### RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data

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**RS Aggarwal Solutions Class 10 Chapter 9**

**Exercise 9A**

**Question 1:**

Table is as given below:

**Question 2:**

We have

Class | Frequency f_{i} | Mid Value x_{i} | f_{i}x_{i} |

0-1010-2020-3030-4040-5050-60 | 7561282 | 51525354555 | 3575150420360110 |

[latex]\sum { { f }_{ i } } =40 [/latex] | [latex]\sum { { f }_{ i } } { x }_{ i }=1150 [/latex] |

**Question 3:**

We have

Class | Frequency f_{i} | Class Mark x_{i} | f_{i}x_{i} |

10 – 2020 – 3030 – 4040 – 5050 – 6060 – 70 | 111520301410 | 152535455565 | 1653757001350770650 |

[latex]\sum { { f }_{ i } } =100 [/latex] | [latex]\sum { { f }_{ i } } { x }_{ i }=4010 [/latex] |

**Question 4:**

We have

Class | Mid value f_{i} | Frequency x_{i} | f_{i}x_{i} |

10 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 80 | 15253545556575 | 68137321 | 9020045531516513075 |

[latex]\sum { { f }_{ i } } =40 [/latex] | [latex]\sum { { f }_{ i } } { x }_{ i }=1430 [/latex] |

**Question 5:**

We have

Class | Frequency f_{i} | Mid value x_{i} | f_{i}x_{i} |

25 – 3535 – 4545 – 5555 – 6565 – 75 | 6108124 | 3040506070 | 180400400720280 |

[latex]\sum { { f }_{ i } } =40 [/latex] | [latex]\sum { { f }_{ i } } { x }_{ i }=1980 [/latex] |

Mean,**Question 6:**

We have

Class | Frequency f_{i} | Mid Value x_{i} | f_{i}x_{i} |

0 – 100100 – 200200 – 300300 – 400400 – 500 | 6915128 | 50150250350450 | 3001350375042003600 |

[latex]\sum { { f }_{ i } } =50 [/latex] | [latex]\sum { { f }_{ i } } { x }_{ i }=13200 [/latex] |

Mean,**Question 7:**

We have

Class | Frequency f_{i} | Mid Value x_{i} | f_{i}x_{i} |

0 – 1010 – 2020 – 3030 – 4040 – 50 | 152035p10 | 515253545 | 7530087535p450 |

[latex]\sum { { f }_{ i } } =80+p [/latex] | [latex]\sum { { f }_{ i } } { x }_{ i }=1700+35p [/latex] |

**Question 8:**

We have

Class | Frequency f_{i} | Mid Value x_{i} | f_{i}x_{i} |

0 – 20 | 17 | 10 | 170 |

20 – 40 | f_{1} | 30 | 30f_{1} |

40 – 60 | 32 | 50 | 1600 |

60 – 80 | 52 -f_{1} | 70 | 3640 – 70f_{1} |

80 – 100 | 19 | 90 | 1710 |

[latex]\sum { { f }_{ i } } =120 [/latex] | [latex]\sum { { f }_{ i } } { x }_{ i }=7120-40{ f }_{ 1 } [/latex] |

**Question 9:**

We have

Class | Frequency f_{i} | Mid Value x_{i} | f_{i}x_{i} |

0 – 20 | 7 | 10 | 70 |

20 – 40 | f_{1} | 30 | 30f_{1} |

40 – 60 | 12 | 50 | 600 |

60 – 80 | f_{2}=18 -f_{1} | 70 | 1260 – 70f_{1} |

80 – 100 | 8 | 90 | 720 |

100 – 120 | 5 | 110 | 550 |

[latex]\sum { { f }_{ i } } =50 [/latex] | [latex]\sum { { f }_{ i } } { x }_{ i }=3200-40{ f }_{ 1 } [/latex] |

**Question 10:**

We have, Let A = 25 be the assumed mean

Marks | Frequency f_{i} | Mid value x_{i} | Deviation d_{i}=(x_{i}-25) | (f_{i}× d_{i}) |

0 – 1010 – 2020 – 3030 – 4040 – 5050 – 60 | 12182720176 | 51525 = A354555 | -20-100102030 | -240-1800200340180 |

[latex]\sum { { f }_{ i } } =100 [/latex] | [latex]\sum { { (f }_{ i } } \times { d }_{ i })=300 [/latex] |

Hence mean = 28.**Question 11:**

A = 100 be the assumed mean, we have

Marks | Frequency f_{i} | Mid value x_{i} | Deviation d_{i}=(x_{i}-100) | (f_{i}× d_{i}) |

0 – 4040 – 8080 – 120120 – 160160 – 200 | 1220353023 | 2060100 = A140180 | -80-4004080 | -960-800012001840 |

[latex]\sum { { f }_{ i } } =120 [/latex] | [latex]\sum { { (f }_{ i } } \times { d }_{ i })=1250 [/latex] |

