BUDGETING

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Budgeting

Budgeting is the process of estimating the availability of resources and then allocating them to various activities of an organization according to a pre-determined priority. In most cases, approval of a budget also means the approval to various spending units to utilize the allocated resources. Budgeting plays a criucial role in the socio-economic development of the nation.

Budget is the annual statement of the outlays and tax revenues of the government of India together with the laws and regulations that approve and support those outlays and tax revenues . The budget has two purposes in general :
1. To finance the activities of the union government
2. To achieve macroeconomic objectives.

The Budget contains the financial statements of the government embodying the estimated receipts and expenditure for one financial year, ie.  it is a proposal of how much money is to be spent on what and how much of it will
be contributed by whom or raised from where during the coming year.

Different types of Budgeting

Economists throughout the globe have classified the budgets into different types based on the process and purpose of the budgets, which are as follows:-

1- The Line Item Budget

line-item budgeting was introduced in some countries in the late 19th centuary. Indeed line item budgeting which is the most common form of budgeting in a large number of countries and suffers from several drawbacks was a major reform initiative then. The line item budget is defined as “the budget in which the individual financial statement items are grouped by cost centers or departments .It shows the comparison between the financial data for the past  accounting or budgeting periods and estimated figures for the current or a future period”In a line-item system, expenditures for the budgeted period are listed according to objects of expenditure, or “line-items.” These line items include detailed ceilings on the amount a unit would spend on salaries, travelling allowances, office expenses, etc. The focus is on ensuring that the agencies
or units do not exceed the ceilings prescribed. A central authority or the Ministry of Finance keeps a watch on the spending of various units to ensure that the ceilings are not violated. The line item budget approach is easy to understand and implement. It also facilitates centralized control and fixing of authority and responsibility of the spending units. Its major disadvantage is that it does not provide enough information to the top levels about the activities and achievements of individual units.

2 – Performance Budgeting

a performance budget reflects the goal/objectives of the organization and spells out performance targets. These targets are sought to be achieved through a strategy. Unit costs are associated with the strategy and allocations are accordingly made for achievement of the objectives. A Performance Budget gives an indication of how the funds spent are expected to give outputs and ultimately the outcomes. However, performance budgeting has a limitation – it is not easy to arrive at standard unit costs especially in social programmes which require a multi-pronged approach.

3- Zero-based Budgeting

The concept of zero-based budgeting was introduced in the 1970s. As the name suggests, every budgeting cycle starts from scratch. Unlike the earlier systems where only incremental changes were made in the allocation, under zero-based budgeting every activity is evaluated each time a budget is made and only if it is established that the activity is necessary, are funds allocated to it. The basic purpose of Zero-based Budgeting is phasing out of programmes/ activities which do not have relevance anymore. However, because of the efforts involved in preparing a zero-based budget and institutional resistance related to personnel issues, no government ever implemented a full zero-based budget, but in modified forms the basic principles of ZBB are often used.

4- Programme Budgeting and Performance Budgeting

Programme budgeting in the shape of planning, programming and budgeting system (PPBS) was introduced in the US Federal Government in the mid-1960s. Its core themes had much in common with earlier strands of performance budgeting.
Programme budgeting aimed at a system in which expenditure would be planned and controlled by the
objective. The basic building block of the system was classification of expenditure into programmes, which meant objective-oriented classification so that programmes with common objectives are considered together.
It aimed at an integrated expenditure management system, in which systematic policy and expenditure planning would be developed and closely integrated with the budget. Thus, it was too ambitious in scope. Neither was adequate preparation time given nor was a stage-by-stage approach adopted. Therefore, this attempt to introduce PPBS in the federal government in USA did not succeed, although the concept of performance budgeting and programme budgeting endured.

 

 

Budgetary Control

Budgetary control refers to how well managers utilize budgets to monitor and control costs and operations in a given accounting period. In other words, budgetary control is a process for managers to set financial and performance goals with budgets, compare the actual results, and adjust performance, as it is needed.

Budgetary control involves the following steps :

(a) The objects are set by preparing budgets.

(b) The business is divided into various responsibility centres for preparing various budgets.

(c) The actual figures are recorded.

(d) The budgeted and actual figures are compared for studying the performance of different cost centres.

(e) If actual performance is less than the budgeted norms, a remedial action is taken immediately.

The main objectives of budgetary control are the follows:

  1. To ensure planning for future by setting up various budgets, the requirements and expected performance of the enterprise are anticipated.
  2. To operate various cost centres and departments with efficiency and economy.
  3. Elimination of wastes and increase in profitability.
  4. To anticipate capital expenditure for future.
  5. To centralise the control system.
  6. Correction of deviations from the established standards.
  7. Fixation of responsibility of various individuals in the organization.

 

Responsibility Accounting

Responsibility accounting is an underlying concept of accounting performance measurement systems. The basic idea is that large diversified organizations are difficult, if not impossible to manage as a single segment, thus they must be decentralized or separated into manageable parts.

These decentralized parts are divided as : 1) revenue centers, 2) cost centers, 3) profit centers and 4) investment centers.

  1. revenue center (a segment that mainly generates revenue with relatively little costs),
  2. costs for a cost center (a segment that generates costs, but no revenue),
  3. a measure of profitability for a profit center (a segment that generates both revenue and costs) and
  4. return on investment (ROI) for an investment center (a segment such as a division of a company where the manager controls the acquisition and utilization of assets, as well as revenue and costs).

 

Advantages:-

  1. It provides a way to manage an organization that would otherwise be unmanageable.
  2. Assigning responsibility to lower level managers allows higher level managers to pursue other activities such as long term planning and policy making.
  3. It also provides a way to motivate lower level managers and workers.
  4. Managers and workers in an individualistic system tend to be motivated by measurements that emphasize their individual performances.

In India the budget is prepared from top to bottom approach and responsible accounting would not only improve the efficiency of Indian budgetary system but also will help in performance analysis.

Social Accounting

Social accounting is concerned with the statistical classification of the activities of human beings and human institutions in ways which help us to understand the operation of the economy as a whole.

Social accounting is the process of communicating the social and environmental effects of organizations’ economic actions to particular interest groups within society and to society at large

The components of social accounting are production, consumption, capital accumulation, government transactions and transactions with the rest of the world.

The uses of social accounting are as follows:

(1) In Classifying Transactions

(2) In Understanding Economic Structure

(3) In Understanding Different Sectors and Flows

(4) In Clarifying Relations between Concepts

(7) In Explaining Movements in GNP

(8) Provide a Picture of the Working of Economy

(9) In Explaining Interdependence of Different Sectors of the Economy

(10) In Estimating Effects of Government Policies

(11) Helpful in Big Business Organisations

(12) Useful for International Purposes

(13) Basis of Economic Models

 

Budgetary Deficit

Budgetary Deficit is the difference between all receipts and expenditure of the government, both revenue and capital. This difference is met by the net addition of the treasury bills issued by the RBI and drawing down of cash balances kept with the RBI. The budgetary deficit was called deficit financing by the government of India. This deficit adds to money supply in the economy and, therefore, it can be a major cause of inflationary rise in prices.

Budgetary Deficit of central government of India was Rs. 2,576 crores in 1980-81, it went up to Rs. 11,347 crores in 1990-91 to Rs. 13,184 crores in 1996-97.

The concept of budgetary deficit has lost its significance after the presentation of the 1997-98 Budget. In this budget, the practice of ad hoc treasury bills as source of finance for government was discontinued. Ad hoc treasury bills are issued by the government and held only by the RBI. They carry a low rate of interest and fund monetized deficit. These bills were replaced by ways and means advance. Budgetary deficit has not figured in union budgets since 1997-98. Since 1997-98, instead of budgetary deficit, Gross Fiscal Deficit (GFD) became the key indicator.

 

Fiscal Deficit

  • The difference between total revenue and total expenditure of the government is termed as fiscal deficit. It is an indication of the total borrowings needed by the government and thus amounts to all the borrowings of the government . While calculating the total revenue, borrowings are not included.
  • The gross fiscal deficit (GFD) is the excess of total expenditure including loans net of recovery over revenue receipts (including external grants) and non-debt capital receipts. The net fiscal deficit is the gross fiscal deficit less net lending of the Central government.
  • Generally fiscal deficit takes place either due to revenue deficit or a major hike in capital expenditure. Capital expenditure is incurred to create long-term assets such as factories, buildings and other development.
  • A deficit is usually financed through borrowing from either the central bank of the country or raising money from capital markets by issuing different instruments like treasury bills and bonds.

 

Revenue Deficit

  • Revenue deficit is concerned with the revenue expenditures and revenue receipts of the government. It refers to excess of revenue expenditure over revenue receipts during the given fiscal year.
  • Revenue Deficit = Revenue Expenditure – Revenue Receipts
  • Revenue deficit signifies that government’s own revenue is insufficient to meet the expenditures on normal functioning of government departments and provisions for various services.
  • In India social expenditure like MNREGA is a revenue expenditure though a part of Plan expenditure.
  • Its targeted to be 2.9% of GPD in the year 2014-15, though the fiscal revenue and budget management act specifies it to be zero by 2008-09

Drain Theory

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  • Dadabhai Naoroji: ‘Poverty in India’ (1876)
  • He claimed that the drain of wealth and capital from the country which started after 1757 was responsible for absence of development in India.
  • Drain was done through trade, industry and finance
  • Two elements of the drain
    • That arising from the remittances by European officials of their savings, and fro their expenditure in England
    • Arising from remittance by non-official Europeans
  • India has to export much more than she imported to meet the requirements of the economic drain
  • In 1880 it amounted to 4.14% of India’s national income
  • Consequences of the Drain
    • Prevented the process of capital formation in India
    • Through the drained wealth, the British established industrial concerns in India owned by British nationals
    • It acted as a drag on economic development

 

