Number System, Problems on Number

Question: There are 45 students in a certain class 2/3 of the students are girls and 1/2 of the girls are blue-eyed. How many blue- eyed girls are there in the class?

Solution:
মোট ছাত্র-ছাত্রীর সংখ্যা = 45 জন
বালিকার সংখ্যা = (45 × 2)/3 জন
= 30 জন

blue- eyed বালিকার সংখ্যা = (30 × 1)/2 জন
= 15 জন

n < 0 ⇒ 2n < 0,
again, - n > 0 and n² = (-n)2 > 0.
Thus, out of the numbers 0, -n, 2n and n²
We find that 2n is the least number here. 
Answer: 2n

Question: If 5 students run a mile in 5 minutes, how much time will 50 students take to run a mile?

Solution:
5 students run a mile in 5 minutes
1 students run a mile in 5 minutes
50 students run a mile in 5 minutes

Question: In a class of 100 students, 55 are taking Biology, 35 are taking Chemistry and 10 are taking both courses. How many students are not enrolled in either course?

Solution:
দেওয়া আছে,
মোট শিক্ষার্থীর সংখ্যা, n(U) = 100
জীববিজ্ঞান নেওয়া শিক্ষার্থীর সংখ্যা, n(B) = 55
রসায়ন নেওয়া শিক্ষার্থীর সংখ্যা, n(C) = 35
উভয় বিষয় নেওয়া শিক্ষার্থীর সংখ্যা, n(B ∩ C) = 10

কমপক্ষে একটি বিষয় নেওয়া শিক্ষার্থীর সংখ্যা, n(B ∪ C) = n(B) + n(C) - n(B ∩ C)
⇒ n(B ∪ C) = 55 + 35 - 10
⇒ n(B ∪ C) = 90 - 10
⇒ n(B ∪ C) = 80

কোনোটিই নেয়নি এমন শিক্ষার্থীর সংখ্যা = n(U) - n(B ∪ C)
= 100 - 80
= 20

সুতরাং, 20 জন শিক্ষার্থী কোনো কোর্সই গ্রহণ করেনি।

Question: When 40% of the first number is added to the second number, the second number becomes 7/5 times the first number. What is the ratio of the first number to the second number?

Solution:
Let the first number = x
and the second number = y.

According to the question,
y + 40% of x = (7/5)x
⇒ y + (40/100) x = (7/5)x
⇒ y + (4/10)x = (7/5)x
⇒ y = (7/5)x - (4/10)x
⇒ y = (14 - 4)x/10
⇒ y = 10x/10
⇒ y = x

Therefore, x : y = x : x = 1 : 1

0.001/(0.1×0.1)
= 0.001/0.01
= 100/1000
= 0.1

When the number is divided by 14 it gives a remainder of 8,
The number = 14N + 8 (14N is divisible by 14)
When same number is divided by 7 it will give remainder 1

LCM of 12, 14, 18 and 22 = 2772
Hence the least number which will be exactly divisible by 12, 14, 18, and 22 = 2772
2772/2 = 1386
1386 is the number which when doubled, we get 2772
Hence, 1386 is the least number which when doubled will be exactly divisible by 12, 14, 18, and 22.

Question: What is the least number which when divided by 3, 5, 6, 8, 10, and 12 leaves a remainder 2 in each case, but when divided by 22 leaves no remainder?

Solution:
The number, when divided by 3, 5, 6, 8, 10, and 12 leaves a remainder of 2. This means the number is 2 more than a multiple of their L.C.M.
LCM of 3, 5, 6, 8, 10, 12 = 120

The required number, N, must be in the form:
N = 120K + 2, where K is a positive integer.

