Question

The sum of two 2 - digit numbers is 132. If their HCF is 11, the numbers are <br />a) 55, 77<br />b) 44, 88<br />c) 33, 99<br />d) 22, 110​

Answer 1

Answer:

$\huge\underline\bold {Answer:}$

Since the HCF is 11, the numbers must be of the form 11a and 11b, where a and b are prime to each other.

11a + 11b = 132

=> a + b = 12

The possible pairs of numbers whose sum is 12 are 1, 11; 2, 10; 3, 9; 4,8; 5, 7; 6,6.

Out of these pairs, the pairs of numbers that are prime to each other are 1, 11 and 5, 7. The pairs which satisfies the given condition is 5, 7. The pairs which satisfies the given condition is 5, 7.

Hence the numbers are 11 × 5 and 11 × 7, i.e., 55 and 77.

Answer 2

Answer:

a) 55&77

Step-by-step explanation:

Let the numbers be x and y.

ATQ,

x + y = 132

As, x and y have 11 as HCF, they r divisible by 11.

x = 11m, y = 11n

so,

11m + 11n = 132

m + n = 132/11

m + n = 12

But, as m and n has no common factor they are co-prime

Values for m and n.

m - 1, n -11

m - 2, n - 10(Not Possible)

m - 3, n - 9(Not Possible)

m - 4, n - 8(Not Possible)

m - 5, n - 7

m - 6, n - 6(Not Possible)

the reverse order is also possible.

so we get that

m = 1,5 and n = 11,7

x = 1*11 =11; n = 11*11=121. (not possible, as x and y are 2 digit nos)

x = 5*11 = 55; n = 7*11 = 77

Numbers are 55 and 77

PLS MARK IT BRAINLIEST AND FOLLOW ME!!!!!!!!