Industrial Transition in India

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  • The process of industrial transition divided into: industrial growth during the 19th century and industrial progress during the 20th century
  • Industrial growth during the 19th century
    • Decline of indigenous industries and the rise of large scale modern industries
    • 1850-55: first cotton mill, first jute mill and the first coal mine established. Railway also introduced.
    • Despite some industrialisation, India was becoming an agricultural colony
    • The thrust to industrialisation came from the British because
      • They had capital
      • They had experience in setting up industries in Britain
      • They had state support
    • British industrialists were interested in making profits rather than economic growth of India
    • Parsis, Jews and Americans were also setting industries
    • No Indian industrialists because
      • Neither the merchants nor the craftsmen took the lead in setting industries
      • While the craftsmen didn’t possess capital, the merchants were happy with trading and money lending activity which was also growing at that time.
    • However, some Parsis, Gujaratis, Marwaris, Jains and Chettiars joined the ranks of industrialists
  • Industrial Growth in the first half of the 20th century
    • Imp events that stimulated industrial growth
      • 1905: Swadeshi Movement
      • First WW
      • Second WW
    • Great stimulus was given to the production of iron and steel, cotton and woollen textiles, leather products, jute.
    • Tariff protection was given to Indian industries between 1924 and 1939. This led to growth and Indian industrialists were able to capture the market and eliminate foreign completion altogether in important fields
    • The increase in industrial output between 1939 and 1945 was about 20 percent
    • After the WW I, the share of the foreign enterprises in India’s major industries began to decline.
  • Causes for the slow growth of private enterprise in India’s industrialisation
    • Inadequacy of entrepreneurial ability
      • Indian industrialists were short-sighted and cared very little for replacement and renovation of machinery
      • Nepotism dictated choice of personnel
      • High profits by high prices rather than high profits by low margins and larger sales
    • Problem of capital and private enterprise
      • Scarce capital
      • Few avenues for the investment of surplus
      • No government loans
      • Absence of financial institutions
      • Banking was not highly developed and was more concerned with commerce rather than industry
    • Private enterprises and the role of government
      • Lack of support from the government
      • Discriminatory tariff policy: one way free-trade
      • Restrictions transfer of capital equipments and machinery from Britain
      • Almost all machinery was imported
    • Despite these difficulties, the Indian indigenous business communities continued to grow, albeit at a slow pace.

Forms and Consequences of Colonial Exploitation

  • Main forms of colonial exploitation
    • Exploitation through trade policies
    • Exploitation through export of British Capital to India
    • Exploitation through finance capital via the Managing agency system
    • Exploitation through the payments for the costs of the British administration
  • Exploitation through trade policies
    • Exp of cultivators to boost indigo export: forced
    • Exp of artisans by compulsory procurement by the Company at low prices: gomastas were the agents of the Company who used to do this
    • Exp through manipulation of export and import duties:
      • Imports of Indian printed cotton fabrics in England were banned
      • Heavy import duties on Indian manufactures and very nominal duties on British manufactures.
      • Discriminating protection was given (to industries that had to face competition from some country other than Britain). This was whittled down, however, by the clause of Imperial Preference under which imports from GB and exports to GB should enjoy the MFN status.
    • Exploitation through export of British Capital to India
      • There were three purposes of these investment (in transport and communication)
        • To build better access systems for exploited India’s natural resources
        • To provide a quick means of communication for maintaining law and order
        • To provide for quicker disbursal of British manufactures throughout the country and that raw materials could be easily procured
      • Fields of FDI
        • Economic overhead and infrastructure like railways, shippings, port, roads, communication
        • For promoting mining of resources
        • Commercial agriculture
        • Investment in consumer goods industries
        • Investments made in machine building, engineering industries and chemicals
      • Forms of investment
        • Direct private foreign investment
        • Sterling loans given to the British Government in India
      • Estimates show that foreign capital increased from 365 mn sterling in 1911 to 1000 mn sterling in 1933.
      • British multinationals were the chief instruments of exploitation and it were they who drained out the wealth of India.
      • These investments show that
        • British were interested in creating economic infrastructure to aid exploitation and resource drain
        • They invested in consumer goods and not in basic and heavy industries to prevent the development of Indian industries
        • Ownership and management of these companies lay in British hands
      • Exploitation through finance capital via the Managing agency system
        • Managing agency system: The British merchants who had earlier set up firms acted as pioneers and promoters in several industries like jute, tea and coal. These persons were called managing agents
        • It may be described as partnerships of companies formed by a group of individuals with strong financial resources and business experience
        • Functions of managing agents
          • To float new concerns
          • Arrange for finance
          • Act as agents for purchase of raw materials
          • Act as agents to market the produce
          • Manage the affairs of the business
        • They were important because they supplied finance to India when it was starved of capital
        • In due course, they started dictating the terms of the industry and business and became exploitative and inefficient
        • They demanded high percentage of profits. When refused they threatened to withdraw their finance
      • Exploitation through payments for the costs of British administration
        • British officers occupied high positions and were paid fabulous remunerations.
        • These expenditures were paid by India
        • They transferred their savings to Britain
        • India had to pay interest on Sterling Loans
        • India has to pay for the war expedition of the Company and later the Crown

Consequences of the exploitation

  • India remained primarily an agricultural economy
  • Handcrafts and industries were ruined
  • Trade disadvantage developed due to the policy of the British
  • Economic infrastructure was developed only to meet the colonial interests
  • Drain of Wealth
  • The net result of the British policies was poverty and stagnation of the Indian economy

India under the British Rule

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The economic consequences of the British rule can be studied under three heads:

  • Decline of Indian Handicrafts and progressive ruralisation of the Indian economy
  • Growth of the new land system and the commercialisation of Indian agriculture
  • Process of industrial transition of India

Decline of Handicrafts

  • While India was an exporter of Handicrafts before the Industrial Revolution, the revolution reversed the character of India’s foreign trade
    • Increase in demand for raw material for British industries
    • Hence, steps were made to crush Indian handcrafts as well as commercialise agriculture to meet the interests of the British industries
  • Principle causes for the decline of Indian handicrafts
    • Disappearance of Princely courts
    • Hostile policy of the East India Company and the British Parliament
    • Competition of machine-made goods
    • The development of new forms and patterns of demand as a result of foreign influence
  • Economic consequences of the decline of handicrafts
    • Increased unemployment
    • Back-to-the-land movement: handicrafts were forced to take up agriculture or become landless labourers. This increased the pressure on land. This trend of growing proportion of the working force on agriculture is described as ‘progressive ruralisation’ or ‘deindustrialisation of India’. Thus, the crisis in handicrafts and industries seriously crippled Indian agriculture.

Land System during 1793-1850

  • 1793: permanent settlement
  • Zamindari, Ryotwari, Mahalwari systems
  • Absentee landlordism emerged
  • The result of the whole change in the land system led to the emergence of subsistence agriculture
  • It helped the concentration of economic power in the hand of absentee landlords and moneylenders in rural India.

Commercialisation of Agriculture (1850-1947)

  • Define: Production of crop for sale rather than for family consumption
  • What distinguished commercial agriculture from normal sales of marketable surplus was that it was a deliberate policy worked up under the pressure from British industries. It was thus forced upon the Indian peasantry.
  • Resistance: Indigo revolution etc
  • Why CA? Industrial Revolution
  • Impact of railways and road transport: Railways and road transport made possible a huge expansion in cash cropping, for national and international markets, and production regimes across the subcontinent were placed in a new context of opportunity
  • Impact of CA
    • Mass movement to commercial agriculture caused decline in food production, increase in prices and famines.
    • Halted the process of industrialisation in India

 

PIPES & CISTERN

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Pipes and Cistern

 

  1. Inlet:

A pipe connected with a tank or a cistern or a reservoir, that fills it, is known as an inlet.

 

Outlet:

A pipe connected with a tank or cistern or reservoir, emptying it, is known as an outlet.

 

  1. If a pipe can fill a tank in xhours, then:
part filled in 1 hour = 1 .
x
  1. If a pipe can empty a tank in yhours, then:
part emptied in 1 hour = 1 .
y
  1. If a pipe can fill a tank in xhours and another pipe can empty the full tank in y hours (where y > x), then on opening both the pipes, then
the net part filled in 1 hour = 1 – 1 .
x y
  1. If a pipe can fill a tank in xhours and another pipe can empty the full tank in y hours (where x > y), then on opening both the pipes, then
the net part emptied in 1 hour = 1 – 1 .
y x

 

 

Questions:

 

Level-I:

 

1. Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P,Q and R respectively. What is the proportion of the solution R in the liquid in the tank after 3 minutes?
A.
5
11
B.
6
11
C.
7
11
D.
8
11

 

2. Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in:
A.
1 13 hours
17
B.
2 8 hours
11
C.
3 9 hours
17
D.
4 1 hours
2

 

3. A pump can fill a tank with water in 2 hours. Because of a leak, it took 2 hours to fill the tank. The leak can drain all the water of the tank in:
A.
4 1 hours
3
B. 7 hours
C. 8 hours
D. 14 hours
 

4.

 Two pipes A and B can fill a cistern in 37 minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the B is turned off after:
A. 5 min.
B. 9 min.
C. 10 min.
D. 15 min.

 

5. A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:
A. 6 hours
B. 10 hours
C. 15 hours
D. 30 hours
 

 

6.

  

 

Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank is:
A. 60 gallons
B. 100 gallons
C. 120 gallons
D. 180 gallons

 

7. A tank is filled in 5 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank?
A. 20 hours
B. 25 hours
C. 35 hours
D. Cannot be determined
E. None of these

 

8. Two pipes A and B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately?
A. 1 hour
B. 2 hours
C. 6 hours
D. 8 hours

 

9. Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both the pipes are used together, then how long will it take to fill the tank?
A. 12 min
B. 15 min
C. 25 min
D. 50 min

 

10. Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?
A. 10 min. 20 sec.
B. 11 min. 45 sec.
C. 12 min. 30 sec.
D. 14 min. 40 sec.
 

 

 

 

11.

  

 

Level-II:

 

One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank in:
A. 81 min.
B. 108 min.
C. 144 min.
D. 192 min.

 

12. A large tanker can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time and A and B fill it together for the other half?
A. 15 min
B. 20 min
C. 27.5 min
D. 30 min

 

13. A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?
A. 3 hrs 15 min
B. 3 hrs 45 min
C. 4 hrs
D. 4 hrs 15 min

 

14. Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in:
A. 6 hours
B.
6 2 hours
3
C. 7 hours
D.
7 1 hours
2

 

15. Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:
A. 10
B. 12
C. 14
D. 16
 

 

 

16.

