Actually best explanation i was depressed before watching this❤

What am I doing wrong here Can anyone help me ? Eq1 1/x+1/y=36 Eq 2 x/3+2y/3=90 =>x=270-2y Replacing 2 in 1 y^2-153y+(270*18)=0 ( firstly this factorisation is too time consuming for a gmat question ) y=135 or y=36 Both of which cannot be true

“Essentially” word is repeated several times in the video’s why???

Thanks,Bhaskar Bhai 🙂

Considering P is the profit Mary got 1/3P ( remuneration ) + 1/4P ( Half of remaining profit ) = 14/24P Mike got 1/6P ( remuneration ) + 1/4P ( Half of remaining profit ) = 10P/24 Mary got 800 more than mike =>(14P/24) -(10P/24)=800 P=4800 Mary got 14P/24 =2800

1. D+E+F= 1280 2. (D+15)+(E+20)+(F+17)=13k+17k+22k D+E+F+52= 52K =>1332=52k k=1332/52 Emma got after addition E+20= 17*1332/52 E= 435+20 = 455.

bonus question answer is $1000 , D?

Sir please make new videos for gmat for 2025-26 batch please sir it would be really helpful for those students who can't afford coaching and wants to pursue MBA

thanks a lot, sir

link ?

B. 360m

Bhaskar bhai tosi great ho ❤

for factorization method if it's not factorized it may waste ur time during the exam so better go directly with the 2nd approach i guess. Thumbs up Sir

2nd approach t Dr, what if in the 3rd one i did 15*2=5*2*3 on trailing zero instead of 15*4

Can you make full video on this with college name and different waivers they offer for application form. It will be great help 😊. Thank you ❤

What about option C? S2 is sufficient when S1 is implied with it right?

Thanks for the video 😊

bonus q answer is D $2800?

Five identical balls are to be separated into -- at most -- 3 groupings. Thus, we need five balls and two separators: OO|OO|O Every arrangement of the elements above represents one way to distribute the 5 balls among three boxes A, B and C: OO|OO|O = A gets 2 balls, B gets 2 balls, C gets 1 ball. OO||OOO = A gets 2 balls, B gets 0 balls, C gets 3 balls. OOOOO|| = A gets all 5 balls. And so on. To count all of the possible distributions, we simply need to count the number of ways to arrange the 7 elements above (the 5 identical balls and the 2 identical separators). The number of ways to arrange 7 elements = 7!. But when an arrangement includes identical elements, we must divide by the number of ways each set of identical arrangements can be arranged. The reason: When the identical elements swap positions, the arrangement doesn't change, reducing the total number of unique arrangements. Here, we must divide by 5! to account for the 5 identical balls and by 2! to account for the 2 identical separators: 7!5!2!=21

Here's how you solve it First of all you have to calculate the total work done by A and B which is 1/60 + 1/120 which is 1/40. Now they have worked together for 24 days right ( because a leaves after 24 days so they have worked together for 24 days) Let's calculate the work done in 24 days which is 24×1/40 which is 3/5. Now remaining work because B will work alone. Remaining work = 1-3/5 which is 2/5 B will complete 2/5 work in 2/5×120 days ( because according to question b can complete one work in 120 days) Calculate this and you will get 48. Now adding 24 days of together and 48 days alone of B we get 72. The answer is 72. Hope you undersstood. Happy preparation.

Could you pls tell us how did you solve the question

Best

Hello sir kindly tell the process of getting application fee waiver of gmat

A. 45

wow great teaching !thank you sir

are these usefull for the one who taking gmat focus edition?

Can you use calculus , the first derivation to find out the x value?? Reply if you are noted ..

Thank you for the encouragement at the very end. Hope all is good for you.

too much boring

Hi, could you please solve the bonus q if possible? When you tell first half and second half for the time, how should i approach. It's a bit confusing.

is this content valid for gmat focus edition too?

smallest percentage difference Nepal and Nigeria , largest is Nigeria and myanmar And Bangladesh receives 32 oercent more than mayanmar

Sir kindly solve this, as I’m getting K as 25.6 and the subsequent answer for Emma what I got is not one in the answer options. I used the same method to solve as the original question

422?

55599

Hi, I wanted to know whether I can access the paid module of wizako through my iPad?

This is remarkable explanation, thank you so much 😮

The probability of hitting the target is 1/4 Therefore, the probability of not hitting the target is = 1 - 1/4 = 3/4 Let E be the event that the man to hit the target twice in four shots, we need to consider the following cases: Case 1: He hits the target in 1st and 2nd shot and does not hit the target in 3rd and 4th shot. P(Case 1) = 1/4 * 1/4 * 3/4 * 3/4 = 9/256 Case 2: He hits the target in 2nd and 3rd shot and does not hit the target in 1st and 4th shot. P(Case 2) = 3/4 * 1/4 * 1/4 * 3/4 = 9/256 Case 3: He hits the target in 3rd shot and 4th shot and does not hit the target in 1st and 2nd shot. P(Case 3) = 3/4 * 3/4 * 1/4 * 1/4 = 9/256 Case 4: He hits the target in 1st shot and 3rd shot and does not hit the target in 2nd and 4th shot. P(Case 4) = 1/4 * 3/4 * 1/4 * 3/4 = 9/256 Case 5: He hits the target in 1st shot and 4th shot and does not hit the target in 2nd and 3rd shot. P(Case 5) = 1/4 * 3/4 * 3/4 * 1/4 = 9/256 Case 6: He hits the target in 2nd shot and 4th shot and does not hit the target in 1st and 3rd shot. P(Case 6) = 3/4 * 1/4 * 3/4 * 1/4 = 9/256 P(E) = 9/256 * 6 = 54/256 OPTION (C)

why have we not discarded 0 since the question says "positive" integers?