Question

please answer this duds ....... ​

Answer 1

Answer:

25 and 58

Step-by-step explanation:

Solving with the help of basic mathematics.

 Tens as well as ones place will always be an integer.

Here,

If product of both ones digits is 40( = 8 * 5 ), then there are only two possible integer which can give 40 as their product so ones( possible ) digit will be 8 and 5.

In the same manner, 10 = 5 * 2, so there are only two possible digits those are 5 and 2( for tens place ).

 So, numbers( by writing them as 10a + b ):

Possible combinations :

  10(2) + 8 = 28  and   10(5) + 5 = 55

          but 55*22 = 1540 ≠ 1450

  10(2) + 5 = 25 and 10(5) + 8 = 58

       25*58 = 1450

 Hence the require numbers are 25 and 58

Answer 2

Let x,y be the numbers.

x.y = 1450

Let x be a + 10b and y be m + 10n in terms of units and tens place.

Given, b.n = 10 = 5*2 or 2*5 ---------------------> 1

Similarly, a.m = 40 = 5*8 or 8*5 ----------------> 2

Since, (a + 10b)(m + 10n) = 1450

a.m + 10an + 10bm + 100bn = 1450

40 + 10 ( an + bm ) + 1000 = 1450

an+bm = 41 -------------------------------------------> 3

Now checking values of 1 and 2 in 3.

CASE 1

b = 5, n = 2, a = 5, m = 8

10 + 40 $\neq$ 41

CASE 2

b = 2, n = 5, a = 5, m = 8

16 + 25 = 41

Hence b = 2, n = 5, a = 5, m = 8,

The numbers are 25 and 58.