Apti Shortcuts

8/2/2019 Apti Shortcuts

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Apti Shortcuts:

Additions Shortcuts

Addition of numbers close to multiples of ten (e.g. 19, 29, 38, 59 etc.)

This technique is useful for Mental calculations

116 + 39 (Here we can write this 39 as 40-1)

= 116 + (40 - 1)

= 116 + 40 - 1

= 156 - 1 (Instead ofadding 39 to 116, we just add 40to 116 (because we can do this without using pen and paper) and later we

subtract one from it)

= 155

Try this.. This is very useful tip while doing calculations.

Now lets try another example.

116 + 97

= 116 + (100 - 3)

= 116 + 100 - 3 (Here, instead of adding 97 to 116, we are just adding a 100 to

116 and then subtracting 3 from it :)

= 216 - 3

= 213
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Addition of decimals

12.5 + 6.25

= (12 + 0.5) + (6 + 0.25)

= 12 + 6 + 0.5 + 0.25 (Here we just added the rounded numbers first and later

we added the decimal numbers :)

= 18 + 0.5 + 0.25

= 18.75

Subtractions Shortcuts

PLEASE READ THE SHORTCUTS IN ADDITIONS FIRST. SO THAT IT

WILL BE EASY FOR YOU TO

UNDERSTAND SUBTRACTION SHORTCUTS

SUBTRACTION BY NUMBERS CLOSE TO 100, 200, 300, 400,

ETC.

250 - 96

= 250 - (100 - 4) (here, instead ofsubtracting 96 from 250, we arejust subtracting 100 from 250 and then adding 4)= 250 - 100 + 4 (Why adding? because the actual amount we haveto subtract from 250 is 96. but we are subtracting 100. Thatmeans, we are subtracting 4 numbers more than we actuallydeserve. so our 250 will feel bad. so we should add that 4 to it:)
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= 150 + 4= 154

250 - 196= 250 - (200 - 4)

= 250 - 200 + 4 (here also same. In order tosubtract 196, we subtract 200 and adding 4)= 50 + 4= 54

Note : We can use this logic for any number. According toour convenience.

Lets see,

216 - 61 (Here i found it difficult to subtract 61 from 216)= 216 - (100 - 39) (So i just decided to subtract 100 to it and later

will subtract the extra 39)
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= 216 - 100 + 39 (Hey, see here. How about writing this 39 as 40 -1

?)= 116 + (40 - 1) (dont be confused. just practice this method and

you will come to know how easy and efficient method it is :)= 156 - 1= 155

Subtraction of decimals47 - 9.9 (How about dividing this 9.9 as 9 + 0.9 ??)= 47 - (9 + 0.9) we can write this as...= 47 - 9 - 0.9

= 38 - 0.9= 37.1

18.3 - 0.8= 18 + 0.3 - 0.8= (18 - 0.8) + 0.3

= 17.2 + 0.3

= 17.5


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Multiplication by "giving and taking"12 x 47 (Here its little difficult for us to calculatethe multiplication of 12 and 47 mentally. so just check for the

ROUNDED number nearer to 47. Yes it is 50. So.....

= 12 x (50 - 3)

= (12 x 50) - (12 x 3) (we have discussed this rule earlier)

= 600 - 36

= 564

Multiplication by 5:

* If we have to multiply a number with 5, just divide the number

with 2 and then multiply the result with 10. Confused? Its very

simple step actually....

428 x 5 (Now just divide the number with 2)

= 428 x 1/2 = 214 (Now multiply it with 10. I mean just add a zero at

the end :P)

= 214 x 10

= 2140 (This is our result)

Whats the logic behind this step?Very simple.* Lets say the number is X.* Now we are dividing the number with 2. so here X becomes X/2.* And then we are multiplying it with 10. So itwill become 10x / 2* Now cancel it with 2. so it becomes 10x / 2 = 5X = 5 multiplied by X.Thats it ;)


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Multiplication by 10 ------------ just move the decimal point one

place to the right

16 x 10

= 160

5.9 = 159

169.93 = 169.3 (Need an explanation for this too??? :P)

Multiplication by 50 ------ divides with 2 and then multiplies by

100

Well, this is also same process as we did for 5. Here we should

add an extra zero. Thats it

18 x 50

= (18/2) = 9

= 9 x 100

= 900

Multiplication by 100 -------- move the decimal point two places to

the right

45 x 100

= 4500

Multiplication by 500-------- divide with two and multiply with1000

21 x 500

= 21/2 x 1000

= 10.5 x 1000


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= 10500

Multiplication by 25 ---------- use the analogy Rs 1 = 4 x 25 Paise

25 x 14 (just divide the 14 as 10+4)

= (25 x 10) + (25 x 4)

= 250 + 100 ---> Rs2.50 + Rs1

= 350

Hey one more thing. Here you can use another technique too.

