Algebra

Question: The sum of fifth and thirteenth term of an arithmetic progression is 28. What is the sum of the first seventeen terms of that progression?

Solution:
In an arithmetic progression.
Let first term = a
Common difference = d

We know,
a_n = a + (n - 1)d
∴ a₅ = a + 4d and a₁₃ = a + 12d

Given that, 
fifth term + thirteenth term = 28
⇒ a₅ + a₁₃ = 28
⇒ a + 4d + a + 12d = 28
⇒ 2a + 16d = 28
⇒ 2(a + 8d) = 28
∴ a + 8d = 14  ........(1)

We need the sum of the first 17 terms.
S₁₇ = (n/2) × [2a + (n - 1)d]
= (17/2) × [2a + 16d]
= 17/2 × 2(a + 8d)
= 17 × (a + 8d)
= 17 × 14
= 238

So the sum of the first seventeen terms is 238.

Question: If a = 0.202 , then the value of

is:

Solution:

সঠিক উত্তর 1.202 হবে,
কারণ (+) যোগ চিহ্ন দিয়ে বের করা রাশির উত্তর নেই।

Suppose
a = 3, b = 2
Option a) a² > b² = (3)² > (2)² = 9 > 4; True
Option b) a² < ab = (3)² < 3×2 = 9 < 6; False
Option c) a - b < 0 = 3 - 2 < 0 = 1 < 0; False
Option d) b + a > 2a = 2 + 3 > 2×3 = 5 > 6; False

2/a + 1/b
= (2b + a)/ab
= 6/4
= 3/2

আমরা জানি, মূলবিন্দুগামী রেখার সমীকরণ y = mx
এখানে, ঢাল m = 3
y = 3x............(i)
এখন (6, y) বিন্দুর জন্য (i) নং হতে পাই,
y = 3x
∴ y = 3 × 6 = 18 [∵ ভূজ = 6]
আবার, (x, 12) বিন্দুর জন্য (i) নং হতে পাই,
y = 3x
⇒ 12 = 3x [∵ কোটি = 12]
∴ x = 12/3 = 4
অতএব, y - x = 18 - 4 = 14

Question: A boy agrees to work at the rate of one Taka on the first day, two Taka on the second day, and four Taka on third day and so on. How much will the boy get if he started working on the 1^st of February and finishes on the 20^th of February?

Solution:
Given that,
1st term, a = 1
Common ratio, r = 2   ; r > 1

We know, 
Sum Sn = a × {(rⁿ - 1)/(r - 1)}
= 1 × {(2²⁰ - 1)/(2 - 1)}. ; [Putting, a = 1, r = 2 and n = 20]
= 2²⁰ - 1

So the boy will get 2²⁰ - 1 Takas if he works from February 1^st to February 20^th.

Question: Which value of x will satisfy the given inequality,
2(x - 4) ≥ 3x - 5 ?

Solution:
Given,
2(x - 4) ≥ 3x - 5
⇒ 2x - 8 ≥ 3x - 5 
⇒ 2x - 3x ≥ -5 + 8
⇒ - x ≥ 3
⇒ x ≤ - 3 [ Multiplying both sides of an inequality by a negative number reverses the inequality sign ]

Question: Three times a whole number is equal to four less than the square of the number. Find the number.

Solution:
Let the number be x.
Then, 3x = x² - 4
⇒ x² - 3x - 4 = 0
⇒ x² - 4x + x - 4 = 0
⇒ x (x - 4) + 1 (x - 4) = 0
⇒ (x - 4) (x + 1) = 0
⇒ x = 4, - 1 
Solving, x = 4 is the only whole number solution.

Players who play at least one sport = 37 + 30 - 21 = 46
∴ Players who play neither cricket nor badminton = 50 - 46 = 4

Question: 


Solution:

x⁴ + x² + 1
= (x²)² + 2x².1 + 1 - x²
= (x² + 1)² - x²
= (x² + x + 1) (x² - x + 1)

Question: Solve: 3(2x - 1) ≥ 4(x + 5), Then what is the solution set?

