Top

# Number System Problems

Here are few Number System Problems, which covers problems on integers , series and their operations like intersections and unions.

 Related Calculators Linear System of Equations Solver Metric System Calculator

## Questions

Question 1:
For given sequence, suggest possible next three terms and find the general term.
7,11,15,...

First term = 7, Common difference = 11 - 7 = 15 - 11 = 4
The three terms form an A.P. :
Next three terms are 15 + 4 = 19, 19 + 4 = 23, 23 + 4 = 27
General term is = a + (n-1) d = 7 + (n-1) 4 = 4n + 3

Question 2:
Let n(H) = 125, N(U) = 75 and n(H $\cap$ U) = 50
Find : n(H $\cup$ U)

Using formula n(H $\cup$ U) = n(H) + n(U) - n(H $\cap$ U)
= 125 + 75 - 50
= 150

Question 3:
At dinner, seven people ordered the special of the day. Three people order from the list of the menu, one person orders only salad. Find the number 'm' (dinners ordered by the people).

The sets are disjoint. (assuming no one orders more than one dinner) m = 7 + 3 + 1 = 10, the total number of dinners ordered.

Question 4: Solve for x; (60 - x) + x + (35 - x) = 80
(60 - x) + x + (35 - x) = 80
=> x = 95 - 80 = 15

Question 5:  Find the unknown value.
1. If n(C $\cup$ T) = 80, n(C) = 60
2. If n(C $\cup$ T) = 80, n(T) = 35

1. Using n(C $\cup$ T) = n(C) + n(T) - n(C $\cap$ T)
80 = 60 + n(T) - n(C $\cap$ T)
n(T) - n(C $\cap$ T) = 80 - 60 = 20

2. Again n(C $\cup$ T) = n(C) + n(T) - n(C $\cap$ T)
80 = n(C) + 35 - n(C $\cap$ T)
n(C) - n(C $\cap$ T) = 80 - 35 = 45

Question 6:
In a class of 80 boys, there are 60 boys who play chess and 35 play table tennis. Use Venn diagrams to show how many boys play both the games? How many play chess only and how many play tennis only?

Let n(C) = number of boys of who play chess
n(T) = number of boys who play table tennis
Let x = number of boys who play both the games
From the above diagram it is clear that,
(60 - x) + x + (35 - x) = 80
x = 95 - 80 = 15

Those who play chess only = 60 - x = 60 - 15 = 45
Those who play table tennis only = 35 - 15 = 20

Question 7:
A survey shows that 71% of Indians like to watch cricket, whereas 64% like to watch hockey. What percentage of Indians like to watch both cricket and hockey? (Assuming that every indian watches at least one of these games)

Let n(C) = percentage of Indians who watch cricket and
n (H) = percentage of Indians who watch hockey.
Then n(C) = 71 n(H) = 64 n(C $\cup$ H) = 100

Using n(C $\cup$ H) = n(C) + n(H) - n(C $\cap$ H)
100 = 71 + 64 - n(C $\cap$ H)
n(C $\cap$ H) = 135 - 100 = 35
Hence 35% of Indians like to watch both cricket and hockey.

Question 8:
In a group of 80 persons, 30 drink Fanta but not Limca and 41 drink Fanta.
1. How many drink Fanta and Limca both?
2. How many drink Limca, but not Fanta?

Let n(F) = number who drink Fanta
n(L) = number who drink Limca
F - L = Set of Persons who drink Fanta but not Limca
L - F = Set of Persons who drink Limca but not Fanta

n(F) = 41 , n(F - L) = 30,  n(F $\cup$ L) = 80
Now,

(i) n(F - L) + n(F $\cap$ L) = n(F)
n(F  $\cap$ L) = n(F) - n(F - L)
= 41 - 30 = 9

9 persons drink Fanta and Limca both.

(ii) n(F $\cup$ L) = n(F) + n(L) - n(F $\cap$ L)
n(L) - n(F $\cap$ L) = n(F $\cup$ L) - n(F)
So, n(L) = n(F $\cup$ L) - n(F) + n(F $\cap$ L)
= 80 - 41 + 9 = 48

n(L - F) + n(F $\cap$ L) = n(L)
n(L - F) = n(L) - n(F $\cap$ L)
= 48 - 9 = 39

39 people drink Limca but not Fanta.

Question 9: In a group of 1500 people, 950 can speak Hindi and 650 can speak both Hindi and Kannada. How many can speak Hindi only? How many can speak Kannada only?

Answer: Let n(H) = number who can speak Hindi
n(K) = number who can speak Kannada
n(H $\cap$ K) = number who can speak both Hindi and Kannada
n(H $\cup$ K) = number who can speak either Hindi and Kannada
To find : n(H) - n(H $\cap$ K) and n(K) - n(H $\cap$ K)
Using n(H $\cap$ K) = n(H) + n(K) - n(H $\cup$ K)
Here n(H) = 950 n(K) = 650 n(H $\cup$ k) = 1500
Number who can speak Hindi only =
n(H) - n(H $\cap$ K) = n(H $\cup$ K) - n(K) = 1500 - 650 = 850
Number who can speak Kannada only =
n(K) - n(H $\cap$ K) = n(H $\cup$ K) - n(H) = 1500 - 950 = 550

Question 10:
Represent the following using Venn diagram
B= {2, 3, 4, 5} and A= {2, 3}

A is a subset of B.

Question 11: Find three numbers in AP whose sum is 9 and sum of their square is 77.

Answer: Let a - d, a, a + d be three numbers in AP.
a - d + a + a + d = 9
3a = 9
a = 3
(a - d)$^2$ + a$^2$ + (a + d)$^2$ = 77
(3 - d)$^2$ + 3$^2$ + (3 + d)$^2$ = 77 (Put a = 3)
9 + d$^2$ - 6d + 9 + 9 + d$^2$ + 6d = 77
27 + 2d$^2$ = 77
2d$^2$ = 50
d = 5

Question 12: If S1, S2, S3 be respectively the sum of n, 2n, 3n terms of GP, prove that:
i) S12 + S22 = S1(S2 + S3)
ii) S1(S3 - S2) = (S2 - S1)2

Question 13: Show that if the positive numbers a,b,c are in AP, then below numbers are also in AP.

Answer: Since a,b,c are in an AP, we have

Now, we shall show that sum of extreme terms is double the middle terms.

Question 14: Using Euclid's division algorithm, find the HCF of the following numbers. 2103, 9945, 9216

Hence HCF of 9945, 9216 and 2103 is 3.

Question: Write the factor tree for the following numbers and write them as the product of primes i) 81 ii) 32 iii) 45 iv) 120 v) 200 vi) 100

Here is a Elementary Number theory Worksheets problem for which they have given tree based explanation step by step

Answer: i) 81 = 3 x 3 x 3 x 3 = 34

ii) 32 =2 x 2 x 2 x 2 x 2 = 25

iii) 45 = 3 x 3 x 5 = 32 x 5

iv) 120 = 2 x 2 x 2 x 3 x 5 = 23 x 3 x 5

v) 200 = 2 x 2 x 2 x 5 x 5 = 23 x 52

vi) 100 =2 x 2 x5 x 5 =22 x 52

 Related Topics Math Help Online Online Math Tutor
*AP and SAT are registered trademarks of the College Board.