Bhagya

MP Board Class 6th Hindi Sugam Bharti Solutions विविध प्रश्नावली 1

प्रश्न 1.
सही जोड़ी बनाइए
1. मंगल – (क) आंदोलन
2. असहयोग – (ख) कामनाएँ
3. राजेन्द्र सिंह – (ग) उर्वरा शक्ति
4. धरती की – (घ) मैगसेसे सम्मान
उत्तर-
1. (ख),
2. (क),
3. (घ),
4. (ग)

प्रश्न 2.
सही विकल्प चुनकर रिक्त स्थानों की पूर्ति कीजिए
1. मैं ……………………………………. तुझे था जब कुंज और वन में। (पुकारता/ढूंढ़ता)
2. भूमि के भीतर जो जल है, उसे हम उनके विधियों द्वारा ……………………………………. कर सकते हैं। (संक्षित/दोहित)
3. निर्मला ……………………………………. सा मुख लिए कमल को साथ लेकर स्कूल चली गई। (उदास/पुलकित)
4. महिमा महान् तू है, गौरव ……………………………………. तू है। (प्रधाननिधान)
5. चौरी-चौरा नामक स्थान में ……………………………………. घटना घटी। (सनसनीपूर्ण/उत्तेजक)
उत्तर-
1. ढूंढता
2. संरक्षित
3. उदास
4. निधान
5. सनसनीपूर्ण।

प्रश्न 3.
निम्नलिखित प्रश्नों के उत्तर एक-एक वाक्य में दीजिए.

(क) आधी छुट्टी के समय कमल ने माँ से निर्मला की कौन-सी शिकायत की?
उत्तर-
कमल ने माँ से शिकायत की कि निर्मला उसे अपने साथ-साथ नहीं चलने देती, पीछे छोड़ देती है।

(ख) कवि जनम-जनम भर किसकी वंदना करने की बात कहता है?
उत्तर-
कवि जनम-जनम भर अपने देश भारत की वंदना करने की बात कहता हैं

(ग) जल स्त्रोतों के सूख जाने पर हम क्या कहने को विवश हो जाएँगे?
उत्तर-
जल स्त्रोतों के सूख जाने पर हमें विवश होकर कहना पड़ेगा-‘सुखी धरती, प्यासे लोग’

(घ) मदनमोहन मालवीय जी ने किन समाचार-पत्रों का संपादन किया?
उत्तर-
मदनमोहन मालवीय जी ने ‘लीडर’ और ‘हिन्दुस्तान’ जैसे पत्रों का संपादन किया?

(ङ) ‘वृक्ष निभाता रिश्ता नाता’ कविता किस कवि ने लिखी है?
उत्तर-
‘वृक्ष निभाता रिश्ता नाता’ शीर्षक कविता के कवि हैं-अज़हर हाशमी।

प्रश्न 4.
निम्नलिखित प्रश्नों के उत्तर तीन से पाँच वाक्यों में दीजिए

(क) भू-जल स्रोतों के अंधाधुंध दोहन से क्या हानियाँ हो रही हैं?
उत्तर-
भू-जल स्रोतों के अंधाधुंध दोहन से धीरे धीरे हम इन स्रोतों को खत्म करते जा रहे हैं। बोरिंग अथवा नल-कूपों के माध्यम से आवश्यकता से अधिक जल निकालकर हम प्राकृतिक जल-भंडार को लगातार खाली करते जा रहे हैं। परिणामस्वरूप जल की विकट समस्या खड़ी हो गई है।

(ख) ‘रामलाल का परिवार’ एकांकी में नई पीढ़ी के किस दायित्व की ओर संकेत किया गया है?
उत्तर-
‘रामलाल का परिवार’ नामक एकांकी में नई पीढ़ी को एक जिम्मेदारी दी गई है। आज के समाज में जो बुराईयाँ, भ्रष्टाचार फैली हैं, या फैलती जा रही हैं, उन्हें खत्म करने की जिम्मेदारी हमारी नई पीढ़ी की हैं यदि वे चाहें तो हमारा समाज बिल्कुल स्वच्छ और निर्मल बन सकता है।

(ग) भाई बहन के प्रेम के किस दृश्य को देखकर सावित्री मंत्र-मुग्ध सी रह गई?
उत्तर-
निम्रला का छोटा भाई कमल मुहर्रम के जुलूस में खो गया है। यों तो निर्मला भाई से हरदम लड़ती रहती थी, किंतु जैसे ही उसे एहसास होता है, कि कमल खो गया है, वह बेचैन सी हो उठी। फूटफूटकर रोने लगी रोते-रोते उसकी आँखें सूज आईं। एक बहन को अपने भाई के लिए इस प्रकार रोते-सिसकते देखकर सावित्री मंत्र-मुग्ध-सी रह गई।

