Tag: maths

Questions Related to maths

If $|a|$ denotes the absolute value of an integer, then which of the following are correct?
1.$|ab| = |a| |b|$
2. $|a+b| \le |a|+|b|$
3. $|a-b| \ge| |a| -|b||$
Select the correct answer using the code given below.

maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

  1. 1 and 2 only

  2. 2 and 3 only

  3. 1 and 3 only

  4. 1, 2 and 3

Show answer
Correct Option: D
Explanation:

Given $\left| a \right| $ is the absolute value of an integer,

From the definition,
$\left| a \right| =a$ if $ a\ge 0$,
$=-a\quad $ is $a\le0$
$\therefore$ $\left| ab \right| =\left| a \right| \left| b \right| $ where $a,b$ are real numbers.
We know that from the triangle inequality sum of any two sides is always greater than the third side,
i.e.,$\left| a+b \right| \le \left| a \right| +\left| b \right| $,
We can also prove by considering 
Absolute part of the difference between any two sides is always less than the third side,
$\Longrightarrow \left| a-b \right| \ge \left| \left| a \right| -\left| b \right|  \right| $

The difference between a two digit number and the number obtained by interchanged the two digits of the number is $9$. What is the difference between the two digits of number.

maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

  1. $3$

  2. $2$

  3. $1$

  4. Cannot be determined

  5. None of these

Show answer
Correct Option: C
Explanation:
Let the unit's digit be $y$ and ten's digit be $x$.

Then, the number $= 10x + y$. When we interchange the digits, the number will be $10y + x$.

Now, it is given that the difference between a two digit number and the number obtained by interchanged the two digits of the number is $9$, therefore, we have:

$(10x+y)−(10y+x)=9\\ \Rightarrow 9x-9y=9\\ \Rightarrow 9(x-y)=9\\ \Rightarrow x-y=\frac { 9 }{ 9 } \\ \Rightarrow x-y=1$

Hence, the difference between the two digits of number is $1$.

What will come in place of the question mark $(?)$ in the following question?
$34.667-15.597-8.491-0.548=?$

maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

  1. $14.403$

  2. $10.031$

  3. $18.301$

  4. $21.043$

  5. None of these

Show answer
Correct Option: B
Explanation:

Let the missing place in the given question be $x$, then we have:


$34.667-15.597-8.491-0.548=x\ \Rightarrow \dfrac { 34667 }{ 1000 } -\dfrac { 15597 }{ 1000 } -\dfrac { 8491 }{ 1000 } -\dfrac { 548 }{ 1000 } =x\quad \quad \quad \quad \quad \left{ \because \quad \dfrac { 1 }{ 10 } =0.1,\dfrac { 1 }{ 100 } =0.01,.... \right} \ \Rightarrow \dfrac { 34667-15597-8491-548 }{ 1000 } =x\ \Rightarrow \dfrac { 34667-(15597+8491+548) }{ 1000 } =x$

$\Rightarrow \dfrac { 34667-24636 }{ 1000 } =x$

$\ \Rightarrow \dfrac { 10031 }{ 1000 } =x\ \Rightarrow x=10.031$

Hence, $34.667-15.597-8.491-0.548=10.031$

Find three consecutive even integers such that the sum of first two integers is same as the sum of third integer and $6$.

maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

  1. $4,6,8$

  2. $6,8,10$

  3. $8,10,12$

  4. $10,12,14$

Show answer
Correct Option: C
Explanation:

Let us say the first even integer be $x$. The second consecutive even integer would be $x+2$ (zit would not be $x+1$ because that would result in an odd integer. The sum of two even integers is even). The third consecutive even integer would be $(x+2)+2$ or $x+4$.


Now, it is given that the sum of first two integers is same as the sum of the third integer and $6$ which means:

$x+(x+2)=(x+4)+6\ \Rightarrow 2x+2=x+10\ \Rightarrow 2x-x=10-2\ \Rightarrow x=8$

Therefore, the first even integer is $8$ then the second integer is $x+2=8+2=10$ and the third integer is $x+4=8+4=12$

Hence, the three consecutive even integers are $8,10,12$.

Subtract  $ - \frac{2}{3}{y^3}-\frac{2}{7}{y^2} - 5$ from $\frac{1}{3}{y^3} + \frac{5}{7}{y^2} - 2$, then the resultant value is . 

maths brackets order operations and algebra using brackets in algebraic expressions order of operations

  1. ${y^3} + {y^2} - 7$

  2. ${y^3} + {y^2} + 3$

  3. ${y^3} + {y^2} - 3$

  4. ${y^3} + {y^2} + 7$

Show answer
Correct Option: A

If r and s are zeroes of the polynomial $t^2-4t+3$, then $\dfrac{1}{r}+\dfrac{1}{s}-2rs+\dfrac{14}{3}$ is equal to

maths brackets order operations and algebra using brackets in algebraic expressions order of operations

