RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F

RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 5 Fractions Ex 5F.

Other Exercises

Find the difference:

Question 1.
Solution:
\(\frac { 5 }{ 8 } -\frac { 1 }{ 8 } \)
= \(\\ \frac { 5-1 }{ 8 } \)
= \(\frac { 4 }{ 8 } \)
= \(\frac { 4\div 4 }{ 8\div 4 } \)
= \(\frac { 1 }{ 2 } \)

Question 2.
Solution:
\(\frac { 7 }{ 12 } -\frac { 5 }{ 12 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 2.1

Question 3.
Solution:
\(4\frac { 3 }{ 7 } -2\frac { 4 }{ 7 } \)
= \(\frac { 31 }{ 7 } -\frac { 18 }{ 7 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 3.1

Question 4.
Solution:
\(\frac { 5 }{ 6 } -\frac { 4 }{ 9 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 4.1

Question 5.
Solution:
\(\frac { 1 }{ 2 } -\frac { 3 }{ 8 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 5.1

Question 6.
Solution:
\(\frac { 5 }{ 8 } -\frac { 7 }{ 12 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 6.1

Question 7.
Solution:
\(2\frac { 7 }{ 9 } -1\frac { 8 }{ 15 } \)
= \(\frac { 25 }{ 9 } -\frac { 23 }{ 15 } \)
(changing into improper fractions)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 7.1

Question 8.
Solution:
\(3\frac { 5 }{ 8 } -2\frac { 5 }{ 12 } \)
= \(\frac { 29 }{ 8 } -\frac { 29 }{ 12 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 8.1

Question 9.
Solution:
\(2\frac { 3 }{ 10 } -1\frac { 7 }{ 15 } \)
= \(\frac { 23 }{ 10 } -\frac { 22 }{ 15 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 9.1

Question 10.
Solution:
\(6\frac { 2 }{ 3 } -3\frac { 3 }{ 4 } \)
= \(\frac { 20 }{ 3 } -\frac { 15 }{ 4 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 10.1

Question 11.
Solution:
\(7-5\frac { 2 }{ 3 } \)
= \(\frac { 7 }{ 1 } -\frac { 17 }{ 3 } \)
(changing into improper fractions)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 11.1

Question 12.
Solution:
\(10-6\frac { 3 }{ 8 } \)
= \(\frac { 10 }{ 1 } -\frac { 51 }{ 8 } \)
(changing into improper fractions)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 12.1

Simpilify

Question 13.
Solution:
\(\frac { 5 }{ 6 } -\frac { 4 }{ 9 } +\frac { 2 }{ 3 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 13.1

Question 14.
Solution:
\(\frac { 5 }{ 8 } +\frac { 3 }{ 4 } -\frac { 7 }{ 12 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 14.1

Question 15.
Solution:
\(2+\frac { 11 }{ 15 } -\frac { 5 }{ 9 } \)
= \(\frac { 90+33-25 }{ 45 } \)
(LCM of 15 and 9 = 45)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 15.1

Question 16.
Solution:
\(5\frac { 3 }{ 4 } -4\frac { 5 }{ 12 } +3\frac { 1 }{ 6 } \)
= \(\frac { 23 }{ 4 } -\frac { 53 }{ 12 } +\frac { 19 }{ 6 } \)
(changing into improper fractions)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 16.1

Question 17.
Solution:
\(2+5\frac { 7 }{ 10 } -3\frac { 14 }{ 15 } \)
= \(\frac { 2 }{ 1 } +\frac { 57 }{ 10 } -\frac { 59 }{ 15 } \)
(changing into improper fractions)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 17.1

Question 18.
Solution:
\(8-3\frac { 1 }{ 2 } -2\frac { 1 }{ 4 } \)
= \(\frac { 8 }{ 1 } -\frac { 7 }{ 2 } -\frac { 9 }{ 4 } \)
(changing into improper fractions)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 18.1

Question 19.
Solution:
\(8\frac { 5 }{ 6 } -3\frac { 3 }{ 8 } +2\frac { 7 }{ 12 } \)
= \(\frac { 53 }{ 6 } -\frac { 27 }{ 8 } +\frac { 31 }{ 12 } \)
(changing into improper fractions)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 19.1

Question 20.
Solution:
\(6\frac { 1 }{ 6 } -5\frac { 1 }{ 5 } +3\frac { 1 }{ 3 } \)
= \(\frac { 37 }{ 6 } -\frac { 26 }{ 5 } +\frac { 10 }{ 3 } \)
(changing into improper fractions)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 20.1

Question 21.
Solution:
\(3+1\frac { 1 }{ 5 } +\frac { 2 }{ 3 } -\frac { 7 }{ 15 } \)
= \(\frac { 3 }{ 1 } +\frac { 6 }{ 5 } +\frac { 2 }{ 3 } -\frac { 7 }{ 15 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 21.1

Question 22.
Solution:
By subtracting \(9 \frac { 2 }{ 3 } \) from 19, we get the required number
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 22.1

Question 23.
Solution:
By subtracting \(6 \frac { 7 }{ 15 } \) from \(8 \frac { 1 }{ 5 } \) we get the required number
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 23.1

Question 24.
Solution:
Sum of \(3 \frac { 5 }{ 9 } \) and \(3 \frac { 1 }{ 3 } \)
= \(\frac { 32 }{ 9 } +\frac { 10 }{ 3 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 24.1

Question 25.
Solution:
\(\\ \frac { 3 }{ 4 } \), \(\\ \frac { 5 }{ 7 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 25.1

Question 26.
Solution:
Milk bought by Mrs. Soni = \(7 \frac { 1 }{ 2 } \) litres
and milk consumed by here = \(5 \frac { 3 }{ 4 } \) litres
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 26.1

Question 27.
Solution:
Total time of film show = \(3 \frac { 1 }{ 3 } \) hours
Total spent on advertisement = \(1 \frac { 3 }{ 4 } \) hours
Duration of the film
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 27.1

Question 28.
Solution:
On a day, rickshaw pullar earned
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 28.1

Question 29.
Solution:
Total length of wire =\(2 \frac { 3 }{ 4 } \)-metres
Length of one piece = \(\\ \frac { 5 }{ 8 } \) metre
Length of the other piece
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5F 29.1

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RS Aggarwal Class 6 Solutions Chapter 18 Circles Ex 18

RS Aggarwal Class 6 Solutions Chapter 18 Circles Ex 18

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 18 Circles Ex 18

Question 1.
Solution:
Method :
RS Aggarwal Class 6 Solutions Chapter 18 Circles Ex 18 Q1.1
Take a point O on the paper as shown in the figure. With the help of the rular, open out compasses in such a way that the distance between the metal point and pencil point is 4 cm. Take the compasses in the same position and put its metal point at O and draw the circle.
Remove the compasses and again open out the compasses in such a way that the distance between the metal point and pencil point is 5.3 cm. Taking O as the centre, draw another circle. Again remove the compasses and similarly draw the third circle with radius 6.2 cm. Then the required circles are as shown in the figure which have radius OA = 4 cm., OB = 5.3 cm. and OC = 6.2 cm.

Question 2.
Solution:
Method : Take a point C on the paper. With the help of the rular, open out the compasses in such a way that the distance between its metal point and pencil point is 4.5 cm. Take the compasses in the same position and put its metal point at C and draw the circle. Mark points P, Q and R as shown in the figure as required.
RS Aggarwal Class 6 Solutions Chapter 18 Circles Ex 18 Q2.1

Question 3.
Solution:
Method : Take a point O on the paper. With the help of the rular, open out the compasses in such a way that the distance between the metal point and pencil point is 4 cm. Take the compasses in the same position and put the metal point at O and draw the circle.
RS Aggarwal Class 6 Solutions Chapter 18 Circles Ex 18 Q3.1
Take A and B any points on the circle and join AB. ThenAB is the chord of the circle. Mark points X and Y on the circle as shown. Then arc AXB and arc AYB are the required minor and major arcs respectively.