(घ) ‘जो संस्कृति अभी तक, दुर्जेय-सी बनी है।’! पंक्ति का भाव स्पष्ट कीजिए।
उत्तर-
भारत की संस्कृति पूरी दुनिया में विख्यात है।। यहाँ के आदर्श एवं अमूल्य विरासत का आज भी चारों तरफ बोलवाला है और दुनिया का कोई भी देश उसका मुकाबला नहीं कर सकता है। यह हमेशा से विश्व गुरु रहा है। इसकी संस्कृति को अपनाना अन्य देशों के लिए टेढ़ी खीर की तरह है।

(ङ) मदनमोहन मालवीय जी बच्चों की आरंभिक शिक्षा की ओर अधिक ध्यान देना क्यों चाहते थे?
उत्तर-
मदनमोहन मालवीय जी बच्चों की आरंभिक शिक्षा की ओर अधिक ध्यान देना चाहते हैं क्योंकि इस उम्र में बच्चों को जो बातें बताई या सिखाई जाती हैं, उन्हें वे हमेशा याद रखते हैं और उसके अनुसार काम करते हैं। अतः बचपन में अच्छे-अच्छे गुणों के संस्कार डालने चाहिए।

प्रश्न 5.
निम्नलिखित शब्दों के मानक रूप लिखिएजनम, भगत, माखी, चरन, धरम
उत्तर-
जन्म, भक्त, मक्खी, चरण, धर्म।

प्रश्न 6.
(क) निम्न शब्दों की वर्तनी शुद्ध कीजिएकठनाई, रिण, साहितयिक, अध्यन, हरयाली, आर्शीवाद
उत्तर-
कठिनाई, ऋण, साहित्यिक, अध्ययन।

(ख) निम्न शब्दों का वाक्यों में प्रयोग कीजिए
जल-संरक्षण, जाति-पांति, बड़े-बूढ़े, कहा-सुनी, पारिवारिक, मुक्तभोगी।
उत्तर-
जल-संरक्षण-हम सबको मिलकर जल-संरक्षण के लिए उपाए करना चाहिए।
जाति-पांति-समाज में आज भी जाति-पांति के भेदभाव. देखने को मिलते हैं।
बड़े-बूढ़े-बड़े-बूढ़ों का आदर करना हमारा परम कर्त्तव्य
कहा-सुनी-देखते-देखते दोनों मित्रों के बीच कहा-सुनी हो गई।
पारिवारिक-पारिवारिक शांति से जीवन खुशहाल होता है।
भुक्तभोगी-श्याम चोट के मामले में भुक्तभोगी है।

प्रश्न 7.
(क) निम्न शब्दों के दो-दो पर्यायवाची शब्द – लिखें
आँख, आकाश, हवा, फूल
उत्तर-
आँख – नेत्र, नयन
आकाश – गगन, आसमान
हवा – वायु, पवन
फूल – सुमन, पुष्प

(ख) निम्नलिखित विलोम शब्दों की सही-जोड़ी बनाएँ
अमृत, उत्थान, गरीब, दानव, अमीर, पतन, देवता, सजीव, विष, निर्जीव
उत्तर-
अमृत – विष
उत्थान – पतन
गरीब – अमीर
देवता – दानव
सजीव – निर्जीव

प्रश्न. 8.
(क) निम्न शब्दों में ‘ता’ प्रत्यय जोड़कर भाववाचक संज्ञा बनाएँ
मित्र, सुन्दर, शत्रु, मधुर, दयालु
उत्तर-
मित्रता, सुन्दरता, शत्रुता, मधुरता, दयालुता

(ख) निम्न उपसर्गों से एक-एक शब्द बाएँ
उप, अप, वि, अनु
उत्तर-
उप – उपहार
अप – अपकार
वि – विकार
अनु – अनुकूल

प्रश्न 9.
भाई-बहन कहानी का सारांश लिखिए।
उत्तर-
पाठ-3 का सारांश देखे।

प्रश्न 10.
‘जल ही जीवन है’ विषय पर 100 शब्दों में एक निबंध लिखिए।
उत्तर-
जल जीवन का स्त्रोत है। हमारे शरीर का भी एक तत्त्व जल है। धरती का अधिकांश भाग जल ही है। जल के बिना धरती का कोई अस्तित्व ही नहीं रह जाएगा। जल से हम फसल सींचते हैं जो हमें जीवन देता है। जल से वाष्पीकरण से नमी होती है। जल एक प्राकृतिक संसाधन है। यह सीमित है। हम इसका दुरुपयोग भी करते हैं। जनसंख्या के बोझ से इसकी खपत बढ़ती जा रही है जिससे इसका स्त्रोत सूखता जा रहा है। जरूरत है इसके स्त्रोत को बचाने का।

MP Board Class 6th Hindi Solutions

MP Board Class 7th Maths Solutions Chapter 4 Simple Equations Ex 4.1

Question 1.
Complete the last column of the table.