Which we have used for multiplication with 5.

Multiplication by 25 ----------- Divide by 4 and multiply by 100

36 x 25

= (36/4) x 100

= 9 x 100

= 900

Multiplication by 11 (if sum of digits is less than 10)

72 x 11

= 7+2 =9, it is Less than 10. so,

= place this term 9 between 7 &2

= 792 (That's the answer)

Multiplication by 11 (if sum of digits is greater than 10)

87 x 11

=> 8 + 7 = 15

because here 15 is greater than 10, first use 5 and then add 1 to the first

term 8,


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which gives you the answer

= 957

Multiplication of numbers ending in 5 with the same first terms

(square of a number)

25 x 25

first term = (2 + 1) x 2 = 6

last term = 25

answer = 625 ---> square of 25

75 x 75

first term = (7 + 1) x 7 = 56

last term = 25

answer = 5625 ---> 75 squared

Shortcuts in Division

There are so many types of divisions are there. Lets have a look.

Divisions by parts -> Imagine you have Rs.874. You have to givethat to your two children.874/2 [We can write this 874 as 800+74 (for our convenience)= 800/2 + 74/2= 400 + 37= 437


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Division using the factors of the divisor: "this is also called asDouble Division"70/14= (70/7)/2 (Because 7 and 2 are the factors of 14)= 10/2= 5

Division using Fractions:132/2= (100/2 + 32/2) ( here we've broken the given fraction intotwo separate fractions)= (50 + 16)= 66

Division by 5 :Note: if you have to divide any number with 5, then divide it by100 and then just multiply by 201400/5= (1400/100) x 20= 14 x 20= 280

Division by 10 (Its very simple, just move the decimal point oneplace to the left)

0.5/10= 0.05

Division by 50 ( Just divide with 100 then multiply by 2)2100/50= (2100/100) x 2= 21 x 2= 42

700/50= (700/100) x 2= 7 x 2= 14

Division by 100 (just move the decimal point two places to the


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left)25/100= 0.25

Division by 500 (just divide with 100 and then multiply with 0.2)17/500= (17/100) x 0.2= 0.17 x 0.2= 0.034

Division by 25 (just divide by 100 and then multiply by 4 )500/25= (500/100) x 4= 5 x 4

= 20

750/25= (750/100) x 4= 7.5 x 2 x 2= 30

Decimals

Some times, you have to convert or express the given percentages in theform of decimals. It is not such a difficult task as we think. Have a look atthe following.

1% = 1/100 = 0.01 (if two zeros are given, just move the decimal pointertwo places left)2% = 2/100 = 0.02 = 1/50 (the simplification of 2/100)3% = 3/100 = 0.034% = 4/100 = 0.04 = 1/255% = 5/100 = 0.05 = 1/20


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6.25% = 6.25/100 = 0.0625 = 1/167% = 7/100 = 0.077.5% = 7.5/100 = 0.07510% = 10/100 = 0.1 = 1/1012.5% = 12.5/100 = 0.125 = 1/820% = 0.2 = 1/521% = 0.2125% = 0.25 = 1/430% = 0.3 = 3/1033.33% = 33.33/100 = 0.3333 = 1/337.5% = 0.375 = 3/840% = 0.4 = 2/550% = 0.5 = 1/260% = 0.6 = 3/5

62.5% = 0.625 = 5/866.66% = 66.66/100 = 2/375% = 0.75 = 3/480% = 0.8 = 4/587.5% = 0.875 = 7/8100% = 1125% = 1.25 = 1 1/4150% = 1.5 = 1 1/2200% = 2

Divisibility Rules

A number is divisible by 2, when its unit digit is either Even or Zero.

A number is divisible by 3, wehen the sum of its digits is divisible by 3.

