Number System, Problems on Number

14 থেকে 34 পর্যন্ত = 3 টি

40 থেকে 49 পর্যন্ত = 11 টি

54 থেকে 94 পর্যন্ত = 5 টি
_____________________________

                   মোট = 19 টি

Question: The square root of (6 + 5√2)(6 - 5√2) is:

Solution:
√{(6 + 5√2)(6 - 5√2)}
= √{6² - (5√2)²}   [∵ (a + b)(a - b) = a² - b²]
= √{36 - (25 × 2)}
= √{36 - 50}
= √(-14)
= √{14(- 1)}
= √14 × √(- 1)
= i√14   [যেখানে i² = - 1]

Let the numbers be x and y.
Then, 2x + 3y = 39 .......(i) and
3x + 2y = 30 .........(ii)
On solving (i) and (ii), we get : x = 6 and y = 9.
larger number = 9

প্রশ্ন:The sum of 3 consecutive integers is less than 75. What is the greatest possible value of the smallest one?

সমাধান:
ধরি,
 সংখ্যা তিনটি যথাক্রমে  x, x + 1, x + 2

প্রশ্নমতে,
x + 2 + x + x + 1 < 75
3x + 3 < 75
3x + 3 - 3 < 75 - 3
3x < 72
x/3 < 72/3
x < 24

ছোট সংখ্যাটি = ( 24 - 1 ) = 23

এখানে 24টি বাস = 36টি গাড়ি
1 টি বাস = 36/24 = 1.5টি গাড়ি
যেহেতু ফেরিটি 18 টি বাস বহন করে ফেলছে, সুতরাং আরো বহন করতে পারবে = 24 - 18 = 6 টি বাস = (6×1.5) টি গাড়ি অর্থাৎ 9 টি গাড়ি।

আমরা জানি,
গ.সা.গু x ল.সা.গু = সংখ্যাদ্বয়ের গুণফল
⇒ 12 x 288 = 96 x অপর সংখ্যা
⇒ অপর সংখ্যা = (12×288)/96
∴ অপর সংখ্যা = 36

LCM of 4 and 3 is 12. So they meet after every 12 day.
12 day after Monday is Saturday.
So, they will meet on Saturday again after the meeting on Monday.

Question: If the sum of five consecutive odd integers is 255, what is the largest number?

Solution:
ধরি, মাঝের সংখ্যাটি = x
সুতরাং, 5টি ক্রমিক বিজোড় সংখ্যা হবে যথাক্রমে: x - 4, x - 2, x, x + 2 এবং x + 4

প্রশ্নমতে,
(x - 4) + (x - 2) + x + (x + 2) + (x + 4) = 255
⇒ 5x = 255
⇒ x = 51

∴মাঝের সংখ্যা, x = 51 
∴ সবচেয়ে বড় সংখ্যা = x + 4 = 51 + 4 = 55

As given the questions these numbers are co-primes, so there is only 1 as their common factor.
It is also given that two products have a middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29;

So the first number is: 551/29 = 19
Third number = 1073/29 = 37

So sum of these numbers is = (19 + 29 + 37) = 85

Question: The least number by which 150 must be multiplied to make it a perfect square is:

Solution:
Prime factorization of 150:
150 = 2 × 3 × 5 × 5
= 2¹ × 3¹ × 5²

Here, the powers of 2 and 3 are odd.
∴ To make it a perfect square, we need to multiply by 2 × 3 = 6.

Since the numbers are co-prime, their HCF = 1

Product of first two numbers = 119
Product of last two numbers = 391

The middle number is common in both of these products. Hence,
if we take HCF of 119 and 391, we get the common middle number.

HCF of 119 and 391 = 17
⇒ Middle Number = 17
First Number = 119/17 = 7
Last Number = 391/17 = 23

Sum of the three numbers = 7 + 17 + 23 = 47.

