Home
Class 12
MATHS
Let X and Y be the sets of all positive ...

Let X and Y be the sets of all positive divisions of400 and 1000 respectively (including 1 and the number). Then `n(X cap Y)` is T, then `(T)/(4)` is-

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we need to find the sets of positive divisors for 400 and 1000, determine their intersection, and then calculate \( \frac{T}{4} \). ### Step 1: Find the positive divisors of 400 To find the positive divisors of 400, we can start by determining its prime factorization. 1. **Prime Factorization of 400**: \[ 400 = 2^4 \times 5^2 \] 2. **Finding Divisors**: The formula for finding the number of divisors from the prime factorization \( p_1^{e_1} \times p_2^{e_2} \) is \( (e_1 + 1)(e_2 + 1) \). - For \( 400 = 2^4 \times 5^2 \): \[ \text{Number of divisors} = (4 + 1)(2 + 1) = 5 \times 3 = 15 \] 3. **List of Divisors**: The positive divisors of 400 are: \[ 1, 2, 4, 5, 8, 10, 16, 20, 25, 40, 50, 80, 100, 200, 400 \] ### Step 2: Find the positive divisors of 1000 Next, we will find the positive divisors of 1000. 1. **Prime Factorization of 1000**: \[ 1000 = 10^3 = (2 \times 5)^3 = 2^3 \times 5^3 \] 2. **Finding Divisors**: - For \( 1000 = 2^3 \times 5^3 \): \[ \text{Number of divisors} = (3 + 1)(3 + 1) = 4 \times 4 = 16 \] 3. **List of Divisors**: The positive divisors of 1000 are: \[ 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500, 1000 \] ### Step 3: Find the intersection of sets X and Y Now, we will find the intersection of the two sets of divisors. 1. **Divisors of 400 (Set X)**: \[ X = \{1, 2, 4, 5, 8, 10, 16, 20, 25, 40, 50, 80, 100, 200, 400\} \] 2. **Divisors of 1000 (Set Y)**: \[ Y = \{1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500, 1000\} \] 3. **Finding Intersection \( X \cap Y \)**: The common elements in both sets are: \[ X \cap Y = \{1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 200\} \] ### Step 4: Count the number of elements in the intersection The number of elements in \( X \cap Y \) is: \[ T = 12 \] ### Step 5: Calculate \( \frac{T}{4} \) Now we will calculate \( \frac{T}{4} \): \[ \frac{T}{4} = \frac{12}{4} = 3 \] ### Final Answer Thus, the final answer is: \[ \frac{T}{4} = 3 \]
Promotional Banner

Topper's Solved these Questions

  • SETS

    AAKASH INSTITUTE ENGLISH|Exercise SECTION-H(Multiple True-False type Questions)|2 Videos
  • SETS

    AAKASH INSTITUTE ENGLISH|Exercise SECTION-I(Aakash Challengers Questions)|4 Videos
  • SETS

    AAKASH INSTITUTE ENGLISH|Exercise SECTION-F(Matrix-Match Type Questions)|1 Videos
  • SEQUENCES AND SERIES

    AAKASH INSTITUTE ENGLISH|Exercise Assignment (SECTION - J) Aakash Challengers|11 Videos
  • STATISTICS

    AAKASH INSTITUTE ENGLISH|Exercise Section-C Assertion-Reason|15 Videos

Similar Questions

Explore conceptually related problems

Let X= the sets of all letters in the word "DELHI" and Y=the sets of all letters in the word "INDIA" Find X cup Y .

If x = 4y + 5 and y = kx + 4 are the lines of regression on X on Y and Y on X respectively, where k is a positive constant, prove that it can't exceed (1)/(4) .

Let X be the set of all positive such that f(x+y) = f(xy) for all x ge 4, y ge 4 . If f(8)= 9 , then f(9) is equal to.

The number of the positive integral solutions (x,y, z) of the equation xyz = 24 is t, then the number of all possible factors of t is

if X and Y are two sets , then X cap (Y cup X)' equals

The x and y coordinates of the particle at any time are x=5t-2t^(2) and y=10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t=2s is:

The x and y coordinates of the particle at any time are x=5t-2t^(2) and y=10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t=2s is:

The x and y coordinates of the particle at any time are x = 5t - 2t^(2) and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at I = 2s is

Consider radioactive decay of A to B with which further decays either to X or Y , lambda_(1), lambda_(2) and lambda_(3) are decay constant for A to B decay, B to X decay and Bto Y decay respectively. At t=0 , the number of nuclei of A,B,X and Y are N_(0), N_(0) zero and zero respectively. N_(1),N_(2),N_(3) and N_(4) are the number of nuclei of A,B,X and Y at any instant t . The net rate of accumulation of B at any instant is

Let R be a relation on the set of all real numbers defined by xRy iff |x-y|leq1/2 Then R is