Algebra

If (x - 1) and (x + 1) are the factors y equation then,

x - 1 = 0
x = 1
Put x = 1 we get,
1 + a - 3 + 2 + b = 0
a + b = 0 ........(i)

x + 1 = 0
x = -1
Put x = -1 we get,
1 - a - 3 - 2 + b = 0
b - a = 4 ........(ii)

(i) - (ii) we get,
a + b - b + a = 0 - 4
2a = -4
a = -2
∴ b = 2
a, b = -2, 2.

Let, first number of that series = a
Common difference = d
So, 4th term = a + 3d and 12th term = a + 11d

ATQ, a + 3d + a + 11d = 20
⇒ 2a + 14d = 20

∴ S₁₅ = 15/2{2a + (15 - 1)d}
= 15/2(2a + 14d)
= 15/2 × 20
= 150

x = a(b - c),
y = b(c - a),
z = c(a - b)

Let,
x/a = b - c = A
y/b = c - a = B
z/c = a - b = C

∴ A + B + C = b - c + c - a + a - b
= 0
A³ + B³ + C³ = 3ABC
= (x/a)³ + (y/b)³ + (z/c)³
= 3 × (x/a) × (y/b) × (z/c)
= 3xyz/abc

Question: If 

Solution:

Question: If one root of x² - (q + 2)x + 15 = 0 is 5, then the value of q is:

Solution:
Given equation: x² - (q + 2)x + 15 = 0

One root is x = 3. Substitute:
(5)² - (q + 2)(5) + 15 = 0
⇒ 25 - 5(q + 2) + 15 = 0
⇒ 25 - 5q - 10 + 15 = 0
⇒ 30 - 5q = 0
⇒ 5q = 30
∴ q = 6 

Question: In a series of 5 consecutive odd numbers if 13 is the 5^th number, what is the 3^rd number in the series?

Solution:
5^th odd number = 13
4^th odd number = 11
3^rd odd number = 9
2^nd odd number = 7
1^st odd number = 5

Adding the given equations:
12a + 3b + 7b - 2a = 10
Or, 10a + 10b = 10
Or, 10(a + b) = 10
Or, a + b = 1
So, average of a and b is 0.5

Question: What should be the value of "P" so that, the expression (9 - 12x + Px²) becomes a perfect square?

Solution:
Given Expression,
(9 - 12x + Px²)
= (3)² - 2 × 3 × 2x + (2x)² + Px² - (2x)²
= (3 - 2x)² + Px² - 4x²

∴ the expression becomes a perfect square if,
Px² - 4x² = 0
⇒ Px² = 4x²
∴ P = 4

মনে করি, মোট ফ্রিসবি এর পরিমাণ x টি
প্রশ্নমতে, 7x = 10000 + 2x
⇒ 7x - 2x = 10000
⇒ 5x = 10000
∴ x = 10000/5 = 2000

4b² + 1/b² = 2
(2b)² + (1/b)² + 4 = 2 + 4
(2b)² + 2×2b×1/b + (1/b)² = 6
(2b + 1/b)² = 6
So (2b + 1/b) = (6)^1/2.
Now 8b³ + 1/b³
= (2b)³ + (1/b)³
= (2b + 1/b)³ - 3×2(2b + 1/b)
= (6^1/2)³ - 6(6)^1/2
= 6(6)^1/2 - 6(6)^1/2
= 0

Question: If (x + 3)² = 64, which of the following can be the value of (x + 2)?

Solution:
Given, (x + 3)² = 64
⇒ (x + 3)² = 8²
∴ x + 3 = ± 8

Case 1: x + 3 = 8
x = 8 - 3 = 5
x + 2 = 5 + 2 = 7

Case 2: x + 3 = - 8
x = - 8 - 3 = - 11
x + 2 = - 11 + 2 = - 9

Possible values of (x + 2) are 7 or - 9.

এখানে,
21, 18, 15, 12,
১ম পদ, a = 21
সাধারন অন্তর, d = 18 - 21 = -3
∴ সমষ্টি = n/2{2a + (n - 1)d}
⇒ 0 = n/2{(2 × 21) + (n - 1)(-3)}
⇒ 0 = n/2(42 - 3n + 3)
⇒ 45n - 3n² = 0
⇒ 3n(n - 15) = 0
⇒ n - 15 = 0
∴ n = 15.