Let, the number be a and b.
When it is reversed and added to itself we get (10a + b) + (10b + a)
= 11a + 11b
= 11(a + b)
We are given,
143 = 11(a + b)
a + b = 143/11
a + b = 13
so the digits are a and 13 - a.
We are given their products as a(13 - a) = 36, which is a quadratic expression.
a(13 - a) = 36
13a - a2 = 36
-a² + 13a - 36 = 0
a² - 13a + 36 = 0
a² - 9a - 4a + 36 = 0
a(a - 9) -4(a - 9) = 0
a - 9)(a - 4) = 0
a = 9 or a = 4
So, the number could be 49 or 94.
Hence the option is 49 then the answer will be 49.

Question: If 0 ≤ x ≤ 4 and y < 6, which of the following cannot be the value of xy?

Solution:
y < 6 হলে y  এর মান 5, 4, 3, 2, 1, 0, - 1,..............
0 ≤ x ≤ 4 হলে x  এর মান 0, 1, 2, 3, 4

এখন
x = 0, y = 1 হলে xy = 0 × 1 = 0
x = 2, y = - 1 হলে xy = 2 × (- 1) = - 2
x = 3, y = 2 হলে xy = 3 × 2 = 6

সঠিক উত্তর: None of these

Let the number be x
As given 65% of x = 4/5 of x -21
So, solving above equation:
65 X x/100= 4 X x/5 - 21
⇒ 65 X x /100 = ( 4x - 105)/5
⇒ (65x) X 5 = (4x- 105) X 100
⇒ 65 x = (4x -105) X 20
⇒ 65 x = 80 x - 2100
⇒ 15x = 2100
∴ x = 140
So, the number is 140.

Question: A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. The fraction obtained is -

Solution:
Let, the fraction is a/b

Now, (a - 1)/b = 1/3
⇒ b = 3a - 3

Again, a/(b + 8) = 1/4
⇒ b + 8 = 4a
⇒ b = 4a - 8

∴ 3a - 3 = 4a - 8
⇒ 4a - 3a = 8 - 3
⇒ a = 5

And, b = (3 × 5) - 3
= 15 - 3
= 12

∴ The fraction = a/b = 5/12 

Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit.
Lets, ten's and unit's be 2x and x respectively
Then, (10 times; 2x + x) - (10x + 2x) = 36
⇒ 9x = 36
⇒ x = 4

∴ Required difference (2x + x) - (2x - x ) = 2x = 4 × 2 = 8
Answer : 8

Let, n =  3

n - 1 = 2
n + 1 = 4
4n + 1 =  13
3n + 1 =  10

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, 18 is not possible.

When Divided by 7,
A = 7x + 4
So, numbers can be: 4, 11, 18, 25, 32, 39, 46, 53…….

Again,
when divided by 11,
A = 11y + 9
So, numbers can be: 9, 20, 31, 42, 53…….

Here, 53 is common.

Now, LCM of 7 & 11 is 77.
So, the numbers pattern is : 77x + 53

Then,
77x ≤ 1000
Or, x ≤ 12.2
x can be 0, 1, 2 ...... 12 = total 13 integers 
It can't be x = 13, then the number will be bigger than 1000

Question: The sum of the digits of two - digit number is 10, while when the digits are reversed, the number decrease by 54. Find the changed number.

Solution:
Let number be (10x + y)

ATQ
(10x + y) - (10y + x) = 54
⇒ 10x - 10y + y - x = 54
⇒ 9x - 9y = 54
⇒ x - y = 6 ................ (1)

Sum of digits,
x + y = 10 .................(2)

(1) + (2)
x - y + x + y = 6 + 10
⇒ 2x = 16
∴ x = 8
Put the value of x in (2)
We get,
x + y = 10
⇒ y = 10 - 8
∴ y = 2

The required number is = (10x + y)
= (10 × 8) + 2
= 82

Changed number = 28 

Let the required number numbers be 33a and 33b
Then, 33a + 33b = 528
⇒a + b = 16
Now, co - primes with sum 16 are (1,15), (3,13), (5,11) and (7,9)
∴ Required numbers are (33 × 1, 33 × 15), (33 × 3, 33 × 13), ( 33 × 5, 33 × 11), (33 × 7, 33 ×9)
The numbers of such pairs are 4

Question: 5/9 of a number equals twenty-five percent of the second number. Second number equals 1/4 of third number. The value of third number is 2960. What is 30% of first number? 

