Average: Math Ka Jugaad

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Average: Matlab, Beech Ka Banda! Average, yaani "औसत," woh magical number hai jo saare numbers ko sum karke unki ginti se divide karne par milta hai. Socho, agar saare numbers ek hi party mein hain, toh average unka sabse balanced dost hai jo sabko khush rakhta hai! Formula: $Average = \frac{Sum~of~all~terms}{Number~of~terms}$ Deviation Method: Jab Data Bada Ho, Toh Jugaad Bada! Jab numbers bade-bade hon, aur unka sum karna headache lage, toh deviation method aata hai rescue karne. Yeh method kehta hai, "Ek assumed average le lo, aur baaki numbers kitne 'tedhe' hain usse, woh dekh lo." Step 1: Ek "Assumed Average" choose karo. Koi bhi number jo tumhe beech ka lagta ho. Isko 'A' bolte hain. Step 2: Har number ka deviation nikalo assumed average se. Deviation = (Number - Assumed Average). Step 3: Saare deviations (positive aur negative) ko sum karo. Isko "Sum of Deviations" bolte hain. Step 4: "Sum of Deviations" ko total number of terms se divide karo. Step 5: Final average = Assumed Average + (Result from Step 4). Ta-da! Example: Numbers 48, 52, 40, 60, 53, 53, 49. Let's assume A = 50. Deviations: (48-50)=-2, (52-50)=+2, (40-50)=-10, (60-50)=+10, (53-50)=+3, (53-50)=+3, (49-50)=-1 Sum of Deviations: $-2+2-10+10+3+3-1 = 5$ Number of terms: 7 Average: $50 + \frac{5}{7} = 50 + 0.714... \approx 50.71$ Kramagat Numbers Ke Funde (Consecutive Numbers) 1. Natural Numbers (1, 2, 3...) Average of first 'n' natural numbers: $\frac{n+1}{2}$ Average of square of first 'n' natural numbers: $\frac{(n+1)(2n+1)}{6}$ Average of cube of first 'n' natural numbers: $\frac{n(n+1)^2}{4}$ 2. Even Numbers (2, 4, 6...) Average of first 'n' even numbers: $n+1$ Average of square of first 'n' even numbers: $\frac{2(n+1)(2n+1)}{3}$ Average of cube of first 'n' even numbers: $2n(n+1)^2$ 3. Odd Numbers (1, 3, 5...) Average of first 'n' odd numbers: $n$ Average of square of first 'n' odd numbers: $\frac{(2n-1)(2n+1)}{3}$ Average of cube of first 'n' odd numbers: $n(2n^2-1)$ 4. General Consecutive Numbers / AP Series Agar numbers ek arithmetic progression (AP) mein hain, toh average hamesha hota hai: $\frac{First~number + Last~number}{2}$ Concept 01: Missing Number Ka Game Agar kuch numbers ka average diya hai, aur usmein se do numbers ko 'a' aur 'b' maan liya, aur unka sum 'k' hai, toh: Sum of all numbers = $n \times Average$ Agar teen numbers $2a, 2b, 2c$ hain aur unka average 40 hai, toh: $2a+2b+2c = 3 \times 40 = 120$ $2(a+b+c) = 120 \implies a+b+c = 60$ Example: Three natural numbers ka average 60 hai. First number $2a$, second $2a-k$, third $2a+k$. Average 60 hai. Sum: $2a + (2a-k) + (2a+k) = 6a$ $6a = 3 \times 60 = 180 \implies a = 30$ Numbers: $60, 60-k, 60+k$. Ismein k ki value pata nahi, par 'a' mil gaya. Concept 02: Average of Averages Jab aapko $n_1$ observations ka average $a_1$ diya ho, $n_2$ observations ka average $a_2$ diya ho, aur aise hi aage. Toh total average nikalne ka tareeka: Overall Average = $\frac{n_1 a_1 + n_2 a_2 + n_3 a_3 + ...}{n_1 + n_2 + n_3 + ...}$ Example: First 12 observations ka average 36 hai. Remaining 9 observations ka average 15 hai. Total 21 observations ka average kya hai? Sum of first 12 = $12 \times 36 = 432$ Sum of remaining 9 = $9 \times 15 = 135$ Total Sum = $432 + 135 = 567$ Overall Average = $\frac{567}{12+9} = \frac{567}{21} = 27$ Concept 03: Jab Data Mein Miscalculation Ho Jaye (When data gets messed up) Kabhi-kabhi data entry mein gadbad ho jaati hai. Jaise, ek number ko galat padh liya. Isko theek kaise karein? Method: $Correct~Average = Old~Average + \frac{Sum~of~corrections}{Total~numbers}$ Where $Sum~of~corrections = (Correct~value_1 - Incorrect~value_1) + (Correct~value_2 - Incorrect~value_2) + ...