Dear Readers,

As all of you may aware that State Bank of India (SBI) has introduced the Computer Aptitude in the Main Examination of its recruitment of Junior Associates
(Customer Support & Sales) & Junior Agricultural Associates (Including Special Drive) in this year. After getting continuous demands from our
readers, today we are presenting first quiz set on Computer Aptitude as similar as SBI Clerk Main 2016 Exam.

**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 %@@,

5 is written as %@% and so on,

5 is written as %@% and so on,

**1. Which of the following will represent 16?**

a) @%%@%

b) %%@%@

c) %@@@@

d) %@@%@

e) None of these

**2. Which of the following will represent 31?**

a) %@%@%

b) %%%%%

c) @@@@@

d) @%@%@

e) None of these

**3. Which of the following will represent 121?**

a) @%@%@%%

b) %%%%%%%

c) @@@@@@@

d) %%%%@@%

e) None of these

**4. Which of the following number will be represented by %@%%%%?**

a) 47

b) 44

c) 45

d) 46

e) None of these

**5. Which of the following number will be represented by %%@%@%%?**

a) 108

b) 127

c) 107

d) 90

e) None of these

### Answers:

1. c) %@@@@

## Solution:

We can write 16 in binary form as:= (1x2^{4})+(0x2^{3})+(0x2^{2})+(0x 2^{1})+(0x2^{0})= %@@@@

2. b) %%%%%

## Solution:

We can write 31 in binary form as:= (1x2^{4})+(1x2^{3})+(1x2^{2})+(1x 2^{1})+(1x2^{0})= %%%%%

3. d) %%%%@@%

## Solution:

We can write 121 in binary form as:= (1x2^{5})+(1x2^{4})+(1x2^{3})+(1x2^{2})+(1x 2^{1})+(1x2^{0})= %%%%@@%

4. a) 47

## Solution:

%@%%%%= (1x2^{5})+ (0x2^{4})+(1x2^{3})+(1x2^{2})+(1x 2^{1})+(1x2^{0})= 32+0+8+4+2+1= 47

5. c) 107

## Solution:

%%@%@%%= (1x2^{6})+(1x2^{5})+(0x2^{4})+(1x2^{3})+(0x2^{2})+(1x 2^{1})+(1x2^{0})= 64+32+0+8+0+2+1= 107

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