Statistics Class 10 Previous Year Questions (Chapter 13)

1. 1

   Step 1: Prepare the table of class marks and products

   2 marks

   | Lifetime (in hours) | \(f_i\) | \(x_i\) | \(f_i x_i\) | | --- | --- | --- | --- | | 0 – 20 | 10 | 10 | 100 | | 20 – 40 | 35 | 30 | 1050 | | 40 – 60 | 50 | 50 | 2500 | | 60 – 80 | 60 | 70 | 4200 | | 80 – 100 | 30 | 90 | 2700 | | 100 – 120 | 15 | 110 | 1650 |

   Why? A class interval has no single value, so NCERT uses its mid-point, the class mark, as the representative value \(x_i\) of every observation in that class.

2. 2

   Step 2: Add the columns

   2 marks

   \(\sum f_i = 10 + 35 + 50 + 60 + 30 + 15 = 200\) and \(\sum f_i x_i = 100 + 1050 + 2500 + 4200 + 2700 + 1650 = 12200\).

   Why? Adding the frequencies gives the total number of observations; adding the products gives the total of all the observation values represented by the class marks.

3. 3

   Step 3: Apply the direct method formula

   1 mark

   The mean lifetime is \(\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{12200}{200} = 61\) hours.

   Why? The direct method formula for the mean of grouped data is \(\bar{x} = \frac{\sum f_i x_i}{\sum f_i}\); substituting the two sums gives the average lifetime.

Marks: 2 + 2 + 1 = 5 · Split as in the CBSE marking scheme, 2025 · Basic 430/1/1 Q35 (the marks split is taken from the 2025 Basic paper, where this question also appeared)

1. 1

   Step 1: Find the class marks

   1 mark

   Take the class mark as x_i for each class: (50+100)/2 = 75, (100+150)/2 = 125, (150+200)/2 = 175, (200+250)/2 = 225, (250+300)/2 = 275, (300+350)/2 = 325, (350+400)/2 = 375.

   Why? NCERT assumes the frequency of each class is centred at its class mark, so the class mark represents all the observations in that class.

2. 2

   Step 2: Find f_i x_i

   1 mark

   f_i x_i: 4×75 = 300, 5×125 = 625, 13×175 = 2275, 20×225 = 4500, 14×275 = 3850, 8×325 = 2600, 4×375 = 1500.

   Why? Multiplying each class mark by its frequency gives the total consumption contributed by the consumers in that class.

3. 3

   Step 3: Total the frequencies

   1 mark

   Σf_i = 4+5+13+20+14+8+4 = 68.

   Why? The sum of the frequencies is the total number of consumers, which is the denominator of the mean.

4. 4

   Step 4: Total f_i x_i

   1 mark

   Σf_i x_i = 300+625+2275+4500+3850+2600+1500 = 15650.

   Why? Adding the products gives the total electricity consumed by all the consumers.

5. 5

   Step 5: Apply the mean formula

   1 mark

   x̄ = Σf_i x_i / Σf_i = 15650/68 ≈ 230.15 units. Therefore, mean monthly consumption = 230.15 units (approx.).

   Why? Substituting the totals into the direct method formula gives the average consumption per consumer.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Split as in the CBSE marking scheme, 2026 · Basic 430/1/2 Q34

1. 1

   Step 1: Use total frequency

   1 mark (suggested)

   Let x be the number of households in the 1250 – 1500 class. Since there are 25 households, 4 + (2x + 1) + 12 + x + 2 = 25.

   Why? The sum of all frequencies in a grouped frequency distribution equals the total number of observations, so adding the five frequencies gives the equation in x.

2. 2

   Step 2: Solve for x

   1 mark (suggested)

   3x + 19 = 25, so 3x = 6, x = 2.

   Why? Collecting like terms and using inverse operations isolates x; the value must make the total of the frequencies 25.

