If `10! =2^p .3^q .5^r .7^s`, then `(A)` `2q = p` `(B)` `pqrs = 64` `(C)` number of divisors of `10!` is `280` `(D)` number of ways of putting `10!` as a product of two natural numbers is `135`

If p:0 is a natural number q:5 is a factor of 10, then

Find the number of divisors of the number N=2^3 .3^5 .5^7 .7^9 which are perfect squares.

The number of even divisor of the number N=12600=2^3 3^2 5^2 7 is a. 72 b. 54 c. 18 d. none of these

If p:4 is an even prime number q:6 is a divisor of 12 and r: The HCF of 4 and 6 is 2, then which of the following is true.

Let z be a complex number satisfying equation z^p=z^(-q),w h e r ep ,q in N ,t h e n (A) if p=q , then number of solutions of equation will be infinite. (B) if p=q , then number of solutions of equation will be finite. (C) if p!=q , then number of solutions of equation will be p+q+1. (D) if p!=q , then number of solutions of equation will be p+qdot

The standard deviation of first 10 natural numbers is 8.25 (b) 6.5 (c) 3.87 (d) 2.87

Match each of the set on the left in the roster form with the same set on the right described in set-builder form: (i) {1, 2, 3, 6} (a) {x : x is a prime number and a divisor of 6} (ii) {2, 3} (b) {x : x is an odd natural number less than 10} (iii) {M,A,T,H,E,I,C,S} (c) {x : x is natural number and divisor of 6} (iv) {1, 3, 5, 7, 9} (d) {x : x is a letter of the word MATHEMATICS}.

Consider the ten numbers a r ,a r^2, a r^3, ,a r^(10)dot If their sum is 18 and the sum of their reciprocals is 6, then the product of these ten numbers is 81 b. 243 c. 343 d. 324

The number of solutions of sum_(r=1)^5cosrx=5 in the interval [0,2pi] is 0 (b) 2 (c) 5 (d) 10

Match the roster form with set builder form. (i) {1,2,3,6} (a) {x : x is prime number and a divisior of 6} (ii) {2,3} (b) {x : x is an odd natural number smaller than 10} (iii) {m,a,t,h,e,I,c,s} (c) {x : x is a natural number and divisor of 6} (iv) {1,3,5,7,9} (d) {x : x is a letter of the word MATHEMATICS}