Question 4.
Solution:
(i) False
(ii) True
(iii) False
(iv) False
(v) True.

Question 5.
Solution:
Steps of construction :
(i) With centre O and radius 3.7 cm, draw a circle.
(ii) Take a point A on the circumference of the circle.
(iii) Join OA.
(iv) At O, draw another radius OB such that ∠AOB = 72° with the help of protractor. Then sector AOB is the required one.
RS Aggarwal Class 6 Solutions Chapter 18 Circles Ex 18 Q5.1

Question 6.
Solution:
(i) > (ii) < (iii) > (iv) >. Ans.

Question 7.
Solution:
(i) Passes through
(ii) at the centre, on the circle
(iii) chord
(iv) arc
(v) sector.

 

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CA Foundation Business Economics Study Material – Determination of Prices

CA Foundation Business Economics Study Material Chapter 4 Price Determination in Different Markets – Determination of Prices

Determination of Equilibrium Price

  • We know that law of demand reveals, if other conditions remain unchanged, more quantity of a commodity is demanded in the market at a lower price and less quantity is demanded at a higher price. Therefore, demand curve slopes downward.
  • Similarly, the law of supply reveals, if other conditions remain unchanged, more quantity of a commodity is supplied in the market at a higher price and less quantity is supplied at a lower price. Therefore, supply curve slopes upward.
  • Demand and supply are the two main factors that determine the price of a commodity in the market. In other words, the price of a commodity is determined by the inter-action of the forces of demand and supply.
  • The price that will come to prevail in the market is one at which quantity demanded equals 1 quantity supplied.
  • This price at which quantity demand equals quantity supplied is called equilibrium price.
  • The quantity demanded and supplied at equilibrium price is called equilibrium quantity.

The process of price determination is illustrated with the help of following imaginary schedule and diagram.

CA Foundation Business Economics Study Material - Determination of Prices

The above table shows that at a price of ₹ 3 per unit, the quantity demanded equals quantity supplied of the commodity. At ₹ 3 two forces of demand and supply are balanced. Thus, ₹ 3 is the equilibrium price and equilibrium quantity at ₹ 3 is 300 units.

CA Foundation Business Economics Study Material - Determination of Prices 1

  • The equilibrium between demand and supply can also be explained graphically as in Fig.
  • In Fig. the market is at equilibrium at point ‘E’, where the demand curve and supply curve intersect each other. Here quantity demanded and supplied, are equal to each other.
  • At point ‘E’, the equilibrium price is ₹ 3 per unit and equilibrium quantity is 300 units.
  • If the price rises to ₹ 4 per unit, the supply rises to 400 units but demand falls to 200 units. Thus, there is excess supply of 200 units in the market.
  • In order to sell off excess supply of 200 units the sellers will compete among themselves and in doing so the price will fall.
    As a result the quantity demand will rise and quantity supplied will fall and becoming equal to each other at the equilibrium price ₹ 3.
  • Similarly, if the price falls to ₹ 2 per unit, the demand rises to 400 units but supply falls to 200 units. Thus, there is excess demand of 200 units in the market.
  • As the price is less there is competition among the buyers to buy more and more. This competition among buyers increases with the entry of new buyers.
  • More demand and less supply and competition among buyers will push up the price.
  • As a result, quantity demanded will fall and quantity supplied will rise and become equal to each other at the equilibrium price of ₹ 3.

Effects of Shifts in Demand and Supply on Equilibrium Price

While determining the equilibrium price, it was assumed that demand and supply conditions were constant. In reality however, the condition of demand and supply change continuously.
Thus, changes in income, taste and preferences, changes in the availability and prices of related goods, etc. brings changes in demand conditions and cause demand curve to shift either to right or left.
In the same way, changes in the technology, changes price of labour, raw materials, etc., changes in the number of firms, etc. brings changes in supply conditions and cause supply curve to shift either to right or left.

(a) Change (shift) in Demand and Supply remaining constant.

CA Foundation Business Economics Study Material - Determination of Prices 2

  • In Fig.- DD and SS are the original demand and supply curves respectively intersecting each other at point E.
  • At point E, the equilibrium price is OP and the demand and supply (ie. equilibrium quantity) are equal at OQ.
  • When the demand increases, the demand curve shifts upwards from DD to D1D1 supply remaining the same.
    As a result, the equilibrium price rises from OP to OP1 and the equilibrium quantity increases from OQ to OQ1 as shown at point E1.
  • When the demand decreases, the demand curve shifts downwards from DD to D2D2, Supply remaining the same.
  • As a result, the equilibrium price falls from OP to OP2 and the equilibrium quantity decreases from OQ to OQ2 as shown at point E2.

(b) Change (shift) in Supply and Demand remaining constant.

CA Foundation Business Economics Study Material - Determination of Prices 3

  • In Fig. – DD and SS are the original demand and supply curves respectively inter-sections each other at point E.
  • At point E, the equilibrium price is OP and the demand and supply (i.e. Equilibrium quantity) are equal at OQ.
  • When the supply increases, the supply curve shifts to the right from SS to S1S1 demand remaining the same.
  • As a result, the equilibrium price falls from OP to OP1 and the equilibrium quantity increases from OQ to OQ1 as shown at point E1.
  • When the supply decreases, the supply curve shifts to the left from SS to S2S2, demand remaining the same.
  • As a result, the equilibrium price rises from OP to OP2 and the equilibrium quantity decreases from OQ to OQ2 as shown at point E2.

Effects of Simultaneous Shifts in Demand and Supply on Equilibrium Price

Sometimes demand and supply conditions may change at the same time changing the equilibrium price and quantity. The changes in both demand and supply simultaneously can be discussed with the help of following diagrams:

CA Foundation Business Economics Study Material - Determination of Prices 4

  • In Fig. – DD and SS are the original demand and supply respectively intersecting each other at point E at which the equilibrium price is OP and the equilibrium quantity is OQ.
  • Fig. (a) shows that the increase in demand is equal to increase in supply. The new curves D1D1 and S1S1 intersect at E1. Therefore, the new equilibrium price is equal to old equilibrium price OP. But equilibrium quantity increases.
  • Fig. (b) shows that the increase in demand is more than increase in supply. The new curves D1D1 and S1S1 intersect each other at point E, which shows that new equilibrium price OP1 is higher than old equilibrium price OP. But equilibrium quantity increases.
  • Fig. (c) shows that the increase in supply is more than increase in demand. The new curves D1D1 and S1S1 intersect each other at point E1 which shows that new equilibrium price OP1 is lower than old equilibrium price OP. But equilibrium quantity increases.

RS Aggarwal Class 6 Solutions Chapter 17 Quadrilaterals Ex 17B

RS Aggarwal Class 6 Solutions Chapter 17 Quadrilaterals Ex 17B

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 17 Quadrilaterals Ex 17B

Other Exercises

Objective questions
Mark against the correct answer in each of the following :

Question 1.
Solution:
(c) ∵ Sum of angles of a quadrilateral is 360°.

Question 2.
Solution:
Sum of 4 angles of a quadrilateral = 360° and three angles of a quadrilateral are 80°, 70° and 120°
∵ Fourth angle = 360° – (80° + 70° + 120°)
= 360° – 270° – 90° (c)

Question 3.
Solution:
Sum of angles of a quadrilateral = 360°
The ratio in there four angles is 3 : 4 : 5 : 6
RS Aggarwal Class 6 Solutions Chapter 17 Quadrilaterals Ex 17B Q3.1

Question 4.
Solution:
(d) Y Quadrilateral having one pair of parallel sides is a trapezium.

Question 5.
Solution:
(d) ∵ Quadrilateral having opposite sides parallel is called a parallelogram.

Question 6.
Solution:
(b) ∵ A trapezium having nonparallel sides equal is called an isosceles trapezium.