Solution: 
second number = 1/4 of third number = 2960/4 
 = 740 

(5/9)first number = 25% of second number 
first number = (1/4) × 740 × (9/5)
= 333

30% of first number = (3/10) × 333
= 99.9 

Question: If (x/y) > 0, which of the following must be true?

Solution:
যেহেতু (x/y) > 0 তাই
x ও y এর দুইটি একই সাথে ধণাত্মক বা ঋণাত্বক হবে। 
আমরা জানি
 দুইটি একই সাথে ধণাত্মক বা ঋণাত্বক সংখ্যার গুণফল ধণাত্মক হয়।
( - x) ( - y) = xy
x × y = xy

i) xy > 0 অবশ্যই ধণাত্মক

Question: The average of a non-zero number and its square is 11 times the number. The number is-

Solution:
Let the number be x (x ≠ 0).

According to the question,
The average of the number and its square is 11 times the number.
⇒ (x + x²)/2 = 11x
⇒ x + x² = 22x
⇒ x² + x - 22x = 0
⇒ x² - 21x = 0
⇒ x(x - 21) = 0
So, x = 0  or  x = 21
But the number is non-zero, so we discard x = 0.

Therefore, the number is 21.

L.C.M. of 3, 4, 5, 6, 8 is 120
Now 120 = 2 × 2 × 2 × 3 × 5
To make it a perfect square, it must be multiplied by 2 × 3 × 5
So, required number
=22×22×32×52
=3600

Question: Which of the following is irrational?

Solution:
একটি সংখ্যা অমূলদ (irrational) হয় যদি এটি p/q আকারে প্রকাশ করা না যায়, যেখানে p এবং q পূর্ণসংখ্যা এবং q ≠ 0।

ক) 0.75 = 75/100 = 3/4 = এটি p/q আকারে প্রকাশ করা যায়, তাই এটি মূলদ সংখ্যা।

খ) √256 = 16 একটি পূর্ণবর্গ সংখ্যা (16² = 256), তাই √256 = 16 একটি মূলদ সংখ্যা।

গ) 2/5 = এটি ইতিমধ্যে p/q আকারে আছে, তাই এটি মূলদ সংখ্যা।

ঘ) √27 = √(9 × 3) = 3√3 একটি পূর্ণবর্গ সংখ্যা নয়, তাই √27 একটি অমূলদ সংখ্যা। এটি p/q আকারে প্রকাশ করা যায় না।

উত্তর: ঘ) √27 একটি অমূলদ (irrational) সংখ্যা।

Question: Bowl S contains only marbles. If (1/4) of the marbles were removed, the bowl would be filled to (1/2) of its capacity. If 100 marbles were added, the bowl would be full. How many marbles are in bowl S?

Solution: 
Let
there are x number of marbles and capacity of the bowl is y marbles
3x/4 = y/2
⇒ y = 3x/2

x + 100 = y 
⇒ x + 100 = 3x/2
⇒ (3x/2) - x = 100 
⇒ x/2 = 100 
∴ x = 200 

Question: An officer was appointed on maximum daily wages on contract money of Tk. 6720. But on being absent for some days, he was paid Tk. 5600. For how many days was he absent?