  

 

How much time will the leak take to empty the full cistern?
I. The cistern is normally filled in 9 hours.
 II. It takes one hour more than the usual time to fill the cistern because of la leak in the bottom.
A. I alone sufficient while II alone not sufficient to answer
B. II alone sufficient while I alone not sufficient to answer
C. Either I or II alone sufficient to answer
D. Both I and II are not sufficient to answer
E. Both I and II are necessary to answer

 

17.
How long will it take to empty the tank if both the inlet pipe A and the outlet pipe B are opened simultaneously?
I. A can fill the tank in 16 minutes.
 II. B can empty the full tank in 8 minutes.
A. I alone sufficient while II alone not sufficient to answer
B. II alone sufficient while I alone not sufficient to answer
C. Either I or II alone sufficient to answer
D. Both I and II are not sufficient to answer
E. Both I and II are necessary to answer

 

18.
If both the pipes are opened, how many hours will be taken to fill the tank?
I. The capacity of the tank is 400 litres.
II. The pipe A fills the tank in 4 hours.
 III. The pipe B fills the tank in 6 hours.
A. Only I and II
B. Only II and III
C. All I, II and III
D. Any two of the three
E. Even with all the three statements, answer cannot be given.

 

 

Answers:

 

Level-I:

 

Answer:1 Option B

 

Explanation:

Part filled by (A + B + C) in 3 minutes = 3 1 + 1 + 1 = 3 x 11 = 11 .
30 20 10 60 20

 

Part filled by C in 3 minutes = 3 .
10

 

 Required ratio = 3 x 20 = 6 .
10 11 11

 

Answer:2 Option C

 

Explanation:

Net part filled in 1 hour 1 + 1 – 1 = 17 .
5 6 12 60

 

 The tank will be full in 60 hours i.e., 3 9 hours.
17 17

 

 

Answer:3 Option D

 

Explanation:

Work done by the leak in 1 hour = 1 – 3 = 1 .
2 7 14

Leak will empty the tank in 14 hrs.

 

 

Answer:4 Option B

 

Explanation:

Let B be turned off after x minutes. Then,

Part filled by (A + B) in x min. + Part filled by A in (30 –x) min. = 1.

 x 2 + 1 + (30 – x). 2 = 1
75 45 75

 

11x + (60 -2x) = 1
225 75

11x + 180 – 6x = 225.

x = 9.

 

 

Answer:5 Option C

 

Explanation:

Suppose, first pipe alone takes x hours to fill the tank .

Then, second and third pipes will take (x -5) and (x – 9) hours respectively to fill the tank.

1 + 1 = 1
x (x – 5) (x – 9)

 

x – 5 + x = 1
x(x – 5) (x – 9)

(2x – 5)(x – 9) = x(x – 5)

x2 – 18x + 45 = 0

(x – 15)(x – 3) = 0

x = 15.    [neglecting x = 3]

 

 

Answer:6 Option C

 

Explanation:

Work done by the waste pipe in 1 minute = 1 – 1 + 1
15 20 24

 

    = 1 – 11
15 120

 

    = – 1 .    [-ve sign means emptying]
40

 

 Volume of 1 part = 3 gallons.
40

Volume of whole = (3 x 40) gallons = 120 gallon

 

 

Answer:7 Option C

 

Explanation:

Suppose pipe A alone takes x hours to fill the tank.

Then, pipes B and C will take x and x hours respectively to fill the tank.
2 4

 

1 + 2 + 4 = 1
x x x 5

 

7 = 1
x 5

x = 35 hrs.

 

Answer:8 Option C

 

Explanation:

Let the cistern be filled by pipe A alone in x hours.

Then, pipe B will fill it in (x + 6) hours.

1 + 1 = 1
x (x + 6) 4

 

x + 6 + x = 1
x(x + 6) 4

x2 – 2x – 24 = 0

(x -6)(x + 4) = 0

x = 6.     [neglecting the negative value of x]

 

 

Answer:9 Option A

 

Explanation:

Part filled by A in 1 min = 1 .
20

 

Part filled by B in 1 min = 1 .
30

 

Part filled by (A + B) in 1 min = 1 + 1 = 1 .
20 30 12

Both pipes can fill the tank in 12 minutes.

 

 

Answer:10 Option D

 

Explanation:

Part filled in 4 minutes = 4 1 + 1 = 7 .
15 20 15

 

Remaining part = 1 – 7 = 8 .
15 15

 

Part filled by B in 1 minute = 1
20

 

1 : 8 :: 1 : x
20 15

 

x = 8 x 1 x 20 = 10 2 min = 10 min. 40 sec.
15 3

The tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec.

 

Level-II:

 

Answer:11 Option C

 

Explanation:

Let the slower pipe alone fill the tank in x minutes.

Then, faster pipe will fill it in x minutes.
3

 

1 + 3 = 1
x x 36

 

4 = 1
x 36

x = 144 min.

 

 

 

Answer:12 Option D

 

Explanation:

Part filled by (A + B) in 1 minute = 1 + 1 = 1 .
60 40 24

Suppose the tank is filled in x minutes.

Then, x 1 + 1 = 1
2 24 40

 

x x 1 = 1
2 15

x = 30 min.

 

Answer:13 Option B

 

Explanation:

Time taken by one tap to fill half of the tank = 3 hrs.

Part filled by the four taps in 1 hour = 4 x 1 = 2 .
6 3

 

Remaining part = 1 – 1 = 1 .
2 2

 

2 : 1 :: 1 : x
3 2

 

 x = 1 x 1 x 3 = 3 hours i.e., 45 mins.
2 2 4

So, total time taken = 3 hrs. 45 mins.

 

Answer:14 Option C

 

Explanation:

(A + B)’s 1 hour’s work = 1 + 1 = 9 = 3 .
12 15 60 20

 

(A + C)’s hour’s work = 1 + 1 = 8 = 2 .
12 20 60 15

 

Part filled in 2 hrs = 3 + 2 = 17 .
20 15 60

 

Part filled in 6 hrs = 3 x 17 = 17 .
60 20

 

Remaining part = 1 – 17 = 3 .
20 20

 

Now, it is the turn of A and B and 3 part is filled by A and B in 1 hour.
20

Total time taken to fill the tank = (6 + 1) hrs = 7 hrs.

 

Answer:15 Option C

 

Explanation:

Part filled in 2 hours = 2 = 1
6 3

 

Remaining part = 1 – 1 = 2 .
3 3

 

 (A + B)’s 7 hour’s work = 2
3

 

(A + B)’s 1 hour’s work = 2
21

C’s 1 hour’s work = { (A + B + C)’s 1 hour’s work } – { (A + B)’s 1 hour’s work }

   = 1 – 2 = 1
6 21 14

C alone can fill the tank in 14 hours.

 

Answer:16 Option E

 

Explanation:

  1. Time taken to fill the cistern without leak = 9 hours.
Part of cistern filled without leak in 1 hour = 1
9
  1. Time taken to fill the cistern in presence of leak = 10 hours.
Net filling in 1 hour = 1
10

 

Work done by leak in 1 hour = 1 – 1 = 1
9 10 90

Leak will empty the full cistern in 90 hours.

Clearly, both I and II are necessary to answer the question.

Correct answer is (E).

 

 

 

 

Answer:17 Option E

 

Explanation:

 I. A’s 1 minute’s filling work = 1
16

 

II. B’s 1 minute’s filling work = 1
8

 

(A + B)’s 1 minute’s emptying work = 1 – 1 = 1
8 16 16

Tank will be emptied in 16 minutes.

Thus, both I and II are necessary to answer the question.

Correct answer is (E).

 

Answer:18 Option B

 

Explanation:

  II. Part of the tank filled by A in 1 hour = 1
4

 

III. Part of the tank filled by B in 1 hour = 1
6

 

(A + B)’s 1 hour’s work = 1 + 1 = 5
4 6 12

 

 A and B will fill the tank in 12 hrs = 2 hrs 24 min.
5

So, II and III are needed.

Correct answer is (B).

SQUARE ROOT & CUBE ROOTS

by

Square Root & Cube Root

 

Step 1: First of all group the number in pairs of 2 starting from the right.

 

Step 2: To get the ten’s place digit, Find the nearest square (equivalent or greater than or less than) to the first grouped pair from left and put the square root of the square.

 

Step 3: To get the unit’s place digit of the square root

 

Remember the following

If number ends in Unit’s place digit of the square root
1 1 or 9(10-1)
4 2 or 8(10-2)
9 3 or 7(10-3)
6 4or 6(10-4)
5 5
0 0

 

Lets see the logic behind this for a better understanding

We know,

12=1

22=4

32=9

42=16

52=25

62=36

72=49

82=64

92=81

102=100

 

Now, observe the unit’s place digit of all the squares.

Do you find anything common?

 

We notice that,

Unit’s place digit of both 12 and 92 is 1.

Unit’s place digit of both 22 and 82 is 4

Unit’s place digit of both 32 and 72 is 9

Unit’s place digit of both 42 and 62 is 6.


Step 4: Multiply the ten’s place digit (found in step 1) with its consecutive number and compare the result obtained with the first pair of the original number from left.

 

Remember,

If first pair of the original number > Result obtained on multiplication then  select the greater number  out of the two numbers as the unit’s place digit of the square root.

 

If firstpair of the original number < the result obtained on multiplication,then select the lesser number out of the two numbers as the unit’s place digit of the square root.

 

 

Let us consider an example to get a better understanding of the method

 

 

Example 1: √784=?

Step 1: We start by grouping the numbers in pairs of two from right as follows

7 84

 

Step 2: To get the ten’s place digit,

We find that nearest square to first group (7) is 4 and √4=2

Therefore ten’s place digit=2

 

Step 3: To get the unit’s place digit,

We notice that the number ends with 4, So the unit’s place digit of the square root should be either 2 or 8(Refer table).

 

Step 4: Multiplying the ten’s place digit of the square root that we arrived at in step 1(2) and its consecutive number(3) we get,

2×3=6
ten’s place digit of original number > Multiplication result
7>6
So we need to select the greater number (8) as the unit’s place digit of the square root.
Unit’s place digit =8

Ans:√784=28

 

 

 

Cube roots of perfect cubes

It may take two-three minutes to find out cube root of a perfect cube by using conventional method. However we can find out cube roots of perfect cubes very fast, say in one-two seconds using Vedic Mathematics.

We need to remember some interesting properties of numbers to do these quick mental calculations which are given below.