Question: If x² - 10x + 25 = 0, then the value of x is:

Solution:
দেওয়া আছে,
x² - 10x + 25 = 0
⇒ x² - 2. x. 5 + 25 = 0
⇒ (x - 5)² = 0
⇒ (x - 5)(x - 5) = 0
∴  x = 5 এবং x = 5 [যেহেতু সমীকরণটি একটি দ্বিঘাত সমীকরণ তাই এর মূল হবে দুইটি]

Question: If a - b = 2 and a² + b² = 20, then the value of a + b will be?

Solution:
We know, (a - b)² = a² - 2ab + b²
⇒ (a - b)² = a² + b² - 2ab ----------------(1)

Given,
a - b = 2
⇒ (a - b)² = 4
Also given, a² + b² = 20

Substitute the values in equation (1): 
20 - 2ab = 4
⇒ 2ab = 16
⇒ ab = 8

Now, (a + b)² = a² + b² + 2ab
= 20 + 16
= 36
∴ a + b = ±6

Players who play at least one sport = 35 + 30 - 18 = 47
Players who play neither golf nor soccer = 50 - 47 = 3

Formula for calculating two overlapping sets:
A + B - both + NOT (A or B) = Total

ATQ,
41 (french) + 22 (german) - 9 (both) + NOT = 78
⇒ 54 + NOT = 78
⇒ NOT = 78 - 54 = 24
⇒ So answer is 24

Question: If a + 1/a = √3, then what is the value of a³⁰ + a²⁴ + a⁶ + 1?

Solution:
Given, a + 1/a = √3
Now,
a³ + 1/a³ = (a + 1/a)³ - 3 . a . (1/a)(a + 1/a)
⇒ a³ + 1/a³ = (√3)³ - 3(√3) [∵ a + 1/a = √3]
⇒ a³ + 1/a³ = 3(√3) - 3(√3)
⇒ a³ + 1/a³ = 0 
⇒ a⁶ + 1 = 0 [Multiplying both sides by a³]

Then,
a³⁰ + a²⁴ + a⁶ + 1
= a²⁴ (a⁶ + 1) + (a⁶ + 1)
= (a²⁴ × 0) + 0
= 0

Question: The sum of seventh and eleventh term of an arithmetic progression is 18. 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 = a + (n - 1)d
∴ a₇ = a + 6d and a₁₁ = a + 10d

Given that, 
Seventh term + eleventh term = 18
⇒ a₇ + a₁₁ = 18
⇒ a + 6d + a + 10d = 18
⇒ 2a + 16d = 18
⇒ 2(a + 8d) = 18
∴ a + 8d = 9  ........(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 × 9
= 153

So the sum of the first seventeen terms is 153.

Question: Solve the inequality: 5(x + 2) - 3 ≤ 2(x - 4) + 17

Solution: 
Given inequality,
5(x + 2) - 3 ≤ 2(x - 4) + 17
⇒ 5x + 10 - 3 ≤ 2x - 8 + 17
⇒ 5x + 7 ≤ 2x + 9
⇒ 5x - 2x ≤ 9 - 7
⇒ 3x ≤ 2
∴ x ≤ 2/3

Question: The slope of a line perpendicular to one with slope 2 is-

Solution:
দেওয়া আছে, 
প্রথম রেখের ঢাল, m₁ = 2

আমরা জানি,
যদি দুটি রেখা পরস্পর লম্ব হলে তাদের ঢালের গুণফল - 1 হয়। 
অর্থাৎ, 
m₁⋅m₂ = - 1
⇒ 2⋅m₂ = - 1
∴ m₂ = - 1/2

Question: Solve the inequality: 4(x - 6) + 4 < 8(x - 4)

Solution: 
Given inequality,
4(x - 6) + 4 < 8(x - 4)
⇒ 4x - 24 + 4 < 8x - 32
⇒ 4x - 20 < 8x - 32
⇒ 4x - 8x < - 32 + 20
⇒ - 4x < - 12
∴ x > 3

Question: In a group of people, 35 speak English, 25 speak French, and 12 speak both English and French. How many people are in the group if everyone speaks at least one of these two languages?