Question 7.
Solution:
(b) ∵ Diagonals of a rhombus bisect each other at right angles.

Question 8.
Solution:
(b) ∵ A square has four equal sides and also diagonals are equal.

Question 9.
Solution:
A quadrilateral having two pairs of equal adjacent sides but unequal opposite angles is called a kite. (c)

Question 10.
Solution:
A regular quadrilateral is a quadrilateral having equal sides and equal angles which is a square. (c)

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RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5E

RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5E

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 5 Fractions Ex 5E.

Other Exercises

Find the sum :

Question 1.
Solution:
\(\frac { 5 }{ 8 } +\frac { 1 }{ 8 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5E 1.1

Question 2.
Solution:
\(\frac { 4 }{ 9 } +\frac { 8 }{ 9 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5E 2.1

Question 3.
Solution:
\(1\frac { 3 }{ 5 } +2\frac { 4 }{ 5 } \)
\(\frac { 8 }{ 5 } +\frac { 14 }{ 5 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5E 3.1

Question 4.
Solution:
\(\frac { 2 }{ 5 } +\frac { 5 }{ 6 } \)
= \(\\ \frac { 4+15 }{ 18 } \) (LCM of 9 and 6 = 18)
= \(\\ \frac { 19 }{ 18 } \)
= \(1 \frac { 1 }{ 18 } \)

Question 5.
Solution:
\(\frac { 7 }{ 12 } +\frac { 9 }{ 16 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5E 5.1

Question 6.
Solution:
\(\frac { 4 }{ 15 } +\frac { 17 }{ 20 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5E 6.1

Question 7.
Solution:
\(2\frac { 3 }{ 4 } +5\frac { 5 }{ 6 } \)
= \(\frac { 11 }{ 4 } +\frac { 35 }{ 6 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5E 7.1

Question 8.
Solution:
\(3\frac { 1 }{ 8 } +1\frac { 5 }{ 12 } \)
= \(\frac { 25 }{ 8 } +\frac { 17 }{ 12 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5E 8.1

Question 9.
Solution:
\(2\frac { 7 }{ 10 } +3\frac { 8 }{ 15 } \)
= \(\frac { 27 }{ 10 } +\frac { 53 }{ 15 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5E 9.1

Question 10.
Solution:
\(3\frac { 2 }{ 3 } +1\frac { 5 }{ 6 } +2 \)
\(\frac { 11 }{ 3 } +\frac { 11 }{ 6 } +\frac { 2 }{ 1 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5E 10.1

Question 11.
Solution:
\(3+1\frac { 4 }{ 15 } +1\frac { 3 }{ 20 } \)
=\(\frac { 3 }{ 1 } +\frac { 19 }{ 15 } +\frac { 23 }{ 20 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5E 11.1

Question 12.
Solution:
\( 3\frac { 1 }{ 3 } +4\frac { 1 }{ 4 } +6\frac { 1 }{ 6 } \)
\(\frac { 10 }{ 3 } +\frac { 17 }{ 4 } +\frac { 37 }{ 6 } \)
(changing into improper fractions)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5E 12.1

Question 13.
Solution:
\(\frac { 2 }{ 3 } +3\frac { 1 }{ 6 } +4\frac { 2 }{ 9 } +2\frac { 5 }{ 18 } \)
\(\frac { 2 }{ 3 } +\frac { 19 }{ 6 } +\frac { 38 }{ 9 } +\frac { 41 }{ 18 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5E 13.1

Question 14.
Solution:
\(2\frac { 1 }{ 3 } +1\frac { 1 }{ 4 } +2\frac { 5 }{ 6 } +3\frac { 7 }{ 12 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5E 14.1

Question 15.
Solution:
\(2+\frac { 3 }{ 4 } +1\frac { 5 }{ 6 } +3\frac { 7 }{ 16 } \)
\(\frac { 2 }{ 1 } +\frac { 3 }{ 4 } +\frac { 13 }{ 8 } +\frac { 55 }{ 16 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5E 15.1

Question 16.
Solution:
Cost of a pencil = Rs. \(3 \frac { 2 }{ 5 } \)
Cost of an eraser = Rs.\(2 \frac { 7 }{ 10 } \)
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5E 16.1

Question 17.
Solution:
Length of cloth for kurta = \(4 \frac { 1 }{ 2 } \) metres
Length of cloth for pyjamas = \(2 \frac { 2 }{ 3 } \) metres
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5E 17.1

Question 18.
Solution:
Distance travelled by Rickshaw = \(4 \frac { 3 }{ 4 } \) km
Distance travelled on foot = \(1 \frac { 1 }{ 2 } \) km
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5E 18.1

Question 19.
Solution:
Weight of empty cylinder = \(16 \frac { 4 }{ 5 } \) kg
Weight of gas filled in it = \(14 \frac { 2 }{ 3 } \) kg
Total. weight of cylinder with gas
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5E 19.1

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NCERT Solutions for Class 10 Science Chapter 7 Control and Coordination

In this chapter 7 Control and Coordination, students will learn about the nervous system of animals, reflex actions, the human brain, how tissues are protected and how nervous tissues cause action, coordination in plants, hormones in animals.

NCERT Solutions for Class 10 Science Chapter 7 Control and Coordination

These Solutions are part of NCERT Solutions for Class 10 Science. Here we have given NCERT Solutions for Class 10 Science Chapter 7 Control and Coordination. Learn Insta provides you the Free PDF download of NCERT Solutions for Class 10 Science (Biology) Chapter 7 – Control and Coordination solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 7 – Control and Coordination Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Questions

In Text Questions

Question 1.
What is the difference between reflex action and walking ?
Answer:

Reflex Walking/Voluntary
1.      Origin. Reflex action is inborn and present in an individual right from birth.

2.      Control. It is automatic. An individual cannot control it.

3.      Intensity. It cannot be changed.

4.      Value. It has survival and protective value.

 It is acquired through learning.

It is under control of the will or brain.

It can be changed.

It has various functions, generally other than survival and protection.

More Resources

Question 2.
What happens at the synapse between two neurons ?
Answer:
NCERT Solutions for Class 10 Science Chapter 7 Control and Coordination image - 1
At synapse, axon terminal is expanded to form presynaptic knob. The dendrite terminal that lies close to it is slightly broadened and depreseed to form post-synaptic depression. A fluid filled narrow space, called synaptic cleft, occurs between the two. When an impulse reaches the presynaptic knob, it stimulates the release of neurotransmitter into synaptic cleft. Neurotransmitter comes in contact with chemoreceptor sites of the membrane of postsynaptic depression. This generates an electrochemical signal or impulse in the dendrite part of second neuron.

Question 3.
Which part of brain maintains posture and equilibrium of the body ?
Answer:
Cerebellum.

Question 4.
How do we detect the smell of an agarbatti (incense stick) ?
Answer:
Burning of an agarbatti emits smoke having very large number of odorant molecules. They enter the nose along with inhaled air. The odorant molecules are trapped in mucus present over olfactory epithelium. Olfactory receptor cells have a number of non-motile olfactory hair containing special protein molecules. Contact between the two forms cyclic AMP that generates an impulse in the receptor cells. Nerve fibres coming out of the cells carry the information to olfactory bulbs which transmit the same to temporal lobes of cerebrum for interpretation.

Question 5.
What is the role of brain in reflex action ? (CCE 2010, 2015)
Answer:
It functions as a relay centre for transferring impulse from sensory to motor neurons in several reflex actions called cerebral reflexes, e.g, closure of eyes exposed to flash of light, salivation at the sight or smell of food. In spinal reflexes it acts as information collecting and evaluation centre without any direct involvement in reflex action.

Question 6.
What are plantohormones ?
Answer:
Phytohormones are chemical substances other than nutrients produced naturally in plants which regulate growth, development, differentiation and a number of physiological processes, e.g., auxin, gibberellins, abscisic acid, cytokinins.

