Approach for my answer I posted above in response to @Atulit123 's stamps question.
From the question we can infer that 2 paise stamps are six times the 1 paise stamps. Also,Total paise=75 (Given)
Now, we do supposition. Supposition works nice. Suppose 10 or such easy figures only. This works many times.
Let us suppose 10 stamps of 1 paise. So, 2 paise stamps must be 6 times the 1 paise stamps. i.e. 60 stamps of 2 paise. => 120 paise that exceeds 75 paise(given).
Let us suppise 5 1paise stamps. So, 2 paisr stamps would be 6 times the 1 paise stamps=>30 stamps of 2paise=>60 paise. So now, 5 1paise stamps+30 stamps of 2 paise=>65 paise.
Remaining=10paise. So, there must have been 2 stamps of 5 paise.
In this way, we get
5 stamps of 1paise
30 stamps of 2paise
2 stamps of 5paise
total=75 paise
Two women were selling marbles in the market place €” one at three for a Rupee and other at two for a Rupee. One day both of then were obliged to return home when each had thirty marbles unsold. They put together the two lots of marbles and handing them over to a friend asked her to sell then at five for 2 Rupees. According to their calculation, after all, 3 for one Rupee and 2 for one Rupee was exactly same as 5 for two Rupees.
Now they were expecting to get 25 Rupees for the marbles, (10 Rupees to first and 15 Rupees to second), as they would have got, if sold separately. But much to their surprise they got only 24 Rupees ( ) times for the entire lot.
Now where did the one Rupee go? CAN YOU EXPLAIN THE MYSTERY?
Now they were expecting to get 25 Rupees for the marbles, (10 Rupees to first and 15 Rupees to second), as they would have got, if sold separately. But much to their surprise they got only 24 Rupees ( ) times for the entire lot.
Now where did the one Rupee go? CAN YOU EXPLAIN THE MYSTERY?
@Atulit123 bhai, calm down, think logically.
There are two Women - let them be A and B
- First lets deal with A
She sells marbles at the rate of three for a rupee
MATLAB - one rupee you get three marbles
so one marble will cost 1/3 rs.
- Second lets deal with B
She sells at the rate of two for on rs.
so one marble from her will cost 1/2 rs.
THIRD WOMAN
SHE HAS to sell 30+30 marbles at the rate of 2 for 5. Matlab price of one marble from third woman is 2/5 rs
you must understand that she is not selling at 1/3rs for 1/2 rs.
so calculating
third woman would give 2/5 * 30 = 12 rs. each.
However, if A and B had sold individually,
they would have got 10 and 15 rs each.
Third woman is selling at an average price lower than A+B ka average.
hope this helps.
@vbhvgupta said:Can you explain the logic....why u took 420
we have to take the lcm of 2 , 3 ,4 ,5 closest to 400
ist it
ist it
@Atulit123 said:Two women were selling marbles in the market place €” one at three for a Rupee and other at two for a Rupee. One day both of then were obliged to return home when each had thirty marbles unsold. They put together the two lots of marbles and handing them over to a friend asked her to sell then at five for 2 Rupees. According to their calculation, after all, 3 for one Rupee and 2 for one Rupee was exactly same as 5 for two Rupees.Now they were expecting to get 25 Rupees for the marbles, (10 Rupees to first and 15 Rupees to second), as they would have got, if sold separately. But much to their surprise they got only 24 Rupees ( ) times for the entire lot.Now where did the one Rupee go? CAN YOU EXPLAIN THE MYSTERY?
60 is being sold for 5 for 2 ==> 60/5*2 = 24
if u want logical explanation, think of it like this
separate them into 2 piles, 30 each, whenever the customer gives u 2 rupees, take out 2 from 1 pile and 3 from other, once u have 20 rs, 1st pile will be over and 2nd one would have 10 left
now u have to sell them at 5 for 2 for which u'll get 4 rs, so total =10*2+4 = 24
so basically, the 3 for 1 deal is decreasing the value of 2 for 1
if u want logical explanation, think of it like this
separate them into 2 piles, 30 each, whenever the customer gives u 2 rupees, take out 2 from 1 pile and 3 from other, once u have 20 rs, 1st pile will be over and 2nd one would have 10 left
now u have to sell them at 5 for 2 for which u'll get 4 rs, so total =10*2+4 = 24
so basically, the 3 for 1 deal is decreasing the value of 2 for 1
@MJ477 said:Find the probability that three points chosen on a circle lie on the same semi-circle ?
1/2 ?
@vbhvgupta said:Find the no of no between 100 to 400 which r divisible by either 2,3,7,5
PLZ SOLVE THIS
minimum value that a can take if x^2 + ax + 900 has
1. Integral roots
2. Negative integral roots
3. Positive integral roots
how many APs are possible such that the first term is 3525 , last term is 6259
and there are atleast...
1. 10 terms
2. 6 terms
3. 13 terms
and there are atleast...
