RRB, SSC, ADRE, Banking, and Other Competitive Exams
Q. If the product of two consecutive numbers is 71 more than their sum, find the numbers.
Soln
Let numbers be consecutive → x and x+1
Product - Sum = 71 → x(x+1)−(x+x+1) = 71
Simplify: x^- x - 72 = 0
Factor: (x−9)(x+8)=0 → x = 9
Ans: Numbers are 9 and 10
Trick: “Product - Sum = given number” → set up x(x+1) - (2x+1)=71and factor immediately.
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
Answer: 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 a^x = b^y = c^z and b^2 = 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, b^2 = ac
⇒ b^2 = 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. In 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.
Let the number of students = x
Then each contributes = ₹x
According to the question,
X × x = 6561
X^2 = 6561
Taking square root on both sides,
X = 81
Ans : 81Nos
Q. If a^3 – b^3 = 513 and a-b = 3 find ab.
Use identity: a^3 - b^3 = (a−b)(a^2+ab+b^2)
So 513 = 3 (a^2+ab+b^2)
⇒ a^2+ab+b^2 = 171
Also (a−b)^2 = a^2−2ab+b^2 = 9
Subtract the second from the first:
(a^2 + ab + b^2) - (a^2 - 2ab + b^2)
⇒ 3ab = 162
⇒ ab = 54
Ans: ab = 54