A number is divisible by 4, when the number formed by the two extreme

right end digits is either divisible by 4 or both these digits are zeroes.

A number is divisible by 5, when its unit digit is either zero or 5.

A number is divisible by 6, when it is divisible bye 2 as well as 3.
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A number is divisible is by 7, if it passes the following Test...

Take the last digit in a number.

Double and subtract the last digit in your number from the rest

of the digits.

Repeat the process for larger numbers.

Take an Example 357 (Double the 7 to get 14. Subtract 14 from

35 to get 21 which is divisible by 7 and we can now say that

357 is divisible by 7.

A number is divisible by 8, when the number formed by its three extreme

right end digits is divisible by 8 or when these last three digits are Zeros.

A number is divisible by 9, when the sum of its digits is divisible by 9.

A number is divisible by 10, when its unit digit is zero.

A number is divisible by 11, when the difference between the sums of the

alternate digits is either zero or divisible by 11.

A number is divisible by 12, when it is divisible by 3 as well as 4. A number is divisible by 13, if sum of 4 times the digit in units place and the

number in the remaining part is multiple of13.

If the difference of5 times the digit in units place and the number in the

remaining part is 0 or multiple of17, then the number is divisible by 17.

If the sum ofdouble the digit in units place of a given number and number

in the remaining part is multiple of19, then the given number is divisible of

19.

LCM and HCF


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Factor :A number is said to be a factor of other when it EXACTLYdivides the other.

o Ex : 6 and 7 are Factors of 42. Multiple : A number is said to be a multiple of another, when it is

Exactly divisible by the othero Ex : 42 is a multiple of 6 and 7

Please re - read these definitions. So that you can get the differencebetween Factor and Multiple.

Prime Number : Prime number is a number which has no factorsexcept itself and Unity.

o Ex :2, 3, 5, 7, 11, 13, 17 etc are prime numbers Composite Number : Composite number is a number which has

other factors besides itself and Unity.o Ex : 14, 15, 16, 18 etc

Co-Prime : Two numbers are said to be Co-Prime (Prime-To-EachOther) when they have no common factors except Unity.

o Note : The Co-Primes need not necessarily be Primes. 15 and 19 15, 17 and 22 are Co-Primes

Common Multiple : A Common Multiple of two or more numbers

is a number which is exactly divisible by each of them.o Ex : 12 is a common multiple of 2, 3, 4 and 6

Least Common Multiple (LCM) : The LCM of two or more givennumbers is the Least Number which is exactly divisible by each ofthem.

o Ex : 20 is the Common Multiple of 2, 4, 5 and 10
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40 is the Common Multiple of 2, 4, 5 and 10 80 is the Common Multiple of 2, 4, 5 and 10, But

Here 20 is the Least Common Multiple of 2, 4, 5,and 10

Highest Common Factor (HCF) :The HCF of two or morenumbers is the Greatest Number which divides each of them Exactly.

o It is also Called Greatest Common Divisor (GCD)o Ex : Find the HCF of 18, 24

Factors of 18 --> 1, 2, 3, 6, 9, 18 Factors of 24 --> 1, 2, 3, 4, 6, 8, 12, 24

Here the Greatest number, which divides themexactly is 6. So 6 is the H.C.F of 18, 24

In the above example they have given very small numbers. So it was easy

for us to find the HCF. What if they ask you to find the HCF for 84 and 540? Will you write the factors to both of them and then find out the Highestnumber? If you are planning to do that , please erase that thought from

your mind :) Because there are several methods to make the process simple

Methods of finding HCF :

HCF by factorization :1. Express each of the given number as the product of Prime

Factors2. Choose common factors3. Find the Product of Lowest Power of these Factors.

This Product is the required HCF of the given Numbers

Ex : Find the HCF of 84, 540


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If you find this method lil confusing, dont worry. There is another method

to find HCF.

HCF by Method of Division :o Consider two different numbers.o Divide the longer number by the smaller one.o Now divide the divisor by the reminder.o Repeat this process of dividing the preceding divisor by the last

reminder obtained, till you get the reminder "0"o The LAST DIVISOR is the HCF of the given TWO numbers

Ex : Find the HCF of 42, 70
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Thats it. Now the answer is 14 :)

BODMAS

In The simplifications of numerical questions, the order of mathematicaloperations to be followed is given in the abbreviated form : "BODMAS",

where

o B stands for 'Bracket'

o Ostands for 'Of'
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o Dstands for 'Division'

o Mstands for 'Multiplication '

o Astands for 'Addition' and

o Sstands for 'Subtraction'.

