Arithmetic aptitude :: Problems on Numbers :: General Questions

• Problems on Numbers - General Questions
• Problems on Numbers - Important Formulas and Facts

Basic Formulae:
1). $(a+b)(a-b)=(a^{2}-b^{2})$.
2). $(a+b)^{2}=(a^{2}+b^{2}+2ab)$.
3). $(a-b)^{2}=(a^2+b^{2}-2ab)$.
4). $(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$.
5). $(a^{3}+b^{3})=(a+b)(a^{2}-ab+b^{2})$.
6). $(a^{3}-b^{3})= (a-b)(a^{2}+ab+b^{2})$.
7). $(a^{3}+b^{3}+c^{3}-3abc)=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ac)$.
8). When $a+b+c = 0$, then $a^{3}+b^{3}+c^{3}=3abc$.

Three numbers which are co-prime to one another are such that the product of the first two is 551 and that of the last two is 1073 .The sum of three numbers is 

Answer:

Explanation:

Let  the numbers be x,y,z which are prime to one another

Xy = 551 yz = 1073

Y = HCF of 551,1073 = 29

$x=\frac{551}{29}=19$

$z=\frac{1073}{29}=37$

Sum of the numbers  = 29+19+37 =85

The difference between the squares of two numbers is 256000 and the sum of the numbers is 1000. The numbers are :

Answer:

Explanation:

Let the numbers be $x$ and $y$.

Then, $x^{2}-y^{2}=256000$ and $x+y=1000$.

On dividing, we get : $x-y= 256$.

Solving $x+y=1000$ and $x-y= 256$, we get $x = 628$ and $y=372$

The difference between two positive integers is 3. If the sum of their squares is 369, then the sum of the numbers is :

Answer:

Explanation:

Let the numbers be $x$ and $(x+3)$.
Then, $x^{2}+(x+3)^{2}= 369 \Leftrightarrow x^{2}+x^{2}+9+6x = 369$
$\Leftrightarrow 2x^{2}+6x-360=0$
$\Leftrightarrow x^{2}+3x-180=0$
$\Leftrightarrow (x+15)(x-12)=0$
$\Leftrightarrow x = 12$
So, the numbers are 12 and 15
$\therefore$ Required sum = $(12+15)= 27$.

The sum of the squares of two numbers is 3341 and the difference of their squares is 891. The numbers are :

Answer:

Explanation:

Let the numbers be $x$ and $y$.

Then, $x^{2}+y^{2}=3341$-----(1) $x^{2}-y^{2}= 891$-----(2)

Adding (1) and (2), we get $2x^{2}=4232$ or $x^{2}=2116$ or $x=46$.

Subtracting (2) from (1), we get : $2y^{2}=2450$ or $y^{2}= 1225$ or $y=35$.

So, the numbers are 35 and 46.

The product of two numbers is 120 and the sum of their squares is 289. The sum of the numbers is :

Answer:

Explanation:

Let the numbers be $x$ and $y$. Then, $xy=120$ and $x^{2}+y^{2}=289$.
$\therefore$ $(x+y)^{2}=x^{2}+y^{2}+2xy=289+240=529$.
$\therefore x+y = \sqrt{529}=23$

• Problems on Numbers - General Questions