Question 7.
How is movement of leaves of Sensitive Plant different from movement of shoot towards light ? (CCE 2015)
Answer:
Movement in the leaves of Sensitive Plant (Mimosa pudica) is haptonastic movement which occurs due to turgor changes in the cells of pulvinules and pulvinus. Movement of a shoot towards light is. phototropic movement that is caused by differential growth.

Question 8.
Give an example of a plant hormone that promotes growth ?
Answer:
Indole 3-acetic acid or IAA (auxin).

Question 9.
How do auxins promote the growth of a tendril around a support ? (CCE 2015)
Answer:
Less auxin occurs on the side of contact as compared to the free side. More growth occurs on the free side.
As a result of more growth on the free side, the tendril coils around the support.

Question 10.
Design an experiment to demonstrate hydrotropism. (CBSE 2010, CCE 2011)
Answer:
Apparatus: Trough with perforated base, saw dust, water, seeds of Pea/Gram, wooden support.
Procedure: Take a trough with perforated base. Fill it with saw dust. Moisten the same. Sow several seeds of Pea or Gram. Place the trough in slanting position by means of a wooden block. Keep the saw dust moist by sprinkling water at intervals. Observe after 2-3 days.
Observation: As the radicles come out of the seeds, they are seen to move towards the perforations. They come out of the pores and hang downwardly for some time under the influence of gravity. However, after some growth they bend back and enter the perforations to reach moist saw dust in complete disregard of gravity (Fig. 2.11).
NCERT Solutions for Class 10 Science Chapter 7 Control and Coordination image - 2
Inference: Bending of radicles back into moist saw dust is hydrotropic movement. It occurs despite being against the force of gravity.

Question 11.
How does chemical coordination take place in animals ?
Answer:
In animals, chemical coordination is achieved through the agency of hormones which function as chemical messengers or informational molecules. Hormones are secreted by ductless glands in response to specific conditions or nervous stimulation. Timing and amount of a hormone released are regulated by feed-back mechanism. After a meal, sugar level of blood rises. It is detected by pancreas. Pancreas responds by producing hormone insulin from (3-cells of islets of Langerhans. Insulin causes glucose to be absorbed by all cells as well as get stored in liver and muscles in the form of glycogen. As the level of glucose falls in blood, insulin secretion is reduced.

Question 12.
Why is the use of iodised salt advisable ? (CCE 2011, 2015)
Answer:
Iodine is essential for synthesis of hormone thyroxine in thyroid gland. Thyroxine controls basal metabolic rate, physical activity, body temperature, heart beat, mental, physical and sexual development besides regulating carbohydrate, protein and fat metabolism. Deficiency of thyroxine disturbs metabolic, physical and mental activities besides causing disorders of simple goitre, cretinism and myxedema. Therefore, it is always advisable ^ to take iodised salt so that there is no deficiency of iodine.

Question 13.
How does our body respond when adrenaline is secreted into blood ?
Answer:
Adrenaline/emergency hormone/triple F hormone

  1. Reduces blood supply to peripheral blood vessels and gastrointestinal tract,
  2. More blood flows to skeletal and heart muscles.
  3. Increases breathing and gives more oxygen to muscles,
  4. Increases heart rate,
  5. Mobilises more glucose to muscles for higher activity.

Question 14.
Why are some patients of diabetes treated by giving injections of insulin ?
Answer:
Diabetes mellitus is of two types, insulin dependent and insulin independent. In insulin dependent diabetes, pancreas is unable to produce required quantity of insulin. As a result blood sugar continues to rise and part of sugar is excreted through urine resulting in diabetes. This is kept under check by regular injection of insulin. Availability of insulin will help the cells to take up glucose while liver and muscles are induced to store excess of glucose as glycogen.

NCERT Chapter End Exercises

Question 1.
Which of the following is a plant hormone ?
(A) Insulin
(B) Thyroxine
(C) Oestrogen
(D) Cytokinin.
Answer:
(D).

Question 2.
The gap between two neurons is called
(A) Dendrite
(B) Synapse
(C) Axon
(D) Impulse.
Answer:
(B).

Question 3.
The brain is responsible for
(A) Thinking
(B) Regulating the heart beat
(C) Balancing
(D) All the above.
Answer:
(D).

Question 4.
What is the function of receptors in our body ? Think of situation where receptors do not work properly. What problems are likely to arise ? (CCE 2011)
Answer:
Receptors are specialised cells, tissues, organs and nerve endings which are able to pick up specific stimuli, e.g., photoreceptors, gustatoreceptors, thermoreceptors, photoreceptors, statoreceptors, tangoreceptors, pain and pressure receptors. Receptors provide sensory input about external and internal environment. Without them, an animal will not able to observe, handle and taste food. It will not be aware of an approaching enemy. The animal may not be able to correct its position and fall down repeatedly if its statoreceptors are damaged. Therefore, the animal will not be able to perform the activities connected with the defective receptors.

Question 5.
Draw the structure of a neuron and explain its functions. (CCE 2011, 2013)
Answer:
NCERT Solutions for Class 10 Science Chapter 7 Control and Coordination image - 3
Functions:

  1. Dendrites. Picking up sensations and transmitting the same to cell body.
  2. Cell Body,
    1. Sustaining structure and function of dendrites and axon,
    2. Functioning as passage¬way for transmission of sensation or impulse to axon.
  3. Axon. Carrying impulse to another neuron, muscle, gland and organ. A single impulse can be transmitted to several structures with the help of axon terminals.

Mechanism of Impulse Transmission:
Impulse is a self propagated electrical current that travels from one end to another of a neuron for the passage of a message.

The pathway is stimulus ——— >dendrite ——- > cell body ——- > axon——– > axon terminal ——– > passage of stimulus.

A stimulus received by a neuron travels through it in the form of an electrical disturbance. During rest the outer surface of a neuron is positively charged while the interior has negative charge. Stimulus causes opening of ion channels which makes the outer surface negatively charged while the interior becomes positively charged. This creates the impulse which moves forward. The posterior region returns to the condition of rest. At the end of the neuron, the impulse is passed on to the next neuron, an organ, muscle or gland in the form of a neurotransmitter. Neurotransmitter is a chemical secreted by axon terminal for transmission of impulse to the next neuron, muscle, gland or organ, e.g., acetylcholine, noradraneline, glutamic acid.

Question 6.
How does phototropism occur in plants ?
Answer:
It is directional growth movement of curvature which occurs in response to unidirectional exposure to light. The region of photoperception is sho.ot apex while the region of response is in the area of elongation. The light effective in phototropic response is blue light. The photoreceptor is a flavoprotein called phototropin. Leaves are essential for producing the response.
Stems generally bend towards the direction of light. They are positively phototropic. Leaves generally come to lie at right angles to light. They are diaphototropic. Roots are either neutral (non-phototropic) or negatively phototropic. Positively phototropic heads of Sunflower perform solar tracking as they move from east to west along the direction of sun.
Phototropic movement is generally caused by increased auxin on the dark side and lesser auxin on the illuminated side. It causes more growth on the dark side of stem causing it to bend towards the source of light. The opposite happens in root where less auxin stimulates growth while higher auxin inhibits growth.
In the plant growing in the open, sunlight is received from above. Auxin diffuses equally on all sides so that the stem does not bend but grows straight vertically.

Question 7.
Which signals will get disrupted in case of spinal cord injury ?
Answer:

  1. Sensory impulses from the area innervated by injured portion,
  2. Transmission of motor impulses through the injured portion,
  3. Reflex action in the area of injury. Sensations and movements are restricted.

Question 8.
How does chemical coordination occur in plants ?
Answer:
Plants produce a number of hormones which control and coordinate their functioning. Amount of hormone depends upon the environment and other stimuli. Its effect is also regulated by its antagonistic * hormone, e.g., auxin and abscisic acid. The effect is enhanced by synergic presence of another hormone, e.g., auxin and gibberellin.

