Q: If p^x = q^y = r^z, where x, y and z are in GP, then consider the following statements: I. p, q and r are in AP. II. p, q and r are in GP. Which of the statements given above is/are correct?

Let p^x = q^y = r^z = k Taking natural logarithm on all sides: x p = y q = z r = k x = k p , y = k q , z = k r Since x, y, and z are in GP, y^2 = xz Substituting the values of x, y, and z: ( k q )^2 = ( k p ) ( k r ) 1 ( q)^2 = 1 p r ( q)^2 = p r This implies that p, q, and r are in GP. Statement II is correct, while Statement I is not necessarily true. Answer: II only

Question 1

If 1- _ 10 2= _ 10 (5^x+4^x+3^x+2^x+1), then what is a value of x ?

Accepted Answer

The given equation is 1 - _ 10 2 = _ 10 (5^x + 4^x + 3^x + 2^x + 1) Simplifying the left hand side: 1 - _ 10 2 = _ 10 10 - _ 10 2 = _ 10 ( 10 2 ) = _ 10 5 Equating this to the right hand side: _ 10 5 = _ 10 (5^x + 4^x + 3^x + 2^x + 1) Since the logarithmic function is one-to-one, we can equate the arguments: 5 = 5^x + 4^x + 3^x + 2^x + 1 5^x + 4^x + 3^x + 2^x = 4 Substituting x = 0 into the equation: 5^0 + 4^0 + 3^0 + 2^0 = 1 + 1 + 1 + 1 = 4 Thus, x = 0 satisfies the equation. Answer: 0

Question 2

What is the sum of the binary numbers (101101101)_2 and (100011)_2?

Accepted Answer

Converting the given binary numbers to decimal: (101101101)_2 = 1 2^8 + 0 2^7 + 1 2^6 + 1 2^5 + 0 2^4 + 1 2^3 + 1 2^2 + 0 2^1 + 1 2^0 (101101101)_2 = 256 + 0 + 64 + 32 + 0 + 8 + 4 + 0 + 1 = 365 (100011)_2 = 1 2^5 + 0 2^4 + 0 2^3 + 0 2^2 + 1 2^1 + 1 2^0 (100011)_2 = 32 + 0 + 0 + 0 + 2 + 1 = 35 Adding the decimal values: Sum = 365 + 35 = 400 Converting 400 back to binary: 400 = 256 + 128 + 16 = 2^8 + 2^7 + 2^4 Therefore, 400 = (110010000)_2 Answer: (110010000)_2

Question 3

For the following two (02) items: Let p = _ j=1 ^ n _ 10 2^j and q = _ j=1 ^ n _ 10 5^j. If p + q = 15, then what is q - p equal to?

Accepted Answer

Given p = _ j=1 ^ n _ 10 2^j and q = _ j=1 ^ n _ 10 5^j. Simplifying p and q: p = _ j=1 ^ n j _ 10 2 = _ 10 2 _ j=1 ^ n j = n(n+1) 2 _ 10 2 q = _ j=1 ^ n j _ 10 5 = _ 10 5 _ j=1 ^ n j = n(n+1) 2 _ 10 5 Adding p and q: p + q = n(n+1) 2 ( _ 10 2 + _ 10 5) = n(n+1) 2 _ 10 (2 5) = n(n+1) 2 _ 10 10 Since _ 10 10 = 1, we get: p + q = n(n+1) 2 Given p + q = 15, we have: n(n+1) 2 = 15 Now, finding q - p: q - p = n(n+1) 2 _ 10 5 - n(n+1) 2 _ 10 2 q - p = n(n+1) 2 ( _ 10 5 - _ 10 2) q - p = 15 _ 10 ( 5 2

Question 4

For the following two (02) items: Let p = _ j=1 ^ n _ 10 2^j and q = _ j=1 ^ n _ 10 5^j. If p + q = 66, then which one of the following is correct?

Accepted Answer

Given p = _ j=1 ^ n _ 10 2^j and q = _ j=1 ^ n _ 10 5^j Adding p and q: p + q = _ j=1 ^ n ( _ 10 2^j + _ 10 5^j) p + q = _ j=1 ^ n _ 10 (2^j 5^j) p + q = _ j=1 ^ n _ 10 (10^j) p + q = _ j=1 ^ n j _ 10 10 Since _ 10 10 = 1, we get: p + q = _ j=1 ^ n j = n(n+1) 2 Given p + q = 66: n(n+1) 2 = 66 n(n+1) = 132 n^2 + n - 132 = 0 (n - 11)(n + 12) = 0 Since n must be a positive integer, n = 11. The value n = 11 satisfies the condition 9 Answer: 9 < n < 12

Question 5

If p^x = q^y = r^z, where x, y and z are in GP, then consider the following statements: I. p, q and r are in AP. II. p, q and r are in GP. Which of the statements given above is/are correct?

Accepted Answer

Let p^x = q^y = r^z = k Taking natural logarithm on all sides: x p = y q = z r = k x = k p , y = k q , z = k r Since x, y, and z are in GP, y^2 = xz Substituting the values of x, y, and z: ( k q )^2 = ( k p ) ( k r ) 1 ( q)^2 = 1 p r ( q)^2 = p r This implies that p, q, and r are in GP. Statement II is correct, while Statement I is not necessarily true. Answer: II only

Basic of Mathematics — Previous Year Questions & Solutions