PSC Notes

SURDS

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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

 

Scarcity of water, methods of conservation-rain water harvesting and watershed management, ground water management

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Scarcity of water

 

Water scarcity is possibly to pose the greatest challenge on account of its increased demand coupled with shrinking supplies due to over utilisation and pollution. Water is a cyclic resource with abundant supplies on the globe. Approximately, 71 per cent of the earth’s surface is covered with it but fresh water constitutes only about 3 per cent of the total water. In fact, a very small proportion of fresh water is effectively available for human use. The availability of fresh water varies over space and time.

According to the United Nation Developement Program ,occurrence of water availability at about 1000 cubic meters per capita per annum is a commonly threshold for water indicating scarcity.

Krishna, Cauvery, Subernarekha, Pennar, Mahi, Sabarmati, Tapi, East Flowing Rivers and West Flowing Rivers of Kutch and Saurashtra including Luni are some of the basins, which fall below the 1000 cubic meter mark- out of which Cauvery, Pennar, Sabarmati and East Flowing rivers and West Flowing Rivers of Kutch and Saurashtra including Luni facing more acute water scarcity with per capita availability of water less than or around 500 cu m.

The need of the hour to change the condition of water scarity are as follows:-

  • The need to change cropping patterns based on scientific advice,
  • use of drip and sprinkler irrigation,
  • fertigation for increasing water use efficiency,
  • community participation, especially women, for better water management
  • Use of treated urban waste water to be used for farming in the adjoining areas
  • desilting of rivers
  • recharging of rivers,
  • check dams and other water storage mechanisms.

Rain water harvesting

 

Rain water harvesting generally means collection of rain water. Its special meaning is a technique of recharging of underground water. In this technique water is made to go underground after collecting rain water locally, without polluting the same.

Rain water harvesting is a low cost and eco-friendly technique for preserving every drop of water by guiding the rain water to bore well, pits and wells. Rainwater harvesting increases water availability, checks the declining ground water table, improves the quality of groundwater through dilution of contaminants like fluoride and nitrates, prevents soil erosion, and flooding and arrests salt water intrusion in coastal areas if used to recharge aquifers.

Rainwater is relatively clean and the quality is usually acceptable for many purposes with little or even no treatment. The physical and chemical properties of rainwater are usually superior to sources of groundwater that may have been subjected to contamination. Rainwater harvesting can co‐exist with and provide a good supplement to other water sources and utility systems, thus relieving pressure on other water sources. Rainwater harvesting provides a water supply buffer for use in times of emergency or breakdown of the public water supply systems, particularly during natural disasters.

Watershed management

 

The term watershed refers to a “contiguous area draining into a single water body or a water course” or “it is a topographical area having a common drainage”. This means that the rainwater falling on an area coming within a ridgeline can be harvested and will flow out of this area thorough single point. Some refer it as a catchment area or river basin.

Watershed management is an efficient management and conservation of surface and groundwater resources. It involves prevention of runoff and storage and recharge of groundwater through various methods like percolation tanks, recharge wells, etc. However, in broad sense watershed management includes conservation, regeneration and judicious use of all resources – natural (like land, water, plants and animals) and human with in a watershed.

Integrated Watershed Management Programme  is to restore the ecological balance by harnessing, conserving and developing degraded natural resources such as soil, vegetative cover and water.  The outcomes are  prevention of soil run-off, regeneration of natural vegetation, rain water harvesting and recharging of the ground water table.  This enables  multi-cropping and the introduction of diverse agro-based activities, which help to provide sustainable livelihoods to the people residing in the watershed area.

The main benefits of watershed management are:-

  1. Supply of water for drinking and irrigation.
    2. Increase in bio-diversity.
    3. Loss of acidity in the soil and free for standing water.
    4. Increase in the agricultural production and productivity.
    5. Decrease in the cutting of forests.
    6. Increase in the standard of living.
    7. Increase in employment.
    8. Increase in personal get together by participation of local people.

Ground water management.

 

Scientific management of ground water resources involves a combination of

  1. A) Supply side measures aimed at increasing extraction of ground water depending on its availability and
  2. B) Demand side measures aimed at controlling, protecting and conserving available resources.

The rainfall occurrence in different parts of India is limited to a period ranging from about 10 to 100 days. The natural recharge to ground water reservoir is restricted to this period only and is not enough to keep pace with the excessive continued exploitation. Since large volumes of rainfall flows out into the sea or get evaporated, artificial recharge has been advocated to supplement the natural recharge.

Ground water resources management requires to focus attention on the judicious utilization of the resources for ensuring their long-term sustainability. Ownership of ground water, need-based allocation pricing of resources, involvement of stake holders in various aspects of planning, execution and monitoring of projects and effective implementation of regulatory measures wherever necessary are the important considerations with regard to demand side ground water management.

 

Major Armed Uprising

by
Wahabi Shah Abdul Aziz & Saiyed Ahmad Raebarelvi. Objective was to reform the Muslim
Movement society& convert ‘Dur-ul-Harb’ (Non-Islamic community) into ‘Dar-ul-Islam’.
  Origianlly the movement was started in Arabia by Muhammad Ibn-Aba-e-Wahid.
  Its main centre was Patna, Sittana (NW province).
Kuka Bhai Ram Singh (Disciple of Bhai Balak Singh). It is also called Namdhari Mission.
Movement Bhai Ram Singh asked his followers to worship cow & run langars, wear white clothes
  & not use any foreign commodity or service. Bhai Ram Singh was deported to Burma.
Santhal Siddhu, Khanhu, Chand & Bhareo (four sons of Chulu Santhal of Raj Mahal district).
Rebellion Under the Permanent Settlement of 1793 the lands of Santhal passed to Zamidars &
1855-56 later to European Indigo planters. 10,000 santhals were killed in this rebellion. After
  this the area was put under the direct control of the Governor General & was named
  Santhal Paragana.

Vasudeo Balwant Phadke was born in Maharashtra. He left the army & became a revolutionary. Later deported to Aden & died in 1883. He may be justly called the father of militant nationalism in India.

 

Kisan Launched by Lala Lajpat Rai & Ajit Singh. The passing of the 1906 Punjab Land
Movements Revenue Act & heavy increase in water tax caused panic. The poem of Banke Dayal,
  ‘Pagri Sambhal O Jatta’ became famous. Lala & Ajit Singh were sentenced to 6 months
  prison. Later the DSP of Layalpur Clough was assassinated. Ajit Singh escaped to
  France while Bhai Parmanand’s house search yielded a book on bomb making.
Moplah Khilafat movement in Malabar incited communal feelings in Muslim peasants directed
Rebellion towards Hindu land holders.

Important Formulas – Mixtures and Alligations

by

 

  1. Alligation

    It is the rule which enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a specified price.

  2. Mean Price

    Mean price is the cost price of a unit quantity of the mixture

 

  1. Suppose a container contains x of liquid from which y units are taken out and replaced by water.After n operations, the quantity of pure liquid = [x(1−y/x)n]

 

  1. Rule of Alligation

    If two ingredients are mixed, then

    (Quantity of cheaper/Quantity of dearer)=(P. of dearer – Mean Price)/(Mean price – C.P. of cheaper)

Cost Price(CP) of a unit quantity
of cheaper (c)		 Cost Price(CP) of a unit quantity
of dearer (d)
   
Mean Price
(m)
(d – m) (m – c)
  1. => (Cheaper quantity) : (Dearer quantity) = (d – m) : (m – c)

 

Solved Examples

Level 1

1.A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?
A. 26 litres B. 29.16 litres
C. 28 litres D. 28.2 litres

 

 

Answer : Option B

Explanation :

Assume that a container contains x of liquid from which y units are taken out and replaced

by water. After n operations, the quantity of pure liquid

=[x(1−y/x)n]

Hence milk now contained by the container = 40(1−4/40)3=40(1−1/10)3

40×9/10×9/10×9/10=(4×9×9×9)/100=29.16

2.A jar full of whiskey contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohols and now the percentage of alcohol was found to be 26%. The quantity of whisky replaced is
A. 43 B. 34
C. 32 D. 23

 

 

Answer : Option D

Explanation :

Concentration of alcohol in 1st Jar = 40%

Concentration of alcohol in 2nd Jar = 19%

After the mixing, Concentration of alcohol in the mixture = 26%

By the rule of alligation,

Concentration of alcohol in 1st Jar Concentration of alcohol in 2nd Jar
40% 19%
Mean
26%
7 14

Hence ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2

 

3.In what ratio should rice at Rs.9.30 per Kg be mixed with rice at Rs. 10.80 per Kg so that the mixture be worth Rs.10 per Kg ?
A. 7 : 8 B. 8 : 7
C. 6 : 7 D. 7 ; 6

 

 

Answer : Option B

 

Explanation :

By the rule of alligation, we have

 

Cost of 1 kg rice of 1st kind Cost of 1 kg rice of 2nd kind
9.3 10.80
Mean Price
10
10.8-10 = .8 10 – 9.3 = .7

Required ratio = .8 : .7 = 8 : 7.

 

 

 

 4.In what ratio must water be mixed with milk costing Rs.12 per litre in order to get a mixture worth of Rs.8 per litre?
A. 1 : 3 B. 2 : 2
C. 1 : 2 D. 3 : 1

 

 

Answer : Option C

 

Explanation :

By the rule of alligation, we have

Cost Price of 1 litre of water Cost Price of 1 litre of milk
0 12
Mean Price
8
12-8=4 8-0=8

Required Ratio = 4 : 8 = 1 : 2

 

 

 

5.In 1 kg mixture of iron and manganese 20% of manganese. How much iron should be added so that the proportion of manganese becomes 10%
A. 1.5 Kg B. 2 Kg
C. .5 Kg D. 1 Kg

 

 

 

 

Answer : Option D

 

Explanation :

By the rule of alligation, we have

Percentage concentration of
manganese in the mixture : 20		 Percentage concentration of
manganese in pure iron : 0
Percentage concentration of manganese in the final mixture
10
10 – 0 = 10 20 – 10 = 10

=> Quantity of the mixture : Quantity of iron = 10 : 10 = 1 : 1

Given that Quantity of the mixture = 1 Kg

Hence Quantity of iron to be added = 1 Kg

 

6.A trader has 1600Kg of sugar. He sells a part at 8% profit and the rest at 12% profit. If he gains 11% on the whole , find the quantity sold at 12%.
A. 1200 Kg B. 1400 Kg
C. 1600 Kg D. 800 Kg

 

 

 

Answer : Option A

 

Explanation :

By the rule of alligation, we have

 

% Profit by selling part1 % Profit by selling part2
8 12
Net % Profit
11
12 – 11 = 1 11 – 8 = 3

=>Quantity of part1 : Quantity of part2 = 1 : 3

Given that total quantity = 1600 Kg

Hence quantity of part2 (quantity sold at 12%) = 1600×3/4=1200

 

 

7.How many litres of water must be added to 16 liters of milk and water contains 10% water to make it 20% water in it
A. 4 litre B. 2 litre
C. 1 litre D. 3 litre

 

 

Answer : Option B

 

Explanation :

By the rule of alligation, we have

% Concentration of water
in pure water : 100		 % Concentration of water
in the given mixture : 10
Mean % concentration
20
20 – 10 = 10 100 – 20 = 80

=> Quantity of water : Quantity of the mixture = 10 : 80 = 1 : 8

Here Quantity of the mixture = 16 litres

=> Quantity of water : 16 = 1 : 8

Quantity of water = 16×1/8=2 litre

 

8.A merchant has 1000 kg of sugar part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The Quantity sold at 18% profit is
A. 300 B. 400
C. 600 D. 500

 

 

Answer : Option C

 

Explanation :

By the rule of alligation,we have

Profit% by selling 1st part Profit% by selling 2nd part
8 18
Net % profit
14
18-14=4 14-8=6

=> Quantity of part1 : Quantity of part2 = 4 : 6 = 2 : 3

Total quantity is given as 1000Kg

So Quantity of part2 (Quantity sold at 18% profit) = 1000×3/5=600Kg

 

Level 2

 

1.Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety of tea in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, what is the price of the third variety per kg ?
A. Rs.182.50 B. Rs.170.5
C. Rs.175.50 D. Rs.180

 

 

Answer : Option C

Explanation :

Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed in the ratio 1 : 1

So their average price = (126+135)2=130.5

Hence let’s consider that the mixture is formed by mixing two varieties of tea.

one at Rs. 130.50 per kg and the other at Rs. x per kg in the ratio 2 : 2, i.e.,

1 : 1. Now let’s find out x.

