PERCENTAGE (%)

A percentage is a ratio expressed in terms of a unit being 100. A percentage is usually denoted by the symbol

“%”.

To express a% as a fraction, divide it by 100 ⇒ a% = a/100

To express a fraction as %, multiply it by 100 ⇒ a/b = [(a/b) × 100] %

x% of y is given by (y × x/100 )

Point to remember for faster Calculation

1 = 100%

1 /2= 50%

1 /3= 33.33%

1/4 = 25%

1/5 = 20%

1 /6= 16.66%

1/7 = 14.28%

1/8 = 12.5%

1/9 = 11.11%

1/10 = 10%

1/11 = 9.09%

1/12 = 8.33%

Some Short tricks based on Condition

If A's income is r% more than B's income, the B's income is less than A's income by

[ r / (100+r)] * 100%

If A‟s income is r% less than B‟s income, the B‟s income is more than A‟s income by

[ r / (100-r)] * 100%

If 'A' is x% of 'C' and 'B' is y% of 'C' then

'A' is (x/y) * 100% of 'B'.

If the sides of the triangle, rectangle, square, circle, rhombus etc is

(i) Increased by x%. Its area is increased by

2x+(x2/100)

(ii)If decreased x%. Its areas is decreased by,

-2x+(x2/100)

If a number „x‟ is successively changed by a%, b%, c%. then final value :

x (1+a/100) (1+b/100) (1+c/100)

The net change after two successive changes of a% and b% is

(𝑎+ 𝑏+𝑎𝑏/100) %

The population of a town is 'P'. It increased by x% during 1st year, increased by y% during 2

nd year and

again increased by z% during 3rd

year. The population after 3 years will be,

P * [(100+x)/100] * [(100+y)/100] * [ (100+z)/100 ]

Rate Change and Change in quantity available for fixed expenditure:

Let original price = x per unit quantity;

x(x+ rate change)= (Expenditure × Rate change) / (change in available quantity)

Here rate change = change in rate per unit quantity .

Mixture problems:

If x% of a quantity is taken by the first person, y% of the remaining quantity is taken by the second person, and

z% of the remaining is taken by the third person and if A is left, then initial quantity was

The same concept we can use, if we add something, then the initial quantity was

Profit & Loss Cost Price-The price at which an article is purchased is called its cost price (C.P.)

Selling Price-The price at which the article is sold is called its selling price (S.P.)

CP = Cost Price

SP = Selling Price

When SP < CP → Loss = CP - SP

When SP > CP → Profit = SP - CP

Note: Loss% and Profit% both are calculated upon CP

Profit% = [Profit/CP] * 100

Loss% = [Loss/CP] * 100

Now, some more general things we can deduce...

(1) When profit% is given, say, 10% profit

Then SP becomes [100+profit%] of CP = [100+Profit%]/100

For 10% profit, SP = 110% of CP = [110/100]*CP = 1.1*CP

Means to say, if 10% profit, then we simply do SP = 1.1*CP

If CP is given, then SP is calculated, and if SP is given then CP is calculated

(CP = SP/1.1)

(2) When loss% is given, say, 10% loss ,Then again, SP becomes [100-loss%] of CP, again same approach.

For 10% loss, SP becomes 90% of CP = 0.9*CP

If CP is given, then SP is calculated, and if SP is given, then CP is calculated

(CP = SP/0.9)

e.g.

If an article is sold at a profit/gain of 30%, then S.P. = 130% of the C.P.

If an article is sold at a loss of 20%, then S.P. = 80% of the C.P.

When there are two successive Profit of x % and y % then the resultant profit per cent is given by

[x + y+ (x × y/100)]

If there is a Profit of x% and loss of y % in a transaction, then the resultant profit or loss% is given by

[x – y - (x × y/100)]

Note- For profit use sign + in previous formula and for loss use – sign.

if resultant come + then there will be overall profit . if it come – then there will be overall loss.

If a cost price of m articles is equal to the selling Price of n articles,

(C.P of m article = S.P. of n article) then Profit percentage

(m - n)/n×100%

If m part is sold at x% profit , n part is sold at y % profit, and p part is sold at z% profit and Rs. r is earned

as overall profit then the value of total consignment

r×100 / (mx +ny +pz)

A man purchases a certain no. of article at m a rupee and the same no. at n a rupee. He mixes them together

and sold them at p a rupee then his gain or loss

[{2mn/(m+n)p} -1]× 100 Note += profit ,- = loss

When a person sells two similar items, one at a gain of say x%, and the other at a loss of x%, then in this

transaction the seller always incurs a loss given by:

(x^2/100)% Marked price = Cost price + Markup

CP --> + Markup --> = MP

Always Remember: Markup is extra price on Cost Price. So, Markup is always calculated on CP

And

%Markup = [Markup/CP]*100

Means to say if 400Rs CP and 100Rs Markup, then MP becomes 500Rs and %Markup becomes [100/400]*100

= 25%

Discount (if SP < MP) = MP - SP i.e. SP = MP - Discount

MP → - Discount → = SP

Always Remember: Discount is deducted from Marked Price. So, Discount is always calculated on MP and

%Discount = [Discount/MP]*100

A single discount equivalent to discount series of x% and y% given by the seller is equal to

(x +y - xy/100)%

If a seller marks his goods at x% above his cost price and allows purchasers a discount of y % for cash,

then overall gain or loss

(x – y –xy/100)%

If a trader professes to sell his goods at cost price, but uses false weights, then

Gain% = {(Error)/(True value - Error)x 100] %

SIMPLE INTEREST AND COMPOUND INTEREST

Principal: - The money borrowed or lent out for certain period is called the principal or the Sum.

