RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.1

RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 6 Algebraic Expressions and Identities Ex 6.1

Other Exercises

Question 1.
Identify the terms, their co-efficients for each of the following expressions.
(i) 7x2yz – 5xy
(ii) x2 + x + 1
(iii)3x2y2 – 5x2y2 + z2 + z2
(iv) 9 – ab + be-ca
(v) \(\frac { a }{ 2 } +\frac { b }{ 2 }\)-ab
(vi)2x – 0.3xy + 0.5y
Solution:
(i) Co-efficient of 7x2yz = 7
co-efficient of -5xy = -5
(ii) Co-efficient of x1 = 1
co-efficient of x = 1
co-efficient of 1 = 1
(iii) Co-efficient of 3x2_y2 = 3
co-efficient of -5x2y2z2 = -5
co-efficient of z2 – 1
(iv) Co-efficient of 9 = 9
co-efficient of -ab = -1
co-efficient of be = 1
co-efficient of -ca = -1
(v) Co-efficient of \(\frac { a }{ 2 } =\frac { 1 }{ 2 }\)
Co-efficient of \(\frac { b }{ 2 } =\frac { 1 }{ 2 }\)
co-efficient of -ab = -1
(vi) co-efficient of 0.2x = 0.2
co-efficient of-0.3xy = -0.3
co-efficient of 0.5y = 0.5

Question 2.
Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any category ?
(i) x+y
(ii) 1000
(iii) x + x2 + x3 + x4
(iv) 7 + a + 5b
(v) 2b – 3 b2    
(vi) 2y – 3y2 +4y3
(vii) 5x – 4y + 3x
(viii) 4a – 15a2
(ix) xy+yz + zt + tx
(x)   pqr
(xi) p2q + pq2       
(xii)  2p + 2 q
Solution:
Monomials are (ii), (x)
Binomials are (i), (v), (viii), (xi), (xii)
Trinomials are (iv), (vi) and (vii)
None of these are (iii) and (ix)

 

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CA Foundation Business Economics Study Material – Business Cycle

CA Foundation Business Economics Study Material Chapter 5 Business Cycle

All countries have gone through fluctuations in economic activities i.e. ups and downs in its economic activities. In other words, every country passes through a pattern where there are period of economic growth, followed by periods of slowing growth and even failing growth. There are periods of prosperity followed by downturns. Thus,

“The Business Cycle is the periodic fluctuations in economic activity measured by change in Real GDP.”

Although these economic fluctuations are recurrent and occur periodically, they are not at regular interval and are not of same length.

Phases of Business Cycle

A business cycle passes through the following four distinct phases:

  1. Expansion/Boom/ Recovery/ Upswing
  2. Peak/Boom/Prosperity
  3. Contraction/Recession/ Downswing
  4. Trough/Depression

The following figure shows the four stages of the business cycle.
CA Foundation Business Economics Study Material Business Cycle

In the figure above the four phases business cycles are shown. The broken line represents long time growth trend or potential GDP. It shows rising trend of growth over a period of time. The figure starts from Trough when the overall economic activities ie. level of production and employment are at the lowest level. With increase in the economic activities the economy moves into Expansion Phase. But expansion phase cannot continue indefinitely, and after reaching Peak, economy starts contracting i.e. Contraction Phase sets in and continue till it reaches the lowest turning point called Trough. Here cycle completes and new cycle starts.

Expansion

In the expansion phase, there is increase in OUTPUT and EMPLOYMENT. Expansion phase is characterized by-

  1. increase in national output,
  2. increase in employment,
  3. increase in aggregate demand,
  4. increase in capital i.e. investments,
  5. increase in consumer spending,
  6. increase in sales, profits, stock prices, & expansion of bank credit.
  7. increase in standard of living.

There is no INVOLUNTARY UNEMPLOYMENT and whatever unemployment exist is only of FRICTIONAL or STRUCTURAL in nature.
The growth ultimately slows down and reaches its peak.

Peak

Peak phase of the cycle is the highest point. The economy is producing at its maximum level. The economy becomes overheated i.e. unsustainable. The expansion phase ends here. The prices of inputs increase, resulting higher cost of production, leading to higher output prices. Higher output price leads to increased cost of living. Fixed income earners and consumers suffer. Economic growth stabilizes at peak an then starts the downswing.

Contraction

In contraction phase. There is fall in OUTPUT and EMPLOYMENT levels. Contraction phase is characterized by—

  1. fall in the level of investments,
  2. fall in the level of production and employment,
  3. fall in the incomes of people,
  4. demand and consumption of both capital goods and consumer goods fall,
  5. bank credit, shrinks as investments fall,
  6. stock prices fall,
  7. firms become pessimistic about future,
  8. there is lot of excess production capacity in industries.

There is large scale involuntary unemployment.

A severe contraction or recession of economic activities pushes the economy into Depression.

Trough and Depression

The lowest level of economic activity is called trough or depression. All economic activity touch the bottom and the phase of trough is reached. Trough is the turning point into expansion. Increased investments lead to increase in consumption. Therefore, industries expand production and start using their idle production capacity and rate of unemployment falls. With this the cycle is complete.

It is very difficult to predict turning points of business cycles. Changes in different economic activities is used to measure the business cycle and to predict in which direction the economy is headed. There are three types of economic indicators, depending on their timing namely—

  1. Leading Indicators,
  2. Lagging Indicators, and
  3. Coincident Indicators

Leading Indicators signal future changes

  • Leading Indicators change before the economy itself changes Le. change prior to large economic adjustments.
    E.g.– changes in stock prices, profit margins and profits, the house market, manufacturing activity, etc. Leading Indicators should be used with caution as they may not be always accurate.
  • Lagging Indicators usually change after the economy as a whole changes i.e. after the real output changes. Lagging Indicators are useful to confirm the business cycle.
    E.g.– unemployment, the consumer price index, interest rates, lending by banks, etc.
  • Coincident Indicators also called Concurrent Indicators occur at about the same time with business cycle movements. They give us idea about current state of economy.
    E.g.    – GDP, inflation, industrial output, personal income, etc.

Features of Business Cycles

  • Business cycles occur periodically. They do not show same regularity, duration and intensity.
  • The length of different phases of business cycles is not definite and hence do not show smoothness and regularity.
  • Business cycles do not bring about changes in one industry or sector but occur simultaneously in all industries and sectors. Further, it passes from one industry to another.
  • Fluctuations take place not only in the level of output but also in other related variables like consumption, employment, investment, interest rates and price level.
  • Cyclical fluctuations affect adversely the consumption of durable goods like capital goods, scooters, cars, houses, refrigerators, etc. Their demand falls. As a result investments become unstable.
    However, consumption of non-durable goods and services does not vary much during different phases of business cycle. .
  • Business cycles causes lot of uncertainty for businessmen and forecasting becomes difficult. Profits fluctuate.
  • Business cycles affect the inventories of goods. During depression inventories start accumulation more than the desired level. This results reduction in the production. When recovery starts, inventories are below the required level.
  • Business cycles are international in character.

Causes of Business Cycles

Business Cycles may occur due to internal and external causes or a combination of both.

Internal Causes (Endogenous Factors): Internal causes of business cycle are those, which are built within the economic system. They are—

1. Fluctuations in Effective Demand:
Fluctuations in economic activities is due to fluctuations in aggregate effective demand. When aggregate demand falls, it results in lower output, income and employment. This causes a downward spiral. Increase in aggregate demand causes conditions of expansion and boom.

2. Fluctuations in Investments:
Investments fluctuate because of changes in profit expectations of entrepreneurs. High investments brings increase in aggregate demand and thus result in upswing and vice versa.

3. Variations in government spending:
Fluctuations in government spending affects the economic activities and results in business fluctuations.