 

Points to remember  for speedy  calculation of cube roots

  1. To calculate cube root of any perfect cube quickly, we need to remember the cubes of 1 to 10 which is given below.
13 = 1
23 = 8
33 = 27
43 = 64
53 = 125
63 = 216
73 = 343
83 = 512
93 = 729
103 = 1000
  1. From the above cubes of 1 to 10, we need to remember an interesting property.
13 = 1 => If last digit of the perfect cube = 1, last digit of the cube root = 1
23 = 8 => If last digit of the perfect cube = 8, last digit of the cube root = 2
33 = 27 => If last digit of the perfect cube = 7, last digit of the cube root = 3
43 = 64 => If last digit of the perfect cube = 4, last digit of the cube root = 4
53 = 125 => If last digit of the perfect cube =5, last digit of the cube root = 5
63 = 216 => If last digit of the perfect cube = 6, last digit of the cube root = 6
73 = 343 => If last digit of the perfect cube = 3, last digit of the cube root = 7
83 = 512 => If last digit of the perfect cube = 2, last digit of the cube root = 8
93 = 729 => If last digit of the perfect cube = 9, last digit of the cube root = 9
103 = 1000 => If last digit of the perfect cube = 0, last digit of the cube root = 0

 

It’s very easy to remember the relations given above because

1 -> 1 (Same numbers)
8 -> 2 (10’s complement of 8 is 2 and 8+2 = 10)
7 -> 3 (10’s complement of 7 is 3 and 7+3 = 10)
4 -> 4 (Same numbers)
5 -> 5 (Same numbers)
6 -> 6 (Same numbers)
3 -> 7 (10’s complement of 3 is 7 and 3+7 = 10)
2 -> 8 (10’s complement of 2 is 8 and 2+8 = 10)
9 -> 9 (Same numbers)
0 -> 0 (Same numbers)

 

Also see
8 ->  2 and 2 ->  8
7 -> 3 and 3-> 7

 

 

 

 

 

Questions

Level-I

1. The cube root of .000216 is:
A. .6
B. .06
C. 77
D. 87

 

 

2.
What should come in place of both x in the equation x = 162 .
128 x
A. 12
B. 14
C. 144
D. 196

 

3. The least perfect square, which is divisible by each of 21, 36 and 66 is:
A. 213444
B. 214344
C. 214434
D. 231444

 

4. 1.5625 = ?
A. 1.05
B. 1.25
C. 1.45
D. 1.55

 

5. If 35 + 125 = 17.88, then what will be the value of 80 + 65 ?
A. 13.41
B. 20.46
C. 21.66
D. 22.35
 

 

6.

  

 

If a = 0.1039, then the value of 4a2 – 4a + 1 + 3a is:
A. 0.1039
B. 0.2078
C. 1.1039
D. 2.1039

 

7.
If x = 3 + 1 and y = 3 – 1 , then the value of (x2 + y2) is:
3 – 1 3 + 1
A. 10
B. 13
C. 14
D. 15

 

8. A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to Rs. 59.29, the number of the member is the group is:
A. 57
B. 67
C. 77
D. 87

 

9. The square root of (7 + 35) (7 – 35) is
A. 5
B. 2
C. 4
D. 35

 

 

 

 

10.
If 5 = 2.236, then the value of 5 – 10 + 125 is equal to:
2 5
A. 5.59
B. 7.826
C. 8.944
D. 10.062

 

 

 

Level-II

 
11.
625 x 14 x 11 is equal to:
11 25 196
A. 5
B. 6
C. 8
D. 11

 

12. 0.0169 x ? = 1.3
A. 10
B. 100
C. 1000
D. None of these

 

13.
3 – 1 2 simplifies to:
3
A.
3
4
B.
4
3
C.
4
3
D. None of these

 

14. How many two-digit numbers satisfy this property.: The last digit (unit’s digit) of the square of the two-digit number is 8 ?
A. 1
B. 2
C. 3
D. None of these

 

15. The square root of 64009 is:
A. 253
B. 347
C. 363
D. 803

 

 

16. √29929 = ?
A. 173
B. 163
C. 196
D. 186

 

 

 

 

 

 

17. √106.09 = ?
A. 10.6
B. 10.5
C. 10.3
D. 10.2
 
 

 

 

18.  ?/√196 = 5
A. 76
B. 72
C. 70
D. 75
 
 

 

Answers

Level-I

 

Answer:1 Option B

 

Explanation:

(.000216)1/3 = 216 1/3
106

 

   = 6 x 6 x 6 1/3
102 x 102 x 102

 

   = 6
102

 

   = 6
100

= 0.06

 

Answer:2 Option A

 

Explanation:

Let x = 162
128 x

Then x2 = 128 x 162

= 64 x 2 x 18 x 9

= 82 x 62 x 32

= 8 x 6 x 3

= 144.

x = 144 = 12.

 

Answer:3 Option A

 

Explanation:

L.C.M. of 21, 36, 66 = 2772.

Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11

To make it a perfect square, it must be multiplied by 7 x 11.

So, required number = 22 x 32 x 72 x 112 = 213444

 

Answer:4 Option B

 

Explanation:

1|1.5625( 1.25

|1

|——-

22| 56

| 44

|——-

245| 1225

| 1225

|——-

|    X

|——-

1.5625 = 1.25.

 

 

Answer:5 Option D

 

Explanation:

35 + 125 = 17.88

35 + 25 x 5 = 17.88

35 + 55 = 17.88

85 = 17.88

5 = 2.235

80 + 65 = 16 x 5 + 65

= 45 + 65

= 105 = (10 x 2.235) = 22.35

 

 

 

Answer:6 Option C

 

Explanation:

4a2 – 4a + 1 + 3a = (1)2 + (2a)2 – 2 x 1 x 2a + 3a

= (1 – 2a)2 + 3a

= (1 – 2a) + 3a

= (1 + a)

= (1 + 0.1039)

= 1.1039

 

Answer:7 Option C

 

Explanation:

x = (3 + 1) x (3 + 1) = (3 + 1)2 = 3 + 1 + 23 = 2 + 3.
(3 – 1) (3 + 1) (3 – 1) 2

 

y = (3 – 1) x (3 – 1) = (3 – 1)2 = 3 + 1 – 23 = 2 – 3.
(3 + 1) (3 – 1) (3 – 1) 2

x2 + y2 = (2 + 3)2 + (2 – 3)2

= 2(4 + 3)

= 14

 

Answer:8 Option C

 

Explanation:

Money collected = (59.29 x 100) paise = 5929 paise.

Number of members = 5929 = 77

 

 

Answer:9 Option B

 

Explanation:

(7 + 35)(7 – 35) = (7)2 – (35)2  = 49 – 45  = 4  = 2

 

 

Answer:10 Option B

 

Explanation:

5 – 10 + 125 = (5)2 – 20 + 25 x 55
2 5 25

 

= 5 – 20 + 50
25

 

= 35 x 5
25 5

 

= 355
10

 

= 7 x 2.236
2

 

= 7 x 1.118

 

= 7.826

 

 

Level-II

Answer:11 Option A

 

Explanation:

Given Expression = 25 x 14 x 11 = 5.
11 5 14

 

 

 

Answer:12 Option B

 

Explanation:

Let 0.0169 x x = 1.3.

Then, 0.0169x = (1.3)2 = 1.69

 x = 1.69 = 100
0.0169

 

 

 

Answer:13 Option C

 

Explanation:

3 – 1 2 = (3)2 + 1 2 – 2 x 3 x 1
3 3 3

 

= 3 + 1 – 2
3

 

= 1 + 1
3

 

= 4
3

 

 

 

Answer:14 Option D

 

Explanation:

A number ending in 8 can never be a perfect square.

 

 

Answer:15 Option A

 

Explanation:

2 |64009( 253      |4      |———-45  |240      |225      |———-503| 1509      |  1509      |———-      |     X      |———-

64009 = 253.

 

 

Answer:16 Option A

 

Explanation:
√29929 = So, √29929 = 173

 

 

Answer:17 Option C

 

Answer:18 Option C

SURDS

by

Surds

A surd is a square root which cannot be reduced to a rational number.

For example,  is not a surd.

However  is a surd.

If you use a calculator, you will see that  and we will need to round the answer correct to a few decimal places. This makes it less accurate.

If it is left as , then the answer has not been rounded, which keeps it exact.

Here are some general rules when simplifying expressions involving surds.

 

 

 

  1. amx an = am + n
am = am – n
an
   
  • (am)n= amn

 

  1. (ab)n= anbn

 

a n = an
b bn
           
  1. a0= 1

 

 

Questions

Level-I

 

1. (17)3.5 x (17)? = 178
A. 2.29
B. 2.75
C. 4.25
D. 4.5

 

2.
If a x – 1 = b x – 3 , then the value of x is:
b a
A.
1
2
B. 1
C. 2
D.
7
2

 

3. Given that 100.48 = x, 100.70 = y and xz = y2, then the value of z is close to:
A. 1.45
B. 1.88
C. 2.9
D. 3.7

 

4. If 5a = 3125, then the value of 5(a – 3) is:
A. 25
B. 125
C. 625
D. 1625

 

5. If 3(x – y) = 27 and 3(x + y) = 243, then x is equal to:
A. 0
B. 2
C. 4
D. 6

 

.6. (256)0.16 x (256)0.09 = ?
A. 4
B. 16
C. 64
D. 256.25

 

7. The value of [(10)150 ÷ (10)146]
A. 1000
B. 10000
C. 100000
D. 106

 

8.
1  + 1 + 1 = ?
1 + x(b – a) + x(c – a) 1 + x(a – b) + x(c – b) 1 + x(b – c) + x(a – c)
A. 0
B. 1
C. xa – b – c
D. None of these

 

9. (25)7.5 x (5)2.5 ÷ (125)1.5 = 5?
A. 8.5
B. 13
C. 16
D. 17.5
E. None of these

 

10. (0.04)-1.5 = ?
A. 25
B. 125
C. 250
D. 625

 

 

Level-II

 

11.
(243)n/5 x 32n + 1 = ?
9n x 3n – 1
A. 1
B. 2
C. 9
D. 3n

 

12.
1 + 1 = ?
1 + a(n – m) 1 + a(m – n)
A. 0
B.
1
2
C. 1
D. am + n

 

13. If m and n are whole numbers such that mn = 121, the value of (m – 1)n + 1 is:
A. 1
B. 10
C. 121
D. 1000

 

14.
xb (b + c – a) . xc (c + a – b) . xa (a + b – c) = ?
xc xa xb
A. xabc
B. 1
C. xab + bc + ca
D. xa + b + c

 

  1. If 5√5 * 53÷ 5-3/2= 5a+2 , the value of a is:
    A. 4
    B. 5
    C. 6
    D. 8
 

16.(132)7 ×(132)? =(132)11.5.