Interest: - Extra money paid for using other money is called interest.

Simple Interest (S.I.):- If the interest on a sum borrowed for certain period is reckoned uniformly, then it is

called simple interest.

Let Principal = P, Rate = r % per annum (p.a.), and Time = t years then

Using this formula we can also find out

Principal or Sum ; Rate % per annum ;

Time (in years) .

Compound Interest: When compound interest is applied, interest is paid on both the original principal and on earned interest.

So for One year Simple interest and Compound interest both are equal.

So Amount at the end of 1st year (or Period) will become the principal for the 2

nd year (or Period) and Amount

at the end of 2nd year (or Period) becomes the Principal of 3rd

year.

Amount = Principal + Interest

A= P (1+r/100) ^n A= Amount, P= Principal, r = Rate %, n = no. of years.

So Compound Interest = [P (1+r/100) ^ n - P]

C.I. = P [(1+r/100) ^ n – 1]

Some Important Condition:-

i) When interest is compounded annually,

Amount =

ii) When interest is compounded half yearly,

Amount =

iii) When interest is compounded Quarterly,

Amount =

iv) When interest is compounded annually but time is in fraction, say 3 year

Amount = (

)

(

)

v) When Rates are different for different years, say r1%, r2%, and r3% for 1st, 2

nd and 3

rd year respectively.

Then,

Amount = . vi) Difference between Compound Interest & Simple interest Concept

a) For Two years

– (

)

b) For Three Year

–

vii) For Two year ratio of Compound Interest to Simple interest

RATIO & PROPORTION / AVERAGE / PROBLEM ON AGES

A Ratio is comparison of two quantities by division.

A Proportion is a statement that two ratio or equivalent.

A Proportion is considered to be true if the ratios on the both side are equivalent.

If a: b = c: d, we write a: b :: c : d and we say that a, b, c, d are in proportion.

a : b :: c : d (b x c) = (a x d) Fourth Proportional:

If a : b = c : d, then d is called the fourth proportional to a, b, c.

Third Proportional: a : b = c : d, then c is called the third proportion to a and b.

Mean Proportional: Mean proportional between a and b is ab.

Duplicate Ratios: Duplicate ratio of (a : b) is (a

2 : b

2).

Sub-duplicate ratio of (a : b) is (a1/2

: b1/2

).

Triplicate ratio of (a : b) is (a3 : b

3).

Sub-triplicate ratio of (a : b) is (a1/3

: b1/3

).

Properties of Proportion: If a/b = c/d then,

b/a= d/c

a/c = b/d

a × d=b × c

Average An average or more accurately an arithmetic mean is, in crude terms, the sum of n different data divided by n.

Averages of a group are defined as the ratio of sum of all the items in the group to the number of items in the

group.

Average = (Sum of all items in the group)/ Number of items in the group

Some Important Concepts:

Average =total of data/No. of data

If the value of each item is increase by the same value a, then the average of the group or items will also

increase by a.

If the value of each item is decreased by the same value a, then the average of the group of items will also

decrease by a.

If the value of each item is multiplied by the same value a, then the average of the group or items will also get

multiplied by a.

If the value of each item is multiplied by the same value a, then the average of the group or items will also get

divided by a.

If we know only the average of the two groups individually, we cannot find out the average of the combined

group of items.

Average of n natural no's = (n+1)/2

Average of even No’s = (n+1)

Average of odd No's = n

Change in the value of the Quantity and its effect on Average

When one/more than one quantity are removed but replaced with same no. of quantities of different value,

Change in the no. of quantities and its effect on Average

+ = if quantities ADDED, - = if quantities removed

Problem on Ages: 1. If the current age is x, then n times the age is n x.

2. If the current age is x, then age n years later/hence = x + n.

3. If the current age is x, then age n years ago = x - n.

4. The ages in a ratio a : b will be ax and bx.

5. If the current age is x, then 1/n of the age is x/n .

Mixture or Allegation: Mixtures are generally two types.

i) Simple mixture:- When two different ingredients are mixed together

ii) Compound mixture:- when two or more simple mixture are mixed together to form another mixture,

it is known as a compound mixture.

Allegation: Allegation means linking; it is the rule that enables us to find the ratio in which two or more

ingredients at the given price must be mixed to produce a mixture of desired price.

Mean Price: The cost of a unit quantity of the mixture is called the mean price.

Rule of Allegation:

If two ingredients are mixed then, 𝑎 𝑓 𝑐 𝑎 𝑒 𝑒

𝑎 𝑓 𝑑𝑒𝑎 𝑒

𝑓 𝑑𝑒𝑎 𝑒 𝑒𝑎 𝑐𝑒

𝑒𝑎 𝑐𝑒 𝑓 𝑐 𝑒𝑎 𝑒

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 = (

)

CONCEPT REVISION QUESTIONS:

1.) The side of a square increases by p%, then find by what percent its area increase does?

A) 2p B) p ^2/100 C) (2p+p/100) ^2

D) 2p+ p^2/100 E) None of these

2.) ‘a’ liters of oil was poured into a tank and it was still e% empty How much oil must be poured into

the tank in order to fill it to the brim?