Order to be followed by in case of brackets is as follows :

o ( ) : Small Brackets,

o { } : Curly Brackets and

o [ ] : Square Brackets

Average

Average of Given Items :

Average =Sum of the given Items

Number of those Items

Average Speed :
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Suppose a Person covers a certain distance at X kmph and

an equal distance at Ykmph.

Then, the average speed for the whole journey will be

(2XY)/(X+Y) Kmph

Lets do some problems on Average (Click Here)

Average

Average of Given Items :

Average =

Sum of the given Items

Number of those Items

Average Speed :
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Suppose a Person covers a certain distance at X kmph and

an equal distance at Ykmph.

Then, the average speed for the whole journey will be(2XY)/(X+Y) Kmph

What is the average of first 21 multiples of 7 ?

1. 492. 1473. 774. Cant be determined5. None of these

2. In a class, the average age of 30 boys is 13 yearsand the average of 20 girls is 12 years. what is the average age of the wholeclass?

1. 12.5 Yrs2. 14.2 Yrs3. 12.3 Yrs
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4. 12.6 Yrs5. None of these

3. The average of seven numbers is 18. The average of first three numbersis 14 and the average of last three numbers is 19. What is the middlenumber?

1. 422. 573. 274. Cant be determined5. None of these

4. The average age of a class of 32 students is 16 yrs. if the teacher's age is

also included, the average increases by one year. Find the age of the teacher

1. 492. 463. 484. Cant say5. NOne of these

5. The average weight of a class of 20 boys was calculated to be 58.4 kgsand it was later found that one weight was misread as 56 kg instead of 65

kg. What is the correct weight?

1. 67.4kg2. 58.85kg3. 57.75kg4. 49.4kg5. None of these

6. The average weight of 8 persons increases by 1.5kg when aperson weighting 65kg is replaced by a new person. What could be the

weight of the new person?

1. 53kg2. 58.85kg3. 75kg4. 77kg5. None of these


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7. The average temperature for Monday, Tuesday and Wednsday is 36.3degrees C. The average temperature for Tuesday, Wednesday and Thursdayis 36.7 degrees C. if Monday's temperature recorded as 39 degrees c, findthe Thursday's temperature ?

1. 40.2 degrees C2. 39.2 degrees C3. 41.4 degrees C4. 40.4 degrees C5. None of these

8. A man spends Rs. 1600 per month on an average for the first threemonths, Rs 1550 for next four months and Rs. 1800 per month for the lastfive months and saves Rs . 5200 a year. What is his average monthly

income?

1. Rs. 20502. Rs. 2103. Rs. 19504. Rs. 22005. None of these

9. There were 45 students in a hostel, if the numbers of students increasedby 7, the expenses of the mess were increased by Rs. 39 per day while the

average expenditure per head diminished by Re.1. What is the originalexpenditure of the mess?

1. Rs. 6242. Rs. 5623. Rs. 5854. Rs. 5985. None of these

10. The average age of father and his two sons is 27 Years. Five years ago,

the average age of the two sons was 12 Years. If the difference between theages of the two sons is four years, what is the present age of the father?

1. 422. 483. 444. 47


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5. None of these

Answers :

1. Required average = 7(1+2+....+21) / 21= (7/21)X ( (21x22) / 2 ) (because sum of first 21 natural

numbers)= 77

2. Total age of 50 students= (30X 13+20 X 12) = 630

Average = 630/50 = 12.6 Years

3. The total of seven numbers = 7X18 = 126The total of first 3 and last 3 numbers is = 3 X 14+3 X 19 = 99So, the middle number is (126 - 99 ) = 27

4. Total age of students is 32X16 = 512 YearsTotal age inclusive of teacher = 33X (16+1) = 561So, Teacher's age is 561-512 = 49 Yrs

There is a shortcut for these type of problems :Teacher's age is 16+(33X1) = 49 Years :)

