Saturday 11 September 2010

Problems on LCM and HCF

☛ Read Basics of LCM and HCF Here, Before starting Practice

1. Find the least number which when increased by 4 is exactly divisible by 8, 16, 24, 30 and 32 ?
a) 480
b) 484
c) 476
d) 472
e) None of these

2. What is the greatest number of five digits which when 3769 is added to it will be exactly divisible by 5, 6 , 10, 12, 15 and 18 ?
a) 4309
b) 99459
c) 100539
d) 99911
e) None of These

3. Find the minimum number of square tiles required to pave the floor of a room of 2m 50cm long and 1m 50cm broad ?
a) 50
b) 750
c) 45
d) 15
e) None of these

4. Five bells toll together at the intervals of 5, 6, 8, 12 and 20 seconds respectively. Find the number of times they toll together in one hour's time (Inclusive of the toll at the beginning)
a) 120
b) 31
c) 30
d) 5
e) None of These

5. A milk man has three different kinds of milk 493liters, 551 liters and 435 liters. Find the minimum number of equal size containers required to store all the milk without mixing.
a) 29
b) 51
c) 58
d) 49
e) None of these

6. The circumference of the front and back wheels of a vehicle are 6 3/14 m and 8 1/18 m respectively. At any given moment, a chalk mark is put on the point of contact of each wheel with the ground. Find the distance traveled by the vehicle so that both the chalk marks are again on the ground at the same time
a) 217.5 m
b) 435 m
c) 412m
d) 419m
e) None of these

7. The LCM of two numbers is 28 times of their HCF. The sum of their LCM and HCF is 1740. If one of the numbers is 420, the other number is
a) 150
b) 225
c) 180
d) 240
e) None of these

8. Two persons A and B walk around a circular track whose radius is 1.4 km. A walks at a speed of 176 meters per minute while B walks at a speed of 110 meters per minute. if they both start at the same time, from the same point and walk in the same direction, at what interval of time would they both be at the same starting point again? (in Hours)
a) 6 2/3
b) 2 1/3
c) 5 1/4
d) 3 2/3
e) None of these

9. Find the least number which when divided by 8, 9, 15, 24, 32 and 36 leaves remainders 3, 4, 10, 19, 27 and 31 respectively?
a) 2880
b) 2885
c) 2974
d) 2875
e) None of these

10. Find the greatest number which when divide 357, 192 and 252 leaves same remainder in each case
a) 45
b) 1
c) 15

d) Cant be determinde
e) None of these

Solutions :

1. LCM of 8, 16, 24, 30 and 32 is 480
So, Required number is 480 - 4 = 476

2. LCM of 5, 6, 10, 12 and 18 is 540
On dividing (99999 + 3769) by 540, the remainder is 88
So, the required number is 99999 - 88 = 99911

3. HCF of 250 cm and 150 cm is 50 cm, which is the side of the tile

So, the required number of tiles = (250 X 150) / (50X50) = 15

4. Time after which all the bells tol together is the LCM of 5, 6, 8, 12 and 20.
i.e., 120 seconds = 20 minutes
The number of times they toll together in one hour = 60/2 = 30 + 1 (beginning tone)
So, the answer is 31

5. As minimum number of containers are required, the size of the container should be maximum and the size is also equal. so size of the container will be HCF of 493, 551 and 435 i.e., 29
So, required number of containers is = (493+551+435) / 29 = 51

6. The required distance is the LCM of 6 3/14 and 8 1/18
LCM of 6 3/14 and 8 1/18 = LCM (87/14, 145/18)
=> LCM(87,145) / HCF (14,18) = 435/2 = 217.5m
7. LCM = 28 HCF
LCM + HCF = 1740
=> 28 HCF + HCF = 1740
HCF = 1740/29 = 60 and LCM = 28X60
if a and b are two numbers, then
LCM of (a & b) X HCF of (a & b) = aXb
28X60X60 = x X 420 = > x = 240
So, the other number is 240

