Swaroop Garg
1 year, 5 months ago

while doing RCs in CAT i attempt 3 RCs in which i manage 10 questions and 7-8 correct, is it correct or need to improve??

Swaroop Garg
0 Likes Posted 1 year, 5 months ago

what is an ideal time for attempting all RCs ,actually i solve 3 rcs in 40-45 minutes..

Nishank Garg
1 year, 5 months ago

px^2-(p-1)x+2p=0, find its range

Manu Jindal
0 Likes Posted 1 year, 5 months ago

Hi Nishank,

To find the range of p, we have to equate discrominant>=0

(p-1)^2-4*2p*p>=0

7p^2+2p-1<=0

(-1-2root2)/7<=p<=(-1+2root2)/7

SHAHEED HOSSAIN HOSSAIN
1 year, 5 months ago

How to increase the speed in number system problem

Praneeth Madhunanthu
1 Likes Posted 1 year, 5 months ago

1) There are quite a few formulae and concepts in Number System. Knowing the concepts beforehand will help you to get to the trick immediately after reading the question. This saves a lot of time which can be used better elsewhere.

2) A lot of practice is required. Practice all types of problems, but concentrate more on remainders, HCF-LCM and base system as questions from these topics can be quite tricky.

3) Don't depend a lot on the given calculator as it takes a lot of time in clicking each button. Improving your calculation speed will significantly help to reduce the amount of time you spend on a question.

Dipinti Phutela
1 year, 5 months ago

Could anyone please explain the last example given at the end of "Introduction to Quadratic equations" video. It is a graph based question asking for the appropriate equation explaining the graph given.

Nishank Garg
0 Likes Posted 1 year, 5 months ago

the graph shows that the equation has 2 positive roots and 1 negative root. So maximum positive roots for any given option should be two, ie no. of sign changes should be at least 2, its not in option 2 ,

Biswajit Barua
1 year, 5 months ago

Q) Ram and Shyam are competing with each other in a friendly community competition in a pool of 50 m length and the race is for 1,000 m. Ram crosses 50 m in 2 min and Shyam in 3 min 15 s. Each time they meet/cross each other, they do handshake’s. How many such handshake’s will happen? (They both start from the same end at the same time.) options: a)18 b)19 c)23 d)20 e)17

Pv Aaditya
0 Likes Posted 1 year, 5 months ago

When the faster swimmer meets (overtakes) the slower swimmer for the first time, the faster swimmer would have covered 2 laps (one lap = 50m) more than the slower swimmer, and both are swimming in the same direction at this point of time. Let this time be t.

So, t/2 = t/3.25 + 2 => t = 10.4 minutes

So, the faster swimmer overtakes the slower swimmer for the first time at 10.4 minutes after the start of the race. Number of handshakes till then = 5 (they cross each other 4 times and meet one time)

So, number of handshakes by t = 31.2 minutes = 15

From this moment, the 5th handshake occurs at t = 41.6 minutes. But the faster swimmer would have stopped swimming by then.

So, number of handshakes from t = 31.2 to t = 40 is 4

Total number of handshakes = 19

Himanshu Jain
1 year, 5 months ago

find the remainder when 7^99 is divided by 101

Pv Aaditya
1 Likes Posted 1 year, 5 months ago

Euler of 101 = 100

So, 7^100 mod 101 = 1

7 * 7^99 mod 101 = 1

Let 7^99 mod 101 = k

=> 7k mod 101 = 1

=> 7k = 101p + 1

p = 2 satisfies for k = 29

So, remaninder = 29

Akshay Kaushal
1 year, 5 months ago

In a kilometer race, A beats B by 1 minute. In the same race, B beats C by 30 seconds. If A beats C by 600 m, find the ratio of speeds of A and B. Please share you approach.

Pv Aaditya
2 Likes Posted 1 year, 5 months ago

A ran 1000 m in the time that C ran 400 m. So, ratio of speeds of A and C is 5:2

Let the ratio of speeds of A, B and C be 5:x:2. So, speeds = 5k, xk and 2k

Time taken by B - Time taken by A = 1000/xk - 1000/5k = 60 seconds

Time taken by C - Time taken by B = 1000/2k - 1000/xk = 30 seconds

Equating for k, we get, 200*(5/x - 1)/60 = 500*(1-2/x)/30 => x = 2.5

So, speed of A : Speed of B = 5 : 2.5 = 2:1

Akshay Kaushal
1 year, 5 months ago

A boy travelled half the distance from his house to school at a speed of 10 km/hr. For exactly half of the remaining time he travelled at 15 km/hr and the rest of the time at 20 km/hr. Find his average speed. Options: A) 13 km/hr B) 140/11 km/hr C) 15 km/hr D) 97/6 km/hr

Pv Aaditya
1 Likes Posted 1 year, 5 months ago

Let the total distance be 2d.

Total time = d/10 + x/15 + (d-x)/20

x/15 = (d-x)/20 => x = 3d/7

Total time = d/10 + 3d/105 + 4d/140 = 11d/70

Average speed = Total distance/Total time = 2d/(11d/70) = 140/11 kmph

135 points

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Hi Swaroop, 80% accuracy is good. It would be great if you can increase the number of attempts. Ideally, if you can attempt all the RCs with 80% accuracy, that should be a very good score.