By the rule of alligation, we can write as

Cost of 1 kg of 1st kind of tea Cost of 1 kg of 2nd kind of tea
130.50 x
Mean Price
153
(x – 153) 22.50

(x – 153) : 22.5 = 1 : 1

=>x – 153 = 22.50

=> x = 153 + 22.5 = 175.5

 

2.A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5?
A. 5litres, 7 litres B. 7litres, 4 litres
C. 6litres, 6 litres D. 4litres, 8 litres

 

 

Answer : Option C

Explanation :

Let cost of 1 litre milk be Rs.1

Milk in 1 litre mix in 1st can = 3/4 litre

Cost Price(CP) of 1 litre mix in 1st can = Rs.3/4

Milk in 1 litre mix in 2nd can = 1/2 litre

Cost Price(CP) of 1 litre mix in 2nd can = Rs.1/2

Milk in 1 litre of the final mix=5/8

Cost Price(CP) of 1 litre final mix = Rs.5/8

=> Mean price = 5/8
By the rule of alligation, we can write as

CP of 1 litre mix in 2nd can CP of 1 litre mix in 1st can
1/2 3/4
Mean Price
5/8
3/4 – 5/8 = 1/8 5/8 – 1/2 = 1/8

=> mix in 2nd can :mix in 1st can = 1/8 : 1/8 = 1:1

ie, from each can, 12×12=6 litre should be taken

 

3.Two vessels A and B contain spirit and water in the ratio 5 : 2 and 7 : 6 respectively. Find the ratio in which these mixture be mixed to obtain a new mixture in vessel C containing spirit and water in the ration 8 : 5 ?
A. 3: 4 B. 4 : 3
C. 9 : 7 D. 7 : 9

 

 

Answer : Option D

 

Explanation :

 

Let Cost Price(CP) of 1 litre spirit be Rs.1

Quantity of spirit in 1 litre mixture from vessel A = 5/7

Cost Price(CP) of 1 litre mixture from vessel A = Rs. 5/7

Quantity of spirit in 1 litre mixture from vessel B = 7/13

Cost Price(CP) of 1 litre mixture from vessel B = Rs. 7/13

Quantity of spirit to be obtained in 1 litre mixture from vessel C = 8/13

Cost Price(CP) of 1 litre mixture from vessel C = Rs. 8/13 = Mean Price

By the rule of alligation, we can write as

 

CP of 1 litre mixture from vessel A CP of 1 litre mixture from vessel B
5/7 7/13
Mean Price
8/13
8/13 – 7/13 = 1/13 5/7 – 8/13 = 9/91

=> Mixture from Vessel A : Mixture from Vessel B = 1/13 : 9/91 = 7 : 9 = Required Ratio

 

4.How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per Kg so that there may be a gain of 10 % by selling the mixture at Rs. 9.24 per Kg ?
A. 60 Kg B. 63 kg
C. 58 Kg D. 56 Kg

 

 

Answer : Option B

Explanation :

Selling Price(SP) of 1 Kg mixture= Rs. 9.24

Profit = 10%

Cost Price(CP) of 1 Kg mixture= 100×SP /(100+Profit%) =100×9.24/(100+10)

=100×9.24/110=92.4/11=Rs.8.4

By the rule of alligation, we have

CP of 1 kg sugar of 1st kind CP of 1 kg sugar of 2nd kind
Rs. 9 Rs. 7
Mean Price
Rs.8.4
8.4 – 7 = 1.4 9 – 8.4 = .6

ie, to get a cost price of 8.4, the sugars of kind1 and kind2 should be mixed in the

ratio 1.4 : .6 = 14 : 6 = 7 : 3

Let x Kg of kind1 sugar is mixed with 27 kg of kind2 sugar

then x : 27 = 7 : 3

⇒x/27=7/3

⇒x=(27×7)/3=63

5.A container contains a mixture of two liquids P and Q in the ratio 7 : 5. When 9 litres of mixture are drawn off and the container is filled with Q, the ratio of P and Q becomes 7 : 9. How many litres of liquid P was contained in the container initially?
A. 23 B. 21
C. 19 D. 17

 

 

Answer : Option B

Explanation :

Let’s initial quantity of P in the container be 7x

and initial quantity of Q in the container be 5x

Now 9 litres of mixture are drawn off from the container

Quantity of P in 9 litres of the mixtures drawn off = 9×7/12=63/12=21/ 4

Quantity of Q in 9 litres of the mixtures drawn off = 9×5/12=45/12=1/54

HenceQuantity of P remains in the mixtures after 9 litres is drawn off =7x−21/4

Quantity of Q remains in the mixtures after 9 litres is drawn off =5x−15/4

Since the container is filled with Q after 9 litres of mixture is drawn off,

Quantity of Q in the mixtures=5x−15/4+9=5x+21/4

Given that the ratio of P and Q becomes 7 : 9

⇒(7x−21/4):(5x+21/4)=7:9

⇒(7x−21/4)(5x+21/4)=7/9

63x−(9×21/4)=35x+(7×2/14)

28x=(16×21/4)

x=(16×21)/(4×28)

litres of P contained in the container initially = 7x=(7×16×21)/(4×28)=(16×21)/(4×4)=21

 

6.A dishonest milkman sells his milk at cost price but he mixes it with water and thereby gains 25%. What is the percentage of water in the mixture?
A. 25% B. 20%
C. 22% D. 24%

 

 

Answer : Option B

Explanation :

Let CP of 1 litre milk = Rs.1

Given that SP of 1 litre mixture = CP of 1 Litre milk = Rs.1

Given than Gain = 25%

Hence CP of 1 litre mixture = 100/(100+Gain%)×SP

=100(100+25)×1

=100/125=4/5

By the rule of alligation, we have

CP of 1 litre milk CP of 1 litre water
1 0
CP of 1 litre mixture
4/5
4/5 – 0 = 4/5 1- 4/5 = 1/5

=> Quantity of milk : Quantity of water = 4/5 : 1/5 = 4 : 1

Hence percentage of water in the mixture = 1/5×100=20%

MENSURATION

by

 

Mensuration is the branch of mathematics which deals with the study of different geometrical shapes, their areas and Volume. In the broadest sense, it is all about the process of measurement. It is based on the use of algebraic equations and geometric calculations to provide measurement data regarding the width, depth and volume of a given object or group of objects

  • Pythagorean Theorem (Pythagoras’ theorem)

In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides

c2 = a2 + b2 where c is the length of the hypotenuse and a and b are the lengths of the other two sides

  • Pi is a mathematical constant which is the ratio of a circle’s circumference to its diameter. It is denoted by π

π≈3.14≈227

  • Geometric Shapes and solids and Important Formulas
Geometric Shapes Description Formulas
Rectangle

l = Length

b = Breadth

d= Length of diagonal

 Area = lb

Perimeter = 2(l + b)

d = √l2+b2
Square

a = Length of a side

d= Length of diagonal

 Area= a*a=1/2*d*d

Perimeter = 4a

d = 2√a
Parallelogram

b and c are sides

b = base

h = height

                                  Area = bh

Perimeter = 2(b + c)
Rhombus

a = length of each side

b = base

h = height

d1, d2 are the diagonal

 Area = bh(Formula 1)

Area = ½*d1*d2 (Formula 2 )

Perimeter = 4a
Triangle

a , b and c are sides

b = base

h = height

 Area = ½*b*h (Formula 1) Area(Formula 2)                         = √S(S−a)(S−b)(S−c              where S is the semiperimeter
S  =(a+b+c)/2 (Formula 2 for area          -Heron’s formula) Perimeter = a + b + c

Radius of incircle of a triangle of area A =AS
where S is the semiperimeter
=(a+b+c)/2
Equilateral Triangle

a = side

 Area = (√3/4)*a*a               Perimeter = 3a

Radius of incircle of an equilateral                                                                  triangle of side a = a/2*√3

Radius of circumcircle of an equilateral triangle
of side a = a/√3

 
Base a is parallel to base b Trapezium(Trapezoid in American English)

h = height

 Area = 12(a+b)h

 
Circle

r = radius

d = diameter

 d = 2r

Area = πr2 = 14πd2

Circumference = 2πr = πd

 
Sector of Circle

r = radius

θ = central angle

 Area  = (θ/360) *π*r*r
Arc Length, s = (θ/180)* π*r

In the radian system for angular measurement,
2π radians = 360°
=> 1 radian = 180°π
=> 1° = π180 radians
Hence,
Angle in Degrees
= Angle in Radians × 180°π
Angle in Radians
= Angle in Degrees × π180°

 
Ellipse

Major axis length = 2a

Minor axis length = 2b

 Area = πab

Perimeter ≈
Rectangular Solid

l = length

w = width

h = height

 Total Surface Area
= 2lw + 2wh + 2hl
= 2(lw + wh + hl)

Volume = lwh
Cube

s = edge

 Total Surface Area = 6s2

Volume = s3
Right Circular Cylinder

h = height

r = radius of base

 Lateral Surface Area
= (2 π r)h

Total Surface Area
= (2 π r)h + 2 (π r2)

Volume = (π r2)h
Pyramid

h = height

B = area of the base

 Total Surface Area = B +                 Sum of  the areas of the triangular sides

Volume = 1/3*B*h
Right Circular Cone

h = height

r = radius of base

 Lateral Surface Area=πrs
where s is the slant height =√r*r+h*h
Total Surface Area
                                =πrs+πr2
Sphere

r = radius

d = diameter

  d = 2r

Surface Area =4πr*r=πd*d

Volume =4/3πr*r*r=16πd*d*d
     
 
 
 

 
 
 
 

 

 

 
 
 
 
 
 

 

  • Important properties of Geometric Shapes
    1. Properties of Triangle
      1. Sum of the angles of a triangle = 180°
      2. Sum of any two sides of a triangle is greater than the third side.
  • The line joining the midpoint of a side of a triangle to the positive vertex is called the median
  1. The median of a triangle divides the triangle into two triangles with equal areas
  2. Centroid is the point where the three medians of a triangle meet.
  3. Centroid divides each median into segments with a 2:1 ratio
  • Area of a triangle formed by joining the midpoints of the sides of a given triangle is one-fourth of the area of the given triangle.
  • An equilateral triangle is a triangle in which all three sides are equal
  1. In an equilateral triangle, all three internal angles are congruent to each other
  2. In an equilateral triangle, all three internal angles are each 60°
  3. An isosceles triangle is a triangle with (at least) two equal sides
  • In isosceles triangle, altitude from vertex bisects the base.

