A trader cheats both his supplier and customer by using faulty weights. When he buys from the supplier, he takes 10% more than the indicated weight. When he sells to his customer, he gives 10% less than the indicated weight. If he sells at cost price (i.e., charges the cost price of the indicated weight), what is his profit percentage?
@maroof10 said:Ram and Shyam are moving upwards on a moving escalator.When Ram takes 2 steps Shyam takes 3 steps in same time. Shyam reaches the top taking 25 steps while Ram takes 20 steps to reach the top.If escalator's movement is ceased,then how many steps are needed to be taken by Shyam to reach the top?
Speed of shyam be 3x
speed of Ram be 2x
Time taken by Shyam=25/2
25+e/x*25/3=20+e/x*10
5=5/3*e/x
e/x=3
Total steps=25+25=50
@meenu05 said:Find the sum of the first 20 terms of the sequence 5,5.5, 5.55,5.555 ...
Its 101.666(17 times)100
the dots are decimal points I guess
@nam2aspire said:hey guys,mine is a very basic question.i just wanna know the algorithm behind euclid's long divison method for finding hcf.i just want to know the logic behind this method of first dividing the bigger number by smallernumber(in case of two numbers) and then dividing the divisor by the coming remainder and so on.......y do we divide the divisor by the remainder(what's the logic behind this!?!) and how come we get hcf by this continuous subtraction and division?? (though hcf comes out by finding factors of the numbers!!!!! )p.s. i know its unnecessary but i just want to understand the concept in detail!! please explain...
Let us take two numbers say 2223 and 3762
now if you will apply long division method you will get hcf as 171
we define hcf as the highest number that could divide the given numbers completely
so in long division we try to find out the largest number in terms of which the two numbers could be expressed
now 3762=2223*1+1539
2223=1539*1+684
1539=684*2+171
684=171*4
now if substituting backward successively
3762=22*171
so the highest common factor of(3762,2223) will be 171
as both can be expressed in its terms completely...hope this helps..:)
now if you will apply long division method you will get hcf as 171
we define hcf as the highest number that could divide the given numbers completely
so in long division we try to find out the largest number in terms of which the two numbers could be expressed
now 3762=2223*1+1539
2223=1539*1+684
1539=684*2+171
684=171*4
now if substituting backward successively
3762=22*171
so the highest common factor of(3762,2223) will be 171
as both can be expressed in its terms completely...hope this helps..:)
@meenu05 said:@staaalinnn answer is 110.5
can you elaborate your approach? if there is any short-cut
@jain4444 said:x is a number such that x^2 + 3x + 9 = 0. What is the value of x^3?
x=-3+/-3rt3*i/(2)
x^3
(-3+/-3rt3*i/(2))^3=27
@meenu05
Shld be 110.5
Here, t(1)= 5/9*(10 - 1)
=> t(2)= 5/9*(10 - 0.1)
=> (3)= 5/9*(10 - 0.01)
.
.
=> t(20) = 5/9*(10 - 10^-19)
Sum= 5/9*[200 - 10/9*(1 - 10^-20)
=> ~ 5/9*(200 - 10/9)
=> ~ 110.49.. or, 110.5..
@meenu05 said:Find the sum of the first 20 terms of the sequence 5,5.5, 5.55,5.555 ...
5+5.5+5.55+.....20 terms
=(5*20) + (5/10 + 55/100 + 555/1000 + .....19 terms)
=100 + 5/9(9/10 + 99/100 +999/1000 + 19 terms)
=100 + 5/9((1-1/10) + (1-1/100) + (1-1/1000) + ... 19 terms)
=100 + 5/9(19-(1/10 + 1/100 + 1/1000 +....19 terms)
=100 + 5/9(19 - (1/10(1-(1/10)^19))/(1-1/10))
=100 + 5/9(19 - 1/9)=110.5
=(5*20) + (5/10 + 55/100 + 555/1000 + .....19 terms)
=100 + 5/9(9/10 + 99/100 +999/1000 + 19 terms)
=100 + 5/9((1-1/10) + (1-1/100) + (1-1/1000) + ... 19 terms)
=100 + 5/9(19-(1/10 + 1/100 + 1/1000 +....19 terms)
=100 + 5/9(19 - (1/10(1-(1/10)^19))/(1-1/10))
=100 + 5/9(19 - 1/9)=110.5
@tmohan02
Shld be 22 2/9 % or 200/9%..
Let Originally 100 kg of article cost 100 /-
Thus, CP of 110 kg of article = 100/-...CP of 1 kg of article = 10/11 /-
Again, SP of 90 kg of article = 100/-...SP of 1 kg of article = 10/9 /-
Thus, Gain% = (10/9 - 10/11)/(10/11)*100 = 200/9 %..
Three cards are drawn from a well shuffled pack of 52 cards. Find the probability that all the three cards are of different suits and there is at least one face card.
1) 1017/1105
2) 1197/5505
3) 129/1105
4) 1197/5525
5) 1017/5505
1) 1017/1105
2) 1197/5505
3) 129/1105
4) 1197/5525
5) 1017/5505
A and M can do a work in 12 days. M and B can do a work in 15 days. if A is twice as good a workman as B. in how many daya will mohan alone can do the work??
@pavimai said:if A= 8888^8888 B= sum of digits of A , C=sum of digits of B , D=sum of digits of C...In this series there will be a point where you will get sum of digits of X=X..Find X
Divide by 9 to check sum of digits
5^8888
Euler number of 9 is 6
5^2divided by 9
7
8888 sum of dgits is 32
32 sum of digits is 7
@vbhvgupta said:A and M can do a work in 12 days. M and B can do a work in 15 days. if A is twice as good a workman as B. in how many daya will mohan alone can do the work??
Let A do in X days and so B does it n 2x days
Now 1/x + 1/m = 1/12 and 1/2x + 1/m = 1/15
Solve and get M can do in 20 days
Now 1/x + 1/m = 1/12 and 1/2x + 1/m = 1/15
Solve and get M can do in 20 days
@vbhvgupta
Shld be 20 days..
Given, W= 12(A + M)...(i) n W = 15(B + M)...(ii) n A=2B
Thus, 4A + 4M = 5B + 5M, or, M=3B, or, A= 2/3*M
Thus, W = 12(5/3*M) => W/M = 20..
@pavimai said:how many integers between 1000 and 10000 have their sum of the digits equal to 9???
1008,1017,---------------------9999
8991/9=999+1=1000
@vbhvgupta said:A and M can do a work in 12 days. M and B can do a work in 15 days. if A is twice as good a workman as B. in how many daya will mohan alone can do the work??
20 days.
A=x
B=2x
first eq 1/m+1/x=1/12
second eq 1/m+1/2x=1/15
solve for x. it comes out to be 30
put this in any one of the eq, M=20.
A=x
B=2x
first eq 1/m+1/x=1/12
second eq 1/m+1/2x=1/15
solve for x. it comes out to be 30
put this in any one of the eq, M=20.
@gnehagarg said:1008,1017,---------------------99998991/9=999+1=1000
aapne bhi meri wali galti kari....sum of digits bola hai digita sum nahi..