1. 10 terms
2. 6 terms
3. 13 terms
@krum said:1/2 ?
Find the probability that three points chosen on a circle lie on the same semi-circle. ?
if we fix a semicircle then 3 cases lag rahe h :
1. 3 on one side2. 2 on one and 3rd on other
3. first on one and rest two on other... what say 1/3 ?
3. first on one and rest two on other... what say 1/3 ?
If ab^2c^3= 27*2^8, then find the minimum value of a+b+c.
a. 12 b. 18 c. 24 d. 16
a. 12 b. 18 c. 24 d. 16
@joyjitpal said:If ab^2c^3= 27*2^8, then find the minimum value of a+b+c.a. 12 b. 18 c. 24 d. 16
4*8^2*3^3
a+b+c=15
a+b+c=15
@joyjitpal said:If ab^2c^3= 27*2^8, then find the minimum value of a+b+c.a. 12 b. 18 c. 24 d. 16
12
c=6, b=4, a=2
@vbhvgupta said:how many APs are possible such that the first term is 3525 , last term is 6259and there are atleast...1. 10 terms2. 6 terms3. 13 terms
2734 = (n-1)*d => 2 APs
in all cases ?
in all cases ?
@Atulit123 said:Please if possible solve this, My friend Shuba works in a post office and she sells stamps. One day a man walked in and kept seventy five paise on the counter and requested, 'Please give me some 2 paise stamps, six times as many as one paisa stamps, and for the rest of the amount give me 5 paise stamps.' The bewildered Shuba thought for a few moments and finally she handed over the exact fulfilment of the order to the man €”with a smile. How would you have handled the situation?
30 1p stamps
5 2p stams
2 5p stamps
let no of 1p stamps= x
then total money n 1p+2p stamps= 13x
only 13*5= 65 leaves 10 which is divisible by 5.
@joyjitpal said:If ab^2c^3= 27*2^8, then find the minimum value of a+b+c.a. 12 b. 18 c. 24 d. 16
(assuming a,b,c as +ve)
a+b+c = a +(b/2+b/2) + (c/3 + c/3 + c/3) = k
applying AM >= GM we get
(k/6)^6 > = a*(b/2)^2 * (c/3)^3
=>(k/6)^6 > = ab^2c^3/(4*27)
=>(k/6)^6 > = 27*2^8/(4*27)
=> (k/6)^6 > = 2^6
=> k > = 12
hence min value = 12
and this occurs at a =2; b =4 and c = 6.
if the condition of a,b,c being +ve is not there then the above solution is not valid. For example if we put a = -2; b= 4 and c = -6 then ab^2c^3 = 27*2^8 is satisfied but the value of a+b+c = -4.
ATDH.
@MJ477 said:Find the probability that three points chosen on a circle lie on the same semi-circle. ? if we fix a semicircle then 3 cases lag rahe h :1. 3 on one side2. 2 on one and 3rd on other3. first on one and rest two on other... what say 1/3 ?
i think its 3/4 , u have OA?
@vbhvgupta said:minimum value that a can take if x^2 + ax + 900 has1. Integral roots2. Negative integral roots3. Positive integral roots
1,3) x^2-901x+900 =>x^2-900x-x+900=(x-900)(x-1) , a=-901
2) x^2+60x+900 =>(x+30)(x+30) , a=60
2) x^2+60x+900 =>(x+30)(x+30) , a=60
@vbhvgupta said:minimum value that a can take if x^2 + ax + 900 has1. Integral roots2. Negative integral roots3. Positive integral roots
900 is the product of roots (i.e it is +ve) so both the roots must of be of the same sign.
say p and q are the roots
=>pq = 900 and we want to minimize (p+q)
so product is fixed and we need to maximize/minimize the sum
1. for minimum value of a we take both the roots +ve (so that a is -ve) and keep both the roots as far as possible such that product is 900.
900 and 1 as roots gives a = -900-1 = -901
2. If we want the roots to be -ve then a is +ve. In this case we try to find two +ve numbers such that product is 900 and they are as close to each other. 30,30 is the pair. So in this case minimum value of a = 30+30 = 60
3. same as solution for 1.
ATDH.
@anytomdickandhary sir
I marked polynomial of x
please solve this XAT 2013 question
f(x)=x^4+x^3+x^2+x+1
find the remainder when f(x^5) is divided by f(x)
options=> 1,4,5,a monomial of x, a polynomial of x
I marked polynomial of x
TF's key says E but CL and TIME keys say C
@sumeet1489 said:@anytomdickandhary sirplease solve this XAT 2013 questionf(x)=x^4+x^3+x^2+x+1find the remainder when f(x^5) is divided by f(x)options=> 1,4,5,a monomial of x, a polynomial of xI marked polynomial of xTF's key says E but CL and TIME keys say C
i checked with x = 2 and 3 .. got 5