3. 3

   Step 3: Mean table with class marks

   2 marks (suggested)

   | Daily Expenditure (in ₹) | Number of households (f_i) | Class mark (x_i) | f_i x_i | | --- | --- | --- | --- | | 500 – 750 | 4 | 625 | 2500 | | 750 – 1000 | 5 | 875 | 4375 | | 1000 – 1250 | 12 | 1125 | 13500 | | 1250 – 1500 | 2 | 1375 | 2750 | | 1500 – 1750 | 2 | 1625 | 3250 | | Total | 25 | | 26375 |

   Why? For grouped data, each class is represented by its class mark, the average of its two limits; the products f_i x_i and their sum give the numerator of the mean formula.

4. 4

   Step 4: Find the mean

   1 mark (suggested)

   x̄ = Σ f_i x_i / Σ f_i = 26375/25 = 1055. Therefore, the mean daily expenditure is ₹1055.

   Why? The direct method formula for the mean divides the sum of the f_i x_i products by the total frequency.

Marks: 1 + 1 + 2 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Identify the modal class

   1 mark (suggested)

   The maximum frequency is 30, so the modal class is 40 – 60.

   Why? In grouped data, the class with the highest frequency is called the modal class; the mode lies inside this class.

2. 2

   Step 2: Substitute into the mode formula

   1 mark (suggested)

   Substitute l as 40, h as 20, f1 as 30, f0 as 25, f2 as 24 in Mode = l + ((f1 − f0)/(2f1 − f0 − f2)) × h.

   Why? The mode formula uses the lower limit of the modal class, the class size, and the frequencies of the modal class and its two neighbouring classes.

3. 3

   Step 3: Calculate the mode

   1 mark (suggested)

   Mode = 40 + (5/11) × 20 = 40 + 100/11 = 49.09. The mode of the distribution is 49.09 marks.

   Why? The fraction gives the correction to be added to the lower limit; this locates the mode inside the modal class.

Marks: 1 + 1 + 1 = 3 · Suggested split (CBSE’s usual pattern for 3-mark questions)

1. 1

   Step 1: Cumulative frequency table

   1 mark (suggested)

   | Class | 100−105 | 105−110 | 110−115 | 115−120 | 120−125 | 125−130 | | Frequency | 6 | 8 | 10 | 4 | 9 | 3 | | Cumulative frequency | 6 | 14 | 24 | 28 | 37 | 40 |

   Why? The cumulative frequency of a class is the sum of the frequencies of all classes up to that class; it shows how many observations lie up to each upper limit.

2. 2

   Step 2: Locate the median class

   1 mark (suggested)

   n = 6 + 8 + 10 + 4 + 9 + 3 = 40, so n/2 = 20. The cumulative frequency 24 is just greater than 20, so the median class is 110−115.

   Why? The median class is the class whose cumulative frequency is just greater than n/2; the median formula places the middle observation inside this class.

3. 3

   Step 3: Median value

   1 mark (suggested)

   Median = 110 + ((20 − 14)/10) × 5 = 110 + 3 = 113.

   Why? Multiplying the fraction by the class size gives the distance of the median from the lower limit of the median class.

4. 4

   Step 4: Modal class and formula

   1 mark (suggested)

   The highest frequency is 10, so the modal class is 110−115. Mode = 110 + ((10 − 8)/(2×10 − 8 − 4)) × 5.

   Why? The modal class has the maximum frequency; the mode formula uses the frequencies of the modal class and its two neighbouring classes.

5. 5

   Step 5: Mode value and result

   1 mark (suggested)

   Mode = 110 + (2/8) × 5 = 110 + 1.25 = 111.25. Hence the median is 113 and the mode is 111.25.

   Why? The formula gives the mode as a point inside the modal class, and the value lies inside the modal class, so the result is sensible.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Total frequency equation

   1 mark (suggested)

   Let x be the number of children with pocket allowance 190 – 210 and y the number with 230 – 250. The known frequencies add to 7 + 6 + 9 + 13 + 5 = 40, so x + y = 64 − 40 = 24.

   Why? The sum of all frequencies equals the total number of children, 64. This gives the first equation in the two unknowns.