4. Money Supply:
According to Hawtrey and Friendman, business cycles relate to fluctuations in money and credit supply. Cheap money policy leads to expansion of money and credit supply resulting in increased economic activities and vice versa. .

5. Monetary and Fiscal Policies:
Monetary and Fiscal Policies also cause business cycles. Expansionary policies, like low interest rates, rates increased government spending and tax cuts boost economic activities. It there is inflation opposite will be done resulting in showing down of economy.

6. Psychological Factors:
According to Pigou, business cycles appear because of the optimistic and pessimistic mood of the business community. It business community is optimistic about future market conditions, they make investments. Here, the expansion phase starts ultimately leading to boom and vice versa.

7. Other Factors:
According to Schumpeter, business cycles occur due to innovations that take place from time to time in economic system. (Innovation Theory)
According to Nicholas Kaldor, the present fluctuations in prices are responsible for fluctuations in output and employment in future (Cobweb Theory)

External Causes (Exogenous Factors):

1. Wars:
During war time, all the available resources are used up for the production of arms and ammunitions. This results in the fall of production of capital and consumer goods. This in turn causes fall in income, profits and employment and contraction in economic activities take place which may lead to depression.

2. Post War Reconstruction:
After war, the level of consumption and investment goes upward. Both the government and individuals are involved in construction. E.g.- houses, roads, bridges, communication, etc. The economy picks up resulting in higher output, employment and income.

3. Technology:
Another cause of business is scientific development leading to improved technology. Adoption of new technology for production of new and better goods and services require huge investments. Increased investments increases employment income and profits this gives boost to the economy.

4. Natural Factors:
Weather cycles causes fluctuations in agricultural output. If in any year, weather is good the output of agriculture sector will increase. This will also increase the demand for industrial goods and vice versa.

5. Population Growth:
If the population growth rate is higher than the economic growth rate, income level will be low. This will result is lower savings and investments and therefore, lower income and employment.

Relevance of Business Cycles in Business Decision Making

  • Understanding the business cycle is important for all types of business enterprises because it affects the demand for their product and in turn their profits.
  • Knowledge of business cycles, its phases and characteristics help the business enterprises to frame appropriate policies. E.g.-New opportunities for investment, employment and production opens up at the time of prosperity. So understanding the economic environment is important white making business decisions.
  • Business managers have to advantageously respond in complex time during the whole business cycle through boom, downswing, recession and recovery to arrive at sound strategic environment.
  • We have seen that business cycles do not affect all the sector uniformly. Some business are more vulnerable white others are not or less vulnerable to changes in business cycle. Businesses like fashion retailers, electrical goods, restaurants, constructors, advertising, foreign tour operators, etc. are directly linked with economic growth. Such business are called cyclical businesses. So during recession such businesses slump and vice versa.
  • The phase of the business cycle is important to decide on entry into the market by a new firm or to decide about launch of a new product.

RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C

RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 7 Decimals Ex 7C.

Other Exercises

Add the following decimals :

Question 1.
Solution:
9.6, 14.8, 37 and 5.9
Converting these decimals into like decimals and then adding 9.6 + 14.8 + 37.0 + 5.9
= 67.3 Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q1.1

Question 2.
Solution:
23.7, 106.94, 68.9 and 29.5
Converting them into like decimals and then adding
23.70 + 106.94 + 68.90 + 29.50
= 229.04 Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q2.1

Question 3.
Solution:
72.8, 7.68, 16.23 and 0.7
Converting them into like decimals and then adding
72.80 + 7.68 + 16.23 + 0.70
= 97.41 Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q3.1

Question 4.
Solution:
18.6, 84.75, 8.345 and 9.7
Converting them into like decimals and then adding
18.600 + 84.750 + 8.345 + 9.700
= 121.395 Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q4.1

Question 5.
Solution:
8.236, 16.064, 63.8 and 27.53
Converting them into like decimals and then adding
8.236 + 16.064 + 63.800 + 27.530
= 115.630 Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q5.1

Question 6.
Solution:
28.9, 19.64, 123.697 and 0.354
Converting them into like decimals and then adding
28.900 + 19.640 + 123.697 + 0.354
= 172.591 Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q6.1

Question 7.
Solution:
4.37, 9.638, 17.007 and 6.8
Converting them into like decimals and then adding
4. 370 + 9.638 + 17.007 + 6.800
= 37.815 Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q7.1

Question 8.
Solution:
14.5, 0.038, 118.573 and 6.84
Converting them into like decimals and then adding
14.500 + 0.038 + 118.573 + 6.840
= 139.951 Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q8.1

Question 9.
Solution:
Earning for the first day = 32.60 rupees
Earning for the second day = 56.80 rupees
Earning for the third day = 72 rupees
Total earning = Rs. 32.60 + Rs. 56.80 + Rs. 72
= Rs. 161.40 Ans.
Working
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q9.1

Question 10.
Solution:
Cost of almirah = Rs. 11025
Cartage = Rs. 172.50
Cost on repair = Rs. 64.80
Total cost = Rs. 11025 + Rs. 172.50 + Rs. 64.80
= Rs. 11262.30 Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q10.1

Question 11.
Solution:
Distance covered by taxi = 36 km 235 m
= 36.235 km
Distance covered by Rickshaw = 4 km 85 m
= 4.085 km
and distance covered on foot
= 1 km 80 m
= 1.080 m
Total distance covered = 36.235 km + 4.085 km + 1.080 km
= 41.400 km
= 41 km 400 m Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q11.1

Question 12.
Solution:
Weight of sugar in a bag = 45 kg 80 g
= 45.080 kg
Mass (weight) of empty bag = 950 g
= 0.950 kg
Total weight of the bag with sugar = 45 kg 80 g + 950 g
= 45.080 kg + 0.950 kg
= 46.030 kg
= 46 kg 30 g Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q12.1

Question 13.
Solution:
Length of cloth for shirt = 2 m 70 cm
= 2.70 m
Length of cloth for pyjamas = 2 m 60 cm
= 2.60 m
Total length of cloth = 2.70 m + 2.60 m
= 5.30 m
= 5 m 30 cm Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q13.1

Question 14.
Solution:
Cloth of salwar = 2 m 5 cm = 2.05 m
Cloth for shirt = 3 m 35 cm = 3.35 m
Total length of cloth = 2.05 m + 3.35 m
= 5.4.0 m
= 5 m 40 cm Ans.
Working :
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7C Q14.1

Hope given RS Aggarwal Solutions Class 6 Chapter 7 Decimals Ex 7C 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 5 Playing With Numbers Ex 5.3

RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3

Other Exercises

Solve each of the following cryptarithms.
Question 1.
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 1
Solution:
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 2
Values of A and B be from 0 to 9 In ten’s digit 3 + A = 9
∴ A = 6 or less.
∴ 7 + B = A = 6 or less
∴ 7 + 9 or 8 = 16 or 15
∴ But it is two digit number
B = 8
Then A = 5
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 3

Question 2.
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 4
Solution:
Values of A and B can be between 0 and 9
In tens digit, A + 3 = 9
∴ A = 9 – 3 = 6 or less than 6
In ones unit B + 7 = A = 6or less
∴ 7 + 9 or 8 = 16 or 15
But it is two digit number
∴ B = 8 and
∴ A = 5
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 5

Question 3.
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 6
Solution:
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 7
Value of A and B can be between 0 and 9 In units place.
1+B = 0 ⇒1+B = 10
∴ B = 10 – 1 = 9
and in tens place
1 + A + 1 = B ⇒ A + 2 = 9
⇒ A = 9 – 2 = 7
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 8

Question 4.
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 9
Solution:
Values of A and.B can be between 0 and 9
In units place, B+1 = 8 ⇒ B = 8-1=7
In tens place A + B= 1 or A + B = 11
⇒ A + 7 = 11 ⇒ A =11-7 = 4

Question 5.
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 10
Solution:
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 11
Values of A and B can be between 0 and 9
In tens place, 2 + A = 0 or 2 + A=10
A = 10-2 = 8
In units place, A + B = 9
⇒ 8 + B = 9 ⇒ B = 9- 8 = 1
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 12

Question 6.
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 13
Solution:
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 14
Values of A and B can be between 0 and 9
In hundreds place,
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 15

Question 7.
Show that cryptarithm 4 x \(\overline { AB } =\overline { CAB }\) does not have any solution.
Solution:
RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 16
It means that 4 x B is a numebr whose units digit is B
Clearly, there is no such digit
Hence the given cryptarithm has no solution.

