Finding HCF by the method of Division

Finding HCF by the method of Division

Step 1: Divide the bigger number with the smaller number.

Step 2: divide the divisor by the remainder

Step 3: divide the remainder by the next remainder, and proceed till no remainder is left.

Example:

Find the HCF of 13281 and 15844

The smaller of the two numbers is 13281, so, we will divide 15844 by 13281

13281 ) 15844 ( 1

               13281

                 2563 ) 13281 ( 5

                             12815

                                 466) 2563 (5

                                         2330

                                           233 ) 466 ( 2

                                                    466

                                                     x x

Therefore, 233 is the HCF of the 2 numbers.

tougher problems in HCF and LCM

                                                                                                                                                                                               

The numbers 11284 and 7655, when divided by a certain number of three digits, leave the same remainder. Find the number and the remainder.

                                                                                                                                                                                               

Answer:

Let, Y be the common remainder. So, we have 11284 – y , 7655 – y as the perfectly divisible numbers.

If 2 numbers are exactly divisible by a certain number, then there difference is also exactly divisible.

So, (11284 – y) – (7655 – y), is exactly divisible.

Or, 3629 is divisible by the required number. Let us find the factors of 3629.

We have, 3629 = 191 x 19

Therefore the 3 digit number is 191.

Now let us divide 11284 by 191 to find the remainder

11284/191 = 59 ¹⁵/₁₉₁.

Therefore, the remainder is 15.

Tougher problems on HCF and LCM

What is the greatest number that will divide 2930 and 3250 and will leave a remainder 7 and 11 respectively?  

    

Answer:

(2930 -7)= 2923 is perfectly divisible.

Also,( 3250 -11) 3239 is perfectly divisible.

Now, let us find the HCF of 2923 and 3239

2923 = 37 x 79

3239 = 41 x 79

Therefore, the HCF is 79.              

Find the greatest number which is such that when 12288, 19139 

and 28200 are divided by it, the remainders are all the same.

Answer:

Let, Y be the common remainder. So, we have 12288 – y , 19139 – y , 28200 – y as the perfectly divisible numbers.

If 2 numbers are exactly divisible by a certain number, then there difference is all exactly divisible.

So, (19139 – y) – (12288 – y), (28200 – y) – (19139 – y) and (28200 – y) – (12288 – y) are exactly divisible.

Or, 6851, 9061 and 15912 are divisible by the required number.

Therefore let us find the HCF of  6851, 9061 and 15912.

6851 = 17 x 13 x 31

9061 = 17 x 13 x 41

15912 = 17 x 13 x 2 x 3 x 3 x 2 x 2

Therefore, HCF = 17 x 13 = 221.

So, the required number is 221.

 

simple equations STD VI

Present ages of Anu and Raj are in the ratio 4:5. Eight years from now the ratio of their ages will be 5:6. Find their present ages.

Answer:

Let the present ages of Anu and Raj be 4x years and 5x years respectively.

After eight years. Anu’s age = (4x + 8) years;

After eight years, Raj’s age = (5x + 8) years.

Therefore, the ratio of their ages after eight years = (4x + 8)/(5x + 8)

But this is given as 5: 6

Therefore, (4x + 8)/(5x + 8) = 5/6

Cross-multiplication gives 6 (4x + 8) = 5 (5x + 8)

Or, 24x + 48 = 25x + 40

Or, 24x + 48 – 40 = 25x

Or, 24x + 8 = 25x

Or, 8 = 25x – 24x

Or, 8 = x

Therefore, Anu’s present age = 4x = 4 × 8 = 32 years

Raj’s present age = 5x = 5 × 8 = 40 years

The ages of Hari and Harry are in the ratio 5:7. Four years from now the ratio of their ages will be 3:4. Find their present ages.

Answer:

Let the present ages of Hari and Harry be 5x years and 7x years respectively.

After eight years. Hari’s age = (5x + 4) years;

After eight years, Harry’s age = (7x + 4) years.

