Latest Elitmus Aptitude Question SOLUTION: The sum of four consecutive two-digit odd numbers , when divided by 10, becomes a perfect square. Which of the following can possibly be one of these four numbers? A. 21 B. 25 C. 4

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(#M40038000) ELITMUS Category question Keep an EYE

The sum of four consecutive two-digit odd numbers , when divided by 10, becomes a perfect square. Which of the following can possibly be one of these four numbers?
A. 21
B. 25
C. 41
D. 67
E. 73

Asked In Elitmus Liti (9 years ago)

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deepika  (9 years ago)

Perfect squares : 4,9,16,25,36,49,64,81......
4=40/10,(40/4=10)(sum of 4 2-digit numbers cannot be 10).
9=90/10,(90/4=22.5) (we cannot get decimas when we sum 4 2-digit nos)
16=160/10,(160/4=40) (so the 2 digit numbers are around 40 , the 2- digit odd numbers around 40 are 37,39,41,43)


37+39+41+43=160
160/10=16 (a perfect square)

Answer: 41


When it comes to 36
36=360/10, (360/4=90)

The odd nums around 90 are 87,89,91,93

87+89+91+93=360

If we take 64 in the same way we get 640/4 = 160 (3-digit number so we cannot proceed more than this square number)

Other than these all the other squares give decimal values when multiples by 10 and then divided by 4

Is this solution Helpfull? Yes (13) | No   

kummari venkatesh  (9 years ago)

ans 41

37+39+41+43=160/10=4*4

Is this solution Helpfull? Yes (6) | No (1)  

Niranjan C  (9 years ago)

its option C .41
How to solve it:
1. think of number which divided by 10 will be a whole number,
2. Look at the option which is nearly equal to 1/4 th of that number.(because we are adding 4 consecutive numbers)

here in this case:
option 1: 21*4=80 (there is no perfect square when divided by 10 near by to 80)
option 2: 25*4=100 (again no perfect square)
option 3: 41*4=164 (there is an option 160 )
160/4=16=4*4.

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ashish mehta  (9 years ago)

sum of 4 terms in ap having cd=2
s=2(2a+6),s/10=a perfect square(ps)
=>2a+6=5*ps
now putting different values of ps,a can be obtained
ps=1,a=-1/2 not possible
ps=4,a=7 not possible
ps=9,a=39/2 not possible
ps=16,a=37...so no may be 37,39,41,43,45
and 41 is in the option too

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Dharmendra  (9 years ago)

Lets number be a,a+2,a+4,a+6 where a is any 2 digit odd no
so a+(a+2)+(a+4)+(a+6)=4a+12
since
4a+12 is divisible by 10 so its unit place digit must be 0(zero) or multiple of 10 i.e.(140,160,170,180)

for 4a+12 becames multiple of 10 ,4a digit must contain 8 on unit place
so only possible value of a is 17,27,37,47
now u can check
17*4+12=80 div 10 =8 not a perfect square
27*4+12=120 div 10 =12 not a perfect square
37*4+12=160 div 10 =16 it is a perfect square

now odd number is a=37,a+2=39,a+4=41,a+6=43 so ans is 41(c) given in option

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roshan  (9 years ago)

A. 21

21 + 22 + 23 + 24 = 90 , 90/10 = 9 [3*3=9]

Is this solution Helpfull? Yes (1) | No (22)  

DIMPLE SINGH  (9 years ago)

@ Roshan
the no. should be consecutiv odd

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ANKIT SRIVASTAVA  (9 years ago)

C. 41
the last 4 one's digit must be 7,9,1,3 such that 0 comes at one's digit,
so by doing it we get 37+39+41+43= 160/10= undroot16=4

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raushan kumar  (9 years ago)

41 is ans 37 39 41 43 = 160/10 =16 is prefect sqr

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saira fatma  (9 years ago)

WE ARE GETTING THE ANSWER via 21 also
21+23+25+27/10=96/10=9
so ans can be 21 also since,9 is also a perfect square

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