Which of the following is a finite decimal or terminating decimal?
\( \frac{1}{24} \)
\( \frac{1}{32} \)
\( \frac{1}{96} \)
\( \frac{1}{48} \)
A fraction represents a terminating decimal if and only if the prime factorization of its denominator contains only the prime numbers 2 and/or 5. We will simplify each given fraction to its lowest terms and then check the prime factors of its denominator.
The fraction is \( \frac{1}{24} \). The denominator is \( 24 \). Prime factorization of \( 24 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3^1 \). Since the prime factorization of the denominator contains a factor of \( 3 \) (which is not \( 2 \) or \( 5 \)), \( \frac{1}{24} \) is a non-terminating decimal.
The fraction is \( \frac{1}{32} \). The denominator is \( 32 \). Prime factorization of \( 32 = 2 \times 2 \times 2 \times 2 \times 2 = 2^5 \). Since the prime factorization of the denominator contains only the prime factor \( 2 \), \( \frac{1}{32} \) is a terminating decimal.
The fraction is \( \frac{1}{96} \). The denominator is \( 96 \). Prime factorization of \( 96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^5 \times 3^1 \). Since the prime factorization of the denominator contains a factor of \( 3 \), \( \frac{1}{96} \) is a non-terminating decimal.
The fraction is \( \frac{1}{48} \). The denominator is \( 48 \). Prime factorization of \( 48 = 2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3^1 \). Since the prime factorization of the denominator contains a factor of \( 3 \), \( \frac{1}{48} \) is a non-terminating decimal.
Correct Option: B) \( \frac{1}{32} \)
We are given two conditions involving the LCM and HCF of two positive numbers: their LCM is three times their HCF, and the product of their LCM and HCF is 867. We need to find the numbers. We will use the fundamental property that for any two positive numbers, the product of the numbers is equal to the product of their LCM and HCF.
Let the two positive numbers be \(a\) and \(b\).
Let HCF be \(H\) and LCM be \(L\).
From the first condition, we are given that the LCM is three times their HCF:
\[ L = 3H \quad \text{(Equation 1)} \]From the second condition, we are given that the product of the LCM and the HCF is 867:
\[ L \times H = 867 \quad \text{(Equation 2)} \]Substitute Equation 1 into Equation 2:
\[ (3H) \times H = 867 \]\[ 3H^2 = 867 \]Solve for \(H^2\):
\[ H^2 = \frac{867}{3} \]\[ H^2 = 289 \]Solve for \(H\):
\[ H = \sqrt{289} \]\[ H = 17 \]Now, substitute the value of \(H\) back into Equation 1 to find \(L\):
\[ L = 3 \times 17 \]\[ L = 51 \]We know that for any two positive numbers \(a\) and \(b\), the product of the numbers is equal to the product of their LCM and HCF:
\[ a \times b = L \times H \]\[ a \times b = 51 \times 17 \]\[ a \times b = 867 \]We also know that the HCF of the two numbers is 17. This means both numbers must be multiples of 17. Let the numbers be \(17x\) and \(17y\), where \(x\) and \(y\) are coprime (their HCF is 1).
The product of the numbers is \(17x \times 17y = 289xy\). We found this product to be 867.
\[ 289xy = 867 \]\[ xy = \frac{867}{289} \]\[ xy = 3 \]Since \(x\) and \(y\) must be coprime integers, the only possible pair for \((x, y)\) (or \((y, x)\)) is \((1, 3)\).
Therefore, the two numbers are:
A) 17, 51 — The numbers are 17 and 51. Their HCF is 17, and their LCM is 51. \(L = 3H\) (\(51 = 3 \times 17\)) and \(L \times H = 51 \times 17 = 867\). This matches all conditions.
The type of image formed by a concave mirror depends on the position of the object relative to its focal point (F) and center of curvature (C). When an object is placed between the pole (P) and the focal point (F) of a concave mirror, the rays of light diverge after reflection, appearing to originate from a point behind the mirror. This results in a specific type of image.
