Hi Puys, Please solve and explain the following question
the ratio of Jims salary for October to his salry for November is 1.5:1.333 and the ratio of salary for November ato December is 2:2.6666. the worker got 40 rs. More for December than for October and received a bonus constituting 40% of salary for 3 months. Find the bonus.( assume that the no. of workdays is same in every month) a) 368.888 b) 152.5555 c) 265.6 d) 222.22 e)none
Oct:Nov = 3/2:4/3 = 9:8 Nov:Dec = 2:8/3 = 6:8
Combining the ratio for three months gives us 27:24:32
Now the difference between Oct and Dec is Rs.40 So, 5x=40 => x=8
Q.67-Lod2-NS-> K is a 3-digit number such tat the ratio of number to the sum of its digits is least. what is the difference between hundreds and the tens digit of K. (a) 9 (b) 8 (c) 7 (d) none of these.
hi i did this through a little bit of error and trial method let x, y, z, be the hundreds, tens and units digit. clearly we want the ratio of 100x+10y+z/x+y+z now we want this ratio to be least . for this two objectives must be fulfilled numerator should be least and denominator should be max the max effect will be when we when the units digit is highest eg 100/1 and 109/10.hence deno is max similiary in order to have less numerator to be min let the hundred digit be 1 now in case of tens digit just check a litle by putting 1 and 9 .thus we see 199 is the no ..
hope this solves some of the doubt
this process wuld nt help if the ratio asked to be highest .
this process wuld nt help if the ratio asked to be highest .
when u consider sets of 10 i.e, suppose 101-109. 101 has highest ratio and 109 has least.similarily 111-119 and so on. so for least consider 109,119..199..299..999.among these the order is like dis.. 199 is least for 109-199.299 is least for 209 to 299 and so on.. s0 199 has least ratio.
now for highest anyways u can see 100, 200,300,400,500,600,700,800,900 all have same ratio and its the max.
Find the 28383rd term of series 12345678910111213....
a 3 b 4 c 9 d 7
plz xplain the steps
1-9 there are 9 terms 10-99 there are 180 terms 100-999 there are 2700 terms till now in total 2700+180+9=2889 terms we need 28383 rd term 28383-2889=25494 25494/4=6373.5 starting frm 1000 6373rd number is 6372 therefore 28383rd number is 2nd digit of 6373 hence the answer is (A) 3
hey can anyone give me the working steps of these questions???? 1)Find the average of first 97 natural numbers? 2)Find the average of first 10 multiples of 3?
hey can anyone give me the working steps of these questions???? 1)Find the average of first 97 natural numbers? 2)Find the average of first 10 multiples of 3?
1) They will form an A.P of c.d =1 Sum of an A.P=(n/2)*(a + l) Average=Sum/n= (1/2)*(a + l) Average=(1/2)*(1+ 97) = 49
2)First term will be 3,last term will be 30,difference=3,n=10 Hence average = (1/2)*(a + l) = (1/2)*(3+30) =33/2
hey can anyone give me the working steps of these questions???? 1)Find the average of first 97 natural numbers? 2)Find the average of first 10 multiples of 3?
First find the sum of all the numbers from 1 to 97
are those books new and what abt the old books market near jnu and iit
u will not find the latest edition there...but still a year or two old version u will find there......but u will many good MBA books...thats guranteed and u can obviously bargain with them.......sorry no idea about those jnu and iit market
thanks everyone for their reply....really appreciated
1)3 yrs ago,average age of A,B & C was 27 and of B & C 5 yrs ago was 20 yrs.What is A's present age?
2)The average salary of all the workers is 95.Avg salary of 15 workers is 525 and the avg salary of rest of the workers is 85.Find the total no. of workers?
3)In a bag,there r 150 coins of Re.1,50p & 25p denominations.If the total value of coins is 150.Find how many rupees can be constituted by 50p coins.
