RRB, SSC, ADRE, Banking, and Other Competitive Exams

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Q. If the product of two consecutive numbers is 71 more than their sum, find the numbers.

Solⁿ

Let the consecutive numbers be x and x+1. The product of the numbers is x(x+1) and their sum is x + (x+1). According to the question:

ক্রমিক সংখ্যাদ্বয় x আৰু x+1 ধৰা হ’ল। সংখ্যাদ্বয়ৰ গুণফল হৈছে x(x+1) আৰু যোগফল হৈছে x+(x+1)। প্ৰশ্ন অনুসৰি:

Ans / উত্তৰ: 9 and 10

Q. If x + y = 9 and xy = 18, find x² + y². Option : (a) 18 (b) 36 (c) 12 (d) 45

Formula: x² + y² = (x+y)² − 2xy

Substitute: = 9² - 2(18) = 81−36 = 45

Ans: 45

Q. If (a + b)(a – b) = 25(a + b), find a – b. Option : (a) 18 (b) 25 (e) 30 (d) 15

Given:

(a+b)(a–b) = 25(a+b)

If a+b ≠ 0 , divide both sides by (a + b):

a – b = 25

Ans: 25

Q. If (y + z) = 26 and (y – 2)² = 225, find the value of z. Option : (a) 10 (b) 5 (c) 9 (d) 7

(y–2)² = 225 ⇒ y – 2 = ±15

Case 1:
If y – 2 = 15, then y = 17.

Z = 26 – 17 = 9

Case 2:
If y – 2 = -15, then y = -13.

Z = 26 - (-13) = 39

From the options (10, 9, 7), the valid one is 9.

Ans: 9

Q. If y² – 15y + 56 = 0, find the value of y.

Solⁿ:
y² – 15y + 56 = 0
⇒ y² – 7y – 8y + 56 = 0
⇒ y(y – 7) – 8(y – 7) = 0
⇒ (y – 7)(y – 8) = 0

So, y – 7 = 0 ⇒ y = 7
or y – 8 = 0 ⇒ y = 8

Ans: y = 7 or 8

Q. If If a^x = b^y = c^z   and b² = ac , Prove that 1/x + 1/z = 2/y.

Solⁿ.

Given: a^x = b^y = c^z  

⇒ a^x = b^y ⇒ a = b^y/x

 ⇒ b^y = c^z   and c = b^y/z

Now, b² = ac

⇒ b² = b^y/x × b^y/z = b^{(y/x + y/z)}

Since bases “b”are the same, equate exponents:

⇒ 2 = y/x + y/z

Divide both sides by y : ⇒ 2/y = 1/x + 1/z

Hence proved: 1/x + 1/z = 2/y

Q. n a class, if each student contributes as many rupees as the total number of students, the total amount collected is ₹6561. Find the number of students. এটা শ্ৰেণীত প্ৰতিজন ছাত্ৰই শ্ৰেণীৰ মুঠ ছাত্ৰসংখ্যাৰ সমান টকা দিলে, মুঠ সংগৃহীত টকা ₹6561 হয়। ছাত্ৰসংখ্যা উলিয়াওক।

Let the number of students be x.

Each student contributes x rupees.

So, x² = 6561

x = √6561​ = 81

So, the number of students is 81.

Ans / উত্তৰ: 81

Q. If a³ – b³ = 513 and a-b = 3 find ab.

Use identity: a³ - b³ = (a−b)(a²+ab+b²)

So 513 = 3 (a²+ab+b²)

⇒ a²+ab+b²= 171

Also (a−b)² = a²−2ab+b² = 9

Subtract the second from the first:

(a² + ab + b²) - (a² - 2ab + b²)

⇒ 3ab = 162

⇒ ab = 54

Ans: ab = 54