Question 9.
What is the need for a system of control and coordination in an organism ?
Answer:
The body of a multicellular organism consists of a number of components and sub-components, each specialised to perform a particular fonction. However, all the components are not required to fonction all the time at the same speed. A system of controls is required to allow them to perform or not to perform, slow down or speed up their working. Further, most activities require the simultaneous or sequential functioning of a number of parts, stopping some and stimulating others. During feeding, eyes locate the food, nose registers its smell, hands pick up the food and take it to mouth, mouth opens to receive the food, teeth and muscles take part in its mastication and saliva moistens it. Tongue perceives its taste. It moves the food below the teeth. Later it pushes the crushed food into pharynx. All this is possible only through a system of coordination.

Question 10.
How are involuntary actions and reflex actions different from each other ?
Answer:
Refex actions are involuntary in nature which are carried out to meet emergencies. However, all involuntary actions are not reflex actions. They fulfill critical life processes, e.g., circulation of blood, movement of food in food pipe.

Question 11.
Compare and contrast nervous and hormonal mechanisms for control and coordination in animals.
Answer:
Nervous system controls and coordinates many body functions as it has a well spread network of neurons. Messages travel very fast, in the form of electrical impulses. However, it has limitations,

  1. Nerve impulses do not reach each and every cell of the body,
  2. The effect of nerve impulse is of short duration,
  3. Nerve impulses cannot pass continuously.

A small gap is required between two impulses. These short-comings are overcome in endocrine system. Here, the stimulated glandular cells secrete chemicals that diffuse throughout the body. Cells have receptors for picking up chemical information. The information can pass persistendy. The passage of information is, of course, slower. It, however, influences all the cells of the target. A multiple effect can also be produced. Adrenaline reduces blood supply to skin and digestive system but increases the same to skeletal or voluntary muscles. There is increase in breathing rate and heart beat. The body becomes ready to deal with an emergency. Further, endocrine system controls and coordinates many processes of the body where nervous system has no role, e.g., cell permeability, cell division, cell growth, cell differentiation, development of sex organs, secondary sex characters and several other activities. Any discrepancy can lead to a disorder, e.g, dwarfism and gigantism, hypothyroidism (simple goitre, cretinism, myxedema), hyperthyroidism (exophthalmia).

Question 12.
What is the difference between the manner in which movement in Sensitive Plant and movement in our legs takes places ?
Answer:

NCERT Solutions for Class 10 Science Chapter 7 Control and Coordination image - 4

Selection Type Questions

Alternate Response Type Questions
(True/False, Right(√)/Wrong (x), Yes/No)

Question 1.
Fore brain is centre of intelligence, control of movements, hearing, smell and sight.
Question 2.
Chewing cud is a movement of growth.
Question 3.
Immediate response to stimulus is shown by Mimosa pudica.
Question 4.
Rise in sugar level in blood stops secretion of insulin by pancreas.
Question 5.
Control and coordination are functions of nervous and endocrine systems.
Question 6.
Stems are positively geotropic while roots are negatively geotropic.
Question 7.
Major part of taste is smell.
Question 8.
I withdrew my hand back from hot plate reflexly.

Matching Type Questions

Question 9.
Match the articles in columns I and II (single matching) :

Column I

Column II

(a)   Cell growth

(b)   Wilting

(c)   Emergency

(d)    Electrical impulses

(i) Abscisic acid

(ii) Nerve conduction

(iii) Adrenaline

(iv) Auxin.

Question 10.
Match the contents of columns I, II and III (double matching) :

Column I Column II Column III

(a)    Thyroid

(b)    Shoot tip

(c)     Receptor

(d)    Motor end plate

 (i)     Reflex arc

(ii)  Neuromuscular junction

(iii) Thyroxine

(iv) Auxin

p.   Iodine

q.  Acetylcholine

r.  Apical dominance

s.  Effector

Question 11.
Name the control — voluntary (V), involuntary (I) and endocrine (E) in the following (Key or Check list Items) :
Action                                         Control
(i) Peristalsis                             ………………..
(ii) Lifting of arms                     ……………….
(iii) Growth                               ……………….
(iv) BMR                                   ………………..

Question 12.
Match each stimulus with appropriate response :

Hormone Dwarfism
(A)		 Cretinism
(B)		 Pregnancy
(C)		 Calcium level (D) Dilute urine (E) Mammary glands
(F)

(i) Thyroxine

(ii) Growth hormone

(iii) Parathormone

(iv)  Prolactin

(v)    Vasopressin

(vi)  Progesterone

Fill In the Blanks

Question 13. A ………………… mechanism regulates the action of hormones.
Question 14. An axon terminal passes the electrical stimulus to a dendrite of next neuron through …………………. reaction.
Question 15. Reflex arc formed in spinal cord also sends information input to ………………… .
Question 16. ……………………….. coordinates the activity of picking up pencil for writing.
Question 17. Positive geotropism of root is due to greater growth on ……………………… side as compared to ………………… side.

Answers:
NCERT Solutions for Class 10 Science Chapter 7 Control and Coordination image - 5

NCERT Solutions for Class 10 Science Chapter 7 – Control and Coordination

Hope given NCERT Solutions for Class 10 Science Chapter 7 are helpful to complete your science homework.

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

CA Foundation Business Economics Study Material – Meaning and Types of Markets

CA Foundation Business Economics Study Material Chapter 4 Price Determination in Different Markets – Meaning and Types of Markets

MEANING OF MARKET

  • In ordinary language, a market refers to a place where the buyers and sellers of a commodity gather and strike bargains.
  • In economics, however, the term “Market” refers to a market for a commodity. E.g. Cloth market; furniture market; etc.
    According to Chapman, “the term market refers not necessarily to a place and always to a commodity and buyers and sellers who are in direct competition with one another”.
  • According to the French economist Cournot, “Market is not any particular place in which things are bought and sold, but the whole of any region in which buyers and sellers are in such free intercourse with each other that the prices of the same goods tend to equality easily and quickly”,

The above mentioned definitions reveals the following features of a market:

  1. A region. A market does not refer to a fixed place. It covers a region, which may be a town, state, country or even world.
  2. Existence of buyers and sellers. Market refers to the network of potential buyers and sellers who may be at different places.
  3. Existence of commodity or service. The exchange transactions between the buyers and sellers can take place only when there is a commodity or service to buy and sell.
  4. Bargaining for a price between potential buyers and sellers.
  5. Knowledge about market conditions. Buyers and sellers are aware of the prices offered or accepted by other buyers and sellers through any means of communication.
  6. One price for a commodity or service at a given time.

Classification of Market:

Markets may be classified on the basis of different criteria. In Economics, generally the classification is made as pointed out in the following chart—

CA Foundation Business Economics Study Material - Meaning and Types of Markets 1

TYPES OF MARKET STRUCTURES

Market can be classified on the basis of area, volume of business, time, status of sellers, regulation and control.
The main types of markets can be summed up as follows:

  1. Perfect Competition:
    • Perfect competition market is one where there are many sellers selling identical products to many buyers at a uniform.
  2. Monopoly:
    • Monopoly market structure is a market situation in which there is a single seller of a commodity selling to many buyers.
    • The commodity has no close substitutes available.
    • A monopolist therefore, has a considerable influence on the price and supply of his commodity.
  3. Monopolistic Competition:
    • Monopolistic competition is a market situation in which there are many sellers selling differentiated goods to many buyers.
  4. Oligopoly:
    Oligopoly is a market situation in which there are few sellers selling either homogeneous or differentiated goods.