Solution:
দেওয়া আছে,
ইংরেজি বলে, n(E) = 35 জন
ফরাসি বলে, n(F) = 25 জন
উভয় ভাষা বলে, n(E ∩ F) = 12 জন

যেহেতু সবাই অন্তত একটি ভাষা বলে, তাই মোট ব্যক্তির সংখ্যা হবে n(E ∪ F)।

আমরা জানি,
n(E ∪ F) = n(E) + n(F) - n(E ∩ F)
⇒ n(E ∪ F) = 35 + 25 - 12
⇒ n(E ∪ F) = 60 - 12
⇒ n(E ∪ F) = 48

∴ ঐ গ্রুপে মোট 48 জন ব্যক্তি আছেন। 

Question: The value of is
(Senior Officer 2022 অনুযায়ী)

Solution:
As x → 0, we know cosx → 1 [cos0 = 1]
So the expression becomes x/1 = x near 0

Applying the limit,

Let us assume that he answered x question correctly. Marks scored by him in x question = 2x
Then, wrong answer would be = 60 – x
Marks lost by him in (60 – x) questions = (60 – x)×1
ATQ,
2x – (60 – x) = 39
Or, 3x = 99
∴ x = 33

Question: In a class of 78 students, 41 are taking French, 22 are taking German, and 9 are taking both courses. How many students are not enrolled in either course?

5 Combine Banks (২০২২ সাল ভিত্তিক) Post Name: Officer Cash/Officer Teller (১০ম গ্রেড) Exam Date: 11.07.2025 Faculty of Business Studies (FBS), DU

Solution:
Total students = 78
Students taking French, n(F) = 41
Students taking German, n(G) = 22
Students taking both French and German, n(F ∩ G) = 9

We know,
n(F ∪ G) = n(F) + n(G) - n(F ∩ G)
n(F ∪ G) = 41 + 22 - 9 = 54

∴ Not enrolled = Total students - n(F ∪ G) = 78 - 54 = 24

Question: The 3rd term of a geometric sequence is 48, and the 6th term is 384. What is the common ratio is?

Solution:
Given, 
The 3rd term of a geometric sequence, a₃ = 48
The 6th term of the same sequence, a₆ = 384

In a geometric sequence we know,
a_n = ar^{n - 1}

So,
a₃ = ar^{3 - 1}= ar² .......(1)
And
a₆ = ar^{6 - 1}= ar⁵........ (2)

Now (2) ÷ (1), 
ar⁵/ar² = 384/48
⇒ r³ = 8 = 2³
∴ r =  2

So the common ratio is 2.

Given, d = -1.5t² + 120 
As, d =10 
∴ 10 = -1.5t² + 120  
⇒ 1.5t² - 110 = 0 
⇒ 1.5t² = 110 
⇒ t² = 110/1.5 
⇒ t = √(110/1.5)
∴  t = 8.56

Question: Tk. 720 was divided among A, B, C, D, E. The sum received by them was in ascending order and in arithmetic progression. E received Tk. 40 more than A. How much did B receive? 

Solution:
Given that,
A + B + C + D + E = Tk. 720
And E - A = 40

We know,
Arithmetic progression,
a, a + d, a + 2d, a + 3d, a + 4d
And nth term = a + (n - 1)d

Let, A receive Tk. a and the difference between each consecutive person be Tk. d.
Amount, E = a + 4d
Amount, A = a

According to the question,
⇒ a + 4d - a = 40
⇒ 4d = 40
∴ d = 10

Also,
a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = 720
⇒ 5a + 10d = 720
⇒ 5a + 10 × 10 = 720
⇒ 5a = 720 - 100
⇒ a = 620/5
∴ a = 124

So, Amount, B = a + d = 124 + 10 = Tk. 134

Question: A sequence of numbers a₁, a₂, a₃, …, a_n is generated by the rule a_{n + 1} = 3a_n. If a₅ - a₄ = 48, then what is the value of a₅?

সমাধান:
প্রদত্ত অনুক্রমের নিয়মটি হলো: a_n+1 = 3a_n
n = 4 বসালে পাই,
a_{4 + 1} = 3a₄
⇒ a₅ = 3a₄ 

প্রশ্নমতে,
a₅ - a₄ = 48
⇒ 3a₄ - a₄ = 48
⇒ (3 - 1)a₄ = 48
⇒ 2a₄ = 48
⇒ a₄ = 48/2
⇒ a₄ = 24

এখন,
a₅ = 3a₄
⇒ a₅ = 3 × 24
⇒ a₅ = 72

অতএব, a₅ এর মান হলো 72

Question: What is the sum of the squares of the digits from 1 to 11?

