Number system

Number system is one of the most important topic for understanding basics of mathematics and quantitative ability.A bunch of questions are asked from Number system in various compitative exams.

In this study note we discuss frequently asked questions on  'Number Systems' with solutions and detailed explanation

Question 1

A positive whole number M less than 100 is represented in base 2 notation, base 3 notation, and base 5 notation. It is found that in all three cases the last digit is 1, while in exactly two out of the three cases the leading digit is 1. Then M equals

(a)  31

(b)  63

(c)  75

(d)   91

Solution:

31 = 5^2+5+1 ; 3^3+3+1; 2^4+2^3+2^2+2+1 

63 =2^5+ 2^4+2^3+2^2+2+1; 2*3^3+3^2; 2*5^2+2*5+3 

75 = 3*5^2 

91 = 2^6+2^4+2^3+2+1 ; 3^4+2^2+1; 3*5^2+3*5+1 

As some of you can see, 91 satisfies all the conditions

Hence, option 4 is the correct answer.
or 
Since  in all three cases the last digit is 1, the number should give remainder 1 when divided individually by 2,3,5 . So the no. may be 31 or 91 . Now 31 in base 2,3 and 5 give first digit as 1 in all the 3 cases while 91 gives exactly two out of three cases the leading digit as 1 . Hence option D .

Question 2

In a certain examination paper, there are n questions. For j = 1, 2 …n, there are ( 2^n–J ) students who answered j or more questions wrongly. If the total number of wrong answers is 4095, then the value of n is ? 

(a)      12

(b)     11

(c)      10

(d)     9

Solution:

There will many probabilitiesfrom getting n ans wrong to getting one ans ...
1+2+4+8+……+2ⁿ=4095
But in this equation some of the student are repeated 1 student got n ans wrong and 2 student got n-1 or more ans wrong in this n-1 and n both are included.
So how will we calculate no of student who exactly got n ans wrong….?
2^(n-j) student got j or more wrong
2^(n-j-1)got j+1or more wrong
So , 2^(n-j) - 2^(n-j-1) students got exactly j wrong.
Sum….
{2^(n+1)-1}- {2^(n)-1}=2^n=4095
Approximately
n=12

Question 3

If 'x' varies inversely as (y² + 1) and equal to 6 when 'y' = 8. Find 'x' when 'y' = 7?

(a)       7.8

(b)     11.3

(c)      12

(d)     9

Solution:

Given that x varies inversely as y² + 1. 
In math form
x ∞ 1/y^2+1
x =k/y^2+1
k = x (y² + 1) .....(1)
Now plug y = 8 and x = 6 into above equation
k = 6 (8² + 1)
k = 6 (64 +1)
k = 6 (65)
k = 390
Now substitute k = 390 into equation (1)
390 = x (y² + 1) .....(2)
Now put y = 7 into the equation 2
390 = x (7² + 1)
390 = x (49 +1)
390 = x (50)
x =390/50
So, our answer is x = 7.8

Question 4

A three-digit number which is being subtracted from another three-digit number consisting the same digit in reverse order gives 594. What is the minimum possible sum of these three-digit numbers?

(a)      76, 670

(b)     156, 750

(c)      104, 698

(d)     107, 701

Solution:

lets assume x,y,z are three digit and y is middle digit in number .

lets assume x>z

100x+10y+Z-(100z+10y+x)=594

99x-99z =594

99(x-z) =594

x-z = 594/99=6

x= z+6

as Z is starting digits in 3 digit number so z must be greater than Zero, so minimum value 0f z must be 1

so x = Z+6 =7, those 3 digit number will be 107 & 701.

OR 

after x-z=6 , you can choose answer by options  as well .

Question 5

Calculate the total number of prime factors in the expression (4)¹¹ x (5)⁵ x (3)² x (13)²

(a)  30

(b)  31

(c)  33

(d)  32

Solution:

(4)¹¹ x (5)⁵ x (3)² x (13)²

(2²)¹¹ x (5)⁵ x (3)² x (13)²

(2)²² x (5)⁵ x (3)² x (13)²

The total number of prime factors = Sum of powers of the expression = 22 + 5 + 2 + 2 = 31 

 Question 6

The denominator of a fraction is 3 more than its numerator. If the numerator is increased by 7 and the denominator is decreased by 2, we obtain 2. The sum of numerator and denominator of the fraction is 

(a)      5

(b)     13

(c)      17

(d)     19

Solution:

Let the numerator be x and the denominator be x + 3 

∴ Fraction =x/(x + 3)

According to the question, 

(x+7)/(x+3-2)=2

⇒ x + 7 = 2x + 2 

⇒ x = 7 – 2 = 5 

∴ Numerator = 5 

And denominator = 5 + 3 = 8 

∴ Sum of numerator and denominator = 8 + 5 = 13 

Question 7

If a positive integer n, divided by 5 has a remainder 2, which of the following must be true?

(a)      n is odd

(b)     n + 1 cannot be a prime number

(c)      (n + 2) divided by 7 has remainder 2

(d)     n + 3 is divisible by 5

Solution:

You can find the integers which when divided by 5 have a remainder 2 by adding 2 to all multiples of 5.

