1) a tank holds 100 gallons of water.it's inlet is 7 inches in diameter and fills the tank at 5 gallons/min.the outlet of tank is twice the diameter of inlet.how many minutes will it take to empty the tank if inlet is shut off,when tank is fill and outlet is opened?(answer: 10 min)
2)a tank of capacity 25l has inlet and outlet tap.if both are opened simultaneously,tank is filled in 5 min.but if outlet flow rate is doubled and taps opened,the tank never gets filled up.what is outlet flow rate?(answer: 6)
3)a tank of capacity 3600m^3 capacity is being filled with water.the delivery of pump discharging the tank is 20% more than delivery of pump filling the same tank.as a result 12 min more tank needs to fill tank than to discharge it.determine delivery of pump discharging tank.(answer:60)
plz share ur approach for such questions..
A sold a table to B at a profit of 20%. B sold the same table to C for Rs. 75 thereby making a profit of 25%. Find the price at which A brought the table from X if it is known that X gained 25% in the transaction ?? please eplain !!! Im getting 60Rs. Ans given Rs. 50
@ChinmayaDave said:Pls explain how?
sum of the numerator = 102 * 50 / 2 = 2550
sum of denominator = n/2 *( 2 + (n-1)2 ) = n^2
==> 2550 / n^2 = 51 / 2
n^2 = 100
n=10
@mitali0390 said:1) a tank holds 100 gallons of water.it's inlet is 7 inches in diameter and fills the tank at 5 gallons/min.the outlet of tank is twice the diameter of inlet.how many minutes will it take to empty the tank if inlet is shut off,when tank is fill and outlet is opened?(answer: 10 min)2)a tank of capacity 25l has inlet and outlet tap.if both are opened simultaneously,tank is filled in 5 min.but if outlet flow rate is doubled and taps opened,the tank never gets filled up.what is outlet flow rate?(answer: 6)3)a tank of capacity 3600m^3 capacity is being filled with water.the delivery of pump discharging the tank is 20% more than delivery of pump filling the same tank.as a result 12 min more tank needs to fill tank than to discharge it.determine delivery of pump discharging tank.(answer:60)plz share ur approach for such questions..
Solutions for 2 and 3 are as follows:
2) Since the tank is being filled up in 5 mins therefore the amount filled in one min = 5 litres. Now the flow of water in the Inlet will be more than 5 litres(some value) and the flow of water from the outlet would be less than 5 litres(some value), only then the average flow rate can be equal to 5.
Second condition is when the flow rate in outlet is doubled the tank is never filled implies that when the rate is doubled it will be equal to the flow rate of the outlet, which we have already concluded to be more than 5..... mind you this cannot be solved by this method without options....
3) let the inlet flow rate be X then the outflow rate would be 6X/5.(20% more it says)... hence the equation would be
3600/X + 3600/(6X/5) = 12...on solving this you get X to be 50 and 6X/5 to be 60...
1) For the first option it simply assumes that the rate of flow is directly proportional to the diameter...
So when diameter is 7 ------ rate is 5
so when diameter is 14 ------ rate is 10....
so 100 gallons/10 = 10 mins..
i don't know how that is done....if somebody could clarify how then it would be of great help....
@ziatron said:A sold a table to B at a profit of 20%. B sold the same table to C for Rs. 75 thereby making a profit of 25%. Find the price at which A brought the table from X if it is known that X gained 25% in the transaction ?? please eplain !!! Im getting 60Rs. Ans given Rs. 50
check for calculation mistake...i am getting 50 as well....straight forward equation of 1st degree
@avaneeshvyas said:Solutions for 2 and 3 are as follows:2) Since the tank is being filled up in 5 mins therefore the amount filled in one min = 5 litres. Now the flow of water in the Inlet will be more than 5 litres(some value) and the flow of water from the outlet would be less than 5 litres(some value), only then the average flow rate can be equal to 5.Second condition is when the flow rate in outlet is doubled the tank is never filled implies that when the rate is doubled it will be equal to the flow rate of the outlet, which we have already concluded to be more than 5..... mind you this cannot be solved by this method without options....3) let the inlet flow rate be X then the outflow rate would be 6X/5.(20% more it says)... hence the equation would be3600/X + 3600/(6X/5) = 12...on solving this you get X to be 50 and 6X/5 to be 60...1) For the first option it simply assumes that the rate of flow is directly proportional to the diameter...So when diameter is 7 ------ rate is 5so when diameter is 14 ------ rate is 10....so 100 gallons/10 = 10 mins..i don't know how that is done....if somebody could clarify how then it would be of great help....