$ Example 1: 10 numbers ka average 40 hai. Ek number 64 ko galati se 46 padh liya. Correct average kya hai? Old Average = 40 Incorrect value = 46, Correct value = 64 Correction needed = $64 - 46 = +18$ (Sum mein 18 add karna hai) Correct Average = $40 + \frac{18}{10} = 40 + 1.8 = 41.8$ Example 2: 'n' numbers ka average 'm' hai. Ek number 'p' ko galati se 'q' padh liya aur ek number 'x' ko galati se 'y' padh liya. Correct average? Sum of Corrections = $(p-q) + (x-y)$ Correct Average = $m + \frac{(p-q) + (x-y)}{n}$ Concept 04: Age Related Average (Umra Ka Hisaab) Jab groups mein log add ya remove hote hain, toh average age kaise badalti hai? Example: Ek class mein 40 students ka average 15 saal hai. Agar 8 new students join karte hain toh average age 0.2 saal badh jaati hai. New students ki average age kya hai? Old total age = $40 \times 15 = 600$ years New total students = $40 + 8 = 48$ New average age = $15 + 0.2 = 15.2$ years New total age = $48 \times 15.2 = 729.6$ years Age of 8 new students = $729.6 - 600 = 129.6$ years Average age of 8 new students = $\frac{129.6}{8} = 16.2$ years Concept 05: Replacement, Inclusion, Exclusion (Aadmi Aaye, Aadmi Gaye!) Jab group mein koi naya member aata hai, ya koi purana member chala jata hai, ya koi replace ho jata hai, toh average kaise change hota hai? 1. Inclusion (Naya Banda Aaya) Example: Ek class mein 50 students ka average weight 45 kg hai. Jab ek naya student aata hai, toh average weight 100g badh jata hai. New student ka weight kya hai? Old total weight = $50 \times 45 = 2250$ kg New number of students = $50+1 = 51$ Increase in average = 100g = 0.1 kg New average weight = $45 + 0.1 = 45.1$ kg New total weight = $51 \times 45.1 = 2299.1$ kg Weight of new student = $2299.1 - 2250 = 49.1$ kg 2. Exclusion (Ek Banda Chala Gaya) Example: 34 students ka average weight 42 kg hai. Agar teacher ka weight bhi include karein toh average 400g badh jata hai. Teacher ka weight kya hai? Total weight of students = $34 \times 42 = 1428$ kg Increase in average = 400g = 0.4 kg New average (students + teacher) = $42 + 0.4 = 42.4$ kg Total number of people = $34 + 1 = 35$ Total weight (students + teacher) = $35 \times 42.4 = 1484$ kg Teacher's weight = $1484 - 1428 = 56$ kg 3. Replacement (Ek Gaya, Doosra Aaya) Example: 25 boys ka average height 1.5 m hai. Agar 5 boys group chhod dete hain aur 5 naye boys join karte hain, toh average height 0.15 m badh jati hai. Naye 5 boys ka average height kya hai? Initial total height of 25 boys = $25 \times 1.5 = 37.5$ m Number of boys remains 25. New average height = $1.5 + 0.15 = 1.65$ m New total height of 25 boys = $25 \times 1.65 = 41.25$ m Increase in total height = $41.25 - 37.5 = 3.75$ m This increase in total height is due to the difference between the sum of heights of the 5 incoming boys and the 5 outgoing boys. Let the sum of heights of 5 outgoing boys be $S_{out}$ and 5 incoming boys be $S_{in}$. So, $S_{in} - S_{out} = 3.75$ m. To find the average height of the new 5 boys, we need $S_{in}$. The problem implies that the 5 boys who left had the initial average height. If the 5 outgoing boys had the initial average height, $S_{out} = 5 \times 1.5 = 7.5$ m. Then, $S_{in} = 3.75 + S_{out} = 3.75 + 7.5 = 11.25$ m. Average height of 5 new boys = $\frac{11.25}{5} = 2.25$ m. Concept 06: Cricket Based (Cricket Ka Khel, Average Ka Mel) Cricket mein runs, wickets ka average kaise nikalte hain. Thoda tricky hota hai, dhyan se dekho! 