Two alloys of iron have different percentage of iron in them.The first one weighs 6kg and second one weighs 12 kg.One piece each of equal weight was cut off from both the alloys and the first piece was alloyed with the second alloy and the second piece alloyed with the first one.As a result,the percentage of iron became the same in the resulting two new alloys.What was the weight of each cut -off peice ?
1)4kg
2)2kg
3)3kg
4)5kg
5)3.5kg solution is 4kg
How to solve this ? Also "percentage of iron became same in resulting alloys" How is this information useful. I would have understood if it was writen " amount of iron was same in new alloys", but instead of amount percentage word is there.
@paridhi11890 said:Elaborate it a bit more... mujhe samajh nhi aaya ye method..
Arey I considered 0 and 7 as the multiples of 7
Now, one obvious case is that the 5th digit is 0 or 7=> _ _ _ _ 0 _ _ _ _ [ See that every consecutive 5 numbers' product is 0] => satisfies the condition, so I took every number of this form + _ _ _ _ 7 _ _ _ _
So, 9C8*8! + ( 9C8*8! - (starting with zero)) = 17*8!
Now, consider this -> 7_ _ _ _ 0 _ _ _
Here, the first five consecutive digits' product has a 7, and rest are 0 -> satisfies
Also, the 7 can move to the right till the 4th digit (I neglected the 5th because that is already covered in the first case)
=> 4*(8C7*7!) = 4*8!
Now, for rest, move the 0 to the right by one place each, and find the cases.Similar thing would happen when we place 0 in the second place and 7 in the 7th place in the number
@nole said:Two alloys of iron have different percentage of iron in them.The first one weighs 6kg
let the first part have x% conc. of iron ....wt. is 6kg
let the second part have y% conc. of iron .....wt. is 12kg
let the cut weight be t kg
now first part + small second part in total have
(6-t)*x/100 + t*y/100 kg of iron ( 6-t kg has x% comp. of iron and t kg has y%(from second part) )
similarly second part has (12-t)*y/100 + t*x/100 kg of iron
now % of iron in first part
= ((6-t)*x/100 + t*y/100) / (6-t+t)-------------------------1
comp. of second part iron is
= ((12-t)*y/100 + t*x/100) / (12-t+t) ----------------------2
1=2 ...on solving we get 12(x-y) = 3t(x-y) => 12=3t => t=4kg(the cut out part)
let the second part have y% conc. of iron .....wt. is 12kg
let the cut weight be t kg
now first part + small second part in total have
(6-t)*x/100 + t*y/100 kg of iron ( 6-t kg has x% comp. of iron and t kg has y%(from second part) )
similarly second part has (12-t)*y/100 + t*x/100 kg of iron
now % of iron in first part
= ((6-t)*x/100 + t*y/100) / (6-t+t)-------------------------1
comp. of second part iron is
= ((12-t)*y/100 + t*x/100) / (12-t+t) ----------------------2
1=2 ...on solving we get 12(x-y) = 3t(x-y) => 12=3t => t=4kg(the cut out part)
@YouMadFellow said:33 * 8! ?Case 1 : _ _ _ _ 0 _ _ _ _ / _ _ _ _ 7 _ _ _ _9C8*8! + (9C8*8! - 8*7!) = 17*8!Case 2: 7_ _ _ _ 0 _ _ _ / _ 7_ _ _ _ 0 _ _ / _ _ 7 _ _ _ _ 0_ / _ _ _7 _ _ _ _04*8! + 3*8! + 2*8! + 8! = 10*8! (7 can change positions towards the right)Case 3: _ 0_ _ _ _ 7 _ _ / _ _ 0 _ _ _ _ 7_ / _ _ _0 _ _ _ _73*8! + 2*8! + 8! = 6 * 8!Total sum = 33* 8! ?Solved with a headache, may be wrong
no OA getting 35*8!
_ _ _ _ 7 _ _ _ _ = 8*8! cases
_ _ _ _ 0 _ _ _ _ = 9C8*8! cases
_ 0 _ _ _ 7 _ _ _ = 8C7*7!*2! * 2 cases
_ _ 0 _ _ 7 _ _ _ = 8C7*7!*2! * 3 cases
_ _ _ 0 _ 7 _ _ _ = 8C7*7!*2! * 4 cases
_ _ _ _ 7 _ _ _ _ = 8*8! cases
_ _ _ _ 0 _ _ _ _ = 9C8*8! cases
_ 0 _ _ _ 7 _ _ _ = 8C7*7!*2! * 2 cases
_ _ 0 _ _ 7 _ _ _ = 8C7*7!*2! * 3 cases
_ _ _ 0 _ 7 _ _ _ = 8C7*7!*2! * 4 cases
7 _ _ _ _ 0 _ _ _ = 8C7*7! (EDIT)
sum = 8*8! + 9*8! + 4*8! + 6*8! + 8*8! + 8! = 36*8!
sum = 8*8! + 9*8! + 4*8! + 6*8! + 8*8! + 8! = 36*8!
