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Mathematics 16 Online
OpenStudy (anonymous):

The least number which when divided by 48,64,90,120 leave the remainders 38,54,80,110 respectively is

OpenStudy (anonymous):

The first solid hint here is that the remainder when divided by 120 is 110. That means whatever it is, it is divisible by 10. I'm checking mentally to see if guess-and-check is an efficient way to solve this.

OpenStudy (anonymous):

Oh, note that 48*5=240=120*2 In fact, most of these numbers have only a few prime factors different from each other.

OpenStudy (anonymous):

ho we can solve this typpe of ques fast

OpenStudy (anonymous):

can u plz explain me properly

OpenStudy (anonymous):

Note the similarities in the numbers. 90*8=720=120*6=48*15 All remainders are 10 less than the number, so we now know that this is 10 less than a multiple of 720

OpenStudy (anonymous):

That just leaves 64, which, come to think of it, also leaves a remainder 10 less than itself.

OpenStudy (anonymous):

64*45=2880=720*4. We now have a number with all of these factors in common. There is, of course, a faster way!

OpenStudy (anonymous):

is there any simple way to solve such problems?

OpenStudy (anonymous):

Once you've noticed that each remainder is 10 less than the number itself, factor each number before you start multiplying.

OpenStudy (anonymous):

can u plz show by this quest the fast method?

OpenStudy (anonymous):

OK. 48, 64, 90, and 120 leave remainders of 38, 54, 80, and 110. Each remainder is 10 less than the original number. This means our final number must be the Least Common Multiple (LCM) of these four numbers minus ten.

OpenStudy (anonymous):

got the ans 2870..is it right? I took the LCM of 48,64,90,120 & then less 10 from it

OpenStudy (anonymous):

\[ 48=2^4*3^1. 64=2^6. 90=2^1*3^2*5^1. 120= 2^3*3^1*5^1.\] Take the highest power of each prime factor, then multiply them: \[2^6 from64, 3^2 from90, and 5^1 from 90 or 120\] \[ 2^6*3^2*5^1=2880 \] That means you're right!

OpenStudy (anonymous):

thankyou

OpenStudy (anonymous):

A,B,C subscribs together Rs 50,000 in a business.A subscribes 4000 more than B,B subscribes 5000more than C.Out of the toal profit of 35000,A recieves?

OpenStudy (anonymous):

A=C+9,000 B=C+5,000 Total=50,000=A+B+C=14,000+3C Solve for A, B, and C, then take that ratio and apply it to 35,000.

OpenStudy (anonymous):

I got equation as 14000+3C...after that?

OpenStudy (anonymous):

Solve for C, then A and B.

OpenStudy (anonymous):

C=14000/3

OpenStudy (anonymous):

No. 50,000=14,000+C

OpenStudy (anonymous):

A=21000 B=17000 C=12000 now instead of 50000 I can put 35000 & solve for ABC?

OpenStudy (anonymous):

A=16000 B=12000 C=7000 is it correct?

OpenStudy (anonymous):

?

OpenStudy (anonymous):

Not quite. You had A,B,C right. Consider the percentages.

OpenStudy (anonymous):

plz help me

OpenStudy (anonymous):

OK. A, B, and C are all out of 50,000. The percentage is just double the number of thousands (x,000/50,000 *100%=2x)

OpenStudy (anonymous):

So A=42%, B=34%, and C=24%. Expressed as a ratio, that's \[0.42:0.34:0.24\] Multiply each of those by 35,000 and you have your answer.

OpenStudy (anonymous):

got it 14700

OpenStudy (anonymous):

I'm going to bed. I'll be back in the morning! Perhaps you should post separate questions on separate posts in future.

OpenStudy (anonymous):

when u will be ol bcz I have some more quest...tell me whr to post the question bcz I can easily understand ur way of explaining.

OpenStudy (anonymous):

You can always message me. I'll be back on some time in the next 24 hours, that's all I know at this point.

OpenStudy (anonymous):

whr to msg u? here only....or I can post u quest & u can give me ans

OpenStudy (anonymous):

I am new to this site I dun know whr to msg?

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