Hope given RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.3 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.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS

More Exercises

Question 1.
In the given figure, O is the centre of the circle. If ∠ABC = 20°, then ∠AOC is equal to
(a) 20°
(b) 40°
(c) 60°
(d) 10°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q1.1
Solution:
In the given figure,
Arc AC subtends ∠AOC at the centre
and ∠ABC at the remaining part of the circle
∠AOC = 2∠ABC = 2 × 20° = 40° (b)

Question 2.
In the given figure, AB is a diameter of the circle. If AC = BC, then ∠CAB is. equal to
(a) 30°
(b) 60°
(c) 90°
(d) 45°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q2.1
Solution:
In the given figure,
AB is the diameter of the circle and AC = BC
∠ACB = 90° (angle in a semi-circle)
AC = BC
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q2.2

Question 3.
In the given figure, if ∠DAB = 60° and ∠ABD = 50° then ∠ACB is equal to
(a) 60°
(b) 50°
(c) 70°
(d) 80°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q3.1
Solution:
In the given figure,
∠DAB = 60°, ∠ABD = 50°
In ∆ADB, ∆ADB = 180° – (60° + 50°)
= 180° – 110° = 70°
∠ACB = ∠ADB
(angles in the same segment) = 70° (c)

Question 4.
In the given figure, O is the centre of the circle. If ∠OAB = 40°, then ∠ACB is equal to
(a) 50°
(b) 40°
(c) 60°
(d) 70°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q4.1
Solution:
In the given figure, O is the centre of the circle.
In ∆OAB,
∠OAB = 40°
But ∠OBA = ∠OAB = 40°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q4.2

Question 5.
ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140°, then ∠BAC is equal to
(a) 80°
(b) 50°
(c) 40°
(d) 30°
Solution:
ABCD is a cyclic quadrilateral,
AB is the diameter of the circle circumscribing it
∠ADC = 140°, ∠BAC = Join AC
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q5.1

Question 6.
In the given figure, O is the centre of the circle. If ∠BAO = 60°, then ∠ADC is equal to
(a) 30°
(b) 45°
(c) 60°
(d) 120°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q6.1
Solution:
In the given figure, O is the centre of the circle ∠BAO = 60°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q6.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q6.3

Question 7.
In the given figure, O is the centre of the circle. If ∠AOB = 90° and ∠ABC = 30°, then ∠CAO is equal to
(a) 30°
(b) 45°
(c) 90°
(d) 60°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q7.1
Solution:
In the given figure, O is the centre of the circle
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q7.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q7.3
∠CAO = 105° – 45° = 60° (d)

Question 8.
In the given figure, O is the centre of a circle. If the length of chord PQ is equal to the radius of the circle, then ∠PRQ is
(a) 60°
(b) 45°
(c) 30°
(d) 15°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q8.1
Solution:
In the given figure, O is the centre of the circle
Chord PQ = radius of the circle
∆OPQ is an equilateral triangle
∴∠POQ = 60°
Arc PQ subtends ∠POQ at the centre and
∴∠PRQ at the remaining part of the circle
∴∠PRQ = \(\\ \frac { 1 }{ 2 } \) ∠POQ = \(\\ \frac { 1 }{ 2 } \) x 60° = 30° (c)

Question 9.
In the given figure, if O is the centre of the circle then the value of x is
(a) 18°
(b) 20°
(c) 24°
(d) 36°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q9.1
Solution:
In the given figure, O is the centre of the circle.
Join OA.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q9.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q9.3

Question 10.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(a) 7 cm
(b) 12 cm
(c) 15 cm
(d) 24.5 cm
Solution:
From Q, length of tangent PQ to the circle = 24 cm
and QO = 25 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q10.1

Question 11.
From a point which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is
(a) 60 cm²
(b) 65 cm²
(c) 30 cm²
(d) 32.5 cm²
Solution:
Let point P is 13 cm from O, the centre of the circle
Radius of the circle (OQ) = 5 cm
PQ and PR are tangents from P to the circle
Join OQ and OR
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q11.1
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q11.2

Question 12.
If angle between two radii of a circle is 130°, the angle between the tangents at the ends of the radii is
(a) 90°
(b) 50°
(c) 70°
(d) 40°
Solution:
Angles between two radii OA and OB = 130°
From A and B, tangents are drawn which meet at P
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q12.1

Question 13.
In the given figure, PQ and PR are tangents from P to a circle with centre O. If ∠POR = 55°, then ∠QPR is
(a) 35°
(b) 55°
(c) 70°
(d) 80°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q13.1
Solution:
In the given figure,
PQ and PR are the tangents to the circle from a point P outside it
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q13.2

Question 14.
If tangents PA and PB from an exterior point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to
(a) 50°
(b) 60°
(c) 70°
(d) 100°
Solution:
Length of tangents PA and PB to the circle from a point P
outside the circle with centre O, and inclined an angle of 80°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q14.1

Question 15.
In the given figure, PA and PB are tangents from point P to a circle with centre O. If the radius of the circle is 5 cm and PA ⊥ PB, then the length OP is equal to
(a) 5 cm
(b) 10 cm
(c) 7.5 cm
(d) 5√2 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q15.1
Solution:
In the given figure,
PA and PB are tangents to the circle with centre O.
Radius of the circle is 5 cm, PA ⊥ PB.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q15.2

Question 16.
At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is
(a) 4 cm
(b) 5 cm
(c) 6 cm
(d) 8 cm
Solution:
AB is the diameter of a circle with radius 5 cm
At A, XAY is a tangent to the circle
CD || XAY at a distance of 8 cm from A
Join OC
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q16.1

Question 17.
If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is tangent to the other is
(a) 3 cm
(b) 6 cm
(c) 9 cm
(d) 1 cm
Solution:
Radii of two concentric circles are 4 cm and 5 cm
AB is a chord of the bigger circle
which is tangent to the smaller circle at C.
Join OA, OC
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q17.1

Question 18.
In the given figure, AB is a chord of the circle such that ∠ACB = 50°. If AT is tangent to the circle at the point A, then ∠BAT is equal to
(a) 65°
(b) 60°
(c) 50°
(d) 40°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q18.1
Solution:
In the given figure, AB is a chord of the circle
such that ∠ACB = 50°
AT is tangent to the circle at A
AT is tangent and AB is a chord
∠ACB = ∠BAT = 50°
(Angles in the alternate segments) (c)

Question 19.
In the given figure, O is the centre of a circle and PQ is a chord. If the tangent PR at P makes an angle of 50° with PQ, then ∠POQ is
(a) 100°
(b) 80°
(c) 90°
(d) 75°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q19.1
Solution:
In the given figure, O is the centre of the circle.
PR is tangent and PQ is chord ∠RPQ = 50°
OP is radius and PR is tangent to the circle
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q19.2