5. Actual total weight is (20X 58.4 - 56 + 65) = 1177 KgsActual average weight is 1177/20 = 58.85kgs

6. Let the average weight of 8 persons be 'x'years and the age of the new person be 'y' years

so, (8x - 65 + y) / 8 = x+1.5so, y = 77kgs Shortcut : 65 + 8X1.5 = 77kgs

7. MOn + Tue + wed Temperature = 3X36.3 = 108.9

Tue + wed Temperature = 108.9 - 39 = 69.9Tue + wed + Thu temperature = 3X36.7 = 110.1So, Thursday's temperature = 110.1 - 69.9 = 40.2 degrees C

8. Total expenditure for the first 3 months = 3X1600 = 4800Total expenditure for 4 months = 4X1550 = 6200

Total expenditure for 5 months = 5X1800 = 9000


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Total expenditure and saving (which is income for one year)= 4800 + 6200 + 9000 + 5200 = Rs. 25200

So, Average monthly income = 25200/12 = Rs. 2100

9. Let the original expenditure be Rs.xOriginal average expenditure = X/45

New average expenditure = (x+39)/52So (x/45) - ((x+39) / 52) = 1 so x = 585

so, original expenditure is Rs 585

10. The total present age of father and two sons is 3S27 = 81 yrsThe total present age of sons is (12+5) X 2 = 34Yearsso, present age of father is 81 - 34 = 47 yrs

If you have any doubts or comments, please feel free to share.Thank you.

Time and Work

Work = Strength X Time=> Strength X Time = Work=> (Strength X Time ) / Work = 1 :)

So, For every Case (Strenth X Time) / Work = 1 [ No matter thevalues]


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So, (SxT)/Wwill be same for all the cases.

I mean st/w = ST/W

Keep this point in your mind.

Now lets see some more important formulas

If Days are Given :o If A can do some work in n days, then he can do 1/n work in

One day

If Work is Given :o If A can do 1/n work in One day, he can finish it in n days

If A is TWICE as good a work man as B, theno The ratio of work done by A and B = 2:1o The ratio of time taken by A and B to finish the work = 1:2

(Please Dont be confused)

Now lets do some problems :

1. 36 men can do a piece of work in 25 days. In how many days can 30 mendo it?
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Here , the work is same.. So,

(S X T) / W = (s X t) / w

= > (S X T) / W = (s X t) / w [ Cancel W for both sides]

Now, substitute the values...= > 36 X 25 = 30 X t

= > (36 X 25) / 30 = t => t = 900/3 = 30

2. 32 men can do a piece of work in 15 days working for 6 hours a day. Inhow many days will 40 men can finish it if they work for 8 hours a day?

32 X 15 X 6 = 40 X d X 8 = 9

3. If 16 men can build a wall of 52 m long in 25 days working for 8 hours aday, in how many days can 64 men build a similar wall of 260m long

working for 10hrs a day?

(16 X 200) / 52 = (64 X ? X 10) / 260 = 25

4. A man engaged 10 laborers to make 320 toys in 5 days. After 3 days hefound that only 120 toys were made. How many additional men should he

engage to finish the work in time?

here, equate the complete work with the remaining work

(10 X 3) / 120 = (10+X) x 2 / 200

= > 10+X = 25 => X = 15

5. A can do a task in 20 days and B can do it in 30 days. In how many days

can they finish it if they work together?

A's one day's work = 1/20B's one day's work = 1/30


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=> Their one day's work = (1/20)+(1/30) = (3+2)/60 = 5/60 =1/12

This is their one day's work TOGETHER.So, obviously the number of days will be = 12

Short Cut : calculate Product/Sum = (20 X 30) / 50 = 12 Thats it

6. A, B and C can do a job in 20 days, 30 days and 60 days respectively. Ifthey work together, in how many days will the work be finished?

(1/20)+(1/30)+(1/60) = (30+20+10)/60 = 6/60 = 1/10

So, the number of days is = 10

7. Two taps A and B can fill a tank in 10 hours and 15 hours respectively. athird tap C can empty the full tank in 12 hours. How many hours will berequired if all of them are opened simultaneously to fill in an empty tankcompletely?

Here, first two are Inlets and the last one is Outlet,

So, (1/10)+(1/15)-(1/12) = (6+4-5)/60 = 5/60 = 1/12

So, our answer is 12

8. A and B can do a job in 12 days. B and C in 15 days and C and A in 20days. In how many days can they finish it if they work TOGETHER?

