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# Miscellaneous

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A positive whole numbers m less than  100 is represented in base 2 notation , base 3 notation, and base 5 notation . It is found that in all three cases the last digit is 1, while in exactly 2 out of 3 cases the leading digit is 1 . Then M equals?

Since in all three cases the last digit is 1, the number should give remainder 1 when divided individually by 2,3,5 . So the no. may be 31 or 91 . Now 31 in base 2,3 and 5 give first digit as 1 in all the 3 cases while 91 gives exactly two out of the three cases the leading digit as 1

In precise it is simply saying that out of 4 option which no. when converted to base 2,3,5 will have last digit 1 for all three base(2,3,5) conversion and any two conversion will have leading digit as 1. let us see for 31(base 2 conversion)-11111 31(base 3 conversion)-1011 31(base 5 convertion0-111 here we see that all there conversion is having last as well as leading digit as 1 so second condition of having any two with leading 1 is not satisfied try with others. 63(base 5 conversion)---(223) it do not satisfy 1st condition so this will be incorrect. 75(base 5 conversion)---(300) it do not satisfy 1st condition so this will be incorrect. 91(base 5 conversion)---(331) it satisfy 1st condition. 91(base 3 conversion)---(10101) it satisfy 1st condition. 91(base 2 conversion)---(1011011) it satisfy 1st condition. so 91 is satisfying second condition also of having any 2 leading digit as 1.

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In a sale, out of 30 buyers, 10 buyers bought 5 or fewer items, 22 buyers bought fewer than 8 items and 14 buyers bought at least 7 items. Find the numbers of buyers who bought exactly 6 items.

8 ??

correct ans:6

I hope this figure helps

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is it 24?

Yes. How did you do it?

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sin36 sin72 sin108 sin144=?

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please provide the solution for all the ques.

1.
_ _ _ _ 0 _ _ _ _ :

for first four digits there are 9C4 ways , for last four digits there are 5C4 ways.
=> 9C4 x 5C4 = 630 numbers

_ _ _ _ 1 _ _ _ _ :
For first four digits select 4 digits from { 2, 3, 4, 5, 6, 7, 8, 9, } in 8C4 ways and last four in 4C4 ways.
=> 8C4 x 4C4 = 70 numbers

Total : 630 + 70 => 700 numbers .

2nd

let AB=a, BC=b, CA=c

now a+b+c=2

and we have to find the max value of ab+bc which can also be written as b(a+c)

from given equation a+c=2-b

b(a+c)= b(2-b)

=2b-b2

for any quadratic eq min value is -D/4a but in this condition since term of b2 is -ve graph will be inverted curve so min value will become maximum value

-D/4a= -4/-4=1 hence option 1st

From 112 to 118 => sum of the product of the digits will be (1 x 1 x 2 )+ (1 x 1 x 4) , (1 x 1 x 6) , ......, (1x 1 x 8)

From 122 to 118 =>sum of the  product of the digits will be (1 x 2 x 2) + (1 x 2 x 4) , (1 x  2 x 6) ,....., (1 x 2 x 8).

From 133 to 118 => sum of the product of the digits will be ( 1 x 3 x 2 ) + ( 1 x 3 x 4 ) + ........+ ( 1 x 3 x 8)

from 112 to 198

sum of all products : (2 + 4 + 6 + 8) x ( 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 ) i.e 20 x 45

and for 212 to 298 sum of all products 20 x 45 x 2 , for 312 to 398 => 20 x 45 x 3 and so on ...

Hence sum of all such products : 20 x 45 x ( 1 + 2 + 3 + ....+ 9) = 40500 .

Last three digits : 500

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Hello sir , kindly share the solution to this problem -

A retailer has 'n' stones of distinct integral weights (in kg) by which he can measure from 1 kg to 11 kg keeping them on either side of the weighing machine. Find the minimum value of 'n'.

