1. How a New Value Shifts an Average
"The average age of 10 people is 25. A new person joins and the average becomes 26. Find the new person's age." Recomputing a full average is slow — a shift-based shortcut is faster.
New person's value = (new average)
+ (number of ORIGINAL people) x (increase in average)
Worked: 10 people avg 25, new average after 1 joins is 26.
New person's age = 26 + 10 x (26 - 25)
= 26 + 10
= 36The intuition: the new average (26) itself accounts for the new person being "one of the group now," but the new person also has to make up for every ORIGINAL person being 1 below the new average — 10 people, each 1 short, is 10 extra that has to come from the new person alone. This same "shift × count" logic works for someone leaving a group too, with the sign flipped.
2. The Alligation Cross
"How much of a 20%-fat milk should be mixed with 5%-fat milk to get a mixture that's 12% fat?" The alligation cross answers any two-component mixture question in one diagram.
Cheaper/weaker value (5) Dearer/stronger value (20)
\ /
\ /
\ /
Mean value (12)
/ \
/ \
(20 - 12) = 8 (12 - 5) = 7
Ratio of quantities (cheaper : dearer) = 8 : 7
Read the cross DIAGONALLY: each quantity's ratio number comes
from the DIFFERENCE on the OPPOSITE side of the mean.So mix 5%-fat and 20%-fat milk in the ratio 8:7 to get a 12%-fat mixture. The cross-diagonal reading is the one detail worth double-checking every time: the quantity of the cheaper item is proportional to the gap on the dearer side, not its own side — mixing this up flips the ratio backward.
3. Alligation Is Just Weighted Average
The alligation cross isn't a separate trick — it's the weighted-average formula, rearranged to solve for the ratio instead of the mean. Seeing this connection makes the cross something you can re-derive, not just something to memorize.
Weighted average: Mean = (q1 x v1 + q2 x v2) / (q1 + q2)
Rearranging to solve for q1/q2 (with v1=5, v2=20, Mean=12):
12(q1 + q2) = 5q1 + 20q2
12q1 + 12q2 = 5q1 + 20q2
7q1 = 8q2
q1/q2 = 8/7
Same 8:7 ratio as the cross -- the cross is a fast shortcut
for exactly this algebra, not a different method.This is worth internalizing because it means the alligation cross generalizes cleanly to averages, percentages, prices, speeds — anything that's fundamentally a weighted average of two values, which is a much broader set of problems than "actual liquid mixtures" alone.
4. Repeated Replacement
A container has pure liquid. Some is removed and replaced with water, repeatedly.
What fraction remains pure after n replacements? This is the one
formula in this chapter that isn't alligation, and is worth having separately.
Remaining pure quantity = Initial x (1 - removed/total)^n
Worked: a 20-liter container is full of milk. 4 liters are
removed and replaced with water, and this is repeated 3 times
total. Find the amount of pure milk remaining.
Remaining = 20 x (1 - 4/20)^3
= 20 x (0.8)^3
= 20 x 0.512
= 10.24 liters
The exponent structure mirrors Week 5's compound interest exactly: each replacement
multiplies the remaining pure quantity by the same fraction, so after
n replacements the fraction remaining is that ratio raised to the
nth power — "compounding," here applied to dilution instead of growth.
5. Practice Problems
30 problems, no calculator — Easy, Medium & Tough
Use the alligation cross for every two-component mixture problem below. Easy problems should take under 30 seconds each; Medium under 60 seconds; Tough under 90 seconds.
Easy (1–10)
- Find the average of 12, 18, 24, 30.
- Find the average of the first 10 natural numbers.
- The average of 5 numbers is 20. Find their sum.
- Find the average of 15, 25, 35.
- The average weight of 4 students is 55kg. Find their total weight.
- Find the average of all even numbers from 2 to 20.
- The average of two numbers is 25. If one number is 30, find the other.
- Find the average speed of a car that travels 100km in 2 hours at constant speed.
- The average marks of a class of 30 students is 60. Find the total marks.
- Find the average of 7, 14, 21, 28, 35.
1) (12+18+24+30)/4=21. 2) (1+2+...+10)/10=5.5. 3) 20×5=100. 4) (15+25+35)/3=25. 5) 55×4=220kg. 6) (2+20)/2=11. 7) 25×2-30=20. 8) 100/2=50 km/h. 9) 60×30=1,800. 10) (7+14+21+28+35)/5=21.
Medium (11–20)
- The average age of 30 students is 12 years. Including the teacher's age, the average becomes 13. Find the teacher's age.
- The average of 6 numbers is 18. If one number is excluded, the average becomes 16. Find the excluded number.
- In what ratio should tea worth Rs. 40/kg be mixed with tea worth Rs. 60/kg to get a mixture worth Rs. 45/kg?
- Find the average of the first 5 multiples of 4.
- A cricketer's average after 10 innings is 40. After the 11th innings, his average becomes 42. Find his score in the 11th innings.
- Find the average of all odd numbers between 1 and 15 (inclusive).
- Two vessels contain milk worth Rs. 20/liter and Rs. 30/liter. In what ratio should they be mixed to get a mixture worth Rs. 24/liter?