A+B = 12B+C = 15C+A = 20

So here, A+B's One day's work = > 1/ (A+B) = 1/12

B+C's one day's work => 1/ (B+C) = 1/15

C+A's one day's work = > 1/(C+A) = 1/20


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Just Add them => 1/( 2A+2B+2C) = 12/60 = 1/5

=> 1/2(A+B+C) = 1/5

=>1/(A+B+C) = 1/10 [this is their one day's workTOGETHER]

So, they can finish it in 10 days :)

9. A and B can do a job in 12 days. B and C can do the same job in 15 days.C and A in 20 days. In how many days can A alone finish the whole task???

A+B = 12

B+C = 15C+A = 20

Here we need A, so take a pair which is NOT HAVING Aand subtract it from the others,

so, A+B-(B+C)+C+A = A+B-B-C+C+A = 2A => ( 60/4 )x2 = 30:)

10. A and B can do a piece of work in 20 days. A alone can do it in 30 days.In how many days can B alone do it?

Per day work of A and B = 1/20

Work done by A= 1/30

So, B's one day work = 1/20 - 1/30 = (3-2)/60 = 1/60=> B's work is 60 Days

Short Cut : Product/diff = 600/10 = 60 :)

Ok, that's all for now friends. Complaints, Comments and Suggestions areWelcome. Have a Great Day. All The Best


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Simple Interest

I need money. I don't have enough to meet my requirement. Then whatshould I Do?

Well I have 3 wonderful options.

1. Robing a Bank : It needs Courage, Effort, Planning and Hard-work.Well I am sure that I don't have any of these.

2. To kidnap you and ask your dad for money : Ummmm, not a goodidea!!! Nexxxxxxxt

3. To borrow the money from my friend. Sounds good, isn't it?o But Nobody gonna give me money if Isay that I'm not going to

return it to him. So, I should promise him that I will return theamount within a specif amount of time (not really :P). And in-order to TEMPT him, I also will promise him that I will return

his money with some ADDITIONAL money. This additionalmoney is called INTEREST.o There are TWO types of Interests are there

Simple Interest : The amount charged by the lender forgiving you his money for a specific amount of time.
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Compound Interest : Here also same. But here thelender calculates the Interest on Interest if the given timeexceeds (dont worry if you are unable to understand whatI am saying. We shall discuss about this in our next post.Now concentrate on Simple Interest)

When money is borrowed at Simple Interest, the interest charged issame irrespective of the period involved.

I mean, if Simple Interest for One year is Rs 1000/-, then for 2 Yearswith teh same rate percent, it will be 2X1000 = 2000/-

So, for 8 years it will be 8 X1000 = 8000.

So, if the Simple Interest (or SI in shorter form) on a certain sum is Rs 600

in 3 yrs, then the SI on that sum for one year will be 600/3 = 200

Now have a look at some formulas :

SI = (P*T*R)/100

Where

SI=Simple Interest P= Principle (the actual money borrowed) T=Time in Years R=Rate of % per annum (The percentage of the Principle, we should

pay as the Interest)

So, Obviously

P = (100I) /TR T = (100I) /PR R = (100I) / PT

Now, how much money we should pay to the Lender?Its the total of the money we have taken from him and the money we shouldpay in the name of Interest.

So, AmountA = P+I The actual amount (Principle) + Interest


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Now lets see some Examples

Click HEREfor Practise Problems onSimple Interest

Simple Interest - Practice Problems

Before Doing Problems, Have a look at the BasicsHERE

1. Find the Simple Interest on Rs 750 in 4 years at 15% perannum

here S.I = PTR/100

= (750 X 4 X 15) / 100 = 4500

2. On what sum of money will the Simple Interest be Rs. 2000 in

5 years at 8% Per Annum?

here they are asking about P,

So, P = 100I / TR
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= (100X2000) / 40 = 5000

3. A man invested Rs. 20,000 at 10% P.A. Rs. 15,000 at 12% P.A.and some money at 15% P.A. If the total annual interest recieved

is Rs. 56,000, find the money invested by him at 15% P.A

here, the total Interest he recieved is Rs. 56,000.

