Latest TCS Aptitude Question SOLUTION: I bought a certain number of marbles at rate of 59 marbles for rupees 2 times M, where M is an integer. I divided these marbles into two parts of equal numbers, one part of which I

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I bought a certain number of marbles at rate of 59 marbles for rupees 2 times M, where M is an integer. I divided these marbles into two parts of equal numbers, one part of which I sold at the rate of 29 marbles for Rs. M, and the other at a rate 30 marbles for Rs. M. I spent and received an integral number of rupees but bought the least possible number of marbles. How many did I buy?

Asked In TCS shubham tiwari (8 years ago)

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prabin  (8 years ago)

Let he bought x marbles.
59 marbles costs = Rs. 2M
Therefore, x marbles costs = Rs. (2M * x) / 59
Since the marble is divided into 2 equal parts so the number x should be an even number.
For first x/2 marbles,
29 marbles s.p. is = Rs. M
Therefore, x/2 marbles s.p. = Rs. (M * x) / 58
For other x/2 marbles,
30 marbles s.p. is = Rs. M
Therefore, x/2 marbles s.p. = Rs. (M * x) / 60
Now we can't equate like [(M * x) / 58] + [(M * x) / 60] = (2M * x) / 59
because (M *x ) will get cancel each side and of course 1/58 + 1/60 is not equal to 2/59
So here we don't need M and we can cancel it. After that we have,
CP = 2x/59
For first x/2 marbles,
SP = x/58
And for other x/2 marbles,
SP = x/60
Now this CP and SP must be an integer (as per question). So we have to find a number x which will be divisible simultaneously by 59, 58 and 60. So we have to find the LCM of 59, 58, 60 which will turn out minimum value as 102660 and it is even as well. So the value of x will be 102660 minimum.

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manikandan  (8 years ago)

x*2M/59 = x/2*M/(29) +x/2* M/(30)
2Mx/59 =xM/58+xM/60..sub for x and m to find the ans frm options.

Is this solution Helpfull? Yes (1) | No (3)  

Shubham  (8 years ago)

let the number of marbles bought be 2k ,since it needs to be a even number since it needs to be divided into two equal parts.
now the money spend =59 *2k/2m=59*k/m,(since 59*k/m is a integer therefore k is multiple of m , let it be m*k1.
the money received =m*(m*k1)/29 +m*(m*k1)/30=(m^2)*(k1*59/(29*30) since these to be integer

((m^2)*k1)/(29*30) should be integer.m^2 cannot be 29 but can be a number which is a factor of 30 .but there is no such number accept 1. therefore k1=29*30=870 and k=m*k1=870
therefore the answer is 2*29*30=1740

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Meera  (4 years ago)

Let he bought 2x marbles.
27 marbles costs = Rs.2M so 1 marble costs = Rs.2M/27
Therefore, x marbles costs = Rs. (2M × x) / 27
Now we calculate the selling prices.
He sold x marbles at the rate of 13 for Rs.M so 1 marble selling price = M/13
x marbles selling price = x × M/13
He sold another x marbles at the rate of 14 for Rs.M so 1 marble selling price = M/14
x marbles selling price = x × M/14
Now 2Mx27,xM13,xM142Mx27,xM13,xM14 are integers.
So x marbles must be divisible by 27, 13, 14. LCM of 4914
So 2x = 9828

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