Presented by: Mr. Partha Sarathi Biswas.

B.E. Computer Sc. & Engg.

Jadavpur University.

The number system that we use in our day-to-day life in each & every field is the decimal number system. Decimal number system has base 10 as it uses 10 digits( 0,1,2,3,4,5,6,7,8 9).

In decimal number system, the successive positions to the left of the decimal point represent units, tens, hundreds, thousands and so on. And fractions can be shown as its equivalent decimal followed by a point.

----- H.C.F. & L.C.M.

----- Base conversion.

----- A.P.,G.P. & H.P.

----- Unit digit.

----- Remainder.

----- No. of zeros at the end.

----- No. of factors or number of divisors.

1. Express the numbers as product of prime factors

12 = 3*22

36 = 32 *22 ◦ = 3*24

The common prime factors are 2 and 3 and the corresponding

least indices are 2 and 1 respectively The product of all the common prime factors with the respective

least indices H.C.F of 12, 36, 48 = 22 *3= 12

Highest Common Factor (H.C.F) H.C.F is the highest common factor or also

known as greatest common divisor, the greatest number which exactly divides all the given numbers.

There are two methods to find H.C.F of given

numbers, they are: ◦ Prime factorization method.

◦ Division Method.

How to find H.C.F of given numbers by prime factorization method? Step 1:- Express the given numbers as product of their prime factors Step 2:- Check for the common prime factors and find least index of

each common prime factor in the given numbers Step 3:- The product of all common prime factors with the respective

least indices is H.C.F of given numbers .

Ex:

H.C.F of 12, 36, 48

How to find H.C.F of two numbers by division method?

Step 1:- Divide higher number with the smaller number .

Step 2:- Divide smaller number with the remainder of above division .

Step 3:- Then second remainder divides first remainder and the process of division continues till remainder is zero.

Step 4:- Last divisor of division is the H.C.F of two numbers .

H.C.F of 12 and 56 12) 56 ( 4 48 8) 12 ( 1 8 4) 8 ( 2 8 0 The last divisor in the above division is 4. Hence, H.C.F of 12 and 56 = 4

L.C.M is least common multiple, the smallest number which is exactly divisible by all the given numbers.

There are two methods to find L.C.M of given numbers, they are: ◦ Prime factorization method.

◦ Division Method.

How to find L.C.M of given numbers by prime factorization method? Step 1:- Express the given numbers as product of their prime factors. Step 2:- Find highest index in all the prime factors of given numbers. Step 3:- The product of all the prime factors with respective highest

indices is the L.C.M of given numbers.

Ex: L.C.M of 14, 42, 36 Express the numbers as product of prime factors. 14 = 2*7 36 = 32 *22 42 = 2*3*7The highest index of 2, 3, 7 are 2, 2, 1 respectively The product of all the prime factors with the respective highest

indices. L.C.M of 14, 36, 42 = 22 *32 *7 = 252

How to find L.C.M of given numbers by division method?

Step 1:- Divide all the numbers with the common prime factors.

Step 2:- The division process continues until there is no common factor to the numbers.

Step 3:- The product of divisors and remaining numbers with no common factor is the L.C.M of given numbers.

Ex: L.C.M of 12, 98, 188

2) 12,98,188

2 ) 6, 49, 94

3, 49, 47

The product of divisors and remaining numbers = 2*2*3*49*47 = 27636

Hence, the L.C.M of 12, 98, 188 = 27636

H.C.F . Of numerator

H.C.F of given fractions = --------------------- L.C.M. Of denominator

L.C.M of numerator

L.C.M of given fractions = ---------------------

H.C.F. Of denominator

Find out H.C.F & L.C.M. 0f 15/16 , 21/22, 33/52

Question 3). Six bells commence tolling together and toll

at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?

A. 4 B. 10 C. 15 D. 16

Question 4).

Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

A. 4

B. 7

C. 9

D. 13

Question 5).

The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

A. 9000

B. 9400

C. 9600

D. 9800

Decimal to Other Base System Other Base System to Decimal Other Base System to Non-Decimal Shortcut method − Binary to Octal Shortcut method − Octal to Binary Shortcut method − Binary to Hexadecimal Shortcut method − Hexadecimal to Binary

Step 1 − Divide the decimal number to be converted by the value of the new base.

Step 2 − Get the remainder from Step 1 as the rightmost digit least significant digit of new base number.

Step 3 − Divide the quotient of the previous divide by the new base.

