M4maths Previous Puzzles

25 Feb 2015 Rank of the word

Rank of the word "NOTICE" in the dictionary of words formed by all the six letters of the word NOTICE.

OPtion

1) 373
2) 385
3) 391
4) 400
5) 432
6) 455
7) 478
8) 501
9) 513
10)618

Solution

Ordered arrangement of letters is
C, E, I, N, O, T
so rank
(720/6)*3 + (120/5)*3 + (24/4)*3 + (6/3)*2 + (2/2)*0 + 1
455

Option 6)

Correct Option: 6 Best Solution (64)

Sanu Panji   9 years AGO

There are 6 letters and all are different : C E I N O T
In dictionary, words starting from C,E & I will come before word NOTICE.
so, total no. of words starting from C = 5!= 120
(we have to arrange 5 letters on five places)
similarly, no. of words starting from E = 5! = 120
and no. of words starting from I = 5! = 120
now, first word starting with letter N will come

similarly, no. of words starting with NC.... = 4!
no. of words starting with NE.... = 4!
no. of words starting with NI.... = 4!
now, words starting from NO will come

no. of words starting with NOC... =3!
no. of words starting with NOE... =3!
no. of words starting with NOI... =3!
now, words starting from NOT will come

no. of words starting with NOTC.. = 2!
no. of words starting with NOTE.. = 2!

then after words starting from NOTI will come and first word will NOTICE

so, total no. of words before NOTICE = 3*5! + 3* 4! + 3 * 3! + 2*2!
= 360 + 72 + 18 + 4
= 454
so, rank of NOTICE in dictionary = 455

Like? Yes (15) | No (5)  

Surit Datta   9 years AGO

1st we arrange the words of NOTICE in alphabet order
like,
C,E,I,N,O,T
Now strike off the first letter N.
Then count the no.of letters before N, and it is equal to 3,which is the coefficient of 5!.

Again strike off the letter O.
Then count the no.of letters before O and it is equal to 3 which is coefficient of 4!

Again strike off the first letter T.
Then count the no.of letters before T and it is equal to 3 which is coeffcient of 3!

Again strike off the first letter I.
Then count the no.of letters before I and it is equal to 2 which is coeffcient of 2!

Again strike off the first letter C.
Then count the no.of letters before C and it is equal to 0 which is coeffcient of 1!

Finally add 1 to the above values to get the rank of the word NOTICE as follows:

3*(5!) + 3*(4!) +3*(3!)+2(2!)+0(1!)+1 = 455
so option 6
that's it.

Like? Yes (6) | No (2)  

Jaspalsingh Parmar   9 years AGO

3*(5!)+ 3*(4!) + 3*(3!) + 2*(2!) +1 = 360+72+18+4+1 = 455

Like? Yes (7) | No (4)  

Meenakshi   9 years AGO

First arrange the given word in alphabetical order
CEINOT
1.search for N.Remove that word from list and see how many letters before N.. 3*(5!)=360
2.search for o.Remove that word from list and count the no of letters before o..3*(4!)=72
3.similarly , 3*(3!)=18
4. 2*(2!)=4
5. 0*(1!)=0
6. 1*(1!)=1
so sum=360+72+18+4+0+1=455
option 6)

Like? Yes (4) | No (1)  

Emanuel Abhishek Murmu   9 years AGO

Arranging the letters in alphabetical order we have, C,E,I,N,O,T

C_ _ _ _ _= 5!
E_ _ _ _ _= 5!
I_ _ _ _ _= 5!

NC_ _ _ _= 4!
NE_ _ _ _= 4!
NI_ _ _ _= 4!

NOC_ _ _= 3!
NOE_ _ _= 3!
NOI_ _ _= 3!

NOTC_ _= 2!
NOTE_ _= 2!

so, number of words before "NOTICE"= (3*5!)+(3*4!)+(3*3!)+(2*2!) = 454

SO, "NOTICE" will be 455th word.

Like? Yes (4) | No (1)  

sandeep ramchandra pawaskar   9 years AGO

option 6 is correct

Like? Yes (3) | No (1)  

srinivas   9 years AGO

The Letter starting with C,E,I,
5!+5!+5!
The letter statinr with
NC,NE,NI=4!+4!+4!
NOC,NOE,NOI=3!+3!+3!
The Letter Starting with:
NOTC,NOTE,=2!+2!
NOTICE=1
SO TOTAL:
5!+5!+5!+4!+4!+4!+3!+3!+3!+2!+2!+1

Like? Yes (4) | No (2)  

SURESH STANAM   9 years AGO

3*5!+3*4!+3*3!+2*2!+1*1!
=360+72+18+4+1=455

Like? Yes (4) | No (2)  

Tasneem Patanwala   9 years AGO

Arrange Notice in alphabetically order
NOTICE-> C,E,I,N,O,T
Now strike off the first letter N.

C,E,I,N,O,T

Then count the no.of letters before N, and it is equal to 3,which is the coefficient of 5!.

Again strike off the first letter O.
C,E,I,N,O,T

Then count the no.of letters before O and it is equal to 3 which is coefficient of 4!

Again strike off the first letter T.

C,E,I,N,O,T

Then count the no.of letters before T and it is equal to 3 which is coeffcient of 3!

Again strike off the first letter I.

C,E,I,N,O,T

Then count the no.of letters before I and it is equal to 2 which is coeffcient of 2!

Again strike off the first letter C.

C,E,I,N,O,T

Then count the no.of letters before C and it is equal to 0 which is coeffcient of 1!

Finally add 1 to the above values to get the rank of the word NOTICE as follows:

3(5!) + 3(4!) +3(3!)+2(2!)+0(1!)+1=455

Like? Yes (1) | No   

praveen kumar   9 years AGO

rank= 3*5! +3*4! +3*3! +2*2! +0*1!+0*0! +1
= 455

Like? Yes (1) | No   

RAKESH ARYA   9 years AGO

The word is NOTICE .
We will arrange it alphabetically , then it be : CEINOT

Now , 3*5!+3*4!+3*3!+2*2!+0*1!+1 = 455

Like? Yes (4) | No (3)  

Renugadevi   9 years AGO

option is 6

Like? Yes (3) | No (2)  

Ritesh   9 years AGO

Write the word in alphabetical order i.e. CEINOT
To write word NOTICE,
For 1st place Of N there are 3 letters C,E &I.
So it will be 3*5!
For next place i.e. NO there are 3 letters before O excluding N, because it is used on 1st place. So 3*4!
Similarly for others and expression would be
3*5!+3*4!+3*3!+2*2!+1=455

Like? Yes (3) | No (2)  

nitin kishore   9 years AGO

total number of words sarting from:
C = 5!
E = 5!
I = 5!
NC = 4!
NE = 4!
NI = 4!
NOC = 3!
NOE = 3!
NOI = 3!
NOTC = 2!
NOTE = 2!
then comes the word NOTICE.
So, rank of the word NOTICE is:
3 *5! + 3*4! +3*3!+5 =455

Like? Yes (1) | No   

sindhura   9 years AGO

3*5! + 3*4! + 3*3! + 2*2! + 0*1! + 0*0! =360+72+18+4+0+0= 454+1= 455 is the answer.

Like? Yes (1) | No   

sujatha   9 years AGO

In alphaetical order C,E,I,N,O,T
starting with letter C - 5!=120 words
starting with letter E - 5!=120 words
starting with letter I - 5!=120 words

starting with letter N- 5!=120
but IN "NOTICE"
start with NC-4!=24,NE-4!=24,NI-4!=24
NOC-3!=6,NOE-3!=6,NOI-3!=6,
for NOTICE-5
total rank=120+120+120+24+24+24+6+6+6+5=455

Like? Yes (1) | No   

Rishikesh Ranjan   9 years AGO

3*5! + 3*4! + 3*3! + 2*2! + 1(itself)
= 355

Like? Yes (1) | No   

Akshaya   9 years AGO

c_ _ _ _ _ = 5!
E _ _ _ _ _ = 5!
I _ _ _ _ _ = 5!
NC_ __ _ _= 4!
NE_ _ __ _= 4!
NI_ _ _ _ = 4!
NOC __ _ _= 3!
NOE __ _ _= 3!
NOI_ _ _= 3!
NOTC _ _= 2!
NOTE _ __= 2!
NOTICE= 0!
Totally 120+120+120+24+24+24+6+6+6+2+2+1 = 455

Like? Yes (1) | No   

arush gupta   9 years AGO

ans 455

Like? Yes (1) | No   

gourav   9 years AGO

5!+5!+5!+4!+4!+4!+3!+3!+3!+2!+2!+2

Like? Yes (1) | No   

Ramesh Bathula   9 years AGO

NOTICE:
arrange the letters in alphabetical order C,E,I,N,O,E
C_=5!=120
E_=5!=120
I_=5!=120
[N]C_=4!=24
[N]E_=4!=24
[N]I_=4!=24
[NO]C_=3!=6
[NO]E_=3!=6
[NO]I_=3!=6
[NOT]C_=2!=2
[NOT]E_=2!=2
[NOTICE]=1

sum the all values=120+120+120+24+24+24+6+6+6+2+2+1=455

Like? Yes (1) | No   

nikhil   9 years AGO

for N 5!*3
for O 4!*3
for T 3!*3
for I 2!*2
+1 for CE

Like? Yes (1) | No   

A M ANKIT KALLURAYA   9 years AGO

guessed

Like? Yes (1) | No   

sheikh shahrukh   9 years AGO

First C,E,I come before N so 3*5!=360
Then NC,NE,NI will come followed by other letters so 3*4!=72
Then NOC,NOE,NOI will come so 3*3!=18
Then NOTC,NOTE will come so 2*2!=4
Then NOTICE will be the next letter therefore
(360+72+18+4)+1=455th letter

Like? Yes (1) | No   

Sangam Prasad   9 years AGO

C E I N O T in alphabetical order( NOTICE)

Now

C - No match = 5! =120

E- No match = 5! =120

I- no match = 5!=120

N- match so we move on to the next position

N C- no match = 4! =24

N E- no match= 4! = 24

N I-no match= 4! = 24

N O- match so we move on to the next position

N O C-no match= 3!= 6

N O E- no match = 3! = 6

N O I - no match = 3!= 6

N O T- match , so next position

N O T C - no match = 2 != 2

N O T E - no match = 2!= 2

N O T I- match next pos

N O T I C - match next

N O T I C E -full match so=1!= 1


now adding all= 120+120+120+24+24+24+6+6+6+2+2+1=455

so option 6)455 ans

Like? Yes (1) | No   

Jitendra Prasad   9 years AGO

C = 5!==120
E = 5!==120
i = 5!==120
NC = 4!==24
NE = 4!==24
NI = 4!==24
NOC = 3!=6
NOE = 3!==6
NOI = 3!==6
NOTC=2!==2
NOTE==2
NOTICE==1
add all= 120+120+120+24+24+24+6+6+6+2+2+1=455

Like? Yes (1) | No   

karthiga   9 years AGO

CETNOT(DICTIONARY ORDER)
if C is fixed remaining can be arranged in 5! (120) ways
if E is fixed remaining can be arranged in 5! (120) ways
if I is fixed remaining can be arranged in 5! (120) ways
totally 360 ways
first letter N fixed
if C is fixed remaining can be arranged in 4! (24) ways
if E is fixed remaining can be arranged in 4! (24) ways
if I is fixed remaining can be arranged in 4! (24) ways
totally (360 +72) 432 ways
second letter NO fixed
like that.....
arranged in (6 +6 +6 = 18) ways
third letter NOT fixed
Remaining can arranged in 2! +2! = 4 ways
totally 454 ways
tourth letter NOTI fixed
454th word is NOTIEC
455th word is NOTICE
so answer is 455

Like? Yes (1) | No   

mohini gupta   9 years AGO

455

Like? Yes (1) | No   

Pranati Kundu   9 years AGO

fix I,C,E in 1st position and the remaining alphabets can be arranged in 5! ways so the total is 3×5!. then fix N in 1st and I,C,E in 2nd position so the remaining alphabets can be arranged in 4! ways and the total is 3×4! then fix N in 1st and O in 2nd position and I,C,E in 3rd position so the remaining alphabets can be arranged in 3! ways and the total is 3×3!.then NOTCIE , NOTCEI, NOTECI, NOTEIC then NOTICE so whole total is 455

Like? Yes (1) | No   

Kavya Bhat   9 years AGO

i guessed.

Like? Yes (1) | No   

Thirupathi Reddy   9 years AGO

NOTICE=CEINOT
=3*5!+3*4!+3*3!+2*2!+0*1!+1
=455

Like? Yes (1) | No   

Arunithi   9 years AGO

rearrange the word alphabetically and find the corresponding coefficints by striking each letter
CEINOT
strike N .no of letters before N = 3(5!)
strike O.no of letters before O = 3(4!)
strike T.no of letters before T =3(3!)
strike I.no of letters before I =2(2!)
strike C.no of letters before C =0(1!)
then add 1 so total rank =455

Like? Yes (1) | No   

ADARSH GAURAV   9 years AGO

455 is the answer.