A) (a × e)/(100-e) B) a/(100-e) C) 100a/(100-e)

D) 100ae / (100-e) E) None of these

3.) A reduction of Rs. 2 per kg enables a man to purchase 4 kg more sugar for Rs. 16, Find the original

Price of sugar? A) 2/kg B) 8/kg C) 16/kg

D) 4/kg E) None of these

4.) A reduction of 25% in the Price of Dairy milk Chocolates, enables a person to buy 5 kg more for Rs.

120, Find the original price of Dairy milk Chocolates/kg?

A) 10/kg B) 12/kg C) 8/kg

D) 6/kg E) None of these

5.) If the price of rice is increased by 7%, then by how much per cent should a housewife reduce her

consumption of sugar, to have no extra expenditure?

A) 7 B) 6.5 C) 8

D) 10 E) None of these

6.) The Daily age of workers increase by 20 % but the no. of hours worked by him per week drops by 20

%. If he was originally getting Rs. 500 per week, how much will he get now? A) Rs. 480

B) Rs. 510

C) Rs. 490

D) Rs. 475

E) None of these

7.) A fruit seller bought a number of apples at 6 for rupee and sold them at 4 for rupee.

I) Find his gain per cent?

II) If his total gain is Rs. 26, find the no. of apples he bought?

A) 25 %, 350 B) 50 %, 312 C) 75%, 300

D) 80 %, 305 E) None of these

8.) A man sold two chairs for Rs. 500 each .On,one he gain 20% and on other he loses 12%. How much

does he gain or lose in the whole transaction?

A) Loss 5% B) Gain 2%

C) Gain 1.5 % D) Loss 1.5% E) None of these

9.) Raman bought a bag with 20 % discount on the original price. He got a profit of Rs. 60 by selling at

150 % of the Price at which he bought. What is the original price of bag? A) 600 B) 300 C) 150

D) 125 E) None of these

10.) The list Price of watch is Rs. 160. A retailer bought the same watch for Rs. 122.40. He got successive

discount one at 10% and the other at a rate which was not legible. What is the second discount rate?

A) 20% B) 15% C) 10%

D) 12.5% E) None of these

11.) A milk man makes a Profit of 20% on the sale of milk .If he were to add 10% water to the milk, by

what % would profit increase?

A) 12 B) 15 C) 10

D) 32 E) None of these

12.) In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price

remains constant, approximately what percentage of the selling price is the profit?

A) 25% B) 50 % C) 70%

D) 75 % E) None of these

13.) Salaries of Ankit and Sumit are in the ratio 2 : 3. If the salary of each is increased by Rs. 4000, the

new ratio becomes 40 : 57. What is Sumit's salary?

A) 40000 B) 42000 C) 38000

D) 45000 E) None of these

14.) In a bag, there are coins of 25 p, 10 p and 5 p in the ratio of 1 : 2 : 3. If there is Rs. 30 in all, how

many 5 p coins are there?

A) 150

B) 200

C) 100

D) 175

E) None of these

15.) In a mixture 60 liters, the ratio of milk and water 2 : 1. If this ratio is to be 1 : 2, then the quantity of

water to be further added is:

A) 45 liter B) 60 liter C) 50 liter

D) 100 liter E) None of these

16.) The age of Chandan’s father is 4 times his age, 5 year ago it was 7 times of the age of his son. At that

time find the Present age of father?

A) 45 year B) 60 year C) 40 year

D) 50 year E) None of these

17.) The ratio of the ages of Reeta & Geeta is 4:5 if the sum of the ages 81 years, find the ratio of their

ages after 9 years?

A) 9:13 B) 36:45 C) 45: 54

D) 33:11 E) None of these

18.) A, B and C start a business with investment in ratio 5:6:8 respectively After one year C withdraw

50% of his capital and A increase his capital by 60%of his investment. After two years, in what ratio

should they earn profit divided among A,B and C respectively?

A) 13:12:12

B) 25: 24:16

C) 12:12:19

D) 16:23:19

E) None of these

19.) A and B enter into a partnership and invest in stock market trading. Their investment initially was

Rs. 50,000 and 45,000 respectively. After 4 months A withdraw half of his capital. At the end of 8

months B withdraw half of his capital and C joins with a capital of Rs. 70,000.What would be the

ratio in the profit?

A) 40:45:28

B) 45:29:54

C) 40:25:17

D) 36:43:49

E) None of these

20.) The Simple interest on a sum of money is 1/9 of the Principal and the no. of years is equal to the rate

%pa. The rate % pa is?

A) 9 %

B) 3.33%

C) 10%

D) 16.67%

E) None of these

21.) A sum of money is accumulating at compound interest at a certain rate of interest. If Simple interest

instead of compound interest were reckoned, the interest for the first two years would be diminished

by Rs. 20 and that for the first three years by Rs. 61. Find the Sum?

A) 8000

B) 6000

C) 10,000

D) 12,000

E) None of these

22.) A certain sum is interested at compound. The interest occurred in the first two years is Rs. 272 and

that in the First three years is Rs. 434. Find the rate %?