A. 3
B. 3.5
C. 4
D. 4.5

 

 

17. (ab)x−2=(ba)x−7. What is the value   of x ?

 

A. 3
B. 4
C. 3.5
D. 4.5

 

 

 

18. (0.04)-2.5 = ?

 

A. 125
B. 25
C. 3125
D. 625

 

 

 

  
 

Answers

Level-I

Answer:1 Option D

 

Explanation:

Let (17)3.5 x (17)x = 178.

Then, (17)3.5 + x = 178.

3.5 + x = 8

x = (8 – 3.5)

x = 4.5

 

Answer:2 Option C

 

Explanation:

Given a x – 1 = b x – 3
b a

 

a x – 1 = a -(x – 3)  = a (3 – x)
b b b

x – 1 = 3 – x

2x = 4

x = 2.

 

 

Answer:3 Option C

 

Explanation:

xz = y2        10(0.48z) = 10(2 x 0.70) = 101.40

0.48z = 1.40

 z = 140 = 35 = 2.9 (approx.)
48 12

 

Answer:4 Option A

 

Explanation:

5a = 3125        5a = 55

a = 5.

5(a – 3) = 5(5 – 3) = 52 = 25.

 

 

Answer:5 Option C

 

Explanation:

3x – y = 27 = 33        x – y = 3 ….(i)

3x + y = 243 = 35        x + y = 5 ….(ii)

On solving (i) and (ii), we get x = 4

 

 

Answer:6 Option A

 

Explanation:

(256)0.16 x (256)0.09 = (256)(0.16 + 0.09)

= (256)0.25

= (256)(25/100)

= (256)(1/4)

= (44)(1/4)

= 44(1/4)

= 41

= 4

Answer:7 Option B

 

Explanation:

(10)150 ÷ (10)146 = 10150
10146

= 10150 – 146

= 104

= 10000.

 

Answer:8 Option B

 

Explanation:

Given Exp. =
1  + 1  + 1
1 + xb + xc
xa xa
1 + xa + xc
xb xb
1 + xb + xa
xc xc

 

   = xa + xb + xc
(xa + xb + xc) (xa + xb + xc) (xa + xb + xc)

 

   = (xa + xb + xc)
(xa + xb + xc)

= 1.

 

Answer:9 Option B

 

Explanation:

Let (25)7.5 x (5)2.5 ÷ (125)1.5 = 5x.

Then, (52)7.5 x (5)2.5 = 5x
(53)1.5

 

5(2 x 7.5) x 52.5 = 5x
5(3 x 1.5)

 

515 x 52.5 = 5x
54.5

5x = 5(15 + 2.5 – 4.5)

5x = 513

x = 13.

 

Answer:10 Option B

 

Explanation:

(0.04)-1.5 = 4 -1.5
100

 

   = 1 -(3/2)
25

= (25)(3/2)

= (52)(3/2)

= (5)2 x (3/2)

= 53

= 125.

 

Level-II

 

Answer:11 Option C

 

Explanation:

Given Expression
= (243)(n/5) x 32n + 1
9n x 3n – 1
= (35)(n/5) x 32n + 1
(32)n x 3n – 1
= (35 x (n/5) x 32n + 1)
(32n x 3n – 1)
= 3n x 32n + 1
32n x 3n – 1
= 3(n + 2n + 1)
3(2n + n – 1)
= 33n + 1
33n – 1
= 3(3n + 1 – 3n + 1)   = 32   = 9.

Answer:12 Option C

 

Explanation:

1 + 1 =
1  + 1
1 + an
am
1 + am
an
1 + a(n – m) 1 + a(m – n)

 

   = am + an
(am + an) (am + an)

 

   = (am + an)
(am + an)

= 1.

 

Answer:13 Option D

 

Explanation:

We know that 112 = 121.

Putting m = 11 and n = 2, we get:

(m – 1)n + 1 = (11 – 1)(2 + 1) = 103 = 1000.

 

Answer:14 Option B

 

Explanation:

Given Exp.  

= x(b – c)(b + c – a) . x(c – a)(c +a – b) . x(a – b)(a + b – c)
= x(b – c)(b + c) – a(b – c)  .  x(c – a)(c + a) – b(c – a)
.  x(a – b)(a + b) – c(a – b)
= x(b2 – c2 + c2 – a2 + a2 – b2)  .   x–a(b – c) – b(c – a) – c(a – b)
= (x0 x x0)
= (1 x 1) = 1.

 

Answer:15 option C

 

Answer:16

Explanation

am.an=am+n

(132)7 × (132)x = (132)11.5

=> 7 + x = 11.5

=> x = 11.5 – 7 = 4.5

 

 

Answer:17

Explanation:

an=1a−n

(ab)x−2=(ba)x−7⇒(ab)x−2=(ab)−(x−7)⇒x−2=−(x−7)⇒x−2=−x+7⇒x−2=−x+7⇒2x=9⇒x=92=4.5

 

Answer:18

Explanation:

a−n=1/an

(0.04)−2.5=(1/.04)2.5=(100/4)2.5=(25)2.5=(52)2.5=(52)(5/2)=55=3125

 

SIMPLIFICATION

by

Simplification

Simplification is one of the most important part of Quantitative Aptitude section of any competitive exam. Today I am sharing all the techniques to solve Simplification questions quickly.

Rules of Simplification

V → Vinculum

B → Remove Brackets – in the order ( ) , { }, [ ]

O → Of
D → Division

M → Multiplication

A → Addition

S → Subtraction

 

Classification

Types Description
Natural Numbers: all counting numbers ( 1,2,3,4,5….∞)
Whole Numbers: natural number + zero( 0,1,2,3,4,5…∞)
Integers: All whole numbers including Negative number + Positive number(∞……-4,-3,-2,-1,0,1,2,3,4,5….∞)
Even & Odd Numbers : All whole number divisible by 2 is Even (0,2,4,6,8,10,12…..∞) and which does not divide by 2 are Odd (1,3,5,7,9,11,13,15,17,19….∞)
Prime Numbers: It can be positive or negative except 1, if the number is not divisible by any number except the number itself.(2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61….∞)
Composite Numbers: Natural numbers which are not prime
Co-Prime: Two natural number a and b are said to be co-prime if their HCF is 1.

 

Divisibility

Numbers IF A Number Examples
Divisible by 2 End with 0,2,4,6,8 are divisible by 2 254,326,3546,4718 all are divisible by 2
Divisible by 3 Sum of its digits  is divisible by 3 375,4251,78123 all are divisible by 3.  [549=5+4+9][5+4+9=18]18 is divisible by 3  hence 549 is divisible by 3.
Divisible by 4 Last two digit divisible by 4 5648 here last 2 digits are 48 which is divisible by 4 hence 5648 is also divisible by 4.
Divisible by 5 Ends with 0 or 5 225 or 330 here last digit digit is 0 or 5 that mean both the numbers are divisible by 5.
Divisible by 6 Divides by Both 2 & 3 4536 here last digit is 6 so it divisible by 2 & sum of its digit (like 4+5+3+6=18) is 18 which is divisible by 3.Hence 4536 is divisible by 6.
Divisible by 8 Last 3 digit divide by 8 746848 here last 3 digit 848 is divisible by 8 hence 746848 is also divisible by 8.
Divisible by 10 End with 0 220,450,1450,8450 all numbers has a last digit zero it means all are divisible by 10.
Divisible by 11 [Sum of its digit in
odd places-Sum of its digits
in even places]= 0 or multiple of 11		 Consider the number 39798847

(Sum of its digits at odd places)-(Sum of its digits at even places)(7+8+9+9)-(4+8+7+3)

(23-12)

23-12=11, which is divisible by 11. So 39798847 is divisible by 11.

Division & Remainder Rules

Suppose we divide 45 by 6

 

hence ,represent it as:

dividend = ( divisor✘quotient ) + remainder

or

divisior= [(dividend)-(remainder] / quotient

could be write it as

x = kq + r where (x = dividend,k = divisor,q = quotient,r = remainder)

 

 

Rules

  1. Modulus of a Real Number:

Modulus of a real number a is defined as

|a| = a, if a > 0
–a, if a < 0

Thus, |5| = 5 and |-5| = -(-5) = 5.

  1. Virnaculum (or Bar):

When an expression contains Virnaculum, before applying the ‘BODMAS’ rule, we simplify the expression under the Virnaculum.

 

 

Example:

On dividing a certain number by 342, we get 47 as remainder. If the same number is divided by 18, what will be the remainder ?

Number = 342k + 47

( 18 ✘19k ) + ( 18 ✘2 ) + 11

18 ✘( 19k + 2 ) +11.

Remainder = 11

 

Sum Rules

(1+2+3+………+n) = 1/2 n(n+1)

(12+22+32+………+n2) = 1/6 n (n+1) (2n+1)

(13+23+33+………+n3) = 1/4 n2 (n+1)2

 

Questions:

Level-I:

 

1. A man has Rs. 480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?
A. 45
B. 60
C. 75
D. 90

 

2. There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is:
A. 20
B. 80
C. 100
D.  

200

 

3. The price of 10 chairs is equal to that of 4 tables. The price of 15 chairs and 2 tables together is Rs. 4000. The total price of 12 chairs and 3 tables is:
A. Rs. 3500
B. Rs. 3750
C. Rs. 3840
D. Rs. 3900

 

4. If a – b = 3 and a2 + b2 = 29, find the value of ab.
A. 10
B. 12
C. 15
D. 18

 

5. The price of 2 sarees and 4 shirts is Rs. 1600. With the same money one can buy 1 saree and 6 shirts. If one wants to buy 12 shirts, how much shall he have to pay ?
A. Rs. 1200
B. Rs. 2400
C. Rs. 4800
D. Cannot be determined
E. None of these

 

 

6.

