**Directions (Q.1- 5): Read the following information and answer the questions that follow:**

In a coded binary language, the symbol for '0' is '@' and for '1' is '%'. There are no other symbols for numbers greater than 1 (one). The numbers greater
than 1 (one) are to be written only by using the two symbols given above. The value of symbol for '1' doubles itself every time it shifts one place to the
left. Study the following examples:

0 is written as @,

1 is written as %,

2 is written as %@,

3 is written as %%,

4 is written as %@@ and so on,

**1. Which of the following numbers will be represented by this code (%@%% + %@ x %@@%)?**

a) 29

b) 31

c) 32

d) 28

e) None of these

**2. Which of the following codes will give 11 as the answer?**

a) %%@%% + %@@@@

b) %@@@@ + %%@%%

c) %%@%% - %@@@@

d) %%@%% x %@@@@

e) %@@@@ - %%@%%

**3. Which of the following is the answer of code (%@%@ - %@@)?**

a) %@@

b) %@%S

c) %%@

d) %%%

e) None of these

**4. Which of the following codes will give the value (7 x 3 - 4 ÷ 5 + 2.4 x 2)?**

a) 52

b) 48

c) 54

d) 50

e) None of these

**5. Which are the Remainder and Quotient when 58 is divided by 5?**

a) %@@%, %@

b) %@%@, %%

c) %@%%, %@

d) %@%%, %%

e) None of these

### Answers:

## 1. a) 29

Solution:(%@%% + %@ x %@@%)= [{(1x2^{3})+(0x2^{2})+(1x2^{1})+(1x2^{0})}+{(1x2^{1})+(0x2^{0})}*{(1x2^{3})+(0x2^{2})+(0x2^{1})+(1x2^{0})}]= [11 + 2 * 9]= [11 + 18]= 29

## 2. c) %%@%% - %@@@@

Solution:%%@%% - %@@@@= (1x2^{4})+(1x2^{3})+(0x2^{2})+(1x2^{1})+(1x2^{0}) - (1x2^{4})+(0x2^{3})+(0x2^{2})+(0x2^{1})+(0x2^{0})= 27 - 16= 11

## 3. c) %%@

Solution:(%@%@ - %@@)= (1x2^{3})+(0x2^{2})+(1x2^{1})+(0x2^{0}) - (1x2^{2})+(0x2^{1})+(0x2^{0})= 10 - 4= 6We can write 6 in binary form as:= (1x2^{2})+(1x2^{1})+(0x2^{0})= %%@

## 4. a) 52

Solution:(7 x 3 - 4 ÷ 5 + 2.4 x 2)= 21 – 0.8 + 4.8= 20.2 + 4.8= 25We can write 25 in binary form as:= (1x2^{3})+(1x2^{2})+(1x2^{1})+(1x2^{0})= %%%%

## 5. d) %@%%, %%

Solution:When 58 is divided by 5Quotient is = 11Remainder = 311 = (1x2^{3})+(0x2^{2})+(1x2^{1})+(1x2^{0})= %@%%3 = (1x2^{1})+(1x2^{0})= %%

**Thanks.**