Solution:
Maximum daily wages of the officers = H.C.F of Tk. 6720 and Tk. 5600

H.C.F of 6720 & 5600 = 1120

Maximum daily wage = Tk. 1120
Total days appointed = 6720 ÷ 1120 = 6 days

Days present = 5600 ÷ 1120 = 5 days
Absent days = 6 − 5 = 1

Required L.C.M = (L.C.M of 3, 6, 9)/(H.C.F of 4, 7, 8)
=18/1 = 18
Answer : 18

Question: If the sum of seven consecutive odd integers is 385, what is the largest number?

Solution:
ধরি, মাঝের সংখ্যাটি = x

সুতরাং, 7টি ক্রমিক বিজোড় সংখ্যা হবে যথাক্রমে: x - 6, x - 4, x - 2, x, x + 2, x + 4 এবং x + 6

প্রশ্নমতে,
(x - 6) + (x - 4) + (x - 2) + x + (x + 2) + (x + 4) + (x + 6) = 385
⇒ 7x = 385
⇒ x = 55

∴ মাঝের সংখ্যা, x = 55

∴ সবচেয়ে বড় সংখ্যা = x + 6 = 55 + 6 = 61

Question: The number n yields a remainder p when divided by 15 and a remainder q when divided by 5. If p = q+10, then which one of the following could be the value of n?

Solution:
প্রদত্ত শর্ত অনুযায়ী, আমাদের এমন একটি সংখ্যা n খুঁজে বের করতে হবে, যাকে 15 দ্বারা ভাগ করলে ভাগশেষ p এবং 5 দ্বারা ভাগ করলে ভাগশেষ q পাওয়া যায়, যেখানে p = q + 10।

• অপশন (ক): 
ধরা যাক n = 24
24 ÷ 15 = 1 এবং ভাগশেষ p = 9।
24 ÷ 5 = 4 এবং ভাগশেষ q = 4।
শর্ত অনুযায়ী, p = q + 10। এখানে 9 ≠ 4 + 10 ; সুতরাং, এই অপশনটি সঠিক নয়।

• অপশন (খ):
ধরা যাক n = 32
32 ÷ 15 = 2 এবং ভাগশেষ p = 2।
32 ÷ 5 = 6 এবং ভাগশেষ q = 2।
শর্ত অনুযায়ী, p = q + 10। এখানে 2 ≠ 2 + 10 ;  সুতরাং, এই অপশনটি সঠিক নয়।

• অপশন (গ): 
ধরা যাক n = 43
43 ÷ 15 = 2 এবং ভাগশেষ p = 13।
43 ÷ 5 = 8 এবং ভাগশেষ q = 3।
শর্ত অনুযায়ী, p = q + 10। এখানে 13 = 3 + 10 = 13, যা সত্য। সুতরাং, এই অপশনটি সঠিক।

• অপশন (ঘ): 
ধরা যাক n = 50
50 ÷ 15 = 3 এবং ভাগশেষ p = 5।
50 ÷ 5 = 10 এবং ভাগশেষ q = 0।
শর্ত অনুযায়ী p = q + 10। এখানে 5 ≠ 0 + 10 ; সুতরাং, এই অপশনটি সঠিক নয়।

প্রশ্ন: -1 হতে কত বিয়োগ করলে বিয়োগফল 0 (শূণ্য) হবে?

সমাধান:
- 1 - ( - 1)
= - 1 + 1
= 0

অর্থাৎ, -1 হতে -1 বিয়োগ করলে বিয়োগফল 0 (শূণ্য) হবে।

Given that, (10a + b) - (10b + a) = 27
⇒ 9a - 9b = 27 
⇒ a - b = 3

Let,
the number be x
then, 63x - 36x = 3834
⇒ 27x = 3834
⇒ x = 3834/27
⇒ x = 142.

(489.1375 x 0.0483 x 1.956) / (0.0873 x 92.581 x 99.749)
≈ 489 x 0.05 x 2 / 0.09 x 93 x 100
= 489/ (9 x 93 x 10)
= (163/279)x (1/10)
= 0.58/ 10
= .058
≈ .06