Therefore, the ratio of their ages after eight years =

(5x + 4)/(7x + 4)

This is given to be 3: 4

Therefore, (5x + 4)/(7x + 4) = 3/4

Cross-multiplication gives  4(5x + 4)=  3(7x + 4)

Or, 20x + 16 = 21x + 12

Or, 16 – 12 = 21x – 20x

Or, 4 = x

Or, x = 4

Therefore, Hari’s present age = 5x = 20 years

Harry’s present age = 7x = 28 years

The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is 3/2. Find the rational number.

Answer:

Let the numerator be y

So, the denominator will be 8 + y

Given, ( y + 17)/ (8 + y -1) = 3/2

Or, 2(y + 17) = 3( 7 + y)

Or, 2y + 34 = 21 + 3y

Or, 34 – 21 = 3y – 2y

Or, y = 13

So, the rational number is = 13/21

simple equations STD VI

Half of a herd of deer are grazing in the field and three fourths of the remaining is playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

Answer:

Let the number of deer be y

No. of deer grazing = y/2

Remaining no. of deer = 1-y/2 = y/2

No. of deer playing = ¾ x y/2 = 3y/8

Given that, y – (y/2 + 3y/8) = 9

Or, y – y(4+3)/8 = 9

Or, y(1 – 7/8) = 9

Or, y x 1/8 = 9

Or, y = 72.

A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.

Answer:

Let the age of the granddaughter be y

So, the age of the grandfather is 10y.

Also given that 10y – y = 54

Or, 9y = 54

Or, y = 6 years

So, the age of the granddaughter is 6, and the age of the grandfather is 60.

Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.

Answer:

Let son’s age by y

So, Aman’s age is 3y

Also given that, 5 x (y-10) = 3y – 10

Or, 5y – 50 = 3y – 10

Or, 5y – 3y = 50 – 10

Or, 2y = 40

Or, y = 40/2 = 20

So, the son’s age is 20 years and Aman’s age is 60 years.

Simple Equations STD VI

Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of his mother’s present age. What are their present ages?

Answer:

Let Shobo’s present age by y.

So, Shobo’s mother’s present age is 6y.

Also given, (y + 5) = 1/3 x 6y

Or, y + 5 = 2y

Or 2y – y = 5

Or, y = 5

So, Shobo’s age is 5.

And Shobo’s mother’s age is 6y = 30. 

There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate Rs100 per meter it will cost the village panchayat Rs 75000 to fence the plot. What are the dimensions of the plot?

Answer:

Let the length of the plot be L

Let the breadth of the plot be B

So, given L:B = 11: 4

Or, 4L = 11B

Or, B = 4L /11

Now perimeter of the rectangular plot is

= 2 ( L + B)

= 2 ( L + 4L/11)  because B = 4L/11

Now given that 100 x 2(L + 4L/11) = 75000

Or, 200(L + 4L /11) = 75000

Or, L (1 + 4/11) = 75000/200

Or, L(15/11) = 375

Or, L = 375 x 11/15 = 25 x 11 = 275

So, the length of the plot is 275 meters

And the breadth of the plot is 4x275/11 = 4 x 25 = 100 meters.

Check:

2(275+100) x 100 = 750x100 = 75000

Simple equations STD VI

A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Answer:

Let the number be y

Therefore the other number is 5y

Also, 2(21 + y) = 5y + 21

Or, 42 + 2y = 5y + 21

Or, 3y = 42 – 21

Or, 3y = 21

Or, y = 7

So, the two numbers are 7 and 35

Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?

Answer:

Let the number in the units place be y

So, the other number is 9-y.

Given, 10(9 – y) + y + 27 = 10y + (9 – y)

Or, 90 – 10y + y + 27 = 10y + 9 – y

Or, 117 – 9y = 9y + 9

Or, 18 y = 117 – 9

Or, y = 108/18 = 6

So, the other number is 9- y = 3

So, the 2 digit number is 36.