Hence, mean = 110.67**Question 12:**

Let the assumed mean be 150, h = 20

Marks | Frequency f_{i} | Mid value x_{i} | Deviation d_{i} = – 150 | (f_{i}× d_{i}) |

100 – 120120 – 140140 – 160160 – 180180 – 200 | 102030155 | 110130150=A170190 | -40-2002040 | -400-4000300200 |

[latex]\sum { { f }_{ i } } =80 [/latex] | [latex]\sum { { (f }_{ i } } \times { d }_{ i })=300 [/latex] |

Hence, Mean = 146.25**Question 13:**

Let A = 50 be the assumed mean, we have

Marks | Frequency f_{i} | Mid value x_{i} | Deviation d_{i}=(x_{i}-50) | f_{i }× d_{i} |

0 – 2020 – 4040 – 6060 – 8080 – 100100 – 120 | 203552443831 | 103050 = A7090110 | -40-200204060 | -800-700088015201860 |

[latex]\sum { { f }_{ i } } =200 [/latex] | [latex]\sum { { (f }_{ i } } \times { d }_{ i })=2760 [/latex] |

**Question 14:**

Marks | Frequency f_{i} | Mid value x_{i} | [latex]{ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h } \right) [/latex] | f_{i }× u_{i} |

0 – 1010 – 2020 – 3030 – 4040 – 5050 – 60 | 12182720176 | 51525 = A354555 | -2-10123 | -24-180203418 |

[latex]\sum { { f }_{ i } } =100 [/latex] | [latex]\sum { { (f }_{ i } } \times { u }_{ i })=30 [/latex] |

We have h = 10 and let assumed mean = 25.

A = 25, h = 10, ∑ f_{i}= 100 and ∑(f_{i}×u_{i})= 30

Hence the mean of given frequency distribution is 28.**Question 15:**

We have h = 4 and let assumed mean be A = 26. We have the table given below:

Marks | Frequency f_{i} | Mid value x_{i} | [latex]{ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h } \right) [/latex] | f_{i }× u_{i} |

4 – 88 – 1212 – 1616 – 2020 – 2424 – 2828 – 3232 – 36 | 21215251812133 | 61014182226 = A3034 | -5-4-3-2-1012 | -10-48-45-50-180136 |

[latex]\sum { { f }_{ i } } =100 [/latex] | [latex]\sum { { (f }_{ i } } \times { u }_{ i })=-152 [/latex] |

A = 26, h = 4, ∑ f_{i}= 100 and ∑(f_{i}×u_{i})= -152

Hence the mean of given frequency distribution is 19.92.**Question 16:**

We have h= 30 and let A = 75 be the assumed mean. we have the table given below:

Marks | Frequency f_{i} | Mid value x_{i} | [latex]{ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h } \right) [/latex] | f_{i }× u_{i} |

0 – 3030 – 6060 – 9090 – 120120 – 150150 – 180 | 122134522011 | 144575 = A105135165 | -2-10123 | -24-210524033 |

[latex]\sum { { f }_{ i } } =150 [/latex] | [latex]\sum { { (f }_{ i } } \times { u }_{ i })=80 [/latex] |

Thus, A = 75, h = 30, ∑ f_{i}= 150 and ∑(f_{i}×u_{i})= 80

Hence, the mean of the given frequency distribution is 91.**Question 17:**

We ahve h = 20 and let A = 70 be the assumed mean. We have the table given below:

Marks | Frequency f_{i} | Mid value x_{i} | [latex]{ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h } \right) [/latex] | f_{i }× u_{i} |

0 – 2020 – 4040 – 6060 – 8080 – 100100 – 120120 – 140 | 1218152526159 | 10305070 = A90110130 | -3-2-10123 | -36-36-150263027 |

[latex]\sum { { f }_{ i } } =150 [/latex] | [latex]\sum { { (f }_{ i } } \times { u }_{ i })=-4 [/latex] |

Thus, A = 70, h = 20, ∑ f_{i}= 120 and ∑(f_{i}×u_{i})= -4

Hence the mean of given frequency distribution is 69.33.**Question 18:**

We have h = 14 and let A = 35 be the assumed mean.

For calculating the mean, we prepare the table given below:

Marks | Frequency f_{i} | Mid value x_{i} | [latex]{ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h } \right) [/latex] | f_{i }× u_{i} |

0 – 1414 – 2828 – 4242 – 5656 – 70 | 721351116 | 72135 = A4963 | -2-1012 | -14-2101132 |

[latex]\sum { { f }_{ i } } =90 [/latex] | [latex]\sum { { (f }_{ i } } \times { u }_{ i })=8 [/latex] |

Thus, A = 35, ∑ f_{i}= 90, h = 14 and ∑(f_{i}×u_{i})= 8

Hence, Mean = 36.24**Question 19:**

Let h = 5 and let A = 22.5 be the assumed mean.

