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 $a^{3}+b^{3}+c^{3}=3abc$.

The difference between 120 percent of number and its three fourth is 50% of that number .What is that number?

Answer:

Explanation:

Let the number be x

120 percent of $x - \frac{3}{4}x=$50 percent of $x$

= >      $\frac{6x}{5}-\frac{3x}{4}=\frac{x}{2}$

= >$\frac{9x}{20}=\frac{x}{2}$

The equation cannot be solved

So Data inadequate 

When a teacher distributed craft papers to students in her class each one got 7 papers and  14 papers remained with the teacher .If two more students would have been in the class ,no paper would have remained undistributed .How many students ,the teacher had in her class?

Answer:

Explanation:

Either the total number of craft papers should be given or in the second case the number of papers got by each student should be given.

Therefore  Data inadequate.

A bridge crosses a river which is 228 meters wide .If one bank of the river holds 1/5^th of the bridge and the other holds 1/6^th of it. What is the total length of the bridge?

Answer:

Explanation:

Let the length of the bridge  = x m

One bank holds = 1/5^th of bridge

Other bank holds  = 1/6^th of bridge

$x-(\frac{x}{5}+\frac{x}{6})=228$

$x-\frac{11x}{30} = 228$

$\frac{19x}{30} = 228$

$x=\frac{228\times30}{19} $

X= 360 metres

Three fifth of the square of certain number is 126.15.What is the number?

Answer:

Explanation:

Let the umber be x

Given $\frac{3}{5}x^{2}=126.15$

$x^{2}=\frac{126.15\times5}{3}$

$x=\sqrt{210.25}$

=14.5

Rs.6500 were divided equally among a certain number of persons. Had there been 15 more persons, each would have got Rs.30 less .Find the original number of person?

Answer:

Explanation:

Let the original number of person be x

Share of each person = 6500/x

If the noumber of person be increased by 15 ,then the new share of each person = $\frac{6500}{x+15}$

Given  $\frac{6500}{x}-\frac{6500}{x+15}=30$

Solve the above equation x =-65,50

X cannot be negative ,x = 50

• Problems on Numbers - General Questions