 A sum of Rs. 1360 has been divided among A, B and C such that A gets  of what B gets and B gets  of what C gets. B’s share is:
A. Rs. 120
B. Rs. 160
C. Rs. 240
D. Rs. 300

 

7. One-third of Rahul’s savings in National Savings Certificate is equal to one-half of his savings in Public Provident Fund. If he has Rs. 1,50,000 as total savings, how much has he saved in Public Provident Fund ?
A. Rs. 30,000
B. Rs. 50,000
C. Rs. 60,000
D. Rs. 90,000

 

8. A fires 5 shots to B’s 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed:
A. 30 birds
B. 60 birds
C. 72 birds
D. 90 birds

 

9. Eight people are planning to share equally the cost of a rental car. If one person withdraws from the arrangement and the others share equally the entire cost of the car, then the share of each of the remaining persons increased by:
A.
1
7
B.
1
8
C.
1
9
D.
7
8

 

10. To fill a tank, 25 buckets of water is required. How many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to two-fifth of its present ?
A. 10
B. 35
C. 62.5
D. Cannot be determined
E. None of these

 

Level-II:

1. In a regular week, there are 5 working days and for each day, the working hours are 8. A man gets Rs. 2.40 per hour for regular work and Rs. 3.20 per hours for overtime. If he earns Rs. 432 in 4 weeks, then how many hours does he work for ?
A. 160
B. 175
C. 180
D. 195

 

12. Free notebooks were distributed equally among children of a class. The number of notebooks each child got was one-eighth of the number of children. Had the number of children been half, each child would have got 16 notebooks. Total how many notebooks were distributed ?
A. 256
B. 432
C. 512
D. 640
E. None of these

 

13. A man has some hens and cows. If the number of heads be 48 and the number of feet equals 140, then the number of hens will be:
A. 22
B. 23
C. 24
D. 26

 

14.
(469 + 174)2 – (469 – 174)2 = ?
(469 x 174)
A. 2
B. 4
C. 295
D. 643

 

15. David gets on the elevator at the 11th floor of a building and rides up at the rate of 57 floors per minute. At the same time, Albert gets on an elevator at the 51st floor of the same building and rides down at the rate of 63 floors per minute. If they continue travelling at these rates, then at which floor will their paths cross ?
A. 19
B. 28
C. 30
D. 37

 

  1. Find the value of 1/(3+1/(3+1/(3-1/3)))
A.) 3/10 B.) 10/3
C.) 27/89 D.) 89/27

  1. Find the value of
A.) 3½ 99; B.) 34/99
C.) 2.131313 D.) 3.141414

 

18.Find the value of

((0.1)3 + (0.6)3 + (0.7)3 − (0.3)(0.6)(0.7))/((0.1)2 + (0.6)2 + (0.7)2 − 0.006 − 0.42 − 0.07)

 

A.) 14/10 B.) 1.35
C.) 13/10 D.) 0

 

 

 

 

  1. Solve(0.76 × 0.76 × 0.76 − 0.008)/(0.76 × 0.76 + 0.76 × 0.2 + 0.04)
A.) 0.56 B.) 0.65
C.) 0.54 D.) 0.45
   
  1. Find the value of
A.) 1.5 B.) -1.5
C.) 1 D.) 0

 

Answers:

Level-I

 

Answer:1 Option D

 

Explanation:

Let number of notes of each denomination be x.

Then x + 5x + 10x = 480

16x = 480

x = 30.

Hence, total number of notes = 3x = 90.

 

 

Answer:2 Option C

 

Explanation:

Let the number of students in rooms A and B be x and y respectively.

Then, x – 10 = y + 10      x – y = 20 …. (i)

and x + 20 = 2(y – 20)      x – 2y = -60 …. (ii)

Solving (i) and (ii) we get: x = 100 , y = 80.

The required answer A = 100.

 

 

Answer:3 Option D

 

Explanation:

Let the cost of a chair and that of a table be Rs. x and Rs. y respectively.

Then, 10x = 4y   or   y = 5 x.
2

15x + 2y = 4000

 15x + 2 x 5 x = 4000
2

20x = 4000

x = 200.

So, y = 5 x 200 = 500.
2

Hence, the cost of 12 chairs and 3 tables = 12x + 3y

= Rs. (2400 + 1500)

= Rs. 3900.

 

 

 

 

 

Answer:4 Option A

 

Explanation:

2ab = (a2 + b2) – (a – b)2

= 29 – 9 = 20

ab = 10.

 

 

 

Answer:5 Option B

 

Explanation:

Let the price of a saree and a shirt be Rs. x and Rs. y respectively.

Then, 2x + 4y = 1600 …. (i)

and x + 6y = 1600 …. (ii)

 

Divide equation (i) by 2, we get the below equation.

 

=> x +  2y =  800. — (iii)

 

Now subtract (iii) from (ii)

 

x +  6y = 1600  (-)

x +  2y =  800

—————-

4y =  800

—————-

 

Therefore, y = 200.

 

Now apply value of y in (iii)

 

=>  x + 2 x 200 = 800

 

=>  x + 400 = 800

 

Therefore x = 400

 

Solving (i) and (ii) we get x = 400, y = 200.

Cost of 12 shirts = Rs. (12 x 200) = Rs. 2400.

 

 

Answer:6 Option C

 

Explanation:

Let C’s share = Rs. x

Then, B’s share = Rs. x ,   A’s share = Rs. 2 x x = Rs. x
4 3 4 6

 

x + x + x = 1360
6 4

 

17x = 1360
12

 

 x = 1360 x 12 = Rs. 960
17

 

Hence, B’s share = Rs. 960 = Rs. 240.

 

 

Answer:7 Option C

 

Explanation:

Let savings in N.S.C and P.P.F. be Rs. x and Rs. (150000 – x) respectively. Then,

1 x = 1 (150000 – x)
3 2

 

x + x = 75000
3 2

 

5x = 75000
6

 

 x = 75000 x 6 = 90000
5

Savings in Public Provident Fund = Rs. (150000 – 90000) = Rs. 60000

 

 

Answer:8 Option A

 

Explanation:

Let the total number of shots be x. Then,

Shots fired by A = 5 x
8

 

Shots fired by B = 3 x
8

 

Killing shots by A = 1 of 5 x = 5 x
3 8 24

 

Shots missed by B = 1 of 3 x = 3 x
2 8 16

 

3x = 27 or x = 27 x 16 = 144.
16 3

 

Birds killed by A = 5x = 5 x 144 = 30.
24 24

 

 

 

 

 

 

 

Answer:9 Option A

 

Explanation:

Original share of 1 person = 1
8

 

New share of 1 person = 1
7

 

Increase = 1 – 1 = 1
7 8 56

 

 Required fraction = (1/56) = 1 x 8 = 1
(1/8) 56 1 7

 

 

Answer:10 Option C

 

Explanation:

Let the capacity of 1 bucket = x.

Then, the capacity of tank = 25x.

New capacity of bucket = 2 x
5

 

 Required number of buckets = 25x
(2x/5)

 

=  25x x  

5
2x

 

= 125
2

= 62.5

 

Level-II:

Answer:11 Option B

 

Explanation:

Suppose the man works overtime for x hours.

Now, working hours in 4 weeks = (5 x 8 x 4) = 160.

160 x 2.40 + x x 3.20 = 432

3.20x = 432 – 384 = 48

x = 15.

Hence, total hours of work = (160 + 15) = 175.

 

Answer:12 Option C

 

Explanation:

Let total number of children be x.

Then, x x 1 x = x x 16     x = 64.
8 2

 

 Number of notebooks = 1 x2 = 1 x 64 x 64 = 512

 

Answer:13 Option D

 

Explanation:

Let the number of hens be x and the number of cows be y.

Then, x + y = 48 …. (i)

and 2x + 4y = 140      x + 2y = 70 …. (ii)

Solving (i) and (ii) we get: x = 26, y = 22.

The required answer = 26.

 

Answer:14 Option B

 

Explanation:

Given exp. = (a + b)2 – (a – b)2
ab

 

   = 4ab
ab

= 4 (where a = 469, b = 174.)

 

Answer:15 Option C

 

Explanation:

Suppose their paths cross after x minutes.

Then, 11 + 57x = 51 – 63x        120x = 40

x = 1
3

 

Number of floors covered by David in (1/3) min. = 1 x 57 = 19.
3

So, their paths cross at (11 +19) i.e., 30th floor.

 

Answer:16 Option ‘C’

Explanation:

1/[3 + (1/(3+1/(3 – 1/3)))]

=> 1/[3 + 1/(3 + 1/(8/3))]

=> 1/[3 + 1/(3 + 3/8)]

=> 1/[3 + 8/27]

=> 1/(89/27)

=> 27/89

 

Answer:17 Option ‘D’

Explanation:

6/9 + 7/9 + 9/9 + 69/99

2/3 + 7/9 + 1 + 69/99

(66 + 77 + 99 + 69)/99

311/99 => 3.141414

 

Answer:18 Option ‘A’

Explanation:

((0.1)3 + (0.6)3 + (0.7)3 − (0.3)(0.6)(0.7))/((0.1)2 + (0.6)2 + (0.7)2 − 0.006 − 0.42 − 0.07)

=> (0.1 + 0.6 + 0.7)3/(0.1 + 0.6 + 0.7)2

=> 0.1 + 0.6 + 0.7 => 1.4 = 14/10

 

Answer:19 Option ‘A’

 

Answer:20 Option ‘D’

11/30 − [1/6 + 1/5 + [7/12 − 7/12]]

11/30 − [1/6 + 1/5 + [0]]

11/30 − [(5 + 6)/30]

11/30 − 11/30 = 0.

PROFIT & LOSS

by

Profit and loss

 

IMPORTANT FACTS

Cost Price:

The price, at which an article is purchased, is called its cost price, abbreviated as C.P.

 

Selling Price:

The price, at which an article is sold, is called its selling prices, abbreviated as S.P.

 

Profit or Gain:

If S.P. is greater than C.P., the seller is said to have a profit or gain.

 

Loss:

If S.P. is less than C.P., the seller is said to have incurred a loss.

