to 199, so their average is 39.8.

Where it fails:

• Averaging the two limits instead of the first and last multiples. $\frac{50 + 200}{2}$ = 125 is the trap in the example above.
• The average of the squares is not the square of the average. The numbers 1 to 25 average 13, and $13^2$ = 169, but their squares average 221.
• The first n even numbers start at 2, so they average n + 1, not n.

Since the first n odd numbers average n, they add up to $n^2$: the first 30 odd numbers add up to 900.

One Joins or Leaves: the Change × Count Idea

One person joins

When to use it: a group of n has an average A. One person joins and the average changes by d.

Fast way: the newcomer has to bring the new average for himself and also top up each of the n old members by d.

$$\text{newcomer} = \text{new average} + n \times d$$

The same answer comes from the old average and the count after the change: newcomer = A + (n + 1) × d.

For example: 24 students have an average weight of 42 kg. A new student joins and the average rises by 0.2 kg. The new student weighs 42.2 + 24 × 0.2 = 47 kg (or 42 + 25 × 0.2, the same thing).

One person leaves

Fast way: the same pairing works. Take the old average with the count after the change. If the average falls when someone leaves, the leaver was above it, so add. If it rises, the leaver was below it, so subtract.

For example: 30 workers have an average wage of ₹640 a day. One leaves and the average falls by ₹4. The 29 who stay are each ₹4 lower, so the leaver earned 640 + 29 × 4 = ₹756.

Several join or leave together

Fast way: total of the newcomers = (number joining) × old average + (count after) × change.

For example: 40 parcels average 35 kg. Eight more are loaded and the average rises by 1.5 kg. The new parcels total 8 × 35 + 48 × 1.5 = 280 + 72 = 352 kg, so they average 44 kg.

Where it fails:

• Mixing the pairs. Old average with the old count (42 + 24 × 0.2 = 46.8) is the wrong option you will usually see. Pair the old average with the count after, or the new average with the count before.
• Direction. A joiner who raises the average is above it. A leaver who raises the average was below it, so you subtract.
• If the average doesn't change at all, the person who joined or left is exactly at the average.

Replacement: New = Old ± Count × Change

One member replaced

When to use it: one member of a group is swapped for someone new, so the count stays the same.

Fast way: the whole change in the total comes from the swap, and the total changes by n × (change in average), where n is the size of the group.

$$\text{new member} = \text{old member} + n \times (\text{change in average})$$

Use + if the average rises and − if it falls.

For example: the average weight of the 12 members of a tug-of-war team goes up by 1.5 kg when a member weighing 57 kg is replaced. The new member weighs 57 + 12 × 1.5 = 75 kg.

Working the other way is just as quick. In a group of 15, a man of 72 kg is replaced by one of 51 kg, so the average falls by $\frac{21}{15}$ = 1.4 kg.

Two members replaced

Fast way: the same rule for the pair: (total of the new pair) = (total of the old pair) ± n × change.

For example: the average age of a team of 15 falls by 1.2 years when two players aged 33 and 29 are replaced by two new players. The new pair add up to 62 − 15 × 1.2 = 44 years, so their average is 22 years.

The same rule covers "the average age falls when an older member is replaced". Nobody has aged; one person has simply been swapped for a younger one, and the gap between them is n × fall.

Where it fails:

• n is the size of the whole group, not 1 or 2. Writing 57 + 1.5 = 58.5 is the trap.
• Replacement keeps the count the same, so there is no n + 1 here.
• "A is replaced by B" means B is the new one. Read which weight the question gives you.
• With two replaced the rule gives the pair's total. Halve it only if the question asks for their average.

Wrong Entries: Difference ÷ Count

When to use it: a value was copied wrongly, and you want the correct average.