Table: Features of major types of markets

Points Market Types
Perfect Competition Monopoly Monopolistic Competition Oligopoly
i. Number of sellers Many One Many Few
ii. Product Homogeneous Unique having no substitutes Differentiated Homogeneous or Differentiated
iii. Selling Cost No Negligible High High
iv. Degree of control over price No Control. Price taker. Full control. Price maker Limited due to product differentiation. Limited
v. Demand (or AR) Curve Horizontal straight line parallel to x-axis Downward sloping Downward sloping Indeterminate
vi. Price elasticity of demand Infinite P = MC Small P > MC Large P > MC Small

CONCEPTS OF TOTAL REVENUE, AVERAGE REVENUE AND MARGINAL REVENUE

Total Revenue: (TR)

  • Total revenue may be defined as the total amount of money received by the firm by selling a certain units of a commodity.
  • It is obtained by multiplying the price per unit of a commodity with the total number of units sold.
  • Total Revenue = Price per unit X Total No. of units sold
    TR = P X Q
  • E.g. A firm sells 100 units of a commodity @ ₹ 15 each, then its total revenue is ₹ 15 X 100 units = ₹ 1,500

Average Revenue: (AR)

  • Average revenue is the revenue per unit of the commodity sold.
  • It is simply the total revenue divided by the number of units of output sold.
    CA Foundation Business Economics Study Material - Meaning and Types of Markets 2
  • E.g. A firm earns total revenue of ₹ 2,000 by the sale of 100 units of a commodity, then its average revenue is ₹ 20 (₹ 2000 -MOO units)
  • By definition average revenue is the price per unit of output. To prove it
    CA Foundation Business Economics Study Material - Meaning and Types of Markets 3

Marginal Revenue (MR):

  • Marginal revenue refers to the addition to total revenue by selling one more unit of a commodity.
  • Marginal revenue may also be defined as the change in total revenue resulting from the sale of one more unit of a commodity
  • E.g. If a firm sells 100 units of a commodity @ ₹ 15 each, its TR is ₹ 1,500. Now, if it increases the sale by ten units i.e. it sells 110 units @ ₹ 14 each, its TR is ₹ 1,540. Thus,
    CA Foundation Business Economics Study Material - Meaning and Types of Markets 4
    Where
    ∆TR is the change in total revenue
    ∆Q is the change in the quantity sold
  • For one unit change – MRn = TRn – TRn-1
    Where
    MRn = Marginal Revenue from ‘n’ units
    TRn = Total Revenue of ‘n’ units
    TRn-1 = Total Revenue from ‘n-1’ units
    n = any give number

MARGINAL REVENUE, AVERAGE REVENUE, TOTAL REVENUE AND ELASTICITY OF DEMAND

The relationship between AR, MR and price elasticity of demand can be examined with the formula —
CA Foundation Business Economics Study Material - Meaning and Types of Markets 5
CA Foundation Business Economics Study Material - Meaning and Types of Markets 6
Figure: The relationship between AR, MR, TR & elasticity of demand.

The above figure reveals the following on a straight line demand curve (or AR curve):

  1. When e > 1, marginal revenue is positive and therefore total revenue is rising,
  2. When e = l, marginal revenue is zero and therefore total revenue is maximum, and
  3. When e < l, marginal revenue is negative and therefore total revenue is falling.

BEHAVIOURAL PRINCIPLES

Principle 1: A firm should not produce at all if its total revenue is either equal to or less than its total variable cost.
Principle 2: It will be profitable for the firm to expand output so long as marginal revenue is more than marginal cost till the point where marginal revenue equals marginal cost.
Also the marginal cost curve should cut its marginal revenue curve from below.

 

RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D

RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 5 Fractions Ex 5D.

Other Exercises

Question 1.
Solution:
(i) Like fraction : Fractions having the same denominators are called like fractions. For examples:
\(\frac { 2 }{ 11 } ,\frac { 3 }{ 11 } ,\frac { 4 }{ 11 } ,\frac { 5 }{ 11 } ,\frac { 8 }{ 11 } \)
(ii) Unlike fraction : Fraction having the different denominators, are called unlike fractions. For examples:
\(\frac { 1 }{ 3 } ,\frac { 4 }{ 7 } ,\frac { 5 }{ 9 } ,\frac { 3 }{ 8 } ,\frac { 7 }{ 11 } \)

Question 2.
Solution:
We know that like fractions have same denominator
Now \(\frac { 3 }{ 5 } ,\frac { 7 }{ 10 } ,\frac { 8 }{ 15 } ,\frac { 11 }{ 30 } \)
LCM of 5, 10, 15 and 30 = 30
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 2.1

Question 3.
Solution:
We know that like fraction have same denominators
\(\frac { 1 }{ 4 } ,\frac { 5 }{ 8 } ,\frac { 7 }{ 12 } ,\frac { 13 }{ 24 } \)
LCM of 4, 8, 12, 24 = 24
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 3.1

Question 4.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 4.1
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 4.2

Question 5.
Solution:
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 5.1
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 5.2

Compare the fractions given below :

Question 6.
Solution:
\(\frac { 4 }{ 5 } and\frac { 5 }{ 7 } \)
LCM of 5 and 7 = 35
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 6.1

Question 7.
Solution:
\(\frac { 3 }{ 8 } and\frac { 5 }{ 6 } \)
LCM of 8 and 6 = 24
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 7.1

Question 8.
Solution:
\(\frac { 7 }{ 11 } and\frac { 6 }{ 7 } \)
LCM of 11 and 7 = 77
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 8.1

Question 9.
Solution:
\(\frac { 5 }{ 6 } and\frac { 9 }{ 11 } \)
LCM of 6 and 11 = 66
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 9.1

Question 10.
Solution:
\(\frac { 2 }{ 3 } and\frac { 4 }{ 9 } \)
LCM of 3 and 9 = 9
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 10.1

Question 11.
Solution:
\(\frac { 6 }{ 13 } and\frac { 3 }{ 4 } \)
LCM of 13 and 4 = 52
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 11.1

Question 12.
Solution:
\(\frac { 3 }{ 4 } and\frac { 5 }{ 6 } \)
LCM of 4 and 6 = 12
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 12.1

Question 13.
Solution:
\(\frac { 5 }{ 8 } and\frac { 7 }{ 12 } \)
LCM of 8 and 12 = 24
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 13.1

Question 14.
Solution:
\(\frac { 4 }{ 9 } and\frac { 5 }{ 6 } \)
LCM of 9 and 6 = 18
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 14.1

Question 15.
Solution:
\(\frac { 4 }{ 5 } and\frac { 7 }{ 10 } \)
LCM of 5 and 10 = 10
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 15.1

Question 16.
Solution:
\(\frac { 7 }{ 8 } and\frac { 9 }{ 10 } \)
LCM of 8 and 10 = 40
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 16.1

Question 17.
Solution:
\(\frac { 11 }{ 12 } and\frac { 13 }{ 15 } \)
LCM of 12 and 15 = 60
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 17.1

Question 18.
Solution:
\(\frac { 1 }{ 2 } ,\frac { 3 }{ 4 } ,\frac { 5 }{ 6 } and\frac { 7 }{ 8 } \)
LCM of 2, 4, 6 and 8 = 24
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 18.1

Question 19.
Solution:
\(\frac { 2 }{ 3 } ,\frac { 5 }{ 6 } ,\frac { 7 }{ 9 } and\frac { 11 }{ 18 } \)
LCM of 3, 6, 9 and 18 = 18
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 19.1

Question 20.
Solution:
\(\frac { 2 }{ 5 } ,\frac { 7 }{ 10 } ,\frac { 11 }{ 15 } and\frac { 17 }{ 30 } \)
LCM of 5, 10, 15 and 30 = 30
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 20.1

Question 21.
Solution:
\(\frac { 3 }{ 4 } ,\frac { 7 }{ 8 } ,\frac { 11 }{ 16 } and\frac { 23 }{ 32 } \)
LCM of 4, 8, 16 and 32 = 32
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 21.1

Arrange the following fractions in the descending order :