8. Circumference of the track is 2πr
= 2 X (22/7) X 1400m = 8800 m

Time taken by A to complete one round
= (8800m) / (176m/min) = (8800X60)/(176) = 3000 sec

Time taken by B to complete one round

= (8800) / (110m/min) = (8800X60) / (110) = 4800 sec

Time they meet together at the starting point is LCM of 3000 and 4800 sec

i.e., 24000 sec = 6 2/3 hours

So, they meet at the starting point after 6 2/3 hours

9. We can observe that the difference between the numbers and their remainders is same

i.e., 8-3 = 9-4 = 15-10 = 24-19 = 32-27 = 36-31 = 5

So, required answer is

LCM (8, 9, 15, 24, 32, 36) - 5

= > 2880 - 5 = 2875

10. Required answer is

HCF (357-192, 192-252, 357-252)

= > HCF (165, 60, 105) = 15

00:52 - By Unknown 0

Friday 10 September 2010

LCM and HCF

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

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 other

Ex : 42 is a multiple of 6 and 7

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

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

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.

Ex : 14, 15, 16, 18 etc

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

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.

Ex : 12 is a common multiple of 2, 3, 4 and 6

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

Ex :

20 is the Common Multiple of 2, 4, 5 and 10
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 more numbers is the Greatest Number which divides each of them Exactly.

It is also Called Greatest Common Divisor (GCD)
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 them exactly 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 Highest number? 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 :

Express each of the given number as the product of Prime Factors
Choose common factors
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

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

HCF by Method of Division :

Consider two different numbers.
Divide the longer number by the smaller one.
Now divide the divisor by the reminder.
Repeat this process of dividing the preceding divisor by the last reminder obtained, till you get the reminder "0"
The LAST DIVISOR is the HCF of the given TWO numbers

Ex : Find the HCF of 42, 70

Thats it. Now the answer is 14 :)

☛ Check some Practice Problems on LCM and HCF Here

00:50 - By Unknown 0

Thursday 9 September 2010

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.
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 of 13.
If the difference of 5 times the digit in units place and the number in the remaining part is 0 or multiple of 17, then the number is divisible by 17.
If the sum of double the digit in units place of a given number and number in the remaining part is multiple of 19, then the given number is divisible of 19.

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00:49 - By Unknown 0

Wednesday 8 September 2010

Shortcut for Multiplication (Base Number Method Part 2)

We've already discussed about the Shortcut Method (Base Number Method) for doing multiplications. In the earlier post we've discussed doing the multiplication by taking the base as 100 (Click Here to read that). I mean, you can do the multiplication for the numbers nearer to the 100. Then what about the remaining numbers? Well, being the Guide 4 Bank Exams we have solutions for all Banking Problems... So, lets have a look at some more approaches of the Base Number Method so that it will be easier for you to multiply any number with any number :)

Find the product of 117 and 88

117 ---> +17

88 ---> -12

____ _____

105 -204 Ans : 10296

Here, note that to take care of -204 of the second part, borrowing a 1 from the first part is not sufficient (because the 100 it becomes when it comes to the second part is not numerically greater than -204). So, we should borrow 3 from 105 (leaving 102 as the first part) which becomes 300 in the second part to which -204 should be added giving us 96. Hence, the product of 117 and 88 is 10296.

Find the product of 997 and 983.

Here, both the numbers are close to 1000. So take 1000 as the base number.

997 ---> -3

983 ---> -17

_____ _____

980 +51 Ans : 980051

The Second part 51 has only two digits whereas the base 1000 has three zeroes. So, 51 will be written as 051. hence the Product is 980051.

Find the product of 297 and 292.

Here, the numbers are not close to any power of 10 but are close to 300 which is a multiple of 100 which itself is a power of 10. So we adopt 300 as a "temporary base". This temporary base is a multiple (or a sub-multiple) of the main base 100. Here, the temporary base 300 = 3 X 100. Then, the procedure of finding out the deviation from the base, getting the cross-totals and the product of the deviations should be done in a manner similar to the previous cases except that the deviations will be taken from the temporary base.