A) 33.33% B) 25% C) 12.5%

D) 16.67% E) None of these

23.) Sarah opened a restaurant, with a initial investment of Rs. 30,000. In that first year, he incurred a

loss of 5%. However, during the second year, he earned the Profit of 10% which in third year rose to

12.5%. Calculate his net profit for the entire period of three years?

A) 5300 B) 5620 C) 6000

D) 6600 E) None of these

24.) A bottle is full of Dettol. 1/3 of it is taken out and then an equal amount of water is poured into bottle

to fill it. This operation is done 4 times. Find the final ratio of Dettol and water in the bottle?

A) 43: 65 B) 17: 43 C) 16:63

D) 16:65 E) None of these

25.) In what ratio must a grocer mix two varieties of rice costing Rs. 15 and Rs. 20 per kg respectively

so as to get a mixture worth Rs. 16.50 kg? A) 8:9 B) 7: 10 C) 7:3

D) 8: 9 E) None of these

26.) A shopkeeper gives 12% additional discount on the discounted price, after giving an initial discount of 20%

on the labeled price of a radio. If the final sale price of the radio is Rs.704, then what is its labeled price? A) 1000 B) 2000 C) 1500

D) 3000 E) None of these

27.) In Arun’s opinion, his weight is greater than 65 kg but less than 72 kg. His brother doesn’t agree with Arun

and he thinks that Arun’s weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight

cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different

probable weights of Arun? A) 67 kg B) 68 kg C) 69 kg.

D) Data inadequate E) None of these

28.) The average weight of 8 person's increases by 2.5 kg when a new person comes in place of one of them

weighing 65 kg. What might be the weight of the new person?

A) 76Kg B) 76.5Kg C) 85Kg

D) Data inadequate E) None of these

29.) A fruit seller had some apples. He sells 40% apples and still has 420 apples, originally he had A) 600 B) 700 C) 800

D) 90 E) None of these

30.) The ratio between the present ages of Meera and Priya is 3 : 4 respectively. Ten years ago the ratio between

their ages was 4 : 7 respectively. What will be Meera’s age after 5 years? A) 23 year B) 24 year C) 25 year

D) Can‟t be determined E) None of these

ANSWERS OF CONCEPT REVISION QUESTIONS:

1) D

2) A

3) D

4) C

5) B

6) A

7) B

8) D

9) C

10) B

11) A

12) C

13) C

14) A

15) B

16) C

17) C

18) A

19) A

20) B

21) A

22) C

23) B

24) D

25) C

26) A

27) A

28) C

29) B

30) A

1. % change in area = p + p + (p*p)/100 = 2p+p^2/100

2. After filling ‘a’ liter still e% empty. So (100-e)% = a liter ;

Let x liter require to fill remain tank.

So full capacity of tank = 100a/(100-e) So x= e% of capacity = 100a/(100-e) * e/100 = ae/(100-e)

3. Expenditure is fixed, let original price be x per kg. So x(x-2) = (16 × 2) / 4 On solving and taking + value in consideration x= 4per kg.

4. Expenditure is fixed but rate decrease by 25%. So quantity increases by 5 kg. Let the original price is x/kg. So x(x-0.25x) = (120 * 0.25x) /5 So x= 8/kg.

5. No change in expenditure. So % reduce consumption = 7 /(100+7) × 100 = 6.5 % approx.

6. There is always loss. So loss % = 20*20 /100 = 4% So New Price = 96% of 500 = 480

7. Gain % = 2/4 * 100 = 50% And Total gain in Quantity / Total apple = unit S.P. – Unit C.P 26/ Total Apple = (1/4- 1/6) Total Apple = 312

8. Overall gain or loss = [100- (2* 120*88) / ( 120+ 88)] %

= 100 – 101.5 = -1.5 So loss 1.5%

9. Let the original Price x Rs. So selling Price = x*80/100 * 150/ 100= 1.2x Difference = 1.2x- 0.8x= 60 So x= 150 Rs.

10. Marked Price = 160 Selling Price = 122.40 160 *90/100 *x / 100 = 122.40 On solving x= 85 So % discount = 100-85 = 15%

11. After adding water total profit = 20 + 10 + 20*10/100 = 32% Increase in Profit = 32-20 = 12%

12. Let C.P = 100 Profit = 320 S.P = 420 New C.P = 125 New Profit = 420- 125 = 295 Profit % of the selling price = 295/420 ×100 = 70% approx

13. 2x+4000 / 3x+4000 = 40/57 On solving, Sumit’s salary = 34, 000 Option E

14. No. of coin * value of coin in rupees = Amount in rupees. So (1*.0.25x) + (2*0.10x)+(3*0.05x) = Rs. 30 On solving x= 50 So no. of coin of 5p= 3x = 150

15. Mixture = 60 liter ; Ratio of milk: water = 2:1 So milk = 40, Water = 20 Water added = x 40/20+x= 1:2, on solving x= 60 liter;

16. Age of father (F)= 4* age of chandan (C) F-5 =7(C-5) On solving both, C= 10; Then age of father = 4C= 40 year

17. 4x+5x= 81; So x=9 Ratio after 9 years= (36+9) : (45+9) = 45:54

18. 1st year ratio = 5:6:8, At 2 nd year = 8:6:4 After two year profit divided in ratio = 13:12:12