Question 22.
Solution:
\(\frac { 3 }{ 4 } ,\frac { 5 }{ 8 } ,\frac { 11 }{ 12 } and\frac { 17 }{ 24 } \)
LCM of 4, 8, 12 and 24 = 24
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 22.1

Question 23.
Solution:
\(\frac { 7 }{ 9 } ,\frac { 5 }{ 12 } ,\frac { 11 }{ 18 } and\frac { 17 }{ 36 } \)
LCM of 9, 12, 18 and 36 = 36
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 23.1

Question 24.
Solution:
\(\frac { 2 }{ 3 } ,\frac { 3 }{ 5 } ,\frac { 7 }{ 10 } and\frac { 8 }{ 15 } \)
LCM of 3, 5, 10 and 15 = 30
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 24.1

Question 25.
Solution:
\(\frac { 5 }{ 7 } ,\frac { 9 }{ 14 } ,\frac { 17 }{ 21 } and\frac { 31 }{ 42 } \)
LCM of 7, 14, 21 and 42 = 42
RS Aggarwal Class 6 Solutions Chapter 5 Fractions Ex 5D 25.1

Question 26.
Solution:
∴ the numerators are equal
∴ The fraction having small denominator is greater than the fraction having large denominator
∴ In descending order, we can write
\(\frac { 1 }{ 12 } ,\frac { 1 }{ 23 } ,\frac { 1 }{ 7 } ,\frac { 1 }{ 9 } ,\frac { 1 }{ 17 } ,\frac { 1 }{ 50 } \)

Question 27.
Solution:
Here, the numerators of all fractions are equal
∴ The fraction having small denominator is greater than the fraction having large denominator
Now in descending order is
\(\frac { 3 }{ 4 } ,\frac { 3 }{ 5 } ,\frac { 3 }{ 7 } ,\frac { 3 }{ 11 } ,\frac { 3 }{ 13 } ,\frac { 3 }{ 17 } \)

Question 28.
Solution:
Lalita read 30 pages of a book containing 100 pages
She read \(\\ \frac { 30 }{ 100 } \) = \(\\ \frac { 3 }{ 10 } \) part of the book and Sarita read \(\\ \frac { 2 }{ 5 } \) of the book
Now in \(\\ \frac { 3 }{ 10 } \) and \(\\ \frac { 2 }{ 5 } \), LCM of 10, 5 = 10
\(\\ \frac { 3 }{ 10 } \) = \(\\ \frac { 3 }{ 10 } \)
\(\\ \frac { 2 }{ 5 } \) = \(\\ \frac { 2\times 2 }{ 5\times 2 } \) = \(\\ \frac { 4 }{ 10 } \)
From above, Sarita read more
as \(\\ \frac { 4 }{ 10 } \) or \(\frac { 2 }{ 5 } >\frac { 3 }{ 10 } \)

Question 29.
Solution:
Rafiq exercised for \(\\ \frac { 2 }{ 3 } \) hour and Rohit exercised for \(\\ \frac { 3 }{ 4 } \) hour
In \(\\ \frac { 2 }{ 3 } \) and \(\\ \frac { 3 }{ 4 } \), LCM of 3 and 4 = 12
\(\\ \frac { 2 }{ 3 } \) = \(\\ \frac { 2\times 4 }{ 3\times 4 } \) = \(\\ \frac { 8 }{ 12 } \)
\(\\ \frac { 3 }{ 4 } \) = \(\\ \frac { 3\times 3 }{ 4\times 3 } \) = \(\\ \frac { 9 }{ 12 } \)
\(\frac { 9 }{ 12 } >\frac { 8 }{ 12 } \)
=> \(\frac { 3 }{ 4 } >\frac { 2 }{ 3 } \)
∴ Rohit exercised more time

Question 30.
Solution:
In VI A, 20 student passed out of 25 or \(\\ \frac { 20 }{ 25 } \) or \(\\ \frac { 4 }{ 5 } \) students passed
But in VI B, 24 out of 30 passed 24 or \(\\ \frac { 24 }{ 30 } \) or \(\\ \frac { 4 }{ 5 } \) students passed
Now \(\\ \frac { 4 }{ 5 } \) = \(\\ \frac { 4 }{ 5 } \)
∴ Both sections gave same result

Hope given RS Aggarwal Solutions Class 6 Chapter 5 Fractions Ex 5D 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.

RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3C

RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3C

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3C.

Other Exercises

Question 1.
Solution:
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3C 1.1
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3C 1.2

Question 2.
Solution:
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3C 2.1

Question 3.
Solution:
(i) 463 – 9
= 464 – 1 – 9
= 464 – 10
= 454
(ii) 5632 – 99
= 5632 – 100 + 1
= 5633 – 100
= 5533
(iii) 8640 – 999
= 8640 – 1000 + 1
= 8641 – 1000
= 7641
(iv) 13006 – 9999
= 13006 – 10000 + 1
= 13007 – 10000
= 12007

Question 4.
Solution:
Smallest number of 7-digits = 1000000
Largest number of 4-digits = 9999
Required difference = (1000000 – 9999)
= 990001

Question 5.
Solution:
Deposit in the beginning = Rs. 136000
Next day withdrew = Rs. 73129
Amount left in the bank account = Rs. 136000 – 73129
= Rs. 62871

Question 6.
Solution:
Amount withdrawn from the bank = Rs. 1,00,000
Cost of TV set = Rs. 38750
Cost of refrigerator = Rs. 23890
Cost of jewellery = Rs. 35560
Total amount spent on her purchase = Rs. (38750 + 23890 + 35560)
= Rs. 98200
Amount left with her – Rs. 1,00,000 – Rs. 98200
= Rs. 1800

Question 7.
Solution:
Population of a town = 110500
Increase in 1 year = 3608
Persons left or died = 8973
The population at the end of year = 110500 + 3608 – 8973
= 114108 – 8973
= 105135

Question 8.
Solution:
(i) We have n + 4 = 9
n = 9 – 4 = 5
n = 5
(ii) n + 35 = 101
Subtracting 35 from both sides
n + 35 – 35 = 101 – 35
=> n = 66
n = 66
(iii) n – 18 = 39
Adding 18 to both sides
n – 18 + 18 = 39 + 18
=> n = 57
n = 57
(iv) n – 20568 = 21403 Adding 20568 to both sides
n – 20568 + 20568 = 21403 + 20568
n = 41971

Hope given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3C 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.

RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1

RD Sharma Class 8 Solutions Chapter 22 Mensuration III (Surface Area and Volume of a Right Circular Cylinder) Ex 22.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1

Other Exercises

Question 1.
Find the curved surface area and total surface area of a cylinder, the diameter of whose base is 7 cm and height is 60 cm.
Solution:
Diameter of the base of cylinder = 7 cm
∴ Radius (r) = \(\frac { 7 }{ 2 }\) cm
Height (h) = 60m
∴ carved surface area = 2πh
= 2 x \(\frac { 22 }{ 7 }\) x \(\frac { 7 }{ 2 }\) x 60cm2 = 1320cm2
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 1
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 2

Question 2.
The curved surface area of a cylindrical road is 132 cm2. Find its length if the radius is 0.35 cm.
Solution:
Curved surface area =132 cm2
Radius (r) = 0.35 cm
Let h be the length of the rod Then 2πrh = 132
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 3

Question 3.
The area of the base of a right circular cylinder is 616 cm2 and its height is 2.5 cm. Find the curved surface area of the cylinder.
Solution:
Let r be the radius of the base of the cylinder, then
Area of the base = πr2
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 4

Question 4.
The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find its curved surface area and total surface area.
Solution:
Height of the cylinder (h) = 15 cm
Circumference of the base = 88 cm
Let r be the radius of the base, the circumference = 2πr
∴ 2πr = 88 cm …(i)
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 5

Question 5.
A rectangular strip 25 cm x 7 cm is rotated about the longer side. Find the total surface area of the solid thus generated.
Solution:
Dimensions of rectangular strip = 25 cm x 7 cm
By rotating the strip along longer side, a solid is formed whose radius = 7 cm
and height = 25 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 6
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 7

Question 6.
A rectangular sheet of paper 44 cm x 20 cm, is rolled along its length to form a cylinder. Find the the total surface area of the cylinder thus generated.
Solution:
By rolling along length wire, we get a cylinder whose circumference of its base = 20 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 8

Question 7.
The radii of two cylinders are in the ratio of 2 : 3 and their heights are in the ratio 5 : 3. Calculate the ratio of their curved surface areas.
Solution:
Ratio in radii of two cylinders = 2:3
and ratio in their heights = 5:3
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 9

Question 8.
The ratio between the curved surface area and the total surface area of a right circular cylinder is 1: 2. Prove that its height and radius are equal.
Solution:
Let r be the radius and h be the height of a right circular cylinder, then Curved surface area = 2πrh
and total surface area = 2πrh x 2πr2 = 2πr (h + r)
But their ratio is 1 : 2
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 10
Hence their radius and height are equal.