2. 2

   Step 2: Class marks and f_i x_i

   1 mark (suggested)

   Using class mark = (upper limit + lower limit)/2: | Pocket allowance (in ₹) | f_i | x_i | f_i x_i | | --- | --- | --- | --- | | 110 – 130 | 7 | 120 | 840 | | 130 – 150 | 6 | 140 | 840 | | 150 – 170 | 9 | 160 | 1440 | | 170 – 190 | 13 | 180 | 2340 | | 190 – 210 | x | 200 | 200x | | 210 – 230 | 5 | 220 | 1100 | | 230 – 250 | y | 240 | 240y | So Σf_i x_i = 6560 + 200x + 240y.

   Why? For grouped data, each class is represented by its class mark, the average of its upper and lower limits. The mean formula needs the product of each class mark and its frequency.

3. 3

   Step 3: Mean equation

   1 mark (suggested)

   Mean = (Σf_i x_i)/(Σf_i) = (6560 + 200x + 240y)/64. The given mean is ₹180, so 6560 + 200x + 240y = 180 × 64 = 11520. Hence 200x + 240y = 11520 − 6560 = 4960. Dividing by 40, 5x + 6y = 124.

   Why? The direct method formula for the mean of grouped data is mean = sum of f_i x_i divided by sum of f_i. Equating it to the given mean gives the second equation in x and y.

4. 4

   Step 4: Solve for x

   1 mark (suggested)

   From x + y = 24, y = 24 − x. Substituting into 5x + 6y = 124 gives 5x + 6(24 − x) = 124, so 6 × 24 = 144 and 144 − x = 124. Hence x = 144 − 124 = 20.

   Why? Substitution solves the two linear equations: use x + y = 24 to write y in terms of x, then substitute into 5x + 6y = 124.

5. 5

   Step 5: Find y and state the result

   1 mark (suggested)

   Then y = 24 − 20 = 4. Hence x = 20 and y = 4.

   Why? Once x is known, the total-frequency equation gives y. The final sentence states the two values asked for.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Find class marks and products

   1 mark (suggested)

   Let \(x_i\) be the class mark of each class. | Class | \(f_i\) | \(x_i\) | \(f_i x_i\) | |---|---|---|---| | 0-10 | 12 | 5 | 60 | | 10-20 | 23 | 15 | 345 | | 20-30 | 34 | 25 | 850 | | 30-40 | 25 | 35 | 875 | | 40-50 | 6 | 45 | 270 |

   Why? The mid-point of a class interval is called its class mark, and it represents all the observations of that class in the mean formula.

2. 2

   Step 2: Add the columns

   1 mark (suggested)

   \(\sum f_i = 12 + 23 + 34 + 25 + 6 = 100\) \(\sum f_i x_i = 60 + 345 + 850 + 875 + 270 = 2400\)

   Why? The mean formula needs the total number of students and the total of the products of class marks with their frequencies.

3. 3

   Step 3: Apply the mean formula

   1 mark (suggested)

   Mean \(= \frac{\sum f_i x_i}{\sum f_i} = \frac{2400}{100} = 24\) The mean marks of the students is 24.

   Why? The mean of grouped data is the sum of the products divided by the total frequency.

Marks: 1 + 1 + 1 = 3 · Suggested split (CBSE’s usual pattern for 3-mark questions)

1. 1

   Step 1: Mean table

   1 mark (suggested)

   Assume the frequency of each class is centred at its class mark. Class mark x_i = (upper limit + lower limit)/2. | Class | f | x_i | f_i x_i | | --- | --- | --- | --- | | 0-20 | 5 | 10 | 50 | | 20-40 | 8 | 30 | 240 | | 40-60 | 10 | 50 | 500 | | 60-80 | 12 | 70 | 840 | | 80-100 | 7 | 90 | 630 | | 100-120 | 8 | 110 | 880 | | Total | 50 | | 3140 |

   Why? The class mark is the mid-point of a class interval. The direct method uses each class mark as the representative value of all observations in that class.

2. 2

   Step 2: Mean by direct method

   1 mark (suggested)

   x̄ = Σf_i x_i / Σf_i = 3140/50 = 62.8.