- The average weight of 8 people increases by 2kg when a new person replaces one weighing 60kg. Find the new person's weight.
- A batsman scores 87 runs in the 17th innings and thus increases his average by 3. Find his average after the 17th innings.
- Find the average of the squares of the first 5 natural numbers.
11) Total with teacher=31×13=403; without=30×12=360; teacher=43. 12) Total=6×18=108; new total=16×5=80; excluded=28. 13) Cross: (60-45):(45-40)=15:5=3:1. 14) Multiples 4,8,12,16,20; avg=(4+20)/2=12. 15) Total after 11=42×11=462; after 10=400; 11th score=62. 16) Odd 1-15: avg=(1+15)/2=8. 17) Cross: (30-24):(24-20)=6:4=3:2. 18) Increase×count=2×8=16; new person=60+16=76kg. 19) Total before=16x; 16x+87=17(x+3) → x=36; average after=39. 20) (1+4+9+16+25)/5=11.
Tough (21–30)
- The average of 11 numbers is 60. The average of the first 6 is 58 and that of the last 6 is 63. Find the 6th number.
- A vessel contains 40 liters of milk. 8 liters are removed and replaced with water, and this is repeated once more (2 replacements total). Find the amount of pure milk remaining.
- The average of 20 numbers is 15. If each number is multiplied by 3, find the new average.
- In what ratio must rice at Rs. 32/kg be mixed with rice at Rs. 24/kg so that, on selling the mixture at Rs. 30/kg, a profit of 10% is made?
- The average of 9 consecutive numbers is 20. Find the largest number.
- A shopkeeper mixes two varieties of rice in the ratio 3:2, costing Rs. 20/kg and Rs. 25/kg respectively, and sells the mixture at Rs. 26/kg. Find his profit percentage.
- The average weight of A, B and C is 45kg. The average weight of A and B is 40kg, and that of B and C is 43kg. Find B's weight.
- A container has milk and water in the ratio 4:1. If 10 liters of the mixture is replaced with pure water, the new ratio becomes 2:1. Find the initial quantity of milk.
- The average of 7 numbers is 18. If one number is removed, the average becomes 17. Find the removed number.
- Find the average of all multiples of 7 between 1 and 100.
21) Total(11)=660; first6+last6=348+378=726 (6th number counted twice); 6th number=726-660=66. 22) 40×(1-8/40)²=40×0.64=25.6 liters. 23) New average=15×3=45. 24) Mean cost=30/1.1=300/11; alligation gives Rs.24-rice:Rs.32-rice=13:9, i.e. Rs.32-rice:Rs.24-rice=9:13. 25) Middle(5th)=20; largest(9th)=24. 26) Mixture cost=(3×20+2×25)/5=22; profit=(26-22)/22×100≈18.18%. 27) A+B+C=135; A+B=80 → C=55; B+C=86 → B=86-55=31kg. 28) Removing 10L keeps ratio 4:1 in what's removed; solve 4x-8=2(x+8) → x=12; initial milk=4x=48 liters. 29) Total=126; new total=102; removed=24. 30) Multiples 7 to 98 (14 terms); avg=(7+98)/2=52.5.
6. Knowledge Check
Four quick questions. Expand each to check your answer.
Q1
In the average-shift shortcut, why is the increase in average multiplied by the ORIGINAL number of people, not the new total?
In the average-shift shortcut, why is the increase in average multiplied by the ORIGINAL number of people, not the new total?
Each of the original people individually falls short of the new, higher average by that increase amount — the new person's value has to compensate for all of those shortfalls simultaneously, one per original person. Since there are exactly as many shortfalls as original people, that's the count that gets multiplied by the per-person shortfall.
Q2
Why does the alligation cross assign the cheaper item's ratio number from the gap on the DEARER side, rather than its own side?
Why does the alligation cross assign the cheaper item's ratio number from the gap on the DEARER side, rather than its own side?
This falls directly out of rearranging the weighted-average equation algebraically (Section 3) — solving for the quantity ratio produces exactly this cross-diagonal pattern, where each item's coefficient in the ratio comes from the other item's distance to the mean, not its own. It's a consequence of the algebra, not an arbitrary convention.
Q3
Why is the alligation cross considered "the same method" as the weighted-average formula rather than a separate technique?
Why is the alligation cross considered "the same method" as the weighted-average formula rather than a separate technique?
Directly rearranging the weighted-average equation to solve for the quantity ratio produces exactly the same result the alligation cross gives — the cross is simply a fast, visual shortcut for that one specific piece of algebra, not an independent formula that happens to give the same answer.
Q4
Why does the repeated-replacement formula raise the retained fraction to the power of n, rather than multiplying by n?
Why does the repeated-replacement formula raise the retained fraction to the power of n, rather than multiplying by n?
Each replacement operates on whatever pure liquid remains AFTER the previous replacement, not on the original full amount — so the same fraction is retained repeatedly, compounding multiplicatively each time rather than being subtracted the same fixed amount each round. This is structurally identical to compound interest's repeated multiplicative growth, just working in the opposite (shrinking) direction.