Fast way: only the total was wrong, and it was wrong by (correct value − wrong value). Spread that over the count.

$$\text{correct average} = \text{wrong average} + \frac{\text{correct} - \text{wrong}}{n}$$

For example: the average of 42 numbers was worked out as 58, but 92 had been read as 29. The total was 63 short, so the correct average is 58 + $\frac{63}{42}$ = 59.5.

Several mistakes: net them first

For example: 25 observations were found to average 46. Two were copied wrongly, 62 as 26 and 41 as 47. The net error is (62 − 26) + (41 − 47) = 36 − 6 = 30, so the correct average is 46 + $\frac{30}{25}$ = 47.2.

Digits reversed

A two-digit number and its reverse always differ by 9 × (difference of the digits). If one number in a list of 12 had its digits reversed and the average went up by 3, the total went up by 36, so the digits differ by 36 ÷ 9 = 4. The number could be 15, 26, 37, 48 or 59 (written as 51, 62, 73, 84 or 95). The question will add one more fact, such as the digit sum, to pin it down.

A value left out

When a value was missed altogether, the count was wrong too, so go back to totals. Suppose 19 numbers averaged 32 and a 20th number, 70, had been left out. The correct average is $\frac{19 \times 32 + 70}{20} = \frac{678}{20}$ = 33.9.

Where it fails:

• Direction. If the correct value is bigger than the one used, the average goes up. Subtracting gives 56.5 in the first example.
• Divide by the full count n, not by 2 or by anything else.
• If a value was left out or counted twice, the count changes as well, and difference ÷ count no longer works on its own. Use totals.

[SSC Tip] In digit-reversal questions the change in the total is always a multiple of 9. An option that would make it anything else can go straight away.

Overlapping Groups and Overlapping Days

A value counted twice

When to use it: the average of all N values is given, along with the average of the first k and of the last m, and the two groups share exactly one value (k + m = N + 1).

Fast way: adding the two group totals counts the shared value twice, so

$$\text{shared value} = \text{first group total} + \text{last group total} - \text{full total}$$

For example: 11 numbers average 46. The first six average 43 and the last six average 50. The sixth number is 258 + 300 − 506 = 52.

It is faster still with differences from the overall average. The first six are 3 below each (−18 in all) and the last six are 4 above each (+24 in all), so the shared value = 46 − 18 + 24 = 52. No three-digit totals at all.

Three numbers from pair averages

If the averages of A and B, B and C, and C and A are given, then A + B + C = the sum of the three pair averages. Each pair total is twice its average, and each letter sits in two pairs, so the 2s cancel. With pair averages 40 (A and B), 50 (B and C) and 45 (C and A), A + B + C = 135, and C = 135 − 80 = 55.

Days: the last day minus the first day

When to use it: averages over sliding windows of days, such as Monday to Wednesday and Tuesday to Thursday.

Fast way: the middle days are in both windows and cancel out, so (last day − first day) = (days in each window) × (difference of the averages).

For example: the average temperature from Monday to Wednesday was 29°C and from Tuesday to Thursday was 31°C. Thursday was 3 × 2 = 6°C warmer than Monday. If Thursday was 35°C, Monday was 29°C.

For a whole week, use the shared-value rule. A week averages 30°C, the first four days 29°C and the last four 32°C. Thursday is in both groups, so Thursday = 30 − 4 + 8 = 34°C.

Where it fails:

• The difference of the averages (2°C) is not the difference of the days. Multiply by the number of days in each window.
• The shared-value rule needs exactly one value in both groups. If the groups share two or more values, you can find only their total.
• Watch the direction the question asks for: Thursday − Monday or Monday − Thursday.

[SSC Tip] First k and last k of 2k − 1 numbers always share the middle one. "First 7 and last 7 of 13" means the 7th, so you know which number you are finding before you start.

Cricket: Batting and Bowling Averages

A batting average is runs per innings, and a bowling average is runs given away per wicket. Both run on the joining rule from earlier in this section.

Runs needed to reach a target average

When to use it: after n innings the average is A, and the batsman wants it to be A + d after the next innings.

