naveen   9 years AGO

let the numbers be x,y, z
x+y+z=xyz-15 & (x-2)+y+z=(x-2)yz+7
by solving them we get yz=12
the y, z may be 2,6 or 3,4:
but 3, 4 are satisfying the above condition from that we get x=2;
hence the product xyz=2*3*4=24
hence option (4) is correct

Like? Yes (14) | No (2)  

Anup Sahay   9 years AGO

if digits are 2,3,4 then their sum =9 and product =24 so the sum is 15 less than their product satisfies and if 2 is subtracted from 1st number the numbers are 0,3,4 so their sum = 7 and product =0 which satisfies the equation
so the product of numbers = 2*3*4=24
option 4)
i made a hit and trial here no equations which took me less time

Like? Yes (6) | No   

ABHINAV PUSHPAM   9 years AGO

The required number is 243

Like? Yes (7) | No (4)  

shibanti sanyal   9 years AGO

the two equations will be: x+y+z=xyz-15 & (x-2)+y+z=(x-2)yz+7.
Solving we get yz=12 & x=2.
Hence, product=2*12=24. ans

Like? Yes (3) | No   

kriti kankariya   9 years AGO

let the three numbers be x,y,z
then,x+y+z=x*y*z - 15
& (x-2)+y+z=(x-2)*y*z + 7
solving these we will get y*z=12
possiblities are {1,12} , {2,6} , {3,4}
taking these {3,4} is satisfying the conditions giving x=2
so x*y*z=2*12=24

Like? Yes (2) | No   

Anubrindha   9 years AGO

24 = 4*3*2
Sum of 4+3+2=9
Sum of these 3 numbers are 15 less than their product

Like? Yes (2) | No   

rahul shaw   9 years AGO

since diff in 1st case is 15 so the value of product should be above 15...
then we came to 24 and then breaking it into factors and calculating...we can get it...
24=4*3*2
sum=9
9+15=24
so option 4

Like? Yes (2) | No   

bhaskar   9 years AGO

xyz=x+y+z+15
(x-2)yz+7=x+y+z-2
yz=12
y=3
z=4
(y=2,z=6 is not possible for both the conditions)
12x=x+22
x=2
product =2*3*4=24

Like? Yes (2) | No (1)  

rajat gupta   9 years AGO

since numbers are 2, 3 and 4 this is the only possible case by hit and trial method

Like? Yes (1) | No   

Arvind kumar   9 years AGO

Product of 2,3 and 4 is 24 which is 15 more than their sum i.e.9

Like? Yes (3) | No (2)  

Deepak Agarwal   9 years AGO

the three digit no. is 234
where 2+3+4=9 & 2*3*4=24 which is 15 more than their sum.
If we subtract 2 from 2 we get 0 so their product will be zero and sum will be 0+3+4=7.
So the product is 2*3*4=24

Like? Yes (1) | No   

Rohit Agrawal   9 years AGO

x+y+z=xyz-15
x+y+z-2=(x-2)yz+7
=>yz=12
y=2or 3 z=6 or 4
3 4 holds for the 1st equatiion
solving x=2
answr is 2x3x4=24

Like? Yes (1) | No   

sudiptaa   9 years AGO

numbers are 2,3,4
Product is 2*3*4 = 24

Like? Yes (1) | No   

praveendas   9 years AGO

three digits are 4,3,2.no is 432..verify it with given conditions

Like? Yes (1) | No   

Mayur jaiswal   9 years AGO

2*3*4=24
Dat is 15 less Dan 2+3+4=9

Like? Yes (1) | No   

siddhi datri   9 years AGO

a+b+c=abc-15
a+b+c-2=abc+7
on solving them, we get
ab=12,then abc=24 satifies the both equation.