Like? Yes (1) | No   

PRANEETH   9 years AGO

3*fact(5)+3*fact(4)+3*fact(3)+4

Like? Yes (1) | No   

Supriya Mondal   9 years AGO

Word starting with C has 5!=120 combinations
Word starting with E has 5!=120 combinations
Word starting with I has 5!=120 combinations
Word starting with NC has 4!=24 combinations
Word starting with NE has 4!=24 combinations
Word starting with NI has 4!=24 combinations
Word starting with NOC has 3!=6 combinations
Word starting with NOE has 3!=6 combinations
Word starting with NOI has 3!=6 combinations
Word starting with NOTC has 2!=2 combinations
Word starting with NOTE has 2!=2 combinations
And, the next word is NOTICE
So,rank of this word is=(120+120+120+24+24+24+6+6+6+2+2+1)=455
So, option 6)455 is the right answer

Like? Yes (1) | No   

sairamesh   9 years AGO

N O T I C E
4 5 6 3 1 2
5! 4! 3! 2! 1! 0!
3 3 3 2 0 0 ====(5!*3)+4!*3)+............(0!*0)+1=455 answer 6 th option

Like? Yes (1) | No   

harvinder   9 years AGO

starting with
C =5! =120
E= 5!=120
I =5!=120
NC=4!=24
NE=4!=24
NI=4!=24
NOC=6
NOE=6
NOI=6
NOTC=2
NOTE=2
NOTICE=1


120+120+120+24+24+24+6+6+6+2+2+1=455

Like? Yes (1) | No   

POOJA SINGH   9 years AGO

3*5!+(3*4!)+(3*3!)+(2*2!)+1

Like? Yes (1) | No   

anusha   9 years AGO

answr is 6

Like? Yes (1) | No   

Gurram Mahendra ranganath   9 years AGO

notice
rank=455
option 6.

Like? Yes (1) | No   

shovan Manna   9 years AGO

there is a shortcut process to calculate that rank
N O T I C E (we have to see no less than that
3 3 3 2 0 0 particular no.)
5! 4! 3! 2! 1! 0!
=(3*5!+3*4!+3*3!+2*2!)+1
=455

Like? Yes (1) | No   

Gopal gupta   9 years AGO

c*****=120
E*****=120
I*****=120
NC****=24
NE****=24
NI****=24
NOC***=6
NOE***=6
NOI***=6
NOTC**=2
NOTE**=2
NOTICE=1
=455

Like? Yes (4) | No (4)  

Harish Reddy Namala   9 years AGO

455 is the rank.

Like? Yes (1) | No (1)  

Dorababu M   9 years AGO

Given word : NOTICE
Alphabetical order : C E I N O T
words starting with C:
C - - - - - =5!=120
words starting with E:
E - - - - - =5!=120
words starting with I:
I - - - - - =5!=120
NC - - - - =4!=24
NE - - - - =4!=24
NI - - - - =4!=24
NOC - - - =3!=6
NOE - - - =3!=6
NOI - - - =3!=6
NOTC - - =2!=2
NOTE - - =2!=2
NOTICE=1
120+120+120+24+24+24+6+6+6+2+2+1 =455

Like? Yes (1) | No (1)  

MRITUNJAY   9 years AGO

this is the answer

Like? Yes (1) | No (1)  

Arvind gupta   9 years AGO

Alphabets in NOTICE are in order
C
E
I
N
O
T

words forming from starting by C,E and I are 5! so total 3*5!
Now keeping fixed N ,words foriming from C,E,I are total 3*4!
now keeping NO fixed
words forming from C,E,I are total 3*3!

now keeping NOT fixed
words forming from C,E are total 2*2!
now keeping NOTI fixed next letter comes C and next one is E
so we found our word NOTICE
so its position is
3*5!+3*4!+3*3!+2*2!+1=455

Like? Yes (1) | No (1)  

KUMAR HARSH   9 years AGO

notice corresponding position in alphabate
456312
now we have to find from first and see how much no. are less then that of first posion no.(4) there are three no. less then 4 ie; 3,2,1
and like this we find all the no less then position
like 4 5 6 3 1 2 corespondin no. of less no. then posional no.
3 3 3 2 0 0 now we have to write from last position 0!,1!,2!,3!,4!,5! and then multiply them with 3, 3, 3 ,2 ,0 ,0 corresponding and then sum all
like 0!*0+1!*0+2!*2+3!*3+4!*3+5!*3=455

Like? Yes (5) | No (5)  

ABHINAV PUSHPAM   9 years AGO

Answer is 454+1 = 455

Like? Yes (1) | No (1)  

Saurabh Kumar Gupta   9 years AGO

the rank will be 455

Like? Yes (1) | No (1)  

sam   9 years AGO

rank of word"NOTICE"=3(5!)+3(4!)+3(3!)+2(2!)+0(1!)+1
ans=455

Like? Yes (1) | No (1)  

RAHUL KUMAR   9 years AGO

Ans is 455.

Like? Yes (1) | No (1)  

KALYAN KUMAR REDDY VALLURU   9 years AGO

Words start with
C-->120=5!
E-->120
I--->120
NC-->24=4!
NE-->24
NI--->24
NOC-->6=3!
NOE-->6
NOI--->6
NOTC-->2=2!
NOTE-->2=2!
NOTICE is the going to occur after all of the above cases in the dictionary.
so, (120*3)+(24*3)+(6*3)+(2*2)=454
NOTICE is the 455th word.

Like? Yes (3) | No (3)  

Abhishek birkh   9 years AGO

NOTICE write in alfabetical order and get ceinot and
first we take N .in CEINOT we see that N is after 3 words than 3 is the factor of !5 and so on and in last we add 1

Like? Yes (1) | No (1)  

Dharmraj Sharma   9 years AGO

using stroke and rank formula
C,E,I,N,O,T
rank=3*5!+3*4!+3*3!+2*2!+0*1!+0*1!+1
rank=455

Like? Yes (1) | No (1)  

Riyas   9 years AGO

Given word:
NOTICE

Arranging alphabetically
CEINOT

consider
first letter N
No of words before N in alpahbet order 3*5!(coefficient)

second letter O
No of words before O in alphabet order 3*4!
Remove N before counting

Likewise look for each alphabet
so finally 3(5!) + 3(4!) + 3(3!) + 2(2!) + 0(1!) +1 = 455

Like? Yes (1) | No (1)  

aman aditya   9 years AGO

3*5!+3*4!+3*3!+2*2!+0*1+=455

Like? Yes (2) | No (2)  

Math Lovers   9 years AGO

NOTICE

Words starts from each letter = 6*5*4*3*2*1 =120
Before N, 3 letters I,C and E are there so after 3*120 = 360 words, words from N starts.

again after 1*3*4*3*2*1 = 72 words, words from NO starts...

again after 1*1*3*3*2*1 = 18 words, words from NOT starts...

again after 1*1*1*2*2*1 = 4 words, words from NOTI starts... which is NOTICE

so rank = 360+72+18+4+1 = 455

Like? Yes (1) | No (1)  

Swathy Murugan   9 years AGO

option is 6

Like? Yes (1) | No (1)  

shahbaz Alam Ansari   9 years AGO

Given word is NOTICE

The letters of the are N,O,T,I,C,E.

Write the alphabetical order of the letters of the given word 'NOTICE ' as C,E,I,N,O,T

Now strike off the first letter N.

C,E,I,N,O,T.

Then count the no.of letters before N, and it is equal to 3,which is the coefficient of 5!.

Again strike off the first letter O.

C,E,I,N,O,T

Then count the no.of letters before O and it is equal to 3 which is coefficient of 4!

Again strike off the first letter T.

C,E,I,N,O,T

Then count the no.of letters before T and it is equal to 3 which is coeffcient of 3!

Again strike off the first letter I.

C,E,I,N,O,T

Then count the no.of letters before I and it is equal to 2 which is coeffcient of 2!

Again strike off the first letter C.

C,E,I,N,O,T

Then count the no.of letters before C and it is equal to 0 which is coeffcient of 1!

Finally add 1 to the above values to get the rank of the word NOTICE as follows:

3(5!) + 3(4!) +3(3!)+2(2!)+0(1!)+1=455

Like? Yes (4) | No (4)  

urmita das   9 years AGO

NOTICE in alphabetical order gives CEINOT
Thus...
letters before N is 3,hence coefficient is 5
letters before O is 3,coefficient is 4
letters before T is 3,coefficient is 3
letters before I is 2,coefficient is 2
letters before C is 0,coefficient is 1
letters before E is 1,coefficient is 0

hence, 3(5!)+3(4!)+3(3!)+2(2!)+0(1!)+1=455

Like? Yes (3) | No (3)  

Ravishankar Karn   9 years AGO

C,E,I comes before N, so word- 5!+5!+5!
NC,NE,NI words-- 4!+4!+4!
NOC,NOE,NOI-- 3!+3!+3!
NOTC,NOTE-- 2!+2!
total word=454.
so the rank is 455

Like? Yes (2) | No (2)  

snehlata   9 years AGO

position= 456312
NOTICE
(no of letters) 333200
less than)

3*5!+3*4!+3*3!+2*2!+1
360+72+18+4+1
=455

Like? Yes (1) | No (1)  

roma gautam   9 years AGO

if we use permutation and combination we can find

Like? Yes (1) | No (1)  

Yathiraj   9 years AGO

Arranging the letters of "NOTICE" alphabetically, it will be C E I N O T
Keeping C as constant find the different combinations of words that can be formed using remaining 5 letters
5! = 120
Let me take each letter alphabetically
Keeping C as constant -> 120 words (5!)
Keeping E as constant -> 120 words (5!)
Keeping I as constant -> 120 words (5!)

Now, I can't take N as constant as NOTICE can appear anywhere between 120 combinations.
Take N and arrange remaining letters alphabetically (C E I O T). Now, take one by one along with N
Keeping N&C as constant -> 24 words (using E I O T... 4!)
Keeping N&E as constant -> 24 words (using C I O T... 4!)
Keeping N&I as constant -> 24 words (using C E O T... 4!)

Now, I can't take N&O as constant as NOTICE can appear anywhere between 24 combinations
Take N&O and arrange remaining letters alphabetically (C E I T). Now, take one by one along with N&O
Keeping N,O,C as constant -> 6 words (using E I T... 3!)
Keeping N,O,E as constant -> 6 words (using C I T... 3!)
Keeping N,O,I as constant -> 6 words (using C E T... 3!)

Now, I can't take N,O,T as constant as NOTICE can appear anywhere between 6 combinations.
Take N,O,T and arrange the remaining letters alphabetically(C E I). Now, take one by one
Keeping N,O,T,C as constant -> 2 words (using E I... 2!)
Keeping N,O,T,E as constant -> 2 words (using C I... 2!)
Now, if you take N,O,T,I as constant the first possible word (alphabetically) is NOTICE

Now, add all the above combinations
(120+120+120) + (24+24+24) + (6+6+6) + 2+2+1
=> 360+72+18+5
=> 455

Rank of the word NOTICE will be 455

Option 6.

Like? Yes (2) | No (2)  

24 Feb 2015 Possibility

x, y and z are three positive integers such that x not= y not= z and 0 < x < y < z < 10; If x+y+z = 15 then number of possible pairs of (x, y, z) will be

OPtion

1) 1
2) 2
3) 3
4) 4
5) 5
6) 6
7) 7
8) 8
9) 9
10)None of these

Solution

Possible pair are
(1,5,9) (1,6,8) (2,4,9) (2,5,8) (2,6,7) (3,4,8) (3,5,7) (4,5,6)

Option 8)

Correct Option: 8 Best Solution (59)

Yathiraj   9 years AGO

since 0 < x < y < z < 10,
x range 1 to 7
y range 2 to 8
z range 3 to 9

considering the above range and with different combination of x, y, z where x+y+z=15, you will get 8 such pairs of (x,y,z).

x:1, y:5, z:9
x:1, y:6, z:8
x:2, y:4, z:9
x:2, y:5, z:8
x:2, y:6, z:7
x:3, y:4, z:8
x:3, y:5, z:7
x:4, y:5, z:6

So, Option 8.