   Why? The mean of grouped data is the sum of the products of frequencies and class marks divided by the total frequency.

3. 3

   Step 3: Cumulative frequency table

   1 mark (suggested)

   | Class | f | cf | | --- | --- | --- | | 0-20 | 5 | 5 | | 20-40 | 8 | 13 | | 40-60 | 10 | 23 | | 60-80 | 12 | 35 | | 80-100 | 7 | 42 | | 100-120 | 8 | 50 | | Total | 50 | |

   Why? The cumulative frequency of a class is the sum of the frequencies of all classes up to that class. It helps locate the middle observation.

4. 4

   Step 4: Median class and formula

   1 mark (suggested)

   n = 50, so n/2 = 25. The class 60-80 has cf = 35, which is greater than and nearest to 25, so it is the median class. Here l = 60, cf = 23, f = 12, h = 20. Median = l + ((n/2 - cf)/f) × h = 60 + ((25 - 23)/12) × 20 = 60 + 10/3 = 190/3.

   Why? The median class is the class whose cumulative frequency is greater than and nearest to n/2. The formula places the median proportionally inside that class.

5. 5

   Step 5: Mean and median

   1 mark (suggested)

   Mean = 62.8 and Median = 190/3 ≈ 63.33.

   Why? The mean and median are the two measures of central tendency asked for. The median divides the data into two equal halves.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Cumulative frequency table

   1 mark (suggested)

   | Monthly consumption (in units) | Number of consumers (f) | Cumulative frequency (cf) | | --- | --- | --- | | 65 – 85 | 7 | 7 | | 85 – 105 | 8 | 15 | | 105 – 125 | 7 | 22 | | 125 – 145 | 20 | 42 | | 145 – 165 | 14 | 56 | | 165 – 185 | 9 | 65 | | 185 – 205 | 5 | 70 |

   Why? The cumulative frequency of a class is found by adding the frequency of that class to the total of the earlier frequencies. It tells how many consumers lie up to the upper limit of each class.

2. 2

   Step 2: Locate the median class

   1 mark (suggested)

   n = 70, so n/2 = 35. The cumulative frequency 42 is just greater than 35, so the median class is 125 – 145.

   Why? The median is the value of the middle observation. n/2 locates the middle position, and the class whose cumulative frequency first reaches or passes this position is the median class.

3. 3

   Step 3: Apply the median formula

   1 mark (suggested)

   Median = l + ((n/2 − cf)/f) × h = 125 + ((35 − 22)/20) × 20.

   Why? Here l is the lower limit of the median class, cf is the cumulative frequency of the class preceding the median class, f is the frequency of the median class, and h is the class size.

4. 4

   Step 4: Calculate the median

   1 mark (suggested)

   = 125 + (13/20) × 20 = 125 + 13 = 138.

   Why? The fraction gives the position of the middle observation inside the median class. Multiplying by the class size locates it that many units above the lower limit.

5. 5

   Step 5: State the result

   1 mark (suggested)

   The median monthly consumption of electricity is 138 units.

   Why? The median divides the data into two equal halves, so about half the consumers use up to this value and half use more.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Cumulative frequency table

   1 mark (suggested)

   | Lifetime (in hours) | Number of lamps (f) | Cumulative frequency (cf) | | --- | --- | --- | | 1500 – 2000 | 10 | 10 | | 2000 – 2500 | 35 | 45 | | 2500 – 3000 | 52 | 97 | | 3000 – 3500 | 61 | 158 | | 3500 – 4000 | 38 | 196 | | 4000 – 4500 | 29 | 225 |

   Why? Adding the frequencies in order gives the number of lamps with lifetime below the upper limit of each class, which is needed to locate the middle observation.

2. 2

   Step 2: Find the median class

   1 mark (suggested)

   n = 225, so n/2 = 112.5. The cumulative frequency 158 is the first one greater than 112.5, so the median class is 3000 – 3500.

