Civil Engineering Mechanical Engineering Chemical Engineering Networking Database Questions Computer Science Basic Electronics Digital Electronics Electronic Devices Circuit Simulation Electrical Enigneering Engineering Mechanics Technical Drawing

# Aptitude - Problems on Ages - Discussion

@ : Home > Aptitude > Problems on Ages > General Questions - Discussion

### Read more:

"The secret to creativity is knowing how to hide your sources."
- Albert Einstein
2.

The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?

 [A]. 4 years [B]. 8 years [C]. 10 years [D]. None of these

Answer: Option A

Explanation:

Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.

Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50

5x = 20

x = 4.

Age of the youngest child = x = 4 years.

 Vincent said: (Thu, Aug 26, 2010 06:47:27 PM) How did we get 5x = 20 ?

 Martin said: (Sat, Sep 11, 2010 07:22:07 PM) x+3+x+6+x+9+x+12=30 5x+30=50 5x=50-30 5x=20 x=4.

 Priya said: (Tue, Sep 28, 2010 02:34:03 AM) How to form this step : x+3+x+6+x+9+x+12 ?

 Gyan said: (Tue, Oct 5, 2010 06:36:00 AM) Jus Add x+0 X+3 X+6 X+9 X+12

 Cherry said: (Thu, Oct 7, 2010 10:00:33 AM) x+3x+6x+9x+12x=50 x(1+3+6+9+12)=50 x=50/31 x=1.64 If we do the same like this what is that x mean?

 Mani Ramani said: (Tue, Oct 26, 2010 03:32:47 AM) Hi cherry.. you just add all the X first to get 5X then add numbers to get 30. Then add 5x+30=50 5X=50-30(shift 30 to right side) 5x=20 X=20/5 X=4

 Nandish said: (Sat, Nov 20, 2010 11:32:15 AM) x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50 how 50 come

 Prasad Baanu said: (Sat, Dec 4, 2010 10:28:40 PM) Hai Nandish..., They have already gave in the question that the sum of ages of those boys are 50:-) so x + (x+3)+ (x+6)+ (x+9)+(x+12) = 50

 Nandan said: (Wed, Feb 2, 2011 02:33:01 AM) If 1st child age x then 3year gap for next child so x+3 5 children = 5x 5x+30=50 5x=50-30 5x=20 x=20/5 x=4 so 4 years

 Raj said: (Tue, Feb 15, 2011 09:41:52 AM) Use arithmetic prog... formula s=n/2[2a+(n-1)d] where s=total sum of numbers,n-count the numbers ,d-diff&a-intial no. so they ask intial no i.e a? s=50,n=5,d=3,a? 50=5/2[2a+12]=>a=4...

 Jitu said: (Sat, Mar 19, 2011 03:59:27 AM) How can I submitt a question?

 Jinju said: (Sun, Mar 27, 2011 07:51:44 AM) We can easily understand if the avg is 8 then youngestg should be less thgan 8. Then after we calculate the validity of remaining values.

 Stella said: (Wed, Jun 15, 2011 01:32:17 AM) Hi cherry, you just see your calculations: x,3x,6x,9x,12x.(wrong) u have to add x,x+3,x+6,x+9,x+12.(this is the correct format) now u have to add and get the result ya.

 Satheesh said: (Mon, Jun 27, 2011 02:27:01 AM) It seems these calculation are merely mathematical wise not logic wise. Whats the relation between X and younger child?

 Gkgokulan said: (Wed, Jul 13, 2011 12:16:05 PM) Its simple logic frnds 4+7+10+13+16=50. Then the answer is 4.

 Ishu Subham said: (Thu, Aug 4, 2011 12:12:16 PM) Let the ages be (x-6), (x-3), (x), (x+3), (x+6). then sum of ages = 5x => 5x = 50 x = 10 youngest = x-6 =4

 Sangeetha said: (Fri, Aug 5, 2011 10:04:45 PM) Thank you ishu subham, very simple and good logic.

 Durgesh said: (Thu, Aug 11, 2011 07:48:08 PM) Avg. of age of 5 children = 50/5 = 10 Now there are 5 no. having sum 50 and avg. as 10 so no. can only be = (10-6),(10=3),(10),(10+3),(10+6) =4 , 7 ,10 ,13 ,16

 Neeraj Dixit said: (Sat, Sep 17, 2011 04:26:57 PM) Just go through answer select 4 first will be 4 than 7 than 10, 13, 16 add all five you will get 50.

