In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 9A Maths book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 9 |

Chapter Name | Mean, Median, Mode of Grouped Data |

Exercise | 9 A |

Category | RS Aggarwal Solutions |

**RS Aggarwal Solutions Class 10 Chapter 9 Mean, Median, Mode of Grouped Data Ex 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 |

∑fi=40 | ∑fixi=1150 |

**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 |

∑fi=100 | ∑fixi=4010 |

**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 |

∑fi=40 | ∑fixi=1430 |

**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 |

∑fi=40 | ∑fixi=1980 |

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 |

∑fi=50 | ∑fixi=13200 |

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 |

∑fi=80+p | ∑fixi=1700+35p |

**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 |

∑fi=120 | ∑fixi=7120−40f1 |

**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 |

∑fi=50 | ∑fixi=3200−40f1 |

**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 |

∑fi=100 | ∑(fi×di)=300 |

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 |

∑fi=120 | ∑(fi×di)=1250 |

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 |

∑fi=80 | ∑(fi×di)=300 |

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 |

∑fi=200 | ∑(fi×di)=2760 |

**Question 14:**

Marks | Frequency f_{i} | Mid value x_{i} | ui=(xi−Ah) | f_{i }× u_{i} |

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

∑fi=100 | ∑(fi×ui)=30 |

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} | ui=(xi−Ah) | 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 |

∑fi=100 | ∑(fi×ui)=−152 |

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} | ui=(xi−Ah) | f_{i }× u_{i} |

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

∑fi=150 | ∑(fi×ui)=80 |

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} | ui=(xi−Ah) | f_{i }× u_{i} |

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

∑fi=150 | ∑(fi×ui)=−4 |

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} | ui=(xi−Ah) | f_{i }× u_{i} |

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

∑fi=90 | ∑(fi×ui)=8 |

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} | ui=(xi−Ah) | f_{i }× u_{i} |

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

∑fi=40 | ∑(fi×ui)=17 |

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} | ui=(xi−Ah) | f_{i }× u_{i} |

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

∑fi=40 | ∑(fi×ui)=2 |

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} | ui=(xi−Ah) | f_{i }× u_{i} |

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

∑fi=120 | ∑(fi×ui)=81 |

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} | ui=(xi−55020) | f_{i }× u_{i} |

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

∑fi=40 | ∑(fi×ui)=−12 |

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} | ui=(xi−425) | 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 |

∑fi=70 | ∑(fi×ui)=−37 |

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} | ui=(xi−29.510) | 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 |

∑fi=80 | ∑(fi×ui)=43 |

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

Hence, Mean = 34.87 years.

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