Question 9.
The curved surface area of a cylinder is 1320 cm2 and its base has diameter 21 cm. Find the height of the cylinder.
Solution:
Curved surface area of a cylinder = 1320 cm2
Diameter of its base (d) = 21 cm 21
Radius (r) = \(\frac { 21 }{ 2 }\) cm
Let h be the height of the cylinder
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 11

Question 10.
The height of a right circular, cylinder is 10.5 m. If three times the sum of the areas of its two circular faces is twice the area of the curved surface area. Find the radius of its base.
Solution:
Height of cylinder = 10.5 m
Let r be the radius and h be the height of a right circular cylinder, then Area of its two circular faces = 2π2
and area of curved surface = 2πrh
Now, according to the condition:
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 12

Question 11.
Find the cost of plastering the inner surface of a well at Rs 9.50 per m2, if it is 21 m deep and diameter of its top is 6 m.
Solution:
Diameter of the top of a cylindrical well = 6m
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 13
∴ Radius (r) = \(\frac { 6 }{ 2 }\) = 3 m
and depth (h) = 21 m
∴ Curved surface area = 2πrh = 2 x \(\frac { 22 }{ 7 }\) x 3 x 21 m2 = 396 m2
Rate of plastering = Rs 9.50 per m2
∴ Total cost of plastering = Rs 9.50 x 396 = Rs 3,762

Question 12.
A cylindrical vessel open at the top has diameter 20 cm and height 14 cm. Find the cost of the tin-plating it on the inside at the rate of 50 paise per hundred square centimetre.
Solution:
Diameter of the vessel = 20 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 14
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 15

Question 13.
The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find the cost of plastering its inner curved surface at Rs 4 per square metre.
Solution:
Diameter of the well = 3.5 m
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 16

Question 14.
The diameter of a roller is 84 cm and its length is T20 cm. It takes 500 complete revolutions moving once over to level a playground. What is the area of the playground ?
Solution:
Diameter of the roller = 84 cm
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 17

Question 15.
Twenty cylindrical pillars of the Parliament House are to be cleaned. If the diameter of each pillar is 0.50 m and height is 4 m, what will be the cost of cleaning them at the rate of Rs 2.50 per square metre ?
Solution:
Diameter of each pillar = 0.50 m
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 18

Question 16.
The total surface area of a hollow cylinder which is open from both sides is 4620 sq. cm, area of the base ring is 115.5 sq. cm and height 7 cm. Find the thickness of the cylinder.
Solution:
Total surface area of a hollow cylinder opened from both sides = 4620 cm2
Area of base ring = 115.5 cm2
Height of cylinder (h) = 7 cm
Let R be the outer radius and r be the inner radius
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 19
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 20

Question 17.
The sum of the radius of the base and height of a solid cylinder is 37 m, if the total surface area of the solid cylinder is 1628 m2, find the circumference of its base.
Solution:
Let r be the radius and h be the height of the solid cylinder, then r + h = 37 m …(i)
Total surface area = 1628
⇒ 2πr (r + h) = 1628
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 21

Question 18.
Find the ratio between the total surface area of a cylinder to is curved surface area,given that its height and radius are 7.5 cm and 3.5 cm.
Solution:
Radius (r) of cylinder = 3.5 cm
and height (h) = 7.5 cm
∴ Curved surface area = 2πrh
and total surface area = 2πr (h + r)
∴ Ratio = 2πr (h + r)- 2πrh = (h + r): h = 7.5 + 3.5 : 7.5
⇒ 11 : 7.5
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 22

Question 19.
A cylindrical vessel, without lid, has to be tin-coated on its both sides. If the radius of the base is 70 cm and its height is 1.4 m, calculate the cost of tin-coating at the rate of Rs 3.50 per 1000 cm2.
Solution:
Radius of the vessel (r) = 70 cm
and height (h) = 1.4 m = 140 cm
∴ Area of inner and outer curved surfaces and bases = 2 x 2πrh + 2πr2
RD Sharma Class 8 Solutions Chapter 22 Mensuration III Ex 22.1 23

 

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RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3B

RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3B

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3B.

Other Exercises

Question 1.
Solution:
(i) 458 + 639 = 639 + 458 (Commulative law)
(ii) 864 + 2006 = 2006 + 864 (Commulative law)
(iii) 1946 + 984 = 984 + 1946 (Commulative law)
(iv) 8063 + 0 = 8063 (Additive property of zero)
(v) 53501 + (574 + 799) = 574 + (53501 + 799) (Associative law)

Question 2.
Solution:
(i) 16509 + 114 = 16623
Check : 16623 – 114 = 16509 which is given.
(ii) 2359 + 548 = 2907
Check: 2907 – 2359 = 548 which is given
(iii) 19753 + 2867 = 22620
Check : 22620 – 19753 = 2867 which is given

Question 3.
Solution:
(1546 + 498) + 3589
= 2044 + 3589
= 5633
and 1546 + (498 + 3589)
= 1546 + 4087
= 5633
Yes, the above two sum are equal.
The property used is associative law of addition.

Question 4.
Solution:
(i) 953 + 707 + 647
= (953 + 647) + 707
(by associative law)
= 1600 + 707
= 2307
(ii) 1983 + 647 + 217 + 353
= (1983 + 217) + (647 + 353)
= 2200 + 1000 = 3200
(iii) 15409 + 278 + 691 + 422
= (15409 + 691) + (278 + 422)
(by associative law)
= 16100 + 700
= 16800
(iv) 3259 + 10001 + 2641 + 9999
= (3259 + 2641) + (10001 + 9999)
(by associative law)
= 5900 + 20000
= 25900
(v) 1 +2 + 3 +4 + 96 + 97 + 98 + 99
= (1 +99) + (2 + 98) + (3 + 97) + (4 + 96)
= (100+ 100) + (100 +100)
= 200 + 200
= 400
(vi) 2 + 3 + 4 + 5 + 45 + 46 + 47 + 48
= (2 + 48) + (3 + 47) + (4 + 46) + (5 + 45)
= (50 + 50) + (50 + 50)
= 100 + 100 = 209

Question 5.
Solution:
(i) 6784 + 9999 = (6784 – 1) + (9999 + 1)
(Adding and subtracting 1)
= 6783 + 10000
= 16783
(ii) 10578 + 99999
(Adding and subtracting 1)
= (10578 – 1) + (99999 + 1)
= 10577 + 100000
= 110577

Question 6.
Solution:
Yes it is true, by the property of associative law of addition.

Question 7.
Solution:
The magic squares given are completed as under:
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3B 7.1
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3B 7.2
In each row/column, the sum = 46

Question 8.
Solution:
(i) The sum of two odd numbers is an odd number (F)
As sum of two odd numbers is alway an even number
(ii) The sum of two even number is an even number (T)
(iii) The sum of an even number and an odd number is an odd number (T)

Hope given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3B 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.