Number System, Problems on Number

Let, the other number be x

We know, HCF×LCM = product of two numbers
Or, 90×15 = 45×x
Or, x = 1350/45 = 30

So, 30 is the other number

Question: Two-fifth of one-third of three-seventh of a number is 15. What is 40 percent of that number?

Solution: 
let the number be x 

(2/5) × (1/3) × (3/7) x = 15 
⇒ x = (15 × 35)/2 

40% of x = {(15 × 35)/2 } × .4
= {(15 × 35)/2} × (4/10)
= 105 

This is a G.P.(general process) in which a = 2, r = 2^2/2 = 2 and n = 9
S_n = a(rⁿ - 1)/(r - 1)
= 2 x (2⁹ - 1)/(2 - 1)
= 2 x (512 - 1)
= 2 x 511
= 1022.

Question: If the average of 'm' numbers is √2n² and the average of 'n' numbers is √2m², what is the average of the combined (m + n) numbers?

Solution:
দেওয়া আছে,
 m সংখ্যার গড় = √2n²
∴ m সংখ্যার সমষ্টি = m × √2n²

n সংখ্যার গড় = √2m²
∴ n সংখ্যার সমষ্টি = n × √2m²

∴ মোট সমষ্টি = m + n = (m × √2n²) + (n × √2m²)
= √2mn² + √2m²n
= √2mn(m + n)

∴ তাদের গড় = মোট সমষ্টি/(m + n)
= √2mn(m + n)/(m + n)
= √2mn

Question: A and B are two positive integers such that AB = 60. Which of the following cannot be the value of A + B?

Solution:
Factor pairs of 60:
(1, 60) → A + B = 61
(2, 30) → A + B = 32
(3, 20) → A + B = 23
(4, 15) → A + B = 19
(5, 12) → A + B = 17
(6, 10) → A + B = 16

So, possible values of A + B are: 61, 32, 23, 19, 17, 16.

Among the options, 20 is not possible.

(a⁰ - 3b⁰)⁵
= (1 - 3)5 
= -2⁵ 
= -32

x + y = 166,
y + z = 170,
z + 2x = 250.

Solving the equation, y = 84.

LCM of 4, 8, 10, 12, 15 and 20 = 120
120 seconds = 2 minutes
Hence all the six bells will ring together in every 2 minutes
Hence, number of times they will ring together in 60 minutes = 1 + (60/2)
= (2+ 60)/2
= 31.

The HCF of a group of numbers will always be a factor of their LCM.
HCF is the product of all common prime factors using the least power of each common prime factor.
LCM is the product of the highest powers of all prime factors.

Let the number be = 627 + 43
= 670
So, the remainder is,

19) 670 (35
      57
   -------
      100
        95
   --------
         5

Let the number be x

ATQ,
60x/100 - 40 = 50
Or, (60x - 4000) /100 = 50
Or, 60x - 4000 = 50 ×100
Or, 60x -4000 = 5000
Or, 60x = 5000 + 4000
Or, 60x = 9000
Or, x = 9000/60
Or, x = 150
Hence, the number is 150

Given, 624A
= 13 × 8 × 6 × A
= 13 × 2³ × 2 × 3 × A
= 13 × 2⁴ × 3 × A

So, to make the number square of an integer, required value of A is 13 × 3 = 39

perimeter of the plot = 2(90+50) = 280m
number of poles =280/5 = 56m

এখানে,
30 দ্বারা বিভাজ্য সংখ্যা = 1000/30
ভাগফল 33 এবং ভাগশেষ 10
সুতরাং 30 দ্বারা বিভাজ্য সংখ্যা = 33 টি
30 ও 16 এর ল, সা, গু = 240
এখন 1000/240 =
ভাগফল 4 এবং ভাগশেষ 40
∴ 30 ও 16 দ্বারা বিভাজ্য সংখ্যা = 4 টি
সুতরাং, (33 - 4) = 29 টি সংখ্যা 30 দ্বারা বিভাজ্য কিন্তু 16 দ্বারা বিভাজ্য নয়।

Let X be the number which is added to 24
85% of X = 0.85X
Now,
24 + 0.85X = X
⇒ 0.15X = 24
∴ X = 24/0.15 = 160

Let the smaller number be x
Then larger number = (x + 1365)
∴ x + 1365 = 6x + 15
⇒ 5x = 1350
⇒ x = 270
∴ Smaller number = 270

Number = (12 x 35)
Correct Quotient = 420 /21 = 20
Answer : 20

Question: A swimming pool maintenance service charges a fixed fee of Tk. 200 plus Tk. 150 per hour for cleaning. If a customer's total budget is Tk. 1,400, what is the maximum number of full hours the technician can work?