Solution:
(i) x + 3 = 0
By putting x = 3
L.H.S. = 3 + 3 = 6 ≠ R.H.S.
Hence, the equation is not satisfied.

(ii) x + 3 = 0
By putting x = 0
L.H.S. = 0 + 3 = 3 ≠ R.H.S.
Hence, the equation is not satisfied.

(iii) x + 3 = 0
By putting x = – 3
L.H.S. = -3 + 3 = 0 = R.H.S.
Hence, the equation is satisfied.

(iv) x – 7 = 1
By putting x = 7
L.H.S. = 7 – 7 = 0 ≠ R.H.S.
Hence, the equation is not satisfied.

(v) x – 7 = 1
By putting x = 8 L.H.S. = 8 – 7 = 1 = R.H.S.
Hence, the equation is satisfied.

(vi) 5x = 25
By putting x = 0
L.H.S. = 5 × 0 = 0 ≠ R.H.S.
Hence, the equation is not satisfied.

(vii) 5x = 25
By putting x = 5
L.H.S. = 5 × 5 = 25 = R.H.S.
Hence, the equation is satisfied,

(viii) 5x = 25 .
By putting x = – 5
L.H.S. = 5 × (- 5) = – 25 ≠ R.H.S.
Hence, the equation is not satisfied.

(ix) \(\frac{m}{3}\) = 2
By putting m = – 6
L.H.S. = \(\frac{-6}{3}\) = -2 ≠ R.H.S.
Hence, the equation is not satisfied.

(x) \(\frac{m}{3}\) = 2
By putting m = 0
L.H.S. = \(\frac{0}{3}\) = 0 ≠ R.H.S.
Hence, the equation is not satisfied.

(xi) \(\frac{m}{3}\) = 3
By putting m = 6
L.H.S. = \(\frac{6}{3}\) = 2 = R.H.S
Hence, the equation is satisfied.

Question 2.
Check whether the value given in the brackets is a solution to the given equation or not:
(a) n + 5 = 19(n = 1)
(b) 7n + 5 = 19(n = -2)
(c) 7n + 5 = 19(n = 2)
(d) 4p – 3 = 13(p = 1)
(e) 4p – 3 = 13(p = -4)
(f) 4p – 3 = 13 (p = 0)
Solution:
(a) n + 5 = 19 (n = 1)
Putting n = 1
L.H.S. = n + 5 = 1 + 5 = 6 ≠ 19 = R.H.S.
As L.H.S. ≠ R.H.S.
Therefore, n = 1 is not a solution of the given equation, n + 5 = 19.

(b) 7n + 5 = 19 (n =- 2)
Putting n = – 2
L.H.S. = 7n + 5 = 7 × (-2) + 5 = -14 + 5
= -9 ≠ 19 = R.H.S.
As L.H.S. ≠ R.H.S.
Therefore, n = -2 is not a solution of the given equation, 7n + 5 = 19.

(c) 7n + 5 = 19 (n = 2)
Putting n = 2
L.HS. = 7n + 5 = 7 × (2) + 5 = 14 + 5 = 19 = R.HS.
As L.H.S. = R.H.S.
Therefore, n = 2 is a solution of the given equation, 7n + 5 = 19.

(d) 4p – 3 = 13(p = 1)
Putting p = 1
L.H.S. = 4p – 3 = (4 × 1) – 3
= 1 × 13 = R.H.S.
As L.H.S. × R.H.S.
Therefore, p = 1 is not a solution of the given equation, 4p – 3 = 13.

(e) 4p – 3 = 13 (p = – 4)
Putting p = – 4
L.H.S. = 4p – 3 = 4 × (- 4) – 3 = -16 – 3
= -19 ≠ 13 = R.H.S.
As L.H.S. ≠ R.H.S.
Therefore, p = – 4 is not a solution of the given equation, 4p – 3 = 13.

(f) 4p – 3 = 13 (p = 0)
Putting p = 0
L.HS. = 4p – 3 = (4 × 0) – 3 = -3 ≠ 13 = RH.S.
As L.H.S. ≠ R.H.S.
Therefore, p = 0 is not a solution of the given equation, 4p – 3 = 13.

Question 3.
Solve the following equations by trial and error method:
(i) 5p + 2 = 17
(ii) 3m -14 = 4
Solution:
(i) 5p + 2 = 17 Putting p = 1
L.H.S. = (5 × 1)+ 2 = 7 ≠ R.H.S.
Putting p = 2
L.H.S. = (5 × 2) + 2 = 10 + 2 = 12 ≠ R.H.S.
Putting p = 3
L.H.S. = (5 × 3) + 2 = 17 = R.H.S.
Hence, p = 3 is a solution of the given equation.