Fast way: the next innings must bring the new average for itself and also d for each of the n old innings.

$$\text{runs needed} = (A + d) + n \times d$$

For example: the average after 19 innings is 41. To make it 44 in the 20th innings he needs 44 + 19 × 3 = 101 runs.

Average after an innings

When to use it: in the (n + 1)th innings he scores R, and the average goes up by d.

Fast way: new average = R − n × d. The runs above the new average were shared out among the n old innings.

For example: he scores 134 in his 31st innings and his average rises by 2.8. The new average is 134 − 30 × 2.8 = 134 − 84 = 50.

Highest and lowest scores

For example: the average of 32 innings is 44. The highest score is 158 more than the lowest. Leaving these two out, the other 30 innings average 41. The two scores together are 1,408 − 1,230 = 178, so the highest is $\frac{178 + 158}{2}$ = 168 (and the lowest is 10).

Bowling average

Lower is better here, because it means fewer runs per wicket.

For example: a bowler's average is 12.4. He takes 5 wickets for 26 runs in a match, and his average falls by 0.4 to 12. At the new average the 5 wickets would "cost" 5 × 12 = 60 runs, but he gave only 26, a saving of 34. That saving is what makes every old wicket 0.4 cheaper, so old wickets = 34 ÷ 0.4 = 85. Check: 85 × 12.4 + 26 = 1,080 = 90 × 12.

Where it fails:

• In "average after an innings" n is the old count (30 for the 31st innings). Using 31 gives 47.2, which is the old average and a ready-made trap.
• If the question mentions not-out innings, the average is runs ÷ times out, not runs ÷ innings.
• A falling bowling average means the bowler did well. Don't read it the other way.

Target Marks and the Minimum Score

Marks needed in the last test

When to use it: the average over some tests is known, and the student wants a target average after one more test.

Fast way: this is the cricket rule again.

$$\text{marks needed} = \text{target} + (\text{tests done}) \times (\text{target} - \text{current average})$$

For example: Meera's average in 4 tests is 58. To make it 63 after the 5th test she needs 63 + 4 × 5 = 83.

The minimum score, using differences

When to use it: a pass needs an average of at least some value, and every score but one is known.

Fast way: take the pass average as the base and add up the differences. The last paper has to cover any shortfall.

For example: a pass needs an average of 40 in four papers. A student scores 35, 42 and 31 in the first three. The differences from 40 are −5, +2 and −9, a shortfall of 12, so the fourth paper needs at least 40 + 12 = 52.

Is the target possible?

If the marks needed come out above the maximum, one test can't do it. Arjun averages 72 in 4 tests, each out of 100. For an average of 80 after the 5th he would need 80 + 4 × 8 = 112, which is impossible. The best he can reach is $\frac{288 + 100}{5}$ = 77.6.

How many perfect tests would he need? He is 4 × 8 = 32 marks short of an 80 average, and each 100 covers 20 of that, so one test isn't enough and 2 tests are.

Where it fails:

• The count is the tests already done (4), not the total (5). 63 + 5 × 5 = 88 is the trap.
• If the papers have different maximum marks, the target is a weighted average. Work with total marks, as in the weighted-average parts below.
• "At least" questions want the smallest score that reaches the target. If marks are whole numbers, round up, never down.

Weighted Average by Distances: the See-Saw

Weighted and combined averages come up often in recent CGL papers, so keep this method ready.

Group sizes from the averages

When to use it: two groups with averages a and b are put together, and the overall average m is known. The question wants the ratio of the group sizes, or one of the sizes.

Fast way: picture a see-saw. The overall average is the balance point, and each group sits at its own average. The heavier group sits closer to the point, so the sizes are in the opposite ratio of the distances.

$$n_1 : n_2 = (b - m) : (m - a)$$

For example: boys average 61 marks, girls average 72, and the class averages 65. The boys are 4 away and the girls are 7 away, so boys : girls = 7 : 4. With 44 students, that is 28 boys and 16 girls. Check: 1,708 + 1,152 = 2,860 = 44 × 65.