Question 20.
In the given figure, PA and PB are tangents to a circle with centre O. If ∠APB = 50°, then ∠OAB is equal to
(a) 25°
(b) 30°
(c) 40°
(d) 50°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q20.1
Solution:
In the given figure,
PA and PB are tangents to the circle with centre O.
∠APB = 50°
But ∠AOB + ∠APB = 180°
∠AOB + 50° = 180°
⇒ ∠AOB = 180° – 50° = 130°
In ∆OAB,
OA = OB (radii of the same circle)
∠OAB = ∠OBA
But ∠OAB + ∠OBA = 180° – ∠AOB
= 180° – 130° = 50°
∠OAB = \(\frac { { 50 }^{ 0 } }{ 2 } \) = 25° (a)

Question 21.
In the given figure, sides BC, CA and AB of ∆ABC touch a circle at point D, E and F respectively. If BD = 4 cm, DC = 3 cm and CA = 8 cm, then the length of side AB is
(a) 12 cm
(b) 11 cm
(c) 10 cm
(d) 9 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q21.1
Solution:
In the given figure,
sides BC, CA and AB of ∆ABC touch a circle at D, E and F respectively.
BD = 4 cm, DC = 3 cm and CA = 8 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q21.2

Question 22.
In the given figure, sides BC, CA and AB of ∆ABC touch a circle at the points P, Q and R respectively. If PC = 5 cm, AR = 4 cm and RB = 6 cm, then the perimeter of ∆ABC is
(a) 60 cm
(b) 45 cm
(c) 30 cm
(d) 15 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q22.1
Solution:
In the given figure, sides BC, CA and AB of ∆ABC
touch a circle at P, Q and R respectively
PC = 5 cm, AR = 4 cm, RB = 6 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q22.2

Question 23.
PQ is a tangent to a circle at point P. Centre of circle is O. If ∆OPQ is an isosceles triangle, then ∠QOP is equal to
(a) 30°
(b) 60°
(c) 45°
(d) 90°
Solution:
PQ is tangent to the circle at point P centre of the circle is O.
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q23.1

Question 24.
In the given figure, PT is a tangent at T to the circle with centre O. If ∠TPO = 25°, then the value of x is
(a) 25°
(b) 65°
(c) 115°
(d) 90°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q24.1
Solution:
In the given figure, PT is the tangent at T to the circle with centre O.
∠TPO = 25°
OT is the radius and TP is the tangent
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q24.2

Question 25.
In the given figure, PA and PB are tangents at ponits A and B respectively to a circle with centre O. If C is a point on the circle and ∠APB = 40°, then ∠ACB is equal to
(a) 80°
(b) 70°
(c) 90°
(d) 140°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q25.1
Solution:
In the given figure,
PA and PB are tangents to the circle at A and B respectively
C is a point on the circle and ∠APB = 40°
But ∠APB + ∠AOB = 180°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q25.2

Question 26.
In the given figure, two circles touch each other at A. BC and AP are common tangents to these circles. If BP = 3.8 cm, then the length of BC is equal to
(a) 7.6 cm
(b) 1.9 cm
(c) 11.4 cm
(d) 5.7 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q26.1
Solution:
In the given figure, two circles touch each other at A.
BC and AP are common tangents to these circles
BP = 3.8 cm
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q26.2

Question 27.
In the given figure, if sides PQ, QR, RS and SP of a quadrilateral PQRS touch a circle at points A, B, C and D respectively, then PD + BQ is equal to
(a) PQ
(b) QR
(c) PS
(d) SR
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q27.1
Solution:
In the given figure,
sides PQ, QR, RS and SP of a quadrilateral PQRS
touch a circle at the points A, B, C and D respectively
PD and PA are the tangents to the circle
∴ PA = PD …(i)
Similarly, QA and QB are the tangents
∴ QA = QB …(ii)
Now PD + BQ = PA + QA = PQ (a)
[From (i) and (ii)]

Question 28.
In the given figure, PQR is a tangent at Q to a circle. If AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to
(a) 20°
(b) 40°
(b) 35°
(d) 45°
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q28.1
Solution:
In the given figure, PQR is a tangent at Q to a circle.
Chord AB || PR and ∠BQR = 70°
BQ is chord and PQR is a tangent
∠BQR = ∠A
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q28.2

Question 29.
Two chords AB and CD of a circle intersect externally at a point P. If PC = 15 cm, CD = 7 cm and AP = 12 cm, then AB is
(a) 2 cm
(b) 4 cm
(c) 6 cm
(d) none of these
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q29.1
Solution:
In the given figure,
two chords AB and CD of a circle intersect externally at P.
PC = 15 cm, CD = 7 cm, AP = 12 cm
Join AC and BD
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q29.2
ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS Q29.3

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS 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.

CA Foundation Business Economics Study Material – Oligopoly

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

OLIGOPOLY

Introduction:

  • ‘Oligo’ means few and ‘Poly’ means seller. Thus, oligopoly refers to the market structure where there are few sellers or firms.
  • They produce and sell such goods which are either differentiated or homogeneous products.
  • Oligopoly is an important form of imperfect/competition.
  • E.g.- Cold drinks industry; automobile industry; Idea; Airtel. Hutch, BSNL mobile services in Nagpur; tea industry; etc.

Types of Oligopoly:

  • Pure or perfect oligopoly occurs when the product is homogeneous in nature, e.g. Aluminum industry.
  • Differentiated or imperfect oligopoly where products are differentiated. E.g. toilet products.
  • Open oligopoly where new firms can enter the market and compete with already existing firm.
  • Closed oligopoly where entry of new firm is restricted.
  • Collusive oligopoly when some firms come together with some common understanding and act in collusion with each other in fixing price and output.
  • Competitive oligopoly where there is no understanding or collusion among the firms.
  • Partial oligopoly where the industry is dominated by one large firm which is looked upon by other firms as the leader of the group. The dominating firm will be the price leader.
  • Full oligopoly where there is absence of price leadership.
  • Syndicated oligopoly where the firms sell their products through a centralized syndicate.
  • Organized oligopoly where the firms organize themselves into a central association for fixing prices, output, quotas, etc.

Characteristics of Oligopoly Market:

Following are the special features of oligopoly market:

1. Interdependence

  • In an oligopoly market, there is interdependence among firms.
  • A firm cannot take independent price and output decisions.
  • This is because each firm treats other firms as rivals.
  • Therefore, it has to consider the possible reaction to its rivals price-output decisions.

2. Importance of advertising and selling costs

  • Due to interdependence, the various firms have to use aggressive and defensive marketing tools to achieve larger market share.
  • For this the firms spend heavily on advertisement, publicity, sales promotion, etc. to attract large number of customers.
  • Firms avoid price-wars but are engaged in non-price competition. E.g.- free set of tea mugs with a packet of Duncan’s Double Diamond Tea.

3. Indeterminate Demand Curve

  • The nature and position of the demand curve of the oligopoly firm cannot be determined.
  • This is because it cannot predict its sales correctly due to indeterminate reaction patterns of rival firms.
  • Demand curve goes on shifting as rivals too change their prices in reaction to price changes by the firm.

4. Group behaviour

  • The theory of oligopoly is a theory of group behaviour.
  • The members of the group may agree to pull together to promote their mutual interest or fight for individual interests or to follow the group leader or not.
  • Thus the behaviour of the members is very uncertain.

Price and output decisions in an Oligopolistic Market:

As seen earlier, an oligopolistic firm does not know how rival firms react to each other decisions. Therefore, it has to be very careful when it makes decision about its price. Rival firms retaliate to price change by an oligopolistic firm. Hence, its demand curve indeterminate. Price and output cannot be fixed. Some of the important oligopoly models are:

  1. Some economists assume that oligopolistic firms make their decisions independently. Therefore, the demand curve becomes definite and hence equilibrium level of output can be determined.
  2. Some believe that oligopolistic can predict the reaction of rivals on the basis of which he makes decisions about price and quantity.
  3. Cornet considers OUTPUT is the firm’s controlled variable and not price.
  4. In a model given by Stackelberg, the leader firm commits to an output before all other firms. The rest of firms follow it and choose their own level of output.
  5. Bertrand model states PRICE is the control variable for firms and therefore each firm sets the price independently.
  6. In order to pursue common interests, oligopolistic enter into enter into agreement and jointly act as monopoly to fix quantity and price.

