Worksheet on Numbering Systems is here. Follow Numbering Systems problems and solutions in the next sections. Get the step-by-step procedure to solve all the problems involving numbers. With the help of the practice material, you can manage your time-saving skills and also problem-solving skills. Follow the next sections to know the important questions and problems.
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Number System Worksheet
Question 1:
Classify the number “0.1666..” terminating decimal or non terminating recurring decimal or on terminating non recurring decimal?
Solution:
As we know that rational numbers are divided into 2 types they are terminating decimal and non-terminating decimal. In the terminating decimal, the denominator = 2n 5m where m,n ∈ +ve integer. Rational numbers also possess non-rational but repeating decimal.
As the given number is 0.1666….
It is the repeating decimal and non-terminating Rational number.
Question 2:
Classify the following numbers as rational and irrational numbers?
- 2 – √5
- (3 + √23) – √23
- 2√7/2√7
- 1/√2
Solution:
The given 1st number is 2 – √5
2 is the rational number and √5 is the irrational number
Rational number – Irrational Number = Irrational Number
Therefore, the number “2 – √5” is an irrational number
The given 2nd number is (3 + √23) – √23
After calculation, the result value is 3 which is a rational number
Therefore, the number (3 + √23) – √23 is a rational number
The given 3rd number is 2√7/2√7
After calculation, the result value is 1 which is a rational number
Therefore, the number “2√7/2√7” is a rational number
The given fourth number is 1/√2
1/√2 is an irrational number
Question 3:
A number is given which when divided by 899 gives a remainder of 63. The same number when divided by 29, leaves a remainder?
Solution:
Given that,
The remainder of the number, when it is divided by 899, is 63.
When the same number is divided by 29, what is the remainder?
Let the number be x.
Here, we will go by a simple method to find the remainder when x is divided by 29.
63 when it is divided by 29, leaves a remainder 5.
So, when x is divided by 29, leaves a remainder of 5.
Question 4:
A number is given such that when its one by fifth part is increased by 4, it is equal to one by the fourth part when it is diminished by 10. What is the number?
Solution:
Given that,
A number whose one by fifth part is increased by 4 is equal to one by the fourth part when diminished by 10.
Let the number be x, the condition is that
(x/5)+4=(x/4)-10
(x/4)-(x/5)=4+10
(x/4)-(x/5)=14
(5x-4x)/20=14
x=280.
Therefore 280 is the number whose one by fifth part when increased by 4 is equal to one by the fourth part when diminished by 10.
Question 5:
47 is added to the product of 71 and unknown numbers. The new number is divided by 7 giving a quotient of 98. Find the multiple of the unknown number?
Solution:
Let the unknown number be x
As given in the question,
47 is added to the product of 71 and an unknown number
Hence, the equation will be as 47 + 71x
Also, given new number is divided by 7 giving a quotient of 98
47 + 71x/7 = 98
As we know,
Dividend = Divisor * Quotient + Remainder
71x + 47 = 7 * 98 + 0 = 686
71x = 686 – 47 = 639
x = 639/71
x = 9
x = 3 * 3
Therefore, the multiple of unknown number is 3
Question 6:
The Sum of the two numbers is 40 and their product is 375. Find the sum of their reciprocal?
Solution:
Let the two numbers be x and y
As given in the question,
x + y = 40
xy = 375
To find the sum of the reciprocal, we calculate
1/x + 1/y
= (y + x)/xy
= 40/375
= 8/75
Therefore, the sum of the reciprocal values = 8/75
Question 7:
The difference between squares of two consecutive numbers is 39. Find the numbers?
Solution:
Let the two consecutive numbers be x and x+1
As given in the question,
The difference between squares of two consecutive numbers is 39, we get the equation as
(x + 1)² – x² = 39
x² + 1 + 2x – x² = 39
2x + 1 = 39
2x = 38
x = 19
Therefore one number x = 19
The other number is x + 1 = 19 + 1 = 20
Therefore, the two consecutive numbers are 19 and 20.
Question 8:
Find the smallest number which is to be added to 1000 when the 45 divides its sum exactly?