One group's average from the overall average

Fast way: the total shortfall of one group must equal the total surplus of the other.

For example: section A has 30 students averaging 55, and section B has 20 students. Together they average 58.6. Section A is 3.6 below on each student, a shortfall of 30 × 3.6 = 108. Section B must make up 108 over 20 students, which is 5.4 each, so B averages 58.6 + 5.4 = 64.

Three or more groups

Differences from any base still work. Groups of 25, 35 and 40 average 60, 70 and 64. Using 64 as the base: 25 × (−4) + 35 × 6 + 40 × 0 = +110, and 110 ÷ 100 = 1.1, so the overall average is 65.1.

Where it fails:

• Flipping the ratio. The bigger group is nearer the overall average, so the 4 goes with the girls. Boys : girls = 4 : 7 is the trap.
• The overall average must lie between the two group averages. An option outside that range is out.
• The rule needs averages. If the question gives total salaries or total marks, turn them into averages first.

Alligation (Chapter 6) is this same see-saw drawn as a cross.

The Average-of-Averages Trap

When it works and when it doesn't

When to use it: the moment two or more group averages are being combined.

Fast way: a plain average of the averages is right only when the groups are the same size. Otherwise weight each average by its group size, or use the see-saw.

For example: section A has 40 students averaging 70, and section B has 60 averaging 60. The plain average, 65, is wrong. B is bigger, so the answer leans toward 60: 60 + 10 × $\frac{40}{100}$ = 64.

Percentages on papers with different maximum marks

Fast way: turn each percentage into marks, add, and divide by the total maximum.

For example: 84% on a 150-mark paper and 64% on a 50-mark paper is 126 + 32 = 158 out of 200, which is 79%, not 74%.

Average price: by quantity or by money

• Different quantities: weight each price by its quantity. 24 kg at ₹35 and 16 kg at ₹50 average ₹41 a kg, not ₹42.50.
• Equal money spent at two prices: you get more of the cheaper one, so the average price is $\frac{2ab}{a + b}$, which is below the plain average. Spending ₹600 on rice at ₹30 a kg and ₹600 at ₹20 a kg buys 20 kg + 30 kg = 50 kg for ₹1,200, which is ₹24 a kg.

Average speed over equal distances works the same way as equal money.

Where it fails:

• The plain average sits in the options of nearly every combined-group question. If the groups differ in size, it is wrong.
• With equal quantities the plain average is right. With equal money it is not. Read which one is fixed.
• A percentage of percentages needs the bases. "Class A passed 80% and class B 60%" says nothing about the combined pass rate until you know the class sizes.

Ages: the Whole Group Grows Together

n years later, n years more on the average

When to use it: the same group, n years ago or n years from now.

Fast way: every member gets n years older, so the average rises by n and the total by (count × n). Nothing else changes.

For example: a family of 5 has an average age of 31 today. Three years ago, with the same five members, the average was 28, and the total was 15 years less.

A child born, or a member who joined since

Fast way: bring the old group to today first, then compare totals. The difference is the newcomer.

For example: four years ago a family of 6 had an average age of 27. A child has been born since, and today the 7 members also average 27. The six are now 6 × 31 = 186 in total, the seven are 7 × 27 = 189, so the child is 3 years old.

A teacher included

Fast way: this is the joining rule.

$$\text{teacher} = \text{new average} + (\text{number of students}) \times \text{increase}$$

For example: 23 students average 13 years. With the teacher included the average goes up by 1.25. The teacher is 14.25 + 23 × 1.25 = 14.25 + 28.75 = 43 years old.