1. Batting Average Batting Average = $\frac{Total~Runs~Scored}{Total~Innings~Played}$ (Not Out innings ko exclude karte hain from denominator for average calculation, or consider them as completed innings but not as 'dismissals' in some contexts. For general average, total innings played is common) Example: Ek batsman ne 12 innings mein kuch runs banaye. 12th inning mein 90 runs banaye, jisse uska average 5 runs badh gaya. Uska average after 12 innings kya hai? Let average before 12th inning be 'x'. Total runs in 11 innings = $11x$. After 12th inning, total runs = $11x + 90$. New average = $x+5$. New total runs = $12(x+5)$. So, $11x + 90 = 12(x+5)$ $11x + 90 = 12x + 60$ $x = 30$. Average after 12 innings = $x+5 = 30+5 = 35$. 2. Bowling Average Bowling Average = $\frac{Total~Runs~Given}{Total~Wickets~Taken}$ Example: Ek bowler ka average 12.4 runs per wicket hai. Last match mein usne 26 runs dekar 5 wickets liye, jisse uska average 0.4 runs improve ho gaya (matlab average kam ho gaya). Ab uske total wickets kitne hain? Let total wickets before last match be 'w'. Total runs given before last match = $12.4w$. Last match mein runs = 26, wickets = 5. New total runs = $12.4w + 26$. New total wickets = $w + 5$. New (improved) average = $12.4 - 0.4 = 12$. So, $\frac{12.4w + 26}{w + 5} = 12$. $12.4w + 26 = 12(w + 5)$ $12.4w + 26 = 12w + 60$. $0.4w = 34 \implies w = \frac{34}{0.4} = 85$. Total wickets after last match = $w + 5 = 85 + 5 = 90$. Concept 07: Age Based (Umra Ka Khel, Thoda Complex!) Jab family members ki ages ka average diya ho, aur kuch saal baad ya kuch saal pehle ka average pucha ho. Example: 5 saal pehle A, B, C, D ka average age 45 saal tha. Ab E bhi join kar gaya, aur paanchon ka average age 49 saal hai. E ki age kya hai? 5 saal pehle, A, B, C, D ka sum of ages = $4 \times 45 = 180$ saal. Aaj, A, B, C, D ki age har ek ki 5 saal badh gayi hai. So, total increase = $4 \times 5 = 20$ saal. Aaj, A, B, C, D ka sum of ages = $180 + 20 = 200$ saal. Aaj, A, B, C, D, E paanchon ka average age 49 saal hai. Aaj, A, B, C, D, E paanchon ka sum of ages = $5 \times 49 = 245$ saal. E ki age = (Sum of 5) - (Sum of 4) = $245 - 200 = 45$ saal. Concept 08: Number Based (Numberon Ka Chakkar) Jab numbers ek series mein ho, aur unme se kuch nikal diye jayein ya add kar diye jayein. Example: 25 natural numbers ka average 67 hai. Agar 1 number ko remove kar dein toh average 3 kam ho jata hai. Removed number kya hai? Total sum of 25 numbers = $25 \times 67 = 1675$. Naye numbers = 24. Naya average = $67 - 3 = 64$. Naya total sum = $24 \times 64 = 1536$. Removed number = Old total sum - New total sum = $1675 - 1536 = 139$. Example: 11 numbers ka average 42 hai. Agar first 6 numbers ka average 40 hai aur last 6 numbers ka average 44 hai. 6th number kya hai? Sum of 11 numbers = $11 \times 42 = 462$. Sum of first 6 numbers = $6 \times 40 = 240$. Sum of last 6 numbers = $6 \times 44 = 264$. Sum of (first 6 + last 6) = $240 + 264 = 504$. Yahan 6th number do baar count ho gaya hai. 6th number = (Sum of first 6 + Sum of last 6) - (Sum of 11 numbers) 6th number = $504 - 462 = 42$.