Solution:
Given the inequality,
3(2x − 1) ≥ 4(x + 5)
⇒ 6x - 3 ≥ 4x + 20
⇒ 6x - 4x - 3 ≥ 20 ; [Subtract 4x from both sides]
⇒ 2x - 3 ≥ 20
⇒ 2x ≥ 23 ; [Add 3 to both sides] 
⇒ x ≥ 23/2 ; [Divide both sides by 2]
∴ x ≥ 11.5

Solution set: In interval notation: [11.5, ∞)

Question: x² + y² + z² = 2(x + z - 1), then the value of x³ + y³ + z³ = ?

Solution:
Given that,
x² + y² + z² = 2(x + z - 1)
⇒ x² + y² + z² = 2x + 2z - 2
⇒ x² + y² + z² = 2x + 2z - 1 - 1
⇒ (x² + 1 - 2x) + y² + (z² + 1 - 2z) = 0
⇒ (x - 1)² + y² + (z - 1)² = 0
We know,
The sum of three squares of real numbers can only be zero if each individual square is zero.
So,
(x - 1)² = 0
∴ x = 1

y² = 0
∴ y = 0
And,
(z - 1)² = 0
∴ z = 1

Substitute the values of x, y, z into the expression,
x³ + y³ + z³
= 1³ + 0 + 1³
= 1 + 1
= 2

3x + 2y = 10 .... (i)
3x - 2y = 8 .... (ii)

(i) + (ii), 6x = 18
Or, x = 3

From, (i), y = 1/2

So, xy = 3.1/2 = 3/2

The series is = 2 + 5 + 8 + .......
Here, a = 2
d = 3
n = 100
So, n^th number in the series is = a + (n - 2)d
= 2 + (100 - 1)3
= 2 + 99×3
= 299
The number of bacteria on 100th day will be 299

দেয়া আছে,
y/x = 3/7 .........(i)
এবং x + 2y = 13 ............(ii)
(ii) নং সমীকরণ হতে পাই
x + 2y = 13
⇒ x = 13 - 2y .........(iii)
x এর মান (i) নং এ বসাই
y/x = 3/7
⇒ 7y = 3x
⇒ 7y = 3(13 - 2y)
⇒ 7y = 39 - 6y
⇒ 13 y = 39
⇒ y = 3
Answer: 3.

Let the smaller number be x and greater number be y
ATQ,
4x = 3y – 5
Or, 3y – 4x = 5 ......(1)
Again,
x + y = 6(y – x) + 6
Or, x + y = 6y – 6x + 6
Or, 7x – 5y = 6 ..... (2)

Multiply equation (1) by 5 and equation (2) by 3 and add them,
15y – 20x = 25
21x – 15y = 18
_____________
           ∴ x = 43

From equation (1),
3y – 4 × 43 = 5
Or, 3y – 172 = 5
Or, 3y = 177
Or, y = 177/3
∴ y = 59

Question:

Solution:
কোনো বর্গ ম্যাট্রিক্সের ট্রেস (Trace) হলো তার প্রধান কর্ণ বরাবর উপাদানগুলির যোগফল।
এখানে ম্যাট্রিক্স A এর প্রধান কর্ণ বরাবর উপাদানগুলি হলো 2, 5, এবং 8.
∴ Trace(A) = 2 + 5 + 8 = 15

যেহেতু, x ও y ঋণাত্মক পূর্ণসংখ্যা
এবং x - y = 1 অর্থাৎ x > y
তাহলে, x = -1 এবং y = -2 হলে xy = (-1)(-2) = 2

Question: Express the following inequality using absolute value notation: 1 < x < 9

Solution:
1 < x < 9
∴ The midpoint = (1 + 9)/2
= 10/2
= 5
Now subtract the midpoint from all parts. then we get,
1 - 5 < x - 5 < 9 - 5
⇒ - 4 < x - 5 < 4
∴ |x - 5| < 4

Question: The equation of a line that passes through the points (1, 5) and (2, 3) is:

Solution:
We know that the equation of a line passes through two points (x1, y1) and (x2 y2) is
(y - y₁)/(x - x₁) = (y₂ - y₁)/(x₂ - x₁)

Now, substitute the values in the formula, we get
(y - 5)/(x - 1) = (3 - 5)/(2 - 1) ; [(x₁, y₁) = (1, 5) and (x₂, y₂) = (2, 3)]
⇒ (y - 5)/(x - 1) = (- 2)/(1)
⇒ y - 5 = - 2(x - 1)
⇒ y - 5 = - 2x + 2
⇒ 2x + y - 5 - 2 = 0
∴ 2x + y - 7 = 0

Therefore, the equation of a line that passes through the points (1, 5) and (2, 3) is 2x + y - 7 = 0.

