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# Official Quant Thread for CAT 2012 [Part 1]

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-...
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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)
IMS Kolkata Academic Head (Quants dept.) CAT, XAT 99.7+ percentiles
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ans 987 * 563 =556581....

pranam bhaiylog ________/\_________
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J.P Morgan | Media Cell | IIM Kozhikode 2014-16 | CAT 13 99.43 | XAT 97.21 |
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Is it 555681......

Did by multiplication rules........
987*563=555681

same here
IIM Amritsar 2016-18 || CAT 15: 95.92 (QA:99.27) || SNAP 15: 97.3455 || XAT 16: 97.376
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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
Three Lives. Three Destinies. One Name. CAT'12 : 95.86%ile
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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
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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......

Number of positive integral solutions of equation

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
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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
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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 take555681

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
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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 = 143/1088
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Smoothest seas do not make tough sailors
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Q) If in a long division sum, the dividend is 380606 and the successive remainders from the first to the last are 434, 125 and 413, then the divisor is ___ ?

ye solution dekha dekha sa lag raha hai

OA is 552

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