Like? Yes (1) | No   

Kethineni Vinodh   9 years AGO

4+3+2=9
4*3*2=24
432-2=430 i.,e 4+3+0=7
4*3*0=0

Like? Yes (1) | No   

Subhendu Pradhan   9 years AGO

a*b*c=15+a+b+c
(a-2)+b+c=7+(a-2)*b*c=7+abc-2bc
bc=12
abc will be multiple of 12
so,b & c may be 4 and 3 respectively
by putting the value of a=1,2,3.... u have to check the 1st or 2nd condition u will find the option

Like? Yes (1) | No   

prasad   9 years AGO

nm

Like? Yes (1) | No   

ADARSH GAURAV   9 years AGO

24 is the answer.........

Like? Yes (1) | No   

Ankit singh   9 years AGO

24

Like? Yes (1) | No   

DURGESH KUMAR   9 years AGO

24=4*3*2 and sum =9, 24-9 =15

Like? Yes (1) | No   

nirmal das   9 years AGO

the numbers are 2,3,4...as 2+3+4=9 and 2*3*4=24 which is 15 more than their sum...when 2-2=0 then product is 0 bt sum is 7....

Like? Yes (1) | No   

Shankar   9 years AGO

a b c=abc-15
(a-2) b c=7 (a-2)bc
Solving we get
bc=6
Then put bc=6 in first equation
a b c=6a-15
b c 15=5a
bc=6,therefore chances for b and c is 6,1 or 2,3
We take 2,3
Thus a=4
Now abc =4*2*3=24

Like? Yes (1) | No   

rakesh sharma   9 years AGO

15+x+y+z=xyz.....(1)
(x-2)+y+z=(x-2)yz+7...(2)

solve both equ. then we find yz=12
xyz=?
12x=?
by answer:
x=1,2,4,5
put x=2
xyz=24
x+y+z=xyz-15 by equ. (1)
x+y+z=24-15=9
so
x+y+z=9
x=2,y=4/3,z=3/4

so answer (4) is correct
xyz=24

Like? Yes (1) | No   

Anshul khandelwal   9 years AGO

a+b+c+15=a*b*c ----(a)
(a-2)+b+c=(a-2)*b*c -----(b)

solving both equation a and b above
bc=12
so value of b and c will either be 4 and 3 or 6 and 2 or 1 and 12.
best suitable value is 4 and 3 because on solving with other value value of a i not coming integral.
so from first equation (a)
a+4+3+15=12a
a=2

hence multiplication of abc = 24

Like? Yes (1) | No   

CHANDAN KUMAR   9 years AGO

let 3 single digit no. be x , y and z.
now,
x+y+z = xyz -15 ......(I)

(x-2)+y+z = (x-2)(yz) + 7
or , x+y+z = xyz -2yz + 9...... (ii)

solving equations I and ii , we get

xyz-15 =xyz-2yz +9
or,
2yz = 24
or,
yz = 12......... (iii)

let y = 3 & z =4 ( from equation iii )

now , equation (I) become
x+3+4 = 12x -15
or,
x+7 =12x -15

or,
11x = 22
or,
x = 2


finally ,
3 digits are
2 ,3, 4 .

so, their product = 2�3�4 =24

correct answer
options 4) 24

Like? Yes (1) | No   

pawan kumar   9 years AGO

a+b+c+15=a*b*c--- (1)
(a-2)+b+c=(a-2)*b*c+7---(2)
solving 2 equations we get ,b*c=12
taking b=3 & c=4 or b=4 & c=3 we get a=2
hence a*b*c=24,

Like? Yes (1) | No   

saurabh kumar   9 years AGO

digit will be 4,3,2
4+3+2=9
4*3*2=24
diff=15
2-2=0
0+3+4=7
4*3*0=0
diff=7

Like? Yes (1) | No   

Mayank Aditya   9 years AGO

Let no. are x,y and z.
x+y+z+15=xyz.....Eqn. 1
x-2+y+z=(x-2)yz+7....Eqn. 2
From Eqn. 1 and Eqn. 2
yz=2
so the possible conditions may be
1) y=2 and z=6(it is false because then x can't be a single digit no.)
2) y=3 and z=4(then x=2)
so, xyz=2*3*4=24