 

  1. Properties of Quadrilaterals
  2. Rectangle
    1. The diagonals of a rectangle are equal and bisect each other
    2. opposite sides of a rectangle are parallel
  • opposite sides of a rectangle are congruent
  1. opposite angles of a rectangle are congruent
  2. All four angles of a rectangle are right angles
  3. The diagonals of a rectangle are congruent
  4. Square
  • All four sides of a square are congruent
  • Opposite sides of a square are parallel
  1. The diagonals of a square are equal
  2. The diagonals of a square bisect each other at right angles
  3. All angles of a square are 90 degrees.
  • A square is a special kind of rectangle where all the sides have equal length
  1. Parallelogram
  • The opposite sides of a parallelogram are equal in length.
  • The opposite angles of a parallelogram are congruent (equal measure).
  1. The diagonals of a parallelogram bisect each other.
  • Each diagonal of a parallelogram divides it into two triangles of the same area
  1. Rhombus
  • All the sides of a rhombus are congruent
  • Opposite sides of a rhombus are parallel.
  • The diagonals of a rhombus bisect each other at right angles
  1. Opposite internal angles of a rhombus are congruent (equal in size)
  • Any two consecutive internal angles of a rhombus are supplementary; i.e. the sum of their angles = 180° (equal in size)
  • If each angle of a rhombus is 90°, it is a square

Other properties of quadrilaterals

  • The sum of the interior angles of a quadrilateral is 360 degrees
  • If a square and a rhombus lie on the same base, area of the square will be greater than area of the rhombus (In the special case when each angle of the rhombus is 90°, rhombus is also a square and therefore areas will be equal)
  • A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
  • Of all the parallelogram of given sides, the parallelogram which is a rectangle has the greatest area.
  • Each diagonal of a parallelogram divides it into two triangles of the same area
  • A square is a rhombus and a rectangle.
  1. Sum of Interior Angles of a polygon
    1. The sum of the interior angles of a polygon = 180(n – 2) degrees where n = number of sides Example 1 : Number of sides of a triangle = 3. Hence, sum of the interior angles of a triangle = 180(3 – 2) = 180 × 1 = 180 ° Example 2 : Number of sides of a quadrilateral = 4. Hence, sum of the interior angles of any quadrilateral = 180(4 – 2) = 180 × 2 = 360.

 

 

Solved Examples

Level 1

  1. An error 2% in excess is made while measuring the side of a square. What is the percentage of error in the calculated area of the square?
  2. 4.04 %
  3. 2.02 %
  4. 4 %
  5. 2 %

Answer : Option A

Explanation :

Error = 2% while measuring the side of a square.

Let the correct value of the side of the square = 100
Then the measured value = (100×(100+2))/100=102 (∵ error 2% in excess)

Correct Value of the area of the square = 100 × 100 = 10000
Calculated Value of the area of the square = 102 × 102 = 10404

Error = 10404 – 10000 = 404
Percentage Error = (Error/Actual Value)×100=(404/10000)×100=4.04%

 

  1. A towel, when bleached, lost 20% of its length and 10% of its breadth. What is the percentage of decrease in area?
  2. 30 %
  3. 28 %
  4. 32 %
  5. 26 %

Answer : Option B

Explanation :
Let original length = 100 and original breadth = 100
Then original area = 100 × 100 = 10000

Lost 20% of length
=> New length =( Original length × (100−20))/100
=(100×80)/100=80

Lost 10% of breadth
=> New breadth= (Original breadth × (100−10))/100
=(100×90)/100=90

New area = 80 × 90 = 7200

Decrease in area
= Original Area – New Area
= 10000 – 7200 = 2800

Percentage of decrease in area
=(Decrease in Area/Original Area)×100=(2800/10000)×100=28%

  1. If the length of a rectangle is halved and its breadth is tripled, what is the percentage change in its area?
  2. 25 % Increase
  3. 25 % Decrease
  4. 50 % Decrease
  5. 50 % Increase

Answer : Option D

Explanation :
Let original length = 100 and original breadth = 100
Then original area = 100 × 100 = 10000

Length of the rectangle is halved
=> New length = (Original length)/2=100/2=50

breadth is tripled
=> New breadth= Original breadth × 3 = 100 × 3 = 300

New area = 50 × 300 = 15000

Increase in area = New Area – Original Area = 15000 – 10000= 5000
Percentage of Increase in area =( Increase in Area/OriginalArea)×100=(5000/10000)×100=50%

  1. The area of a rectangle plot is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the plot?
  2. 14 metres
  3. 20 metres
  4. 18 metres
  5. 12 metres

Answer : Option B

Explanation:

lb = 460 m2 ——(Equation 1)

Let the breadth = b
Then length, l =( b×(100+15))/100=115b/100——(Equation 2)

From Equation 1 and Equation 2,
115b/100×b=460b2=46000/115=400⇒b=√400=20 m

 

  1. If a square and a rhombus stand on the same base, then what is the ratio of the areas of the square and the rhombus?
  2. equal to ½
  3. equal to ¾
  4. greater than 1
  5. equal to 1

Answer : Option C

Explanation :

If a square and a rhombus lie on the same base, area of the square will be greater than area of the rhombus (In the special case when each angle of the rhombus is 90°, rhombus is also a square and therefore areas will be equal)

 

Hence greater than 1 is the more suitable choice from the given list

================================================================
Note : Proof

Consider a square and rhombus standing on the same base ‘a’. All the sides of a square are of equal length. Similarly all the sides of a rhombus are also of equal length. Since both the square and rhombus stands on the same base ‘a’,

Length of each side of the square = a
Length of each side of the rhombus = a

Area of the sqaure = a2 …(1)

From the diagram, sin θ = h/a
=> h = a sin θ

Area of the rhombus = ah = a × a sin θ = a2 sin θ …(2)

From (1) and (2)

Area of the square/Area of the rhombus= a2 /a2sinθ=1/sinθ

Since 0° < θ < 90°, 0 < sin θ < 1. Therefore, area of the square is greater than that of rhombus, provided both stands on same base.

(Note that, when each angle of the rhombus is 90°, rhombus is also a square (can be considered as special case) and in that case, areas will be equal.

 

  1. The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m, find out the area of the field.
  2. 37500 m2
  3. 30500 m2
  4. 32500 m2
  5. 40000 m2

Answer : Option A

Explanation :

Given that breadth of a rectangular field is 60% of its length
⇒b=(60/100)* l =(3/5)* l

perimeter of the field = 800 m
=> 2 (l + b) = 800
⇒2(l+(3/5)* l)=800⇒l+(3/5)* l =400⇒(8/5)* l =400⇒l/5=50⇒l=5×50=250 m

b = (3/5)* l =(3×250)/5=3×50=150 m

Area = lb = 250×150=37500 m2

 

  1. What is the percentage increase in the area of a rectangle, if each of its sides is increased by 20%?
  2. 45%
  3. 44%
  4. 40%
  5. 42%

Answer : Option B

Explanation :
Let original length = 100 and original breadth = 100
Then original area = 100 × 100 = 10000

Increase in 20% of length.
=> New length = (Original length ×(100+20))/100=(100×120)/100=120

Increase in 20% of breadth
=> New breadth= (Original breadth × (100+20))/100=(100×120)/100=120

New area = 120 × 120 = 14400

Increase in area = New Area – Original Area = 14400 – 10000 = 4400
Percentage increase in area =( Increase in Area /OriginalArea)×100=(4400/10000)×100=44%

  1. What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?
  2. 814
  3. 802
  4. 836
  5. 900

Answer : Option A

Explanation :

l = 15 m 17 cm = 1517 cm
b = 9 m 2 cm = 902 cm
Area = 1517 × 902 cm2

Now we need to find out HCF(Highest Common Factor) of 1517 and 902.
Let’s find out the HCF using long division method for quicker results

902)  1517  (1

 

-902

—————–

615)  902  (1

 

  • 615

————–

 

287)  615 (2

 

-574

—————–

 

41)  287  (7

 

-287

————

0
————

Hence, HCF of 1517 and 902 = 41

Hence, side length of largest square tile we can take = 41 cm
Area of each square tile = 41 × 41 cm2

Number of tiles required = (1517×902)/(41×41)=37×22=407×2=814

 

Level 2

  1. A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, find out the area of the parking space in square feet?
  2. 126 sq. ft.
  3. 64 sq. ft.
  4. 100 sq. ft.
  5. 102 sq. ft.

Answer : Option A

Explanation :

Let l = 9 ft.

Then l + 2b = 37
=> 2b = 37 – l = 37 – 9 = 28
=> b = 282 = 14 ft.

Area = lb = 9 × 14 = 126 sq. ft.

 

  1. A large field of 700 hectares is divided into two parts. The difference of the areas of the two parts is one-fifth of the average of the two areas. What is the area of the smaller part in hectares?
  2. 400
  3. 365
  4. 385
  5. 315

Answer : Option D

Explanation :

Let the areas of the parts be x hectares and (700 – x) hectares.

Difference of the areas of the two parts = x – (700 – x) = 2x – 700

one-fifth of the average of the two areas = 15[x+(700−x)]2
=15×7002=3505=70

Given that difference of the areas of the two parts = one-fifth of the average of the two areas
=> 2x – 700 = 70
=> 2x = 770
⇒x=7702=385

Hence, area of smaller part = (700 – x) = (700 – 385) = 315 hectares.

 

  1. The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq.cm. What is the length of the rectangle?
  2. 18 cm
  3. 16 cm
  4. 40 cm
  5. 20 cm

Answer : Option C

Explanation :

Let breadth = x cm
Then length = 2x cm
Area = lb = x × 2x = 2x2

New length = (2x – 5)
New breadth = (x + 5)
New Area = lb = (2x – 5)(x + 5)

But given that new area = initial area + 75 sq.cm.
=> (2x – 5)(x + 5) = 2x2 + 75
=> 2x2 + 10x – 5x – 25 = 2x2 + 75
=> 5x – 25 = 75
=> 5x = 75 + 25 = 100
=> x = 1005 = 20 cm

Length = 2x = 2 × 20 = 40cm

 

  1. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then what is the area of the park (in sq. m)?
  2. 142000
  3. 112800
  4. 142500
  5. 153600

Answer : Option D

Explanation :

l : b = 3 : 2 —-(Equation 1)

Perimeter of the rectangular park
= Distance travelled by the man at the speed of 12 km/hr in 8 minutes
= speed × time = 12×860     (∵ 8 minute = 860 hour)
= 85 km = 85 × 1000 m = 1600 m

Perimeter = 2(l + b)

=> 2(l + b) = 1600
=> l + b = 16002 = 800 m —-(Equation 2)

From (Equation 1) and (Equation 2)
l = 800 × 35 = 480 m
b = 800 × 25 = 320 m (Or b = 800 – 480 = 320m)

Area = lb = 480 × 320 = 153600 m2

 

  1. It is decided to construct a 2 metre broad pathway around a rectangular plot on the inside. If the area of the plots is 96 sq.m. and the rate of construction is Rs. 50 per square metre., what will be the total cost of the construction?
  2. Rs.3500
  3. Rs. 4200
  4. Insufficient Data
  5. Rs. 4400

Answer : Option C

Explanation :
Let length and width of the rectangular plot be l and b respectively
Total area of the rectangular plot = 96 sq.m.
=> lb = 96

Width of the pathway = 2 m
Length of the remaining area in the plot = (l – 4)
breadth of the remaining area in the plot = (b – 4)
Area of the remaining area in the plot = (l – 4)(b – 4)

Area of the pathway
= Total area of the rectangular plot – remaining area in the plot
= 96 – [(l – 4)(b – 4)] = 96 – [lb – 4l – 4b + 16] = 96 – [96 – 4l – 4b + 16] = 96 – 96 + 4l + 4b – 16
= 4l + 4b – 16
= 4(l + b) – 16

We do not know the values of l and b and hence area of the pathway cannot be found out. So we cannot determine total cost of the construction.

