Classical

Numbers & Algebra - Numbers and Algebra MCQ

6:  

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

 
 

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

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7:  

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

A.

10H < g! < H

B.

H < g! < 10H

C.

10H < g! < 10

D.

10H < g! <10H

 
 

Option: D

Explanation :

Choose k, so that 10H = gλ= 10100k
k = H/100
10H = g H/100 >gg > g!
On the other hand, 10H = 10 x 10g = 10g+1 < g! (most factors are much bigger than 10)
10H < g! < 10H

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8:  

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

 
 

Option: A

Explanation :

(GM)2= HM x AM
HM = (8x8)/ 5 = 64/5 = 12.8

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9:  

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

 
 

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.

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10:  

When x + y + z = 9 and xy + yz + zx = 11, then x3 - y3 - z3 - 3xyz equals

A.

384

B.

192

C.

432

D.

48

 
 

Option: C

Explanation :

x3 - y3 - z3 - 3xyz = (x+y+z)(x2+y2+z2-xy-yz-zx) = 9[(x + y + z)2 - 3(xy + yz + zx)] = 9(81 - 33) = 432.

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