   Why? The middle observation lies in the first class whose cumulative frequency reaches or passes n/2, because that is where the ordered data crosses the halfway point.

3. 3

   Step 3: Apply the median formula

   1 mark (suggested)

   Median = l + ((n/2 − cf)/f) × h = 3000 + ((112.5 − 97)/61) × 500.

   Why? The formula interpolates inside the median class: it starts at the lower limit and adds the fraction of the class width needed to reach the middle observation.

4. 4

   Step 4: Calculate the median

   1 mark (suggested)

   = 3000 + (15.5/61) × 500 = 3000 + 7750/61 ≈ 3127.05.

   Why? The term (n/2 − cf)/f gives the fraction of the class width that must be added to the lower limit to reach the middle observation.

5. 5

   Step 5: State the answer

   1 mark (suggested)

   The median lifetime of a lamp is 3127.05 hours.

   Why? The median divides the ordered data into two equal halves, so it is the required measure of central tendency.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Form the mean table

   1 mark (suggested)

   Using class marks x = (upper limit + lower limit)/2: | Height (in cm) | f | x | fx | | 120 – 130 | 2 | 125 | 250 | | 130 – 140 | 8 | 135 | 1080 | | 140 – 150 | 12 | 145 | 1740 | | 150 – 160 | 20 | 155 | 3100 | | 160 – 170 | 8 | 165 | 1320 | | Total | 50 | | 7490 |

   Why? Each class is represented by its class mark, the average of its upper and lower limits. Multiplying the class mark by the frequency gives that class's contribution to the total.

2. 2

   Step 2: Apply the direct method for the mean

   1 mark (suggested)

   Mean = (sum of fx)/(sum of f) = 7490/50 = 149.8

   Why? The direct method divides the sum of the products of class marks and frequencies by the total frequency. This gives the mean of grouped data.

3. 3

   Step 3: Identify the modal class values

   1 mark (suggested)

   The highest frequency is 20, so the modal class is 150 – 160. Its lower limit l is 150, its class size h = 160 − 150 = 10, the modal frequency f1 is 20, the preceding frequency f0 is 12 and the succeeding frequency f2 is 8.

   Why? The modal class is the class interval with the maximum frequency. The mode formula uses its lower limit, its class size, and the frequencies of the class and its two neighbours.

4. 4

   Step 4: Substitute in the mode formula

   1 mark (suggested)

   Mode = l + ((f1 − f0)/(2f1 − f0 − f2)) × h. Here f1 − f0 = 20 − 12 = 8 and 2f1 − f0 − f2 = 2 × 20 − 12 − 8 = 20.

   Why? The mode formula compares how much the modal frequency exceeds the preceding frequency with the total of the two excesses on either side, then scales this by the class size.

5. 5

   Step 5: Evaluate the mode and state both results

   1 mark (suggested)

   Mode = l + ((f1 − f0)/(2f1 − f0 − f2)) × h = 150 + (8/20) × 10 = 150 + 4 = 154. Therefore, the mean height is 149.8 cm and the mode height is 154 cm.

   Why? Substituting the identified values into the mode formula gives the value inside the modal class. Both measures together answer the question.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Complete the cumulative frequency table

   1 mark (suggested)

   | Monthly consumption (in units) | Number of families (f) | Cumulative frequency (cf) | | --- | --- | --- | | 130-140 | 5 | 5 | | 140-150 | 9 | 14 | | 150-160 | 17 | 31 | | 160-170 | 28 | 59 | | 170-180 | 24 | 83 | | 180-190 | 10 | 93 | | 190-200 | 7 | 100 |

   Why? The cumulative frequency of a class is found by adding all the frequencies up to that class; it tells how many families consume up to the upper limit of that class.

2. 2

   Step 2: Find n/2

   1 mark (suggested)

   n = 5 + 9 + 17 + 28 + 24 + 10 + 7 = 100, so n/2 = 50.

   Why? The median of grouped data is the value of the middle observation, whose position is n/2.

3. 3

   Step 3: Identify the median class

   1 mark (suggested)

   The cumulative frequency just greater than 50 is 59, so the median class is 160-170.