 

  1. A circle is inscribed in an equilateral triangle of side 24 cm, touching its sides. What is the area of the remaining portion of the triangle?
  2. 144√3−48π cm2
  3. 121√3−36π cm2
  4. 144√3−36π cm2
  5. 121√3−48π cm2

Answer : Option A

Explanation :
Area of an equilateral triangle = (3/√4)*a *a where a is length of one side of the equilateral triangle
Area of the equilateral Δ ABC = (3/√4)*a *a = (3/√4)*24*24=144√3 cm2⋯ (1)

Area of a triangle = 12bhwhere b is the base and h is the height of the triangle
Let r = radius of the inscribed circle. Then
Area of Δ ABC
= Area of Δ OBC + Area of Δ OCA + area of Δ OAB
= (½ × r × BC) + (½ × r × CA) + (½ × r × AB)
= ½ × r × (BC + CA + AB)
= ½ x r x (24 + 24 + 24)
= ½ x r x 72 = 36r cm2 —-(2)

From (1) and (2),
144√3=36r⇒r=144√3/36=4√3−−−−(3)

Area of a circle = πr2 where = radius of the circle
From (3), the area of the inscribed circle = πr2=π(4√3)* (4√3)=48π⋯(4)

Hence, area of the remaining portion of the triangle
= Area of Δ ABC – Area of inscribed circle
144√3−48π cm2

 

  1. What will be the length of the longest rod which can be placed in a box of 80 cm length, 40 cm breadth and 60 cm height?
  2. √11600 cm
  3. √14400 cm
  4. √10000 cm
  5. √12040 cm

Answer : Option A

Explanation :
The longest road which can fit into the box will have one end at A and other end at G (or any other similar diagonal).
Hence the length of the longest rod = AG

Initially let’s find out AC. Consider the right angled triangle ABC

AC2 = AB2 + BC2 = 402 + 802 = 1600 + 6400 = 8000
⇒AC = √8000 cm

Consider the right angled triangle ACG

AG2 = AC2 + CG2
(√8000) 2+602=8000+3600=11600
=> AG = √11600 cm
=> Length of the longest rod = √11600cm

 

  1. A rectangular plot measuring 90 metres by 50 metres needs to be enclosed by wire fencing such that poles of the fence will be kept 5 metres apart. How many poles will be needed?
  2. 30
  3. 44
  4. 56
  5. 60

Answer : Option C

Explanation :

Perimeter of a rectangle = 2(l + b)
where l is the length and b is the breadth of the rectangle

Length of the wire fencing = perimeter = 2(90 + 50) = 280 metres
Two poles will be kept 5 metres apart. Also remember that the poles will be placed along the perimeter of the rectangular plot, not in a single straight line which is very important.
Hence number of poles required = 280/5 = 56

Important Formulas – Percentage

by

 

  • Percentage

    Percent means for every 100

    So, when percent is calculated for any value, it means that we calculate the value for every 100 of the reference value.

    percent is denoted by the symbol %. For example, x percent is denoted by x%

  • x%=x/100

    Example : 25%=25/100=1/4

  • To express x/y as a percent,we have x/y=(x/y×100)%

    Example : 1/4=(1/4×100)%=25%

  • If the price of a commodity increases by R%, the reduction in consumptionso as not to increase the expenditure = [R/(100+R)×100]%
  • If the price of a commodity decreases by R%, the increase in consumptionso as not to decrease the expenditure = [R/(100−R)×100]%
  • If the population of a town = P and it increases at the rate of R% per annum, thenPopulation after n years = P((1+R)/100))n
  • If the population of a town = P and it increases at the rate of R% per annum, thenPopulation before n years = P((1+R)/100))n
  • If the present value of a machine = P and it depreciates at the rate of R% per annum,

ThenValue of the machine after n years = P((1-R)/100))n

  • If the present value of a machine = P and it depreciates at the rate of R% per annum,

ThenValue of the machine before n years = P((1-R)/100))n

 

Solved Examples

Level 1

1.    If A = x% of y and B = y% of x, then which of the following is true?
A. None of these B. A is smaller than B.
C. Relationship between A and B cannot be determined. D. If x is smaller than y, then A is greater than B.
E. A is greater than B.

 

   

Answer : Option A

Explanation :

A = xy/100 ………….(Equation 1)

B = yx/100……………..(Equation 2)

From these equations, it is clear that A = B

 

 

2.If 20% of a = b, then b% of 20 is the same as:
A. None of these B. 10% of a
C. 4% of a D. 20% of a

 

Answer :Option C

Explanation :

20% of a = b

=> b = 20a/100

b% of 20 = 20b/100=(20a/100) × 20/100

=(20×20×a)/(100×100)=4a/100 = 4% of a

 

3.Two numbers A and B are such that the sum of 5% of A and 4% of B is two-third of the sum of 6% of A and 8% of B. Find the ratio of A : B.
A. 2 : 1 B. 1 : 2
C. 1 : 1 D. 4 : 3

 

 

Answer :Option D

Explanation :

5% of A + 4% of B = 2/3(6% of A + 8% of B)

5A/100+4B/100=2/3(6A/100+8B/100)

⇒5A+4B=2/3(6A+8B)

⇒15A+12B=12A+16B

⇒3A=4B

⇒AB=43⇒A:B=4:3

4.The population of a town increased from 1,75,000 to 2,62,500 in a decade. What is the average percent increase of population per year?
A. 4% B. 6%
C. 5% D. 50%

 

Answer :Option C

Explanation :

Increase in the population in 10 years = 2,62,500 – 1,75,000 = 87500

% increase in the population in 10 years = (87500/175000)×100=8750/175=50%

Average % increase of population per year = 50%/10=5%

 

5.Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?
A. 57% B. 50%
C. 52% D. 60%

 

Answer :Option A

Explanation :

Votes received by the winning candidate = 11628

Total votes = 1136 + 7636 + 11628 = 20400

Required percentage = (11628/20400)×100=11628/204=2907/51=969/17=57%

 

6.A fruit seller had some oranges. He sells 40% oranges and still has 420 oranges. How many oranges he had originally?
A. 420 B. 700
C. 220 D. 400

 

Answer :Option B

Explanation :

He sells 40% of oranges and still there are 420 oranges remaining

=> 60% of oranges = 420

⇒(60×Total Oranges)/100=420

⇒Total Oranges/100=7

⇒ Total Oranges = 7×100=700

7.A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?
A. 499/11 % B. 45 %
C. 500/11 % D. 489/11 %

 

Answer :Option C

Explanation :

Total runs scored = 110

Total runs scored from boundaries and sixes = 3 x 4 + 8 x 6 = 60

Total runs scored by running between the wickets = 110 – 60 = 50

Required % = (50/110)×100=500/11%

 

8.What percentage of numbers from 1 to 70 have 1 or 9 in the unit’s digit?
A. 2023% B. 20%
C. 21% D. 2223%

 

 

Answer :Option B

Explanation :

Total numbers = 70

Total numbers in 1 to 70 which has 1 in the unit digit = 7

Total numbers in 1 to 70 which has 9 in the unit digit = 7

Total numbers in 1 to 70 which has 1 or 9 in the unit digit = 7 + 7 = 14

Required percentage = (14/70)×100=140/7=20%

 

Level 2

 

1.In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, what was the number of valid votes that the other candidate got?
A. 2800 B. 2700
C. 2100 D. 2500

 

Answer :Option B

Explanation :

Total number of votes = 7500

Given that 20% of Percentage votes were invalid

=> Valid votes = 80%

Total valid votes = (7500×80)/100

1st candidate got 55% of the total valid votes.

Hence the 2nd candidate should have got 45% of the total valid votes
=> Valid votes that 2nd candidate got = (total valid votes ×45)/100

=7500×(80/100)×(45/100)=75×(4/5)×45=75×4×9=300×9=2700

 

2.In a competitive examination in State A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 80 more candidates got selected than A. What was the number of candidates appeared from each State?
A. 8200 B. 7500
C. 7000 D. 8000

 

Answer :Option D

Explanation :

State A and State B had an equal number of candidates appeared.

In state A, 6% candidates got selected from the total appeared candidates

In state B, 7% candidates got selected from the total appeared candidates

But in State B, 80 more candidates got selected than State A

From these, it is clear that 1% of the total appeared candidates in State B = 80

=> total appeared candidates in State B = 80 x 100 = 8000

=> total appeared candidates in State A = total appeared candidates in State B = 8000

 

3.In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is 2/3 of the number of students of 8 years of age which is 48. What is the total number of students in the school?
A. 100 B. 102
C. 110 D. 90

 

 

Answer :Option A

Explanation :

Let the total number of students = x

Given that 20% of students are below 8 years of age

then The number of students above or equal to 8 years of age = 80% of x —–(Equation 1)

Given that number of students of 8 years of age = 48 —–(Equation 2)

Given that number of students above 8 years of age = 2/3 of number of students of 8 years of age

=>number of students above 8 years of age = (2/3)×48=32—–(Equation 3)

From Equation 1,Equation 2 and Equation 3,
80% of x = 48 + 32 = 80

⇒80x/100=80

⇒x100=1⇒x=100

4.In an examination, 5% of the applicants were found ineligible and 85% of the eligible candidates belonged to the general category. If 4275 eligible candidates belonged to other categories, then how many candidates applied for the examination?
A. 28000 B. 30000
C. 32000 D. 33000

 

Answer :Option B

Explanation :

Let the number of candidates applied for the examination = x

Given that 5% of the applicants were found ineligible.

It means that 95% of the applicants were eligible (∴ 100% – 5% = 95%)

Hence total eligible candidates = 95x/100

Given that 85% of the eligible candidates belonged to the general category

It means 15% of the eligible candidates belonged to other categories(∴ 100% – 85% = 15%)
Hence Total eligible candidates belonged to other categories

=(total eligible candidates×15)/100=(95x/100)×(15/100)

=(95x×15)/(100×100)

Given that Total eligible candidates belonged to other categories = 4275

⇒(95x×15)/(100×100)=4275

⇒(19x×15)/(100×100)=855

⇒(19x×3)/(100×100)=171

⇒(x×3)/(100×100)=9

⇒x/(100×100)=3

⇒x=3×100×100=30000

 

5.A student multiplied a number by 3/5 instead of 5/3.What is the percentage error in the calculation?
A. 64% B. 32%
C. 34% D. 42%

 

Answer :Option A

Explanation :

Let the number = 1

Then, ideally he should have multiplied 1 by 5/3.

