Aptitude - Problems on Numbers - Discussion

I have assumed that X ,Y, Z are the three numbers and their positions are 100X+10Y+Z
now;

According to the que;
X+Y+Z=10---> (1)
and
X+Z=Y -------> (2).

From eq 1 and eq 2.

We get;
Y=5 ----> (3)
from eq2 and eq 3.
we get;
X+Y=5----->(4)
now from here, it is sure that number would have 5 digit in its middle.

a/c to que
(100X+ 10Y+Z) +99= 100Z+10Y+X
99= 99Z-99X
1= Z-X------------> (5)

from eq 4 and eq 5;
we get X=2.
Z=3.

So the number is 253 the answer.

Let the three digit numbers be;

100x + 10y + z ---> eq(1)
x + y+ z = 10 (given) ---> eq(2)
y = x + z (given) ---> place this in above eq(2).

i.e x + (x+z) +z = 10.
=> x + z = 10/2 = 5 ---> eq(3)

The number will be increased by 99 if its digits are reversed (Given)
Therefore, (100x +10y +z) + 99 = 100z + 10y + x.
=> 99z = 99x + 99.
=> z - x = 1 ---> eq(4).

from eq(3) and eq(4) we get

x = 2 and z = 3 .and we already know y = 5.

Therefore the number is 100x + 10y + z = 100*2 + 10*5 + 3 = 253.

It's quite simple first see it says the middle digit is equal to the sum of the other digits so in the options (option elimination method).

145 isn't the answer because 1+5 isn't equal to 4
370 isn't the answer because 3+0 isn't equal to 7.
253 and 352 one among these will be the answer. Because 2+3 or 3+2 will be equal to 5
Next, it's given the number is increased by 99 if the digits are reversed so it's obviously 253 because of 253+99=352.

Hi.

Hence the number => X, Y, Z.

Given X+Z = Y AND.

Also Given X+Y+Z = 10.

Then => 2Y = 10.

=> Y = 10/2 = 5.

Here is asked the number increase by 99, when its Reverse please stress the word Reverse.

If let say 1_?_4 then Reverse is 4_?_1.

Let us see difference is 297. But here asked increase 99.

The difference is (352 AND 253).

That's why the number is 253?

The sum of 3 digits = a + b + c.
a + b + c = 10( is given).

Middle num is = sum of the other two.

b = (a + c).
then a + b + c = 10,
(a + c) + b = 10,
b + b = 10,
2b = 10,
b = 5.

So, if a = 3, b = 5, c = 2 then abc = 352.
or a = 2, b = 5, c = 3 then abc= 253.

1) 352 - 253 = 99.
2) 253 - 352 =-99.

In the problem, they gave 99 only so the first step is right.

Let the number be 100x +10y +z.
Now x+y+z = 10 (given),
also y=x+z (given),

--> y+(x+z) = y+y = 2y = 10.
--> y=5.

Now 5= 1+4 or 2+3 .
also z>x (using the third given statement)
So the number can be 154 or 253.
reverse the numbers and subtract,
451-154 = 297,
352- 253 = 99.
Therefore, here the answer can only be 253 i.e option B.

Let me explain guys,

x+y+z=10 (given) --> 1
y(MIDDLE DIGIT)= x+z (given) -->2
substitute equation 2 in 1.
y+y =10,
2y=10,
y=5.
(100x+50+z) - (100z+50+x) = 99 (difference between the number and the reversed one)
100x+50+z-100z-50-x = 99,
99x-99z= 99,
99(x-z) =99,
x-z=1
w.k.t x+z=5
By simultaneous equation x=3;
3+z=5
z=2;
352 - 253 =99.

Lets Suppose three digits are a,b,c
Than a+b+c=10
But b=a+c (given)
a+c+b=10
(a+c)+b=10
b+b=10
b=10/2=5
so the middle digit is 5
so a=3or2 and c=3or2 (3+2=5)
thats why the number is either 253 or 352
But the actual no is 99 less than the reverse no.
352-253=99

: . so the no is 253

Here, conclusion of 352 and 253 is confusing for some people.

Let me explain:

Read last lines REVERSING DIGITS INCREASES BY 99 (gives the answer of the question).

So the present number small according to this statement.

And 253 is smaller than 352 so 253 is the correct answer.

Another way to solve this is,

Let the digits be x,y,z.

Given-
x+y+z=10.

And
y = x + z.

Then=> y+y=2y=10=>y=5
Number is either 253 or 352.
The number is increased by 99,when reversed.
Therefore the answer is 253.