19. Share of A= 50k*4 + 25k*8 = 400k Share of B= 45*8+22.5*4= 450k Share of C= 70k*4= 280k ratio of A:B:C = 40:45:28

20. R=(100/9)^1/2 =3.33% 21. P(r/100)^2= 40

P(r/100)^2 ×(300+r)/100 =61 On Solving P= 8000

22. P[(1+r/100)^2 -1]= 272 P[(1+r/100)^3 -1]= 434 On solving r= 12.5%

23. 1st year loss = 1500

Price after 1st year = 28500,

2nd

year profit = 2850

Price after 2nd

year =31350

Profit in 3rd

year = 3918.75

So net profit during 3 years = 3918.75+ 2850 –

1500= 5268.75

So option E is true

24. Final ratio of Dettol and water = (

)

{ } =

16: 65 25. Ratio= (20- 16.5) / (16.5-15) = 7:3 26. Let labeled price is x

So 704= x*80/100*88/100 On solving x= 1000

27. Go through statements, it’s clear that average of probable weight of Arun is 67 kg.

28. Let new person weight = y and avg of previous 8 person = x So 8x+y-65=8(x+2.5) on solving y= 85kg

29. 60% = 420 So 100% = 700

30. 3x-10 / 4x-10 = 4/7 on solving x=6 so meera present age = 18 years Meera age after 5 years = 23 year .

Time, Speed & Distance

The Speed of a moving body is the Distance travelled by it in unit Time.

So

Distance travelled = Speed × Time

Total Time taken to cover some distance = Distance / Speed

Speed is either measured in Kilometer/ hour or meter/ second

To convert Kilometer/ hour in meter/second,

To convert meter/second in Kilometer/hour,

If a car covers a certain distance at x km/hr and an equal distance at y km/hr, the average speed of the whole journey

2xy/(x + y) km/hr

Speed and time are inversely proportional (when distance is constant)

Speed ∝1/ Time (When Distance is constant)

If the ratio of the speeds of A and B is a : b, then the ratio of the times taken by them to cover the same distance is 1/a:1/b or b : a Concept of Relative Speed: Case1: Two bodies are moving in opposite directions at speed V1 & V2 respectively. The relative speed is defined as

Vr =V1+V2 Case2: Two bodies are moving in same directions at speed V1 & V2 respectively. The relative speed is defined as

Vr =|V1–V2|

Concept of Trains

The basic concept for train related problem is Speed = Distance / time. but we should kept in mind these discussed points below.

(i) When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object.

(ii) The distance to be covered when crossing an object, whenever trains crosses an object will be equal to:

Length of the train + Length of the object

NOTE- When train is crossing a stationary object (with length) like bridge, platform, and then its Length is added to the length of train to get required length.

When train is crossing a pole, tree, man etc.. then their length is neglect with respect to train, Here only length of train is considered.

Condition:

When Train crosses single object:

(Let the speed of train is st & length of train Lt)

1. Train Crosses a stationary object (without length like tree, man, pole etc..)

So time taken by train to cross the object =

=

2. Train Crosses a stationary object of Length L

So time taken by train to cross the object =

=

3. Train crosses a moving object of length L with speed sl in the same direction of train

So time taken by train to cross the object =

4. Train crosses a moving object of length L with Speed Sl in the opposite direction of train

So time taken by train to cross the object =

When two train crossing each other in both directions:

Let length of one train = L ; Length of Second train = L2 They are crossing each other in opposite direction in t1 sec and same direction in t2 sec respectively, Then, Speed of faster train = (L1 + L2) /2 [1/t1 + 1/t2]

Speed of slower train = (L1+ L2) / 2 [1/t1 – 1/t2]

If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then:

(A's speed) : (B's speed) = (b : a)^1/2

Boat & Streams

Downstream/Upstream:

In water, the direction along the stream is called downstream. and, the direction against the stream is called upstream.

If the speed of a boat in still water is u km/hr and the speed of the stream is v km/hr, then:

If the speed downstream is a km/hr and the speed upstream is b km/hr, then:

Speed in still water = 1

(a + b) km/hr. 2

Rate of stream = 1

(a - b) km/hr. 2

Time & Work

Concept

If A can do work in n days, then 1 day work of A = 1/n and vice versa.

If A is thrice as good a workman as B, then: Ratio of work done by A and B = 3 : 1.

Ratio of times taken by A and B to finish a work = 1 : 3. Man – Day – Work If M1 men can do W1 work in D1 days working H1 hours per day and M2 men can do W2 work in D2 days working H2 hours per day (when all men work at same rate)

If A can do a piece of work in p days and b can do in q days, then A and B together can complete the same work in

pq/(p+q) days

Speed downstream = (u + v) km/hr.

Speed upstream = (u - v) km/hr.

M1 D1 H1 / W1 = M2 D2 H2 / W2

Pipe and Cistern

The pipe and cistern problem can be done on the concept of positive work and negative work. The pipe is used to fill tank or something reservoir etc. mainly pipe are two types

Inlet pipe: it is used to fill the tank.

Outlet pipe: it is used to empty the tank

So Inlet pipe work taken as Positive and Outlet pipe work taken as negative .

1. If a pipe can fill a tank in x hours, then:

part filled in 1 hour = 1

. x

2. If a pipe can empty a tank in y hours, then:

part emptied in 1 hour = 1

. y

3. If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where y > x), then on opening both the pipes, then

the net part filled in 1 hour =

1 -

1

. x y

4. If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where x > y), then on opening both the pipes, then

the net part emptied in 1 hour =

1 -

1

. y x

The same concept can be applied for one, two, three and more pipes.