(ii) 3m – 14 = 4
Putting m = 4,
L.H.S. = (3 × 4) – 14 = -2 ≠ R.H.S.
Putting m = 5,
L.H.S. = (3 × 5) – 14 = 1 ≠ R.H.S.
Putting m = 6,
L.H.S. = (3 × 6) – 14 = 18 -14 = 4 = R.H.S.
Hence, m = 6 is a solution of the given equation.

Question 4.
Write equations for the following statements:
(i) The sum of numbers x and 4 is 9.
(ii) 2 subtracted from y is 8.
(iii) Ten times a is 70.
(iv) The number b divided by 5 gives 6.
(v) Three-fourth of f is 15.
(vi) Seven times m plus 7 gives 77.
(vii) One-fourth of a number x minus 4 gives 4.
(viii) If you take away 6 from 6 times y, you get 60.
(ix) If you add 3 to one-third of z, you get 30.
Solution:
(i) x + 4 = 9
(ii) y – 2 = 8
(iii) 10a = 70
(iv) \(\frac{b}{5}\) = 6
(v) \(\frac{3}{4} t\) = 15
(vi) Seven times of m is 7m.
∴ 7m + 7 = 77
(vii) One-fourth of a number x is \(\frac{x}{4}\).

(viii) Six times of y is 6y.
∴ 6y – y = 60

(ix) One-third of z is \(\frac{x}{4}\) .

Question 5.
Write the following equations in statement forms:
(i) p + 4 = 15
(ii) m – 7 = 3
(iii) 2m = 7
(iv) \(\frac{m}{5}\) = 3
(v) \(\frac{3 m}{5}\) = 6
(vi) 3p + 4 = 25
(viii) \(\frac{p}{2}\) + 2 = 8
Solution:
(i) The sum of p and 4 is 15.
(ii) 7 subtracted from m is 3.
(iii) Twice of a number m is 7.
(iv) One-fifth of a number m is 3.
(v) Three-fifth of a number m is 6.
(vi) Three times of a number p, when added to 4, gives 25.
(vii) 2 subtracted from four times of a number p is 18.
(viii) 2 added to half of a number p gives 8.

Question 6.
Set up an equation in the following cases:
(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. (Take m to be the number of Parmit’s marbles.)
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.)
(iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87, (Take the lowest score to be l.)
(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees.)
Solution:
(i) Let Parmit has in marbles.
Number of marbles Irfan has = 5 × Number of marbles Parmit has +7
∴ 5 × m + 7 = 37 ⇒ 5m + 7 = 37

(ii) Let Laxmi be i years old.
Laxmi’s father’s age 3 × Laxmi’s age +4
∴ 49 = 3xy + 4 ⇒ 3y + 4 = 49

(iii) Let the lowest marks be 1.
Highest marks = 2 × Lowest marks +7
∴ 87 = 2 × l + 7 ⇒ 2l + 7 = 87

(iv) An isosceles triangle has two of its angles of equal measure.
Let base angle be b.
Vertex angle =2 × Base angle = 2b
Sum of all interior angles of a triangle 180°
∴ b + b + 2b = 180° ⇒ 4b = 180°

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 3 Data Handling Ex 3.4

Question 1.
Tell whether the following is certain to happen, impossible, can happen but not certain.
(i) You are older today than yesterday.
(ii) A tossed coin will land heads up.
(iii) A die when tossed shall land up with 8 on top.
(iv) The next traffic light seen will be green.
(v) Tomorrow will be a cloudy day.
Solution:
(i) Certain to happen
(ii) Can happen but not certain
(iii) Impossible as there are only six faces of a die marked as 1, 2, 3, 4, 5 and 6 on it.
(iv) Can happen but not certain
(v) Can happen but not certain

Question 2.
There are 6 marbles in a box with numbers from 1 to 6 marked on each of them.
(i) What is the probability of drawing a marble with number 21
(ii) What is the probability of drawing a marble with number 5?
Solution:
Probability of an event

(ii) Probability of drawing a marble with number 5 = \(\frac{1}{6}\).

Question 3.
A coin is flipped to decide which team starts the game. What is the probability that your team will start?
Solution:
A coin has two faces – Head and Tail. One team can opt either Head or Tail. Probability of an event

Probability (our team will start first) = \(\frac{1}{2}\)

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 3 Data Handling Ex 3.3

Question 1.
Use the given bar graph to answer the following questions.
(a) Which is the most popular pet?
(b) How many students have dog as a pet?

Solution:
(a) Since, the bar representing number of students for cats is the tallest, so cat is the most popular pet.
(b) The number of students having dog as a pet is 8.