So we have n = 7 , 12, 17, 22 etc.

From this series we can see that n does not have to be odd.

Also n + 1 can be a prime because, for example, 12 + 1 = 13

And (n + 2) / 7 has a remainder 2 in some cases but not all.

Remember the question asks us for what MUST be true, and we see that none of the statements are true in all cases.

However, adding 3 to any of the values of n will always give a multiple of 5.

Hence, option 4 is the correct answer.

Question 8.

The difference between the digits of a two digit number is 5. Also the original number is 18 more than two times the number obtained by reversing its digits. Find the original number.

(a)      94

(b)     61

(c)      72

(d)     49

Solution:

Let number is 10x+y
Then x-y = 5 or y-x = 5
Now given that, 10x+y = 2(10y+x) + 18
Solve, 8x – 19y = 18
Now solve: 8x – 19y = 18 and x-y = 5.
In this y = 2, x = 7
And also solve; 8x – 19y = 18 and y-x = 5.
In this y come to be negative which is not possible so discard this
So number is 10*7 + 2 = 72

Question 9

The sum of twice the square of a number and 7 times the number equals 15. What is the number?

(a)      1/2

(b)     3/4

(c)      3/2

(d)     5/3

Solution:

2x^2 + 7x = 15
2x^2 + 7x – 15 = 0
2x^2 + 10x-3x – 15 = 0
2x(x+5) - 3(x+5) = 0
(x+5)(2x-3) = 0
x = -5 or x = 3/2

Question 10

In a two number, the product of the digits is 8. when 18 is added to the number,then the digits are reversed.What is the number ?

(a)      81

(b)     42

(c)      24

(d)     18

Solution:

xy = 8,
Also, 10x + y + 18 = 10y + x = 9(x + y) = -18
x - y = -2

or y-x = 2

xy=8 
or y = 8/x 

putting this value in equation :

8/x - x = 2 

(8-x²) /x = 2 

8-x² = 2x 

or 

-x² +8 - 2x = 0

or 

x² + 2x -8 =0

x²+4x-2x-8 =0

x(x+4) -2(x+4)=0

(x+4) (x-2)=0 

i.e x=2 

now x.y=8 

y=4

number will be 10x+y =10x2+4 = 24 
option -c 


 

Question 11

Two numbers are such that their product, their sum and their difference are in the ratio of 6 : 2 : 1 respectively .Their sum is :

(a)      16

(b)     20

(c)      24

(d)     30

Solution:

Let the two number be x and y, then
xy : (x + y) : (x - y) = 6 : 2 : 1
Now, x - y /x + y = 1/2
2x - 2y = x + y
x = 3y
x + y = 3y + y = 4y & xy = 3y * y = 3y²
Now, x + y /xy = 2/6
4y/3y² = 2/6
y = 3 x 4/3 = 4
then, x = 3 x 4 = 12
Hence, x + y = 12 + 4 = 16

Question 12

Sum of eight consecutive odd numbers is 656. Average of four consecutive even numbers is 87. What is the sum of the largest even number and largest odd number?

(a)      171

(b)     191

(c)      101

(d)     179

Solution:

odd numbers — x-8, x-6, x-4, x-2, x, x+2, x+4, x+6
x-8 + x-6 + x-4 + x-2 + x + x+2 + x+4 + x+6 = 656
8x – 8 =656
x = 83
Even numbers — y-2, y, y+2, y+4
4y + 4 = 87 * 4
y = 86
sum of the largest even number and odd number = 89 + 90 = 179

Question 13

The difference between the digits of a two digit number is 5. Also the original number is 18 more than two times the number obtained by reversing its digits. Find the original number.

(a)      94

(b)     61

(c)      72

(d)     49

Solution:

Let number is 10x+y
Then x-y = 5 or y-x = 5
Now given that, 10x+y = 2(10y+x) + 18
Solve, 8x – 19y = 18
Now solve: 8x – 19y = 18 and x-y = 5. In this y = 2, x = 7
And also solve; 8x – 19y = 18 and y-x = 5.
In this y come to be negative which is not possible so discard this
So number is 10*7 + 2 = 72

Question 14

One number is 3 more than another and the sum of their square is 269.Find the number.

(a)      10

(b)     12

(c)      15

(d)     20

Solution:

x is one number
x+3 the other
x^2 +x^2 +6x+9 ; (that is (x+3)(x+3) =269
2x^2+6x -260=0
x^2+3x-130=0 , dividing by 2.
(x+13)(x-10)=0
one number is 10, the other is 13
squared 10=100
square 13=169
sum is 269

Question 15

A two digit number is such that the sum of the digits is 11.When the number with the same digits is reversed is subtracted from this number, the difference is 9.What is the number?

(a)      23

(b)     24

(c)      65

(d)     14

Solution:

let the number be written as xy
x-ten’s place
y units place
x+y=11…………………1
(10x+y)-(10y+x)=9
10x+y-10y-x=9
9x-9y=9
x-y=1………………….2
Add (1) & (2)
2x=12
x=6
y=5
the number is 65