Classical

Numbers & Algebra - Numbers and Algebra MCQ

6:  

 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: C

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


7:  

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


8:  

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.


9:  

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.


10:  

HCF and LCM of two numbers is given. It is possible to ind out the two numbers uniquely if
I. either sum or difference between the two numbers is known.
II. HCF of two numbers = LCM of two numbers.
III. (LCM/HCF)= Prime number.

 

A.

I and II only

B.

II  only

C.

II and III only

D.

I, II and III

 
 

Option: D

Explanation :

Let HCF be h and LCM be l.
I.  Let, numbers be ah and bh. Then abh = and (a + b)h = m
=> (a — b)h = n
Using these ah and bh can be uniquely determined. Thus, I is true II.
II.  If HCF = LCM, then two numbers are equal and same as HCF or LCM. Thus, II is true.
III. LCM/HCF = prime i.e. l/h =P Then one of the numbers is equal to h and other is equal to E. Thus, III is true.




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