Like? Yes (7) | No (1)  

Santosh Kumar   9 years AGO

if x=1then possible (y,z)={5,9 & 6,8}
if x=2then possible (y,z)={4,9 & 5,8 &6,7}
if x=3then possible (y,z)={4,8 & 5,7}
if x=4then possible (y,z)={5,6 }
so total=8

Like? Yes (4) | No (1)  

SAKTHEESVARI S   9 years AGO

the only possibilities are, x,y,z are
1,5,9; 1,6,8; 2 4 9; 2 5 8; 2 6 7; 3 4 8; 3 5 7; 4 5 6;
totally 8 possibilities

Like? Yes (3) | No   

Supriya Mondal   9 years AGO

There are 8 possible pairs and they are-
(1,5,9)
(1,6,8)
(2,4,9)
(2,5,8)
(2,6,7)
(3,4,8)
(3,5,7)
(4,5,6)
so, option 8)8 is the right answer

Like? Yes (3) | No   

Kaju Gupta   9 years AGO

(x,y,z)
(1,5,9)
(2,4,9)
(1,6,8)
(2,5,8)
(3,4,8)
(2,6,7)
(3,5,7)
(4,5,6)

Like? Yes (3) | No (1)  

kunal   9 years AGO

(1,5,9),(1,6,8),(2,4,9),(2,5,8),(2,6,7),(3,4,8),(3,5,7),(4,5,6)

Like? Yes (2) | No   

A.SOMA SUNDARI   9 years AGO

the is only possibilities are
1+5+9
1+6+8
2+4+9
2+5+8
2+6+7
3+4+8
3+5+7
4+5+6

Like? Yes (3) | No (1)  

praveen kumar   9 years AGO

8 pairsare possible

Like? Yes (1) | No   

KUMAR HARSH   9 years AGO

cobination is(4,5,6),(3,5,7),(3,4,8),(2,4,9),(2,5,8),(2,6,7),(1,5,9),(1,6,8)

Like? Yes (2) | No (1)  

Sravanthi   9 years AGO

1 5 9
1 6 8
2 5 8
2 4 9
2 6 7
2 6 7
3 5 7
3 4 8
4 5 6

Like? Yes (2) | No (1)  

Dharmraj Sharma   9 years AGO

for 0

Like? Yes (2) | No (1)  

harvinder   9 years AGO

1,5,9
1,6,8
2,6,7
2,5,8
2,4,9
3,5,7
3,4,8
4,5,6

Like? Yes (2) | No (1)  

MD SOHAIL AKHTER   9 years AGO

951,942,861,852,843,762,753,654

Like? Yes (2) | No (1)  

Dorababu M   9 years AGO

possible pairs are
1,5,9
1,6,8
2,4,9
2,5,8
2,6,7
3,4,8
3,5,7
4,5,6

Like? Yes (2) | No (1)  

anusha   9 years AGO

ans is 8

Like? Yes (2) | No (1)  

RAKESH ARYA   9 years AGO

1. 1,5,9
2. 1,6,8
3. 2,4,9
4. 2,5,8
5. 2,6,7
6. 3,4,8
7. 3,5,7
8. 4,5,6

Like? Yes (2) | No (1)  

jaganmohan rao .pola   9 years AGO

8

Like? Yes (2) | No (1)  

nitin kishore   9 years AGO

The pairs which satisy all the conditions are:-
(1,5,9), (1,6,8), (2,4,9),(2,5,8),(2,6,7),(3,4,8),(3,5,7),(4,5,6)

Like? Yes (1) | No   

shovan Manna   9 years AGO

let x=1,then y and z must be y=5,z=9,
y=6,z=8
when x=2,then y and z must be y=5,z=8
y=6,z=7,y=4 and z =9
when x=3 then y and z must be y=4,z =8
y=5,z=7
when x=4 then y and z must be y=5,z =6

Like? Yes (1) | No   

SANDHYA KALIGITHI   9 years AGO

(1,5,9), (2,4,9),
(1,6,8), (2,5,8), (3,4,8),
(2,6,7), (3,5,7),
(4,5,6)

Like? Yes (1) | No   

Modadugu vishnu kumar   9 years AGO

According to the given condition 0

Like? Yes (1) | No   

POOJA SINGH   9 years AGO

all possible values of x y z that satisfy the condition x + y + z equals to 15 is (1,5,9)(1,6,8)(2,6,7)(2,4,9)(2,5,8)(3,4,8)(3,5,7)(4,5,6). so answer is 8

Like? Yes (1) | No   

Pranati Kundu   9 years AGO

(1,5,9)(1,6,8)(2,6,7)(2,4,9)(2,5,8)(3,4,8)(3,5,7)(4,5,6) r the 8 pairs

Like? Yes (1) | No   

Priyanka Shete   9 years AGO

Possible pairs are-
(1,5,9),
(2,5,8),
(3,5,7),
(4,5,6),
(2,4,9),
(3,5,8),
(2,6,7),
(1,6,8).

Like? Yes (1) | No   

Revathi Perapu   9 years AGO

possible pairs are 8

Like? Yes (1) | No   

Jaspalsingh Parmar   9 years AGO

Option 8 is correct

Like? Yes (1) | No   

KALYAN KUMAR REDDY VALLURU   9 years AGO

starting with 1-2
starting with 2-3
starting with 3-2
starting with 4-1
total 8

Like? Yes (1) | No   

Preethi   9 years AGO

8 possible pairs are 1,5,9
2 ,4,9
1,6,8
2,5,8
3,4,8
2,6,7
3,5,7
4,5,6

Like? Yes (1) | No   

sindhura   9 years AGO

Given that x not= y not= z and 0 < x < y < z < 10; also x+y+z=15
Among x, y and z, z is greater but less than 10, so z=9, 8, 7, 6 only for remaining values given condition violates.
So the combinations are (2,4,9),(1,5,9),(1,6,8),(2,5,8),(3,4,8),(2,6,7),(3,5,7),(4,5,6)
Answer is 8 combinations.

Like? Yes (1) | No   

PIYUSH KUMAR   9 years AGO

1+5+9=15
1+6+8=15
2+4+9=15
2+5+8=15
2+6+7=15
3+4+8=15
3+5+7=15
4+5+6=15

Like? Yes (1) | No   

Arvind gupta   9 years AGO

Answer is 8 combinations
you can run this "C" program
#include
int main(){
int x,y,z,max=10;
for(x=1;x

Like? Yes (1) | No   

Sambar Naresh   9 years AGO

8 psbl pairs are 1,5,9 1,6,8 2,4,9 2,5,8 2,6,7 3,4,8 3,5,7 4,5,6

Like? Yes (1) | No   

Swathy Murugan   9 years AGO

Total combinations available is 8

Like? Yes (1) | No   

Thirupathi Reddy   9 years AGO

The possible sets are (1,5,9) (1,6,8)(2,4,9)(2,5,8)(2,6,7)(3,4,8)(3,5,7)(4,5,6)

Like? Yes (1) | No   

E. UDAY REDDY   9 years AGO

(1,5,9) ->(1,6,8)
(2,5,8)->(2,4,9)->(2,6,7)
(3,5,7)->(3,4,8)
(4,5,6)

Like? Yes (1) | No   

Divya jain   9 years AGO

possible values for x,y,z
(2,4,9)
(1,5,9)
(2,5,8)
(1,6,8)
(3,4,8)
(2,6,7)
((3,5,7)
(4,5,6)

Like? Yes (1) | No   

sandeep ramchandra pawaskar   9 years AGO

9+2+4=15
9+1+5=15
8+4+3=15
8+2+5=15
8+6+1=15
7+5+3=15
7+6+2=15
6+5+4=15

Like? Yes (1) | No   

ayushi   9 years AGO

pairs are :
(1,5,9);(1,6,8);
(2,4,9);(2,5,8);(2,6,7);
(3,4,8);(3,5,7);
(4,5,6);

Like? Yes (1) | No   

Paras Jain   9 years AGO

Ans => 8

Like? Yes (1) | No   

Shanmugapriya.G   9 years AGO

possible combination is 8.
do it by trial method.

Like? Yes (1) | No   

rajkumar   9 years AGO

according to given conditions
possible pairs are
(1,6,8)
(1,5,9)
(2,4,9)
(2,5,8)
(3,4,8)
(3,5,7)
(4,5,6)
(2,6,7)

Like? Yes (1) | No   

NANGINEDI RAVITEJA   9 years AGO

for z=9 it as 942,951
z=8 it as 852,843,861
Z=7 then 753 ,762, z=6 then 654 as z,y,x values as possible

Like? Yes (1) | No   

Ramesh Bathula   9 years AGO

the chances are
(1,5,9) (1,6,8) (2,4,9) (2,5,8) (2,6,7) (3,4,8) (3,5,7) (4,5,6)

Like? Yes (1) | No   

ganesh babu   9 years AGO

(1,5,9)
(2,4,9)
(1,6,8)
(2,5,8)
(3,4,8)
(2,6,7)
(3,5,7)
(4,5,6)

Like? Yes (1) | No   

Ravi   9 years AGO

(1,5,9),(1,6,8),(2,4,9),(2,5,8),(2,6,7),(3,4,8),(3,5,7),(4,5,6)

Like? Yes (1) | No   

mubarak   9 years AGO

1,6,8
1,5,9
2,4,9
2,5,8
2,6,7
3,4,8
3,5,7
4,5,6

Like? Yes (1) | No   

nagaprasad   9 years AGO

cause

Like? Yes (1) | No   

sudhakar   9 years AGO

1)1+5+9=15
2)1+6+8=15
3)2+4+9=15
4)2+5+6=15
5)2+6+7=15
6)3+5+7=15
7)3+4+8=15
8)4+5+6=15
so answer is option 8

Like? Yes (1) | No   

Surit Datta   9 years AGO

the possible pairs are
(1,5,9)
(1,6,8)
(2,5,8)
(2,4,9)
(2,6,7)
(3,5,7)
(3,4,8)
(4,5,6)
so option 8
that's it.

Like? Yes (1) | No   

Vipul Kumar Paliwal   9 years AGO

with 9...only 2 possibilities...ie.9+5+1 and 9+4+2
with 8...8+6+1..8+5+2...8+4+3
with 7..7+6+2....7+5+3
with 6....6+5+4

Like? Yes (1) | No   

MRITUNJAY   9 years AGO

(1,5,9)
(1,6,8)
(2,6,7)
(2,4,9)
(2,5,8)
(3,4,8)
(3,5,7)
(4,5,6)

Like? Yes (1) | No   

RAHUL KUMAR   9 years AGO

159
168
249
258
267
348
357
456
Option 8 is r8.

Like? Yes (1) | No   

haneesha   9 years AGO

the combinations are (1,5,9) (1,6,8) (2,6,7) (2,5,8) (2,4,9) (3,5,7) (3,4,8)(4,6,5)

Like? Yes (1) | No   

ABHINAV PUSHPAM   9 years AGO

159
168
249
258
267
348
357
456
These are possible sets
So option 8)

Like? Yes (1) | No   

T.Aravinth kumar   9 years AGO

only 8 possible pairs

Like? Yes (1) | No   

nikhil   9 years AGO

159,168
249,258,267
348,357
456

Like? Yes (1) | No   

Anil Beck   9 years AGO

According to condition
x must be in between 1,2,3,4
y must be in between 4,5,6
z must be in between 6,7,8,9
So possible values are
x-y-z

1-5-9
1-6-8
2-4-9
2-5-8
2-6-7
3-4-8
3-5-7
4-5-6

Like? Yes (1) | No   

SARAVANAN S K   9 years AGO

1+5+9=15
2+5+8=15
3+5+7=15
4+5+6=15
1+6+8=15
2+7+6=15
3+8+4=15
2+4+9=15

Like? Yes (1) | No   

Harish Reddy Namala   9 years AGO

The pairs are (1,5,9), (1,6,8), (2,4,9), (2,5,8), (2,6,7), (3,4,8) ,(3,5,7),(4,5,6)

Like? Yes (1) | No (1)  

23 Feb 2015 Sum of integer

Sum of a positive integer and inverse of another positive integer is 37/4, sum of integers will be

OPtion

1) 3
2) 6
3) 7
4) 11
5) 13
6) 15
7) 16
8) 17
9) 20
10)23

Solution

37/4 = 9 + 1/4
so first integer = 9
Second integer = 4
Sum = 13

Option 5)

Correct Option: 5 Best Solution (76)

Surit Datta   9 years AGO

one number is 9
another number is 4
so 9 + 1/4 = 37/4
so option 5
that's it.

Like? Yes (10) | No (3)  

sheikh shahrukh   9 years AGO

x+1/y=37/4
or
(xy+1)/y=37/4
so only x=9 and y=4 satisfy this......
so x+y=13
13 is the answer

Like? Yes (6) | No (3)  

Supriya Mondal   9 years AGO

One integer is 9
And another integer is 4
Inverse of the second integer is 1/4
So sum of these=9+(1/4)=37/4
So, ans will be (9+4)=13
So, option 5)13 is the right answer

Like? Yes (7) | No (5)  

Saurabh Kumar Gupta   9 years AGO

37/4 = 9 + 1/4
so the +ve integers are 9 and 4
so required sum = 9+4=13

Like? Yes (4) | No (3)  

nitin kishore   9 years AGO

Let the first integer be X and the second integer be Y.
Then,
X+(1/Y) =37/4
((X*Y)+1)/Y =37/4
It will satisfy the above equation when the values of X and Y are 9 and 4 respectively.
so, the sum of the integers will be, 9+4=13

Like? Yes (1) | No   

Piyush Kumar   9 years AGO

let x and y is the two integers
hence
x+1/y =37/4
this shows that y must be 4,since x and y is an integer
hence x+1/4=37/4
x=9
so sum of integers will be 13
option 5

Like? Yes (2) | No (1)  

ABHINAV PUSHPAM   9 years AGO

9+1/4=37/4
So required sum of intger is 9+4=12
Option 5)

Like? Yes (2) | No (1)  

praveen kumar   9 years AGO

let 1st integerbe x and second integer be y
and their sum is 37/4

means the second integer must be 4
so 1st integer =37/4 - 1/4 =9
so sum of the two integer = 9+4=13

Like? Yes (1) | No   

Harish Reddy Namala   9 years AGO

9 and 4 are the two positive integers.

