The sum of two numbers is 2800 and 7 is their HCF, how many pairs of such numbers are possible?
a)80 b)160 c)60 d)100
Answers
Answered by
0
Step-by-step explanation:
option a is the correct answer
Attachments:
Answered by
0
There are 15 such pairs.
Given:
Sum of numbers = 2800
HCF = 7
To find: Number of such pairs
Solution:
Let the numbers be x and y
We are given that,
x + y = 2800
Also, 7 is the HCF of x and y
⇒ x and y are divisible by 7
⇒ let x = 7a and y = 7b, where a, b ∈ N
⇒ 7a + 7b = 2800 [as given]
⇒ 7(a + b) = 2800
⇒ a + b = 400
⇒ possible values of a and b such that a, and b are co-prime and sum is 400:
3 + 397
11+389
17+383
41+359
47+353
53+347
83+317
89+311
107+293
131+269
137+263
149+251
167+233
173+227
179+211
⇒ There are 15 such pairs.
∴ The correct answer is 15.
SPJ5
Similar questions