Hence the correct result was 1 x (5/3) = (5/3)

By mistake, he multiplied 1 by 3/5.

Hence the result with the error = 1 x (3/5) = (3/5)

Error = 5/3−3/5=(25−9)/15=16/15

percentage error = (Error/True Value)×100={(16/15)/(5/3)}×100

=(16×3×100)/(15×5)=(16×100)/(5×5)=16×4=64%

 

6.The price of a car is Rs. 3,25,000. It was insured to 85% of its price. The car was damaged completely in an accident and the insurance company paid 90% of the insurance. What was the difference between the price of the car and the amount received ?
A. Rs. 76,375 B. Rs. 34,000
C. Rs. 82,150 D. Rs. 70,000

 

Answer :Option A

Explanation :

Price of the car = Rs.3,25,000

Car insured to 85% of its price

=>Insured price=(325000×85)/100

Insurance company paid 90% of the insurance
⇒Amount paid by Insurance company =(Insured price×90)/100

=325000×(85/100)×(90/100)=325×85×9=Rs.248625

Difference between the price of the car and the amount received

= Rs.325000 – Rs.248625 = Rs.76375

 

7.If the price of petrol increases by 25% and Benson intends to spend only an additional 15% on petrol, by how much % will he reduce the quantity of petrol purchased?
A. 8% B. 7%
C. 10% D. 6%

 

 

Answer :Option A

Explanation :

Assume that the initial price of 1 Litre petrol = Rs.100 ,Benson spends Rs.100 for petrol,

such that Benson buys 1 litre of petrol

After the increase by 25%, price of 1 Litre petrol = (100×(100+25))/100=Rs.125

Since Benson spends additional 15% on petrol,

amount spent by Benson = (100×(100+15))/100=Rs.115

Hence Quantity of petrol that he can purchase = 115/125 Litre

Quantity of petrol reduced = (1−115/125) Litre

Percentage Quantity of reduction = ((1−115/125))/1×100=(10/125)/×100=(10/5)×4=2×4=8%

8.30% of the men are more than 25 years old and 80% of the men are less than or equal to 50 years old. 20% of all men play football. If 20% of the men above the age of 50 play football, what percentage of the football players are less than or equal to 50 years?
A. 60% B. 70%
C. 80% D. 90%

 

Answer :Option C

Explanation :

Let total number of men = 100

Then

80 men are less than or equal to 50 years old

(Since 80% of the men are less than or equal to 50 years old)

=> 20 men are above 50 years old (Since we assumed total number of men as 100)

20% of the men above the age of 50 play football

⇒Number of men above the age of 50 who play football = (20×20)/100=4

Number of men who play football = 20 (Since 20% of all men play football)
Percentage of men who play football above the age of 50 = (4/20)×100=20%

=>Percentage of men who play football less than or equal to the age 50 = 100%−20%=80%

FRACTIONS

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Fractions
Any unit can be divided into any numbers of equal parts, one or more of this parts is called fraction of that unit. e.g. one-forth (1/4), one-third (1/3), three-seventh (3/7) etc.

The lower part indicates the number of equal parts into which the unit is divided, is called denominator. The upper part, which indicates the number of parts taken from the fraction is called the numerator. The numerator and the denominator of a fraction are called its terms.

  • A fraction is unity, when its numerator and denominator are equal.
  • A fraction is equal to zero if its numerator is zero.
  • The denominator of a fraction can never be zero.
  • The value of a fraction is not altered by multiplying or dividing the numerator and the denominator by the same number.e.g. 2/3 = 2/6 = 8/12 = (2/4)/(3/4)
  • When there is no common factor between numerator and denominator it is called in its lowest terms.e.g. 15/25 = 3/5
  • When a fraction is reduced to its lowest term, its numerator and denominator are prime to each other.
  • When the numerator and denominator are divided by its HCF, fraction reduces to its lowest term.


Proper fraction: A fraction in which numerator is less than the denominator. e.g. 1/4, 3/4, 11/12 etc.

 

Improper Fraction:  A fraction in which numerator is equal to or more than the denominator. e.g. 5/4, 7/4, 13/12 etc.

 

Like fraction: Fractions in which denominators are same is called like fractions.

e.g. 1/12, 5/12, 7/12, 13/12 etc.

 

Unlike fraction: Fractions in which denominators are not same is called, unlike fractions.

e.g. 1/12, 5/7, 7/9 13/11 etc.

 

Compound Fraction: Fraction of a fraction is called a compound fraction.

e.g. 1/2 of 3/4 is a compound fraction.

 

Complex Fractions: Fractions in which numerator or denominator or both are fractions, are called complex fractions.

 

Continued fraction: Fraction that contain additional fraction is called continued fraction.

e.g.

 

 

 

Rule: To simplify a continued fraction, begin from the bottom and move upwards.

 

Decimal Fractions: Fractions in which denominators are 10 or multiples of 10 is called, decimal fractions. e.g. 1/10, 3/100, 2221/10000 etc.

 

Recurring Decimal: If in a decimal fraction a digit or a set of digits is repeated continuously, then such a number is called a recurring decimal. It is expressed by putting a dot or bar over the digits. e.g.

 

 

Pure recurring decimal: A decimal fraction in which all the figures after the decimal point is repeated is called a pure recurring decimal.

 

Mixed recurring decimal: A decimal fraction in which only some of the figures after the decimal point is repeated is called a mixed recurring decimal.

 

Conversion of recurring decimal into proper fraction: 

CASE I: Pure recurring decimal

 

Write the repeated digit only once in the numerator and put as many nines as in the denominator as the number of repeating figures. e.g.

 

CASE II: Mixed recurring decimal

In the numerator, take the difference between the number formed by all the digits after the decimal point and that formed by the digits which are not repeated. In the denominator, take the number formed as many nines as there are repeating digits followed by as many zeros as is the number of non-repeating digits. e.g.

 
Questions

Level-I

 

1.
Evaluate : (2.39)2 – (1.61)2
2.39 – 1.61
A. 2
B. 4
C. 6
D. 8

 

2. What decimal of an hour is a second ?
A. .0025
B. .0256
C. .00027
D. .000126

 

 

3.
The value of (0.96)3 – (0.1)3 is:
(0.96)2 + 0.096 + (0.1)2
A. 0.86
B. 0.95
C. 0.97
D. 1.06

 

 

4.
The value of 0.1 x 0.1 x 0.1 + 0.02 x 0.02 x 0.02 is:
0.2 x 0.2 x 0.2 + 0.04 x 0.04 x 0.04
A. 0.0125
B. 0.125
C. 0.25
D. 0.5

 

5. If 2994 ÷ 14.5 = 172, then 29.94 ÷ 1.45 = ?
A. 0.172
B. 1.72
C. 17.2
D. 172
 

 

 

6.

  

 

 

When 0.232323….. is converted into a fraction, then the result is:
A.
1
5
B.
2
9
C.
23
99
D.
23
100

 

7.
.009 = .01
?
A. .0009
B. .09
C. .9
D. 9

 

8. The expression (11.98 x 11.98 + 11.98 x x + 0.02 x 0.02) will be a perfect square for x equal to:
A. 0.02
B. 0.2
C. 0.04
D. 0.4

 

9.
(0.1667)(0.8333)(0.3333) is approximately equal to:
(0.2222)(0.6667)(0.1250)
A. 2
B. 2.40
C. 2.43
D. 2.50
   

 

10. 3889 + 12.952 – ? = 3854.002
A. 47.095
B. 47.752
C. 47.932
D. 47.95
 

 

 

 

 

 

11.

  

 

 

Level-II

 

 

0.04 x 0.0162 is equal to:
A. 6.48 x 10-3
B. 6.48 x 10-4
C. 6.48 x 10-5
D. 6.48 x 10-6

 

12.
4.2 x 4.2 – 1.9 x 1.9 is equal to:
2.3 x 6.1
A. 0.5
B. 1.0
C. 20
D. 22

 

 

13.
If 144 = 14.4 , then the value of x is:
0.144 x
A. 0.0144
B. 1.44
C. 14.4
D. 144

 

 

 

14. The price of commodity X increases by 40 paise every year, while the price of commodity Y increases by 15 paise every year. If in 2001, the price of commodity X was Rs. 4.20 and that of Y was Rs. 6.30, in which year commodity X will cost 40 paise more than the commodity Y ?
A. 2010
B. 2011
C. 2012
D. 2013

 

 

15.

  

Which of the following are in descending order of their value ?
A.
1 , 2 , 3 , 4 , 5 , 6
3 5 7 5 6 7
B.
1 , 2 , 3 , 4 , 5 , 6
3 5 5 7 6 7
C.
1 , 2 , 3 , 4 , 5 , 6
3 5 5 6 7 7
D.
6 , 5 , 4 , 3 , 2 , 1
7 6 5 7 5 3
 

 

16.

  

Which of the following fractions is greater than 3 and less than 5 ?
4 6
A.
1
2
B.
2
3
C.
4
5
D.
9
10

 

17. The rational number for recurring decimal 0.125125…. is:
A.
63
487
B.
119
993
C.
125
999
D. None of these

 

18. 617 + 6.017 + 0.617 + 6.0017 = ?
A. 6.2963
B. 62.965
C. 629.6357
D. None of these

 

 

19.
The value of 489.1375 x 0.0483 x 1.956 is closest to:
0.0873 x 92.581 x 99.749
A. 0.006
B. 0.06
C. 0.6
D. 6

 

20. 0.002 x 0.5 = ?
A. 0.0001
B. 0.001
C. 0.01
D. 0.1

 

 

 

 

 

Answers

Level-I

Answer:1 Option B

 

Explanation:

Given Expression = a2 – b2 = (a + b)(a – b) = (a + b) = (2.39 + 1.61) = 4.
a – b (a – b)

 


Answer:2 Option C

 

Explanation:

Required decimal = 1 = 1 = .00027
60 x 60 3600

 

 

Answer:3 Option A

 

Explanation:

Given expression
= (0.96)3 – (0.1)3
(0.96)2 + (0.96 x 0.1) + (0.1)2
= a3 – b3
a2 + ab + b2
= (a – b)  
= (0.96 – 0.1)  
= 0.86

Answer:4 Option B

 

Explanation:

Given expression = (0.1)3 + (0.02)3 = 1 = 0.125
23 [(0.1)3 + (0.02)3] 8

 

 

 

 

Answer:5 Option C

 

Explanation:

29.94 = 299.4
1.45 14.5

 

= 2994 x 1 [ Here, Substitute 172 in the place of 2994/14.5 ]
14.5 10

 

= 172
10

= 17.2

 

 

Answer:6 Option C

 

Explanation:

0.232323… = 0.23 = 23
99

 

Answer:7 Option C

 

Explanation:

Let .009 = .01;     Then x = .009 = .9 = .9
x .01 1

 

 

Answer:8 Option C

 

Explanation:

Given expression = (11.98)2 + (0.02)2 + 11.98 x x.

