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Parin Sir > Quantitative Aptitude (Maths) > H. C. F. & L. C. M. OF NUMBERS > Examples > Examples 1 to 10

Examples 1 to 10

June 2, 2020
Category: Examples

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(1) Two numbers are in the ratio of 15 : 7. If their H.C.F. is 12, find the numbers.

Soln. Let the required numbers be 15x and 7x.

Then, their H.C.F. is x. So, x = 12.

$\therefore$ The numbers are (15 $\times$ 12 and 7 $\times$ 12) i.e., 180 and 84.

(15, 7 ) = 1 and H.C.F. of two numbers is 12.

$\therefore$ The numbers are 15 $\times$ 12 = 180 and 7 $\times$ 12 = 84.

(2) Two numbers are in the ratio of 12 : 9. If their H.C.F. is 15, find the numbers.

Soln. 12 : 9 = 4 : 3

(4, 3 ) = 1 and H.C.F. of two numbers is 15

The numbers are 4 $\times$ 15 = 60 and 3 $\times$ 15 = 45.

(3) The H.C.F. of two numbers is 28 and their L.C.M. is 3500. If one of the numbers is 140, find the other.

Soln. Other number = $\frac{HCF \times LCM}{Given - Number}$ = $\frac{28 \times 3500}{140}$ = 700.

To find the pair of numbers when only H.C.F. and L.C.M. are given.

Step : 1 Find $\frac{L.C.M.}{H.C.F.}$ = k

Step : 2 Now, factorize k in different possible cases.

Step : 3 Now, select the factors which are co-primes.

Step : 4 Multiply both the factors by that H.C.F..

Thus, we get the pairs of required numbers which follow the above condition ( in statement ).

(4) Find the pair of numbers whose H.C.F. = 143 and L.C.M. = 120120.

Soln. Step : 1 $\frac{L.C.M.}{H.C.F.}$ = $\frac{120120}{143}$ = 840

Step : 2 840 = 1 $\times$ 840 (true) As ( 1, 840 ) $\ne$ 1

= 2 $\times$ 420 (false) As ( 2, 420 ) = 2 $\ne$ 1

= 3 $\times$ 280 (true) As ( 3, 280 ) = 1

= 4 $\times$ 210 (false) As ( 4, 210 ) = 2 $\ne$ 1

= 5 $\times$ 168 (true) As ( 5, 168 ) = 1

= 6 $\times$ 140 (false) As ( 6, 140 ) = 2 $\ne$ 1

= 7 $\times$ 120 (true) As ( 7, 120 ) = 1

= 8 $\times$ 105 (true) As ( 8, 105 ) = 1

= 10 $\times$ 84 (false) As ( 10, 84 ) = 2 $\ne$ 1

= 12 $\times$ 70 (false) As ( 12, 70 ) = 2 $\ne$ 1

= 14 $\times$ 60 (false) As ( 14, 60 ) = 2 $\ne$ 1

= 15 $\times$ 56 (true) As ( 15, 56 ) = 1

= 20 $\times$ 42 (false) As ( 20, 22 ) = 2 $\ne$ 1

= 21 $\times$ 40 (true) As ( 21, 40 ) = 1

= 24 $\times$ 35 (true) As ( 24, 35 ) = 1

= 28 $\times$ 30 (false) As ( 28, 30 ) = 2 $\ne$ 1

Step : 3 Hence, number of pairs whose H.C.F. = 143 & L.C.M. = 120120 are 8.

Step : 4 Thus, the required nos. are

1 $\times$ 143 = 143 & 840 $\times$ 143 = 120120

3 $\times$ 143 = 429 & 280 $\times$ 143 = 40040

5 $\times$ 143 = 715 & 168 $\times$ 143 = 24024

7 $\times$ 143 = 1001 & 120 $\times$ 143 = 17160

8 $\times$ 143 = 1144 & 105 $\times$ 143 = 15015

15 $\times$ 143 = 2145 & 56 $\times$ 143 = 8008

21 $\times$ 143 = 3003 & 40 $\times$ 143 = 5720

24 $\times$ 143 = 3432 & 35 $\times$ 143 = 5005

(5) Find the pair of numbers whose H.C.F. = 5 and L.C.M. = 300.

Soln. Step : 1 $\frac{L.C.M.}{H.C.F.}$ = $\frac{300}{5}$ = 60

Step : 2

60 = 1 $\times$ 60 (true) As ( 1, 60 ) = 1

= 2 $\times$ 30 (false) As ( 2, 30 ) = 2 $\ne$ 1

= 3 $\times$ 20 (true) As ( 3, 20 ) = 1

= 4 $\times$ 15 (true) As ( 4, 15 ) = 1

= 5 $\times$ 12 (true) As ( 5, 12 ) = 1

= 6 $\times$ 10 (false) As ( 6, 10 ) = 2 $\ne$ 1

Step : 3 Hence, number of pairs whose H.C.F. = 5 & L.C.M. = 300 are 4.

Step : 4 Thus, the required nos. are

1 $\times$ 5 = 5 & 60 $\times$ 5 = 300

3 $\times$ 5 = 15 & 20 $\times$ 5 = 100

4 $\times$ 5 = 20 & 15 $\times$ 5 = 75

5 $\times$ 5 = 25 & 12 $\times$ 5 = 60

(6) Find the largest number which can exactly divide 513, 261 & 477.

Soln. Required number = H.C.F. of 513, 261 & 477 = 9.

(7) Find the greatest possible length which can be used to measure exactly the lengths 15 m, 10 m 92 cm & 4 m 44 cm.

Soln. 15 m = 1500 cm , 10 m 92 cm = 1092 cm , 4 m 44 cm = 444 cm.

Now, ( 1500, 1092, 444 ) = 12.

$\therefore$ Greatest possible length = 12 cm.

(8) Greatest number which can divide 281, 701, 509 leaving the same remainder 5 in each case, is :

Soln. Required number = H.C.F. of ( 281 – 5 ), ( 701 – 5 ), ( 509 – 5 )

= H.C.F. of 276, 696, 504

= 12

(9) Greatest number which can divide 352, 316, 496 leaving the same remainder 10 in each case, is :

Soln. Required number = H.C.F. of ( 352 – 10 ), ( 316 – 10 ), ( 496 – 10 )

= H.C.F. of 342, 306, 486

= 18

(10) Greatest number which can divide 150, 179, 279 leaving respectively 3, 4, 6 as remainders, is :

Soln. Required number = H.C.F. of ( 150 – 3 ), ( 179 – 4 ), ( 279 – 6 )

= H.C.F. of 147, 175, 273

= 7

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