A) Virtual — When an object is placed between the pole and the focal point (F) of a concave mirror, the reflected rays diverge and appear to intersect behind the mirror, forming a virtual image. This image is also erect and magnified.
To solve this calendar problem, we need to calculate the total number of odd days between the given date (8 December 2007) and the target date (12 August 2008). Odd days are the remainder when the total number of days is divided by 7. We must also account for leap years.
December has 31 days. Days remaining in December 2007 = \(31 - 8 = 23\) days.
Total days = \(23 (\text{Dec 2007}) + 31 (\text{Jan 2008}) + 29 (\text{Feb 2008}) + 31 (\text{Mar 2008}) + 30 (\text{Apr 2008}) + 31 (\text{May 2008}) + 30 (\text{Jun 2008}) + 31 (\text{Jul 2008}) + 12 (\text{Aug 2008})\)
Total days = \(23 + 31 + 29 + 31 + 30 + 31 + 30 + 31 + 12 = 248\) days.
Odd days = Total days \(\pmod{7}\)
\[ 248 \div 7 = 35 \text{ with a remainder of } 3 \]So, there are 3 odd days.
Given 8 December 2007 was a Saturday. We need to add 3 odd days to Saturday.
Therefore, 12 August 2008 was a Tuesday.
A) Tuesday
Find the value of x2 - y2
1. If x + y = 2
2. If x - y = 6
Only 1 is sufficient
Neither 1 nor 2 is sufficient
Only 2 is sufficient
Both 1 and 2 together are sufficient
The problem asks us to find the value of \(x^2 - y^2\) given two statements. We need to determine if statement 1 alone, statement 2 alone, or both statements together are sufficient to find the value.
Analyze Statement 1:
Statement 1 says: \(x + y = 2\).
If we only have this statement, we know the value of \((x+y)\). However, we do not know the value of \((x-y)\). Therefore, we cannot find the value of \(x^2 - y^2\) using only statement 1.
Analyze Statement 2:
Statement 2 says: \(x - y = 6\).
If we only have this statement, we know the value of \((x-y)\). However, we do not know the value of \((x+y)\). Therefore, we cannot find the value of \(x^2 - y^2\) using only statement 2.
Analyze Statements 1 and 2 together:
If we use both statements together, we have:
Now, we can substitute these values into the identity \(x^2 - y^2 = (x + y)(x - y)\):
\[ x^2 - y^2 = (2)(6) \]
\[ x^2 - y^2 = 12 \]
Since we can find a unique value for \(x^2 - y^2\) using both statements together, both statements are sufficient.
D) Both 1 and 2 together are sufficient
The question asks to identify the option that is different or mismatched from the others. This is a classification-based question where we need to find a common characteristic among three options and identify the one that does not share that characteristic.
C) Heart — Hand, Nose, and Eyes are all external parts of the human body or sense organs. The Heart, however, is an internal organ. Therefore, 'Heart' is the mismatched option.
The question asks to identify the statement that is NOT true regarding the effect of increasing temperature on the distribution of molecular speeds in a gas. This relates to the Maxwell-Boltzmann distribution of molecular speeds.
D) The fraction of molecules increases with the most possible speed. This statement is incorrect. While the most probable speed itself increases with temperature, the fraction of molecules possessing that specific speed actually decreases because the distribution curve broadens and flattens.
The question asks to identify the specific coast in Kerala known for its backwaters and coconut trees. This requires knowledge of the geographical divisions of the Indian coastline.
Correct Option: B) The Malabar coast in Kerala is a beautiful backwater lined with coconut trees. This region is renowned for its unique backwater ecosystem, including lakes, lagoons, and canals, which are integral to the local culture and economy, and are indeed heavily lined with coconut trees.
The question asks to identify the second Indian recipient of the Ramon Magsaysay Award in 2018, given that Sonam Wangchuk was one of them. This requires knowledge of the awardees for that specific year.
Correct Option: A) Bharat Vatwani
To find the sum of decimal numbers, align the decimal points vertically and then add the numbers column by column, starting from the rightmost digit.
Correct Option: D) 4.9995