N divisible by 22 ⇒ (120K + 2) ÷ 22 has remainder 0

120K + 2 = 22 × m ⇒ Check for smallest K

Try K = 2 ⇒ 120 × 2 + 2 = 242, divisible by 22

Hence, required number = 242

If the remainder is same in each case and remainder is not given,
HCF of the differences of the numbers is the required greatest number.
34 - 24 = 10
34 - 28 = 6
28 - 24 = 4
Hence, the greatest number which divides 24, 28 and 34 and gives the same remainder
= HCF of 10, 6, 4
= 2

Question: The difference between two positive numbers is 8 and the difference of their squares is 160. What is the smallest number?

Solution:
Let the numbers be x and (x + 8)

According to the question,
(x + 8)² - x² = 160
⇒ x² + 16x + 64 - x² = 160
⇒ 16x + 64 = 160
⇒ 16x = 96
⇒ x = 6

∴ The smallest number is = 6

Greatest number of four digits = 9999
LCM of 15, 25, 40 and 75 = 600
9999 ÷ 600 = 16, remainder = 399
Hence, greatest number of four digits which is divisible by 15, 25, 40 and 75 = 9999 - 399 = 9600

p < 1
⇒ 1/p > 1
⇒ 2/p > 2
2/p - p > 2 - p > 0
[∵ p < 1]
Hence, (2/p - p) is a positive number.

Question: The sum of five consecutive multiples of 3 is 165. What is the largest number?

Solution:
ধরি, পাঁচটি ক্রমিক 3 এর গুণিতক যথাক্রমে x, (x + 3), (x + 6), (x + 9) এবং (x + 12).

প্রশ্নমতে,
x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 165
⇒ 5x + 30 = 165
⇒ 5x = 165 - 30
⇒ 5x = 135
⇒ x = 135/5
⇒ x = 27

∴ বৃহত্তম সংখ্যা = x + 12
= 27 + 12 = 39

Question: If x² is odd, what will x² - x be?

Solution:
যেহেতু x² বিজোড় তাই x ও বিজোড় হবে। 

এখন,
x² - x
= x(x - 1)
= (x - 1)x
∴ (x - 1) এবং x দুইটি ক্রমিক সংখ্যা।

x বিজোড় সংখ্যা হলে (x - 1) অবশ্যই জোড় সংখ্যা হবে।
কারণ দুইটি ক্রমিক সংখ্যার মধ্যে একটি বিজোড় হলে অন্যটি জোড় হবে।

সুতরাং, x ও (x - 1) এর গুনফল,
= x(x - 1)
= x² - x, একটি জোড় সংখ্যা। [জোড় × বিজোড় = জোড়]

x^-3 - 0.0001 = 0
বা, 1/x³ = 0.001
বা, 1/x³ = 1/10³
বা, x³ = 10³
বা, x = 10
∴ x² = 10² = 101

Question: If the n^th term of an arithmetic progression is 5n + 2, then what is the common difference?

Solution:
The n^th term of an arithmetic progression is T_n = 5n + 2
n = 1 then, T₁ = 5 × 1 + 2 = 7
n = 2 then, T₂ = 5 × 2 + 2 = 12
n = 3 then, T₃ = 5 × 3 + 2 = 17
n = 4 then, T₄ = 5 × 4 + 2 = 22
............................

Common difference,
T₂ - T₁ = 12 - 7 = 5
T₄ - T₃ = 22 - 17 = 5

∴ The common difference is 5.

Question: What will come at the place of question mark? 7, 18, 51, 150, ?

Solution:
1st term: 7
2nd term: 18 = 7 × 3 - 3
3rd term: 51 = 18 × 3 - 3
4th term: 150 = 51 × 3 - 3
5th term: 447 = 150 × 3 - 3

Question: If k is a positive integer, what is the smallest possible value of k such that 1080 × k is the square of an integer?