 

IMPORTANT FORMULAE

  1. Gain = (S.P.) – (C.P.)
  2. Loss = (C.P.) – (S.P.)
  3. Loss or gain is always reckoned on C.P.
  4. Gain Percentage: (Gain %)
    Gain % = Gain x 100
C.P.
  1. Loss Percentage: (Loss %)
    Loss % = Loss x 100
C.P.
  1. Selling Price: (S.P.)
    SP = (100 + Gain %) x C.P
100
         
  1. Selling Price: (S.P.)
    SP = (100 – Loss %) x C.P.
100
  1. Cost Price: (C.P.)
    C.P. = 100 x S.P.
(100 + Gain %)
  1. Cost Price: (C.P.)
    C.P. = 100 x S.P.
(100 – Loss %)
  1. If an article is sold at a gain of say 35%, then S.P. = 135% of C.P.
  2. If an article is sold at a loss of say, 35% then S.P. = 65% of C.P.
  3. When a person sells two similar items, one at a gain of say x%, and the other at a loss of x%, then the seller always incurs a loss given by:
    Loss % = Common Loss and Gain % 2 = x 2 .
10 10
  1. If a trader professes to sell his goods at cost price, but uses false weights, then
    Gain % = Error x 100 %.
(True Value) – (Error)

 

Questions:

Level-I:

 

 

1. Alfred buys an old scooter for Rs. 4700 and spends Rs. 800 on its repairs. If he sells the scooter for Rs. 5800, his gain percent is:
A.
4 4 %
7
B.
5 5 %
11
C. 10%
D. 12%

 

2. The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of xis:
A. 15
B. 16
C. 18
D. 25

 

3. If selling price is doubled, the profit triples. Find the profit percent.
A.
66 2
3
B. 100
C.
105 1
3
D. 120

 

4. In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit?
A. 30%
B. 70%
C. 100%
D. 250%

 

 

5. A vendor bought toffees at 6 for a rupee. How many for a rupee must he sell to gain 20%?
A. 3
B. 4
C. 5
D. 6

 

6. The percentage profit earned by selling an article for Rs. 1920 is equal to the percentage loss incurred by selling the same article for Rs. 1280. At what price should the article be sold to make 25% profit?
A. Rs. 2000
B. Rs. 2200
C. Rs. 2400
D. Data inadequate

 

7. A shopkeeper expects a gain of 22.5% on his cost price. If in a week, his sale was of Rs. 392, what was his profit?
A. Rs. 18.20
B. Rs. 70
C. Rs. 72
D. Rs. 88.25

 

8. A man buys a cycle for Rs. 1400 and sells it at a loss of 15%. What is the selling price of the cycle?
A. Rs. 1090
B. Rs. 1160
C. Rs. 1190
D. Rs. 1202

 

9. Sam purchased 20 dozens of toys at the rate of Rs. 375 per dozen. He sold each one of them at the rate of Rs. 33. What was his percentage profit?
A. 3.5
B. 4.5
C. 5.6
D. 6.5

 

10. Some articles were bought at 6 articles for Rs. 5 and sold at 5 articles for Rs. 6. Gain percent is:
A. 30%
B.
33 1 %
3
C. 35%
D. 44%
 

 

 

 

 

 

11.

  

 

 

Level-II:

 

 

On selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls. The cost price of a ball is:
A. Rs. 45
B. Rs. 50
C. Rs. 55
D. Rs. 60

 

 

12. When a plot is sold for Rs. 18,700, the owner loses 15%. At what price must that plot be sold in order to gain 15%?
A. Rs. 21,000
B. Rs. 22,500
C. Rs. 25,300
D. Rs. 25,800

 

13. 100 oranges are bought at the rate of Rs. 350 and sold at the rate of Rs. 48 per dozen. The percentage of profit or loss is:
A.
14 2 % gain
7
B. 15% gain
C.
14 2 % loss
7
D. 15 % loss

 

14. A shopkeeper sells one transistor for Rs. 840 at a gain of 20% and another for Rs. 960 at a loss of 4%. His total gain or loss percent is:
A.
5 15 % loss
17
B.
5 15 % gain
17
C.
6 2 % gain
3
D. None of these

 

 

15. A trader mixes 26 kg of rice at Rs. 20 per kg with 30 kg of rice of other variety at Rs. 36 per kg and sells the mixture at Rs. 30 per kg. His profit percent is:
A. No profit, no loss
B. 5%
C. 8%
D. 10%
E. None of these

 

  1. A man buys an article for Rs. 27.50 and sells it for Rs 28.60. Find his gain percent
  2. 1%
  3. 2%
  4. 3%
  5. 4%

 

 

  1. A TV is purchased at Rs. 5000 and sold at Rs. 4000, find the lost percent.
  2. 10%
  3. 20%
  4. 25%
  5. 28%

 

 

  1. In terms of percentage profit, which among following the best transaction.
    1. P. 36, Profit 17
    2. P. 50, Profit 24
    3. P. 40, Profit 19
    4. P. 60, Profit 29

 

 

 

 

Answer:1 Option B

 

Explanation:

Cost Price (C.P.) = Rs. (4700 + 800) = Rs. 5500.

Selling Price (S.P.) = Rs. 5800.

Gain = (S.P.) – (C.P.) = Rs.(5800 – 5500) = Rs. 300.

Gain % = 300 x 100 % = 5 5 %
5500 11

 

Answer:2 Option B

 

Explanation:

Let C.P. of each article be Re. 1 C.P. of x articles = Rs. x.

S.P. of x articles = Rs. 20.

Profit = Rs. (20 – x).

20 – x x 100 = 25
x

2000 – 100x = 25x

125x = 2000

x = 16.

 

 

Answer:3 Option B

 

Explanation:

Let C.P. be Rs. x and S.P. be Rs. y.

Then, 3(y – x) = (2y – x)    y = 2x.

Profit = Rs. (y – x) = Rs. (2x – x) = Rs. x.

 Profit % = x x 100 % = 100%

 

 

Answer:4 Option B

 

Explanation:

Let C.P.= Rs. 100. Then, Profit = Rs. 320, S.P. = Rs. 420.

New C.P. = 125% of Rs. 100 = Rs. 125

New S.P. = Rs. 420.

Profit = Rs. (420 – 125) = Rs. 295.

 Required percentage = 295 x 100 % = 1475 % = 70% (approximately).
420 21

 

 

Answer:5 Option C

 

Explanation:

C.P. of 6 toffees = Re. 1

S.P. of 6 toffees = 120% of Re. 1 = Rs. 6
5

 

For Rs. 6 , toffees sold = 6.
5

 

For Re. 1, toffees sold = 6 x 5 = 5.
6

 

 

Answer:6 Option A

 

Explanation:

Let C.P. be Rs. x.

Then, 1920 – x x 100 = x – 1280 x 100
x x

1920 – x = x – 1280

2x = 3200

x = 1600

 Required S.P. = 125% of Rs. 1600 = Rs. 125 x 1600 = Rs 2000.
100

 

 

Answer:7 Option C

 

Explanation:

C.P. = Rs. 100 x 392 = Rs. 1000 x 392 = Rs. 320
122.5 1225

Profit = Rs. (392 – 320) = Rs. 72.

 

Answer:8 Option C

 

Explanation:

S.P. = 85% of Rs. 1400 = Rs. 85 x 1400 = Rs. 1190
100

 

 

 

Answer:9 Option C

 

Explanation:

Cost Price of 1 toy = Rs. 375 = Rs. 31.25
12

Selling Price of 1 toy = Rs. 33

So, Gain = Rs. (33 – 31.25) = Rs. 1.75

 Profit % = 1.75 x 100 % = 28 % = 5.6%
31.25 5

 

 

 

Answer:10 Option D

 

Explanation:

Suppose, number of articles bought = L.C.M. of 6 and 5 = 30.

C.P. of 30 articles = Rs. 5 x 30 = Rs. 25.
6

 

S.P. of 30 articles = Rs. 6 x 30 = Rs. 36.
5

 

 Gain % = 11 x 100 % = 44%.
25

 

 

Answer:11 Option D

 

Explanation:

(C.P. of 17 balls) – (S.P. of 17 balls) = (C.P. of 5 balls)

C.P. of 12 balls = S.P. of 17 balls = Rs.720.

 C.P. of 1 ball = Rs. 720 = Rs. 60.
12

 

 

Answer:12 Option C

 

Explanation:

85 : 18700 = 115 : x

 x = 18700 x 115 = 25300.
85

Hence, S.P. = Rs. 25,300.

 

Answer:13 Option A

 

Explanation:

C.P. of 1 orange = Rs. 350 = Rs. 3.50
100

 

S.P. of 1 orange = Rs. 48 = Rs. 4
12

 

 Gain% = 0.50 x 100 % = 100 % = 14 2 %
3.50 7 7

 

 

 

Answer:14 Option B

 

Explanation:

C.P. of 1st transistor = Rs. 100 x 840 = Rs. 700.
120

 

C.P. of 2nd transistor = Rs. 100 x 960 = Rs. 1000
96

So, total C.P. = Rs. (700 + 1000) = Rs. 1700.

Total S.P. = Rs. (840 + 960) = Rs. 1800.

 Gain % = 100 x 100 % = 5 15 %
1700 17

 

 

 

Answer:15 Option B

 

Explanation:

C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600.

S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680.

 Gain = 80 x 100 % = 5%.
1600

 

Answer:16 Option D

 

Explanation:

So we have C.P. = 27.50
S.P. = 28.60

Gain = 28.60 – 27.50 = Rs. 1.10

Gain%=(Gain/Cost∗100)%=(1.10/27.50∗100)%=4%

 

 

 

 

Answer:17 Option B

 

Explanation:

We know, C.P. = 5000
S.P. = 4000
Loss = 5000 – 4000 = 1000
Loss%=(Loss/Cost∗100)%=(1000/5000∗100)%=20%

 

 

Answer:18 Option D

 

Explanation:

Hint: Calculate profit percent as

Profit% = (profit/cost) * 100

AGE PROBLEMS

by

Age Problems

 

Important Formulas on “Problems on Ages”:

 

  1. If the current age is x, then ntimes the age is nx.
  2. If the current age is x, then age nyears later/hence = x+ n.
  3. If the current age is x, then age nyears ago = x– n.
  4. The ages in a ratio a: bwill be ax and bx.
5. If the current age is x, then 1 of the age is x .
n n

Example:

A problem with one variable: How old is Al?

Many single-variable algebra word problems have to do with the relations between different people’s ages. For example:

Al’s father is 45. He is 15 years older than twice Al’s age. How old is Al?

We can begin by assigning a variable to what we’re asked to find. Here this is Al’s age, so let Al’s age = x.

We also know from the information given in the problem that 45 is 15 more than twice Al’s age. How can we translate this from words into mathematical symbols? What is twice Al’s age?