Like? Yes (2) | No (1)  

RAHUL KUMAR   9 years AGO

9+1/4=37/4
so, 9+4=13
option 5 is r8.

Like? Yes (2) | No (1)  

Gurram Mahendra ranganath   9 years AGO

X+(1/Y)=37/4
=>Y(37-4X)=4
Y=4,X=9 will satisfy the condition.
=>X+Y=4+9=13

Like? Yes (2) | No (1)  

Jaspalsingh Parmar   9 years AGO

option 5 is correct

Like? Yes (1) | No   

MRITUNJAY   9 years AGO

x+1/y=37/4
xy+1/y=37/4
y=4
xy+1=37
xy=36
x=9
sum of integer x+y=9+4=13
option 5

Like? Yes (1) | No   

YUGANDHARA RAO    9 years AGO

by converting 37/4 into mixed conversion i.e., 37/4=9*1/4 . then 9+4=13

Like? Yes (1) | No   

sandeep ramchandra pawaskar   9 years AGO

option 5 is correct

Like? Yes (1) | No   

Devendra Marghade   9 years AGO

If two integers are x & y, then
x+ (1/y) =37/4
4xy + 4 =37y or y(37 - 4x) =4
Above condit satisfies for x=9, y=4
So sum of the integers are x+y=9+4=13
Ans 5) 13

Like? Yes (1) | No   

shovan Manna   9 years AGO

x+1/y=(xy+1)/y
now we can write 37/4 as (4*9+1)/4
so by comparing we get ,
y=4,and x=9. so the sum of the no. will be (9+4)=13

Like? Yes (1) | No   

Karan   9 years AGO

9+1/4= 37/4
sum of integers =9+4=13

Like? Yes (1) | No   

Ritesh   9 years AGO

X+1/Y=37/4
(Xy+1)/Y=37/4
On comparing both side
We have,
Y=4
And X = 9
Therefore sum of integers is 9+4=13

Like? Yes (1) | No   

arush gupta   9 years AGO

13 ANS

Like? Yes (1) | No   

sivaramakrishna   9 years AGO

37/4= 9 1/4
So my integers are 9,4
sum=9+4=13

Like? Yes (1) | No   

KUMAR HARSH   9 years AGO

x=9,y=4
given x+1/y=37/4
i e;9+1/4=37/4
x+y=9+4=13

Like? Yes (1) | No   

chirag   9 years AGO

one number is 9 another is 4 inverse of another number is 1/4
so 9+1/4=37/4
and sum will be 9+4

Like? Yes (1) | No   

SAKTHEESVARI S   9 years AGO

let,positive integer is x; inverse of another positive integer is 1/y; sum of both is, x+(1/y)= 37/4;
by solving, we get,(xy+1)/y=37/4; i.e, y=4;
xy+1=37=>xy=36;x=9;
sum of integer is,x+y=9+4=13

Like? Yes (1) | No   

A.SOMA SUNDARI   9 years AGO

Let x be the one integer;(1/y) be the inverse of another integer
then x+(1/y)=37/4;
(xy+1)/y=37/4;
(i.e) y should be equal to 4 then,the value of x is 9
9+(1/4)=37/4
therefore the sum of integers is 9+4=13

Like? Yes (1) | No   

Ravi Konaraddi   9 years AGO

Integers are 9 nd 4

Like? Yes (1) | No   

Sambar Naresh   9 years AGO

9+4=13
so 9+1/4=36+1/4=37/4
so 9+4=13 is the answer...

Like? Yes (1) | No   

NANGINEDI RAVITEJA   9 years AGO

x+1/y=37/4.
9+1/4=37/4.
so x+y=9+4=13.

Like? Yes (1) | No   

Shanmugapriya.G   9 years AGO

the two number will be 4 and 9.
so the sum is 13.

Like? Yes (1) | No   

Yathiraj   9 years AGO

a + (1/b) = (37/4)
37/4 = 9.25
0.25 can be written as (1/4) i.e. 9 + (1/4)
a + (1/b) = 9 + (1/4)
Hence, a = 9 and b=4
a+b = 9+4 = 13

Option 5

Like? Yes (1) | No   

sindhura   9 years AGO

let the two integers be a and b.
given that a+1/b=37/4,so if a=9 and b=4 then this condition is satisfied so a+b=9+4=13 is the answer

Like? Yes (1) | No   

sompon temtheerakij   9 years AGO

x+1/y = 37/4
9 + 1/4 = 37/4
x+y = 9+4 = 13

Like? Yes (1) | No   

MUNISH YADAV   9 years AGO

9 + 1/4 = 37/4

Like? Yes (1) | No   

Priyanka Shete   9 years AGO

let first int b x n other is y, then
x+(1/y)=37/4
(xy+1)/y=37/4
assume y=4, then
x*4+1=37
so, x=9
and sum of integers i.e. x+y=4+9=13
hence ans is 5)13

Like? Yes (1) | No   

Revathi Perapu   9 years AGO

assume x=9 and y=4
then 9+1/4=37/4
hence 9+4=13

Like? Yes (1) | No   

Arvind gupta   9 years AGO

as both the numbers are integers and let x and y
now x+1/y=37/4
so the y must be 4 for the sum with denominator to be 4 as y and x are integers
hence y=4
therefor x+1/4=37/4
=> x=9
hence x+y=13
9+1/4=37/4
therefor
9+4=13

Like? Yes (1) | No   

ganesh babu   9 years AGO

x+1/4=37/4
4x+1=37
4x=36
x=9
sum of integers 9+4=13

Like? Yes (1) | No   

shahbaz Alam Ansari   9 years AGO

X+1/4=37/4
Therefore,
X=9
Two positive integers are
9 and 4
So, 9+4=13
Ans= 13

Like? Yes (1) | No   

mayank mittal   9 years AGO

let an integer be x and another be y.
x+1/y=37/4
let y=4,put in eqn and we get x=9
x+y=13

Like? Yes (1) | No   

ADARSH GAURAV   9 years AGO

9+4=13

Like? Yes (1) | No   

Debashis Dey   9 years AGO

x+1/y=37/4
(xy+1)/y=37/4
clearly , y=4 and xy+1=37 or,4x+1=37 or,x=9 therefore sum of integers =4+9=13

Like? Yes (1) | No   

SUSHANT KUMAR   9 years AGO

x+1/y=37/4
so (xy+1)/y=37/4
so y=4

hence x= 9

therefore x+y=9+4=13

Like? Yes (1) | No   

Priyanka Rao   9 years AGO

37/4=9 + 1/4
9 + 4 = 13

Like? Yes (1) | No   

Sravanthi   9 years AGO

9+(1/4)

Like? Yes (1) | No   

Rishikesh Ranjan   9 years AGO

x+1/y = 37/4;
so x=9 ,y=4

Like? Yes (1) | No   

Padmanabha Reddy   9 years AGO

[First number=9 second number=4]
9+1/4=37/4
answer may be [9+4]=[13]

Like? Yes (1) | No   

urmita das   9 years AGO

the sum given is 37/4 =9.25=9+1/4
the first positive integer is 9
the second is 4
sum of integers=9+4=13

Like? Yes (1) | No   

Thirupathi Reddy   9 years AGO

x+1/y=37/4
(xy+1)/y=37/4
i.e y=4
then x=9
Sum=9+4=13

Like? Yes (1) | No   

soudip   9 years AGO

(4*9 + 1) / 4
= 9 + (1/4)

Therefore, ans: 9+4= 13(5)

Like? Yes (1) | No   

Dharmraj Sharma   9 years AGO

x+1/y=37/4
(xy+1)/y=37/4
y=4
then x=9
so
x+y=13

Like? Yes (1) | No   

mubarak   9 years AGO

x+(1/y)=37/4
so,if we can add any 2 no's one will be normal and another one is inverse we get inverse term denominator. Eg: 6+1/2=13/2.
similarly I take Y=4 because the answer is 37/4.
X+(1/4)=37/4
X=(37-1)/4=9
so X=9; Y=4
X+Y=9+4=13. so the option 5 is correct.

Like? Yes (1) | No   

Shintala Divya   9 years AGO

a+(1/b)=37/4
a=9,b=4
a+b=36

Like? Yes (1) | No   

Kondeti Harish Kumar Rayal   9 years AGO

9+(1/4)=37/4
9+4=13

Like? Yes (1) | No   

pravallika   9 years AGO

9+1/4=37/4 and sum of integers=9+4=13

Like? Yes (1) | No   

sourav pal   9 years AGO

First no. Is 9
And another 4
9+ 1/4 = 37/4
So sum of integers is 9+4=13

Like? Yes (1) | No   

Jitendra Kumar   9 years AGO

9+(1/4)=37/4

Like? Yes (1) | No   

Arunithi   9 years AGO

given x+1/y=37/4
so (xy+1)/y=37/4 ie)y=4
if y=4 then x=9
so x+y=13

Like? Yes (1) | No   

Bathula Tejesh   9 years AGO

x+(1/y)=(37/4)
(xy+1)/y=(37/4)
therefore the denominator y=4
then to become the value of (xy+1)=34. we know the value of y is 4
therefore ((9*4)+1)=37
there the integers are 4,9
the sum of the integers is 4+9=13
hence option:-5 is correct

Like? Yes (1) | No   

Madhumoy Gope   9 years AGO

37/4=(36+1)/4=9+1/4=x+1/y
so,x=9 & y=4
The sum is= 9+4=13
So,the correct option is (5)

Like? Yes (1) | No   

sairamesh   9 years AGO

x+1/y=37/4 then x=8 and y=4 so 8+4+1=13 ..., 13 is the answer

Like? Yes (1) | No   

nishi   9 years AGO

because it could be only 9+1/4=37/4
so the answer is 9+4=13

Like? Yes (1) | No   

Dorababu M   9 years AGO

9+1/4=37/4
then 9+4=13

Like? Yes (1) | No   

Pranati Kundu   9 years AGO

First no. is 9
And another no. 4 
9+1/4 = 37/4
sum of integers are 9+4=13

Like? Yes (1) | No   

Pankaj kumar   9 years AGO

let a >0; a=9 & another integer b=4 because LCM must be 4 when we take inverse of it & add with a , we get 37/4 . I.e. a+1/b = 9+1/4=37/4. now add a+b = 9+4 =13

Like? Yes (1) | No   

SUNITEE GARIMA   9 years AGO

Ffirst number is 9
and another number is 4
9+1/4=37/4
sum of integers=9+4=13

Like? Yes (1) | No   

M.Ruhi parveen   9 years AGO

we can take the inverse of positive integer as 1/4 so we get the 1st positive integer as 9
so 9+4 is 13

Like? Yes (1) | No   

p.venkata kalyan   9 years AGO

X=9.75
Y=3
=13

Like? Yes (2) | No (1)  

Udhaya kumar   9 years AGO

x+1/y = 37/4
xy+1 = 37
y = 4
x = 9
4+9 = 13....

Like? Yes (2) | No (1)  

nikhil   9 years AGO

a = 9
b = 4

Like? Yes (1) | No   

Jitendra Prasad   9 years AGO

First no. Is 9
And another 4
9+ 1/4 = 37/4
So sum of integers is 9+4=13

Like? Yes (1) | No   

Paras Jain   9 years AGO

9 + 1/4 = 37/4 so 9+4 =13 Ans.

Like? Yes (1) | No   

A M ANKIT KALLURAYA   9 years AGO

first no=9 second no=4

Like? Yes (1) | No   

MD SOHAIL AKHTER   9 years AGO

9+(1/4)=37/4
so sum of +ve integers= 9+4=13

Like? Yes (2) | No (2)  

Sanu Panji   9 years AGO

x+1/y=37/4
=9+1/4

so x=9, y= 4
x+y=13

Like? Yes (3) | No (3)  

Math Lovers   9 years AGO

37/4 = 9+ 1/4
=> numbers are 9 and 4
sum =13

Like? Yes (2) | No (2)  

Swathy Murugan   9 years AGO

9+(1/4)=37/4
9+4=13

Like? Yes (2) | No (2)  

20 Feb 2015 Sum of Number

An integer x is selected in such a way that
x + x^(1/2) + x^(1/4) = 276
Then sum of all digits of the number
y = 100000 x^(1/4) + 1000 x^(1/2) + x will be

OPtion

1) 16
2) 19
3) 22
4) 24
5) 26
6) 29
7) 32
8) 35
9) 36
10)39

Solution

x + x^(1/2) + x^(1/4) = 276 Integer (i)
So x^(1/4) should be integer
Possibilities for x are 1, 16, 81, 256, 625 ...
Possibility x = 256 can be checked easily
and after checking we get x = 256 so
y = 100000(4) + 1000(16) + 256
= 416256
Sum = 24
Option 4)

Correct Option: 4 Best Solution (34)

Saurabh Kumar Gupta   9 years AGO

x + x^(1/2) + x^(1/4) = 276
let t = x^(1/4) so x = t^4
now t^4 + t^2 + t = 276
so t = 4 fits in the equation
so x = 4^4 = 256
now y = 100000 x^(1/4) + 1000 x^(1/2) + x
y = 100000*4 + 1000*16 + 256 = 416256
so required sum of digits = 4+1+6+2+5+6 = 24

Like? Yes (13) | No (1)  

Supriya Mondal   9 years AGO

From the question we get that x=256
After putting the value of x in the next equation we get
Y=416256
So sum of digit=4+1+6+2+5+6=24
So,option 4)24 is the right answer

Like? Yes (3) | No   

Yathiraj   9 years AGO

x = 256
y = 100000 x^(1/4) + 1000 x^(1/2) + x = 416256
4+1+6+2+5+6 = 24

Like? Yes (2) | No   

RAKESH   9 years AGO

x = 256
find y and sum of digits = 24

Like? Yes (2) | No   

s.durgadevi   9 years AGO

x+x^(1/2)+x^(1/4)=276
1+1/2+1/4=7/4
=(276)^4/7
=24.8
therefore ans is option 4 -24

Like? Yes (2) | No   

JANARTHANAN RAM   9 years AGO

x=256;
y=416256;
so ans 4+1+6+2+5+6=24;

Like? Yes (2) | No (1)  

MRITUNJAY   9 years AGO

x+x^1/2+x^1/4=276
256+16+4=276

y=100000*4+1000*16+256=416256

sum of digit is 24.
option 4.