Question 2.
Read the given bar graph which shows the number of books sold by a bookstore during five consecutive years and answer the following questions:

(i) About how many books were sold in 1989? 1990? 1992?
(ii) In which year were about 475 books sold? About 225 books sold?
(iii) In which years were fewer than 250 books sold?
(iv) Can you explain how you would estimate the number of books sold in 1989?
Solution:
(i) In 1989, 175 books were sold. In 1990, 475 books were sold. In 1992, 225 books were sold.
(ii) From the graph, it can be concluded that in the year 1990 about 475 books were sold and in the year 1992 about 225 books were sold.
(iii) From the graph, it can be concluded that in the year 1989 and 1992, the number of books sold were less than 250.
(iv) From the graph, it can be concluded that the number of books sold in the year 1989 is about 1 and \(\frac{3}{4}\) th part of 1 cm.
We know that the scale is taken as 1 cm = 100 books.
∴ Number of books sold in 1989

Therefore, about 175 books were sold in the year 1989.

Question 3.
Number of children in six different classes are given below. Represent the data on a bar graph.

(a) How would you choose a scale?
(b) Answer the following questions:
(i) Which class has the maximum number of children? And the minimum?
(ii) Find the ratio of students of class sixth to the students of class eighth.
Solution:

(a) We will choose a scale as 1 unit = 10 children because we can represent a more
clear difference between the number of students of class 7^th and that of class 9^th by this scale.
(b) (i) Since, the bar representing the number of children for class fifth is the tallest. So, there are maximum number of children in class fifth. Similarly, the bar representing the number of children for class tenth is the smallest. So, there are minimum number of children in class tenth.
(ii) The number of students in class sixth is 120 and the number of students in class eighth is 100. Therefore, the ratio of students of class sixth to the students 5.
of class eighth \(=\frac{120}{100}=\frac{6}{5}\) i.e., 6 : 5

Question 4.
The performance of a student in 1^st Term and 2^nd Term is given. Draw a double bar graph choosing appropriate scale and answer the following:

(i) In which subject, has the child improved his performance the most?
(ii) In which subject is the improvement the least?
(iii) Has the performance gone down in any subject?
Solution:
A double bar graph for the given data is as follows.

(i) There was a maximum increase in the marks obtained in Maths. Therefore, the child has improved his performance the most in Maths.
(ii) From the graph, it can be concluded that the improvement was the least in
S. Science.
(iii) From the graph, it can be observed that the performance in Hindi has gone down.

Question 5.
Consider this data collected from a survey of a colony.

(i) Draw a double bar graph choosing an appropriate scale. What do you infer from the bar graph?
(ii) Which sport is most popular?
(iii) Which is more preferred, watching or participating in sports?
Solution:
(i) A double bar graph for the given data is as follows:

The double bar graph represents the number of people who like watching and participating in different sports.
(ii) From the bar graph, it can be observed that the bar representing the number of people who like watching and participating in cricket is the tallest among all the bars. Hence, cricket is the most popular sport.
(iii) The bars representing watching sport are longer than the bars representing participating in sport. Hence, watching different types of sports is more preferred than participating in the sports.

Question 6.
Take the data giving the minimum and the maximum temperature of various cities given in the table. Plot a double bar graph using the data and answer the following:

(i) Which city has the largest difference in the minimum and maximum temperature on the given date?
(ii) Which is the hottest city and which is the coldest city?
(iii) Name two cities where maximum temperature of one was less than the minimum temperature of the other.
(iv) Name the city which has the least difference between its minimum and the maximum temperature.
Solution:
A double bar graph for the given data is constructed as follows.

(i) From the graph, it can be concluded that Jammu has the largest difference in its minimum and maximum temperature on 20.6.2006.
(ii) From the graph, it can be concluded that Jammu is the hottest city and Bangalore is the coldest city.
(iii) Bangalore and Jaipur, Bangalore and Ahmedabad.
For Bangalore, the maximum temperature was 28°C, while minimum temperature of both cities, Ahmedabad and Jaipur, was 29°C.
(iv) From the graph, it can be concluded that the city which has least difference between its minimum and maximum temperature is Mumbai.

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 3 Data Handling Ex 3.2

Question 1.
The scores in Mathematics test (out of 25) of 15 students are as follows:
19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20
Find the mode and median of this data. Are they same?
Solution:
Arranging the given scores in ascending order, we have 5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25
Mode of given data is that value of observation which occurs for the most number of times i.e., 20. 5.
Median of the given data is the middle observation when the data is arranged in ascending order i.e., 8^th term = 20
Hence, mode of data = median of data.

Question 2.
The runs scored in a cricket match by 11 players are as follows:
6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15
Find the mean, mode and median of this data. Are the three same?
Solution:
Arranging the given scores in an ascending order, we have

As 15 occurs for the most number of times.
∴ Mode = 15
Median = Middle term = 15
∴ All three mean, mode and median are not same.

Question 3.
The weights (in kg.) of 15 students of a class are: 38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47
(i) Find the mode and median of this data.
(ii) Is there more than one mode?
Solution:
Arranging the given weights in ascending order, we have 32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45,47, 50
(i) Mode = 38 and 43
Median = Middle term = 40
(ii) Yes, there are 2 modes for the given data i.e., 38 and 43.