  1. 0

  2. 1

  3. 2

  4. -1

Show answer
Correct Option: A
Explanation:
Quadratic Equation
If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^2+bx+c=0$
then $\alpha + \beta=\dfrac{-b}{a}$ and $\alpha\beta=\dfrac{c}{a}$

$f(t)=t^2-4t+3$
$r+s=4$
$rs=3$
Now,
$\dfrac{1}{r}+\dfrac{1}{s}-2rs+\dfrac{14}{3}=\dfrac{r+s}{rs}-2rs+\dfrac{14}{3}$

$=\dfrac{4}{3}-6+\dfrac{14}{3}=\dfrac{4-18+14}{3}=0$

State whether True or False.

Simplify: $(3y+4z)(3y-4z)+(2y+7z)(y+z) $.
The answer is $11y^2+9yz-9z^2$.

maths brackets order operations and algebra using brackets in algebraic expressions order of operations

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

$Multiplying\quad (3y+4z)(3y-4z),\quad (2y+7z)(y+z),\ =-16z^{ 2 }+12yz-12yz+9y^{ 2 }\quad +\quad 7z^{ 2 }+7yz+2yz+2y^{ 2 }\ =-9z^{ 2 }+9yz+11y^{ 2 }$

Simplify $(a + b) (c -d) + (a-  b) (c + d) + 2 (ac + bd)$

maths brackets order operations and algebra using brackets in algebraic expressions order of operations

  1. $4ac$

  2. $4ac - 4bd$

  3. $4bd$

  4. $4ac+4bd$

Show answer
Correct Option: A
Explanation:

$(a+b)(c-d)+(a-b)(c+d)+2(ac+bd)$
$=a(c-d)+b(c-d)+a(c+d)-b(c+d)+2(ac+bd)$
$=ac-ad+bc-bd+ac+ad-bc-db+2ac+2bd$
$=4ac$

Simplify: $(x + y)(x^2 -xy + y^2)$

maths brackets order operations and algebra using brackets in algebraic expressions order of operations

  1. $x^3 - y^3$

  2. $x^3 + y^3$

  3. $x^3 + y^3 + 3ab$

  4. $x^3 + y^3-3ab$

Show answer
Correct Option: B
Explanation:

$(x + y)(x^2 - xy + y^2)$
$=x(x^2 - xy + y^2)+y(x^2 - xy + y^2)$
$=x^3-x^2y+xy^2+x^2y^-xy^2+y^3$
$=x^3+y^3$

The simplified form of the expression given below is :$\dfrac{\dfrac{y^4-x^4}{x(x+y)}-\dfrac{y^3}{x}}{y^2-xy+x^2}$

maths brackets order operations and algebra using brackets in algebraic expressions order of operations

  1. $1$

  2. $0$

  3. $-1$

  4. $2$

Show answer
Correct Option: C
Explanation:

The given expression $\dfrac { \dfrac { { y }^{ 4 }-{ x }^{ 4 } }{ x(x+y) } -\dfrac { { y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } }$ can be simplified as follows:

 
$\dfrac { \dfrac { { y }^{ 4 }-{ x }^{ 4 } }{ x(x+y) } -\dfrac { { y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } } \ =\dfrac { \dfrac { ({ y }^{ 2 })^{ 2 }-({ x }^{ 2 })^{ 2 } }{ x(x+y) } -\dfrac { { y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } } \ =\dfrac { \dfrac { ({ y }^{ 2 }-{ x }^{ 2 })({ y }^{ 2 }+{ x }^{ 2 }) }{ x(x+y) } -\dfrac { { y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } } \quad \quad \quad \quad \quad \quad \quad \left( \because \quad a^{ 2 }-b^{ 2 }=(a+b)(a-b) \right)$
$=\dfrac { \dfrac { ({ y }+x)(y-x)({ y }^{ 2 }+{ x }^{ 2 }) }{ x(x+y) } -\dfrac { { y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } } \ =\dfrac { \dfrac { (y-x)({ y }^{ 2 }+{ x }^{ 2 }) }{ x } -\dfrac { { y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } } \ =\dfrac { \dfrac { { y }^{ 3 }-xy^{ 2 }+yx^{ 2 }-{ x }^{ 3 }-{ y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } }$
$=\dfrac { \dfrac { -x(y^{ 2 }-xy+{ x }^{ 2 }) }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } } \ =-\dfrac { y^{ 2 }-xy+{ x }^{ 2 } }{ { y }^{ 2 }-xy+{ x }^{ 2 } } \ =-1$

Hence, $\dfrac { \dfrac { { y }^{ 4 }-{ x }^{ 4 } }{ x(x+y) } -\dfrac { { y }^{ 3 } }{ x }  }{ { y }^{ 2 }-xy+{ x }^{ 2 } }=-1$