About this discussion

- © 2015-2016
- Careers
- Advertise
- Contact Us

Hi puys, CAT 2011 season is finally over :: :: Let start a new season of Quants thread for CAT 2012.. :: :: Here are few links of Quants thread for CAT 2011 season... Part-8 http://www.pagalguy.com/forum/quanti...at-...

Follow this discussion to get notified of latest updates.

Nice! Share it on Facebook so that your friends can join you here.

Order by:
New Posts

Page 7 of 981

There are three equal containers that are completely filled with different water-alcohol mixtures with

water and alcohol in the ratio 2 : 3, 3 : 4 and 4 : 5 respectively. They are emptied into a bigger

container. What fraction of the mixture in the bigger container should be replaced by water so that

the resulting mixture has equal quantities of water and alcohol?

(a)43/945 (b)143/945 (c)43/544 (d)143/1088

resulting mixture has water:alcohol in the ratio = 401:544

let's say, for every 945 ltr of the mixture, x litres of mixture is replaced by x litres of water. (we are required to find x/945)

the water of the mixture remaining = 401 - 401x/945 + x

the alcohol remaining = 544 - 544x/945

these should be equal,

equating them, you would get:

x/945 = 143/1088 (ANS)

ans 987 * 563 =556581....

pranam bhaiylog ________/\_________

- 2 Likes

Commenting on this post has been disabled.

Is it 555681......

Did by multiplication rules........

987*563=555681

same here

Commenting on this post has been disabled.

Arijit Debnath
@bs0409
4,418

Q.) A boy multiplies 987 by a certain number and obtains 559981 as his answer. If in the answer, both 9s are wrong but the other digits are correct, then the correct answer will be:a) 553681.......b) 555181........c) 555681.........d) 556581

Is it 555681......

Did by multiplication rules........

987*563=555681

- 1 Like

Commenting on this post has been disabled.

Jai Mehra
@jaikishan.n
1,521

O.A. for the previous question is given as 34 but I also think that it should be 33 only

Q.) A contract is to be completed in 50 days and 105 men were set to work, each working 8 hours a day. After 25 days, 2/5th of the work is finished. How many additional men be employed so that the work may be completed on time, each man now working 9 hours a day?

(1) 34..........(2) 36............(3) 35.............(4) 37

Ans is 35

Assume x additional men to be needed.

25 *(3/5)*8*105 = (2/5)*(x+105)*25*9

by solving you will get x=35

why did you take 25 instead of 50 on the LHS and RHS mei why did you multiply by 2/5

Commenting on this post has been disabled.

@HeavenSent
1,409

Aisi approach to koshti bhai ki hoti thi ;)@Heavensent bhai......ye batao ki x^10 ki hi power kyu dekhi generally to hum x^20 ki power dekhtey hain......

x+2y+3z+4w = n

= Coeff. of x^[n - (1+2+3+4) ] in [ (1-x)^-1 * (1-x)^-1 * (1-x)^-1 * (1-x^4)^-1 ]

Hence x^[20 - (1+2+3+4)] = x^10

Koshti bhai :

- 4 Likes

Commenting on this post has been disabled.

@akalola
19

2nd one

the no. shud be divisible by 3 and 7 (987 is divisible by 3 and 7)

so d should be the answer

I think ans would be 555681 = 987 * 563

- 1 Like

Commenting on this post has been disabled.

Jai Mehra
@jaikishan.n
1,521

O.A. for the previous question is given as 34 but I also think that it should be 33 only

Q.) A boy multiplies 987 by a certain number and obtains 559981 as his answer. If in the answer, both 9s are wrong but the other digits are correct, then the correct answer will be:

a) 553681.......b) 555181........c) 555681.........d) 556581

2)my take

987*x = 55aa81

x = 55aa81/987

Denominator is slightly less than 1000, numerator is around 550 thousand.

so we know x is a three digit number starting with 5.

Product ends with 1 so x ends with 3.

x = 5b3 and we need to find b

987*5b3 and perform normal multiplication

first row we get 2961 now 7*b must end in 2 to give (6+2) = 8 in the product.

b has to be 6. so x= 563

563*987 = 555681

- 1 Like

Commenting on this post has been disabled.

Varun Tyagi
@varun.tyagi
14,220

There are three equal containers that are completely filled with different water-alcohol mixtures with

water and alcohol in the ratio 2 : 3, 3 : 4 and 4 : 5 respectively. They are emptied into a bigger

container. What fraction of the mixture in the bigger container should be replaced by water so that

the resulting mixture has equal quantities of water and alcohol?

(a)43/945 (b)143/945 (c)43/544 (d)143/1088

.................A..................B............. ......C

Water.........2...................3............... ...4

Alcohol.......3...................4............... ...5

Ratio in A = 2/3

Ratio in B = 3/4

Ratio in C = 4/5

L.C.M.(5,7, 9) = 315

Let 315 L be the capacity of each container

Ratio in A = 2*63/3*63 = 126/189

Ratio in B = 3*45/4*45 = 135/180

Ratio in C = 4*35/5*35 = 140/175

When all are mixed in the bigger tanker then we can find the amount of each liquid

Amount of Water = 126+135+140 = 401

Amount of Alcohol = 189+180+175 = 544

Amount of solution = 315*3 = 945

Final amount of water = Final amount of Alcohol = 945/2

Quantity of water to be increased to 472.5

=> Addition of 71.5 water and removal of 71.5 of Alcohol

Effectively an addition of (71.5+71.5) 143 L of water in 945+143 = 1088 L Solution

=> Fraction =

- 2 Likes

Commenting on this post has been disabled.

@HeavenSent
1,409

ye solution dekha dekha sa lag raha hai

OA is 552

-----------------------------------------------------------------------

Commenting on this post has been disabled.

When you follow a discussion, you receive notifications about new posts and comments. You can unfollow a discussion anytime, or turn off notifications for it.

377 people follow this discussion.- PG Android App is live on the Play Store!
- CAT • 11:30 AM, 06 Oct '15

- PG CAT Mock Series
- CAT • Sunday, 11 Oct

0Comments »new comments