For the given expression to be a perfect square, we must have

11.98 x x = 2 x 11.98 x 0.02 or x   = 0.04

 

Answer:9 Option D

 

Explanation:

Given expression
= (0.3333) x (0.1667)(0.8333)
(0.2222) (0.6667)(0.1250)
= 3333 x
1 x 5
6 6
2222
2 x 125
3 1000
= 3 x 1 x 5 x 3 x 8
2 6 6 2
= 5
2
= 2.50

 

Answer:10 Option D

 

Explanation:

Let 3889 + 12.952 – x = 3854.002.

Then x = (3889 + 12.952) – 3854.002

= 3901.952 – 3854.002

= 47.95.

 

Level-II

Answer:11 Option B

 

Explanation:

4 x 162 = 648. Sum of decimal places = 6.
So, 0.04 x 0.0162 = 0.000648 = 6.48 x 10-4

 

Answer:12 Option B

 

Explanation:

Given Expression = (a2 – b2) = (a2 – b2) = 1.
(a + b)(a – b) (a2 – b2)

 

 

Answer:13 Option A

 

Explanation:

144 = 14.4
0.144 x

 

144 x 1000 = 14.4
144 x

 

 x = 14.4 = 0.0144
1000

 

 

Answer:14 Option B

 

Explanation:

Suppose commodity X will cost 40 paise more than Y after z years.

Then, (4.20 + 0.40z) – (6.30 + 0.15z) = 0.40

0.25z = 0.40 + 2.10

 z = 2.50 = 250 = 10.
0.25 25

X will cost 40 paise more than Y 10 years after 2001 i.e., 2011.

 

 

 

Answer:15 Option D

Answer:16 Option C

 

Explanation:

3 = 0.75, 5 = 0.833, 1 = 0.5, 2 = 0.66, 4 = 0.8, 9 = 0.9.
4 6 2 3 5 10

Clearly, 0.8 lies between 0.75 and 0.833.

4 lies between 3 and 5 .
5 4 6

 

 

 

Answer:17 Option C

 

Explanation:

0.125125… = 0.125 = 125
999

 

 

Answer:18 Option C

 

Explanation:

617.00

6.017

0.617

+  6.0017

——–

629.6357

———

 

Answer:19 Option B

 

Explanation:

489.1375 x 0.0483 x 1.956 489 x 0.05 x 2
0.0873 x 92.581 x 99.749 0.09 x 93 x 100

 

= 489
9 x 93 x 10

 

= 163 x 1
279 10

 

= 0.58
10

= 0.058  0.06.

 

Answer:20 Option B

 

Explanation:

2 x 5 = 10.

Sum of decimal places = 4

0.002 x 0.5 = 0.001

FUNCTIONS OF MANAGEMENT- DIRECTION (Communication,Supervision,Motivation,Leadership)

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Directing is concerned with instructing, guiding, supervising and inspiring people in the organisation to achieve its objectives. It is the process of telling people what to do and seeing that they do it in the best possiblemanner.

 

Elements in Directing: The four essential elements in Directing are :

 

  1. Communication
  2. Supervision
  3. Motivation
  4. Leadership

 

 

  1. COMMUNICATION

 

Communication is a basic organisational function, which refers to the process by which a person (known as sender) transmits information or messages to another person (known as receiver). The purpose of communication in organisations is to convey orders, instructions, or information so as to bring desired changes in the performance and other attitude of employees. In an organisation, supervisors transmit information to subordinates. Proper communication results in clarity and securing the cooperation of subordinates. Faulty communication may create problems due to misunderstanding between the superior and subordinates. The subordinates must correctly understand the message conveyed to them.

 

Communication Cycle :-

 

Sender—> Message—>Encoding—>Channel/Medium—>Transmission of message—> Receiving & Decoding—>Response & feedback—> Receiver.

 

Classification of Communication :-

 

  • On the basis of Organizational Structure:

 

  • Formal and Informal Communication

The path through which information flows is called channel of communication. In every organisation we have both formal and informal channels. The paths of communication which are based on relationship established formally by management are the formal channels.

For example, The Collector of the district communicates a decision to the SDM who may then issue orders or instructions to the Tahsildaar.

 

Communication, which takes place on the basis of informal or social relations among staff, is   called informal communication.

For example, any sharing of information between a police inspector and an accountant, as they happen to be friends or so. Mostly informal channels are used due to friendly interaction of members of an organisation. In fact, it may be purely personal or related to organisational matters.

 

 

 

  • On the basis of Direction

 

  • Upward: When employees make any request, appeal, report, suggest or communicate ideas to the superior, the flow of communication is upward i.e., from bottom to top. For instance, when a typist drops a suggestion in the suggestion box, or a foreman reports breakdown of machinery to the factory manager, the flow of communication is upward. Upward communication encourages employees to participate actively in the operations of their department. They get encouraged and their sense of responsibility increases when they are heard by their supervisors about problems affecting the jobs.

 

  • Downward: When communication is made from superiors down the hierarchy it is called a downward communication. For instance, when superiors issue orders and instructions to subordinates, it is known as downward communication. When the General Manager orders supervisors to work overtime, the flow of communication is downward i.e., from top to bottom. Similarly, communication of work assignments, notices, requests for performance, etc. through bulletin boards, memos, reports, speeches, meetings, etc, are all forms of downward communication.

 

 

  • Horizontal: Communication can also be amongst members at the same level in the organisation. For instance, production manager may communicate the production plan to the sales manager. This is known as horizontal flow of communication. Here, the communications among people of the same rank and status. Such communication facilitates coordination of activities that are interdependent.

 

  • Diagonal: when communication is not made between people who are in the same department nor at the same level of organisational hierarchy, it is called diagonal communication. For example, cost accountant may request for reports from sales representatives not the sales manager for the purpose of distribution cost analysis. This type of communication does take place under special circumstances.

 

  • On the basis of Mode of Expression

 

  • Verbal and Non verbalCommunication : On the basis of the mode used, communication may be verbal or non-verbal. While communicating, managers may talk to their subordinates either face to face or on telephone or they may send letters, issue notices, or memos. These are all verbal communication. Thus, the verbal modes of communication may be oral and written. Face to face communication, as in interviews, meetings and seminars, are examples of oral communication. Issuing orders and instructions on telephone or through an inter-communication system is also oral communication. The written modes of communication include letters, circulars, notices and memos. Sometimes verbal communication is supported by non-verbal communication such as facial expressions and body gestures. For example – wave of hand, a smile or a frown etc. This is also termed as the gestural communication

 

 

 

 

 

 

 

 

 

 

  1. SUPERVISION

 

It is the duty of the manager to see that they perform the work as per instructions. Managers play the role of supervisors and ensure that the work is done as per the instructions and the plans. Supervisors clarify all instructions and guide employees to work as a team in co-operation with others.

Though supervision is required at all levels of management, it is of great importance at the operational level i.e., at the level of first line supervisor. Managers at this level devote maximum time in supervising the work of subordinates. Though the top or middle level managers also supervise the work of their subordinate managers, but it is the first line supervisors who are in direct and constant touch with operatives i.e., workers in the factory and clerical staff in the office. Thus, they are directly responsible for getting the work done through most of the employees in an organisation.

 

Functions of a Supervisor

 

A supervisor works at the lowest level of management like all other managers he performs the functions of planning, organising, directing and controlling with respect to his own subordinates and department. A major part of his time is devoted in directing and controlling the activities of his subordinates. He also coordinates the activities of his subordinates by integrating the same with the activities of other departments of the enterprise. Besides he performs certain special functions which have been described below:

 

  1. Link between Top Management and Workers: A supervisor works as a link between managers working at higher levels and workers. He conveys the decision of the higher level managers to the workers and also communicates the performance of the workers to the higher level management through different performance reports. He also communicates the grievances, feelings of demands etc. of the workers to the higher level management.

 

  1. Creating Ideal Atmosphere: Being an important link between the operatives and the management a supervisor is expected to create an ideal atmosphere for work in the organisation by correctly communicating the ideas, wishes and decisions of the higher level management to the workers.

 

  1. Guiding the Workers: For obtaining best results the supervisor assigns jobs to the workers keeping in mind their ability and aptitude for work. He makes them available the necessary tools and equipments, raw materials etc. for proper execution of the jobs. He also guides the worker properly to ensure that the job is done with perfection and accuracy.

 

  1. Quality Output : A supervisor has to ensure quality output through constant watch on the performance of workers. He ensures that the performance of the worker takes place as per the plans. This results into study flow of output.

 

  1. Feedback: A supervisor keeps on watching the performance of his subordinates and identifies their strengths and weaknesses. He gives the feedback about this to the workers with the object to further improve the performance of the workers in future.

 

  1. Suggest Training Programmes: A supervisor identifies the areas in which the workers require training and accordingly suggests training programmes that should be organised for them.

 

 

 

  1. MOTIVATION

 

Motivation is one of the important elements of directing.

It is a force that inspires a person at work to intensify his willingness to use the best of his capability for achievement of specified objectives. It may be in the form of incentives like financial (such as bonus, commission etc.) or, non-financial (such as appreciation, growth etc.), or it could be positive or negative. Basically, motivations directed towards goals and prompt people to act.

 

The importance of motivation lies in converting this ability to work into willingness to work. Performance depends on ability as well as willingness; and willingness depends on motivation. Thus, motivation is a key element in directing people to do the job.

 

Each employee has some needs of his own that he wants to fulfil. While directing, it is essential to ensure that any of the unfulfilled need of the individual is being taken care of.

Maslow’s Hierarchy of Needs:-

 

According to Maslow, an individual has many needs and their order can be determined. If a person satisfies his first need, then he thinks about his next need. After satisfying the second need, he tries to satisfy third need and so on. So needs are the motivators. Maslow has given hierarchy of needs in the following ways :

 

  1. Physiological Needs: These needs include need for food, shelter and clothing.

 

  1. 2. Safety and Security Needs: Once physiological needs are fulfilled then the people start thinking about their safety. Safety needs include need for physical safety and economic safety. Physical safety means safety from accidents, disease etc. Economic safety refers to safety of livelihood.

 

  1. Social Needs: Man is a social animal. He wants to live in the society honourably. Therefore, he wants friends and relatives with whom he can share his joys and sorrows. Social needs include need for love, affection, friendship etc.
  2. 4. Esteem Needs: These are the need for respect and recognition. Esteem needs are also known as Ego needs.

 

  1. Self ActualisationNeeds : Self actualisation needs are concerned with becoming what a person is capable of becoming. These needs include need for growth, self-fulfilment etc.

 

 

Financial and Non-financial Hierarchy Theory

 

Monetary / Financial incentives are directly related with money. Non-financial incentives are not directly related with money. Following are the financial incentives:

 

  1. Pay and Allowances: Salary is the basic monetary incentive of every employee. Salary includes basic pay, dearness allowance etc.

 

  1. Bonus: Bonus means the payment to employees in addition to their regular remuneration. Bonus is provided in the form of cash, free trips to resorts or foreign countries etc.

 

  1. Commission: In sales department, sales persons get commission on the basis of their sales.

 

  1. Retirement Benefit: Every employee is concerned about his future after retirement. Some retirement benefits are Provident fund, Pension, Gratuity etc.

 

  1. Perquisites: Rent free accommodation, car allowance, facility of a servant etc.are called as perquisites.

 

Non-financial Incentives: Besides the financial incentives there is certain non financial incentive that motivates the employees. The important non-financial incentive is given below:

 

  1. Career Advancement Opportunity : Appropriate skill development programmes will encourage employees to show improved performance.