Like? Yes (2) | No (1)  

praveen kumar   9 years AGO

from the given condition x+x^1/2+x^1/4=276
the value of x comes out= 256

now according to question,
100000 (x=256)^1/4+1000 (x=256)^1/2+(x=256)
=416256

so the sum of all digits,4+1+6+2+5+6=24

Like? Yes (1) | No   

ABHINAV PUSHPAM   9 years AGO

256+16+4=276
Y=416256
So answer will be 24
Option 4)

Like? Yes (2) | No (1)  

KUMAR HARSH   9 years AGO

take x=256 which satisfy the condition x + x^(1/2) + x^(1/4) = 276
so y=100000*4+1000*16+256=416256
sum of digit=4+1+6+2+5+6

Like? Yes (1) | No   

Math Lovers   9 years AGO

x + x^(1/2) + x^(1/4) = 276
=> x= 4
=> y = 416256
sum of digit = 24

Like? Yes (1) | No   

Revathi Perapu   9 years AGO

x=256,
Then 100000 x^(1/4) + 1000 x^(1/2) + x=416256
sum of the digits are 24

Like? Yes (1) | No   

modem mounika   9 years AGO

X = 256; y= 416256
then 4+1+6+2+5+6 =24

Like? Yes (1) | No   

Debashis Dey   9 years AGO

simplifying x + x^(1/2) + x^(1/4) = 276 we get values for x^1/4=4 ,x=256, and x^1/2=16 putting this values in equation y we get a number 56256 sum of digits of this number is 24

Like? Yes (1) | No   

sandeep ramchandra pawaskar   9 years AGO

option 4 is correct

Like? Yes (1) | No   

SAKTHEESVARI S   9 years AGO

256+(256)^1/2+(256)^1/4=256+16+4=276;
so,here x=256; substituting x value in y eqn, we get 416256 as the answer,so the sum of all digit of the number is=4+1+6+2+5+6=24;

Like? Yes (1) | No   

Ajinkya Kulkarni   9 years AGO

x=256

Like? Yes (1) | No   

Amit Nath   9 years AGO

option 4 is absolutely correct

Like? Yes (1) | No   

valli   9 years AGO

x=256
so y=416256

Like? Yes (1) | No   

madhu   9 years AGO

choose x= 256 then we get
y=100000(4)+ 1000(16)+256=401856
then compute the sum of digits Finally we get, 24
so option 4 is the right choose....

Like? Yes (1) | No   

Testing Account   9 years AGO

x=256
y=416256
then sum of digits=24

Like? Yes (1) | No   

APPIKONDA SANTOSH NAIDU   9 years AGO

24

Like? Yes (1) | No   

mohini gupta   9 years AGO

x=256
then 416256
adding 24

Like? Yes (1) | No   

Aman   9 years AGO

x + x^(1/2) + x^(1/4) = 276
x^(1/4) = 4

y = 100000 x^(1/4) + 1000 x^(1/2) + x = 100000 *4 + 1000 *16 + 256

sum of digits = 4+1+6+2+5+6 = 24

Like? Yes (2) | No (1)  

Thirupathi Reddy   9 years AGO

here x=256
so the result=4+1+6+2+5+6=24

Like? Yes (1) | No   

Chiranjeevi   9 years AGO

Take x=256
so 256+16+4=276 given condition satisfied
y = 100000*4 + 1000*16 + 256=416256
sum of digits of y = 4+1+6+2+5+6 = 24

Like? Yes (1) | No   

RAHUL KUMAR   9 years AGO

ans is 24

X = 256
Y= 100000*4+1000*16+256
Y= 400000+16000+256
Sum of all digits = 4+1+6+2+5+6= 24
Option 4 is correct.

Like? Yes (1) | No   

nikhil   9 years AGO

x=256

Like? Yes (1) | No   

Ritesh   9 years AGO

X=256
256+16+4=276
Y=4*100000+16*1000+256
Y=416256
24

Like? Yes (1) | No   

Priyanka Shete   9 years AGO

take x=256, as 4^4=256 and nearly=276,
then 256+16+4=276.
so, y=400000+16000+256=416256
sum of all digits of y=24
so ans is 4)24

Like? Yes (1) | No   

Preethi   9 years AGO

no is 256
so on solving we get 24

Like? Yes (1) | No   

Harish Reddy Namala   9 years AGO

x would be 4 and the number would be 416256.

Like? Yes (1) | No   

rashmi Aggarwal   9 years AGO

by observation x=256
reqd no is 416256
y=4+1+6+2+5+6=24

Like? Yes (2) | No (2)  

Jaspalsingh Parmar   9 years AGO

Option 4 is correct

Like? Yes (1) | No (2)  

19 Feb 2015 Three positive integers

x, y and z are three positive integers such that x not= y not= z and 0 < x < y < z < 10
If x+y+z = 15 then maximum possible value of xy + yz + zx will be

OPtion

1) 74
2) 78
3) 82
4) 89
5) 112
6) 126
7) 148
8) 156
9) 164
10)178

Solution

x+y+z = 15
So avg value = 5
That pair will give the maximum value for which all value will be closed to 5, that possibility is (4, 5, 6) satisfying all conditions
So 4.5 + 5.6 + 6.4 = 20+30+24 = 74
(Here note that if x=y=z then (5, 5, 5) would be the best answer = 74)

Correct Option: 1 Best Solution (71)

Surit Datta   9 years AGO

the possible values of x , y , z are 4 , 5 , 6 respectively
to get the highest xz + yz + xy value
so according to the above value
the summation will be.
(20 + 30 + 24) = 74
so option 1
that's it.

Like? Yes (11) | No (4)  

Saurabh Kumar Gupta   9 years AGO

x+y+z=15
xy+yz+zx will be maximum only when x, y, z will be closest to each other.
so the possible set of values will be 4, 5, 6
so 4*5+5*6+4*6=74

Like? Yes (5) | No (1)  

Divya jain   9 years AGO

x,y and z can be
(1,5,9)
(1,6,8)
(2,4,9)
(2,5,8)
(2,6,7)
(3,4,8)
(3,5,7)
(4,5,6)

max value from (4,5,6) i.e 74 ans

Like? Yes (4) | No   

Kanhaiya Jha   9 years AGO

4,5,6 as x,y,z respectively

Like? Yes (4) | No (1)  

praveen kumar   9 years AGO

as z is the largest nummber then y and smallest is x
as x+y+z=15 and z

Like? Yes (2) | No   

MRITUNJAY   9 years AGO

after satisfying the condition
numbers are 4,5,6
sum is 4*5+5*6+6*4=74
option 1.

Like? Yes (3) | No (1)  

arush gupta   9 years AGO

74

Like? Yes (1) | No   

Supriya Mondal   9 years AGO

Here value of x, y and z will be 4, 5 and 6 respectively.
That satisfy the above 3 conditions that is,
i)x not= y not = z
ii)0

Like? Yes (3) | No (2)  

priyanka    9 years AGO

4,5,6
20+30+24=74
x+Y+Z= 15

Like? Yes (1) | No   

sompon temtheerakij   9 years AGO

x not= y not= z
0 < x < y < z < 10
x+y+z = 15
so, x=1,2,3,4
xy + yz + zx will be max. --> x=4 , y=5 , z=6
xy + yz + zx = 74

Like? Yes (1) | No   

ramakrishnateja   9 years AGO

x not +y, y not =z and 0

Like? Yes (1) | No   

RAKESH DEY   9 years AGO

x=5,y=4,z=6

Like? Yes (1) | No   

modem mounika   9 years AGO

x=4;y=5;z=6

Like? Yes (1) | No   

jaganmohan rao .pola   9 years AGO

74

Like? Yes (1) | No   

APPIKONDA SANTOSH NAIDU   9 years AGO

74

Like? Yes (1) | No   

sandeep ramchandra pawaskar   9 years AGO

option 1 is correct

Like? Yes (1) | No   

KUMAR HARSH   9 years AGO

given 0

Like? Yes (1) | No   

jalaj   9 years AGO

max value of xy+yz+xz=4*5+5*6+6*4=74

Like? Yes (1) | No   

madhu   9 years AGO

Maximum possible case Will arise at x,y,z are 6,5,4 respectively...
then obviously x+y+z=15 & xy+yz+zx=74
so option 1 is the correct answer...

Like? Yes (1) | No   

Kaju Gupta   9 years AGO

x=4, y=5, z=6
xy+yz+zx = 74

Like? Yes (1) | No   

nitin kishore   9 years AGO

as our equation is xy+yz+zx
to maximize the term as well as the value of equation '
i have to take
x=4,y=5,z=6
then xy+yz+zx=74.....this is the maximum value of the given equation.

Like? Yes (1) | No   

RAHUL KUMAR   9 years AGO

Ans is 74

Like? Yes (1) | No   

sona sharma   9 years AGO

4+5+6=15
so x=4 , y=5 ,z=6
and 4*5+5*6+6*4=74
Thus the answer is 74

Like? Yes (1) | No   

AMIT SAURAV   9 years AGO

x+y+z=15;since we have to assign value of x,y,z so that it's sum must be equal to 15 and all x,y,z must have thier greatest value in order of z>y>x.hence,
=>x=4;y=5;z=6;
=>4*5+5*6+4*6
=74 ans.

Like? Yes (1) | No   

Abhishek Rana   9 years AGO

Bcoz multipication of higher no. Is higher and the combination we could get between 0 to 10 acc. To conditions provided best combination is x=4 , y=5,z=6

Like? Yes (1) | No   

Rahul Mahadik   9 years AGO

x=4
y=5
z=6

Like? Yes (1) | No   

rakhi kumari   9 years AGO

(x,y,z)=4,5,6
xy + yz + zx =74

Like? Yes (1) | No   

ganesh babu   9 years AGO

X=4, Y=5, Z=6
xy+yz+zx=74

Like? Yes (1) | No   

Koustabh Chatterjee   9 years AGO

x=4,y=5,z=6
so it will be 74

Like? Yes (1) | No   

Shruity Govil   9 years AGO

x=5, y=6, z=4
5*6+6*4+4*5=74

Like? Yes (1) | No   

Revathi Perapu   9 years AGO

x=4;y=5;z=6
then xy+yz+zx=20+30+24=74

Like? Yes (1) | No   

A M ANKIT KALLURAYA   9 years AGO

x=4 y=5 z=6

Like? Yes (1) | No   

Mukul Dixit   9 years AGO

answer is 1). 74
Because values of x,y and z will be 4,5 and 6 respectively.

Like? Yes (1) | No   

Sambar Naresh   9 years AGO

If x+y+z=15
There will 3 possible values
i) 3+4+8=15
but xy + yz + zx = 68
ii) 3+5+7=15
but xy + yz + zx = 71
iii) 4+5+6=15
xy + yz + zx = 74
so, options 1 is correct.

Like? Yes (1) | No   

tejomayi   9 years AGO

X=4,y=5,z=6

Like? Yes (1) | No   

Alka Sinha   9 years AGO

we have 3 equation :
X != Y != Z
0 < x < y < z < 10
X + Y + Z = 15

Best combination of number to satisfied this condition is 6, 5 and 4.
Som Answer is 74.

Like? Yes (1) | No   

Naveen Pandey   9 years AGO

take the middle value i.e. 15/3=5
so, 5+6+4=15
so the answer will be 74

Like? Yes (1) | No   

Rajesh R   9 years AGO

4+5+6=15
4*5+5*6+4*6=74

Like? Yes (1) | No   

Prudhvi Kalavakunta   9 years AGO

for values 4,5,6 n according to the formula (a+b+c)^2 , maximum possible value for xy+yz+zx is 74

Like? Yes (1) | No   

surya   9 years AGO

x=4,y=5,z=6
x+y+z=15
xy+yz+zx=74

Like? Yes (1) | No   

Shahrukh   9 years AGO

4+5+6=15
so
4*5+5*6+6*4=74

by hit nd trial method

Like? Yes (1) | No   

Harish Reddy Namala   9 years AGO

Tried with maximum sets of values that x,y,z can have. From the sets of values the product can reach maximum upto 74. The minimum value for this is 59.