Step 4 − Record the remainder from Step 3 as the next digit to the left of the new base number.

Repeat Steps 3 and 4, getting remainders from right to left, until the quotient becomes zero in Step 3.

The last remainder thus obtained will be the Most Significant Digit MSD

of the new base number.

Example − Decimal Number: 2910 Step Operation Result Remainder Step 1 29/ 2 14 1 Step2 14/2 7 0 Step 3 7 / 2 3 1 Step 4 3 / 2 1 1 Step 5 1 / 2 0 1 As mentioned in Steps 2 and 4, the remainders have to be arranged

in the reverse order so that the first remainder becomes the Least Significant Digit LSD and the last remainder becomes the Most Significant Digit MSD.

Decimal Number − 2910 = Binary Number − 111012.

Step 1 − Determine the column positional value of each digit this depends on the position of the digit and the base of the number system.

Step 2 − Multiply the obtained column values inStep1 by the

digits in the corresponding columns. Step 3 − Sum the products calculated in Step 2. The total is the

equivalent value in decimal. Example Binary Number − 111012 Calculating Decimal Equivalent −

111012 = ((1 × 24) + (1 × 23) + (1 × 22) + (0 × 21) + (1

×20))

Binary Number − 111012 = Decimal Number − 2910

Step 1 − Convert the original number to a decimal number base10.

Step 2 − Convert the decimal number so

obtained to the new base number.

Example Octal Number − 258 = Decimal - ((2 × 81) + (5 ×

80))10

Octal Number − 258 = Decimal Number − 2110

Decimal Number − 2110 = Binary Number − 101012

Octal Number − 258 = Binary Number − 101012

3B16 = 3×161+11×160 = 48+11 = 59

E7A916 = 14×163+7×162+10×161+9×160 = 57344+1792+160+9 = 59305

Hex to Dec Dec to Hex

A man ate 100 bananas in five days, each day eating 6 more than the previous day. How many bananas did he eat on the first day? A. 7 B. 8 C. 9 D. 10

Question 2.

The difference between a number and its four-fifth is 25. What is the number?

1. 75 2. 100 3. 125 4. 150 5. None of these

Question 3.

A student got thrice as many marks in Arithmetic as in English. If the total of his marks in the two subjects is 128, how many marks did he get in English?

1. 32 2. 64 3. 96 4. 16 5. None of these

Two radios and one calculator can cost Rs.3300, while two calculators and one radio cost Rs.5400. What is the price of a radio?

1. Rs. 2500 2. Rs. 700 3. Rs. 400 4. Rs. 1100 5. None of these

Question 5.

On selling 17 balls at Rs. 720/-, there is a loss equal to the cost price of 5 balls. The cost price of a ball in rupees is:

1. 45 2. 50 3. 55 4. 60 5. None of these

A man spends half of his salary on household expenses, 1/4th for rent, 1/5th for travel expenses, the man deposits the rest in a bank. If his monthly deposits in the bank amount 50, what is his monthly salary ?

(a) Rs.500 (b) Rs.1500 (c) Rs.1000 (d) Rs. 900

Question 7.

The sum of the squares of three consecutive natural numbers is 2030, then the middle number is:

1. 24 2. 26 3. 25 4. 27 5. None of these

Question 8.

? ÷ √ 0.81 = 300

1. 2700 2. 270 3. 33.33 4. 540 5. None of these

Question 9 .

Find the value of ( 0.75 * 0.75 * 0.75 - 0.001 ) / ( 0.75 * 0.75 - 0.075 + 0.01)?

a) 0.845 b) 1.908 c) 2.312 d) 0.001

The difference between squares of two consecutive even numbers is always divisible by:

1. 3 2. 4 3. 5 4. 2 5. None of these

Question 11.

What is the smallest number by which 2880 must be divided in order to make it into a perfect square ?

(a) 3 (b) 4 (c) 5 (d) 6

Question 12. A box contains 90 mts each of 100 gms and 100

bolts each of 150 gms. If the entire box weighs 35.5 kg., then the weight of the empty box is :

10 kg 10.5 kg 11 kg 11.5 kg None of the above

Question 13.

If the sum of n terms of two series of A.P are in the ratio 5n+4:9n+6 .find the ratio of their 13th terms

(a) 129/231 (b) 1/2 (c) 23/15 (d) None of the above.

Question 14.

The ratio between two numbers is 5:7 and their product is 560. What is the difference between these two numbers?

A. 8 B. 12 C. 6 D. 21