Number System, Problems on Number

Let the two parts be x and (50 - x).
Then,
1/x + 1/(50 - x) = 1/12
(50 - x + x)/x(50 - x) = 1/12
x2 - 50x + 600 = 0
⇒ (x - 30)(x - 20) = 0
⇒ x = 30 or x = 20.
So, the parts are 30 and 20.

Question: Three times the middle of three consecutive even integers is 10 more than the smallest. What is the largest integer?

Solution: 
Let the three consecutive even integers are,
x – 2, x, x + 2. (where x = the middle integer)

According to the question,
Three times the middle is 10 more than the smallest.
∴ 3 × middle = smallest + 10
⇒ 3x = (x - 2) + 10
⇒ 3x = x - 2 + 10
⇒ 3x = x + 8
⇒ 3x - x = 8
⇒ 2x = 8
∴ x = 4

∴ The largest integer is x + 2 = 4 + 2 = 6.

Product of two numbers = Product of their HCF and LCM.
Let one number = x
25 × x = 5 × 150
⇒ x = (5 × 150)/25
⇒ x = 30.

Let the numbers be x and y
According to question,(x>y)
x²−y²=45
(x+y)(x−y)=45
Make factor of 45
15×3
9×5
45×1
Total 3pairs
These pairs gives the value of x and y which satisfy the given condition.

Let the number be x
Then,
⇒ (x+x²)/2 = 5x
⇒ x²− 9x = 0
⇒ x (x−9) = 0
⇒ x = 0 or, x = 9
So, the number is 9.

Let, the number is 3x and 4x
Then 3x + 4x = 420
Or, 7x = 420
Or, x = 60
So, the greater number is 4 × 60 = 240

Quantity of blood donated in 2 years
= (350 × 3) ml
= 1050 ml
= 1.05 litres
∴ Quantity of blood donated in 6 years
=( 1.05/2 ×6)
= 3.15 litres

Question: What should come in place of both the question marks in the equation ?

Solution:

According to question,
√0.00005746
⇒ √5746100000000
⇒ 75.8/10000
⇒ 0.00758

Question: Two positive numbers are in the ratio 3 : 2. The product of their HCF and LCM is 3456. Find the sum of both the numbers.

Solution:
Let two numbers are 3a and 2a.

We know,
HCF × LCM = 1st no. × 2nd no.
⇒ 3456 = 3a × 2a
⇒ 3456 = 6a²
⇒ a² = 576
⇒ a = 24 

∴ Sum of both the numbers = 3a + 2a = 5a = 5 × 24 = 120

Product of three different positive integer is 6 = 1 × 2 × 3
Twice the sum of the integers is = 2(1 + 2 + 3) = 12 

Since the numbers are co-prime, they contain only 1 as the common factor.

Also, the given two products 551 and 1073 have the middle number in common.

So, middle number = H.C.F. of 551 and 1073 = 29

First number = 551/29 = 19
Third number = 1073/29 = 37

Required sum = (19 + 29 + 37) = 85

Given that, a_n = n + 2ⁿ⁻¹ 

Thus:
a₆ = 6 + 2⁶⁻¹ = 38;
a₅ = 5 + 2⁵⁻¹ = 21;

∴ The difference is = 38 - 21 = 17.

Question: If 0.24 ÷ q² = 6, then q equals:

Solution:
0.24 ÷ q² = 6
⇒ q² = 0.24/6
⇒ q² = 0.04
⇒ q = √0.04
⇒ q = 0.2

∴ The value of q is 0.2.

Question: A man deposits certain amount in his bank account. After a few days. he withdraws half of the money deposited and deposits Tk. 500 more. If he has a balance of Tk. 2000 in his bank account, find the amount deposited initially.

Solution:
ধরি,
প্রথমে সে বিনিয়োগ করে ক টাকা 

শর্তমতে,
ক - (ক/২) + ৫০০ = ২০০০
⇒ ক - (ক/২) = ১৫০০
⇒ ক/২ = ১৫০০
∴ ক = ৩০০০

Let the two numbers be are a and b
∴ a + b = 14 .......(i)
a - b = 10 .......(ii)
by adding equation (i) and (ii) we get
2a = 24
∴ a = 12 and b = 2
∴ product of these two numbers = 12 × 2 = 24.