Simple equations STD VI

Arjun is twice as old as Shriya. Five years ago his age was three times Shriya’s age. Find their present ages.

Answer:

Let us take Shriya’s present age to be x years.

Then Arjun’s present age would be 2x years.

Shriya’s age five years ago was (x – 5) years.

Arjun’s age five years ago was (2x – 5) years.

It is given that Arjun’s age five years ago was three times Shriya’s age.

Thus, 2x – 5 = 3(x – 5)

Or, 2x – 5 = 3x – 15

Or, 15 – 5 = 3x – 2x

Or, 10 = x

So, Shriya’s present age = x = 10 years.

Therefore, Arjun’s present age = 2x = 2 × 10 = 20 years.

The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. What can be the original number?

Answer:

Let us take the two digit number such that the digit in the units place is b. The digit in the tens place differs from b by 3. Let us take it as b + 3. So the two-digit number is,          

10 (b + 3) + b = 10b + 30 + b = 11b + 30.

With interchange of digits, the resulting two-digit number will be,

10b + (b + 3) = 11b + 3

If we add these two two-digit numbers, their sum is

(11b + 30) + (11b + 3) = 11b + 11b + 30 + 3 = 22b + 33

It is given that the sum is 143. Therefore, 22b + 33 = 143

Or, 22b = 143 – 33

Or, 22b = 110

Or, b = 110/22

Or, b =5

The units digit is 5 and therefore the tens digit is 5 + 3, which is 8. The number is 85.

Check: On interchange of digits the number we get is

58. The sum of 85 and 58 is 143 as given.

Simple equations STD VI

The organizers of an essay competition decide that a winner in the competition gets a prize of $ 100 and a participant who does not win gets a prize of $ 25. The total prize money distributed is $ 3,000. Find the number of winners, if the total number of participants is 63.

Answer:

Let the number of winners be y

So, number of students who did not win = 63 – y

Given that, 100y + (63 – y )25 = 3000

Or, 100y + 1575 – 25y = 3000

Or, 75y = 3000 – 1575

Or, y = 1425/75

Or, y = 19.

I have a total of $ 300  in notes of denomination $ 1, $ 2 and $ 5. The number of $ 2  notes is 3 times the number of $ 5 notes. The total number of notes is 160. How many notes  of each denomination are with me?

Answer:

Let the number of $5 denomination note be y

So, number of $2 denomination note is 3y

So, number of $1 denomination note = 160 – y – 3y = 160 -4y

Given, 5y + 2x3y +1x (160 – 4y) = 300

Or, 5y +6y + 160 – 4y = 300

Or, 7y = 140

Or, y = 20

The number of $5 denomination note is 20

So, number of $2 denomination note is 3y = 60

So, number of $1 denomination note = 160 - 4y = 80

Simple equations STD VI

Lakshmi is a cashier in a bank. She has currency notes of denominations Rs 100, Rs 50 and Rs 10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is Rs 4,00,000. How many notes of each denomination does she have?

Answer:

Let total number of notes be y

So, fraction of notes of Rs 100 denomination = 2/ (2+3+5) = 2/ 10 = 0.2

So, fraction of notes of Rs 50 denomination = 3/ (2+3+5) = 3/ 10 = 0.3

So, fraction of notes of Rs 100 denomination = 5/ (2+3+5) = 5/ 10 = 0.5

Therefore, 0.2y x 100 + 0.3y x 50 + 0.5y x 10 = 400000

Or 20y +15y + 5y = 400000

Or, 40y = 400000

Or, y = 400000/40 = 10000

So, 100 denomination notes = 0.2 x 10000 = 2000

So, 50 denomination notes = 0.3 x 10000 = 3000

So, 10 denomination notes = 0.5 x 10000 = 5000

Check:

2000 x 100 = 200000

3000 x 50 = 150000

5000 x 10 = 50000

SO, total = 200000 + 150000 + 50000 = 400000