Price Leadership:

A large or dominant firm may be surrounded by many small firms. The dominant firm takes the lead to set the price taking into account of the small firms. Dominant firm may adopt any one of the following strategies—

  1. ‘Live and let live’ strategy where dominant firm accepts the presence of small firms and set the price. This is called price-leadership,
  2. In another strategy, the price leader sets the price in such a way that it allows some profits to the follower firms.
  3. Barometric price leadership where an old, experienced, respectful, largest acts as a leader and sets the price. It makes changes in price which are beneficial from all firm’s and industry’s view point. Price charged by leader is accepted by follower firms.

Kinked Demand Curve:

  • In many oligopolistic industries there is price rigidity or stability.
  • The prices remains sticky or inflexible for a long time.
  • Oligopolists do not change the price even if economic conditions change.
  • Out of many theories explaining price rigidity, the theory of kinked demand curve hypothesis given by American economist Paul M. Sweezy is most popular.
  • According to kinked demand curve 4 hypothesis, the demand curve faced by an oligopolist have a ‘Kink’ at the prevailing price level.
  • A kink is formed at the prevailing price because —
    – the portion of the demand curve above the prevailing price is elastic, and
    – the portion of the demand curve below the prevailing price is inelastic

Consider the following figure.
CA Foundation Business Economics Study Material Oligopoly 1

  • In the fig., OP is the prevailing price at which the firm is producing and selling OQ output.
  • At prevailing price OP, the upper portion of demand curve dK is elastic and lower portion of demand curve KD is inelastic.
  • This difference in elasticities is due to the assumption of particular reactions by kinked demand curve theory.

The assumed reaction pattern are –

  1. If the oligopolist raises the price above the prevailing price OP, he fears that none of his rivals will follow him.
    – Therefore, he will loose customers to them and there will be substantial fall in his sales.
    – Thus, the demand with respect to price rise above the prevailing price is highly elastic as indicated by the upper portion of demand curve dK.
    – The oligopolist will therefore, stick to the prevailing prices.
  2. If the oligopolist reduces the price below the prevailing price OP to increase his sales, his rivals too will quickly reduce the price.
    – This is because the rivals fear that their customers will get diverted to price cutting oligopolist’s product.
    – Thus, the price cutting oligopolist will not be able to increase his sales very much.
    – Hence, the demand with respect to price reduction below the prevailing price is inelastic as indicated by the lower portion of demand curve KD.
    – The oligopolist will therefore, stick to the prevailing prices.
    – Each oligopolist will, thus, stick to the prevailing price realising no gain in changing the price.
    – A kink will, therefore, be formed at the prevailing price which remains rigid or sticky or stable at this level.

Other Important Market Forms:

  1. Duopoly in which there are only TWO firms in the market. It is subset of oligopoly.
  2. Monopoly is a market where there is a single buyer. It is generally in factor market.
  3. Oligopsony market where there are small number of large buyers in factor market.
  4. Bilateral monopoly market where there is a single buyer and a single seller. It is mix of monopoly and monopsony markets

RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.2

RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.2

Other Exercises

Question 1.
Given that the number  \(\overline{35 a 64}\) is divisible by 3, where a is a digit, what are the possible volues of a ?
Solution:
The number \(\overline{35 a 64}\) is divisible by 3
∵The sum of its digits will also be divisible by 3
∴ 3 + 5 + a + b + 4 is divisible by 3
⇒ 18 + a is divisible by 3
⇒ a is divisible by 3 (∵ 18 is divisible by 3)
∴ Values of a can be, 0, 3, 6, 9

Question 2.
If x is a digit such that the number \(\overline { 18×71 }\) is divisible by 3,’ find possible values of x.
Solution:
∵ The number \(\overline { 18×71 }\)
is divisible by 3
∴ The sum of its digits will also be divisible by 3
⇒ l + 8+ x + 7 + 1 is divisible by 3
⇒ 17 + x is divisible by 3
The sum greater than 17, can be 18, 21, 24, 27…………
∴ x can be 1, 4, 7 which are divisible by 3.

Question 3.
If is a digit of the number \(\overline { 66784x }\) such that it is divisible by 9, find the possible values of x.
Solution:
∵ The number 66784 x is divisible by 9
∴ The sum of its digits will also be divisible by 9
⇒ 6+6+7+8+4+x is divisible by 9
⇒ 31 + x is divisible by 9
Sum greater than 31, are 36, 45, 54………
which are divisible by 9
∴ Values of x can be 5 on 9
∴ x = 5

Question 4.
Given that the number \(\overline { 67 y 19 }\) is divisible by 9, where y is a digit, what are the possible values of y ?
Solution:
∵ The number \(\overline { 67 y 19 }\) is divisible by 9
∴The sum of its digits will also be divisible by 9
⇒ 6 + 7+ y+ 1+ 9 is divisible by 9
⇒ 23 + y is divisible by 9
∴ The numbers greater than 23 are 27, 36, 45,……..
Which are divisible by 9
∴y = A

Question 5.
If \(\overline { 3 x 2 }\) is a multiple of 11, where .v is a digit, what is the value of * ?
Solution:
∵ The number \(\overline { 3 x 2 }\) is multiple of 11
∴ It is divisible by 11
∴ Difference of the sum of its alternate digits is zero or multiple of 11
∴ Difference of (2 + 3) and * is zero or multiple of 11
⇒ If x – (2 + 3) = 0 ⇒ x-5 = 0
Then x = 5

Question 6.
If \(\overline { 98125 x 2 }\) is a number with x as its tens digits such that it is divisible by 4. Find all the possible values of x.
Solution:
∵ The number \(\overline { 98125 x 2 }\) is divisible by 4
∴ The number formed by tens digit and units digit will also be divisible by 4
∴ \(\overline { x2 }\) is divisible by 4
∴ Possible number can be 12, 32, 52, 72, 92
∴ Value of x will be 1,3, 5, 7, 9

Question 7.
If x denotes the digit at hundreds place of the number \(\overline { 67 x 19 }\) such that the
number is divisible by 11. Find all possible values of x.
Solution:
∵ The number \(\overline { 67 x 19 }\) is divisible by 11
∴ The difference of the sums its alternate digits will be 0 or divisible by 11
∴ Difference of (9 + x + 6) and (1 + 7) is zero or divisible by 11
⇒ 15+x-8 = 0, or multiple of 11,
7 + x = 0 ⇒ x = -7, which is not possible
∴ 7 + x = 11, 7 + x = 22 etc.
⇒ x=11-7 = 4, x = 22 – 7
⇒ x = 15 which is not a digit
∴ x = 4

Question 8.
Find the remainder when 981547 is divided by 5. Do this without doing actual division.
Solution:
A number is divisible by 5 if its units digit is 0 or 5
But in number 981547, units digit is 7
∴ Dividing the number by 5,
Then remainder will be 7 – 5 = 2

Question 9.
Find the remainder when 51439786 is divided by 3. Do this without performing actual division.
Solution:
In the number 51439786, sum of digits is 5 + 1+ 4 + 3 + 9 + 7 + 8 + 6 = 43 and the given number is divided by 3.
∴ The sum of digits must by divisible by 3
∴ Dividing 43 by 3, the remainder will be = 1
Hence remainder = 1

