JEE MathsStatisticsVisual Solution
Visual SolutionHard
Question
If the mean and variance of eight numbers 3, 7, 9, 12, 14, 20, xx and yy are 10 and 25 respectively, then xyx \cdot y is equal to:
Solution Path
From mean: x+y=15x+y = 15. From variance: x2+y2=121x^2+y^2 = 121. Using (x+y)2=x2+y2+2xy(x+y)^2 = x^2+y^2+2xy: xy=52xy = 52.
01Question Setup
1/4
Given eight numbers 3, 7, 9, 12, 14, 20, xx, yy with mean 10 and variance 25. Find xyx \cdot y.
Find xyx \cdot y
02Use Mean Condition
2/4
Sum = 8×10=808 \times 10 = 80. Known sum = 3+7+9+12+14+20=653+7+9+12+14+20 = 65. So x+y=15x + y = 15.
x+y=15x + y = 15
03Use Variance ConditionKEY INSIGHT
3/4
Variance = xi28100=25\frac{\sum x_i^2}{8} - 100 = 25, so xi2=1000\sum x_i^2 = 1000. Known sum of squares = 879. Therefore x2+y2=121x^2 + y^2 = 121.
x2+y2=121x^2 + y^2 = 121
04Final Answer
4/4
Using (x+y)2=x2+y2+2xy(x+y)^2 = x^2 + y^2 + 2xy: 225=121+2xy225 = 121 + 2xy, so xy=52xy = 52.
xy=52\boxed{xy = 52}
Concepts from this question2 concepts unlocked

Variance Shortcut Formula

EASY

Variance = E(X²) - [E(X)]². Compute the mean of squares minus the square of the mean. Avoids computing individual deviations.

σ2=fixi2N(fixiN)2\sigma^2 = \frac{\sum f_i x_i^2}{N} - \left(\frac{\sum f_i x_i}{N}\right)^2

Nearly every JEE variance question is faster with the shortcut. The direct formula with (xᵢ - mean)² takes longer and has more room for arithmetic errors.

Variance computationSD from raw dataGrouped data variance
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Effect of Linear Transformation on Mean and Variance

STANDARD

If yᵢ = axᵢ + b, then mean(y) = a * mean(x) + b, and Var(y) = a² * Var(x). Adding a constant does not change variance. Multiplying by a constant scales variance by a².

yˉ=axˉ+b,σy2=a2σx2\bar{y} = a\bar{x} + b, \quad \sigma_y^2 = a^2 \sigma_x^2

JEE frequently asks 'if 5 is added to each observation' or 'if each value is doubled'. This single rule answers all such questions instantly.

Change of origin/scaleEffect of adding constantEffect of multiplying constant
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