Number System-Test 1

Answer: Option C

Explanation

Let the original two digit number 'd' be 10x+y

According to the question, half of 'd' exceeds one third of 'd' by the sum of the digits in 'd'

In other words,

$$\begin{gathered} \frac{10x+y}{2}-\frac{10x+y}{3}=x+y \\ \Rightarrow 3(10x+y)-2(10x+y)=6(x+y) \\ \Rightarrow 30x+3y-20x-2y=6(x+y) \\ \Rightarrow 10x+y=6(x+y) \\ \Rightarrow 4x=5y \\ \Rightarrow \frac{x}{y}=\frac{5}{4} \end{gathered}$$

∴ Sum of digits (x+y) is 9.

Hence, option C is correct.

Answer: Option C

Explanation

Let the two digit number be (10x+y)

then according to the question, 10x+y=k(x+y)

the number obtained by interchanging the digits is 10y+x

10y+x=11y+11x−10x−y 

⇒ 11(x+y)−k(x+y)

⇒ (11−k)(x+y)

⇒ 11−k

Hence, option C is correct.

Answer: Option C

Explanation

let the two digit number be 10x+y

and the number obtained by interchanging the position of the digits be 10y+x

According to the question,

(10x+y)−(10y+x)= 9

⇒ 9x−9y=9

⇒ 9(x−y)=9

⇒ (x−y)=1

Hence, option C is correct.

Answer: Option B

Explanation

let the original number be 10x+y

Now, by interchanging the position of the digits of the number, we get 10y+x

Accoding to the question,

(10x+y)−(10y+x)=27

$$\begin{gathered} \Rightarrow 9x-9y=27 \\ \Rightarrow x-y=3 \\ \Rightarrow \frac{y}{x}=\frac{1}{2} \\ \Rightarrow x=2y \end{gathered}$$

If y=3, x=6

Then, (10×6)+3 = 63

Hence option B is correct.

Answer: Option E

Explanation

let the two digits of the number be x and y

Then according to the question,

x+y=15---Eq 1
x−y=3---Eq 2

On solving equations 1 and 2, we get

2x=18
x=9

Inserting the value of x in the equation 10x+y=15, we get the value of y as 6

Now the praduct of the two digits is x×y

⇒(9x6)=54

Hence option E is correct.