The ratio between a two-digit number and the sum of the digits of that number is 3:1. If the digit in the unit’s place is 5 more than digit at ten’s place, what is the number?

a)
17

b)
27

c)
36

d)
34

e)
None of these

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

Question

The ratio between a two-digit number and the sum of the digits of that number is 3:1. If the digit in the unit’s place is 5 more than digit at ten’s place, what is the number?

a)
17
b)
27
c)
36
d)
34
e)
None of these

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

Answer

's place is three times the digit in the tens place, what is the number?

Let's call the tens digit 'x' and the units digit 'y'.

We know that the ratio of the number to the sum of its digits is 3:1, so we can set up the equation:

(number)/(x+y) = 3/1

Simplifying, we get:

number = 3(x+y)

We also know that the units digit is three times the tens digit, or:

y = 3x

Now we can substitute y with 3x in our first equation:

3(x+3x) = number

Simplifying, we get:

10x = number

So the two-digit number is simply 10 times the tens digit.

To find the value of the tens digit, we can use the fact that the sum of the digits is 1/3 of the number:

x + y = (1/3)(10x)

Simplifying, we get:

4x = y

Now we can substitute 4x for y in our sum equation:

x + 4x = (1/3)(10x)

Simplifying, we get:

x = 6

So the tens digit is 6, and the units digit is 3 times that, or 18.

Therefore, the number is 60 + 18 = 78.

Answer

Place is a prime number, what is the number?

Let the tens digit be $t$ and the units digit be $u$. Then the number is $10t+u$. We are given that $$\frac{10t+u}{t+u}=3.$$Multiplying both sides by $t+u$ gives $10t+u=3(t+u)$. Expanding the right side gives $10t+u=3t+3u$, so $7t=2u$.

Since $u$ is a prime number, the only possibilities are $u=2$ or $u=7$. If $u=2$, then $t=1$, but $12$ is not divisible by $3$, so this is not a solution. If $u=7$, then $t=2$, and $27$ is divisible by $3$, so this is the number we seek.

Therefore, the number is $\boxed{27}$.

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