Like? Yes (1) | No   

A.SOMA SUNDARI   9 years AGO

In the above ques they have given 0

Like? Yes (1) | No   

sneha   9 years AGO

0

Like? Yes (1) | No   

s.durgadevi   9 years AGO

x=6,y=4,z=5
xy+yz+zx=74

Like? Yes (1) | No   

ASIM KUMAR HALDAR   9 years AGO

4,5,6 is maximum possibility so answer is 74

Like? Yes (1) | No   

SUSHANT KUMAR   9 years AGO

x= 4 y =5 z =6

so xy +yz +zx =74

Like? Yes (1) | No   

POOJA SINGH   9 years AGO

all possible values of x y z that satisfy the condition x + y + z equals to 15 is (1,5,9)(1,6,8)(2,6,7)(2,4,9)(2,5,8)(3,4,8)(3,5,7)(4,5,6)
out of this only (4, 5, 6 )satisfy the given question so the maximum value of is 74.

Like? Yes (1) | No   

Dorababu M   9 years AGO

sharing the 15 among 3 parts 5,5,5-->but x!=y!=z then one difference among these 3 variables we get better solution 4,5,6
then 20+30+24=74 it is the maximum possible value..........

Like? Yes (1) | No   

ADARSH GAURAV   9 years AGO

74 is the answer.

Like? Yes (1) | No   

naresh kumar   9 years AGO

let x=4,y=5,z=6 1st condition satisfied ,and x+Y+z=15=4+5+6=15,so xy+yz+zx=4*5+5*6+6*4=74

Like? Yes (1) | No   

kishan kumar   9 years AGO

only best combination is 4 5 6
so sum of multiple of two no is 78 only

Like? Yes (1) | No   

Kashish Aneja   9 years AGO

for product to be maximum x=y=z
but as they are not same therefore they must be near as possible to 5 so
x=4;y=5;z=6
hence max value is 74

Like? Yes (1) | No   

RAKESH ARYA   9 years AGO

here the max. value of x,y and z =4,5 ,6 resp.
so we have xy+yz+xz=74 max.

Like? Yes (1) | No   

shahbaz Alam Ansari   9 years AGO

As ( x+y+z)=15
So req values to be 15 according to question will be
9+5+1
9+4+2
9+1+4
8+6+1
8+5+2
8+4+3
7+2+6
7+3+5
6+4+5
All other values will be repeated for the values of x, y,z
So acc to question
Value for xy+yz+zx from above value of x,y,z are
59
62
49
62
66
74
68
71
74
So maximum possible value is 74

Like? Yes (1) | No   

pravallika   9 years AGO

4,5,6 are the numbers with which we get maximum value for the function xy+yz+zx

Like? Yes (1) | No   

Pinaki Ghosh   9 years AGO

4+5+6 = 15
4*5 + 5*6 + 6*4 = 74

Like? Yes (1) | No   

phani krishna   9 years AGO

maximum value is applicable when 4,5,6 are applicable as x,y,z as the product is cyclic. so value is (4*5+5*6+6*4)

Like? Yes (1) | No   

Shripad Kulkarni   9 years AGO

by taking the random values of x y z as per condition given..
x=4 y=5 z=6
so ans. is 74

Like? Yes (1) | No   

Debashis Dey   9 years AGO

taking x=4 y=5 z=6 we get xy+yz+zx=74 satisfying all given conditions

Like? Yes (1) | No   

surya    9 years AGO

x=4,y=5,z=6
20+30+24=74

Like? Yes (1) | No   

Vipul Kumar Paliwal   9 years AGO

the max. value of z can be 6...then y=5 and x=4...we get x+y+z=15...and
xy+yz+zx=4.5+5.6+6.4= 20+30+24= 74

Like? Yes (1) | No   

Abhishek Sumit   9 years AGO

x=4,y=5,z=6
xy+yz+zx=74

Like? Yes (2) | No (2)  

Jaspalsingh Parmar   9 years AGO

Option 1 is correct

Like? Yes (1) | No (1)  

p.venkata kalyan   9 years AGO

take the x=5,y=4,z=6

Like? Yes (1) | No (1)  

Sanu Panji   9 years AGO

x,y,z= 4,5,6
xy+yz+zx=74

Like? Yes (1) | No (1)  

Yathiraj   9 years AGO

x = 4
y = 5
z = 6
xy+yx+zx: 74

------ PL/SQL Program ----------
0

Like? Yes (2) | No (2)  

ABHINAV PUSHPAM   9 years AGO

No. Is 6,5,4

Like? Yes (1) | No (1)  

sravani   9 years AGO

4,5,6 gives the maximum product...I,e. 74

Like? Yes (1) | No (1)  

salman idrisi   9 years AGO

possible value of x,y and z are 4,5 and 6 in oder to minimize x^2+y^2+z^2 so as to maximize xy+yz+xz formula (x+y+z)^2=X^2+y^2+z^2-2(xy+yz+xz)

Like? Yes (1) | No (1)  

RAKESH   9 years AGO

xy + yz + zx < = 75 for any values

for 6,5,4 its 74

Like? Yes (1) | No (2)  

18 Feb 2015 Non-negative Integers

x/y = z + p/q, where x, y, z, p and q are non-negative integers, then what conclusion can be drawn out of the given options

OPtion

1) y and q will be equal
2) x and p will be equal
3) y and z will be equal
4) x and q will be equal
5) x will be less than y
6) p will be less than q
7) y and q may be equal
8) x and z can't be equal
9) Data are insufficient
10)None of these

Solution

Since all are integers (equal or different) so many possibilities can arise
only 7th option will be a right choice.
Example
14/12 = 1 + 2/12 (i)
14/12 = 1 + 1/6 (ii)

In first equation y = q, in second equation y not= q

Option 7)

Correct Option: 7 Best Solution (14)

Jitendra Prasad   9 years AGO

Since all are integers (equal or different) so many possibilities can arise only 7th option will be a possibility
Example
14/12 = 1 + 2/12 (i)
14/12 = 1 + 1/6 (ii)

In first equal y = q, in second equation y not= q
SO ANS IS 7

Like? Yes (2) | No   

Dharmraj Sharma   9 years AGO

for
z=0
and
x=p
so
y=q

Like? Yes (2) | No   

Abhishek birkh   9 years AGO

if z=0,
p=x
and y=q because here condition is may be

Like? Yes (2) | No   

RAKESH   9 years AGO

y and q may be equal

Like? Yes (2) | No   

A M ANKIT KALLURAYA   9 years AGO

guessed

Like? Yes (2) | No   

p.venkata kalyan   9 years AGO

qx=zqy+py,

by analysing with the given choices except 7 remaining may or may not be equal whereas coming to the choice 7 it defines with the equation.so that answer is 7

Like? Yes (2) | No (1)  

KUMAR HARSH   9 years AGO

here x,y,z,p,z>=0
but y & q naver will be 0 so y&q>0
hance y&q may be equal
let us take x=0,z=0,p=0 we get equation is balanced

Like? Yes (1) | No   

Ritesh R. Navthar   9 years AGO

just i think

Like? Yes (1) | No   

rajendra prasad   9 years AGO

it's obvious that y and q may be equal and still we can get the correct solution set.

Like? Yes (1) | No   

Praveen Kumar Soni   9 years AGO

for the two parts to be equal ,there is a need for denominator to be equal.

Like? Yes (1) | No   

Sravanthi   9 years AGO

3/2=1+1/2

Like? Yes (1) | No   

nitin kishore   9 years AGO

X/Y=(ZQ+P)/Q
IF Y=Q,Then
suppose
Y=1
then X=Z+P
so it's value totaly depend on Z+P
not on Y and Q.

Like? Yes (1) | No   

Tushar Agarwal   9 years AGO

bcz y=q in denominator

Like? Yes (1) | No   

diana jena   9 years AGO

its true

Like? Yes (1) | No   

17 Feb 2015 Age Problem

The sum of ages of Mac and Jabi is 26 times the difference of their ages. Maximum possible difference of the ages is 10 years. How many possibilities for ages of the two persons are possible. Ages are taken in integers.

OPtion

1) 1
2) 2
3) 3
4) 4
5) 5
6) 6
7) 7
8) 8
9) 9
10)0

Solution

According to given condition
x + y = 26 (x - y)
25 x = 27 y
Possible Solution is (27n, 25n)
n = 1, 2, 3, 4, 5 are possible so 5 solution

Option 5)

Correct Option: 5 Best Solution (51)

Supriya Mondal   9 years AGO

There are 5 possibilities that satisfy above conditions
They are-
(135,125)
(108,100)
(81,75)
(54,50)
(27,25)

Like? Yes (11) | No (1)  

Gurram Mahendra ranganath   9 years AGO

M+J=26(M-J)
by solving we get,M=27*J / 25
hence,possible values of J are:25,50,75,100,125
hence,ans is 5.

Like? Yes (8) | No (1)  

Yathiraj   9 years AGO

From the puzzle statement, (M + J) = 26(M - J)
J = (25/27)M
For J to be integer, M should be multiples of 27.
Mac: 27, Jabi: 25
Mac: 54, Jabi: 50
Mac: 81, Jabi: 75
Mac: 108, Jabi: 100
Mac: 135, Jabi: 125
No. of possibilities: 5

Like? Yes (4) | No   

Subhadip Samanta   9 years AGO

suppose the age of Mac=x and age of Jabi=y
as per the question (x+y)=26(x-y)
and max(x-y)=10
also the values of x and y should be integer.
to satisfy these conditions (x-y) should be a even no.
(x-y) should not be 0 because in that case both x and y should be 0
the possible values of (x-y) should be 2,4,6,8,10
for these values there should be 5 combinations of x and y
so that bpth x and y are integers.
So the ans is option 5)

Like? Yes (4) | No (1)  

Dharmraj Sharma   9 years AGO

x=(27/25)*y
so for x to be integer y must be multiple of 25.
so possible pairs
(x,y)
(27,25)
(54,50)
(81,75)
(108,100)
(135,125)

Like? Yes (2) | No   

surya   9 years AGO

m+j=26(m-j)
said max possible difference is 10
so for only even values of (m-j) the ages may be possible . so no of possiblities is 5.

Like? Yes (2) | No   

Abhishek Sumit   9 years AGO

possible ages of two person={27,25},{54,50},{81,75},{108,100},{135,125}

Like? Yes (2) | No   

Surit Datta   9 years AGO

five pairs satisfies the above condition
which are
135,125
108,100
81,75
54,50
27,25
so option 5
that's it

Like? Yes (2) | No   

b.subbarao   9 years AGO

let mac and jabi ages are be =x and y' as per the data
x+y=26(x-y) so 25x-27y=0 27+25=26*2=52,------(1)
54+50=26(4)=104--------------(2)
81+75=26(6)=156--------------(3)
108+100=26(8)=208-----------(4)
135+125=26*10=260----------(5)

Like? Yes (2) | No   

A.SOMA SUNDARI   9 years AGO

If the diff of two ages are in odd no then there is no possibilities for the sum of two ages are equal to 26 times of the diff of the ages.
So the even num upto 10 is 2,4,6,8,10
therfore the no of possibilities is 5

Like? Yes (2) | No   

SAKTHEESVARI S   9 years AGO

difference of their age must be an even num,then only the condition is satisfied, the max possibility of their age difference is 10, so all the resultant possibilities are 2,4,6,8,10,totally 5 possibilities.

Like? Yes (2) | No   

Nidadavolu.Suneetha   9 years AGO

The maximum difference is 10.so the diff. in b/w 0 to 10.
for example take diff. is 1.then (age of mac+age of jabi)=26(1)
so the diff. b/w two of them is 1 but it not possible.If diff. is 2 sum of ages is 2(26)=52 bcoz take one person age is 25&another is 27.It satisfies the diff.
The diff. 2,4,6,8,10 satisfies the condition.So max. no of possibilities of their ages are 5

Like? Yes (2) | No   

ABHINAV PUSHPAM   9 years AGO

5 is asnwer.
=> x+y=26(x-y)
=> 25x=27y
=> x/y= 27/25
difference is 2. so possible values could be 27,28 or 54,50 or 81,75 or 108,100 and 135,125
So, answer ie 5.
option 5)

Like? Yes (1) | No   

RAKESH   9 years AGO

M+J = 26*(M-J)

M-J

Like? Yes (1) | No   

Saurabh Kumar Gupta   9 years AGO

Let M > J
(M+J) = 26*(M-J)
Now (M-J) can take values between 1 to 10
Out of these only 2, 4, 6, 8, 10 satisfy the condition, i.e. for eg. if their age difference is 2 years then M+J = 52 and so M & J can be 27 and 25 respectively.
But if the difference is 3 then M+J = 78, now we cant get their ages with difference of 3 years.
So total possible value pairs = 5

Like? Yes (1) | No   

praveen kumar   9 years AGO

5 such ages can be

Like? Yes (1) | No   

sompon temtheerakij   9 years AGO

If diff. 1, ages of the two persons = 24.5+25.5
If diff. 2, ages of the two persons = 25 + 27
If diff. 3, ages of the two persons = 37.5+40.5
If diff. 4, ages of the two persons = 50+54
If diff. 5, ages of the two persons = 62.5+67.5
If diff. 6, ages of the two persons = 75+81
If diff. 7, ages of the two persons = 87.5+94.5
If diff. 8, ages of the two persons = 100+108
If diff. 9, ages of the two persons = 112.5+121.5
If diff. 10, ages of the two persons = 125+135
Ages are taken in integers , possibilities for ages of the two persons are 5.

