Question

Find three consecutive numbers such that if they are divided by 10, 17, and 26 respectively, the sum of their quotients will be 10.
(Hint: Let the consecutive numbers = x , x + 1, x + 2 , then x/10 + (x+1)/17 +(x+2)/26 =10

Answer 1

Let X, (X+1), (X+2) are 3 consecutive numbers.

X/10 + (X+1)/17+ (X+2)/26 =10

L.C.M = 4420

(442(X) +260(X+1) +170(X+2))/4420=10

(442x+260x+260+170x+340)= 4420×10

(442x+260x+170x+260+340)=44200

872x + 600= 44200

872x= 44200- 600

872x= 43600

X= 43600/872

X= 50
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Ist no. (X)= 50
2nd no.(x+1)= 50+1= 51
3rd no.(x+2)= 50+2= 52
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The 3 consecutive numbers are 50, 51, 52
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Hope this will help you...

Answer 2

Solution :-

Let the three consecutive numbers be x, x + 1 and x + 2 respectively.

Then, according to the question.

⇒ x/10 + (x + 1)/17 + (x + 2)/26 = 10

Taking L.C.M. of the denominators and then solving it.

L.C.M. of 10, 17 and 26 is 2210

x/10  +  (x + 1)/17  +  (x + 2)/26  =  10

221x + 130x + 130 + 85x + 170  = 10 
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                 2210

⇒ (221x + 130x + 85x + 130 + 170)/2210 = 10

⇒ (436x + 300)/2210 = 10

⇒ 436x + 300 = 2210*10

⇒ 436x = 22100 - 300

⇒ 436x = 21800

⇒ x = 21800/436

⇒ x = 50

Now, the first number is

x = 50

Second number is (x + 1) = 50 + 1 = 51

Third number is (x + 2) = 50 + 2 = 52

Three consecutive numbers are 50, 51 and 52 respectively.

Answer.