Question 10.
Find the remainder, without performing actual division when 798 is divided by 11.
Solution:
Let n = 798 = a multiple of 11 + [7 + 8 – 9] 798 = a multiple of 11 + 6
∴ Remainder = 6

Question 11.
Without performing actual division, find the remainder when 928174653 is divided by 11.
Solution:
Let n = 928174653
= A multiple of 11+(9 + 8 + 7 + 6 + 3)-(2 + 1+4 + 5)
= A multiple of 11 + 33 – 12
= A multiple of 11 + 21
= A multiple of 11 + 11 + 10
= A multiple of 11 + 10
∴ Remainder =10

Question 12.
Given an example of a number which is divisible by :
(i) 2 but not by 4.
(ii) 3 but not by 6.
(iii) 4 but not by 8.
(iv) both 4 and 8 but not 32.
Solution:
(i) 2 but not by 4
A number is divisible by 2 if units do given is even but it is divisible by 4 if the number formed by tens digit and ones digit is divisible by 4.
∴ The number can be 222, 342 etc.
(ii) 3 but not by 6
A number is divisible by 3 if the sum of its digits is divisible by 3
But a number is divisible by 6, if it is divided by 2 and 3 both
∴ The numbers can be 333, 201 etc.
(iii) 4 but not by 8
A number is divisible by 4 if the number formed by the tens digit and ones digit is divisible by 4 but a number is divisible by 8, if the number formed by hundreds digit, tens digit and ones digit is divisible by 8.
∴ The number can be 244, 1356 etc.
(iv) Both 4 and 8 but not by 32
A number in which the number formed by the hundreds, tens and one’s digit, is divisible by 8 is divisible by 8. It will also divisible by 4 also.
But a number when is divisible by, 4 and 8 both is not necessarily divisible by 32 e.g., 328, 5400 etc.

Question 13.
Which of the following statements are true ?
(i) If a number is divisible by 3, it must be divisible by 9.
(ii) If a number is divisible by 9, it must be divisible by 3.
(iii) If a number is divisible by 4, it must be divisible by 8.
(iv) If a number is divisible by 8, it must be divisible by 4.
(v) A number is divisible by 18, if it is divisible by both 3 and 6.
(vi) If a number is divisible by both 9 and 10, it must be divisible by 90.
(vii) If a number exactly divides the sum of two numbers, it must exactly divide the numbers separately.
(viii) If a number divides three numbers exactly, it must divide their sum exactly.
(ix) If two numbers are co-priirie, at least one of them must be a prime number.
(x) The sum of two consecutive odd numbers is always divisible by 4.
Solution:
(i) False, it is not necessarily that it must divide by 9.
(ii) Trae.
(iii) False, it is not necessarily that it must divide by 8.
(iv) True.
(v) False, it must be divisible by 9 and 2 both.
(vi) True.
(vii) False, it is not necessarily.
(viii)True.
(ix) False. It is not necessarily.
(x) True.

Hope given RD Sharma Class 8 Solutions Chapter 5 Playing With Numbers Ex 5.2 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 7 Decimals Ex 7B

RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 7 Decimals Ex 7B.

Other Exercises

Convert each of the following into a fraction in its simplest form :

Question 1.
Solution:
.9 = \(\\ \frac { 9 }{ 10 } \)

Question 2.
Solution:
0.6
= \(\\ \frac { 6 }{ 10 } \)
= \(\frac { 6\div 2 }{ 10\div 2 }\)
= \(\\ \frac { 3 }{ 5 } \)
(Dividing by 2, the HCF of 6, 10)

Question 3.
Solution:
.08
= \(\\ \frac { 8 }{ 100 } \)
= \(\frac { 8\div 4 }{ 100\div 4 }\)
= \(\\ \frac { 2 }{ 25 } \)
(Dividing by 4, the HCF of 7, 100)

Question 4.
Solution:
0.15
= \(\\ \frac { 15 }{ 100 } \)
= \(\frac { 15\div 5 }{ 100\div 5 }\)
= \(\\ \frac { 3 }{ 20 } \)
(Dividing by 5, the HCF of 15, 100)

Question 5.
Solution:
0.48
= \(\\ \frac { 48 }{ 100 } \)
= \(\frac { 48\div 4 }{ 100\div 4 }\)
= \(\\ \frac { 12 }{ 25 } \)
(Dividing by 4, the HCF of 48, 100)

Question 6.
Solution:
0.53
= \(\\ \frac { 53 }{ 1000 } \)

Question 7.
Solution:
= \(\\ \frac { 125 }{ 1000 } \)
= \(\frac { 125\div 125 }{ 1000\div 125 }\)
= \(\\ \frac { 1 }{ 8 } \)
(Dividing by 125, the HCF of 125, 1000)

Question 8.
Solution:
.224
= \(\\ \frac { 224 }{ 1000 } \)
= \(\frac { 224\div 8 }{ 1000\div 8 }\)
= \(\\ \frac { 28 }{ 125 } \)
(Dividing by 8, the HCF of 224, 1000)

Convert each of the following as a mixed fraction

Question 9.
Solution:
6.4
= \(\\ \frac { 64 }{ 10 } \)
= \(\frac { 64\div 2 }{ 10\div 2 }\)
= \(\\ \frac { 32 }{ 6 } \)
= \(6 \frac { 2 }{ 5 } \)
(Dividing by 2, the HCF of 64, 10)

Question 10.
Solution:
16.5
= \(\\ \frac { 165 }{ 10 } \)
= \(\frac { 165\div 5 }{ 10\div 5 }\)
= \(\\ \frac { 33 }{ 2 } \)
= \(16 \frac { 1 }{ 2 } \)
(Dividing by 5, the HCF of 165, 10)

Question 11.
Solution:
8.36
= \(\\ \frac { 836 }{ 100 } \)
= \(\frac { 836\div 4 }{ 100\div 4 }\)
= \(\\ \frac { 209 }{ 25 } \)
= \(8 \frac { 9 }{ 25 } \)
(Dividing by 4, the HCF of 836, 100)

Question 12.
Solution:
4.275
= \(\\ \frac { 4275 }{ 1000 } \)
= \(\frac { 4275\div 25 }{ 1000\div 25 }\)
= \(\\ \frac { 171 }{ 40 } \)
= \(4 \frac { 11 }{ 40 } \)
(Dividing by 25 )

Question 13.
Solution:
25.06
= \(\\ \frac { 2506 }{ 100 } \)
= \(\frac { 2506\div 2 }{ 100\div 2 }\)
= \(\\ \frac { 1253 }{ 50 } \)
= \(25 \frac { 3 }{ 50 } \)
(Dividing by 2 )

Question 14.
Solution:
7.004
= \(\\ \frac { 7004 }{ 1000 } \)
= \(\frac { 7004\div 4 }{ 1000\div 4 }\)
= \(\\ \frac { 1751 }{ 250 } \)
= \(7 \frac { 1 }{ 250 } \)
(Dividing by 4)

Question 15.
Solution:
2.052
= \(\\ \frac { 2052 }{ 1000 } \)
= \(\frac { 2052\div 4 }{ 1000\div 4 }\)
= \(\\ \frac { 513 }{ 250 } \)
= \(2 \frac { 13 }{ 250 } \)
(Dividing by 4)

Question 16.
Solution:
3.108
= \(\\ \frac { 3108 }{ 1000 } \)
= \(\frac { 3108\div 4 }{ 1000\div 4 }\)
= \(\\ \frac { 777 }{ 250 } \)
= \(3 \frac { 27 }{ 250 } \)
(Dividing by 4)

Question 17.
Solution:
\(\\ \frac { 23 }{ 10 } \)
= 2.3
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q17.1

Question 18.
Solution:
\(\\ \frac { 167 }{ 100 } \)
= 1.67
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q18.1