Question: The range of f(x) = 1/(x + 1) is:

২০২২ সাল ভিত্তিক সমন্বিত ৮ ব্যাংক ও ১ আর্থিক প্রতিষ্ঠান পদের নাম: অফিসার (জেনারেল)
Solution:
দেওয়া আছে,
f(x) = 1/(x + 1)
⇒ y = 1/(x + 1)
⇒1/y = x + 1
⇒ x = (1/y) - 1
⇒ x = (1 - y)/y

∴ f^-1(x) = y = (1 - x)/x
x এর মান 0 ব্যতীত যেকোনো বাস্তব সংখ্যা হবে। কারণ x এর মান 0 হলে ফাংশনটি অসঙ্গায়িত হবে।

অতএব, নির্ণেয় রেঞ্জ: R\{0}

Question: Which of the following is equivalent to the pair of inequalities 2x - 5 ≤ 7 and 3x + 4 > 10?

Solution:
Solve the first inequality,
2x - 5 ≤ 7 
⇒ 2x ≤ 7 + 5
⇒ 2x ≤ 12
∴ x ≤ 6
And,
Solve the second inequality,
3x + 4 > 10 
⇒ 3x > 10 - 4
⇒ 3x > 6
∴ x > 2

∴ We get 2 < x ≤ 6

Question: In a class, 25 students play football, 15 students play cricket, and 5 students play both. 10 students play neither football nor cricket. What is the total number of students in the class?

Solution:
Let the number of students who play football = 25
Number of students who play cricket = 15
Number of students who play both football and cricket = 5
Number of students who play neither = 10

Number of students who play football or cricket:
n(F ∪ C) = n(F) + n(C) − n(F ∩ C)
n(F ∪ C) = 25 + 15 − 5 = 35

Add the students who play neither to get the total number of students:
Total students = n(F ∪ C) + neither = 35 + 10 = 45

Question: a = 2b = 3c and abc = 36, then find the value of c.

Solution:
Given,
a = 2b = 3c
∴ a = 3c
and b = 3c/2

Now, abc = 36
⇒ 3c . (3c/2) . c = 36
⇒ 9c³/2 = 36
⇒ 9c³ = 72
⇒ c³ = 72/9
⇒ c³ = 8
⇒ c³ = 2³
∴ c = 2

Question: In an examination, 65% students passed in Mathematics and 60% students in English, 40% passed in both these subjects. If 90 students failed in Mathematics and English both, then what is the total number of students?

Solution: 
Given that, 
P(M) = 65%
P(E) = 60%
P(M ∩ E) = 40%

We know,
P(M U E) = P(M) + P(E) - P(M ∩ E) 
= 65% + 60% - 40% = 85%
∴ P(M U E) = 85%
∴ passed students = 85%
∴ Failed students = 100% - 85% = 15%

Now,
We are given that 90 students failed in both subjects, which represents 15% of the total students. Let N be the total number of students.
⇒ 15% of N = 90 
⇒ (15/100)N = 90
⇒ N = (90 × 100)/15
∴ N = 600

∴ The total number of students is 600.