Solution:
Given,
Fixed service fee = 200 Tk
Charge per hour = 150 Tk
Total budget = 1,400 Tk

Let h = number of full hours the technician works.

According to the condition,
200 + 150h ≤ 1,400
⇒ 150h ≤ 1,400 - 200
⇒ 150h ≤ 1,200
⇒ h ≤ 1,200/150
∴ h ≤ 8

Therefore, the technician can work a maximum of 8 full hours.

Question: If m is an odd integer, which of the following must be an even integer?

সমাধান:
ধরি, m = 3 (একটি বিজোড় সংখ্যা)

ক) m² + m = 3³ + 3 = 12 → জোড় 
খ) 2m + 1 = 2(3) + 1 = 6 + 1 = 7 → বিজোড়
গ) 5m - 2 = 5(3) - 2 = 15 - 2 = 13 → বিজোড় 
ঘ) m³ + 2 = 3³ + 2 = 27 + 2 = 29 → বিজোড়

Question: A and B are two positive integers such that AB = 72. Which of the following cannot be the value of A + B?

Solution:
Factor pairs of 72:
(1, 72) → A + B = 73
(2, 36) → A + B = 38
(3, 24) → A + B = 27
(4, 18) → A + B = 22
(6, 12) → A + B = 18
(8, 9) → A + B = 17

So, possible values of A + B are: 73, 38, 27, 22, 18, 17.

Among the options, 25 cannot be the value of A + B.

Question: The average of a positive natural number and its cube is 13 times the number. The number is -

Solution:
Let the number be a positive natural number = x (x ≠ 0).

ATQ,
The average of the number and its cube is 13 times the number.
⇒ (x + x³)/2 = 13x
⇒ x + x³ = 26x
⇒ x³ + x - 26x = 0
⇒ x³ - 25x = 0
⇒ x(x² - 25) = 0
⇒ x(x - 5)(x + 5) = 0

So, x = 0 or x = 5 or x = - 5
Therefore, the positive natural number is 5.

Let us consider the options
a) 11 - 2 = 9 and 9/3 = 3; 11 - 4 = 7 and 7/7 = 1
b) 22 - 2 = 20 but 20/3 = 6.67
c) 18 - 2 = 16 but 16/3 = 5.33
d) 32 - 2 = 30 and 30/3 = 10; 32 - 4 = 28 and 28/7 = 4
As the requirement is the least possible value, answer will be 11

The series is: 3 + 0² = 3, 3 + 1² = 4, 4 + 2² = 8, 8 + 3² = 17, 17 + 4³ = 33, 33 + 5² = 58    

Question: If the sum of three consecutive odd integers is 117, what is the largest number?

Solution:
Let the three consecutive odd integers be:
x, x + 2, x + 4

Then:
x + (x + 2) + (x + 4) = 117
⇒ 3x + 6 = 117
⇒ 3x = 117 − 6
⇒ 3x = 111
⇒ x = 111 ÷ 3
⇒ x = 37

So the numbers are: 37, 39, 41 (Largest)

∴ The largest number is 41.

Find LCM of 2, 3 and 5 is 30. So now divide all options.
Option that will have remainder 0 is the answer.

Question: If X ∈ N and 31 < x < 37, and x is a prime number, then which of the following represents the list form of the set of such numbers?

Solution:
The natural numbers between 31 and 37 are:
32, 33, 34, 35, 36

Now, check which of these are prime:
32: divisible by 2 → not prime
33: divisible by 3 and 11 → not prime
34: divisible by 2 → not prime
35: divisible by 5 and 7 → not prime
36: divisible by 2, 3, etc. → not prime

So, there are no prime numbers between 31 and 37.

Therefore, the correct answer is the empty set: { }

H.C.F of two prime numbers is 1.
Product of numbers = (1 × 161) = 161.
Let the numbers be a and b.
Then, ab = 161.
Now, co - primes with product 161 are (1, 161) and (7, 23).
Since x and y are prime numbers and x > y, we have x = 23 and y = 7
∴ 3y -x = (3 × 7) - 23 = -2
Answer is : -2

ধরি, A brand এর কলম সংখ্যা x টি
B brand এর কলম সংখ্যা (8 - x) টি
ATQ,
200x + (8 - x)×100 = 1200
⇒ 200x + 800 - 100x = 1200
⇒ 100x = 1200 - 800
⇒ x = 400/100
⇒ x = 4