Quant by Arun Sharma

MBA CAT 2024
5006
Q12. find the remainder when (10^3 + 9^3)^752 is divided by 12^3
  1. 729

  1. 1000
  2. 752
  3. 1

sol:


10^3 + 9^3 =1729


and we know 12^3=1728


so (1729)^752 / 1728 = R(1)
5007
Q11. find the remainder when 7^99 is divided by 2400
  1. 1
  2. 343
  3. 49
  4. 7

SOL:


(7^4)^48 * 7 = (2401)^24 *7^3 / 2400=R(343)

@ZIDDIARMAN,VYOMCONFUSED AND MDE4IIM THANKS FOR CORRECTING ME.:)

@ MDE4IIM-WHAT A WAY TO EXPRESS YOUR DISAGREEMENT:(
5008
Q11. find the remainder when 7^99 is divided by 2400
  1. 1
  2. 343
  3. 49
  4. 7

SOL:


(7^2)^48 * 7 = (49)^48 / 48 =R(1)


1*7/100 =R(7)


SINCE ANY NO. DIVIDED BY 100 IS THE NO. ITSELF.

i think the answer is wrong....

it should be 343...

7*7*7*7=2401...

so 2401/2400=1

99/4=3 remainder ... hence remainder should be 343
5009
This question is from # system. Question number 12 LOD 2

denotes the greatest integer value just below x and {x} its fractional value. The sum of ^3 and {x}^2 is -7.91. Find x.

Though the book says -1.7 , the justification is not very clear can some one please help me understand, How?

as far as is concerned is fine greatest integer value below -1.7 is -2 and ^3 is -8 but how come {x}^2 is equal to .09 cause 0.7^2=0.49

Please explain

Dude
fractional part of a number cannot be negative..
so in case of -1.7 = -1+ -1+ 1+ .7= -2 + .3
-2^3 =-8 .3^2= .09
=> -8+.09 =-7.91.....
5010
i think the answer is wrong....

it should be 49...

7*7*7*7=2401...

so 2401/2400=1

98/4=2 remainder ... hence remainder should be 49


bhai i think we will take 99/4

which will give 3 as remainder...so it will be 343
5011
i think the answer is wrong....

it should be 49...

7*7*7*7=2401...

so 2401/2400=1

98/4=2 remainder ... hence remainder should be 49


i think it sud be 99/4 -> 3 remainder
=> 343 sud be the answer
5012
Q4 if we divide a 2 digit no by the sum of its digits we get 4 as a quotient and 3 as a remainder. Now, if we divide that 2-digit no. by the product of its digits, we get3 as quotient and 5 as remainder. Find the no?

SOL:
10X+Y = 4(X+Y) + 3------------(1)

10X+Y = 3XY +5--------(2)

SOLVING (1)
WE GET 2X = Y+1

NOW PUT IT INTO (2)

5Y+5+Y = 3/2*(Y+1)Y + 5

Y=3

AND X=2

SO THE REQUIRED NO =23
5013
Q1. The greatest no. which will divide 4003,4126 and 4249:
  1. 43
  2. 41
  3. 45
  4. None of these



is it none of these???


4003 not divisible by neither 45 nor 43.
4003 is also not divisible 41.

mine also ...none of these
5014
Q13. how many values of x and y are such that 4x + 7y = 3, while x
  1. 144
  2. 141
  3. 143
  4. 142

SOL:


FIRST VALUE NEGATIVE VALUE OF X SATISFYING THE EQUATION ;- X=-1 WHERE Y=1


2ND X=-8 AND Y=5


3RD X=-15 AND Y =9


AND SO ON UPTO.......
...........


72ND VALUE


SIMILARLY FOR POSITIVE TERMS THERE WUD BE


72 VALUES ALSO


SO REQUIRED ANS==144 VALUES


CORRECT ME I AM WRONG.
5015

Q. M is a two digit number which has the property that : the product of factorials of it's digits > Sum of factorials of its digits. How many values of M exist?

Ans : 63

Can any1 explain this ???

5016
Q. M is a two digit number which has the property that : the product of factorials of it's digits > Sum of factorials of its digits. How many values of M exist?

Ans : 63

Can any1 explain this ???


sol:
1! * 1! 1! * 2! 1! * 3! 1! * 4! so as we can see from 10 to 20 ,THERE IS NO NUMBER which is holding the above rule - so ZERO NOS.
FROM 21 TO 30 - 23 TO 29 - 7 NOS.
31 TO 40 - 32 TO 39 - 8 NOS.
41 TO 50 - 42 TO 49 - 8 NOS.
51 TO 60 - 52 TO 59 - 8 NOS.
61 TO 70 - 62 TO 69 - 8 NOS.
71 TO 79 - 72 TO 79 - 8 NOS.
81 TO 89 - 82 TO 89 - 8 NOS.
91 TO 99 - 92 TO 99 - 8 NOS.

NO ADD ALL -63 NOS.:)
5017
Q10. x and y are 2 positive integers . Then what will be the sum of coefficients of the expansion of the expression (x+y)^44?
  1. 2^43

  1. (2^43) + 1

  1. 2^44

  1. (2^44) - 1

is it 2^44
5018
that's why this thread has been made.

GO AHEAD


sirjee !

how are you guys posting the questions? pure qstns type karne honge? or if i give qstn number wil do??

Thanks !! \m/
5019

Subscribing............

5020
Q10. x and y are 2 positive integers . Then what will be the sum of coefficients of the expansion of the expression (x+y)^44?
  1. 2^43

  1. (2^43) + 1

  1. 2^44

  1. (2^44) - 1

is it 2^44-1????????


Hello Rakesh

I think it should be 2^44.
Put x=1 and Y=1 and you will get the result.....
5021
sirjee !

how are you guys posting the questions?
? or if i give qstn number wil do??

Thanks !! \m/


yes buddy:)
5022

how will you come across all the digit, hit and trial ?

5023

actually in book the answer was given 41, that's why I got confused. Ans says HCF of 3 no. so 41. I think it is wrong answer

Q1. The greatest no. which will divide 4003,4126 and 4249:
  1. 43
  2. 41
  3. 45
  4. None of these



is it none of these???


4003 not divisible by neither 45 nor 43.
4003 is also not divisible 41.
5024
Q3. A 2 digit no. is thrice as large as the sum of its digits, and the sq. of that sum is equal to the trebled required no.. Find the no?
sol:
x+10y = 3(x+y) --------(1)

(x+y)^2 = (x+10y)*3 ----------------(2)

{(x+10y)/3}^2 = 3*(x+10y)

x+10y = 27..................required answer



no option was given
5025
Dude
fractional part of a number cannot be negative..
so in case of -1.7 = -1+ -1+ 1+ .7= -2 + .3
-2^3 =-8 .3^2= .09
=> -8+.09 =-7.91.....


i think you wanted to say

-1 + 1 - 1 - 0.7 = -2 + 0.3