Given,
(a - 1/a) = 2
∴ a³ - 1/a³ = (a - 1/a)³ + 3.a.1/a(a - 1/a)
= 2³ + 3.2
= 8 + 6 = 14

Question: What is the slope of a line perpendicular to the line whose equation is 10x - y = 3?
(Officer Cash 2022 অনুযায়ী)

Solution:
সরল রেখার সাধারণ সমীকরণ,
y = mx + c ......(1) (এখানেm = ঢাল)

যদি কোনো রেখার ঢাল হয় m, তবে তার লম্ব (perpendicular) রেখার ঢাল হবে,
m' = - (1/m)

এখন,
10x - y = 3
y = 10x + 3
(1) নং এর সাথে তুলনা করে পাই,
m = 10

∴ লম্ব (perpendicular) রেখার ঢাল হবে, m' = - (1/10)

According to the given,
x(x + 1) = 360
⇒ x² + x = 360
⇒x² + x - 360 = 0
⇒x² + 18x - 17x - 360 = 0
⇒(x +18)(x - 17) = 0
x ≠ - 18,
∴ x = 17

One positive interger 17.
Other positive interger 17 + 1 = 18.

x² + x - 360 = 0 is the required quardratic equation. 

Question: In a geometric progression, the 4^th term is 16 and the 7^th term is 128. Find the 10^th term.

Solution:
Let the first term = a
Common ratio = r
We know,
n-term = ar^{n - 1}

Then,
4^th term, ar³ = 16  ........(1)  
7th term, ar⁶ = 128  ........(2)

Now, divide equation (2) by equation (1) then we get,
(ar⁶)/(ar³) = 128/16  
⇒ r³ = 8  
⇒ r³ = 2³  
∴ r = 2  

Then substitute r = 2 into equation (1) 
a.(2)³ = 16   
⇒ a × 8 = 16  
∴ a = 2

Now, 10^th term
= ar⁹  
= 2 × 2⁹ 
= 2 × 2⁹  
= 2¹⁰ 
= 1024

∴ The 10th term is 1024

Question: How many terms of the arithmetic progression 2, 7, 12,... should be taken so that their sum equals 354?

solution:
Given arithmetic progression: 2, 7, 12, …
First term, a = 2
Common difference, d = 5

We know, 
Sum of first n terms, S_n = (n/2) × [2a + (n - 1)d]

ATQ,
(n/2) × [2 × 2 + (n - 1) × 5] = 354
⇒ (n/2) × [4 + 5n - 5] = 354
⇒ (n/2) × (5n - 1) = 354
⇒ n(5n - 1) = 708
⇒ 5n² - n - 708 = 0
⇒ 5n² - 60n + 59n - 708 = 0
⇒ 5n(n - 12) + 59(n - 12) = 0
⇒ (n - 12)(5n + 59) = 0
Now, n - 12 = 0
∴ n = 12
Or
5n + 59 = 0
∴ n = - 59/5 ; [not possible, n must be positive]

∴ 12 terms of the arithmetic progression must be taken to result in a sum of 354. 

Question: What will come at the place of question mark ?
8, 28, 116, 584, ?

Solution:
1^st term = 8
2^nd term = (8 × 3) + 4 = 28 
3^rd term = (28 × 4) + 4 = 116 
4^th term = (116 × 5) + 4 = 584 
5^th term = (584 × 6) + 4 = 3508

​প্রশ্ন: If x + (1/x) = 2, The value of x⁴⁹⁹⁹ + x⁵⁰⁰⁰  is:
​
​সমাধান:
দেয়া আছে, 
​x + (1/x) = 2
⇒ (x²+ 1)/x = 2
⇒ x²+ 1 = 2x
⇒ x²- 2x + 1 = 0
⇒ (x - 1)²= 0 
⇒ x - 1 = 0
∴ x = 1
​
​ x⁴⁹⁹⁹ + x^5000
​= 1 + 1
​= 2

Question: If

Solution:

Question: If 4x + 5y = 140 and 4x / 5y = 2 / 5, then find y - x.
 
Solution:
We are given:
⇒ 4x / 5y = 2 / 5

We simplify:
x / y = (5 / 4) × (2 / 5)
⇒ x / y = 2 / 4 = 1 / 2
∴ x = y / 2

Also given:
4x + 5y = 140

Substitute x = y / 2:
⇒ 4 × (y / 2) + 5y = 140
⇒ (4y / 2) + 5y = 140
⇒ (2y + 5y) = 140
⇒ 7y  = 140
⇒ y = 20

Then x = y / 2 = 10
So, 20 - 10 = 10