A three-digit no when multiplied by 2, gives a four-digit no with at least one digit as 2. How many such three-digit nos are possible ??1) 176 2) 200 3) 576 4) 324 5) NOT
This should be 176.
My approach: instead of concentrating on the 3-digit number, let's focus on the 4-digit number. Now the 4 digit number should be an even number (since 2* 3-digit number!) between 1000 and 1998. There will be 500 such numbers. We need those which have at least one 2, so let's subtract those with no 2, which can be written in 9 x 9 x 4 = 324 ways using Fundamental Principle of Counting. Hence numbers with at least one 2 are 500 – 324 = 176.
A three-digit no when multiplied by 2, gives a four-digit no with at least one digit as 2. How many such three-digit nos are possible ??1) 176 2) 200 3) 576 4) 324 5) NOT
there is number which is minimum is 500 and maximum number is 999
500*2=1000 and 999*2=1998
I am checking numbers between 1000 and 1998 having 2
A and M can do a work in 12 days. M and B can do a work in 15 days. if A is twice as good a workman as B. in how many daya will mohan alone can do the work??
An approach for those who don't like fractions :D
Take LCM of 12, 15 which is 60. Assume 60 pieces of work. A and M do 5 pieces, B and M do 4 so A - B = 1 but A = 2B so B = 1, A = 2, M = 3 and so M alone will do in 60/3 = 20 days. regards scrabbler
here's my explanation. see 4 digit number will be formed starting from 500 - 999 which means maximum of 500 numbers are possible.. now, we need a 2 in any one of the places / all the places.. for this to happen, we must have 1 or 6 in that 3 digit number. .. 600-649 will always have 2, they'll be of the form 12xx... total 50 such numbers also.. like 501 *2 = 1002.. or 501,506,510,511,512,513,514,516,521,526,531,536,541,546,551,556,560,561,566,571,576,581,586,591,596... total 25 numbers. this will be true for 500,700, 800, 900 as well. so 25*4= 100 before we had 50. so total 150 till now. now we'll check for 650-699 651,656,660,661,666,671,676,681,686,691,696 11 numbers.. total 161 nmbrs i got.. :( what am i missing?
Not correct! Ratio of sides is proportional to sine of the corresponding angles!I have no idea of the correct solution approach though - working on it...but not seeing anything (don't know trigo formulae at all!)regardsscrabbler
yess... apply either sine or cosine formua
a/sinA=b/SinB= c/sinC
or cosA=(b^2+c^2- a^2) /2bc ... n so on.
i tried applying cosine formula
A = cos inverse of 3/4 and B = cos inverse (3/4)^2
here's my explanation.see 4 digit number will be formed starting from 500 - 999which means maximum of 500 numbers are possible..now, we need a 2 in any one of the places / all the places..for this to happen, we must have 1 or 6 in that 3 digit number. .. 600-649 will always have 2, they'll be of the form 12xx... total 50 such numbersalso..like 501 *2 = 1002..or 501,506,510,511,512,513,514,516,521,526,531,536,541,546,551,556,560,561,566,571,576,581,586,591,596...
Here you missed 562, 563, 564. Ditto in 700s, 800s, 900s => 3 * 4 = 12 more numbers.
total 25 numbers. this will be true for 500,700, 800, 900 as well.so 25*4= 100before we had 50.so total 150 till now.now we'll check for 650-699651,656,660,661,666,671,676,681,686,691,696
Here you missed 562, 563, 564. Ditto in 700s, 800s, 900s => 3 * 4 = 12 more numbers.Here you missed 662, 663, 664. Another 3.161 + 15 = 176.regardsscrabbler
84^79 mod 100 84^79 mod 4 = 0 84^79 mod 25 = 9^79 mod 25 = 3^158 mod 25 = ((3^3)^52 * 9) mod 25 = (2^52 * 9) mod 25 = (2^12 * 9) mod 25 = (4096 * 9) mod 25 = (21 * 9) mod 25 = 189 mod 25 = 14 mod 25
Tips to Crack Divisions1. 145/182= Step1- add 10% to numerator and denominator 159/200=79.5 % OA is 79.6 %2. 532/745 adding 1% to numerator and denominator 537/750(approx) adding 1/3rd (537+179)/1000 716/1000=71.6% OA-71.5Please comment Giving more examplesHappy to solve them using the above Method
Not correct! Ratio of sides is proportional to sine of the corresponding angles!