Permutation and Combination:

Factorial Notation:

Let n be a positive integer. Then, factorial n, denoted n! is defined as:

n! = n(n - 1)(n - 2) ... 3.2.1.

Examples: i. We define 0! = 1.

ii. 4! = (4 x 3 x 2 x 1) = 24. iii. 5! = (5 x 4 x 3 x 2 x 1) = 120.

Permutations:

The different arrangements of a given number of things by taking some or all at a time, are called permutations.

Examples:

i) All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).

ii) All permutations made with the letters a, b, c taking all at a time are: ( abc, acb, bac, bca, cab, cba)

Number of Permutations:

Number of all permutations of n things, taken r at a time, is given by:

nPr = n(n - 1)(n - 2) ... (n - r + 1) = n!

(n - r)!

Examples:

a. 6P2 = (6 x 5) = 30. b. 7P3 = (7 x 6 x 5) = 210. c. Number of all permutations of n things, taken all at a time = n!

An Important Result:

If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind;p3 are alike of third kind and so on and pr are alike of rth kind, such that (p1 + p2 + ... pr) = n.

Then, number of permutations of these n objects is = n!

(p1!).(p2)!.....(pr!)

Combinations:

Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.

Examples:

1. Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA. Note: AB and BA represent the same selection.

2. All the combinations formed by a, b, c taking ab, bc, ca. 3. The only combination that can be formed of three letters a, b, c taken all at a time is abc. 4. Various groups of 2 out of four persons A, B, C, D are:

AB, AC, AD, BC, BD, CD.

Note that ab ba are two different permutations but they represent the same combination.

Number of Combinations:

The number of all combinations of n things, taken r at a time is:

nCr = n!

= n(n - 1)(n - 2) ... to r factors

. (r!)(n - r)! r!

Note:

I) nCn = 1 and nC0 = 1. II) nCr = nC(n - r)

Probability

Experiment: An operation which can produce some well-defined outcomes is called an experiment.

Random Experiment: An experiment in which all possible outcomes are know and the exact output cannot be predicted in advance, is called a random experiment.

Examples:

i. Rolling an unbiased dice. ii. Tossing a fair coin.

iii. Drawing a card from a pack of well-shuffled cards. iv. Picking up a ball of certain colour from a bag containing balls of different

colours.

Details:

i) When we throw a coin, then either a Head (H) or a Tail (T) appears. ii) A dice is a solid cube, having 6 faces, marked 1, 2, 3, 4, 5, 6 respectively. When we

throw a die, the outcome is the number that appears on its upper face. iii) A pack of cards has 52 cards.

It has 13 cards of each suit; name Spades, Clubs, Hearts and Diamonds. Cards of spades and clubs are black cards. Cards of hearts and diamonds are red cards. There are 4 honors of each unit.

These are Kings, Queens and Jacks. These are all called face cards.

Sample Space:

When we perform an experiment, then the set S of all possible outcomes is called the sample space.

Examples:

1. In tossing a coin, S = {H, T} 2. If two coins are tossed, the S = {HH, HT, TH, TT}. 3. In rolling a dice, we have, S = {1, 2, 3, 4, 5, 6}.

Event: Any subset of a sample space is called an event.

Probability of Occurrence of an Event:

Let S be the sample and let E be an event.

Then, E S.

P(E) = n(E)

. n(S)

Results on Probability:

1) P(S) = 1 2) 0 P (E) 1 3) P( ) = 0 4) For any events A and B we have : P(A B) = P(A) + P(B) - P(A B) 5) If A denotes (not-A), then P(A) = 1 - P(A).

Mensuration

Rectangle

Area of rectangle = length (l) * breadth (b) = lb

Perimeter of rectangle = 2( l + b)

Where l= length of the rectangle, b= breadth of rectangle

Square

Area = a* a = a2

Perimeter = 4a

Where a= side of Square

Rhombus

Area = ½ * Product of Diagonals

Perimeter = 4* length of Side = 4l

Circle

Let r = radius, d = diameter of circle.

Area = π * radius2 = π r2 = ¼π d2

Circumference of Circle = 2πr

Radius r= d/2

Cylinder

Volume of cylinder = πr2h

Total surface area of cylinder = 2πr(r + h)

Curved Surface Area = 2πrh

Where r = radius of base h= height of cylinder

Sphere

Volume of sphere= 4/3 π r3 = 1/6 π d3

Surface Area of sphere = 4 π r2 = πd2

Where r = radius of sphere d = diameter of sphere

Hemisphere

Volume of hemisphere = 2/3 π r3

Surface area of hemisphere = 3π r2

Where r= radius of hemisphere

Cone

Volume of right circular cone = 1/3 π r2h

Area of base of a cone = π r2

Curved Surface Area of Cone = π r l

Total Surface area of cone = πr(r+l)

Where r= radius of base, l = lateral height of cone , h = height of cone

Lateral height of Cone l = {(h2+r2)}1/2

Area of Sector of Circle = π r2 * θ/ 360

Where θ = measure of the angle of the sector, r = radius of sector

Length of an arc = 2π r* θ/360

Cube

Volume of Cube = l * l * l = l3

Length of Diagonal of Cube = √3 l

Where l= side of cube

Cuboid

Volume of cuboid = l *b* h

Length of Diagonal of Cuboid = (l2+b2+h2)1/2

(Where l = length b= breadth h = height)