Question 4.
Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14
Solution:
Arranging the given data in an ascending order, we have
12, 12, 13, 13, 14, 14, 14, 16, 19
14 occurs for the most number of times
∴ Mode = 14
Median = Middle observation = 14

Question 5.
Tell whether the statement is true or false:
(i) The mode is always one of the numbers in a data.
(ii) The mean is one of the numbers in a data.
(iii) The median is always one of the numbers in a data.
(iv) The data 6,4,3,8,9,12,13,9 has mean 9.
Solution:
(i) True, as mode of a given data is that value of observation which occurs for the most number of times. Therefore, it is one of the observations given in the data.
(ii) False, because mean may or may not be one of the numbers in the data.
(iii) False, because median may or may not be one of the numbers in the data
(iv) False, because

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 3 Data Handling Ex 3.1

Question 1.
Find the range of heights of any ten students of your class.
Solution:
Let the heights (in cm) of 10 students of our class be 125, 129, 131, 132, 134, 136, 139, 142, 144, 146
Highest observation = 146 cm
Lowest observation =125 cm
Range = Highest observation – Lowest observation
= (146 – 125) cm = 21 cm

Question 2.
Organise the following marks in a class assessment, in a tabular form. 4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7
Solution:

(i) Highest number = 9
(ii) Lowest number = 1
(iii) Range = (9 – 1) = 8
(iv) Sum of all the observations =4 + 6 + 7 + 5 + 3 + 5 + 4 + 5 + 2 + 6 + 2 + 5 + 1 + 9 + 6 + 5 + 8 + 4 + 6 + 7 = 100
Total number of observations = 20
Arithmetic mean

Question 3.
Find the mean of the first five whole numbers.
Solution:
First five whole numbers are 0, 1, 2, 3 and 4.

Hence, the mean of first five whole numbers is 2.

Question 4.
A cricketer scores the following runs in eight innings: 58, 76,40, 35,46,45,0,100.
Find the mean score.
Solution:
Runs scored by the cricketer in eight innings are 58, 76, 40, 35, 46, 45, 0 and 100.

Therefore, mean score is 50.

Question 5.
Following table shows the points of each player scored in four games:

Now answer the following questions:
(i) Find the mean to determine A’s average number of points scored per game.
(ii) To find the mean number of points per game for C, would you divide the total points by 3 or by 4? Why?
(iii) B played in all the four games. How would you find the mean?
(iv) Who is the best performer?
Solution:
(i) A’s average number of points

(ii) To find the mean number of points per game for C, we will divide the total points by 3 because C played 3 games.

(iii) Mean of B’s Score

(iv) The best performer will have the greatest average among all. Now we can observe that the average of A is 12.5 which is more than that of B and C. Therefore, A is the best performer.

Question 6.
The marks (out of 100) obtained by a group of students in a Science test are 85, 76, 90, 85, 39, 48, 56, 95, 81 and 75. Find the:
(i) Highest and the lowest marks obtained by the students.
(ii) Range of the marks obtained.
(iii) Mean marks obtained by the group.
Solution:
The marks obtained by the group of students in a Science test can be arranged in ascending order as follows.
39, 48, 56, 75, 76, 81, 85, 85, 90, 95
(i) Highest marks = 95 Lowest marks = 39
(ii) Range = 95 – 39 = 56
(iii) Mean marks

Question 7.
The enrolment in a school during six consecutive years was as follows:
1555, 1670, 1750, 2013, 2540, 2820
Find the mean enrolment of the school for this period.
Solution:
Mean enrolment

Question 8.
The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:

(i) Find the range of the rainfall in the above data.
(ii) Find the mean rainfall for the week.
(iii) On how many days was the rainfall less than the mean rainfall?
Solution:
(i) Range = (20.5 – 0.0) mm = 20.5 mm
(ii) Mean rainfall

(iii) For 5 days (i.e., Monday, Wednesday, Thursday, Saturday, Sunday), the rainfall was less than the mean rainfall.

Question 9.
The heights of 10 girls were measured in cm and the results are as follows:
135, 150, 139, 128, 151, 132, 146, 149, 143, 141.
(i) What is the height of the tallest girl?
(ii) What is the height of the shortest girl?
(iii) What is the range of the data?
(iv) What is the mean height of the girls?
(v) How many girls have heights more than the mean height?
Solution:
Arranging the heights of 10 girls in an ascending order;
128, 132, 135, 139, 141, 143, 146, 149, 150, 151
(i) Height of the tallest girl = 151 cm
(ii) Height of the shortest girl = 128 cm
(iii) Range = (151 – 128) cm = 23 cm
(iv) Mean height

(v) The heights of 5 girls are greater than the mean height (i.e., 141.4 cm) and these heights are 143 cm, 146 cm, 149 cm, 150 cm and 151 cm.

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 2 Fractions and Decimals Ex 2.7

Question 1.
Find:
(i) 0.4 ÷ 2
(ii) 0.35 ÷ 5
(iii) 2.48 ÷ 4
(iv) 65.4 ÷ 6
(v) 651.2 ÷ 4
(vii) 3.96 ÷ 4
(viii) 0.80 ÷ 7
Solution:

Question 2.
Find:
(i) 4.8 ÷ 10
(ii) 52.5 ÷ 10
(iii) 0.7 ÷ 10
(iv) 33.1 ÷ 10
(v) 272.23 ÷ 10
(vi) 0.56 ÷ 10
(vii) 3.97 ÷ 10
Solution:
We know that when a decimal number is divided by a multiple of 10 (i.e., 10,100,1000, etc.), the decimal point will be shifted to the left by as many places as there are zeros.
(i) 4.8 ÷ 10 = 0.48
(ii) 52.5 ÷ 10 = 5.25
(iii) 0.7 ÷ 10 = 0.07
(iv) 33.1 ÷ 10 = 3.31
(v) 272.23 ÷ 10 = 27.223
(vi) 0.56 ÷ 10 = 0.056
(vii) 3.97 ÷10 = 0.397

Question 3.
Find:
(i) 2.7 ÷ 100
(ii) 0.3 ÷ 100
(iii) 0.78 ÷ 100
(iv) 432.6 ÷ 100
(v) 23.6 ÷ 100
(vi) 98.53 ÷ 100
Solution:
We know that when a decimal number is divided by a multiple of 10 (i.e., 10, 100, 1000, etc.), the decimal point will be shifted to the left by as many places as there are zeros.
(i) 2.7 ÷ 100 = 0.027
(ii) 0.3 ÷ 100 = 0.003
(iii) 0.78 ÷ 100 = 0.0078
(iv) 432.6 ÷ 100 = 4.326
(v) 23.6 ÷ 100 = 0.236
(vi) 98.53 ÷ 100 = 0.9853

Question 4.
Find:
(i) 7.9 ÷ 1000
(ii) 26.3 ÷ 1000
(iii) 38.53 ÷ 1000
(v) 0.5 ÷ 1000
Solution:
We know that when a decimal number is divided by a multiple of 10 (i.e., 10, 100, 1000, etc.), the decimal point will be shifted to the left by as many places as there are zeros.
(i) 7.9 ÷ 1000 = 0.0079
(ii) 26.3 ÷ 1000 = 0.0263
(iii) 38.53 ÷ 1000 = 0.03853
(iv) 128.9 ÷ 1000 = 0.1289
(v) 0.5 ÷ 1000 = 0.0005

Question 5.
Find:
(i) 7 ÷ 3.5
(ii) 36 ÷ 2
(iii) 3.25 ÷ 0.5
(iv) 30.94 ÷ 0.7
(v) 0.5 ÷ 0.25
(vi) 7.75 ÷ 0.25
(vii) 76.5 ÷ 0.15
(viii) 37.8 ÷ 1.4
(ix) 2.73 ÷ 1.3
Solution:

Question 6.
A vehicle covers a distance of 43.2 km in 2.4 litres of petrol How much distance wilt it cover in one litre of petrol?
Solution:
Distance covered in 2.4 Litres of petrol = 43.2 km
∴ Distance covered in 1 litre of petrol = (43.2 ÷ 2.4) km

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 2 Fractions and Decimals Ex 2.6

Question 1.
Find:
(i) 0.2 × 6
(ii) 8 × 4.6
(iii) 2.71 × 5
(iv) 20.1 × 4
(v) 0.05 × 7
(vi) 211.02 × 4
(vii) 2 × 0.86
Solution:

Question 2.
Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm.
Solution:
Length = 5.7 cm and breadth = 3 cm
Area = Length × Breadth = (5.7 × 3) cm² = 17.1 cm²

Question 3.
Find:
(i) 1.3 × 10
(ii) 36.8 × 10
(iii) 153.7 × 10
(iv) 168.07 × 10
(v) 31.1 × 100
(vi) 156.1 × 100
(vii) 3.62 × 100
(viii) 43.07 × 100
(ix) 0.5 × 10
(x) 0.08 × 10
(xi) 0.9 × 100
(xii) 0.03 × 1000
Solution:
We know that when a decimal number is multiplied by 10, 100, 1000, the decimal point in the product is shifted to the right by as many places as there are zeros.
(i) 1.3 × 10 = 13
(ii) 36.8 × 10 = 368
(iii) 153.7 × 10 = 1537
(iv) 168.07 × 10 = 1680.7
(v) 31.1 × 100 = 3110
(vi) 156.1 × 100 = 15610
(vii) 3.62 × 100 = 362
(viii) 43.07 × 100 = 4307
(ix) 0.5 × 10 = 5
(x) 0.08 × 10 = 0.8
(xi) 0.9 × 100 = 90
(xii) 0.03 × 1000 = 30

