Quant by Arun Sharma

MBA CAT 2024
5806
@jeetuachtani said:Sum to the n terms of series 1+3+7+15+......... ?
The general term or Kth term of the series is [ 2^K -1 ] where K ranges from 1 to N.
Sum of series = Sigma [ 2^K -1 ]
= Sigma[2^K] - N
= (Sum of GP with common ratio 2 and first term 2 ) - N
You can continue from here to get the result as given in options.
5807

Hi pgites


can anyone help me in Q-12 of LOD1 for the chapter Profit and Loss,m not able to understand the question....
5808
@UnCharismatic :pg no219
5809
Could anyone please help me in some of these questions of L.O.D-I Q.42,Q.55,Q.70,Q.71,Q75,Q.76,Q.78,Q.94,Q.95 In case of Q.42 i got the answer 6, but book shows 1..In case of 94, i got20,80
5810
Hi Swati,
LOD I of which chapter?
5811
@UnCharismatic For the LOD 1 question no 12 of profit and loss :
The question is very basic. They have basically given the ratio of the cost price of the shirt and the trouser and the sum of the cost price. U just have to find the individual cost.
Let the cp of the shirt and trouser be S and T respectively.
Given S+T=371
S/T = 112/100 (S is 12 % more than T)
T= 371 * 100/212 = 175
5812

A man buys a spirit at rs 600/l, adds water to it n then sells it at rs 750/l. what is the ratio of the spirit's wt to the wt of the water if his prfit in the deal is 37.5%?

5813

Lod 1 of chapter 1

5814
@webby Profit earned per litre = 37.5 * 600/100 = 225
Difference between SP and CP per litre = 750 - 600 = 150
let the 100 litre be the amount of spirit bought.
total profit = 100*225 = 22500
profit earned from the difference in SP and CP for 100 litres = 150*100 = 15000
Profit not accounted for with the differne in SP and CP = 22500-15000=7500
To get 7500 we need (7500/750) 10 litres of spirit sold at SP.
This extra 10 litre is the quantity of water added to the spirit.
Ratio of spirit wt to wt of water = 100 : 10
5815
@Swati1503
42 : 2^1000/3 = (-1)^1000/3 = 1

55 : here we are talking about real no since its not mentioned that the nos are integers. so the upper limit and lower limit values of x n y are alomost equal to their respective max n min limits.
upper limit (x+y) almost = 4+3
lower limit (x-y)
5816
@Swati1503

42 : 2^1000/3 = (-1)^1000/3 = 1

55 : here the nos are real nos not integers (since it is not mentioned)

70 : x/3 + x/3+x/5 = 1

71 : the first 9 numbers occupy 9 digits : digits left = 120 창€“ 9 = 111


The remaining nos occupy 2 digits : 2n

n = no of numbers after 9. The actual nos at the last position = n+9 = 55+9 = 64

the last 3 digit = 646 (since one more digit was left 67 will cm next n we take the first digit 6)

646/8 = 6
75. 1/(m-1) = n

M^(1/m-1) for this the max value is whn m = 2 : 2^(1/2-1)=2^1=2

As m gets bigger the value gets smaller and smaller

76 : as explained in the answer section

78: as explained in the answer section

94. 20 = 2^2 * 5 : no of factors = 3*2=6

80= 2^4*5 : no of factors = 5*2=10

40=2^3*5 : no of factors = 4*2=8 :::: 40,80 has 2 as the difference in the no of factors

95. 3-1 = 2 : 5-3=2 : 6-4=2 : 7-5=2 :

The required no is a (multiple of LCM of 3,5,6,7

5817
@webby said: A man buys a spirit at rs 600/l, adds water to it n then sells it at rs 750/l. what is the ratio of the spirit's wt to the wt of the water if his prfit in the deal is 37.5%?
10:1 .... Allegation method ....
5818

(36472)^123! * (34767)^76!


i knw that the powers of 2 have digits ending with 2,4,8,6 but which is to be selected and when and how????
5819
@harharmahadev12 wht exactly do u wnt? the last digit? ur question is not clear...if its the last digit...i think :
123! = 1*2*3*4*....=4*6*...=4n
similarly 76! = 4n
6(36472)^123! * (34767)^76 =>6*1=6
(last digit of 2^4n=6 & last digit of 7^4n =1)
hope m rght
5820

yes indeed the question was to find the units digit...thanks @keb
plz cud explain a bit more abt how 76!=4n and 126!=4n??


5821
@harharmahadev12 i explained in my earlier post...if u look carefully u will c...
76!=1*2*3*4*5*....*76 : 4n similarly 126!=4n
5822
Find the sum
1/1*5 + 1/5*9 + 1/9*13 .... + 1/221*225
5823
@AbhiCAT2012 said:
Find the sum
1/1*5 + 1/5*9 + 1/9*13 .... + 1/221*225
1/4(1/1-1/5)+1/4(1/5-1/9)+.....1/4(1/221-1/225)
1/4(1-1/225)
56/225
5824

number plates are issued in following combination .. each has three letters followed by three nos but constrain is numbers has to be smaller or equal than the preceding number for ex ,,132 is wrong 1is wrong it should be 4 or 3.. find no of possibilities

5825
if x(y+z-x) / log x = y(z+x-y) / log y = z(x+y-z) / log z

then

1) xy=yz=zx

2) x^y y^x = z^y y^z = x^z z^x

3) xyz=1

4) none of these