Solution: 
আমরা জানি,
n সংখ্যক ক্রমিক সংখ্যার বর্গের যোগফল, Sn = [n(n + 1)(2n + 1)]/6
= [11(11 + 1)(22 + 1)]/6
= (11 × 12 × 23)/6 
= 22 × 23 
= 506

Question: If x² - √(7). x + 1 = 0 then, x² + 1/x² = ?

Solution:
দেওয়া আছে,
x² - √7x + 1 = 0
⇒ x² + 1 = √7x  
⇒ x²/x + 1/x = √7x/x [উভয় পক্ষকে x দ্বারা ভাগ]
⇒ x + 1/x = √7

এখন,
x² + 1/x²
= (x + 1/x)² - 2 . x . 1/x
= (√7)² - 2
= 7 - 2
= 5

∴ x² + 1/x² এর মান 5।

(3 + 11 + 7 + 9 + 15 + 13 + 8 + 19 + 17 + 21 + 14 + x)/12 = 12 
⇒ (137 + x)/12 = 12
⇒ x = 144 - 137
∴ x = 7

Question: If y = 5, then what is the value of 20y√(y³ - y²)?

Solution:
Given,
y = 5

∴ 20y√(y³ - y²)
= 20. 5. √(53 - 52)
= 100√(125 - 25)
= 100√(100)
= 100 × 10
= 1000

(0.0015×10^m)/(0.03×10^k) = 5×10⁷
⇒ 0.15×10^m-k / 3 = 5×10⁷
⇒ 15×10^m-k / 3×100 = 5×10⁷ 
⇒ 10^m-k/10² = 10⁷ 
⇒ 10^m-k =10⁷ × 10² = 10⁹
∴ m - k = 9

Question: Which of the following equation of the line passing through the points (7, 4) and (5, 1)?

Solution:
Given that, 
Two points are, (7, 4) and (5, 1)

We know,
Slope, m = (y₂ - y₁)/(x₂ - x₁)
= (4 - 1)/(7 - 5)
∴ m = 3/2

So the slope is 3/2.

Now check which option has slope 3/2 and passes through one of the points (we’ll use point (5, 1)),
খ) 2y = 3x - 13
 ⇒ y = (3/2)x - (13/2)
slope, m = 3/2

Similar answer for point (7, 4)

∴ Correct Answer: খ) 2y = 3x - 13

a² − b² = 20
Or, (a + b) (a - b) = 20
Or, a - b = 20/(a + b) = 20/5 = 4

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)  
7^th 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: If x² + yz + zx + xy is divided by x + z, the result is-

Solution:
x² + yz + zx + xy
= x² + xy + zx + yz
= x(x + y) + z(x + y)
= (x + y)(x + z)

∴ divided by (x + z) the result = (x + y)(x + z)/(x +z) = (x + y)

Question: The sum of the squares of three numbers is 120, and the sum of their products taken two at a time is 140. What is the sum of the three numbers?

Solution:
Let the numbers be a, b, c.
Given:
a2+ b2 + c2 = 120, 
ab + bc + ca = 140,
⇒ (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 120 + 2 × 140 = 120 + 280 = 400

∴ a + b + c = √400 = 20

Pillar above the water is 12 feet = 1 - (1/2 + 1/3) = 1/6
So, the length of the pillar is = 12 × 6 = 72 feet

Question: If log_x (64/125) = - 3, then x = ?
(Janata RC 2022 অনুযায়ী) 

Solution:
log_x (64/125)  = - 3
⇒ x^{- 3} = 64/125
⇒ x^{- 3} = (4/5)³
⇒ x^{- 3} = 1/(5/4)³
⇒ x^{- 3} = (5/4)^{- 3}
⇒ x = 5/4

- 2 + (-2) - {-(2)} - 2
= - 2 - 2 + 2 - 2
= - 4

Question: The sum of the squares of three numbers is 123, and the sum of their products taken two at a time is 119. What is the sum of the three numbers?

Solution:
Let the three numbers be a, b, and c
Then,
a² + b² + c² = 123
ab + bc + ca = 119

Now,
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
⇒ (a + b + c)² = 123 + (2 × 119) 
⇒ (a + b + c)² = 361
⇒ (a + b + c) = √361
∴ (a + b + c) = 19