297 ---> -3

292 ---> -8

____ ____

289 +24 Ans : 86724

We've got the first part of the answer as 289 and the second part of the answer as 24. But before we put these two parts together to get the final result, one more step is involved. The first part of the answer is not the final figure. This is an intermediate stage of the first part. This first part should be multiplied by the same figure with which the power of 10 is multiplied to get the temporary base. In this case, we multiplied 100 (which is the power of 10) by 3 to get the temporary base 300. So, the intermediate stage figure of the first part (289) will also have to be multiplied by 3 to get the final figure for the first part. Hence the first part wil be 867 ( = 3 x 289). Now putting the first and the second parts together , the product of 297 and 292 is 86724 (here remember that the product of the deviations should still have as many digits as the number of zeroes in the base, in the above case it is TWO. because 100 has TWO zeroes).

Thats all for now friends. Tomorrow we shall discuss another Shortcut Method. Happy Reading :)

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00:47 - By Unknown 0

Tuesday 7 September 2010

Shortcut for Multiplication (Base Number Method)

We've already discussed some short cut tips for Multiplications. You can check that post Here. Now we shall discuss another Simple and Useful method for Multiplication. In this method we take a base (according to the given numbers) and do the multiplication according to that base. Assume that you have to multiply 97 with 92. The power of 10 to which these two numbers are close is 100. We call this 100 as the base. Now write these two numbers with the difference from the base.

97 ------> -3 (because 97 is obtained as 100 - 3)

92 ------> -8 (because 92 is obtained as 100 -8)

Then take the sum of the two numbers (including their signs) along either one of the two diagonals (see, it will be same for both cases).

Have a look at the above example,

The diagonal sum is 97 - 8 = 92 - 3 = 89

This will form the first part of the Answer.

The second part of the answer is the product (taken along with the sign) of the difference from the power of 10 written for the two numbers. In the above example it is the product of -3 and -8 which is 24.

So, putting these two parts (89 and 24) together one next to the other, the answer is 8924, i.e., the product of 97 and 92 is 8924.

Note : Here keep in mind that the product of the two deviations should have as many digits as the number of zeros in the base. For Example, in this case the product of -8 and -3 has 2 digits which is the same as the number of zeroes in 100.

Lets have a look at another example so that you can understand the above method completely.

Find the product of 113 and 118

Here, both the numbers are greater than 100 and the base here is 100. Taking the difference of the two numbers 113 and 118 from the base, we get +13 and +18 and write them as below.

113 ----- > + 13

118 ----- > +18

_____ _______

131 234 Ans : 13334

The first part of the answer is the cross-total of 113 and +18 which is 131. The second part of the answer, i.e., the product of the deviations (+13 and +18) is equal to 234. But we said there should be as many digits in this product as the number of zeroes in the base (which is 100 here). Since the base has two zeroes, the second part of the answer should also have two digits. Since 234 has three digits, we should retain two digits 4 and 3 and carry forward the third digit 2 to the first part of the answer. hence, the first part of the answer now becomes 133 and the second part is 34. The product of 113 and 118 is thus equal to 13334.

If you want to know the multiplication tips by using the base numbers other than 100 (like 200, 300, 1000, 50 etc...) click Here.

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Thats all for now friends... Tomorrow we shall discuss another shortcut method. Happy Reading :)

Multiplication Tricks and Short Cut Methods for Bank Exams Clerk Po

00:45 - By Unknown 0

Monday 6 September 2010

Shortcuts Maths

Friends, Here you can find Aptitude Shortcuts which are very useful for all the Competitive Exams. We have started with the basic Shortcuts for Aptitude and explained how to save time by improving your Calculation Speed while calculating mathematical problems. The best thing is we have provided aptitude in pdf files when ever it is possible so that you can download and read. We can proudly say that these posts are very useful for those people who are searching for Aptitude for Bank Exams, Aptitude for Cat, SSC and ofcourse each and every Competitive Exam which contains Aptitude Section in it. Thats all for now friends, its upto you to decide how these maths shortcuts and tricks works. All The Best and Happy Reading :)