Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
If the mean of the data
| Class | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 |
| Frequency | 2 | k | 28 | 54 | k+1 | 5 |
is 21, then k is one of the roots of the equation :
(1) $2x^2 - 23x - 10 = 0$
(2) $4x^2 - 35x + 24 = 0$
(3) $2x^2 - 19x - 10 = 0$
(4) $2x^2 - 35x + 98 = 0$
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Answer: Option (3)
Key idea: Mean of grouped data and finding equation from roots
The mean and variance of n observations are 8 and 16, respectively. If the sum of the first $(n-1)$ observations is 48 and the sum of squares of the first $(n-1)$ observations is 496, then the value of n is:
(1) 21
(2) 16
(3) 13
(4) 7
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Answer: Option (4)
Key idea: Mean and Variance formulas with algebraic manipulation
Use mean formula: $\sum x_i = n \times 8 = 8n$. Since $\sum_{i=1}^{n-1} x_i = 48$, we get $x_n = 8n - 48$
Suppose that the mean and median of the non-negative numbers $21, 8, 17, a, 51, 103, b, 13, 67, (a > b)$, are 40 and 21, respectively. If the mean deviation about the median is 26, then $2a$ is equal to:
(1) 109
(2) 117
(3) 161
(4) 131
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
For 10 observations $x_1, x_2, ..., x_{10}$, if $\sum_{i=1}^{10}(x_i + 2)^2 = 180$ and $\sum_{j=1}^{10}(x_j - 1)^2 = 90$, then their standard deviation is:
(1) 2
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Answer: Option (3)
Key idea: Standard Deviation using algebraic expansion of summations
Expand $\sum(x_i + 2)^2 = 180$ to get $\sum x_i^2 + 4\sum x_i + 40 = 180$, so $\sum x_i^2 + 4\sum x_i = 140$
Expand $\sum(x_j - 1)^2 = 90$ to get $\sum x_j^2 - 2\sum x_j + 10 = 90$, so $\sum x_j^2 - 2\sum x_j = 80$
Subtract the two equations: $(\sum x_i^2 + 4\sum x_i) - (\sum x_i^2 - 2\sum x_i) = 60$, giving $6\sum x_i = 60$, so $\sum x_i = 10$
The mean deviation about the mean for the data
| $x_i$ | 5 | 7 | 9 | 10 | 12 | 15 |
|-------|---|---|---|----|----|----|
| $f_i$ | 8 | 6 | 2 | 2 | 2 | 6 |
is equal to:
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Answer: 44/13
Key idea: Mean Deviation about Mean for Grouped Data
A variable X takes values 0, 0, 2, 6, 12, 20, ..., n(n-1) with frequencies $^nC_0$, $^nC_1$, $^nC_2$, $^nC_3$, $^nC_4$, $^nC_5$, ..., $^nC_n$, respectively. If the mean of this data is 60, then its median is:
(1) 56
(2) 42
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Answer: Option (1)
Key idea: Mean and Median of Discrete Distribution with Binomial Coefficients
Identify that the $k$-th value is $k(k-1)$ with frequency $^nC_k$
Calculate mean using $\sum k(k-1) \cdot ^nC_k = n(n-1) \cdot 2^{n-2}$ and total frequency $2^n$
Set up equation: $\frac{n(n-1)}{4} = 60$, giving $n(n-1) = 240$
Solve $n^2 - n - 240 = 0$ to get $n = 16$
Total frequency is $2^{16} = 65536$. By symmetry, $\sum_{k=0}^{7} ^{16}C_k = 32768$
The median position is at $32768.5$, between values at $k=7$ (value $42$) and $k=8$ (value $56$)
The median is $56$ (the value at the $(n/2 + 1)$-th position)
A data consists of 20 observations $x_1, x_2,..., x_{20}$. If $\sum_{i=1}^{20}(x_i+5)^2 = 2500$ and $\sum_{i=1}^{20}(x_i-5)^2 = 100$, then the ratio of mean to standard deviation of this data is:
(1) $2:1$
(2) $3:1$
(3) $3:2$
(4) $4:1$
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Answer: Option (2)
Key idea: Mean and Standard Deviation using algebraic manipulation of summations
Let the mean and the variance of seven observations $2, 4, \alpha, 8, \beta, 12, 14$, $\alpha < \beta$, be $8$ and $16$ respectively. Then the quadratic equation whose roots are $3\alpha + 2$ and $2\beta + 1$ is :
(1) $x^2 - 35x + 306 = 0$
(2) $x^2 - 41x + 420 = 0$
(3) $x^2 - 45x + 506 = 0$
(4) $x^2 - 37x + 342 = 0$
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Answer: Option (2)
Key idea: Mean and Variance of data, forming quadratic equations from roots
Use mean condition: $\frac{2+4+\alpha+8+\beta+12+14}{7} = 8$ gives $\alpha + \beta = 16$
A set of four observations has mean 1 and variance 13. Another set of six observations has mean 2 and variance 1. Then, the variance of all these 10 observations is equal to:
