CBSE Class 7

CBSE Class 7

NCERT Solutions for Class 7 Science Chapter 2 Nutrition in Animals

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NCERT Solutions for Class 7 Science Chapter 2 Nutrition in Animals are part of NCERT Solutions for Class 7 Science. Here we have given NCERT Solutions for Class 7 Science Chapter 2 Nutrition in Animals.

BoardCBSE
TextbookNCERT
ClassClass 7
SubjectScience
ChapterChapter 2
Chapter NameNutrition in Animals
Number of Questions Solved13
CategoryNCERT Solutions

NCERT Solutions for Class 7 Science Chapter 2 Nutrition in Animals

Question 1.
Fill in the blanks:

  1. The main steps of nutrition in humans are ……., ………., ………., ………, and ………….
  2. The largest gland in the human body is ………..
  3. The stomach releases hydrochloric acid and ……… juices which act on the food.
  4. The inner wall of the small intestine has many finger-like outgrowths called ……….
  5. Amoeba digests its food in the ………..

Answer:

  1. ingestion, digestion, absorption, assimilation, and egestion.
  2. liver
  3. digestive
  4. villi
  5. food vacuole.

Question 2.
Mark ‘T’ if the statement is true and ‘F’ if it is false:

  1. Digestion of starch starts in the stomach. (T/F)
  2. The tongue helps in mixing food with saliva. (T/F)
  3. The gall bladder temporarily stores bile. (T/F)
  4. The ruminants bring back swallowed grass into their mouth and chew it for some time. (T/F)

Answer:

  1. F
  2. T
  3. T
  4. T

Question 3.
Tick (√) mark the correct answer in each of the following:
(a) Fat is completely digested in the
(i) stomach
(ii) mouth
(iii) small intestine
(iv) large intestine

(b) Water from the undigested food is absorbed mainly in the
(i) stomach
(ii) food pipe
(iii) small intestine
(iv) large intestine
Answer:
(a) (iii) Small intestine
(b) (iv) Large intestine.

Question 4.
Match the items of Column I with those given in Column II:

Column IColumn II
Food componentsProduct(s) of digestion
CarbohydratesFatty acids and glycerol
ProteinsSugar
FatsAmino acids

Answer:

Column I

Column II

Food componentsProduct(s) of digestion
CarbohydratesSugar
ProteinsAmino acids
FatsFatty acids and glycerol

Question 5.
What are villi? What are their location and function?
Answer:
The inner walls of the small intestine have thousands of finger-like outgrowths. These are called villi. Villi are located in the small intestine. The villi increase the surface area for absorption of the digested food. Each villus has a network of thin and small blood vessels close to its surface. The surface of the villi absorbs the digested food materials.

Question 6.
Where is the bile produced? Which component of the food does it help to digest?
Answer:
Bile is produced in the liver. The bile juice is stored in a sac called the gall bladder. Bile plays an important role in the digestion of fats.

Question 7.
Name the type of carbohydrate that can be digested by ruminants but not by humans. Give the reason also.
Answer:
Cellulose is a type of carbohydrate that can be digested by ruminants but not by humans. Ruminants have a large sac-like structure called rumen which is present in between the small intestine and large intestine. The cellulose is digested here by the action of certain bacteria which are not present in humans.

Question 8.
Why do we get instant energy from glucose?
Answer:
Because glucose can easily breakdown in the cell with the help of oxygen and gives carbon dioxide, water, and energy.

Question 9.
Which part of the digestive canal is involved in:

  1. absorption of food ……..
  2. chewing of food …….
  3. the killing of bacteria …….
  4. complete digestion of food ………
  5. formation of faeces ……..

Answer:

  1. Small intestine
  2. Mouth
  3. Stomach
  4. Small intestine
  5. Large intestine.

Question 10.
Write one similarity and one difference between the nutrition in amoeba and human beings.
Answer:

  1. Similarity: Both amoeba and humans use digestive juices to digest food.
  2. Difference: Human needs to chew food, whereas, an amoeba, there is no chewing.

Question 11.
Match the items of Column I with suitable items in Column II

Column IColumn II
(a) Salivary gland(i) Bile juice secretion
(b) Stomach(ii) Storage of undigested food
(c) Liver(iii) Saliva secretion
(d) Rectum(iv) Acid release
(e) Small intestine(v) Digestion is completed
(f) Large intestine(vi) Absorption of water
(Vii) Release of faeces

Answer:

Column IColumn II
(a) Salivary gland(iii) Saliva secretion
(b) Stomach(iv) Acid release
(c) Liver(i) Bile juice secretion
(d) Rectum(ii) Storage of undigested food
(e) Small intestine(v) Digestion is completed
(f) Large intestine(vi) Absorption of water

 Question 12.
Label figure of the digestive system.
Answer:
NCERT Solutions for Class 7 Science Chapter 2 Nutrition in Animals Q.12
Fig. Human digestive system

Question 13.
Can we survive only on raw, leafy vegetables/grass? Discuss.
Answer:
No. Because to live a healthy life, we need a complete balance of all nutrients. Raw leafy vegetables/grass may have cellulose which can not be digested by us. So, we cannot survive only on raw, leafy vegetables/grass.

We hope the NCERT Solutions for Class 7 Science Chapter 2 Nutrition in Animals help you. If you have any query regarding NCERT Solutions for Class 7 Science Chapter 2 Nutrition in Animals, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Science Chapter 1 Nutrition in Plants

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NCERT Solutions for Class 7 Science Chapter 1 Nutrition in Plants are part of NCERT Solutions for Class 7 Science. Here we have given NCERT Solutions for Class 7 Science Chapter 1 Nutrition in Plants.

BoardCBSE
TextbookNCERT
ClassClass 7
SubjectScience
ChapterChapter 1
Chapter NameNutrition in Plants
Number of Questions Solved13
CategoryNCERT Solutions

NCERT Solutions for Class 7 Science Chapter 1 Nutrition in Plants

Question 1.
Why do organisms need to take food?
Answer:
Organisms need to take food to

  1. get the energy to do work.
  2.  build up the body.
  3. repair damages in the body.
  4.  maintain the functions of the body.

Question 2.
Distinguish between a parasite and a saprotroph.
Answer:

ParasiteSaprotroph

1. A parasite takes the food from the organism on which it survives.

1. They secrete digestive juices on the dead and decaying matter and convert it into a solution

2. They feed on a living organism.2. They feed on dead and decaying matter.
3. The organism on which it survives is called host.3. They do not feed on a living organism.
4. It deprives the host of valuable nutrients.4. There is no host at all.

 Question 3.
How would you test the presence of starch in leaves?
Answer:

  1. Take two healthy green potted plants of the same kind in order to remove all the starch from the leaves.
  2.  Keep one in the darkroom (or in a black box) for 72 hours and the other in the sunlight.
  3.  Now, take one leaf from each of the plants.
  4. Put few drops of iodine solution on each of the leaves.
  5.  The leaf kept in the sunlight will turn blue-black due to the presence of starch.
  6. The leaf kept in the dark will not turn blue-black because of the absence of starch.

Question 4.
Give a brief description of the process of synthesis of food in green plants.
Answer:
The leaves of a plant have a green pigment called chlorophyll. In the presence of sunlight, they use carbon dioxide and water to synthesize carbohydrates.
Carbon dioxide + Water \(\xrightarrow [ Sunlight ]{ Chlorophyll } \) Carbohydrade + Water + Oxygen
During the process, oxygen is released. The carbohydrates ultimately get converted into starch.
Carbon dioxide from the air is taken through stomata. Water and minerals are absorbed by the roots and transported to the leaves.

Question 5.
Show with the help of a sketch that plants are the ultimate source of food.
Answer:
NCERT Solutions for Class 7 Science Chapter 1 Nutrition in Plants Q.5

Fig. Plants capture solar energy by a unique process called photosynthesis

Question 6.
Fill in the blanks:

  1. Green plants are called …….. since they synthesize their own food.
  2. The food synthesized by the plants is stored as ………
  3. In photosynthesis solar energy is captured by the pigment called ……….
  4. During photosynthesis plants take in ……… and release ……….

Answer:

  1. autotrophs
  2. starch
  3. chlorophyll
  4. carbon dioxide, oxygen

Question 7.
Name the following:

  1. A parasitic plant with a yellow, slender, and branched stem.
  2. A plant that is partially autotrophic.
  3. The pores through which leave exchange gases.

Answer:

  1. Cuscuta
  2. Insectivorous plants
  3. Stomata

Question 8.
Tick the correct answer:
(a) Amarbel is an example of:
(i) Autotroph
(ii) Parasite
(iii) Saprotroph
(iv) Host
(b) The plant which traps and feeds on insects is:
(i) Cuscuta
(ii) China rose
(iii) Pitcher plant
(iv) Rose
Answer:
(ii) Parasite
(iii) Pitcher plant

Question 9.
Match the items given in Column I with those in Column II.

Column IColumn II
ChlorophyllBacteria
NitrogenHeterotrophs
CuscutaPitcher plant
AnimalsLeaf
InsectsParasite

Answer:

Column IColumn II
ChlorophyllLeaf
NitrogenBacteria
CuscutaParasite
AnimalsHeterotrophs
InsectsPitcher plant

 Question 10.
Mark T if the statement is true and ‘F’ if it is false:
(i) Carbon dioxide is released during photosynthesis. (T/F)
(ii) Plants which synthesize their food are called saprotrophs. (T/F)
(iii) The product of photosynthesis is not a protein. (T/F)
(iv) Solar energy is converted into chemical energy during photosynthesis. (T/F)
Answer:
(i) F
(ii) F
(iii) T
(iv) T

Question 11.
Choose the correct option from the following:
Which part of the plant takes in carbon dioxide from the air for photosynthesis?
(i) Root hair
(ii) Stomata
(iii) Leaf veins
(iv) Petals
Answer:
(ii) Stomata

Question 12.
Choose the correct option from the following:
Plants take carbon dioxide from the atmosphere mainly through their:
(i) roots
(ii) stem
(iii) flowers
(iv) leaves
Answer:
(iv) leaves

Question 13.
Why do farmers grow many fruits and vegetable crops inside large greenhouses? What are the advantages to the farmers?
Answer:
Most of the crops require a lot of nitrogen to make protein. After the harvest, the soil becomes deficient in nitrogen. Though nitrogen gas is available in the air, plants cannot use it directly. They need nitrogen in a soluble form. The bacterium called Rhizobium can take atmospheric nitrogen and convert it into soluble nitrogenous components. Rhizobium is present in the roots of some fruits and vegetables and legumes plants which provides nitrogen to them. By crop rotation, farmers increase the nitrogenous compounds in soil. So there is no need to add nitrogenous fertilizers to the soil in which leguminous plants are grown. By this practice, farmers provide good quality crops and save money.

We hope the NCERT Solutions for Class 7 Science Chapter 1 Nutrition in Plants help you. If you have any query regarding NCERT Solutions for Class 7 Science Chapter 1 Nutrition in Plants, drop a comment below and we will get back to you at the earliest.

RS Aggarwal Class 7 Solutions Chapter 23 Probability Ex 23

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RS Aggarwal Class 7 Solutions Chapter 23 Probability Ex 23

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 23 Probability Ex 23.

Question 1.
Solution:
(i) Here, total number of trials = 300
Number of heads got = 136.
P(E) = \(\frac { 136 }{ 300 }\) = \(\frac { 34 }{ 75 }\)
(ii) Total number of trials = 300
Number of tails got = 164
P(E) = \(\frac { 164 }{ 300 }\) = \(\frac { 41 }{ 75 }\)

Question 2.
Solution:
Number times, the two coins were tossed = 200
Number of times got two heads = 58
Number of times got one head = 83
and number of times got no head = 59
(i) Probability of getting 2 heads : P(E) = \(\frac { 58 }{ 200 }\) = \(\frac { 29 }{ 100 }\)
(ii) Probability of getting one head : P(E) = \(\frac { 83 }{ 200 }\)
(iii) Probability of getting no head : P(E) = \(\frac { 59 }{ 200 }\)

Question 3.
Solution:
Number of times, a dice was thrown = 100
(i) Number of times got 3 = 18
Probability will be
P(E) = \(\frac { 18 }{ 100 }\) = \(\frac { 9 }{ 50 }\)
(ii) Number of times got 6 = 9
Probability will be
P(E) = \(\frac { 9 }{ 100 }\)
iii) Number of times got 4 = 15
Probability will be
P(E) = \(\frac { 15 }{ 100 }\) = \(\frac { 3 }{ 20 }\)
(iv) Number of times got 1 = 21
Probability will be
P(E) = \(\frac { 21 }{ 100 }\)

Question 4.
Solution:
Total number of ladies = 100
Number of ladies also like coffee = 36.
Number of ladies who dislike coffee = 64
(i) Probability of lady who like coffee :
P(E) = \(\frac { 36 }{ 100 }\) = \(\frac { 9 }{ 25 }\)
(ii) Probability of lady who dislikes coffee:
P(E) = \(\frac { 64 }{ 100 }\) = \(\frac { 16 }{ 25 }\)

 

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RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22

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RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 22 Bar Graphs Ex 22.

Question 1.
Solution:
(i) Draw a horizontal line OX and a vertical line OY on the graph representing x-axis and y-axis.
(ii) Along OX, mark subjects and along y-axis, mark, number of marks
(iii) Take one division = 10 marks.
Now we shall draw bar graph as shown.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 1

Question 2.
Solution:
(i) Draw a horizontal line OX and a vertical line OY on the graph paper. These two lines represent x-axis and y-axis respectively.
(ii) Along OX, write sports and along OY, number of students choosing on division equal to 10 students.
(iii) Now draw the bars of various heights according to the no. of students as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 2

Question 3.
Solution:
(i) Draw a horizontal line OX and a vertical line OY. These represent x-axis and y-axis respectively on the graph paper.
(ii) Along OX, write years and along OY, no. of students choosing one division = 200 students.
(iii) Draw bars of various heights according to the number of students given.
This is the required bar graph as shown.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 3

Question 4.
Solution:
(i) Draw a horizontal line OX and a vertical line OY which represent x-axis and y-axis respectively on the graph.
(ii) Along OX, write years and along OY, no. of scooters
(iii) Choose 1 division = 300
(iv) Now draw bars of different heights according to given data as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 4

Question 5.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along OX write countries and along OY, take Birth rate per thousand.
(iii) Choose 1 division = 10.
(iv) Now draw bars of different heights according to the given data as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 5

Question 6.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis on the graph paper.
(ii) Along x-axis write states and along y-axis population in lakhs.
(iii) Choose one division = 200 (Lakhs)
(iv) Draw bars of different heights according to the data given as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 6

Question 7.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis write years and along y-axis Interest in thousand cores rupees.
(iii) Choose one division = 20 thousand crore rupees.
(iv) Draw bars of different heights according to the given data as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 7

Question 8.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis, write city and along y-axis the distance (in km).
(iii) Choose one division = 200 km.
(iv) Draw bars of different heights according to the given data as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 8

Question 9.
Solution:
(i) Draw a horizontal line OX and a vertical line OY represent x-axis and y-axis respectively on the graph paper.
(i) Along x-axis write countries and along y-axis life expectancy (in years)
(ii) Choose one division = 10 years
(iv) Draw bars of different heights according to the given data as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 9

Question 10.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis write years and along y-axis imports in thousand crores rupees.
(iii) Choose one division = 50 thousand crore rupees.
(iv) Draw bars of different heights according to the given data as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 10

Question 11.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis write months and along y-axis average rainfall in cm.
(iii) Choose one small division = 5cm.
(iv) Draw different bars of different heights according to the given data as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 11

Question 12.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis write Brand and along y-axis, percentage of buyers.
(iii) Choose one division = 5% of buyers.
(iv) Draw bars of different heights according to the given data, as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 12

Question 13.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis, write week and along y-axis. Rate per 10gm in rupees.
(iii) Choose and division = 1000
(iv) Draw bars of different heights according to the data given as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 13

Question 14.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis, write mode of transport and on the y-axis is number of students.
(iii) Choose one division =100 students
(iv) Draw bars of different heights according to given data as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 14

Question 15.
Solution:
(i) The bar graph shows the marks obtained by a student in different subjects.
(ii) The student is very good in Mathematics.
(iii) The student is very poor in Hindi.
(iv) Average marks
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 15

Question 16.
Solution:
(i) The bar graph shows the number of members in each of the 85 families.
(ii) 40 families have 3 number each.
(iii) Number of people living alone is nil.
(iv) The families having 3 members each is most common.

Question 17.
Solution:
(i) The highest peak is Mount Everest whose heighest is 8800 m.
(ii) The required ratio between the highest peak and the next heighest peak = 8800 : 8200 or 44 : 41
(iii) Arranging the heights of peaks in descending order are : 8800 m, 8200 m, 8000 m, 7500 m, 6000 m

Question 18.
Solution:
(i) We can draw the double bar graph in the following steps :
(a) On a graph paper, draw a horizontal line OX and a vertical line OY, representing the x-axis and the y-axis respectively.
(b) Along OX, write the names of the subjects taken at appropriate uniform gaps.
(c) Choose the scale : 1 division = 10
(d) The heights of various pairs of bars in terms of the number of small divisions are :
Mon. 350 and 200; Tues. 400 and 450; Wed. 500 and 300; Thurs. 450 and 250; Fri. 550 and 100 and Sat. 450 and 50.
(e) On the x-axis, draw pairs of bars of equal width and of heights shown in step 4 at the points marked in step 2.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 16
(ii) The number of readers in the library was maximum on TUESDAY.
(iii) Total Number of magazine readers in a week = 200 + 450 + 300 + 250 + 100 + 50 = 1350
Mean Number of readers per day = \(\frac { 1350 }{ 6 }\) = 225

Question 19.
Solution:
(i) We can draw the double bar graph in the following steps :
(a) On a graph paper, draw a horizontal line OX and a vertical line OY, representing the x-axis and the y-axis respectively.
(b) Along OX, write the names of the subjects taken at appropriate uniform gaps.
(c) Choose the scale: 1 division = 4
(d) The heights of various pairs of bars in terms of the number of small divisions are:
VI 95 and 92; VII 90 and 85; VIII 82 and 78; IX 75 and 69; X 68 and 62.
(e) On the x-axis, draw pairs of bars of equal width and of heights shown in step 4 at the points marked in step 2.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 17
(ii) For class VII, total Number of Students = 90
Number of students present = 85
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 18

Question 20.
Solution:
(i) We can draw the double bar graph in the following steps :
(a) On a graph paper, draw a horizontal line OX and a vertical line OY, representing the x-axis and the y-axis respectively.
(b) Along OX, write the names of the subjects taken at appropriate uniform gaps.
(c) Choose the scale: 1 division = 10
(d) The heights of various pairs of bars in terms of the number of small divisions are:
January 2 and 1.5 ; February 3.25 and 3; March 4 and 3.5; April 4.5 and 6; May 7.75 and 5.5; June 8 and 6.5.
(e) On the x-axis, draw pairs of bars of equal width and of heights shown in step 4 at the points marked in step 2.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 19
(ii) June
(iii) January

Question 21.
Solution:
(i) We can draw the double bar graph in the following steps :
(a) On a graph paper, draw a horizontal line OX and a vertical line OY, representing the x-axis and the y’-axis respectively.
(b) Along OX, write the names of the subjects taken at appropriate uniform gaps.
(c) Choose the scale: 1 division = 10
(d) The heights of various pairs of bars in terms of the number of small divisions are:
Town A 640000 and 750000; Town B 830000 and 920000; Town C 460000 and 630000; Town D 290000 and 320000
(e) On the x-axis, draw pairs of bars of equal width and of heights shown in step 4 at the points marked in step 2.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 20
(ii) Town B
(iii) Town D

 

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RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21C

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RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data (Mean, Median and Mode) Ex 21C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21C.

Other Exercises

Question 1.
Solution:
(i) Arranging in ascending order :
4, 6, 7, 8, 8, 8, 8, 10, 11, 15
We see that 8 occurs maximum times
Mode = 8
(ii) Arranging in ascending order :
18, 21, 23, 27, 27, 27, 27, 27, 36, 39, 40
We see that 27 occurs maximum times
Mode = 27

Question 2.
Solution:
Arranging in ascending order :
28, 31, 32, 32, 32, 32, 34, 36, 38, 40, 41.
We see that 32 occurs maximum times
Mode = 32 years

Question 3.
Solution:
We prepare the table as given below:
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21C 1
Here, number of terms = 45, which is odd
Median = \(\frac { n + 1 }{ 2 }\) th term = \(\frac { 45 + 1 }{ 2 }\) = \(\frac { 46 }{ 2 }\) th term
= 23th term = 450
Now, mode = 3(median) – 2(mean)
= 3 x 450 – 2 x 470
= 1350 – 940
= 410

Question 4.
Solution:
We prepare the table as given below:
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21C 2
Here, number of terms (N) = 41, which is odd
Median = \(\frac { n + 1 }{ 2 }\) th term = \(\frac { 41 + 1 }{ 2 }\) th term
= \(\frac { 42 }{ 2 }\) = 21 th term = 22 {value of 18 to 29 = 22}
Mode = 3 (median) – 2 (mean)
= 3 x 22 – 2 x 21.92 = 66 – 43.84 = 22.16

Question 5.
Solution:
We prepare the table as given below:
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21C 3
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21C 4

Hope given RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.