Numbers & Algebra - Numbers and Algebra MCQ

If x = b + c, y = c a, z = a b, then
x2 + y2 + z2 - 2xy - 2xz + 2yz is equal to

A.	a + b + c
B.	4b²
C.	abc
D.	a²+ b²
	Answer Report Discuss
	Option: B Explanation : x²+ y² + z²-2xy - 2xz + 2yz = (x - y - z)² =(b+c-c+a-a+b)²= 4b² Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

An Egyptian fraction has a numerator equal to 1, and its denominator is a positive integer. What is the maximum number of different Egyptian fraction such that their sum is equal to 1, and their denominators are equal to 10 or less?

A.	3
B.	5
C.	7
D.	9
	Answer Report Discuss
	Option: A Explanation : We ignore 1/7 and 1/9 because no sum of other denominator.numbers is going to give 7ths or 9ths in the denominator Also, 1/5 and 1/10 are not enough to add up to anything (1/10, 2/10 and 3/10 are going to leave tenths left over no matter what else you add) What's left is 1/2, 1/3, 1/4, 1/6, 1/8. Sum total of these is 11/8. So we need all of them except 3/8, which means 1/2+1/3+1/6. Which is the only way to do this with egyptian fractions whose denominators are 10 or less. Hence maximum number of Egyptian fractions needed is 3 Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

If g = 10100 and H = 10g, then in which interval does g! = 1 . 2 .3 ........ 10100 lie?

A.	10H < g! < H
B.	H < g! < 10^H
C.	10H < g! < 10
D.	10H < g! <10^H
	Answer Report Discuss
	Option: D Explanation : Choose k, so that 10^H = g^λ= 10^100k k = H/100 10^H = g^H/100>g^g > g! On the other hand, 10H = 10 x 10^g= 10^g+1 < g! (most factors are much bigger than 10) 10H < g! < 10^H Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

What is the harmonic mean of two numbers whose geometric mean and arithmetic mean is 8 and 5 respectively?

A.	12.8
B.	12
C.	13.5
D.	14.6
	Answer Report Discuss
	Option: A Explanation : (GM)²= HM x AM HM = (8x8)/ 5 = 64/5 = 12.8 Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

10:

Anil wants to divide Rs.100 into a number of bags so that one can ask for any amount between Rs.1 and Rs.100, he can give the proper amount by giving certain number of these bags without taking out the amount from them. What is the minimum number of bags he will require if each bag has whole number of rupees?

A.	5
B.	6
C.	7
D.	8
	Answer Report Discuss
	Option: C Explanation : If Anil has to give 1 rupee he needs a bag with Re.1. For 2 rupees he had two bags with Re.1 each or Rs.2. bag. To have minimum bags, he has a bag with Rs. 2. Now with the two bags he can give Rs.3. So next he will require a bag with Rs.4. With these three he can give Rs.5. Rs.6 and Rs.7 and next bag will be one containing Rs.8 and so on. Thus he would have bags with Re.1. 2. 4. 8. 16. 32. Sum of which is 63 and remaining 37 can be put in the last bag. So total number of bags is 7. Click on Discuss to view users comments. Write your comments here: Updating your Comments.. Updating your Comments..

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