Solution:
Let the smallest number be x
As given in the question,
The smallest number is added to 1000 when 45 divides its sum exactly, we get the equation as
= 1000 + x /45
Suppose that we multiply the random numbers and check the solutions which gives
45 * 10 = 450
45 * 20 = 900
45 * 21 = 945
45 * 22 = 990
45 * 23 = 1035
This is the nearest number to 1000
Hence, 1035 – 1000 = 35
Therefore 35 is the smallest number which is to be added to 1000 when the 45 divides its sum exactly
Question 9:
The Sum of both the numbers is 13 and the sum of their squares is 89. Find the product of two numbers?
Solution:
Let the two numbers be x and y
As given in the question
The sum of both the numbers = 13
ie., x + y = 13
Sum of their squares = 89
x² + y² = 89
As we know that
(x+y)² = x² + y² + 2xy
(13)² = 89 + 2xy
169 = 89 + 2xy
2xy = 169 – 89
2xy = 80
xy = 40
Therefore, the product of two numbers is 40
Question 10:
The sum of squares of two numbers is 40. The square of their difference is 20. Find the product of two numbers?
Solution:
Let the two numbers be x and y
As given in the question,
The sum of squares of two numbers is 40
x² + y² = 40
(x – y)² = 20
As we know that
(x – y)² = x² + y² + 2xy
20 = 40 – 2xy
-20 = – 2xy
xy = 20/2
xy = 10
Therefore, The product of two numbers is 10.
Question 11:
A cricket player so far has scored 7,849 runs in the test matches. He wishes to complete 10,000 runs. Find the number of runs he needs to complete 10,000 runs?
Solution:
Total number of runs = 7,849
Target run = 10,000
Number of runs needed = 10,000 – 7,849
= 2,151
Therefore, he needs 2,151 runs to complete 10,000 runs
Hence, the final solution is 2,151 runs
Question 12:
A man has 1,57,184 with him. He placed an order for purchasing 80 articles at Rs.125 each. How much money will remain with him after the purchase?
Solution:
Total amount of money = Rs. 1,57,184
Cost of 1 article = Rs. 125
Cost of 80 articles = Rs 125 * 80
=Rs. 10,000
Amount of money left after the purchase = 1,57,184 – 10,000
= Rs. 1,47,184
Therefore, Rs. 1,47,184 money will remain with him after the purchase.
Question 13:
The sum of the two numbers is 72. The sum of the same two numbers is 48. Find the numbers.
Solution:
Let the two numbers be x and y.
Assume that x is big and y is small.
According to the given condition,
x+y=72. Let this be equation1.
x-y=48. Let this be equation 2.
Add equation 1 and equation2. You will get 2x=120.
Then x=60.
Put the value of x in equation 1.
Now, we will get the value of y.
y=12.
Therefore, the sum of 60 and 12 is 72. The difference between 60 and 12 is 48.
Question 14:
One number is four times the other. Six times the smaller number plus half the larger number equals 212. Find those numbers.
Solution:
Let the two numbers be x and y.
Given the first condition is on number is four times the other.
x=4y
Also given,
Six times the smaller number plus half the larger number equals 212, then we get
6y + 1/2x = 212
6y + 1/2(4y) = 212
8y = 212
y = 26.5
As the equation is x=4y
Substitute the value of y in the above equation
x = 4(26.5)
x = 106
Therefore the numbers are 106 and 26.5
Question 15:
The difference between the two numbers is 45. Three times the larger number less five times the smaller number equals 75. What are the two numbers?
Solution:
Let the two numbers be x and y
As given in the question,
The difference between the two numbers is 45, we get the equation as
x – y = 45
Three times the larger number less five times the smaller number equals 75, we get the equation as
3x – 5y = 75
Now, we have to solve both the equations
Multiply the first equation with -3, we get the equation as
-3x + 3y = -135
By solving both the equations, we get
-2y = -60
y = 30
Now, substitute the value of y in the equation x – y = 45
x – 30 = 45
x = 75
Therefore, the numbers are 75 and 30