Question 19.
Solution:
\(\\ \frac { 1589 }{ 100 } \)
= 15.89
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q19.1

Question 20.
Solution:
\(\\ \frac { 5413 }{ 1000 } \)
= 5.413
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q20.1

Question 21.
Solution:
\(\\ \frac { 21415 }{ 1000 } \)
= 21.415
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q21.1

Question 22.
Solution:
\(\\ \frac { 25 }{ 4 } \)
= 6.25
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q22.1

Question 23.
Solution:
\(3 \frac { 3 }{ 5 } \)
= \(\\ \frac { 3\times 5+3 }{ 5 } \)
= \(\\ \frac { 15+3 }{ 5 } \)
= \(\\ \frac { 18 }{ 5 } \)
= 3.6
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q23.1

Question 24.
Solution:
\(1 \frac { 4 }{ 25 } \)
= \(\\ \frac { 1\times 25+4 }{ 25 } \)
= \(\\ \frac { 25+4 }{ 25 } \)
= \(\\ \frac { 29 }{ 25 } \)
= 1.16
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q24.1

Question 25.
Solution:
\(5 \frac { 17 }{ 50 } \)
= \(\\ \frac { 5\times 50+17 }{ 50 } \)
= \(\\ \frac { 250+17 }{ 50 } \)
= \(\\ \frac { 267 }{ 50 } \)
= 5.34
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q25.1

Question 26.
Solution:
\(12 \frac { 3 }{ 8 } \)
= \(\\ \frac { 12\times 8+3 }{ 8 } \)
= \(\\ \frac { 96+3 }{ 8 } \)
= \(\\ \frac { 99 }{ 8 } \)
= 12.375
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q26.1

Question 27.
Solution:
\(2 \frac { 19 }{ 40 } \)
= \(\\ \frac { 2\times 40+19 }{ 40 } \)
= \(\\ \frac { 80+19 }{ 40 } \)
= \(\\ \frac { 99 }{ 40 } \)
= 2.475
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q27.1

Question 28.
Solution:
\(\\ \frac { 19 }{ 20 } \)
= 0.95
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q28.1

Question 29.
Solution:
\(\\ \frac { 37 }{ 50 } \)
= 0.74
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q29.1

Question 30.
Solution:
\(\\ \frac { 107 }{ 250 } \)
= 0.428
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q30.1

Question 31.
Solution:
\(\\ \frac { 3 }{ 40 } \)
= 0.075
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q31.1

Question 32.
Solution:
\(\\ \frac { 7 }{ 8 } \)
= 0.875
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q32.1

Question 33.
Solution:
(i) 8 kg 640 g in kilograms
= \(8 \frac { 640 }{ 1000 } \) kg
= 8.640kg
(ii) 9 kg 37 g in kilograms
= \(9 \frac { 37 }{ 1000 } \) kg
= 9.037 kg.
(iii) 6 kg 8 g in kilograms
= \(6 \frac { 8 }{ 1000 } \) kg
= 6.008 kg Ans.

Question 34.
Solution:
(i) 4 km 365 m in kilometres
= \(4 \frac { 365 }{ 1000 } \) km
= 4.365 km
(ii) 5 km 87 m in kilometres
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q34.1

Question 35.
Solution:
(i) 15 kg 850 g in kilograms
= \(15 \frac { 850 }{ 1000 } \) kg
= 15.850 kg
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q35.1

Question 36.
Solution:
(i) Rs. 18 and 25 paise in rupees
= \(18 \frac { 25 }{ 100 } \)
= 18.25 rupees
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7B Q36.1

Hope given RS Aggarwal Solutions Class 6 Chapter 7 Decimals Ex 7B are helpful to complete your math homework.

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CA Foundation Business Economics Study Material – Imperfect Competition : Monopolistic Competition

CA Foundation Business Economics Study Material Chapter 4 Price Determination in Different Markets – Imperfect Competition : Monopolistic Competition

IMPERFECT COMPETITION : MONOPOLISTIC COMPETITION

Introduction

  • We have studied two models that represent the two extremes of market structures namely perfect competition and monopoly.
  • The two extremes of market structures are not seen in real world.
  • In reality we find only imperfect competition which fall between the two extremes of perfect competition and monopoly.
  • The two main forms of imperfect competition are —
    – Monopolistic Competition and
    – Oligopoly

Meaning and features of Monopolistic Competition

  • As the name implies, monopolistic competition is a blend of competitive market and monopoly elements.
  • There is competition because of large number of firms with easy entry into the industry selling similar product.
  • The monopoly element is due to the fact that firms produce differentiated products. The products are similar but not identical.
  • This gives an individual firm some degree of monopoly of its own differentiated product.
  • E.g. MIT and APTECH supply similar products, but not identical.
  • Similarly, bathing soaps, detergents, shoes, shampoos, tooth pastes, mineral water, fitness and health centers, readymade garments, etc. all operate in a monopolistic competitive market.

The characteristics of monopolistic competitive market can be summed up as follows:

  1. Large number of buyers and sellers
    • There are large number of firms.
      – So each individual firms can not influence the market.
      – Each individual firm share relatively small fraction of the total market.
    • The number of buyers is also very large and so single buyer cannot influence the market by demanding more or less.
  2. Product Differentiation
    • The product produced by various firms are not identical but are somewhat different from each other but are close substitutes of each other.
    • Therefore, the products are differentiated by brand names. E.g. – Colgate, Close-Up, Pepsodent, etc.
    • Brand loyalty of customers gives rise to an element of monopoly to the firm.
  3. Freedom of entry and exit
    • New firms are free to enter into the market and existing firms are free to quit the market.
  4. Non-Price Competition
    • Firms under monopolistic competitive market do not compete with each other on the basis of price of product.
    • They compete with each other through advertisements, better product development, better after sales services, etc.
    • Thus, firms incur heavy expenditure on publicity advertisement, etc.

Short Run Equilibrium of a Firm in Monopolistic Competition. (Price-Output Equilibrium)

  • Each firm in a monopolistic competitive market is a price maker and determines the price of its own product.
  • As many close substitutes for the product are available in the market, the demand curve (average revenue curve) for the product of individual firm is relatively more elastic.

The conditions of equilibrium of a firm are same as they are in perfect competition and monopoly i.e.

  1. MR = MC, and
  2. MC curve cuts the MR curve from below.

The following figures show the equilibrium conditions and price-output determination of a firm under monopolistic competition.

When a firm in a monopolistic competition is in the short run equilibrium, it may find itself in the following situations —

  1. Firm will earn SUPER NORMAL PROFITS if its AR > AC;
  2. Firm will earn NORMAL PROFITS if its AR = AC; and
  3. Firm will suffer LOSSES if its AR < AC

1. Super Normal Profits (AR > AC):
CA Foundation Business Economics Study Material Imperfect Competition Monopolistic Competition 1
CA Foundation Business Economics Study Material Imperfect Competition Monopolistic Competition 2
The firm will earn NORMAL PROFITS if AC curve is tangent to AR curve i.e. when AR=AC

2. Losses (AR < AC):
CA Foundation Business Economics Study Material Imperfect Competition Monopolistic Competition 3
The firm may continue to produce even if incurring losses if its AR ≥ AVC.

Long Run Equilibrium of a Firm in Monopolistic Competition

  • If the firms in a monopolistic competitive market earn super normal profits, it attracts new firms to enter the industry.
  • With the entry of new firms market will be shared by more firms.
  • As a result, profits per firm will go on falling.
  • This will go on till super normal profits are wiped out and all the firms earn only normal profits.