Solution:
আমরা জানি, একটি সংখ্যা পূর্ণবর্গ হতে হলে এর মৌলিক গুণনীয়কের ঘাতসমূহ জোড় সংখ্যা হতে হবে।

1080 = 2 × 2 × 2 × 3 × 3 × 3 × 5
= 2³ × 3³ × 5

1080k = 2³ × 3³ × 5 × k

এখন,
এখন k এর মান 2 × 3 × 5 = 30  হলে, 1080k একটি পূর্ণবর্গ সংখ্যা হবে।
1080 × 30 = (2³ × 3³ × 5) × (2 × 3 × 5)
= 2⁴ × 3⁴ × 5²

যেহেতু এই গুণফলের সব মৌলিক উৎপাদকের ঘাত জোড়, তাই এটি একটি পূর্ণবর্গ সংখ্যা।
সুতরাং, k = 30 হলে 1080 × k পূর্ণবর্গ সংখ্যা হয়।

The pattern is followed by
155 - 4 = 151
151 - 7 = 144 {7 = 4 + 3}
144 - 12 = 132 {12 = 7 + 5}
132 - 19 = 113 { 19 = 12 + 7}
113 - 28 = 85 { 28 = 19 + 9}
Hence ? = 85.

Question: If one fifth of one sixth of a number is 10, then what is 2/5 of the number?

Solution:
ধরি, সংখ্যাটি = x

প্রশ্নমতে,
(1/5) × (1/6) × (x) = 10
⇒ (1/30)x = 10
⇒ x = 10 × 30
∴ x = 300

∴ সংখ্যাটি = 300

এখন,
সংখ্যাটির 2/5 অংশ = (2/5) × 300
= 600/5
= 120

According to question,
3a+4b /3a−4b = 3c+4d/ 3c−4d
⇒ 3a/4b = 3c/4d
⇒ad=bc

Question: A gardener planted trees in rows and columns such that the number of rows is five more than the number of columns. If the total number of rows and columns is 105, find the number of trees.

Solution:
Let the number of columns = x.
Then, number of rows = x + 5

According to the question: x + (x + 5) = 105
⇒ 2x + 5 = 105
⇒ 2x = 100
⇒ x = 50

Number of rows = x + 5 = 55

Total number of trees = rows × columns = 55 × 50 = 2750

HCF of 84, 144 and 36 is 12
So, 12 is the largest number that divides 84, 144 or 36 without any remainder

F = (BD×TD)/(BD−TD)
= (200 × 100)/ (200−100)
= 200−100/ 100
= Tk.200

It is a G.P (general process) with r i.e;
2¹, 2², 2³,...
If number of term is n.Then,
2× 2^n-1 = 1024
2^n-1 = 512
2^n-1 = 2⁹
n -1 = 9
n = 10
Answer : 10

Question: The difference between two numbers is 840. When the larger number is divided by the smaller, the quotient is 6 and the remainder is 10. What is the smaller number?

Solution:
Given that,
The difference of two numbers = 840
Quotient when the larger number is divided by the smaller number = 6
Remainder when the larger number is divided by the smaller number = 10

Now, Let the smaller number be x.
Larger number = 6x + 10

ATQ,
⇒ 6x + 10 - x = 840
⇒ 5x + 10 = 840
⇒ 5x = 840 - 10
⇒ 5x = 830
⇒ x = 830/5
⇒ x = 166

So the smaller number is 166.

প্রশ্ন: যদি একটি ধনাত্মক পূর্ণসংখ্যা n কে 18 দ্বারা ভাগ করলে ভাগশেষ 7 থাকে, তাহলে n কে 6 দ্বারা ভাগ করলে ভাগশেষ কত হবে?

সমাধান:
এখানে, ভাজ্য = n
ভাজক = 18
ভাগশেষ = 7
ধরি, ভাগফল = q

ভাজ্য = (ভাজক × ভাগফল) + ভাগশেষ 
∴ n = 18q + 7
⇒ n = (18q + 6) + 1
⇒ n = 6(3q + 1) + 1

যেহেতু 6(3q + 1) সংখ্যা, 6 দ্বারা নিঃশেষে বিভাজ্য হবে, তাই ভাগশেষ থাকবে শুধু 1।

∴ n কে 6 দ্বারা ভাগ করলে ভাগশেষ হবে = 1