In a two digit positive number, the digit at the units place is equal to the square of the digit in ten’s place and the difference between the number and the number obtained by interchanging the digits is 54. What is 40% of the original number?

a)
15.6

b)
37.2

c)
39.4

d)
24.8

e)
None of the Above

Correct answer is option 'A'. Can you explain this answer?

Question

In a two digit positive number, the digit at the units place is equal to the square of the digit in ten’s place and the difference between the number and the number obtained by interchanging the digits is 54. What is 40% of the original number?

a)
15.6
b)
37.2
c)
39.4
d)
24.8
e)
None of the Above

Correct answer is option 'A'. Can you explain this answer?

Answer

We can simply consider 3 numbers 11, 24, 39. So only 39 satisfies the condition so 40 percent becomes 15.6

Answer

The number at unit place may be 9 or 4.
then tens place number would be 3 or 2
In this way we get two numbers
39 and 24
Now we are given the difference of the numbers when we change these numbers position. and get the difference 54.
By putting only 39 and it's changed position number 93 gives the difference of 54.
And 40% of 39 is 15.6

Answer

Answer

's place. What is the number?

Let's call the tens digit 'x' and the units digit 'y'. We know that:

- y = x^2 (the units digit is equal to the square of the tens digit)

We also know that the number is two digits, so it must be between 10 and 99. We can combine these two pieces of information to form an equation:

- 10x + y = the number

Substituting y = x^2, we get:

- 10x + x^2 = the number

Now we need to find values of x and y that satisfy this equation and also fit the criteria of being a two-digit positive number. We can start by trying different values of x:

- If x = 1, then y = 1^2 = 1, so the number is 10 + 1 = 11, which is a two-digit positive number.
- If x = 2, then y = 2^2 = 4, so the number is 20 + 4 = 24, which is a two-digit positive number.
- If x = 3, then y = 3^2 = 9, so the number is 30 + 9 = 39, which is a two-digit positive number.
- If x = 4, then y = 4^2 = 16, which is not a valid digit for the units place, so we can stop here.

Therefore, the only possible two-digit positive numbers that satisfy the given conditions are 11, 24, and 39.

Answer

Two digit Nur is for example =ab
then a=3, b=9
3×3= 9=a also b=9 so we assume ab=39
inter changes this Nur's 39 and 93
different between this two Nur's 93-39=54
40%of 39 = ?
40/100=2/5×39=15.6

Answer

's place. What is the number?

Let's call the digit in the ten's place 'x'. Then the digit in the units place is x^2.

So the number is 10x + x^2.

For example, if x = 3, then the number is 10(3) + 3^2 = 30 + 9 = 39.

We can also see that the number must be less than 100, since it is a two-digit number. And since the units digit is the square of the ten's digit, it must be one of the digits 0, 1, 4, 9.

Using these restrictions, we can try out different values of x until we find a number that works.

If x = 0, then the number is 10(0) + 0^2 = 0, which is not a positive two-digit number.

If x = 1, then the number is 10(1) + 1^2 = 11, which works.

If x = 2, then the number is 10(2) + 2^2 = 24, which works.

If x = 3, then the number is 10(3) + 3^2 = 39, which works.

If x = 4, then the number is 10(4) + 4^2 = 56, which works.

If x = 5, then the number is 10(5) + 5^2 = 75, which works.

If x = 6, then the number is 10(6) + 6^2 = 96, which works.

If x = 7, then the number is 10(7) + 7^2 = 119, which is not a two-digit number.

If x = 8, then the number is 10(8) + 8^2 = 144, which is not a two-digit number.

If x = 9, then the number is 10(9) + 9^2 = 171, which is not a two-digit number.

So the possible numbers are 11, 24, 39, 56, and 75.

Answer

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