Answer: A If the number is divisible by 5 and 11 it must be divisible by 55. The numbers are less than 660. Hence, dividing 659 by 55 gives the number of multiples of 55 = 11(ignoring fraction part). The 11 multiples of 55 which are less than 560, but of these 11 multiples some can be multiples of 3. The numbers of such ,multiples is the quotient of 11 by 3. Quotient of 11/3 = 3. Out of 11 multiples of 55. 3 are multiples of 3. Hence, numbers less than 660 and divisible by 5 and 11 but not by 3 = 11-3 =8.
Q. No. 8:
The two positive integers 'p' and 'q' satisfy (p+q)/t = HCF(p,q). Which of the following two numbers sum up to 't'?
Answer: D Let HCF(p,q) = H For two numbers A and B, which are co primes to each other. we can represent p and q as p = H*A and q= H*B So that (p+q)/t = HCF(p,q) = H => t = (A+B) Hence t cab be represented as the sum of any two numbers which are co prime to each other.
Q. No. 9:
What is the unit's digit of the number 6256 - 4256?
Answer: B Since the exponents are even , we can apply the property that, If 'x' is even ax - bx is always divisible by a+b.6256 - 4256 is always divisible by 10.Now any number multiplied by 10 gives the last digit as 'zero'.Alternative:- The last digit of both the number are same as '6'Thus after subtracting the unit's digit be '0'.
Q. No. 10:
The least common multiple of two natural numbers a and b, is 399. What is the minimum possible sum of the digits of the number a(given a > b)?
Answer: B Factorize 399 = 3*7*19 The possible pairs are (57,7), (21,19), (399,1), (133,3) The least possible sum is given when a =21 and b =19 And the sum is 2+1 = 3.
Q. No. 11:
Suppose n is an integer such that the sum of digits on n is 2, and 1010 < n*10n. The number of different values of n is
Answer: A We have, I).1010 < n < 1011 II).Sum of the digits for 'n' = 2 Clearly- (n)min = 10000000001 (1 followed by 9 zeros and finally 1) Obviously, we can form 10 such numbers by shifting '1' by one place from right to left again and again. Again, there is another possibility for 'n' n = 20000000000 So finally : No. of different values for n = 10 + 1 = 11
Q. No. 12:
Four digits of the number 29138576 are omitted so that the result is as large as possible. The largest omitted digit is
Answer: D We have to omit four digits in such a manner that we get the largest possible number as the result. We will omit 2, 1, 3 and 5 and the result will be 9876. The largest omitted digit will be 5.