Like? Yes (1) | No   

Aakashdeep Madaan   9 years AGO

let three numbers be a,b,c then:
a+b+c=abc-15 and (a-2)+b+c=7+(a-2)bc
and after solving these two equations we get the value of bc=12
now, there could only be two possible values for b and c in this case which are 3,4 or 6,12 as the numbers are only single digit numbers
now, let b=3 , c=4 then put these values in equation1 and find the value of a as 2
but when we put b=6 c=2 then we get a as a decimal number , so we ignore that case
hence, a=2 b=3 c=4
and their product is 2*3*4=24

Like? Yes (1) | No   

Pesari Udayasree   9 years AGO

let x,y,z are the no.'s
then given x+y+z=xyz-15 and (x-2)+y+2=(x-2)yz+7
by trail and error x,y,z are 2,3,4
therefore 2*3*4=24

Like? Yes (1) | No   

vivek   9 years AGO

432 is the no.
so 4+3+2=9 and 4*3*2=24
so, product - sum =15
now, subtract 2 from the number, we get
430
sum of new no. 4+3+0=7
product of new no.= 4*3*0=0
so sum of the new no. is 7 more than product of the no.
hence, required product of given no. is 24

Like? Yes (2) | No (1)  

Moutam Vishnu Murthy   9 years AGO

Ans: 24 (2*3*4)
As (2*3*4)-15=2+3+4
(2-2)*3*4=0

Like? Yes (1) | No   

Deepanshu Chaturvedi   9 years AGO

the equation is valid for digits 2,3,4. therefore, product of digits is 24.

Like? Yes (1) | No   

Parthasarathi.S   9 years AGO

(2+3+4)+15=2*3*4=24
(0+3+4)-7=0*3*4=0
xyz=24

Like? Yes (1) | No   

Rahul Hirve   9 years AGO

X=2,
yz=12;
y=3,
z=4;

Like? Yes (1) | No   

vaibhavi lodhiya   9 years AGO

2*3*4=12,
& 2+3+4=9
24-9=15

Like? Yes (1) | No   

AARTI OJHA   9 years AGO

product of 2 nos is 12
so the product of 3 nos must be a multiple of 12
after solving we are getting the first no as 2
thus, the answer is 24

Like? Yes (1) | No   

Shrey Agrawal   9 years AGO

X+y+z=xyz-15....(1)
X-2+y+z=(x-2)yz+7....(2)
Substituting value of equation (1) in eqn(2)
Yz=12
Now..since x,y and z are single digit number.therefore possible combinations for y*z are 4*3 or 6*2...
Taking first possibility and substituting the value in eqn(1)..weget x=2.
Hence x*y*z=24

Like? Yes (1) | No   

sumit kumar   9 years AGO

let no is a,b and c.
a/c to question a+b+c=abc-15
a+b+c-abc=-15 ........(1)
(a-2)+b+c-(a-2)bc=7
a+b+c-2-abc+2bc=7
a+b+c-abc+2bc=9 .........(2)
from eq. (1) and (2)
bc=12=4*3
b=4,c=3
put the value in eq. (2)
a+4+3-12a+24=9
-11a=9-31=-22
a=2
prouct of no is abc=24

Like? Yes (1) | No   

vikash kumar das   9 years AGO

As, x+y+z=xyz-15 ......(i)
now in second case if we subtract 2 from the first no x then a/q
(x-2)+y+z=(x-2)yz+7 ...(ii)
On solving (i) nd (ii) we get yz=12
since x,y and z are single digit
therefore value of y and z should be (3,4),(4,3),(6,2),(2,6)
if we put y=3 and z=4 in eqn-(i) we get x=2 as single digit while (6,2) doesn't satisfy both eqn.
therefore the product is 2*3*4=24.