1. Amit walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is: A) 45 B) 50 C) 55 D) 60 E) None of these

2. A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:: A) 100 km/hr B) 105 km/hr C) 150 km/hr D) 200 km/hr E) None of these

3. Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?: A) 10 B) 15 C) 20 D) 25 E) None of these

4. In a flight of 600 km, an airplane was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is:: A) 45 minutes B) 50 minutes C) 55 minutes D) 60 minutes E) None of these

5. A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. The average speed in km/hour for the first 320 km of the tour is: A) 35.11 B) 55.71 C) 71.11 D) 66.67 E) None of these

6. Prashant is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.? A) 10 kmph B) 15 kmph C) 20 kmph D) 25 kmph E) None of these

7. It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the cars is: A) 3:4 B) 4:5

C) 7:9 D) 8:11 E) None of these

8. A train overtakes two persons walking along a railway track. The first one walks at 4.5 km/hr. The other one walks at 5.4 km/hr. The train needs 8.4 and 8.5 seconds respectively to overtake them. What is the speed of the train if both the persons are walking in the same direction as the train?: A) 80 kmph B) 81 kmph C) 85kmph D) 90kmph E) None of these

9. Two, trains, one from Kolkata to Delhi and the other from Delhi to Kolkata, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is: A) 4:3 B) 3:4 C) 2:3 D) 3:2 E) None of these

10. Two trains of equal lengths take 10 seconds and 15 seconds respectively to cross a telegraph post. If the length of each train be 120 meters, in what time (in seconds) will they cross each other travelling in opposite direction? A) 10 B) 12 C) 30 D) 25 E) None of these

11. A boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water current respectively? A) 8:3 B) 3:8 C) 7:9 D) 9:12 E) None of these

12. A boatman goes 2 km against the current of the stream in 1 hour and goes 1 km along the current in 10 minutes. How long will it take to go 5 km in stationary water? A) 1hour B) 1 hour 15 minutes C) 2 hour D) 2hour 15 minutes E) None of these

13. Speed of a boat in standing water is 9 kmph and the speed of the stream is 1.5 kmph. A man rows to a place at a distance of 105 km and comes back to the starting point. The total time taken by him is? A) 20 hour B) 21 hour C) 23 hour D) 24 hour E) None of these

14. A man rows to a place 48 km distant and come back in 14 hours. He finds that he can row 4 km with the stream in the same time as 3 km against the stream. The rate of the stream is: A) 2km/hr B) 3 km/hr C) 2.5 km/hr D) 4 km/hr E) None of these

15. A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day: A) 15 days B) 20 days C) 25 days D) 30 days E) None of these

16. A is thrice as good as workman as B and therefore is able to finish a job in 80 days less than B. Working together, they can do it in? A) 20 days B) 25 days C) 30 days D) 40 days E) None of these

17. A and B can complete a work in 15 days and 10 days respectively. They started doing the work together but after 2 days B had to leave and A alone completed the remaining work. The whole work was completed in : A) 10 days B) 12 days C) 20 days D) 25 days E) None of these

18. A and B together can do a piece of work in 30 days. A having worked for 16 days, B finishes the remaining work alone in 44 days. In how many days shall B finish the whole work alone? A) 40 days B) 50 days C) 55 days D) 60 days E) None of these

19. A machine A can print one lakh books in 8 hours, machine B can print the same number of books in 10 hours while machine C can print them in 12 hours. All the machines are started at 9 A.M. while machine A is closed at 11 A.M. and the remaining two machines complete work. Approximately at what time will the work (to print one lakh books) be finished? A) 1:05 PM B) 1:30 PM C) 11:35 AM D) 2 PM E) None of these

20. In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?

A) 124045 B) 20890 C) 133156 D) 120960 E) None of these

21. How many 4-letter words with can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed? A) 400 B) 4050 C) 5040 D) 5773 E) None of these

22. In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there? A) 156 B) 209 C) 193 D) 245 E) None of these

23. In a bag, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green? A) 3/91 B) 1/3 C) 3/7 D) 7/15 E) None of these

24. One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)? A) 3/13 B) 1/13 C) 7/52 D) 9/13 E) None of these

25. Three taps A,B and C can fill a tank in 20,30and 40 minutes respectively. All the taps are opened simultaneously and after 5 minutes tap A was closed and then after 6 minutes tab B was closed .At the moment a leak developed which can empty the full tank in 60 minutes. What is the total time taken for the completely full? A) 15 minutes B) 24 minutes C) 30 minutes D) 48 minutes E) None of these

26. There are three taps A, B, and C. A takes thrice as much time as B and C together to fill the tank. B takes twice as much time as A and C to fill the tank. In how much time can the Tap C fill the tank individually, if they would require 10 hours to fill the tank, when opened simultaneously? A) 12 hour B) 48 hour C) 60 hour D) 24 hour E) None of these

27. Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P,Q and R respectively. What is the proportion of the solution R in the liquid in the tank after 3 minutes? A) 5/11 B) 7/11 C) 9/11 D) 3/11 E) None of these

28. The length of a rectangular plot is 20 meters more than its breadth. If the cost of fencing the plot at 26.50 per meter is Rs. 5300, what is the length of the plot in meters? A) 40 B) 50 C) 60 D) Data inadequate E) None of these

29. A Blanket, when washed, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is? A) 25 B) 28 C) 35 D) 40 E) None of these

30. A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?? A) 3m B) 2m C) 1m D) 5m E) None of these

1.