Average age at the birth of the youngest

Take the youngest out, then go back that many years. A family of 5 averages 24 today and the youngest is 4. The other four total 120 − 4 = 116 today, and 4 years ago they totalled 116 − 16 = 100, so their average then was 25.

Where it fails:

• Anyone born or married into the family since then was not in the old total. Add the years only to the people who were there.
• In the teacher rule, start from the new average and use the number of students, not students + 1.
• An older member replaced by a younger one lowers the average without anyone ageing. That is the replacement rule, not this one.

Option Elimination

Range, whole numbers and direction

When to use it: first, before any working. Sometimes it settles the question on its own.

Fast way: test the options against these four checks.

• Range. An average always lies between the smallest and largest values. A combined average lies between the group averages, and nearer the bigger group.
• Whole totals. If the values are whole numbers (people, runs, marks), then average × count must be a whole number too. Seven students can't average 46.3 marks, because 7 × 46.3 = 324.1.
• Direction. A newcomer who raises the average is above it. A replacement that lowers the average is lighter or younger. A batsman whose average rises scored more than the new average.
• Consecutive numbers. An odd count of consecutive whole numbers has a whole-number average. An even count has an average ending in .5.

For example: in a class of 45, boys average 57 marks and girls 66, and the class averages 60. How many girls are there: (a) 30 (b) 15 (c) 25 (d) 20?

60 is nearer 57, so the boys are the bigger group and the girls are fewer than half of 45. That leaves (b) 15 and (d) 20. The see-saw gives boys : girls = 6 : 3 = 2 : 1, so there are (b) 15 girls. Check: 1,710 + 990 = 2,700 = 45 × 60.

Where it fails: when all four options pass the checks, you still need the method. Money and weights can come in decimals, so the whole-total check is strongest for people, runs and marks.

[SSC Tip] A trap option is usually the result of one common slip: the plain average of averages, the old count in place of the new one, or a flipped ratio. If your answer matches an option that one of those slips would give, recheck that one step before you mark it.

Solved Examples

Every answer below uses one of the quick methods above, with a rough time after it.

Question 1

A shop sold 318, 324, 309, 331, 315, 327, 312 and 320 umbrellas on eight days of the monsoon. Find the average number sold per day.

Answer. I take 320 as the base. The differences are −2, +4, −11, +11, −5, +7, −8 and 0, and they add up to −4. Spread over 8 days that is −0.5, so the average is 320 − 0.5 = 319.5.

Time check: about 20 seconds. Adding eight three-digit numbers first takes twice as long and invites a slip.

Question 2

The average of six consecutive even numbers is 47. Find the difference between the square of the largest and the square of the smallest.

Answer. Six even numbers with a step of 2, so the smallest is 47 − 5 = 42 and the largest is 47 + 5 = 52. The difference of squares is (52 − 42)(52 + 42) = 10 × 94 = 940. I didn't square anything, because largest + smallest is just 2 × 47.

Time check: about 15 seconds.

Question 3

Find the difference between the average of the cubes and the average of the squares of the first 11 natural numbers.

Answer. Cubes: $\frac{11 \times 12^2}{4} = \frac{1584}{4}$ = 396. Squares: $\frac{12 \times 23}{6}$ = 46. The difference is 350.

Time check: about 20 seconds, if the two formulas are in your head.

Question 4

The average monthly salary of 20 employees is ₹24,000. Five new employees join and the average falls by ₹600. What is the average salary of the new employees?

Answer. Old average with the count after the change: the newcomers' total = 5 × 24,000 − 25 × 600 = 1,20,000 − 15,000 = 1,05,000. So they average ₹21,000 each.

Time check: about 20 seconds.

Question 5

The average weight of 16 boxes is 34 kg. One box is taken out, and the average of the rest becomes 33.2 kg. How heavy was the box that was taken out?

Answer. The average fell by 0.8 kg when the box left, so the box was above average. The 15 boxes that stay are each 0.8 lower, so the box weighed 34 + 15 × 0.8 = 46 kg.