0.125125... = 0.125 = 125/999

Question: The number of multiples of 4 between 10 and 250 is:

Solution:
10 এবং 250 এর মধ্যে 4-এর প্রথম গুণিতক হলো 12 (যেহেতু 12 > 10)
10 এবং 250 এর মধ্যে 4-এর শেষ গুণিতক হলো 248 (যেহেতু 248 < 250)

এখন, এটি একটি সমান্তর ধারা, যার প্রথম পদ (a) = 12, শেষ পদ (p) = 248 এবং সাধারণ অন্তর (d) = 4।

আমরা জানি,
পদসংখ্যা (n) = {(শেষ পদ - প্রথম পদ)/সাধারণ অন্তর} + 1
= {(248 - 12)/4} + 1
= {236 / 4} + 1
= 59 + 1
∴ n = 60

Question: A number is tripled, then 7 is subtracted from it. If the result is then doubled, it becomes 58. What is the number?

Solution:
Let,
the number be x

ATQ,
2(3x - 7) = 58
⇒ 6x - 14 = 58
⇒ 6x = 58 + 14
⇒ 6x = 72
⇒ x = 72/6 
∴ x = 12

So the number is 12.

Question: What is the average of 0.36, 4.6, 0.64, and 2.42? 

Solution:
Average = (0.36 + 4.6 + 0.64 + 2.4)/4
= 8.00/4
= 2.00

ATQ,
dx + 11 = 2dx + 48 
⇒ dx = 37
So, the divisor is 37

Greatest number of four - digits is 9999.
L.C.M. of 4, 7 and 13 = 364
On dividing 9999 by 364, the remainder obtained is 171.
∴ Greatest number of 4 - digits divisible by 4, 7 and 13 = (9999 - 171)
= 9828.
Hence, required number = (9828 + 3)
= 9831.

Let the ten's digit be x.
Then, the unit's digit = 2x + 1.
[10x + (2x + 1)] - [{10 (2x + 1) + x} - {10x + (2x + 1)}] = 1
⇒ (12x + 1) - (9x + 9) = 1
⇒ 3x = 9, x = 3.
So, ten's digit = 3 and unit's digit = 7.
Hence, original number = 37.

First, identify the number that is multiple of 3 more than 100.
That type of number is 102.
So, 102//3 = 34.
Second, we have to identify the number that is multiple of 3 but nearest less than 198.
Now, 198/3 = 66.
Hence, the answer is (66 - 34) + 1 = 33

Question: If p and q are positive integers and (p + q) is an even number, then (p² + q²) will be always divisible by-

Solution:
In this problem, put any even positive value for both p and q,

For example m = 6 and n = 4
∴ (p² + q²) = (6² + 4²)
= 36 + 16
= 52 is always divisible by 2.

Question: If the nth term of an arithmetic progression is 5n + 3, then what is the common difference?

Solution:
The nth term of an arithmetic progression is T_n = 5n + 3

n = 1 then, T₁ = 5 × 1 + 3 = 8
n = 2 then, T₂ = 5 × 2 + 3 = 13
n = 3 then, T₃ = 5 × 3 + 3 = 18
n = 4 then, T₄ = 5 × 4 + 3 = 23

Common difference, T₂ - T₁ = 13 - 8 = 5
T₄ - T₃ = 23 - 18 = 5

∴ The common difference is 5.

Let, the number y
According to the question, 
y² + 12y = - 27
or, y² + 12y + 27 = 0
or, y² + 3y + 9y + 27 = 0
or, y(y + 3) + 9(y + 3) = 0
or, (y + 3)(y + 9) = 0
or, y = - 3, - 9
The smaller possible value of this number = - 9

Note that bigger number with negative value is actually the smaller number.

HCF of the two numbers = 23
Since HCF will always be a factor of LCM, 23 is a factor of the LCM.
Given that other two factors in the LCM are 13 and 14.
Hence factors of the LCM are 23, 13, 14
So, numbers can be taken as (23 × 13) and (23 × 14)
= 299 and 322
Hence, largest number = 322.

Let the number is = 100x

Now after 15% of increase = 100x + 15% of 100x = 115x
Now 25% decrease = 115x - 25% of 115x = 115x - 28.75x = 86.25x

Actual decreased = 100x - 86.25x = 13.75x

According to the question,
13.75x = 22
⇒ x = 22/13.75
⇒ 100x = (22/13.75) × 100
⇒ 100x = 160.

∴ Original number = 160.