Aptitude - Problems on Numbers - Discussion

Please tell me why we do not take 10x+y here. Please sing 10x+y.

If option A is correct according to naresh why not C, as it is same to a when we add up ?

I don't get this step.

Sum of digits = x + (x + 2) = 2x + 2.

Why we added x instead of 10x to (x+2).

@Venkat.

We want the product of digit (ie. 24) and its sum (ie 2+4) = 144.
And for option C number is 42 doesn't satisfy the condition.

@Shikha.

Hey if we add 10x to (x+2) we will get that number. And when we add x to (x+2) we get the sum so. Hope I clarify a bit.

Let the ten's digit be x.

Then, unit's digit = x + 2.

for Example: Take Number 56.

Here unit digit is 6, ten's digit is 5, Then Number = 50(since its 10th digit)+6 = 56.

Hence in the above case: Number = 10x+ (x+2) = 11x+2.

The remaining same explained by indiabix.

How did 11x2+13x-70 produces (x-2)(11x+35) factors?

Multiply a term(11) and c term (-70) = -770.

Find two numbers that when multiplied equal the number (-770) and add up to be the b term (13).

The numbers be -22 and 35.

Now write this as,

11x2-22x+35x-70 = 0.

11x (x-2) +35 (x-2) = 0.

The factors are x-2 and 11x-35.

If I'm wrong, please explain me.

First mutiple 11* -70 which gives product = -770 and sum = 13.

Next factorise 770 number ,u will get 2*5*7*11...in this series try to make combination such that sum should be 13 and product should be -770.

The combination which we get is by trial and error, is 5*7=35 and 2*11=22, hence we choose 35 and -22 to satisfy p=-770 and sum=13,

Next substitute in equation,
11x^2+35x-22x+4=0.
x(11x+35)-2(11x+35)=0.

Hence,
(x-2)(11x+35) = 0.
Hence we get x = 2 and y = x+2 = 4.

Find Two digit number which is 4 times of digits sum and 2 times of digits multiplication.