 

  1. Status: Status means the rank of a person in a organisation. The rank is linked with authority, responsibility and other extra benefits. Everybody has a wish to be in high rank. Therefore an employee can be motivated by placing him in higher rank.

 

  1. Employee Recognition Programmes: Every employee wants to be considered as an important part of the organisation. Work of an organisation should be distributed in such a way that every employee feels that his work is yield and he is capable to do that work. This motivates the worker and he works hard and in a responsible manner.

 

  1. Employee Participation: It means involving employee in decision making especially when decisions are related to workers.

 

  1. Organisation Climate: It means the relationship between superior and subordinates. Employees can put their best if healthy climate exist in an organisation. It is important to remember that the needs and desires of people change. Once their basic needs are satisfied, other needs arise. Managers have thus, to understand the needs and desires of subordinates and decide how to motivate them. The knowledge of the different types of need enables a manager to adopt different ways to motivate individuals depending upon which need is unsatisfied for the individual. For example, a person whose physiological needs are not fulfilled may be motivated to work with a promise of increase in pay, whereas another person may be motivated if he is given a very challenging job to perform regardless of the pay.

 

 

 

  1. LEADERSHIP

 

Leadership is the process, which influences the people and inspires them to willingly accomplish the organisational objectives. The main purpose of managerial leadership isto gets willing cooperation of the workgroup to achieve the goals.

Leadership is the ability to persuade and motivate others to work in desired way for achieving the goals. Thus, a person who is able to influence others and make them follow his instructions is called a leader.

Leadership and Management are two separate concepts.

Leadership exists in both formal and informal organization but Management operates in formal organization.

 

Leadership Styles :

 

  • Autocratic or Authoritarian Style : 2 types

Pure autocrative or negative Leader : Dictator & makes all decisions by himself.

Benevolent autocrat or Positive Leader : Reward power to influence subordinate and welfare of subordinates.

 

  • Participative Leaders : Decentralise authority, Such leaders involve subordinates in decision-making process.

 

  • Free-rein or Laissez – faire Style : Leaders uses his power very little, gives high degree of freedom to his subordinates in their operation. Aids subordinates in performing their job.

 

 

  • Paternalistic Leadership : It is authoritarian by Nature. Heavily work-centred but has consideration for subordinates.

 

Leadership Qualities: – In order to be successful, a leader must possess certain qualities. A good leader should be professionally competent, intelligent, analytical and he/she should have a sense of fair play, including honesty, sincerity, integrity, and sense of responsibility. He must possess initiative, perseverance, be diligent and realistic in his outlook. He must also be able to communicate his subordinates effectively. Human relation skills are must for any leader. Earlier, it was believed that the success or effectiveness of a leader depends upon his personal traits or characteristics, like physical appearance, intelligence, self-confidence, alertness, and initiative.

RELATIVE SPEEED AND TRAIN QUESTIONS

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Speed has no sense of direction unlike the velocity. Relative speed is the speed of one object as observed from another moving object. Questions on train are the classic examples of relative speed and in all these questions it is assumed that trains move parallel to each other – whether in the same direction or the opposite direction. Thus, we shall see how the relative speed is calculated and using it we come to know the time taken by the trains to cross each other and some other like aspects.

Important Formulas – Problems on Trains

  1. x km/hr = (x×5)/18 m/s

 

  1. y m/s = (y×18)/5 km/hr

 

  1. Speed = distance/time, that is, s = d/t

 

  1. velocity = displacement/time, that is, v = d/t

 

  1. Time taken by a train x meters long to pass a pole or standing man or a post
    = Time taken by the train to travel x meters.

 

  1. Time taken by a train x meters long to pass an object of length y meters

= Time taken by the train to travel (x + y) metres.

 

  1. Suppose two trains or two objects are moving in the same direction at v1 m/s and v2 m/s where v1 > v2,

then their relative speed = (v1 – v2) m/s

 

  1. Suppose two trains or two objects are moving in opposite directions at v1 m/s and v2 m/s ,

then their relative speed = (v1+ v2) m/s

 

  1. Assume two trains of length x metres and y metres are moving in opposite directions at v1 m/s and v2 m/s, Then

The time taken by the trains to cross each other = (x+y) / (v1+v2) seconds

 

  1. Assume two trains of length x metres and y metres are moving in the same direction at at v1 m/s and v2 m/s where v1 > v2, Then

The time taken by the faster train to cross the slower train = (x+y) / (v1-v2) seconds

 

  1. Assume that two trains (objects) start from two points P and Q towards each other at the same time and after crossing they take p and q seconds to reach Q and P respectively. Then,

A’s speed: B’s speed = √q: √p

 

 

Solved Examples

Level 1

1.A train is running at a speed of 40 km/hr and it crosses a post in 18 seconds. What is the length of the train?
A. 190 metres B. 160 metres
C. 200 metres

Answer : Option C

 D. 120 metres

 

Explanation :

Speed of the train, v = 40 km/hr = 40000/3600 m/s = 400/36 m/s

Time taken to cross, t = 18 s

Distance Covered, d = vt = (400/36)× 18 = 200 m

Distance covered is equal to the length of the train = 200 m

2.A train having a length of 240 m passes a post in 24 seconds. How long will it take to pass a platform having a length of 650 m?
A. 120 sec B. 99 s
C. 89 s D. 80 s

 

Answer : Option C

Explanation :

v = 240/24 (where v is the speed of the train) = 10 m/s

t = (240+650)/10 = 89 seconds

3.Two trains having length of 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions (on parallel tracks). The time which they take to cross each other, is
A. 10.8 s B. 12 s
C. 9.8 s D. 8 s

 

Answer : Option A

Explanation :

Distance = 140+160 = 300 m

Relative speed = 60+40 = 100 km/hr = (100×10)/36 m/s

Time = distance/speed = 300 / (100×10)/36 = 300×36 / 1000 = 3×36/10 = 10.8 s

4.A train moves past a post and a platform 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train?
A. 79.2 km/hr B. 69 km/hr
C. 74 km/hr D. 61 km/hr

 

Answer : Option A

Explanation :

Let x is the length of the train and v is the speed

Time taken to move the post = 8 s

=> x/v = 8

=> x = 8v — (1)

Time taken to cross the platform 264 m long = 20 s

(x+264)/v = 20

=> x + 264 = 20v —(2)

Substituting equation 1 in equation 2, we get

8v +264 = 20v

=> v = 264/12 = 22 m/s

= 22×36/10 km/hr = 79.2 km/hr

5.Two trains, one from P to Q and the other from Q to P, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is
A. 2 : 3 B. 2 :1
C. 4 : 3 D. 3 : 2

 

Answer : Option C

Explanation :

Ratio of their speeds = Speed of first train : Speed of second train

= √16: √ 9

= 4:3

 6.Train having a length of 270 meter is running at the speed of 120 kmph . It crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?
A. 320 m B. 190 m
C. 210 m D. 230 m

 

Answer : Option D

Explanation :

Relative speed = 120+80 = 200 kmph = 200×10/36 m/s = 500/9 m/s

time = 9s

Total distance covered = 270 + x where x is the length of other train

(270+x)/9 = 500/9

=> 270+x = 500

=> x = 500-270 = 230 meter

7.Two stations P and Q are 110 km apart on a straight track. One train starts from P at 7 a.m. and travels towards Q at 20 kmph. Another train starts from Q at 8 a.m. and travels towards P at a speed of 25 kmph. At what time will they meet?
A. 10.30 a.m B. 10 a.m.
C. 9.10 a.m. D. 11 a.m.

 

Answer : Option B

Explanation :

Assume both trains meet after x hours after 7 am

Distance covered by train starting from P in x hours = 20x km

Distance covered by train starting from Q in (x-1) hours = 25(x-1)

Total distance = 110

=> 20x + 25(x-1) = 110

=> 45x = 135

=> x= 3 Means, they meet after 3 hours after 7 am, ie, they meet at 10 am

8.Two trains are running in opposite directions in the same speed. The length of each train is 120 meter. If they cross each other in 12 seconds, the speed of each train (in km/hr) is
A. 42 B. 36
C. 28 D. 20

 

Answer : Option B

Explanation :

Distance covered = 120+120 = 240 m

Time = 12 s

Let the speed of each train = v. Then relative speed = v+v = 2v

2v = distance/time = 240/12 = 20 m/s

Speed of each train = v = 20/2 = 10 m/s

= 10×36/10 km/hr = 36 km/hr

 

Level 2

1.A train, 130 meters long travels at a speed of 45 km/hr crosses a bridge in 30 seconds. The length of the bridge is
A. 270 m B. 245 m
C. 235 m D. 220 m

 

Answer : Option B

Explanation :

Assume the length of the bridge = x meter

Total distance covered = 130+x meter

total time taken = 30s

speed = Total distance covered /total time taken = (130+x)/30 m/s

=> 45 × (10/36) = (130+x)/30

=> 45 × 10 × 30 /36 = 130+x

=> 45 × 10 × 10 / 12 = 130+x

=> 15 × 10 × 10 / 4 = 130+x

=> 15 × 25 = 130+x = 375

=> x = 375-130 =245

2.A train has a length of 150 meters. It is passing a man who is moving at 2 km/hr in the same direction of the train, in 3 seconds. Find out the speed of the train.
A. 182 km/hr B. 180 km/hr
C. 152 km/hr D. 169 km/hr

 

Answer : Option A

Explanation :

Length of the train, l = 150m

Speed of the man, Vm= 2 km/hr

Relative speed, Vr = total distance/time = (150/3) m/s = (150/3) × (18/5) = 180 km/hr

Relative Speed = Speed of train, Vt – Speed of man (As both are moving in the same direction)

=> 180 = Vt – 2 => Vt = 180 + 2 = 182 km/hr

3.Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively. If they cross each other in 23 seconds, what is the ratio of their speeds?
A. Insufficient data B. 3 : 1
C. 1 : 3 D. 3 : 2

 

Answer : Option D

Explanation :

Let the speed of the trains be x and y respectively

length of train1 = 27x

length of train2 = 17y

Relative speed= x+ y

Time taken to cross each other = 23 s

=> (27x + 17 y)/(x+y) = 23 => (27x + 17 y)/ = 23(x+y)

=> 4x = 6y => x/y = 6/4 = 3/2

4.A jogger is running at 9 kmph alongside a railway track in 240 meters ahead of the engine of a 120 meters long train . The train is running at 45 kmph in the same direction. How much time does it take for the train to pass the jogger?
A. 46 B. 36
C. 18 D. 22

 

Answer : Option B

Explanation :

Distance to be covered = 240+ 120 = 360 m

Relative speed = 36 km/hr = 36×10/36 = 10 m/s

Time = distance/speed = 360/10 = 36 seconds

5.A train passes a platform in 36 seconds. The same train passes a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, The length of the platform is
A. None of these B. 280 meter
C. 240 meter D. 200 meter

 

Answer : Option C

Explanation :

Speed of the train = 54 km/hr = (54×10)/36 m/s = 15 m/s

Length of the train = speed × time taken to cross the man = 15×20 = 300 m

Let the length of the platform = L

Time taken to cross the platform = (300+L)/15

=> (300+L)/15 = 36

=> 300+L = 15×36 = 540 => L = 540-300 = 240 meter

6.A train overtakes two persons who are walking in the same direction to that of the train at 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. What is the length of the train?
A. 62 m B. 54 m
C. 50 m D. 55 m