@nole said:Two alloys of iron have different percentage of iron in them.The first one weighs 6kg and second one weighs 12 kg.One piece each of equal weight was cut off from both the alloys and the first piece was alloyed with the second alloy and the second piece alloyed with the first one.As a result,the percentage of iron became the same in the resulting two new alloys.What was the weight of each cut -off peice ?1)4kg2)2kg3)3kg4)5kg5)3.5kg solution is 4kgHow to solve this ? Also "percentage of iron became same in resulting alloys" How is this information useful. I would have understood if it was writen " amount of iron was same in new alloys", but instead of amount percentage word is there.
let % be 'a' and 'b'
so amount of iron in first = 6a
amount of iron in second=12b
let x kg be cut out from both
amount left in 1st = a(6-x)
amount left in 2nd = b(12-x)
==> 2*[a(6-x)+bx]=b(12-x)+ax
==> 12a-2ax+2bx=12b-bx+ax
==> 12(a-b)-3x(a-b)=0
==> x=4
so amount of iron in first = 6a
amount of iron in second=12b
let x kg be cut out from both
amount left in 1st = a(6-x)
amount left in 2nd = b(12-x)
==> 2*[a(6-x)+bx]=b(12-x)+ax
==> 12a-2ax+2bx=12b-bx+ax
==> 12(a-b)-3x(a-b)=0
==> x=4
@jain4444 said:no OA getting 35*8!_ _ _ _ 7 _ _ _ _ = 8*8! cases _ _ _ _ 0 _ _ _ _ = 9C8*8! cases _ 0 _ _ _ 7 _ _ _ = 8C7*7!*2! * 2 cases _ _ 0 _ _ 7 _ _ _ = 8C7*7!*2! * 3 cases _ _ _ 0 _ 7 _ _ _ = 8C7*7!*2! * 4 casessum = 8*8! + 9*8! + 4*8! + 6*8! + 8*8! = 35*8!
can you explain the bold part ? Also, have you considered 7_ _ _ _ 0 _ _ _ ?
@YouMadFellow said:can you explain the bold part ? Also, have you considered 7_ _ _ _ 0 _ _ _ ?
_ 0 _ _ _ 7 _ _ _
_ 0 _ _ _ _ 7 _ _
these are the 2 cases and same like in further cases
yeah i missed that one case
@joyjitpal said:Answer question nos. 96 €“ 97 based on the followinginformation.From a group of 545 contenders, a party has to select a leader. Even after holding a series of meetings, the politicians and the general body failed to reach a consensus. It was then proposed that all 545 contenders be given a number from 1 to 545. Then they will be asked to stand on a podium in a circular arrangement, and counting would start from the contender numbered 97. One of the contending politicians, Mr. Chanaya, was quite proficient in calculations and could correctly figure out the exact position. He was the last person remaining in the circle. Sensing foul play the politicians decided to repeat the game. However, this time, instead of removing every alternate person, they agree on removing every 300th person from the circle. All other rules were kept intact. Mr. Chanaya did some quick calculations and found that for a group of 542 people the right position to become a leader would be 437. What is the right position for the wholegroup of 545 as per the modified rule?A. 3B. 194C. 249D. 437E. 543
Solution for the 2nd question:-
http://pagalguy.com/forums/quantitative-ability-and-di/official-quant-thread-cat-2011-part-1-t-60879/p-2389624/r-2415144
The weight of the three heaps of gold are in the ratio 5:6:7.By what fractions of themselves must the first two be increased so that the ratio of weights may be changed to 7:6:5.
answer is 24/25,2/5
But why the first two should be changed.The second one is having the same ratio in both the cases anyways.
