A man and a boy working together can finish a piece of work in 24 days. If for the last 6 days, the man alone works, then the work finished in 26 days. In how many days can the boy alone finish the work??

statement1: They can finish work in 24DStatement2: They worked for 20D together, last 6D man worked alone. So 4D work done by Boy is equal to 2D work done by man.So, their capacities might be 2units and 1unit for man and boy respectively.For 24D, we have 24*3=72 units of work.So for boy it will take 72 days to finish the work, as he do 1 unit of work on everyday

1/24 = 1/M + 1/B5/6th of the work is completed in 20 days. Man compltes 1/6th work in 6 remaining days. So, man will take 36 days to complete work alone. M = 36. Subsituting, B = 72 days.

For which of the following values of K, does the equation |x2 – 12x + 32| = K have exactly four roots? a) 0 b)2 c)4 d)6

Ans. b)2|x2 - 12x + 32| = K can be written as x2 - 12x + 32 = +/- K For 4 solutions, the value of B^2 - 4AC should be positive for both +/- K. Only valid in case of 2. K = 0 has only 2 real rootsK = 4 has only 3 real rootsK = 8 has 2 real and 2 imaginary roots

how to solve quantitative ability questions?

Hi Sanjana, There are some practices that you need to follow to improve your performance in quant.The following blog on quantitative aptitude is very well written and mentions all the practices needed to be followed.Link: https://cracku.in/blog/6-important-tips-quant-cat-2016/Thanks,Varun.

find the three natural no. whose sum is 30

take no as x,x+1,x+2 then given x+x+1+x+2= 30 then by caluvlation x= 9 then no is 9,10,11

I am scoring 30-50 marks in mocks. how can I get 100+ marks in CAT?

answer those question only for which you are sure of focus on accuracy not on no. of question to be attempted and see the change

how many four digits odd no. can be formed such that every 3 in the no, is followed by a 6?

Hi Gaurav,There can be a maximum of one 3 in the four digit number.The three cases1)3 6 _ _2)_ 36_ 3)_ _ _ _ In the first case 36(9*4) numbers are possibleIn the second case 32(8*2) numbers as possibleIn the third case 2592(8*9*9*4) numbers possibleIn total the number of cases = 2592+68 = 2660Thanks,Varun.

raju has forgotten his six digit id no. he remembers the following the first two digits are either 1,5 or 2,6 the no. is even and 6 appears twice.if raju uses trial and error process to find his ID no. at the most, how many trials does he need to suceed

Hi Gaurav, 1: 1, 5, _, _, _, _ There are two cases possible. If the unit digit is 6 and if the unit digit is not 6. (i) If the unit digit is 6 1, 5, _, _, _, 6 6 can be placed in any of the three places, and the remaining can be arranged in 9*9 ways. Therefore, the total number of ways = 3*9*9 = 243 (ii) If the unit digit is not 6 1, 5, _, _, _, _ The last digit will have 4 possibilities. (0,2,4,8) For the remaining digits, 2 digits will be 6, and the other can be any number other than 6. Therefore, the total number of ways = 3*9*4 = 108 2. 2, 6, _, _, _, _ (i) If the last digit is 6 2, 6, _, _, _, 6 Total number of ways = 9*9*9 = 729 (ii) If the last digit is not 6 2, 6, _, _, _, _ The last digit has 4 possibilities, and the remaining numbers can be arranged in 3*9*9 ways. Therefore, the total number of ways = 4*3*9*9 = 972 Maximum number of trials required = 108 + 243 + 729 + 972 = 2052 Hope this helps!

Hi Gaurav,Let's consider the 2 cases1: 1 5 _ _ _ _ Now, unit digit can be 0/2/4/6/8.If unit digit is not 6, it can be selected in 4 ways. But, number must have 2 6s. The third number required can be chosen in 10 ways. So, the total ways will be 3!/2! (arranging 3 digits) X 10 X 4 = 120If unit digit is 6, other 3 numbers must have exactly 1 '6'. The other 2 numbers can be selected in 10 X 10 = 100 ways. So, number of arrangements will be 3! X 100 = 600So, this case must have 720 different arrangements.You can repeat the same process for case 2 as well which is 2 6 _ _ _ _

Akhilesh Singh sir please help me to solve the question posted in the previous question please Akhilesh sir please

Hi Sam,It would be great if you post the question or the a link to the question. So that we can help you better.Thanks

find the value of 72540/5 + 72540/25 +72540/125 & the ans is given on my book is 17967.6 how to solve Akhilesh Singh sir please help me to solve this the process that U have said once is not matching with the ans in my booklet and if the ans in my booklet is wrong then please let me know

hi sam the easiest mathod to solve this question. take 5 as 10/2 and 25 as 100/4 and 125 as. 1000/8 just multiply the with 2,4 and 8 respectively and then put decimal.