 

Answer : Option C

Explanation :

Let x is the length of the train in meter and v is its speed in kmph

x/9 = (v-2) (10/36) — (1)

x/10 = (v-4) (10/36) — (2)

Dividing equation 1 with equation 2

10/9 = (v-2)/(v-4) => 10v – 40 = 9v – 18 => v = 22

Substituting in equation 1, x/9 = 200/36 => x = 9×200/36 = 50 m

7.A train is traveling at 48 kmph. It crosses another train having half of its length, traveling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. What is the length of the platform?
A. 500 m B. 360 m
C. 480 m D. 400 m

 

Answer : Option D

Explanation :

Speed of train1 = 48 kmph

Let the length of train1 = 2x meter

Speed of train2 = 42 kmph

Length of train 2 = x meter (because it is half of train1’s length)

Distance = 2x + x = 3x

Relative speed= 48+42 = 90 kmph = 90×10/36 m/s = 25 m/s

Time = 12 s

Distance/time = speed => 3x/12 = 25

=> x = 25×12/3 = 100 meter

Length of the first train = 2x = 200 meter

Time taken to cross the platform= 45 s

Speed of train1 = 48 kmph = 480/36 = 40/3 m/s

Distance = 200 + y where y is the length of the platform

=> 200 + y = 45×40/3 = 600

=> y = 400 meter

8.A train, 800 meter long is running with a speed of 78 km/hr. It crosses a tunnel in 1 minute. What is the length of the tunnel (in meters)?
A. 440 m B. 500 m
C. 260 m D. 430 m

 

Answer : Option B

Explanation :

Distance = 800+x meter where x is the length of the tunnel

Time = 1 minute = 60 seconds

Speed = 78 km/hr = 78×10/36 m/s = 130/6 = 65/3 m/s

Distance/time = speed

(800+x)/60 = 65/3 => 800+x = 20×65 = 1300

=> x = 1300 – 800 = 500 meter

9.Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. If the fast train completely passes a man sitting in the slower train in 5 seconds, the length of the fast train is :
A. 19 m B. 2779 m
C. 1329 m D. 33 m

 

Answer : Option B

Explanation :

Relative speed = 40-20 = 20 km/hr = 200/36 m/s = 100/18 m/s

Time = 5 s

Distance = speed × time = (100/18) × 5 = 500/18 m = 250/9 = 2779 m = length of the fast train

Tropical and temperate cyclones

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The atmospheric disturbances which involve a closed circulation about a low pressure centre,
anticlockwise in the northern atmosphere and clockwise in the southern hemisphere are called
cyclones. They fall into the following two broad categories: (a) Extra-tropical or Temperate and (b) tropical cyclones.

(a) Temperate Cyclones
Temperate cyclones are formed along a front in mid-latitudes between 35° and 65° N and S. They blow from west to east and are more pronounced in winter season.Temperate cyclones are mainly observed in Atlantic Ocean and North West Europe . They are generally extensive having a thickness of 9 to 11 kilometers and with 1040-1920 km short and long diametres respectively. Each such cyclone alternates with a high pressure anticyclone. The weather associated with the cyclone is drizzling rain and of cloudy nature for number of days. The anticyclone weather is sunny, calm and of cold waves.
(b) Tropical Cyclones
Tropical cyclones are formed along the zone of confluence of north-east and south-east trade winds. This zone is known as the Inter Tropical Convergence Zone (ITCZ). Cyclones generally occur in Mexico, South-Western and North Pacific Ocean, North Indian Ocean and South Pacific Ocean. These cyclones differ from temperate cyclones in many ways. There are no clear warm and cold
fronts as temperature seldom differs in Inter Tropical Convergence Zone. They do not have well-defined pattern of winds and are energised by convectional currents within them. Generally, these are shallow depressions and the velocity of winds is weak. These are not accompanied by anticyclones. The arrangement of isobars is almost circular. These are not extensive and have the diametres of 160-640km. However, a few of them become very violent and cause destruction in the regions of their influence. They are called hurricanes in the Carribean Sea, typhoons in the China, Japan and phillipines,

 

 

 

 

 

 Evaporation and Condensation: dew, frost, fog, mist and cloud, rainfall types

 Evaporation 

 

Evaporation is the process of which water changes from its liquid state to gaseous form. This process takes place at all places, at all times and at all temperatures except at dew point or when the air is saturated. The rate of evaporation is affected by several factors. Important among them are as under:
(i) Accessibility of water bodies :-The rate of evaporation is higher over the oceans than on the continents.
(ii) Temperature :-when the temperature of an air is high, it is capable of holding more moisture in its body than at a low temperature. It is because of this that the rate of evaporation is more in summers than in winters. That is why wet clothes dry faster in summers than in winters.
(iii) Air moisture :-If the relative humidity of a sample of air is high, it is capable of holding less moisture. On the other hand if the relative humidity is less, it can take more moisture. Hence, the rate of evaporation will be high. Aridity or dryness of the air also increases the rate of evaporation. During rainy days, wet clothes take more time to dry owing to the high percentage of moisture content in the air, than on dry days.
(iv) Wind :-If there is no wind, the air which overlies a water surface will get saturated through evaporation. This evaporation will cease once saturation point is reached. However, if there is wind, it will blow that saturated or nearly saturated air away from the evaporating surface and replace it with air of lower humidity. This allows evaporation to continue as long as the wind keep blowing saturated air away and bring drier air.
(v) Cloud cover :-The cloud cover prevents solar radiation and thus influences the air temperatures at a place. This way, it indirectly controls the process of evaporation.

Condensation

Condensation the process by which water vapor (gas) in the atmosphere turns into water (liquid state). It is the opposite of evaporation. This stage is very important because it is the cloud formation stage. Cool temperatures are essential for condensation to happen, because as long as the temperature in the atmosphere is high, it can hold the water vapor and delay condensation.

When a gas is cooled sufficiently or, in many cases, when the pressure on the gas is increased sufficiently, the forces of attraction between molecules prevent them from moving apart, and the gas condenses to either a liquid or a solid.

  • Example: Water vapor condenses and forms liquid water (sweat) on the outside of a cold glass or can.
  • Example: Liquid carbon dioxide forms at the high pressure inside a CO2 fire extinguisher.

The temperature of the air falls in two ways. Firstly, cooling occurs around very small particles of freely floating air when it comes in contact with some colder object. Secondly, loss in air temperature takes place on a massive scale due to rising of air to higher altitudes. The condensation takes place around the smoke, salt and dust particles which attract water vapour to condense around them. They are called hygroscopic nuclei. When the relative humidity of an air is high, a slight cooling is required to bring the temperature down below dew point. But when the relative humidity is low and the temperature of the air is high, a lot of cooling of the air will be necessary to bring the temperature down below dew point. Thus, condensation is directly related to the relative humidity and the rate of cooling.

here are four types of condensation and the worst period for such problems is September to May:-

  1. Surface condensation. This is the most familiar type of condensation, taking the form of water on window panes, cold wall surfaces and tiles.
  2. Interstitial condensation. This is condensation forming between walls or within the building structure.
  3. Reverse condensation. This is also called “Summer condensation”. If rains drenches a wall and strong sunlight then dries it, the heat can actually force water vapour into the wall. When it then meets an insulated surface, it forms condensation at that barrier.
  4. Radiation condensation. This is sometimes called “clear night condensation“. If there is a sudden temperature drop at night, it can cause condensation on the underside of roof coverings, for example: often this drips onto the insulation quilting and causes a distinctive mottled effect upon the quilting.

 

Dew, Frost, Fog, Mist and Cloud

Dew: When the atmospheric moisture is condensed and deposited in the form of water droplets on cooler surface of solid objects such as grass blades, leaves of plants and trees and stones, it is termed as dew. Condensation in dew form occurs when there is clear sky, little or no wind, high relative humidity and cold long nights. These conditions lead to greater terrestrial radiation and the solid objects become cold enough to bring the temperature of air down below dew point. In this process the extra moisture of the air gets deposited on these objects. Dew is formed when dew point is above freezing point. Dew formation can be seen if the water is poured into a glass from the bottle kept in a refrigerator. The outer cold surface of the glass brings the temperature of the air in contact with the surface down below dew point and extra moisture gets deposited on the outer wall of the glass.
Frost: When the dew point is below freezing point, under above mentioned conditions, the condensation of extra moisture takes place in the form of very minute particles of ice crystals. It is called frost. In this process, the air moisture condenses directly in the form of tiny crystal of ice. This form of condensation is disastrous for standing crops such as potato, peas, pulses, grams, etc. It also creates problems for road transport system.
Mist and Fog: When condensation takes place in the air near the earth’s surface in the form of tiny droplets of water hanging and floating in the air, it is called mist. In mist the visibility is more than one kilometer and less than two kilometers. But when the visibility is reduced to less than one kilometer, it is called fog. Ideal conditions for the formation of mist and fog are clear sky, calm and cold winter nights.
Cloud: Clouds are visible aggregates of water droplets, ice particles, or a mixture of both along with varying amounts of dust particles. A typical cloud contains billions of droplets having diameters on the or- der 060.01 to 0.02 mm; yet liquid or solid water accounts for less than 10 parts per million of the cloud volume. Clouds are generally classified on the basis of their general form or appearance and alti- tude.

Rainfall types.

Precipitation or Rainfall is defined as water in liquid or solid forms falling to the earth. It happens when continuous condensation in the body of air helps the water droplets or ice crystals to grow in size and weight that the air cannot hold them and as a result these starts falling on the ground under the force of gravity.

Different types of Rainfall are:-

  • Convectional Rainfall :-Excessive heating of the earth’s surface in tropical region results in the vertical air currents. These currents, lift the warm moist air to higher strata of atmosphere. When-the temperature of such a humid air starts falling below dew point continuously, clouds are formed. These clouds cause heavy rainfall which is associated with lightning and thunder. This type of rainfall is called conventional rainfall. It is very common in equatorial region where it is a daily phenomenon in the afternoon
    (b) Orographic or Relief Rainfall :-Orographic rainfall on formed where air rises and cools because of a topographic barrier. When their temperature fall below dew point, clouds are formed. These clouds cause widespread rain on the windward slopes of the mountain range. This type of rain is called orographic rainfall. However when these winds cross over the mountain range and descend along the leeward slopes, they get warm and cause little rain. Region lying on the leeward side of the mountain receiving little rain is called rainshadow area (see figure 12.4). A famous example of orographic rainfall is Cherrapunji on the southern margin of the Khasi Hills in Meghalaya India.
    (c) Convergence or Cyclonic Rainfall:-Convergence rainfall, produced where air currents converge and rise. In tropical regions where opposing air currents have comparable temperatures, the lifting is more or less vertical and is usually accompanied by con- vention. Convectioned activity frequently occurs along fronts where the temperature of the air masses concerned are quite different. Mixing of air along the front also probably contributes to condensation and therefore to the frontal rainfall. When two large air masses of different densities and temperature meet, the warmer moist air mass is lifted above the colder one. When this happens, the rising warm air mass condenses to form clouds which cause extensive down pour. This rainfall is associated with thunder and lightning. ‘This type of rainfall is also called frontal rainfall. This type of rainfall is associated with both warm and cold fronts, (fig. 12.5) It is gener- ally steady and may persist for a whole day or even longer.

 

 

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