Question 4.
A two-wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol?
Solution:
Distance covered in 1 litre of petrol = 55.3 km
Distance covered in 10 litres of petrol = (10 × 55.3) km = 553 km

Question 5.
Find:
(i) 2.5 × 0.3
(ii) 0.1 × 51.7
(iii) 0.2 × 316.8
(iv) 1.3 × 3.1
(v) 0.5 × 0.05
(vii) 1.07 × 0.02
(viii) 10.05 × 1.05
(ix) 101.01 × 0.01
(x) 100.01 × 1.1
Solution:

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 2 Fractions and Decimals Ex 2.5

Question 1.
Which is greater?
(i) 0.5 or 0.05
(ii) 0.7 or 0.5
(iii) 7 or 0.7
(iv) 1.37 or 1.49
(v) 2.03 or 2.30
(vi) 0.8 or 0.88.
Solution:
(i) 0.5 or 0.05
Converting these decimal numbers into like fractions, we obtain

(ii) 0.7 or 0.5
Converting these decimal numbers into like fractions, we obtain

(iii) 7 or 0.7
Converting these decimal numbers into like fractions, we obtain

(iv) 1.37 or 1.49
Converting these decimal numbers into like fractions, we obtain

(v) 2.03 or 2.30
Converting these decimal numbers into like fractions, we obtain

(vi) 0.8 or 0.88
Converting these decimal numbers into like fractions, we obtain

Question 2.
Express as rupees using decimals:
(i) 7 paise
(ii) 7 rupees 7 paise
(iii) 77 rupees 77 paise
(iv) 50 paise
(v) 235 paise.
Solution:
There are 100 paise in 1 rupee. Therefore, if we want to convert paise into rupees, we have to divide paise by 100.

Question 3.
(i) Express 5 cm in metre and kilometre.
(ii) Express 35 mm in cm, m and km.
Solution:

Question 4.
Express in kg:
(i) 200 g
(ii) 3470 g
(iii) 4 kg 8g
Solution:

Question 5.
Write the following decimal numbers in the expanded form:
(i) 20.03
(ii) 2.03
(iii) 200.03
(iv) 2.034
Solution:

Question 6.
Write the place value of 2 in the following decimal numbers:
(i) 2.56
(ii) 21.37
(iii) 10.25
(iv) 9.42
(v) 63.352.
Solution:

∴ Place value of 2 in 2.56 is ones.

∴ Place value of 2 in 21.37 is tens.

∴ Place value of 2 in 10.25 is tenths.

∴ Place value of 2 in 9.42 is hundredths.

∴ Place value of 2 in 63.352 is thousandths.

Question 7.
Dinesh went from place A to place Band from there to place C. A is 7.5 km from B and B is
12.7 km from C. Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. Who travelled more and by how much?

Solution:
Distance travelled by Dinesh = AB + BC = (7.5 + 12.7) km = 20.2 km
Distance travelled by Ayub = AD + DC
= (9.3+ 11.8) km = 21.1 km
Difference = (21.1 – 20.2) km = 0.9 km
Hence, Ayub travelled 0.9 km more distance than Dinesh.

Question 8.
Shyama bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits?
Solution:
Total fruits bought by Shyama = 5 kg 300 g + 3 kg 250 g = 8 kg 550 g

Hence, Sarala bought more fruits.

Question 9.
How much less is 28 km than 42.6 km?
Solution:
(42.6 – 28.0) km = 14.6 km
Hence, 28 km is less than 42.6 km by 14.6 km.

MP Board Class 7th Maths Solutions

MP Board Class 7th Maths Solutions Chapter 2 Fractions and Decimals Ex 2.4

Question 1.
Find:

Solution:

Question 2.
Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.

Solution:
(i) Reciprocal of \(\frac{3}{7}=\frac{7}{3}\). It is an improper fraction.
(ii) Reciprocal of \(\frac{5}{8}=\frac{8}{5}\). It is an improper fraction.
(iii) Reciprocal of \(\frac{9}{7}=\frac{7}{9}\). It is a proper fraction.
(iv) Reciprocal of \(\frac{6}{5}=\frac{5}{6}\). It is a proper fraction.
(v) Reciprocal of \(\frac{12}{7}=\frac{7}{12}\). It is a proper fraction.
(vi) Reciprocal of \(\frac{1}{8}\) = 8. It is a whole fraction.
(vii) Reciprocal of \(\frac{1}{11}\) = 11. It is a whole fraction.

Question 3.

Question 4.
Find:

Solution:
(i) \(\frac{2}{5}÷\frac{1}{2}=\frac{2}{5} \times 2=\frac{4}{5}\)

MP Board Class 7th Maths Solutions