; 4x= 200 ; x= 50km,

2. Let speed of car x km/hr 75/x-75/1.5x = 12.5/60

Solving x= 120 km/hr

3. Stoppage time per hour = 9/54*60 = 10 min.

4. 600/x- 600/1.5x =200

So x= 1 hour = 60 min.

5. Average Speed =2*64*80/(80+64)= 71.11 km/hr

6. Let speed is x km /hour

x/10 – x/15 = 2 so x= 60 ;

So time = 60/10 = 6 hour so Prashant has started @ 8am

So reach at 1 pm he should travel with = 60/5 = 12 km/hr

7. Let speed of train is x and speed of car = y

120/x+480/y= 8; so 1/x+4/y= 1/15

200/x+400/y=25/3; so 1/x+2/y= 1/24;

So y=80; x=60

So ratio of train that of Car =60:80 = 3:4

8. Let speed of train =x

4.5 km/hr = 5/4 m/s and 5.4 km/hr = 3/2 m/s

Length of train equal

So (x-5/4)* 8.4 = (x- 3/2)8.5

On solving x=22.5m/sec= 81 km/hr

9. Ratio of speed of A: Speed to B = √16:√9 = 4:3

10. Let speed of first train S1 and 2nd

train S2

120/s1=10 ; s1=12m/s same as s2=8m/s

Time taken if they cross each other= (120+120)/(12+8)= 12 sec.

11. 8 hour 48 min = 44/5 hour

Ratio of speed to boat to water ={(44/5+4)/2}:{(44/5-4)/2}= 8:3

12. Speed downstream = 1/10*60 = 6km/hr

Rate upstream = 2km/hr

Speed in still water = (6+2)/2 = 4km/hr

Required time = 5/4 hour = 1 hour 15 min.

13. Time = 105/(9-1.5)+ 105/(9+1.5)

= 14+10 = 24 hour

14. Let speed of man = x km/hr and speed of stream = y km/hr

So 4/(x+y)= 3/(x-y)

And 48/(x+y) + 48/(x-y)= 14

On solving y = 1km/hr

15. A,B and C one day work =(1/20+1/30+1/60) = 1/10

A 2 day work = 2*1/20 = 1/10

Work done in 3 days = 1/10+1/10= 1/5

So work completed = 15 days.

16. Let A taken time x then taken time by B= 3x

Diff. 3x-x= 80 days x= 40 days

If they work together then work /day =1/40+1/120 =4/120

So work completed in = 120/4 =30 days.

17. A and B together 1 day work =1/15+1/10 =5/30=1/6

2 day both work = 1/3

Remain work =2/3

Left work done in (2/3)÷ (1/15) =10 days

So total day = 10+2 = 12 days

18. Let A’s one day work =x and B’s one day work = y

X+y=1/30 and 16x+44y =1

On solving x=1/60 y=1/60

So time required by B to complete work = 60 Days.

19. A,B and C I hour work =1/8+1/10+1/12= 37/120

2 hour work = 37/60

Remain work 1-37/60 =23/60

B+C one hour work = 1/10+1/12=11/60

So 23/60 work done in 23/60÷ 11/60 hour = approx 2 hour 05 min

So time =11am+ 2hour 5min = 1hour 5 min.

20. No. of ways = {8! /(2! * 2!)}×{4!/2!}= 10080 *12 =120960

21. Required no. of words = 10p4 = 10*9*8*7 = 5040

22. For at least one boy required no. of way =(6C1*4C3)+(6C2*4C2)+(6C3*4C1)+(6C4) =209

23. Total no. of balls = 8+7+6 = 21

Probability to chose neither red nor green ball = 7/21= 1/3

24. Probability = 12/52 =3/13

25. Portion of tank filled in 5 min. = 5[1/20+1/30+1/40]=13/24

Portion of tank filled by Band C in next 6 min = 6[1/30+1/40]=7/20

Remain part =1-[13/24 +7/20] =13/120

Time taken by C considering leak also = 13/120 ÷ [1/40 -1/60] = 13 min

Total time =5+6+13= 24 min.

26. 3(portion filled by A in 1 hour) = portion filled by B and C in 1 hour

Add portion of A filled in 1 hour both sides;

4(portion filled by A in 1 hour) = portion filled by A,B and C in 1 hour = 1/10

So portion filled by A in 1 hour = 1/40

P alone can fill = 40 hour

Similarly found Q alone can fill =30 hour

Time taken by R = 1÷[1/10-(1/40+1/30)] = 24 hour

27. Part filled by A+B+C in 3 min = 3(1/20+1/30+1/10) = 11/20

Part filled by C in 10 min = 3/10

Required ratio = 3/10:11/20 = 6:11

28. Let length = l breadth = b

So l=b+20

2(l+b) = 5300/26.50

On solving l= 60 meter

29. % change = -20+-10+2 = -28 So decrease 28% in area

30. Area of Park = 2400 sq m

Area of lawn = 2109 sq m

Area of crossroad = 2400 -2109 = 291 sq m

Let width = x meter, then

60x +40 x- x2

= 291

X= 3 meter