Well, Al’s age is x, so twice Al’s age is 2x, and 15 more than twice Al’s age is 15 + 2x.That equals 45, right? Now we have an equation in terms of one variable that we can solve for x: 45 = 15 + 2x.

original statement of the problem: 45 = 15 + 2x
subtract 15 from each side: 30 = 2x
divide both sides by 2: 15 = x

Since x is Al’s age and x = 15, this means that Al is 15 years old.

It’s always a good idea to check our answer:

twice Al’s age is 2 x 15: 30
15 more than 30 is 15 + 30: 45

This should be the age of Al’s father, and it is.

 

 

Questions:

Level-I:

 

1. Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit’s age. After further 8 years, how many times would he be of Ronit’s age?
A. 2 times
B.
2 1 times
2
C.
2 3 times
4
D. 3 times

 

2. The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
A. 4 years
B. 8 years
C. 10 years
D. None of these

 

3. A father said to his son, “I was as old as you are at the present at the time of your birth”. If the father’s age is 38 years now, the son’s age five years back was:
A. 14 years
B. 19 years
C. 33 years
D. 38 years

 

4. A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, the how old is B?
A. 7
B. 8
C. 9
D. 10
E. 11

 

5. Present ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Anand’s present age in years?
A. 24
B. 27
C. 40
D. Cannot be determined
E. None of these

 

6. A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of his son is:
A. 14 years
B. 18 years
C. 20 years
D. 22 years

 

7. Six years ago, the ratio of the ages of Kunal and Sagar was 6 : 5. Four years hence, the ratio of their ages will be 11 : 10. What is Sagar’s age at present?
A. 16 years
B. 18 years
C. 20 years
D. Cannot be determined
E. None of these

 

8. The sum of the present ages of a father and his son is 60 years. Six years ago, father’s age was five times the age of the son. After 6 years, son’s age will be:
A. 12 years
B. 14 years
C. 18 years
D. 20 years

 

9. At present, the ratio between the ages of Arun and Deepak is 4 : 3. After 6 years, Arun’s age will be 26 years. What is the age of Deepak at present ?
A. 12 years
B. 15 years
C. 19 and half
D. 21 years

 

10. Sachin is younger than Rahul by 7 years. If their ages are in the respective ratio of 7 : 9, how old is Sachin?
A. 16 years
B. 18 years
C. 28 years
D. 24.5 years
E. None of these

 

 

 

 

 

 

 

 

11.

  

Level-II:

 

 

 

 

The present ages of three persons in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years).
A. 8, 20, 28
B. 16, 28, 36
C. 20, 35, 45
D. None of these

 

12. Ayesha’s father was 38 years of age when she was born while her mother was 36 years old when her brother four years younger to her was born. What is the difference between the ages of her parents?
A. 2 years
B. 4 years
C. 6 years
D. 8 years

 

13. A person’s present age is two-fifth of the age of his mother. After 8 years, he will be one-half of the age of his mother. How old is the mother at present?
A. 32 years
B. 36 years
C. 40 years
D. 48 years

 

14. Q is as much younger than R as he is older than T. If the sum of the ages of R and T is 50 years, what is definitely the difference between R and Q’s age?
A. 1 year
B. 2 years
C. 25 years
D. Data inadequate
E. None of these

 

15. The age of father 10 years ago was thrice the age of his son. Ten years hence, father’s age will be twice that of his son. The ratio of their present ages is:
A. 5 : 2
B. 7 : 3
C. 9 : 2
D. 13 : 4

 

16.
What is Sonia’s present age?
I. Sonia’s present age is five times Deepak’s present age.
 II. Five years ago her age was twenty-five times Deepak’s age at that time.
A. I alone sufficient while II alone not sufficient to answer
B. II alone sufficient while I alone not sufficient to answer
C. Either I or II alone sufficient to answer
D. Both I and II are not sufficient to answer
E. Both I and II are necessary to answer

 

17.
Average age of employees working in a department is 30 years. In the next year, ten workers will retire. What will be the average age in the next year?
I. Retirement age is 60 years.
 II. There are 50 employees in the department.
A. I alone sufficient while II alone not sufficient to answer
B. II alone sufficient while I alone not sufficient to answer
C. Either I or II alone sufficient to answer
D. Both I and II are not sufficient to answer
E. Both I and II are necessary to answer

 

 

18.
Divya is twice as old as Shruti. What is the difference in their ages?
I. Five years hence, the ratio of their ages would be 9 : 5.
 II. Ten years back, the ratio of their ages was 3 : 1.
A. I alone sufficient while II alone not sufficient to answer
B. II alone sufficient while I alone not sufficient to answer
C. Either I or II alone sufficient to answer
D. Both I and II are not sufficient to answer
E. Both I and II are necessary to answer

 

 

 

Answers:

Level-I:

 

Answer:1 Option A

 

Explanation:

Let Ronit’s present age be x years. Then, father’s present age =(x + 3x) years = 4x years.

(4x + 8) = 5 (x + 8)
2

8x + 16 = 5x + 40

3x = 24

x = 8.

Hence, required ratio = (4x + 16) = 48 = 2.
(x + 16) 24

 

 

Answer:2 Option A

 

Explanation:

Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.

Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50

5x = 20

x = 4.

Age of the youngest child = x = 4 years.

 

 

 

Answer:3 Option A

 

Explanation:

Let the son’s present age be x years. Then, (38 – x) = x

2x = 38.

x = 19.

Son’s age 5 years back (19 – 5) = 14 years.

 

Answer:4 Option D

 

Explanation:

Let C’s age be x years. Then, B’s age = 2x years. A’s age = (2x + 2) years.

(2x + 2) + 2x + x = 27

5x = 25

x = 5.

Hence, B’s age = 2x = 10 years.

 

Answer:5 Option A

 

Explanation:

Let the present ages of Sameer and Anand be 5x years and 4x years respectively.

Then, 5x + 3 = 11
4x + 3 9

9(5x + 3) = 11(4x + 3)

45x + 27 = 44x + 33

45x – 44x = 33 – 27

x = 6.

Anand’s present age = 4x = 24 years.

 

Answer:6 Option D

 

Explanation:

Let the son’s present age be x years. Then, man’s present age = (x + 24) years.

(x + 24) + 2 = 2(x + 2)

x + 26 = 2x + 4

x = 22.

 

Answer:7 Option A

 

Explanation:

Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.

Then, (6x + 6) + 4 = 11
(5x + 6) + 4 10

10(6x + 10) = 11(5x + 10)

5x = 10

x = 2.

Sagar’s present age = (5x + 6) = 16 years.

 

Answer:8 Option D

 

Explanation:

Let the present ages of son and father be x and (60 –x) years respectively.

Then, (60 – x) – 6 = 5(x – 6)

54 – x = 5x – 30

6x = 84

x = 14.

Son’s age after 6 years = (x+ 6) = 20 years..

 

Answer:9 Option B

 

Explanation:

Let the present ages of Arun and Deepak be 4x years and 3x years respectively. Then,

4x + 6 = 26        4x = 20

x = 5.

Deepak’s age = 3x = 15 years.

 

Answer:10 Option D

 

Explanation:

Let Rahul’s age be x years.

Then, Sachin’s age = (x – 7) years.

x – 7 = 7
x 9

9x – 63 = 7x

2x = 63

x = 31.5

Hence, Sachin’s age =(x – 7) = 24.5 years.

 

Answer:11 Option B

 

Explanation:

Let their present ages be 4x, 7x and 9x years respectively.

Then, (4x – 8) + (7x – 8) + (9x – 8) = 56

20x = 80

x = 4.

Their present ages are 4x = 16 years, 7x = 28 years and 9x = 36 years respectively.

 

Answer:12 Option C

 

Explanation:

Mother’s age when Ayesha’s brother was born = 36 years.

Father’s age when Ayesha’s brother was born = (38 + 4) years = 42 years.

Required difference = (42 – 36) years = 6 years.

 

Answer:13 Option C

 

Explanation:

Let the mother’s present age be x years.

Then, the person’s present age = 2 x years.
5

 

2 x + 8 = 1 (x + 8)
5 2

2(2x + 40) = 5(x + 8)

x = 40.

 

Answer:14 Option D

 

Explanation:

Given that:

1. The difference of age b/w R and Q = The difference of age b/w Q and T.

2. Sum of age of R and T is 50 i.e. (R + T) = 50.

Question: R – Q = ?.

Explanation:

R – Q = Q – T

(R + T) = 2Q

Now given that, (R + T) = 50

So, 50 = 2Q and therefore Q = 25.

Question is (R – Q) = ?

Here we know the value(age) of Q (25), but we don’t know the age of R.

Therefore, (R-Q) cannot be determined.

 

Answer:15 Option B

 

Explanation:

Let the ages of father and son 10 years ago be 3x and x years respectively.

Then, (3x + 10) + 10 = 2[(x + 10) + 10]

3x + 20 = 2x + 40

x = 20.

Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.

 

 

Answer:16  Option E

 

Explanation:

 I. S = 5D     D = S ….(i)
5
  1. S – 5 = 25 (D – 5)    S = 25D – 120 ….(ii)
Using (i) in (ii), we get S = 25 x S – 120
5

4S = 120.

S = 30.

Thus, I and II both together give the answer. So, correct answer is (E).

 

Answer:17 Option E

 

Explanation:

  1. Retirement age is 60 years.
  2. There are 50 employees in the department.

Average age of 50 employees = 30 years.

Total age of 50 employees = (50 x 30) years = 1500 years.

Number of employees next year = 40.

Total age of 40 employees next year (1500 + 40 – 60 x 10) = 940.

Average age next year = 940 years = 23 1 years.
40 2

Thus, I and II together give the answer. So, correct answer is (E).

 

Answer:18   Option C

 

Explanation:

Let Divya’s present age be D years and Shruti’s present age b S years

Then, D = 2 x S        D – 2S = 0 ….(i)

 I. D + 5 = 9 ….(ii)
S + 5 5

 

II. D – 10 = 3 ….(iii)
S – 10 1

From (ii), we get : 5D + 25 = 9S + 45        5D – 9S = 20 ….(iv)

From (iii), we get : D – 10 = 3S – 30        D – 3S = -20 ….(v)

Thus, from (i) and (ii), we get the answer.

Also, from (i) and (iii), we get the answer.

I alone as well as II alone give the answer. Hence, the correct answer is (C).