Find two numbers such that one third of greater exceeds one half on the lesser by 1, and one fifth o

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04 Jun 2013 Problem

Find two numbers such that one third of greater exceeds one half on the lesser by 1, and one fifth of the greater added to one sixth of the lesser equals one half of the lesser.

OPtion

1) 15,9
2) 60,36
3) 30,18
4) 20,10
5) 50,25
6) 32,16
7) 28,12
8) 32,18
9) 46,26
10) None of these

Solution

Now, according to question
(1/3)x=(1/2)y+1 (let larger no x, smallest no y)
and (1/5)x+(1/6)y=(1/2)y
On solving (i) and (ii), we get x=30 and y =18
option 3

Correct Option: 3 Best Solution (11)

Dr.Dipin Singh   11 years AGO

If x ( greater) and y ( lesser) are the numbers

then
x/5 +y/6= y/2
3x=5y .. (1)
also x/3 = y/2+1 or 2x=3y+6 .... (2)

solving (1) and (2), we get
x=30,y=18 ... option 3

Like? Yes (3) | No (1)  

Vinay Sipani    11 years AGO

(x/3)=(y/2)+1
=> 2x - 3y = 6 ----------(1)

(x/5)+(y/6)=(y/2)
=> 6x - 15y = 0 ----------(2)

solving (1) and (2)
=> x = 30 , y = 18

Like? Yes (3) | No (2)  

Adarsha Upadhya   11 years AGO

let 2 no\'s x and y
consider x is greater and y is smaller
according to the problem
one third of greater exceeds one half on the lesser by 1
written as 1/3x-1/2y=1 -----1

one fifth of the greater added to one sixth of the lesser equals one half of the lesser
written as 1/5x+1/6y=1/2y -----2

on solving 2 we get x=5y/3;

substitute same in eqn 1
we get y=18 and x=30

Like? Yes (2) | No (1)  

vigneshwari   11 years AGO

Let X and Y be two given greater and lesser numbers respectively.
From given, we have
X/3 - Y/2 = 1 => 2X - 3Y = 6 .......(1)
And, X/5 + Y/6 = Y/2
=> X/5 - Y/3 = 0 => 3X - 5Y = 0 ......(2)
Solving (1) and (2), we would have
X = 30 and Y = 18.

Like? Yes (1) | No (1)  

vinod   11 years AGO

(x/3)=(y/2)+1
(x/5)+(y/6)=(y/2) verify options

Like? Yes (1) | No (1)  

enthura90   11 years AGO

g/3-l/2=6
g/5+l/6=l/2

Like? Yes (1) | No (1)  

Shoaib   11 years AGO

30*1/3=10
18*1/2=9

10-9=1

30*1/5=6
18*1/6=3

6+3=9

Like? Yes (1) | No (1)  

ajitha   11 years AGO

Assume x and y are the numbers
Then y/3=1+x/2
and y/5+x/6=x/2
By solving the above eq'ns bothe umbers are 30 and 18

Like? Yes (2) | No (2)  

Anil Kumar Bharti   11 years AGO

Ans 3) 30,18
Exp:
let gt no=x, ls no=y
x/3-y/2=1 ---(I)
x/5-y/3=0 ---(II)
Solve eqn I & II
x=30, y=18

Like? Yes (1) | No (1)  

ravishankar   11 years AGO

trial method are 1/3of 30= 10 and 1/2 of 18=9 implies 10 >9 by 1 first condition goes right and 1/5 of 30=6 1/6 of 18=3 when added equals 9 which is one half of lesser 18=9 thus option 3 goes right
(or)
1/3of G=(1/2 of L )+ 1... (1)
1/5 of G + 1/6 of L = 1/2 of L ... (2) on soloving equations we get 30,18..

Like? Yes (1) | No (1)  

Balakrishnan M   11 years AGO

Two Numbers are 30,18.
Condition 1:
one third of greater = (1/3)(30)=10
one half on the lesser=(1/2)(18)=9
Hence one third of greater exceeds one half on the lesser by 1. (10 =9+1)
Condition 2:
one fifth of the greater = (1/5)(30)=6
one sixth of the lesser = (1/6)(18)=3
one half of the lesser = (1/2)(18)=9
Hence one fifth of the greater added to one sixth of the lesser equals one half of the lesser.
6+3=9.

Like? Yes (1) | No (1)