Like? Yes (1) | No   

Santosh Kumar   9 years AGO

(x - y)26 = (x + y)
25x - 26y =x + y
25x - 27y = 0 ................eqn 1
x - y = 1 to 10 (given) .........eqn 2
if x-y=1
we don't get integer age
we get integer only if
x - y = (2,4,6,8,10)
so, obtion 5

Like? Yes (1) | No   

Himanshu   9 years AGO

Let age of mac= x and age of jabi = y
According to question, sum of their ages=26(difference between their ages)
x+y=26(x-y)
now let diff =1
x-y=1 and x+y =26
solving x =13.5 and y = 12.5
but ages are taken as integers, so this can't be the right one,
now diff = 2
x+y = 52 and x-y=2
on solving x=26 y =25
similarly the maximum possible diff can be 10
taking different values of difference of their ages upto 10, ages of x and y which are integers are considered and no. comes out to be 5.
So 5 is the right option

Like? Yes (1) | No   

Divya jain   9 years AGO

mac+jabi=26(mac-jabi)
mac-jabi 125,135
for 8-> 108,100
6-> 81,75
4-> 50,54
2-> 25,27

Like? Yes (1) | No   

Ajinkya Kulkarni   9 years AGO

5

Like? Yes (1) | No   

ramakrishnateja   9 years AGO

given that the m+j=26(m-j). also given that the diff is max of 10 yrs. so, it should be 1,2,.....upto 10. where ages shouldnot be integers frm these diff.
so let us take diff=1. here 27j=25m.so m:j=27:25 hence m=27x and j=25x. so diff is 2x=1 hence x=1/2 where ages will be 27/2 and 25/2. which is wrong. so 1,3,5,7,9 are omitted. take diff as 2,4,6,8,10. then for diff =2 then 2x=2 x=1. so ages will be m=27,25. so sum =52 which is 26(2)

Like? Yes (1) | No   

purba mondal   9 years AGO

x+y=26(x-y)
if x-y=1,x+y=26,and x=13 y=13 x-y=0; x=14,y=12,x-y=2 so x-y!=1 not possible
if x-y=2,x+y=52 and x=27,y=25,x-y=2 possible
then same as x-y=3 not possible
x-y=4 is possible as x=54,y=50
....
then for x-y=even then possible,as max(x-y)=10 then
x-y=2,4,6,8,10 possible
5 possibilities

Like? Yes (1) | No   

Krishna Gandhi   9 years AGO

M+J=26(M-J)
M-J=1,2,3,….,10
So Case-1: M-J=1
M+J=26*1=26
M-J=1
M+j=26
-----------
2M=27;
M=13.5 ; which isn’t a integer
So, M-J=1 isn’t possible

Case-2: M-J=2
M+J=26*2=52
M-J=2
M+J=52
-----------
2M=54; M=27; J=25
So, ages are in integers hence M-J=2 is possible

Similarly, solve for M-J=3,4,…,10
We get the possible ages in integer for M-J=2,4,6,8,10
Ans: So there are 5 possibilities (i.e. M-J= 2,4,6,8,10)

Like? Yes (1) | No   

ADARSH GAURAV   9 years AGO

5 is the answer.

Like? Yes (1) | No   

Alka Sinha   9 years AGO

Max Possible diff of ages = 10
So, Possibilities of differences is from 1 to 10

Let Mac age = X
Jabi age = Y
Equation : X + Y = 26(X-Y)

Case 1: Let diff of ages = 10
X + 10 + X = 26(X+10-X)
X = 125
Y = 135

When we check other possibilities then it will satisfy the equation only for age diff 10, 8, 6, 4, 2.

So, Answer is 5.

Like? Yes (1) | No   

ANKIT RAMAN   9 years AGO

possible combinations are (25,27),(50,54),(75,81),(100,108),(125,135)

Like? Yes (1) | No   

Nupur   9 years AGO

All the even numbered difference can be mapped with ages.

Like? Yes (1) | No   

ganesh babu   9 years AGO

Only possible ages
(27,25)
(54,50)
(81,75)
(108,100)
(135,125)

Like? Yes (1) | No   

Sabitha.R   9 years AGO

x+y=26(x-y)
25x=27y
x/y= 27/25

Like? Yes (1) | No   

mohini gupta   9 years AGO

only cases where x-y is even

Like? Yes (1) | No   

karthik   9 years AGO

fjhjv

Like? Yes (1) | No   

Devendra Marghade   9 years AGO

Possible ages are total 5
(25,27) , (50,54) , (75,81), (100,108), (125,135)
Ans: 5

Like? Yes (1) | No   

lakshmi aparna   9 years AGO

Let the ages of mac and jabi be M, J.
M+J = 26 (M-J); M-J

Like? Yes (1) | No   

A M ANKIT KALLURAYA   9 years AGO

x+y=26(x-y)
25x=27y
x/y= 27/25
difference is 2. so possible values could be 27,28 or 54,50 or 81,75 or 108,100 and 135,125

Like? Yes (1) | No   

Bhavana.K   9 years AGO

guess

Like? Yes (1) | No   

vishal kumar   9 years AGO

=> x+y=26(x-y)
=> 25x=27y
=> x/y= 27/25
difference is 2. so possible values could be 27,28 or 54,50 or 81,75 or 108,100 and 135,125
So, answer ie 5

Like? Yes (1) | No   

Sneha   9 years AGO

=> x+y=26(x-y)
=> 25x=27y
=> x/y= 27/25
difference is 2. so possible values could be 27,28 or 54,50 or 81,75 or 108,100 and 135,125
So, answer ie 5

Like? Yes (1) | No   

Namrata Gupta   9 years AGO

ans is 5
x+y=26(x-y)
25x=27y
x/y=27/25
so possible sol is 5.

Like? Yes (1) | No   

Jitendra Prasad   9 years AGO

5 is the answer.
=> x+y=26(x-y)
=> 25x=27y
=> x/y= 27/25
difference is 2. so possible values could be 27,28 or 54,50 or 81,75 or 108,100 and 135,125
So, answer ie 5

Like? Yes (1) | No   

POOJA SINGH   9 years AGO

x+y=26(x-y)
25x=27y
x/y= 27/25
difference is 2. so possible values could be 27,28 or 54,50 or 81,75 or 108,100 and 135,125So, answer is 5

Like? Yes (1) | No   

neethu   9 years AGO

5

Like? Yes (1) | No   

Sourabh Prasad   9 years AGO

x+y=26(x-y)
25x=27y
x/y= 27/25
difference is 2.
So all possible values could be 27,28 or 54,50 or 81,75 or 108,100 and 135,125

Like? Yes (1) | No   

Paras Jain   9 years AGO

its 5 Ans.

Like? Yes (1) | No   

MRITUNJAY   9 years AGO

This is answer

Like? Yes (1) | No   

tasho k   9 years AGO

Trial and error...
25, 27
50,54
81, 75
100, 108
135, 125

Like? Yes (1) | No   

Dolly Singh   9 years AGO

Since their age difference can be max up to 10

Let say age of Max is x

age of Jabi is y

then

A To P, x + y = 26(x-y)

and x - y

Like? Yes (1) | No   

arush gupta   9 years AGO

5 is the answerx+y=26(x-y)=> 25x=27y=> x/y= 27/25difference is 2. so possible values could be 27,28 or 54,50 or 81,75 or 108,100 and 135,125So, answer is 5

Like? Yes (1) | No   

Jaspalsingh Parmar   9 years AGO

Option 6 is correct

Like? Yes (2) | No (2)  

Anshul Garg   9 years AGO

M + J = 26( M - J )
And M-J

Like? Yes (1) | No (1)  

RAHUL KUMAR   9 years AGO

ans is 5

Like? Yes (1) | No (2)  

16 Feb 2015 Sequence

What is the next number of the following sequence
13, 34, 74, 170, 290, ..............

OPtion

1) 359
2) 389
3) 421
4) 458
5) 471
6) 497
7) 505
8) 530
9) 555
10)595

Solution

13 = 4 + 9 = 2^2 + 3^2
34 = 9+25 = 3^2 + 5^2
74 = 25 + 49 = 5^2 + 7^2
170 = 49 + 121 = 7^2 + 11^2
290 = 121 + 169 = 11^2 + 13^2
Similarly
458 = 169+289 = 13^2 + 17^2

Option 4)
2, 3, 5, 7, 11, 13, 17 ... are prime numbers

Correct Option: 4 Best Solution (44)

Supriya Mondal   9 years AGO

Here, the numbers are addition of two consecutive prime number's square.
1st term=2^2+3^2=13
2nd term=3^2+5^2=34
3rd term=5^2+7^2=74
4th term=7^2+11^2=170
5th term=11^2+13^2=290
So, next term will be=13^2+17^2=458
So, option 4)458 is the right answer

Like? Yes (12) | No (1)  

Santhoshkumar   9 years AGO

Prime number square,
2^2+3^3=13
5^2+3^2=34
7^2+5^2=74
11^2+7^2=170
13^2+11^2=290
17^2+13^2=458

Like? Yes (6) | No   

RAKESH   9 years AGO

2^2 + 3^2 = 13
3^2 + 5^2 = 34
5^2 + 7^2 = 74
7^2 + 11^2 = 170
11^2 + 13^2 = 290
13^2 + 17^2 = 458

Like? Yes (4) | No   

Surit Datta   9 years AGO

2^2 + 3^2 = 13
3^2 + 5^2 = 34
5^2 + 7^2 = 74
7^2 + 11^2 = 170
11^2 + 13^2 = 290
13^2 + 17^2 = 458
the series is summation of two consecutive prime number
starting from 2
so option 4
that's it.

Like? Yes (2) | No   

JANARTHANAN RAM   9 years AGO

prime no^2+(next prime no^2)
2^2+3^2=13;
3^2+5^2=34;
'..
..
13^2+17^2=458

Like? Yes (2) | No   

sunilbaba    9 years AGO

s

Like? Yes (1) | No   

Chandu Reddy   9 years AGO

2^2+3^2=13,3^2+5^2=34,5^2+7^2=74,7^2+11^2=170, 11^2+13^2=290,13^2+17^2=458

Like? Yes (1) | No   

Anshul Garg   9 years AGO

13 = 4 + 9 = 2^2 + 3^2
34 = 9 + 25 = 3^2 + 5^2
74 = 25 + 49 = 5^2 + 7^2
170 = 49 + 121 = 7^2 + 11^2
290 = 121 + 169 = 11^2 + 13^2

From the above, we can conclude that the number is addition of square of lager prime number in previous and the square of next prime number.

And, Hence the next number will be 169 + 289 = 458

Like? Yes (1) | No   

Jaspalsingh Parmar   9 years AGO

Option 4 is correct

Like? Yes (1) | No   

mahi reddy   9 years AGO

2^2+3^2=13,
3^2+5^2=34,
5^2+7^2=74,
7^2+11^2=170,
11^2+13^2=290,
13^2+17^2=458

Like? Yes (1) | No   

Testing Account   9 years AGO

2*2 3*2,3*2 5*2,5*2 7*2,7*2 11*2, 11*2 13*2,13*2
17*2=458

Like? Yes (1) | No   

praveen kumar   9 years AGO

17^2+13^2

Like? Yes (1) | No   

Sravanthi   9 years AGO

2^2+3^2=13
3^2+5^2=34
sum of squres of primary numbers

Like? Yes (1) | No   

ABHINAV PUSHPAM   9 years AGO

2^2 + 3^2 = 13
3^2 + 5^2 = 34
5^2 + 7^2 = 74
7^2 + 11^2 = 170
11^2 + 13^2 = 290
13^2 + 17^2 = 458
So, option 4) is correct

Like? Yes (1) | No   

MRITUNJAY   9 years AGO

2^2+3^2=13
3^2+5^2=34
5^2+7^2=74
7^2+11^2=170
11^2+13^2=290
13^2+17^2=458

option 4.

Like? Yes (1) | No   

RAHUL KUMAR   9 years AGO

Ans is 458

2^2 + 3^2 = 13
3^2 + 5^2 = 34
5^2 + 7^2 = 74
7^2 + 11^2 = 170
11^2 + 13^2 = 290
13^2 + 17^2 = 458 .........ANS

Like? Yes (1) | No   

Abhishek Sumit   9 years AGO

2^2 + 3^2 = 13
3^2 + 5^2 = 34
5^2 + 7^2 = 74
7^2 + 11^2 = 170
11^2 + 13^2 = 290
13^2 + 17^2 = 458

Like? Yes (1) | No   

Dharmraj Sharma   9 years AGO

2^2 + 3^2 =13
3^2 + 5^2 =34
5^2 + 7^2 =74
7^2 + 11^2 =170
11^2 + 13^2=290
13^2 + 17^2=458

Like? Yes (1) | No   

Abhishek birkh   9 years AGO

2^2+3^2,3^2+5^2,5^2+7^2

Like? Yes (1) | No   

ramakrishnateja   9 years AGO

the given numbers are the sum of squares of the prime numbers. they are as follows:
13=3^2+2^2
34=5^2+3^2
74=7^2+5^2
170=11^2+7^2
290=13^2+11^2
so the next number is 458=17^2+13^2

Like? Yes (1) | No   

Subhadip Samanta   9 years AGO

2^2+3^2=13
3^2+5^2=34
5^2+7^2=74
7^2+11^2=170
11^2+13^2=290
the next prime no of 13 is 17
:The next term in the series is(13^2+17^2)=458 (option 4)

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Rabindra Prasad Modi   9 years AGO

sum of two consecutive prime numbers everytime

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jaganmohan rao .pola   9 years AGO

prime numbers: 2,3,5,7,11,13,17,...............

