Quantitative Aptitude - Numbers and Algebra
- 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
7. If g = 10100 and H = 10g, then in which interval does g! = 1 . 2 .3 ........ 10100 lie?
- 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
- 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.
10. When x + y + z = 9 and xy + yz + zx = 11, then x3 - y3 - z3 - 3xyz equals
- Option : C
- Explanation :
x3Â - y3Â - z3Â - 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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