   Why? The median is the middle observation; the class whose cumulative frequency first crosses n/2 contains that middle observation.

4. 4

   Step 4: Apply the median formula

   1 mark (suggested)

   Median = l + ((n/2 - cf)/f) × h = 160 + ((50 - 31)/28) × 10.

   Why? The median formula places the median inside the median class by adding the appropriate fraction of the class size to the lower limit.

5. 5

   Step 5: State the median

   1 mark (suggested)

   Median = 160 + 190/28 = 166.8 units. The median monthly consumption of electricity is 166.8 units.

   Why? The value lies inside the median class 160-170, so the interpolation is consistent.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Prepare the mean table

   1 mark (suggested)

   Class marks: 5, 15, 25, 35, 45, 55, 65, 75. Products fᵢxᵢ: 35, 210, 325, 420, 900, 605, 975, 600.

   Why? For grouped data, each class is represented by its class mark, the average of its two limits. Multiplying each class mark by its frequency gives the contribution of that class to the total.

2. 2

   Step 2: Mean by the direct method

   1 mark (suggested)

   Σ fᵢ = 7 + 14 + 13 + 12 + 20 + 11 + 15 + 8 = 100, Σ fᵢxᵢ = 35 + 210 + 325 + 420 + 900 + 605 + 975 + 600 = 4070, so x̄ = 4070/100 = 40.7.

   Why? The mean of grouped data is the sum of all products fᵢxᵢ divided by the total number of observations, Σ fᵢ.

3. 3

   Step 3: Cumulative frequency table

   1 mark (suggested)

   Cumulative frequencies: 7, 7 + 14 = 21, 21 + 13 = 34, 34 + 12 = 46, 46 + 20 = 66, 66 + 11 = 77, 77 + 15 = 92, 92 + 8 = 100.

   Why? Cumulative frequency is the running total of frequencies. It shows how many observations lie up to the upper limit of each class.

4. 4

   Step 4: Median by the median formula

   1 mark (suggested)

   Since total frequency is 100, n/2 = 50. The class 40–50 has cumulative frequency 66, the first cumulative frequency greater than 50, so it is the median class. Median = 40 + ((50 − 46)/20) × 10 = 40 + 2 = 42.

   Why? The median class is the class whose cumulative frequency first exceeds n/2. The formula places the middle observation inside that class using the cumulative frequency of the preceding class.

5. 5

   Step 5: Final answer

   1 mark (suggested)

   The mean number of cars is 40.7 and the median number of cars is 42.

   Why? These are the two required measures of central tendency of the given data.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Cumulative frequency table

   2 marks (suggested)

   n = 30. | Weight (in kg) | Number of students (f) | Cumulative frequency (cf) | | --- | --- | --- | | 40-45 | 2 | 2 | | 45-50 | 3 | 5 | | 50-55 | 8 | 13 | | 55-60 | 6 | 19 | | 60-65 | 6 | 25 | | 65-70 | 3 | 28 | | 70-75 | 2 | 30 |

   Why? Cumulative frequency of a class is the total frequency of that class and all classes before it. It tells how many students weigh less than the upper limit of that class.

2. 2

   Step 2: Median class and formula

   2 marks (suggested)

   n/2 = 30/2 = 15. The cumulative frequency 19 of the class 55-60 is greater than and nearest to 15, so the median class is 55-60. Here l = 55, cf = 13, f = 6, h = 5. Median = 55 + ((15 - 13)/6) × 5.

   Why? The median class is the class whose cumulative frequency is just greater than n/2. The formula needs the lower limit of this class, the cumulative frequency of the class before it, its frequency, and the class size.

3. 3

   Step 3: Median value and answer

   1 mark (suggested)

   Median = 55 + ((15 - 13)/6) × 5 = 55 + 10/6 = 56.67. The median weight of the students is 56.67 kg.

   Why? Substituting the values in the median formula and simplifying gives the median weight. The unit is kg because the weights are given in kg.

Marks: 2 + 2 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)