Didnt use any logic....just mentally visualized the mutliplication grid and then the answer was obvious...... ought to be a much more elegant way of doing it....but solving via visualization takes only a handful of seconds..so why not!
@anitito123 said:The surface of the water of a swimming pool is a rectangle 26m long and 10m wide and the depth of the water increases uniformly from 1.6m at one end to 4.4m at the other end .Volume (meter cube) of the pool is . options :364, 390, 780, 1560Please mention the approach.
is answer 780
A person starts walking from a point P at 2am and reaches Q at 5am on the same day.Another person starts walking from Q at 4am and reaches P at 9am on the same day.They will cross each other at :
options: 4:22 am, 4:35am, 4:37.5 am, 4:42.5 am
options: 4:22 am, 4:35am, 4:37.5 am, 4:42.5 am
@anitito123 said:A person starts walking from a point P at 2am and reaches Q at 5am on the same day.Another person starts walking from Q at 4am and reaches P at 9am on the same day.They will cross each other at :options: 4:22 am, 4:35am, 4:37.5 am, 4:42.5 am
4 : 37.5
d = S1 * 3 ; d = S2 * 5 ;
at 4 pm person from P would be at D = d/3 * 2 = 2d/3..now two persons are at a distance of d/3..both walk in opp directions ==> t = d/3 / ( d/3 + d/5) = 5/8 hour = 5/8 * 60 = 37.5 minutes...meeting time : 4 : 37.5
d = S1 * 3 ; d = S2 * 5 ;
at 4 pm person from P would be at D = d/3 * 2 = 2d/3..now two persons are at a distance of d/3..both walk in opp directions ==> t = d/3 / ( d/3 + d/5) = 5/8 hour = 5/8 * 60 = 37.5 minutes...meeting time : 4 : 37.5
@anitito123
since options are far apart approx values le lo:
root 2~= 1.3
root 3~= 1.7
root 6= 1.7*1.3 ~= 2.38
numerator becomes ~= 7.56
similarly root(2+ root3) ~= root(3.7) ~= 1.9
denomainator is slightly more than 5.7
now look at the options....and all of them get eliminated except 4/3
since options are far apart approx values le lo:
root 2~= 1.3
root 3~= 1.7
root 6= 1.7*1.3 ~= 2.38
numerator becomes ~= 7.56
similarly root(2+ root3) ~= root(3.7) ~= 1.9
denomainator is slightly more than 5.7
now look at the options....and all of them get eliminated except 4/3
@anitito123 said:solve:2(root2+root6)/(3*root(2+root3))Options: 4/3, 1, 16/9, 2/3
4/3?
take square and then square root of eqn
rt((2(rt2 + rt6)/3*rt(2 + rt3))^2)
rt(4(8 + 4*rt3)/9(2 + rt3)
rt(16/9)
4/3
@anitito123
for the 101 question i came up with a much more elegant logic of solving it than visualization.
take 101. calculate only the non zero digits and not the exact value, it comes out to be 3.
x 101
similarly 10101, comes out to be 5.
x 10101
So now we can generalize the logic as, number of non zero digits in such a sequence = 2*(number of 1's) -1.
hence 2*5-1 = 9.
for the 101 question i came up with a much more elegant logic of solving it than visualization.
take 101. calculate only the non zero digits and not the exact value, it comes out to be 3.
x 101
similarly 10101, comes out to be 5.
x 10101
So now we can generalize the logic as, number of non zero digits in such a sequence = 2*(number of 1's) -1.
hence 2*5-1 = 9.
@anitito123 said:A person starts walking from a point P at 2am and reaches Q at 5am on the same day.Another person starts walking from Q at 4am and reaches P at 9am on the same day.They will cross each other at :options: 4:22 am, 4:35am, 4:37.5 am, 4:42.5 am
It will be 4.37.5 am
Let distance be lcm of 3(5-2) and 5(9-4) =>15 km. So speed of person starting from P is 5 and person starting from q is 3.So by the time person Q starts person P already travelled 10km. remaining is 5 which is travelled by both.
Let distance be lcm of 3(5-2) and 5(9-4) =>15 km. So speed of person starting from P is 5 and person starting from q is 3.So by the time person Q starts person P already travelled 10km. remaining is 5 which is travelled by both.
So meeting time is 4+5/8*60=4.37.5 am
@anitito123
I have found, in questions involving irrational numbers, dont go by the conventional methods and take approx values instead, saves a LOT of time....:)
I have found, in questions involving irrational numbers, dont go by the conventional methods and take approx values instead, saves a LOT of time....:)
@anitito123 said:A person starts walking from a point P at 2am and reaches Q at 5am on the same day.Another person starts walking from Q at 4am and reaches P at 9am on the same day.They will cross each other at :options: 4:22 am, 4:35am, 4:37.5 am, 4:42.5 am
A covers D in 3 hrs...so in 2hrs it will cover---->2*D/3
remaining dist ---->D-2*D/3 = D/3
meeting time = 4+D/3/(D/3+D/5) = 4+5/8*60 = 4:37.5?
@anantn said:@anitito123I have found, in questions involving irrational numbers, dont go by the conventional methods and take approx values instead, saves a LOT of time....
yeah...i too prefer approximations until and unless the options are too close...even in cat used approximation in a log ques :)
@anitito123
Yeah but in cat, you would most probably have a NONE of these in the mix.....so approximations wouldnt work in the question you gave above
Yeah but in cat, you would most probably have a NONE of these in the mix.....so approximations wouldnt work in the question you gave above
@mailtoankit said:A covers D in 3 hrs...so in 2hrs it will cover---->2*D/3remaining dist ---->D-2*D/3 = D/3meeting time = 4+D/3/(D/3+D/5) = 4+5/8*60 = 4:37.5?
correct!
@anitito123 said:A person starts walking from a point P at 2am and reaches Q at 5am on the same day.Another person starts walking from Q at 4am and reaches P at 9am on the same day.They will cross each other at :options: 4:22 am, 4:35am, 4:37.5 am, 4:42.5 am
4:37.5 am.
assume total distance & solve.
assume total distance & solve.
A 5 rupee coin with diameter 2 cm and width .5cm, is tossed into the air. What is the probability of it landing vertically on its thinner side.
@anantn said:@anitito123Yeah but in cat, you would most probably have a NONE of these in the mix.....so approximations wouldnt work in the question you gave above
yeah that's why it's best if u know the conventional method...generally its better to do these approximations if u feel ur attempts are very less and preferably there is no 'none of these' option as u rightly pointed out.
In a class of 45 students, 30 are boys and remaining are girls. If the mean weight of girls is 45 kg and that of boys is 52 kg, then the mean weight of entire class (in kg) is nearest to
(1) 48 (2) 49 (3) 50 (4) 51
(1) 48 (2) 49 (3) 50 (4) 51