For calculating the mean, we prepare the table given below:

Marks | Frequency f_{i} | Mid value x_{i} | [latex]{ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h } \right) [/latex] | f_{i }× u_{i} |

10 – 1515 – 2020 – 2525 – 3030 – 3535 – 40 | 5681263 | 12.517.522.5 = A27.532.537.5 | -2-10123 | -10-6012129 |

[latex]\sum { { f }_{ i } } =40 [/latex] | [latex]\sum { { (f }_{ i } } \times { u }_{ i })=17 [/latex] |

Thus, A = 22.5 and h = 5

∑ f_{i}= 40 and ∑(f_{i}×u_{i})= 17

Hence the mean of given frequency distribution is 24.625.**Question 20:**

We have h = 6 and let assume mean A = 33. For calculating the mean we prepare the table.

Age | Frequency f_{i} | Mid value x_{i} | [latex]{ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h } \right) [/latex] | f_{i }× u_{i} |

18 – 2424 – 3030 – 3636 – 4242 – 4848 – 54 | 6812842 | 212733 = A394551 | -2-10123 | -12-80886 |

[latex]\sum { { f }_{ i } } =40 [/latex] | [latex]\sum { { (f }_{ i } } \times { u }_{ i })=2 [/latex] |

Thus, A = 33, h = 6, ∑ f_{i}= 40 and ∑(f_{i}×u_{i})=2

Hence, Mean = 33.3 years**Question 21:**

We have h = 6 and let assumed mean A = 99. For calculating the mean we prepare the table:

Class | f_{i} | x_{i} | [latex]{ u }_{ i }=\left( \frac { { x }_{ i }-A }{ h } \right) [/latex] | f_{i }× u_{i} |

84 – 9090 – 9696 – 102102 – 108108 – 114114 – 120 | 152220182025 | 879399=A105111117 | -2-10123 | -30-220184075 |

[latex]\sum { { f }_{ i } } =120 [/latex] | [latex]\sum { { (f }_{ i } } \times { u }_{ i })=81 [/latex] |

Thus, A = 99, h = 6 and ∑ f_{i}= 120, ∑(f_{i}×u_{i})=2

Hence, Mean = 103.05.**Question 22:**

Let h = 20 and assume mean = 550, we prepare the table given below:

Age | Frequency f_{i} | Mid value x_{i} | [latex]{ u }_{ i }=\left( \frac { { x }_{ i }-550 }{ 20 } \right) [/latex] | f_{i }× u_{i} |

500 – 520520 – 540540 – 560560 – 580580 – 600600 – 620 | 1495435 | 510530550 = A570590610 | -2-10123 | -27-904615 |

[latex]\sum { { f }_{ i } } =40 [/latex] | [latex]\sum { { (f }_{ i } } \times { u }_{ i })=-12 [/latex] |

Thus, A = 550, h = 20, and ∑ f_{i}= 40, ∑(f_{i}×u_{i})=-12

Hence the mean of the frequency distribution is 544.**Question 23:**

The given series is an inclusive series, making it an exclusive series, we have

Class | Frequency f_{i} | Mid value x_{i} | [latex]{ u }_{ i }=\left( \frac { { x }_{ i }-42 }{ 5 } \right) [/latex] | f_{i }× u_{i} |

24.5 – 29.529.5 – 34.534.5 – 39.539.5 – 44.544.5 – 49.549.5 – 54.554.5 – 59.5 | 4142216653 | 27323742 = A475257 | -3-2-10123 | -12-28-2206109 |

[latex]\sum { { f }_{ i } } =70 [/latex] | [latex]\sum { { (f }_{ i } } \times { u }_{ i })=-37 [/latex] |

Thus, A = 42, h = 5, ∑ f_{i}= 70 and ∑(f_{i}×u_{i})=-37

Hence, Mean = 39.36 years.**Question 24:**

The given series is an inclusive series making it an exclusive series, we get

class | Frequency f_{i} | Mid value x_{i} | [latex]{ u }_{ i }=\left( \frac { { x }_{ i }-29.5 }{ 10 } \right) [/latex] | f_{i }× u_{i} |

4.5 – 14.514.5 – 24.524.5 – 34.534.5 – 44.544.5 – 54.554.5 – 64.5 | 6112123145 | 9.519.529.5=A39.549.559.5 | -2-10123 | -12-110232815 |

[latex]\sum { { f }_{ i } } =80 [/latex] | [latex]\sum { { (f }_{ i } } \times { u }_{ i })=43 [/latex] |

Thus, A = 29.5, h = 10, ∑ f_{i}= 80 and ∑(f_{i}×u_{i})=43

Hence, Mean = 34.87 years.

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