**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 will represent 35?**

a) #$$#$$

b) $$#$##

c) $###$$

d) $##$#$

e) None of these

**2. Which of the following number will be represented by the value of (#$# x $#$$)?**

a) 20

b) 22

c) 21

d) 25

e) None of these

**3. Which of the following will represent the value of [(17-6) x (5+4)]?**

a) $$$##$$

b) $#$###$

c) #$$##$#

d) $$###$$

e) None of these

**4. Which of the following number will be represented by $$#$##?**

a) 52

b) 48

c) 54

d) 50

e) None of these

**5. Which of the following will represent the value of 5**

^{3}?
a) $#$##$$

b) $#$$$$#

c) ##$$$#$

d) $#$#$$#

e) $#$$$$$

### Answers:

## 1. c) $###$$

Solution:We can write 35 in binary form as:= (1x2^{5})+( (0x2^{4})+(0x2^{3})+(0x2^{2})+(1x2^{1})+(1x2^{0})= $###$$

## 2. b) 22

Solution:(#$# x $#$$)= [(0x2^{2})+(1x2^{1})+(0x2^{0}) x (1x2^{3})+(0x2^{2})+(1x2^{1})+(1x2^{0})]= [(0+2+0) x (8+0+2+1)]= 2 x 11= 22

## 3. d) $$###$$

Solution:[(17-6) x (5+4)]= 11 x 9= 99= (1x2^{6})+(1x2^{5})+( (0x2^{4})+(0x2^{3})+(0x2^{2})+(1x2^{1})+(1x2^{0})= $$###$$

## 4. a) 52

Solution:$$#$##= (1x2^{5})+(1x2^{4})+(0x2^{3})+(1x2^{2})+(0x2^{1})+(0x2^{0})= 32+16+0+4+0+0= 52

## 5. e) $#$$$$$

Solution:5^{3}= 125= (1x2^{6})+(0x2^{5})+( (1x2^{4})+(1x2^{3})+(1x2^{2})+(1x2^{1})+(1x2^{0})= $#$$$$$

*Thanks.*

*Smart Prep Kit for Banking Exams by Ramandeep Singh - Download here*
## 0 comments:

## Post a Comment

Thanks for commenting. It's very difficult to answer every query here, it's better to post your query on IBPSToday.com