ব্যাংক নিয়োগ প্রস্তুতি ⎯ লং কোর্স

Question: If 1 < p < 4 and 2 < q < 6, which of the following best describes p - q?

Solution: দেয়া আছে:
1 < p < 4 -------------(1)
2 < q < 6 -------------(2)

এখন, আমরা p - q এর সীমা বের করতে চাই।

(2)⇒
2 < q < 6
⇒ - 2 > -q > - 6 (যদি -1 দ্বারা গুণ করি)
⇒ - 6 < -q < - 2 ----------(3)

(1) এবং (3) যোগ করি,
⇒ (1 + (- 6 )) < p - q < (4 + (- 2))
⇒ - 5 < p - q < 2

Question: If X ∈ N and 17 < x < 23, and x is a prime number, then which of the following represents the list form of the set of such numbers?

Solution:
দেয়া আছে:
X ∈ N and 17 < x < 23

List all natural numbers between 17 and 23
⇒ 18, 19, 20, 21, 22

∴ Identify the prime numbers among them
⇒ 18 → divisible by 2; not prime.
⇒ 19 → prime.
⇒ 20 → divisible by 2; not prime.
⇒ 21 → divisible by 3 and 7; not prime.
⇒ 22 → divisible by 2; not prime.

∴ List of prime numbers in this range
{19}.

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 

Question: If x² + y² = 50 and xy = 21, what is the value of (x - y)²?
 
Solution:
We are given:
x² + y² = 50
xy = 21

Use the identity:
(x - y)² = x² + y² - 2xy

Substitute the values:
⇒ (x - y)² = x² + y² - 2xy
⇒ (x - y)² = 50 - 2 × 21
⇒ (x - y)² = 50 - 42
∴ (x - y)² = 8

Question: If |2x - 3| ≤ 9, then which of the following intervals represents all possible values of the expression 5x + 7?

Solution:
Start with the given inequality:
|2x - 3| ≤ 9

Rewrite as a compound inequality:
- 9 ≤ 2x - 3 ≤ 9
⇒ - 9 + 3 ≤ 2x - 3 + 3 ≤ 9 + 3
⇒ - 6 ≤ 2x ≤ 12
⇒ - 3 ≤ x ≤ 6

Now, find the range of 5x + 7:
Multiply the interval by 5:
5(- 3) ≤ 5x ≤ 5(6) 
⇒ - 15 ≤ 5x ≤ 30
⇒ - 15 + 7 ≤ 5x + 7 ≤ 30 + 7 
⇒ - 8 ≤ 5x + 7 ≤ 37

So, all possible values of 5x + 7 lie in the interval: [ - 8, 37 ]

Question: In a class, 30 students play basketball, 20 students play volleyball, and 8 students play both. 12 students play neither basketball nor volleyball. What is the total number of students in the class?

Solution:
Let the number of students who play basketball = 30
Number of students who play volleyball = 20
Number of students who play both basketball and volleyball = 8
Number of students who play neither = 12

First, calculate the number of students who play basketball or volleyball:
n(B ∪ V) = n(B) + n(V) − n(B ∩ V)
n(B ∪ V) = 30 + 20 − 8 = 42

Now, add the students who play neither sport to get total students:
Total students = n(B ∪ V) + neither

Total students = 42 + 12 = 54

Question: If p and q are the roots of the equation 3x²− 7x + 2 = 0, then what is the value of (1/p) + (1/q)?
 
Solution:

3x²− 7x + 2 = 0
⇒ 3x²- 6x - x + 2 = 0
⇒ 3x (x - 2) - 1 (x - 2) = 0
⇒ (3x - 1)(x - 2) = 0
⇒ x = 1/3 = p
∴ x = 2 = q

Now,
1/p + 1/q
= 1/(1/3) + 1/2
= 3 + 1/2
= 7/2

Question: If x = 2 + √3 and y = 2 - √3, find the value of (x² + y²)²

Solution:
We are given:
x = 2 + √3, y = 2 - √3

Now,
x + y
= (2 + √3) + (2 - √3)
= 4

And,
⇒ xy
= (2 + √3)(2 - √3)
= 2² - (√3)²
= 4 - 3 = 1

We know,
x² + y² = (x + y)² - 2xy
⇒ x² + y² = (4)² - 2(1) [Substitute the values]
⇒ x² + y² = 16 - 2
⇒ x² + y² = 14

∴ (x² + y²)² = 14² = 196

Question: Nine times a whole number is equal to five less than twice the square of the number. Find the number?

Solution: Let the required whole number be x.

According to the question,
9x = 2x² - 5
⇒ 2x² - 9x - 5 = 0
⇒(x - 5)(2x + 1) = 0
⇒ x - 5 = 0 or 2x + 1 = 0
⇒ x = 5 or x = - 1/2

Since x is supposed to be a whole number, the answer, i.e., the required whole number is 5.

Question: If dividing P(x) = 4x³- 7x² + bx - 5 by (x - 3) results the remainder 10, then find the value of b.

Solution:
According to the Remainder Theorem, if a polynomial P(x) is divided by (x - c), then the remainder = P(c).
Here divisor is (x - 3).
So remainder = P(3).

Now,
P(3) = 4(3)³ - 7(3)² + b(3) - 5

= 4 × 27 - 7 × 9 + 3b - 5

= 108 - 63 + 3b - 5

= 40 + 3b

According to the question, the remainder is 10.
So, 40 + 3b = 10

⇒ 3b = 10 - 40
⇒ 3b = - 30
⇒ b = - 10

Question: Find the product of two consecutive numbers if three times the first number is 5 more than twice the second number.

Solution:
Let the numbers be a and a + 1.

According to the question:
3 × (first number) = 2 × (second number) + 5
⇒ 3a = 2(a + 1) + 5
⇒ 3a = 2a + 2 + 5
⇒ 3a = 2a + 7
⇒ 3a - 2a = 7
⇒ a = 7

∴ The numbers are 7 and 8.
Product = 7 × 8 = 56

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

Solution:
(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

Question: If (x + 5)² = 64, which of the following can be the value of (x + 2)?

Solution:
Given, (x + 5)² = 64
⇒ (x + 5)² = 8²
∴ x + 5 = ± 8

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

Case 2: x + 5 = - 8
x = -8 - 5 = -13
x + 2 = -13 + 2 = - 11

Possible values of (x + 2) are 5 or - 11.

Correct Answer: ক) 5

Question: If x² + yz + zx + xy is divided by x + y, 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 + y) the result = (x + y)(x + z)/(x +y) = (x + z)

Question: If one root of x² - (q + 2)x + 15 = 0 is 3, then the value of q is:

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

One root is x = 3. Substitute:
(3)² - (q + 2)(3) + 15 = 0
⇒ 9 - 3(q + 2) + 15 = 0
⇒ 9 - 3q - 6 + 15 = 0
⇒ 18 - 3q = 0
⇒ 3q = 18
∴ q = 6