CA Foundation Business Economics Study Material Imperfect Competition Monopolistic Competition 4

  • In the long run firms in a monopolistic competitive market just earn NORMAL PROFITS.
  • Firms operate at sub-optimal level as shown by point ‘R’ where the falling portion AC curve is tangent to AR curve.
  • In other words firms do not operate at the minimum point of LAC curve ‘L’.
  • Therefore, production capacity equal to QQ, remains idle or unused called excess capacity.
  • This implies that in monopolistic competitive market —
  • Firms are not of optimum size and each firm has excess production capacity
  • The firm can expand its output from Q to Q, and reduce its average cost.
  • But it will not do so because to sell more it will have to reduce its average revenue even more than average costs.
  • Hence, firms will operate at sub-optimal level only in the long run.

RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A

RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 7 Decimals Ex 7A.

Other Exercises

Question 1.
Solution:
(i)Fifty eight point six three = 58.63
(ii)One hundred twenty four point four two five = 124.425
(iii)Seven point seven six = 7.76
(iv)Nineteen point eight = 19.8
(v)Four hundred four point zero four four = 404.044
(vi)Point one seven three = 173
(v)Point zero one five = .015 Ans.

Question 2.
Solution:
(i) 14.83
Place value of 1 = 10,
Place value of 4 = 4,
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A Q2.1
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A Q2.2
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A Q2.3

Question 3.
Solution:
(i) 67.83 = (6 x 10) + (7 x 1)
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A Q3.1

Question 4.
Solution:
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A Q4.1

Question 5.
Solution:
(i) 7.5, 64.23, 0.074 = 7.500, 64.230, 0.074
(Here, at the most 0.074 has 3 places)
(ii) 0.6, 5.937, 2.36, 4.2 = 0.600, 5.937, 2.360, 4.200
(Here, 5.937 has at most 3 places)
(iii) 1.6, 0.07, 3.58, 2.9 = 1.60, 0.07, 3.58, 2. 90
(Here, at the most are two places)
(iv) 2.5. 0.63, 14.08, 1.637 = 2.500, 0.630. 14.080, 1.637 Ans.
(Here, at the most are three places)

Question 6.
Solution:
Making like decimals where ever it is necessary,
RS Aggarwal Class 6 Solutions Chapter 7 Decimals Ex 7A Q6.1

Question 7.
Solution:
First of all making them in like decimals,
(i) 5.8, 7.2, 5.69, 7.14, 5.06
=> 5.80, 7.20, 5.69, 7.14, 5.06
Arranging in ascending order,
5:06 <5.69 <5.80 <7.14 <7.20
=> 5.06 < 5.69 < 5.8 < 7.14 < 7.2 Ans.
(ii) 0.6, 6.6, 6.06, 66.6, 0.06
=>0.60, 6.60, 6.06, 66.60, 0.06
Arranging in ascending order,
0.06 < 0.60 < 6.06 < 6.60 < 66.60
=> 0.06 < 0.6 < 6.06 < 6.6 < 66.6 Ans.
(iii) 6.54, 6.45, 6.4, 6.5, 6.05
=> 6.54, 6.45, 6.4, 6.5, 6.05
Arranging in ascending order,
6. 05 < 6.40 < 6.45 < 6.50 < 6.54
=> 6.05 < 6.4 < 6.45 < 6.5 < 6.54 Ans.
(iv) 3.3,3.303, 3.033, 0.33, 3.003
=> 3.300, 3.303, 3.033, 0.330, 3.003
Arranging in descending order,
0.330 < 3.003 < 3.033 < 3.300 < 3.303
=> 0.33 < 3.003 < 3.033 < 3.3 < 3.303 Ans.

Question 8.
Solution:
Making them in like decimals and them comparing
(i) 7.3, 8.73, 73.03, 7.33, 8.073
=> 7.300, 8.730, 73.030, 7.330, 8.073
Arranging in descending order
73.030 > 8.730 > 8.073 > 7.330 > 7.300
=> 73.03 > 8.73 > 8.073 > 7.33 > 7.3 Ans.
(ii) 3.3, 3.03, 30.3, 30.03, 3.003
=> 3.300, 3.030, 30.300, 30.030, 3.003
Arranging in descending order
30.300> 30.030 >3.300 >3.030 > 3.003
=> 30.3 > 30.03 > 3.3 > 3.03 > 3.003 Ans.
(iii) 2.7, 7.2, 2.27, 2.72, 2.02, 2.007
=> 2.700, 7.200, 2.270, 2.720, 2.020, 2.007
Arranging in descending order
7. 200 > 2.720 > 2.700 > 2.270 > 2.020 > 2.007
=> 7.2 > 2.72 > 2.7 > 2.27 > 2.02 > 2.007 Ans.
(iv) 8.88, 8.088, 88.8, 88.08, ,8.008
=> 8.880, 8.088, 88.800, 88.080, 8.008
Arranging in descending order,
88.800 > 88.080 > 8.880 > 8.088 > 8.008
=> 88.8 > 88.08 > 8.88 > 8.088 > 8.008

 

Hope given RS Aggarwal Solutions Class 6 Chapter 7 Decimals Ex 7A 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 6 Simplification Ex 6B

RS Aggarwal Class 6 Solutions Chapter 6 Simplification Ex 6B

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

Other Exercises

OBJECTIVE QUESTIONS
Mark against the correct answer in each of the following :

Question 1.
Solution:
(c) 8 + 4 ÷ 2 x 5
= 8 + 4 x \(\\ \frac { 1 }{ 2 } \) x 5
= 8 + 10
= 18.

Question 2.
Solution:
(b) 54 ÷ 3 of 6 + 9 = 54 ÷ 18 + 9
= 54 x \(\\ \frac { 1 }{ 18 } \) + 9
= 3 + 9
= 12.

Question 3.
Solution:
(b) 13 – (12 – 6 ÷ 3)
= 13 – (12 – 2)
= 13 – 10
= 3

Question 4.
Solution:
(a) 1001 ÷ 11 of 13
= 1001 ÷ 143
= 1001 x \(\\ \frac { 1 }{ 143 } \)
= 7.

Question 5.
Solution:
(b) 133 + 28 ÷ 7 – 8 x 2
= 133 + 4 – 16
= 137 – 16
= 121.

Question 6.
Solution:
(a) 3640 – 14 ÷ 7 x 2
= 3640 – 2 x 2
= 3640 – 4
= 3636.

Question 7.
Solution:
(b) 100 x 10 – 100 + 2000 ÷ 100
= 1000 – 100 + 20
= 920.

Question 8.
Solution:
(b) \(27-\left[ 18-\left\{ 16-\left( 5-\overline { 4-1 } \right) \right\} \right] \)
= 27 – [18 – {16 – (5 – 4 + 1)}]
= 27 – [18 – {16 – 5 + 4 – 1}]
= 27 – [18 – 16 + 5 – 4 + 1]
= 27 – 18 + 16 – 5 + 4 – 1
= 23

Question 9.
Solution:
\(32-\left[ 48\div \left\{ 36-\left( 27-\overline { 16-9 } \right) \right\} \right] \)
= 32 – [48 ÷ {36 – (27 – 16 + 9)}]
= 32 – [48 ÷ {36 – 27 + 16 – 9}]
= 32 – [48 ÷ 16]
= 32 – 3
= 29

Question 10.
Solution:
(a) 8 – [28 ÷ {34 – (36 – 18 ÷ 9 x 8)}]
\(8-\left[ 28\div \left\{ 34-\left( 36-18\times \frac { 1 }{ 9 } \times 8 \right) \right\} \right] \)
= 8 – [28 ÷ {34 – (36 – 16)}]
= 8 – [28 ÷ {34 – 20}]
= 8 – {28 ÷ 14}
= 8 – 2
= 6.

Hope given RS Aggarwal Solutions Class 6 Chapter 6 Simplification Ex 6B 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.