So, find the total interest and try to equate it with 56,000

So, the total interest = [ (PTR)/100 ] [ here Sigma symbol ( )

represents the Total]

= 200X10+150X12+ (P/100)X15 = 5600

3800+ (3P/20) = 5600

= 3P/20 = 1800 => P = 12000/-

4. On a sum of money the rate of interest is 5% Per Annum for the first 3 years,

6% Per Annum for the next 4 years, and 8% Per Annum for the next years

beyond the first 7 Years. If the interest obtained in 12 Years is Rs. 3,950, Find

the Sum?

5% X 3 + 6% X 4 + 8% X 5

= 15% + 24% + 40% = 79%

=> (79/100) P = 3950

=> P = (39500/79) = 5000


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5. A sum was put at 5% at a certain rate for 5 Years. Had it been put at 3% Per

Annum higher rate, it would have fetched Rs. 900 more. Find the Sum?

P = 100I / TR [Note : Here I is Additional Interest and R is additional Rate]

= (100 X 90) / (5X3)

= 6000

6. On a certain Sum of Money the Simple Interest in 2 years at 10% P.A is Rs.

125, what would be the S.I if the rate of Interest will be 12% P.A in 3 Years?

Here the Principle is same for both the cases. I mean p = P

=> 100i / tr = 100I / TR

= 125 / (2 x 10) = I / (3 x 12)

= > I = 225

7. The S.I on Rs 6400 at 12 1/2 % per annum is Rs. 2000. Find the Period ?

T = 100I / PR = (100 X 2000 X 2) / (6400 X 25) = 2 1/2 years

8. The S.I on Rs. 6000 in 3 Years and 4 Months is Rs. 3000. Find the rate

percent Per Annum ?

[ here T = 3 4/12 = 3 1/3 = 10/3 years]

I = PTR / 100 => R = 100I / PT

=> (100 X 3000 X 3) / (6000 X 10 ) = 15

9. Manish took a loan of Rs. 4000 at S.I. After 2 Years he cleared the loan by

paying Rs. 5600. Find the Rate % P.A?

[ Here Manish took Rs. 4000 and Payed 5600. So the Interest will be 5600-

4000 = 1600 ]


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So, R = 100I / PT

= (100 X 1600) / (400 X 2)

= 20

10. A lent Rs. 25000 to B for 4 years and Rs. 40,000 to C for 3 1/2 years and got

Rs. 24,000 S.I from both B and C. Find the rate PCPA

[ (PTR)/100 ] = 250 X 4R = 400 X (7/2) R = 2400R

=> R = 10

Compound Interest

the interest which is calculated NOT ONLY on the original principal, but also on

the INTEREST earned in previous period is called Compound Interest.

Confused?

Ok, assume that you gave me Rs. 1000 with 10% interest. First year, the

interest will be

(10/100)X1000 = 100 [Same as the Simple Interest]

But in the 2nd year, the interest will be,

(10/100) X (1000+100) = 110

The difference between Simple Interest and Compound Interest :


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S.I and C.I are same for the first year on the same sum and at the same rate

percent.

In S.I, interest is calculated on the original Principal only (for all the years).But from 2nd year, the C.I is calculated not only on the original principal,but also on the interest of the first year.

On the same sum and at the same rate of Interest, the C.I of the 2nd year is always

more than the C.I of the 1st years and their difference is equal to the interest on the

interest of the first year.

Amount = P 1 +R T

100

C.I = Amt - Principal

Where,

A = Amount

P = Principal

R = Rate of Interest

T = Time Period
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Example : Assume that Shivani borrows 3,000 at Rate 10% by C.I.

The C.I on any principal becomes effective only after the 1st year is

complete. Which means, that at the end of the first year. So, at the first year, the

C.I is equal to the S.I of the Principal

so, S.I = PTR/100 = (3000 X 1 X 10) / 100 = 300 at the end of the first

year.

NOw, if the sum borrowed with C.I, the above S.I is added to the Principal (So,

in the above case, the principal becomes 3000+300 = 3300 ) and then if you

calculate S.I on this amount, with the same S.I formula, you can get C.I for thesecond year.

P = 3300 R = 10 T = 1 Year

C.I for the 2nd year = PTR / 100 = (3300 X 1 X 10) / 100

C.I = 330

The total C.I = 300 (at the end of the 1st year) + 330 (at the end of the 2nd year)

= 630

The same C.I can also be calculated by the above mentioned C.I formula.

C.I = Amt - Principal =


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