 Vineshgujjari said: (Sun, Sep 25, 2011 09:50:23 PM) First of all, Raj Thank you very much. But I want to ask you that is this formula will ramain for all the sums of this type i.e. problems of ages. Please reply me asap.

 Vineshgujjari said: (Sun, Sep 25, 2011 09:57:24 PM) Since given 5 children=50 let one children be x given interval 3 each so x+(x+3)+(x+6)+(x+9)+(x+12)=50 5x+30=50 x=20/5 x=4..

 Kumuda said: (Wed, Sep 28, 2011 12:37:11 PM) Younger means the person who born at last, then who is the last one, according to problem the person x+12. Submit X value in x+12=16. Can you please any one clear my doubt ?

 Srujan said: (Wed, Oct 12, 2011 06:31:26 PM) @kumuda sub x value in all then u will get 4,7,10,13,16. So among them d least age is 4 i.e younger one.

 Sudheer said: (Mon, Oct 24, 2011 12:00:18 PM) Let x+(x-3)+(x-6)+(x-9)+(x-12)=50 5x-30=50 5x=80 x=16 is this correct or not? if it is wrong then explain the reason..

 Prince said: (Wed, Nov 16, 2011 02:23:59 PM) For all who don't understand here is the simple explanation. Let the last child ( 5th ) born be x years old, 3 years before the 4th child was born so his age has to be x+3, now the 3rd child was born 3 years before the 4th child. so the age of 3rd child is 3+3 ie 6+x. so on..we get x ( last child ) + x+3 ( a child before x )+ x+6 + x+9 + x+12 =50. so x+x+x+x+x + 3+6+9+12 =50 5x + 30 =50 5x = 50-30 5x = 20 x = 4. where x is the youngest child. Explanation in Hindi: :P jo last baccha hua uski age x hai...jo usse 3 saal pehle paida hua hoga toh uski age to x se 3 saal badi ho hogi na...so x+3, so usse 3 saal pehle hua hoga uski age x se 6 saal badi and so on.

 Udhay Chandran said: (Sun, Jan 15, 2012 12:57:45 AM) Lets assume that the first elder son age is X, then the difference in the age of 3 years among the consecutive child summing to the total age of 50 would be follow.Here (X-3) corresponds to 2nd elder child age and (X-6) corresponds to 3rd elder child age and so on x+(x-3)+(x-6)+(x-9)+(x-12)=50 ----> (1) 5x-30=50 5x=80 x=16 The (x-12) in equation 1 corresponds to 5th child age and by substituting X = 16 & (X-12) = Y in equation 1 we ill get 5th child age as below 16 + 13 + 10 + 7 + Y = 50 46 + y =50 Y = 50-46 Y = 4

 Mitali said: (Sat, Mar 31, 2012 04:32:35 PM) I have a other way to solve it. Avg of 50 = 50/5 = 10. Hence, the middle No. In the series i.e. x+6=10 (middle no is the avg value of the series). So x = 10-6= 4.

 Em0 said: (Sat, Jun 2, 2012 01:38:19 PM) I have an other way to solve it. When 1st baby came and he became 3years old than other baby came. When 2nd baby became 3years old than 3rd baby came and 1st baby became 6years old like it 3x4 = 12, so 1st baby became 12 years old than 5th baby came 1st baby became 15 years old. Like, 1st 15 2nd 12 3rd 9 4th 6 5th 3, 3+6+9+12+15=45 thair is 5 years. So give one year. It will be. 4+7+10+13+16=50.

 Siddhartha Roy Nandi said: (Sat, Jul 7, 2012 12:03:13 PM) The Problem can be solved using Arithmetic Progression: S = n/2{2a+(n-1)d} a = youngest age(We have to calculate this) n = Total no. of people(5) d = Difference of Ages in the progression(3) S = Summation of their ages(50)

 Aarthi said: (Mon, Dec 3, 2012 07:58:31 PM) X+12+x+9+x+6+x+3+x=50. 5x+30=50. (x+6) =10. X=4.

 Sairam said: (Sat, Feb 9, 2013 11:04:02 AM) (x+2),(x+4),(x+6),(x+8),(x+10). Total: x+2+x+4+x+6+x+8+x+10 = 50. 5x+30=50 5x=20 x=4.

 Write your comments here: ...

© 2008-2013 by IndiaBIX™ Technologies. All Rights Reserved | Copyright | Terms of Use & Privacy Policy

Contact us: info@indiabix.com     Follow us on twitter!