Sorry for the type for the solution of 3rd the equ. is
3600/X - 3600/(6X/5) = 12
anyone have pdf for logarithms,plz share here...
Kindly solve these problems..................
1) Brass is an alloy of copper and zinc. Bronze is an alloy containing 80% of copper, 45 of zinc and 16% of tin. A fused mass of brass and bronze is found to contain 74% of copper,16% of zinc and 10% of tin. The ratio of copper to zinc in brass is:
a) 64% and 36% b) 335 and 67% c) 505 and 75% d)35% and 65% e) none of these
2) Concentrations of three wines A, B and C 10% , 20% and 30% respectively. They are mixed in the ratio 2:3:x resulting in a 23% concentration solution find x
a) 7 b) 6 c)5 d)4 e) 3
What is the remainder when (1!)^3+(2!)^3+(3!)^3+...................(1152!)^3 is divided by 1152?
plz sove
@Revenous said:What is the remainder when (1!)^3+(2!)^3+(3!)^3+...................(1152!)^3 is divided by 1152?plz sove
1152=2^7*3^2
all nos greater than 4!^3 will be div div by 1152 as they have enough 2 , 3 ....
(1+8+216) mod 1152====225
@Rshmi05 said:Kindly solve these problems..................1) Brass is an alloy of copper and zinc. Bronze is an alloy containing 80% of copper, 45 of zinc and 16% of tin. A fused mass of brass and bronze is found to contain 74% of copper,16% of zinc and 10% of tin. The ratio of copper to zinc in brass is:a) 64% and 36% b) 335 and 67% c) 505 and 75% d)35% and 65% e) none of these2) Concentrations of three wines A, B and C 10% , 20% and 30% respectively. They are mixed in the ratio 2:3:x resulting in a 23% concentration solution find xa) 7 b) 6 c)5 d)4 e) 3
2.=> 0.2+ 0.6+ 0.3x = 0.23 (5+x)
solve to get x=5
1.=> suppose 100 gms liya hamne bronze ki quantity aur quantity of brass be X,,,,abb jo tin
hai 10% ka naye alloy me vo sirf broze ki vajah se hi aya hoga......
so 10/100*(100+x)=16/100*100
x=60....
now now total copper is 74/100*160 in new alloy..
80+a/100*60=74/100*160
solve to get a=64 { a is the percentage of copper in Brass)
so Ans .....copper =64% and zinc=36%
@abhijit111 : Hi Abhijit,
My name is Nishath....This is my first time for CAT, I am beginning things from scratch...Is Arun Sharma's book the best for Quant??? can you suggest which is the latest edition of this book please?
@abhijit111 Hi Abhijit.... i am pretty new to CAT & this will be my first attempt, can you please tell me which is the latest edition for Arun Sharma's Quant please?
Q) if one root of the eqn (I-m)x^2 + Ix +1=0 is double of the other and is real,find the greatest value of m
a)9/8 b)8/7 c)8/6 d)7/5 e)5/7
if a,b,c,d,e are distinct natural numbers.a+b+c+d+e=55 what is the minimum and maximum value of a^2+b^2+c^2+d^2+e^2?
@vabhuna said:if a,b,c,d,e are distinct natural numbers.a+b+c+d+e=55 what is the minimum and maximum value of a^2+b^2+c^2+d^2+e^2?
min value at a=9,b=10,c=11,d=12,e=13
max value at a=1,b=2,c=3,d=4,e=45
@rayus
as 35 questions have to be unique thus one has to make 35x25 questions for 25 sets .Now for the rest 15 questions out of 50 in one set the person will have to put those 15 questions in second paper also(as we have to have exactly 35 unique).so this way we will have 25/2 i.e 12 pairs plus one single set of DI .That means last pair will take the single section means now there will be 11 pairs of sections having same 15 questions and one group having 3 test papers.This lead to minimum questions = 15x12 + 35x25(previously calculated)...enjoy.
plz solve this question...
in a factory making radio active substances ,it was considered that the three uranium cubes togrther can be hazardous.so the company authorities decided to have the stack of uranium interspreaded by a lead cube.but a new worker ,who was unaware of the rules arranged the cubes the way he wanted .what is the number of hazardous combination of uranium in a stack of 5?