1 kg weight is must

then for 2 kg weight we can have =1+1 or 2 0r 3-1

but 3-1 is best of them because then we can get 3 and 3+1=4 also from them

so for 1 to 4 we get two weights

for 5kg we can have 5kg  or 4+1 or 7-3 or8-3 or 9-3

we can take either of the cases 8-3 or 9-3 to get all the weights from 5 to 11

6= 8-(3-1) , 7=8-1,  9=8+1 ,10=8+(3-1) ,11=8+3

so we used 1kg ,3kg ,8 or 9 kg

no. of weight used =3 (answer)

Yes , 3 is the answer. Thank you

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N is a smallest such positive integer such that when It’s left most digit is removed, then remaining number is 1/37th of N. Find sum of digits of N.

13

14

15

16

(abcde.....)/37 = bcde....

abcde.... = 37 × bcde....

a × 10ⁿ + bcde..... = 37bcde

a × 10ⁿ = 36bcde.....

Smallest value of a which makes LHS a multiple of 36 is 9.

So 9 × 10ⁿ = 36 bcde...

bcde... = 900/36

bcd...= 27

abcd..... = 927

Sum of the digits : 9 +2 + 7 = 16.

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n is a positive integer such that: 1 + 2 + 22 + 23 + … + 2n = 22n – 49. Find n.

6

5

4

3

its a G.P

sum of n terms of G.P =(arn-1)/(r-1)

here a=1 ,r=2

22n – 49= 2n -1

2n (3)=48

2n =16

n=4

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1

2

4

none  ?

2

while doing reciprocal in inequality sign changes

7/3<x/(x+5)<17/3

3/7> (x+5)/x > 3/17

3/7> 1+5/x > 3/17

subtracting 1 from each side

-4/7>5/x>-14/17

by doing reciprocal

-7/4<x/5<-17/14

-35/4<x<-85/14

-8.7<x<-6.01

hence x can take only two integer values i.e -7 and -8

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Medha and Megha are sisters . Madhvi is Medha's daughter and 12 years younger than her aunt . Medha is twice as old as Madhvi . Four years ago, Medha was the same age as Megha is now , and Megha was twice as old as her niece. How old is Madhavi ?

correct ans: 30

@utkarsh-garg25

I think the question has error.

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A bag contains 10 identical balls. All except one ball are of 10gm each. We can use a spring-balance, where we can hang the object in attached pan to know its weight. How many least number of times the spring-balance must be used to identify the faulty ball?

3 ?

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There exists a 10 digit number such that the 1st digit from left represents the number of 0's in the number, the2nd digit from left represents the number of 1's occurring in the number and so on until the 10th digit represents the number of 9's in the number.

The sum of the digits of the number is:

8

9

10

19

for this 10 digit one thing is for sure there will be more no. of zeros than any other no. so we will start by putting 10 zeroes but for that we need to keep no. 10 at first digit from left which is not possible

we will keep 9 on first and lets see

9000000000 (but we also need to fill no. of 9 at 9th place so this case is not possible)

8_ _ _ _ _ _ _ _ _ now we can keep 8 zeroes anywhere of the 9 places vacant but we have to keep 1 at 8th place because no, of "8" is 1 and if we do that we have to keep no. 1 at 2nd place so this case is also not possible

7 _ _ _ _ _ _ _ _ _ for this case we need 7 zeroes and we need 1 at 7th place and 1 at 2nd place also because 1 has appeared in 7th place so we get this 7100001000 but since 1 has occured two times we need to place 2 at the third place so this case is also not possible

6 _ _ _ _ _ _ _ _ _ for this case we need 6 zeroes and then we will keep 1 at 6th place and keep 1 at 2nd place and we will still left with one vacant place like this

61 _ 0010000 now since 1 has occured two times we need to write 2 at 2nd place so we rewrite as

62 _ 0010000 but now this no. suggests that 1 has been written two times and also that 2 has been written one time so we can rewrite as