Like? Yes (1) | No   

Kalpana   9 years AGO

the numbers are 2,3,4
2 is the first number when subtracted it gives 4+3=7
so their product is 24

Like? Yes (1) | No   

Ankush dhadwal   9 years AGO

here yz=12after solving the two equation.then next step is to use hit and trail by assuming values to be 4&3 or 6&2.then substituting the both value onr by one and at last u will find that one with 4&3 on substituting we will get x=2

Like? Yes (1) | No   

SK INTAJ MOHAMMAD   9 years AGO

x+y+z=xyz-15.....(1)
and (x-2)+y+z=(x-2)yz+7;.......(2)
solving two equations we get yz=6;
here 6 is possible if y=2,z=3 or y=3, z=2;
puttting these above values in the equation (1) we get x=4;
so xyz=x*yz=4*6=24...... ans(4)..

Like? Yes (1) | No   

Aman   9 years AGO

From the question we know
X+Y+Z=X*Y*Z-15
and
(X-2)+Y+Z=(X-2)*Y*Z+7
now
on solving we get
Y*Z=12
so
either
4*3 is Y*Z
OR
6*2 is Y*Z
now 6*2 is not possible because it does't satisfies second equation so
4*3*2 is X*Y*Z
SO
24 IS THE X*Y*Z

Like? Yes (1) | No   

himanshu jaiswal   9 years AGO

let 3 no be x ,y,z
x+y+z=xyz-15
(x-2)+y+z=xyz+7
from here yz=12
if y=4 z=3
then we have x=2 which satisfy both the equation
so the product=2*4*3=24

Like? Yes (1) | No   

Ashish Kumar Singh   9 years AGO

let's assume three single digit number is x y z
given that (x+y+z)+15=xyz
and ((x-2)+y+z)-7=xyz
For Example if x=2,y=3,and z=4
then 9+15=24
and (0+3+4)-7=0
so the product of these single digit number is 24

Like? Yes (1) | No   

TONI PRIYA   9 years AGO

X*Y*Z=X+Y+Z+15
(X-2)*Y*Z+7=(X-2)+Y+Z
From these 2 eq. we get y*z=12
by hit and trial we conclude that y=3,z=4
after that when we put x=2 then it satisfy the 2 eq.

Like? Yes (1) | No   

Poornima Bharadvaj   9 years AGO

let the nos be 2,3,4 and their sum is 9 and product is 24 and it is 15 lessthan thier product and product of 3 num is 24

Like? Yes (1) | No   

ramniwas   9 years AGO

the numbers are 2,3,4
1 condition : sum=product-15
i.e 2+3+4=2*3*4-15 => 9=24-15 => 9=9
2 condition : subtract two from first i.e 2-2 =0
sum=product+7
0+3+4=0*3*4+7
7=7
TRUE
since ANS will be 2*3*4 =24

Like? Yes (1) | No   

Amit chaturvedi   9 years AGO

digits are 2,3,4 which proves both case, here we will consider 2 as a first number
2+3+4=9 and 2*3*4=24
sum is 15 less than their product
and
0+3+4=7 and 0*3*4=0
sum is 7 more than their product

Like? Yes (1) | No   

premchand meena   9 years AGO

x+y+z =xyz-15
(x-2)+y+z= (x-2)yz+7
solve both equ..
we get
yz = 12
so x+ y+z= 12x- 15
11x = y+z+15
so we know y+z+15 is multiple of 11
22, 33 ,44
possible y, z
12, 1
6,2
3,4
4,3
2,6
1, 12
takes which setisfy the equation
y+z = 7
so x = 0

Like? Yes (1) | No   

priyanka burnwal   9 years AGO

factor of 24 is 2,3,4
product of 2,3,4=24
and their sum is 9
and when 2 is substracted from 1st digit
sum =7
and product =0

Like? Yes (1) | No   

SAGNIK DATTA   9 years AGO

the given numbers are 2,3 and 4 and all the conditions are satisfied

Like? Yes (2) | No (2)  

UDIT   9 years AGO

24=2x3x4
24-(2+3+4)
24-9=15
here, sum is 15 less than product of three numbers
2 is decreased from 2 i.e.2-2=0
thus, 0x3x4=0
and the new sum is 0+3+4=7
so sum is greater than product by 7.
so product of three numbers is 24(option 4)

Like? Yes (1) | No (1)  

aman aditya   9 years AGO

three numbers are 4,3,2

Like? Yes (1) | No (1)