Question

The one of two numbers is 25 and the H.C.F. and L.C.M. of these numbers are 5 and
120 respectively, then, find the sum of the reciprocals of the numbers?
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Answer 1

Answer:

hey mate here is the answer

Step-by-step explanation:

Let a and b are two numbers.

a/c to question,

a + b = 55 ......(1)

HCF {a, b} = 5

so, we can write a = 5k and b = 5l, where k and l are co-prime numbers.

putting a and b in equation (1),

5k + 5l = 55 => k + l = 11 .....(2)

again, LCM{a, b} = 120

so, 5 × k × l = 120

k × l = 24 ........(3)

from equations (2) and (3),

k × (11 - k) = 24

or, 11k - k² = 24

or, k² - 11k + 24 = 0

or, k² - 3k - 8k + 24 = 0

or, (k - 3)(k - 8) = 0

hence, k = 3 and 8

putting in equation (2), l = 8 and 3

hence, if k = 3 then, l = 8

and if k = 8 then, l = 3

choose anyone of them,

k = 3, and l = 8

then, a = 5k = 15. b = 5l = 40

hence, numbers are 15 and 40

now, sum of reciprocal of the numbers = 1/15 + 1/40 = (8 + 3)/120 = 11/120

hope this helps and plzzzzz mark me as brainlist

Answer 2

GIVEN:

• ONE OF THE NUMBER IS 25
• HCF AND LCM OF THESE NUMBERS IS 5 & 120

TO FIND:

• SUM OF RECIPROCALS OF THOSE NUMBERS

SOLUTION:

WE HAVE A RELATION BETWEEN LCM ,HCF AND THE NUMBERS i.e,

HCF × LCM = PRODUCT OF THE NUMBERS

LET THE OTHER NUMBER BE X

=> 5 × 120 = 25 × X

=> X = 5 × 120 / 25 = 24

∴ THE OTHER NUMBER IS 24

SUM OF THEIR RECIPROCALS

= 1/25 + 1/24

= 25 + 24 /600

= 49/600

HOPE IT HELPS !!!!