2^2+3^2 = 13
3^2+5^2 = 34
5^2+7^2 = 74
7^2+11^2 = 170
11^2+13^2 = 290
13^2+17^2 = 458

Ans:458

Like? Yes (1) | No   

KUMAR HARSH   9 years AGO

because all are the addition of two squared number like
2^2+3^2=13
3^2+5^2=34
5^2+7^2=74
7^2+11^2=170
11^2+13^2=290
so next is by option 13^2+17^2=458

Like? Yes (1) | No   

vishal kumar   9 years AGO

2^2 + 3^2 = 13
3^2 + 5^2 = 34
5^2 + 7^2 = 74
7^2 + 11^2 = 170
11^2 + 13^2 = 290
13^2 + 17^2 = 458 ....... this is the answer

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Ajinkya Kulkarni   9 years AGO

sum of squares of adjacent prime numbers

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Jitendra Prasad   9 years AGO

2^2+3^2=13
3^2+5^2=34
5^2+7^2=74
7^2+11^2=170
11^2+13^2=290
13^2+17^2=458

Like? Yes (1) | No   

ADARSH GAURAV   9 years AGO

458 is the answer

Like? Yes (1) | No   

b.subbarao   9 years AGO

2^2+3^2=13
3^2+5^2=34
5^2+7^2=74
7^2+11^2=170
11^2+13^2=290 so 13^+17^2=458

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Saurabh Soni   9 years AGO

half of the numbers nd thn subtract them..The ans is thn doubled to get d next no.

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Neha Giri   9 years AGO

Here all are two successive prime no squares addition, i.e.
13=2^2+3^2
34=3^2+4^2
74=5^2+7^2
170=7^2+11^2
290=11^2+13^2
so,next will be
13^2+17^2=458.

Like? Yes (1) | No   

Aishwariya    9 years AGO

2^2+3^2=13
3^2+5^2=34
5^2+7^2=74
7^2+11^2=170
11^2+13^2=290
13^2+17^2=458

Like? Yes (1) | No   

A M ANKIT KALLURAYA   9 years AGO

each number is the sum of square of prime numbers.
2^2+3^2=13
3^2+5^2=34
5^2+7^2=74
7^2+11^2=170
11^2+13^2=290
13^2+17^2=458

Like? Yes (1) | No   

Tark Patel   9 years AGO

-> 13 , 34 , 74 , 170 , 290

-> (2^2 + 3^2) , (3^2 + 5^2) , (5^2 + 7^2) , (7^2 + 11^2) , (11^2 + 13^2) so next will be = (13^2 + 17^2)

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pravallika   9 years AGO

3^2+2^2;5^2+3^2;7^2+5^2;11^2+7^2;13^2+11^2;17^2+13^2=13,34,74,170,290,458(sum of squares of primenumber and its preceeding prime)

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tasho k   9 years AGO

Addition of squares of consecutive prime numbers....

Square of 13 + square of 17 = 169+289=458

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sandeep kumar   9 years AGO

3^2+2^2=13

5^2+3^2=34

similarly ..

next prime no.^2+previous prime no.^2 = ANS...................



ANS = 458

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DHEERAJ KUMAR   9 years AGO

we are adding sq. of continuous two prime numbers like 2sq+3sq, 3sq+5sq, 5sq+7sq., 7sq+11sq, 11sq+13sq and finally 13sq+17sq= 458

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POOJA SINGH   9 years AGO

2^2 +3^2=13
3^2 +5^2=34
5^2 + 7^2=74
7^2+11^2=170
11^2+13^2=290
13^2+17^2=458

Like? Yes (1) | No   

arush gupta   9 years AGO

answer is 458
square and add 2 consequent prime numbers 2 3 5 7 11 13 17
2^2+3^2 =13
3^2+5^2=34
5^2+7^2=74
7^2+11^2=170
11^2+13^2=290
13^2+17^2=458

Like? Yes (1) | No   

Mohak Jain   9 years AGO

consecutive two prime nos squared and added.
eg:
2^2+3^2, 3^2+5^2,....... 17^2+13^2

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nitya   9 years AGO

2^2+3^2=13
3^2+5^2=34
5^2+7^2=74
7^2+11^2=170
11^2+13^2=290
13^2+17^2=458

Like? Yes (1) | No   

Akshat Kapoor   9 years AGO

answer is 458
square and add 2 consequent prime numbers 2 3 5 7 11 13 17
2^2+3^2 =13
3^2+5^2=34
5^2+7^2=74
7^2+11^2=170
11^2+13^2=290
13^2+17^2=458

Like? Yes (1) | No   

Paras Jain   9 years AGO

its 458..

Like? Yes (1) | No   

13 Feb 2015 Possible value of integer

The cube of sum of a positive integer and its square is a five digit number. What will be the sum of cubes of sum of integer and its square for all possible value of integer.

OPtion

1) 96462
2) 101088
3) 108652
4) 113764
5) 119766
6) 124760
7) 128642
8) 132496
9) 136468
10)142444

Solution

Let the integer is x then according to given condition
(x+x^2)^3 = Five digit number
for x = 4; (x+x^2)^3 = Four digit number
for x = 7; (x+x^2)^3 = Six digit number
so x = 5 and 6
fox x = 5; (x+x^2)^3 = 27000
for x = 6; (x+x^2)^3 = 74088
Sum = 101088

Option 2)

Correct Option: 2 Best Solution (33)

Surit Datta   9 years AGO

let the integer be a
we have to calculate (a + a^2) such that it become a 5 digit number.
only possible for 5 and 6
(5 + 5^2) = 27000
(6 + 6^2) = 74088
so there sum is 27000 + 74088
which is 101088
so option 2
that's it.

Like? Yes (12) | No (5)  

Supriya Mondal   9 years AGO

Let the number is a
So,according to the question we can write that
(a+a^2)^3=5 digit number
After calculation we get that only 2 values satisfy this condition. They are 5 and 6
So ans will be=(5+5^2)^3+(6+6^2)^3
=(30)^3+(42)^3
=27000+74088
=101088
So,option 2)101088 is the right answer

Like? Yes (8) | No (3)  

g lokesh   9 years AGO

(x+x2)3=fivedigit number,
substituting x=1,2,3......
condition satisfies only for x=5,6
x=5,value is 27000
x=6,value is 74088
sum=101088

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praveen kumar   9 years AGO

5 and 6 are only two integers which fits under 1st condition of given problem.
accordind to Q
(5+25)^3+(6+36)^3=101088

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arush gupta   9 years AGO

ans 101088

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ADARSH GAURAV   9 years AGO

101088 is the answer

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madhu   9 years AGO

[x+x^2]^3=27000 for x=5
and [x+x^2]^3=74088 for x=6
so 101088 is the answer we get after adding both...

Like? Yes (1) | No   

Saurabh Kumar Gupta   9 years AGO

The minimum 5-digit perfect cube is: 22^3 = 10648
The maximum 5-digit perfect cube is: 46^3 = 97336
So out of numbers between 22-46, only 30(5+25) and 42(6+36) can be represented as sum of a +ve integer and its square.
So, the required answer is: 30^3 + 42^3 = 101088

Like? Yes (3) | No (3)  

Vibha   9 years AGO

(x+x^2)^3
by hit and trial
x= 5,6
hence
result = 27000 + 74088
= 101088 ans

Like? Yes (1) | No (1)  

shubham mishra   9 years AGO

there are two possible positive integer.
5,6
now
((5+5^2)^3) +((6+6^2)^3)=101088

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MRITUNJAY   9 years AGO

(X+x^2)^3=five digit number

x= 5 and 6 satisfy the condition
(5+25)^3=27000
(6+36)^3=74088
sum of these two are 74088+27000
=101088

option 2.

Like? Yes (1) | No (1)  

RAHUL KUMAR   9 years AGO

5+5^2 = 30^3=27000
6+6^2=42^3=74088
27000+74088=101088
option 2 is r8.

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Abhishek birkh   9 years AGO

(x+x^2)^3=5digit
22

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ABHINAV PUSHPAM   9 years AGO

Option 2 is answer

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Dharmraj Sharma   9 years AGO

for equation (x+x^2)^3=5 digit number
range of (x+x^2)
22

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Sanu Panji   9 years AGO

x=5, 6
(5+5^2)^3=27000
(6+6^2)^3=74088
sum= 101088

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Jaspalsingh Parmar   9 years AGO

option 8 is correct

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Preethi   9 years AGO

5 and 6 are the numbers that satisfy given condition

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Dimple Kundanani   9 years AGO

Ans is B)

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JEETENDER KALACHAND KEWLANI   9 years AGO

acc to question
(x(1+x))^3=5 digit no.

if x=3......12^3=1628
if x=4....20^3=8000
if x=5....30^3=27000
if x=6....42^3=74088
if x=7....56^3=175616

hence we conclude that acc to ques x=5 and 6 are only valid candidates...hence sum of these will be
30^3 + 42^3=101088

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A M ANKIT KALLURAYA   9 years AGO

(X+x^2)^3=five digit number

x= 5 and 6 satisfy the condition
(5+25)^3=27000
(6+36)^3=74088
sum of these two are 74088+27000
=101088

option 2.

Like? Yes (1) | No (1)  

Ajinkya Kulkarni   9 years AGO

(x+x^2)^3=5 digit no

so only numbers from 22 and 46 have cube in 5 digits ....
21 has cube of 4 digit and 47 has cube of 6 digit ...

(5+25)^3+(6+36)^3=101088

Like? Yes (4) | No (4)  

Sunil Chavan   9 years AGO

let the +ve integer is X and its square X^2.
according to problem defn i.e (X+X^2)^3 = (five digit number)
now go on put 1,2,3,4,5,6,7,......n in the above eqn
for X=5 we get 27000(five digit num).......(a)
and for X=6 we get 74088(five digit num).....(b)
ADD eqn (a) and (b)
we will get 101088 option(2)

Like? Yes (1) | No (1)  

RAKESH   9 years AGO

(x+x^2)^3= five digit no.

x=5 & 6
30^3 + 42^3 = 101088

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kaushik mishra   9 years AGO

Let the no be n
So according to given condition
(n+n^2)^3=five digit number
Number satisfying these conditions are n=5,6 as after 6 for eg 7 onwards will result into 6 digit number so putting n=5
(5+25)^3=27000
Putting n=6
(6+36)^3=74088
Adding these two results we get 27000+74088=101088
Which is the answer...

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Budhaditya Bhattacharya   9 years AGO

(X+x^2)^3=five digit number

x= 5 and 6 satisfy the condition
(5+25)^3=27000
(6+36)^3=74088
sum of these two are 74088+27000
=101088

Like? Yes (1) | No (1)  

Paras Jain   9 years AGO

its 101088 Ans.
for x=5 and x=6 [(x+x^2)]^3

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Jitendra Prasad   9 years AGO

(X+x^2)^3=five digit number

x= 5 and 6 satisfy the condition
(5+25)^3=27000
(6+36)^3=74088
sum of these two are 74088+27000
=101088

option 2.

Like? Yes (1) | No (1)  

Devendra Kumar   9 years AGO

(x+x^2)^3=five digit number

x= 5 and 6 satisfy the condition
(5+25)^3=27000
(6+36)^3=74088
sum of these two are 74088+27000
=101088

option 2 is correct

Like? Yes (1) | No (1)  

Abhik Das   9 years AGO

only The no 5 and 6 satisfies the condition.so the sum will be=[(5^2+5)^3+(6^2+6)^3]=[27000+74088]=101088

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JANARTHANAN RAM   9 years AGO

possible cmntn is 5,6.
min 5 dgt no=10648 ....(22^3)
max 5 dgt no 97336.....(46^3)
sum of a positive integer and its square should be btwen 22 & 46
so we hve only 5+5^2=30...& 6+6^2=42;
so ans is 30^3+42^3=101088;

Like? Yes (2) | No (3)  

KUMAR HARSH   9 years AGO

here let integer is 5 and 6
then in first statement (n[i]+n[i]^2)3
for n[1]=5is 27000
for n[2]=6 is 74088
and in second statement [sum of [(n[i]+n[i]^2)3]]
these=27000+74088=101088

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Alka Sinha   9 years AGO

only 5 and 6 falls in this category.
(5 + 5^2)^3 = 27000
(6 + 6^2)^3 = 74088

Sum = 101088

Like? Yes (1) | No (2)