@nole said:The weight of the three heaps of gold are in the ratio 5:6:7.By what fractions of themselves must the first two be increased so that the ratio of weights may be changed to 7:6:5.answer is 24/25,2/5But why the first two should be changed.The second one is having the same ratio in both the cases anyways.
let heaps be 5x , 6x , 7x
so let first is increased by m fraction so = 5x(m+1)
and second by n fraction = 6x{n+1)
by new ratios
5x(m+1) / 6x(n+1) = 7/6
7n - 5m = -2 ---------------------------(i)
6x(n+1) / 7x = 6/5
n = 2/5
sub in (i)
m = 24/25
@nole said:Two alloys of iron have different percentage of iron in them.The first one weighs 6kg and second one weighs 12 kg.One piece each of equal weight was cut off from both the alloys and the first piece was alloyed with the second alloy and the second piece alloyed with the first one.As a result,the percentage of iron became the same in the resulting two new alloys.What was the weight of each cut -off peice ?1)4kg2)2kg3)3kg4)5kg5)3.5kg solution is 4kgHow to solve this ? Also "percentage of iron became same in resulting alloys" How is this information useful. I would have understood if it was writen " amount of iron was same in new alloys", but instead of amount percentage word is there.
X/(6-X)=12-X/X...
x=4
@mailtoankit can u explain using words how you formed that equation ? I said using "words" so that next time i can imagine it and directly use the equation for such sort of questions :)
@nole said:The weight of the three heaps of gold are in the ratio 5:6:7.By what fractions of themselves must the first two be increased so that the ratio of weights may be changed to 7:6:5.answer is 24/25,2/5But why the first two should be changed.The second one is having the same ratio in both the cases anyways.
7x=5y
=> y=7x/5
5x+a=7y
=> a=24/25x
6x+b=6y
=> b=12/5x
if the ratio of third were same, second wouldn't have changed, as it has decreased, both 1st and 2nd will increase
=> y=7x/5
5x+a=7y
=> a=24/25x
6x+b=6y
=> b=12/5x
if the ratio of third were same, second wouldn't have changed, as it has decreased, both 1st and 2nd will increase
A person lent out some money for 1year at 6% per annum simple interest and after 18months he again lent out the same money at a simple interest of 24% per annum. In both cases he got Rs.4704. What was the amount that he lent out if interest is paid half yearly?a rs 4000 b rs 4200 c rs 4400 d rs 3600
@nole said:The weight of the three heaps of gold are in the ratio 5:6:7.By what fractions of themselves must the first two be increased so that the ratio of weights may be changed to 7:6:5.answer is 24/25,2/5But why the first two should be changed.The second one is having the same ratio in both the cases anyways.
7 in the first ratio becomes 5 in the next
So take the first ratio as
5*5 : 6*5 : 7*5 = 25:30:35
In the second ratio we get
7*7:6*7:5*7 = 49:42:35
So fractional change in the first heap = 24/25
Fractional change in second heap = 12/30 = 2/5
So take the first ratio as
5*5 : 6*5 : 7*5 = 25:30:35
In the second ratio we get
7*7:6*7:5*7 = 49:42:35
So fractional change in the first heap = 24/25
Fractional change in second heap = 12/30 = 2/5
Four identical bags are distributed among four boys. If each boy can get any number of bags then what is the probability that no boy gets more than two bags?(a)18/35(b)2/7(c)19/35(d)16/35
@Aman.Malhotra said:Four identical bags are distributed among four boys. If each boy can get any number of bags then what is the probability that no boy gets more than two bags?(a)18/35(b)2/7(c)19/35(d)16/35
total ways = 4c1+4c2*2+4c2+4c3*3+4c4 = 4+12+6+12+1 = 35
4,0 - 4c1
3,1 - 4c2*2
2,2 - 4c2
1,1,2 - 4c3*3
1,1,1,1 - 4c4
favorable - 4c2+4c3*3+4c4 = 6+12+1 = 19
19/35
4,0 - 4c1
3,1 - 4c2*2
2,2 - 4c2
1,1,2 - 4c3*3
1,1,1,1 - 4c4
favorable - 4c2+4c3*3+4c4 = 6+12+1 = 19
19/35
A teacher asks one of her students to divide a 30-digit number by 11. The number consists of six consecutive 1's, then six consecutive 2's, and likewise six 3's, six 4's and six 7's in that order from left to right. The student inserts a three-digit number between the last 4 and the first 7 by mistake and finds the resulting number to be divisible by 11. Find the number of possible values of the three-digit number.
82
92
81
82
82
92
81
82
@Aman.Malhotra said:Four identical bags are distributed among four boys. If each boy can get any number of bags then what is the probability that no boy gets more than two bags?(a)18/35(b)2/7(c)19/35(d)16/35
2 2 0 0 = 4!/2!*2! = 6
1 2 1 0 = 4!/2! = 12
1 1 1 1 = 1
favorable cases = 19
total cases
a + b + c + d = 4
=> 7C3 = 35
prob. = 19/35