Time check: about 10 seconds.

Question 6

The average weight of 14 men goes up by 1.5 kg when two of them, weighing 61 kg and 66 kg, are replaced by two new men. What is the average weight of the two new men?

Answer. The old pair weighs 127 kg. The total went up by 14 × 1.5 = 21 kg, so the new pair weighs 148 kg, and their average is 74 kg.

Time check: about 15 seconds.

Question 7

The average of 15 numbers was found to be 41. Later it turned out that 74 had been written as 47 and 35 had been written as 53. Find the correct average.

Answer. Net error: (74 − 47) + (35 − 53) = 27 − 18 = 9, so the total was 9 short. Spread over 15 numbers that is 0.6, and the correct average is 41.6.

Time check: about 15 seconds.

Question 8

The average of 13 results is 60. The average of the first 7 is 57 and the average of the last 7 is 64. Find the 7th result.

Answer. I use differences from 60. The first 7 are 3 below each (−21) and the last 7 are 4 above each (+28). The 7th result is in both, so it is 60 − 21 + 28 = 67.

Time check: about 15 seconds. The totals route (399 + 448 − 780) gives the same answer with bigger numbers.

Question 9

The average temperature from Monday to Thursday was 37°C, and from Tuesday to Friday it was 35.25°C. Friday's temperature was $\frac{5}{6}$ of Monday's. Find Friday's temperature.

Answer. Tuesday to Thursday are in both groups, so Monday − Friday = 4 × 1.75 = 7°C. Friday is $\frac{5}{6}$ of Monday, so that gap of 7 is $\frac{1}{6}$ of Monday. Monday was 42°C and Friday was 35°C.

Time check: about 20 seconds.

Question 10

A batsman's average after 21 innings is 44. How many runs must he score in the 22nd innings to raise his average to 46?

Answer. Runs needed = new average + old innings × rise = 46 + 21 × 2 = 88.

Time check: about 10 seconds.

Question 11

A bowler's average is 11 runs per wicket. In his next match he takes 4 wickets for 29 runs, and his average falls to 10.7. How many wickets had he taken before this match?

Answer. At the new average, 4 wickets would "cost" 4 × 10.7 = 42.8 runs. He gave only 29, which saves 13.8. That saving makes every old wicket 0.3 cheaper, so old wickets = 13.8 ÷ 0.3 = 46. Check: 46 × 11 + 29 = 535, and 535 ÷ 50 = 10.7.

Time check: about 25 seconds.

Question 12

Riya's average in 6 tests is 71. What must she score in the 7th test to make her average 73?

Answer. Target + tests done × gap = 73 + 6 × 2 = 85.

Time check: about 10 seconds.

Question 13

In a factory, the average monthly salary of all the workers is ₹21,600. The 9 technicians average ₹28,000 and the rest average ₹19,200. How many workers are there in all?

Answer. Distances from ₹21,600: the technicians are 6,400 above and the rest are 2,400 below. The sizes go the other way, so technicians : rest = 2,400 : 6,400 = 3 : 8. With 9 technicians, 1 unit = 3, so the rest number 24 and the total is 33.

Time check: about 25 seconds.

Question 14

The average age of 27 students is 12 years. When the teacher's age is included, the average goes up by 1.25 years. How old is the teacher?

Answer. Teacher = new average + students × increase = 13.25 + 27 × 1.25 = 13.25 + 33.75 = 47 years.

Time check: about 15 seconds.

Question 15

Three sections have 25, 40 and 55 students, with average marks of 64, 53 and 60. Find the average of all the students together.

Answer. I use 60 as the base: 25 × (+4) + 40 × (−7) + 55 × 0 = 100 − 280 = −180. There are 120 students, so the average is 60 − 1.5 = 58.5. The plain average of 64, 53 and 60 is 59, which is close but wrong.

Time check: about 25 seconds.