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Questions Related to maths

The difference between a discount of $60\%$ on Rs. $500$ and two successive discounts of $36\%$ and $4\%$ on the same amount is __________.

maths money math different discounts successive discounts problems on discounts

  1. $0$

  2. Rs. $2$

  3. Rs. $1.93$

  4. Rs. $7.20$

Show answer
Correct Option: D
Explanation:

Discount of $60\%$ on $500 = 0.6\times 500=300$

Two successive discounts of $36\% $ and $4\%$.
$= (1-0.36)\times 500(1-0.04) = 307.2$
Difference in both the discounts is $307.2 - 300= 7.2$.

On a Rs.$10,000$ order a merchant has a choice among three successive discounts of $20\%, 20\% and 10\%$ and three successive discounts of $40\%, 5\% and 5\%$. By choosing the best offer, he can save:

maths money math different discounts successive discounts problems on discounts

  1. nothing at all

  2. Rs.$400$

  3. Rs.$330$

  4. Rs.$345$

  5. Rs.$360$

Show answer
Correct Option: D
Explanation:

Since a single discount D, equal to three successive discounts $D _1, D _2 and D _3$ is $D = D _1 + D _2 + D _2 - D _1D _2 - D _2D _3 - D _3D _1 + D _1D _2D _3$, then the choices are
$0.20 + 0.20 + 0.10 - 0.04 - 0.02 - 0.02 + 0.004 = 0.424$ and $0.40 + 0.05 + 0.05 - 0.02 - 0.02 - 0.0025 + 0.001 = 0.4585.$
The saving is $0.0345.10,000 = 345 rupees$

Applied to a bill for Rs. $10,000$ the difference between a discount of $40\%$ and two successive discounts of $36\%$ and $4\%$, expressed in rupees, is

maths money math different discounts successive discounts problems on discounts

  1. $0$

  2. $1440$

  3. $2560$

  4. $4000$

  5. $416$

Show answer
Correct Option: B
Explanation:

$40$% of $Rs. 10,000$ is $Rs. 4,000; 36$% of $Rs. 10,000$ is $Rs. 3,600; 4$% of $(Rs. 10,000 - Rs. 3,600)$ is $Rs. 256. Rs. 3,600 + Rs. 256 = Rs. 3,856$;
$\therefore$ the difference is $Rs. 4,000 - Rs. 3,856 = Rs. 144$; or two successive discounts of $36$% and $4$% are equivalent to one discount of $38.56$%.

∴ Percentage difference = $40$ – $38.56$ = $1.44%$
Difference between discount = $1.44%$ of $100000$
=$\dfrac{1.44\times10000}{100}=1440Rs$

Four dealers advertise the same list price for a TV set. Which one of the following discount series is more advantageous to the customer?

maths money math different discounts successive discounts problems on discounts

  1. 25% and 8%

  2. 22% and 8%

  3. 25% and 9%

  4. 25% and 10%

Show answer
Correct Option: D
Explanation:
Let the price of TV be $x$
A. $25\%$ and $8\%$ successive discounts
$\Rightarrow$ Net price $= \dfrac{(100-25)}{100} \times \dfrac{(100-8)}{100} x$
                      $= 0.69x$

B. $22\%$ and $8\%$ successive discounts
$\Rightarrow$ Net price $= \dfrac{(100-22)}{100} \times \dfrac{(100-8)}{100} x$
                      $= 0.7176x$

C. $25\%$ and $9\%$ successive discounts
$\Rightarrow$ Net price $= \dfrac{75}{100} \times \dfrac{91}{100}x$
                     $= 0.6825x$

D. $25\%$ and $10\%$ successive discounts
$\Rightarrow$ Net price $= \dfrac{75}{100} \times \dfrac{90}{100}x$
                     $= 0.675x$
$\therefore$ D is the most advantageous discount series for the customer.

Which discount series is profitable to the buyer $25\%, 12\%, 3\%$ or $18\%, 17\%, 5\%$?

maths money math different discounts successive discounts problems on discounts

  1. First

  2. Second

  3. Both

  4. Neither first nor second

Show answer
Correct Option: A
Explanation:

Let the sum be $x$.

Case 1:
Sum after 1st discount$=x-.25\times x$
                                       $=0.75\times x$
Remaining sum after 2nd discount $=0.75x-0.12\times 0.75x$
                                                          $=0.66x$
Remaining sum after 3rd discount$=0.66x-0.03\times 0.66x$
                                                          $=0.6402x$
$\therefore $Overall discount$=x-0.6402x$                 
                              $=0.3598x$                        
$\therefore$ Overall discount in %$=0.3598\times 100$
                                       $=35.98$%

Case 2:
Sum after 1st discount$=x-.18\times x$
                                       $=0.82\times x$
Remaining sum after 2nd discount $=0.82x-0.17\times 0.82x$
                                                          $=0.6806x$
Remaining sum after 3rd discount$=0.6806x-0.05\times 0.6806x$
                                                          $=0.64657x$
$\therefore $Overall discount$=x-0.64657x$                 
                              $=0.35343x$                        
$\therefore$ Overall discount in %$=0.35343\times 100$
                                       $=35.343$%
Thus, the first case is more profitable.

For any non-singular matrix A, $ \displaystyle A^{-1} $ =

maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. $|A|adj A$

  2. $\dfrac{1}{|A| adj A}$

  3. $\dfrac{adj A}{|A|}$

  4. None of the above

Show answer
Correct Option: C
Explanation:

Singular matrix is square matrix whose determinant is equal to Zero.

So, non-singular matrix $A$ is a matrix whose determinant is non-zero.
$\implies$ inverse of $A$ i.e. $A^{-1}$ exist.
$\therefore A^{-1}=\dfrac{adj A}{|A|}$ .... $[|A|\neq 0]$

Say true or false:
Let A, B be two matrices, such that $AB = A$ and $BA = B$, then $A^2 = A$ and $B^2 = B$.

maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

$AB=A\Rightarrow B\left( AB \right) =B\left( A \right) \Rightarrow BAB=BA\Rightarrow BB=B\Rightarrow { B }^{ 2 }=B\ BA=B\Rightarrow A\left( BA \right) =A\left( B \right) \Rightarrow ABA=AB\Rightarrow AA=A\Rightarrow { A }^{ 2 }=A$

If $AB=A$ and $BA=B$, where $A$ and $B$ are square matrices, then 

maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. ${ B }^{ 2 }=B$ and ${ A }^{ 2 }=A$

  2. ${ B }^{ 2 }=A$ and ${ A }^{ 2 }=B$

  3. $AB=BA$

  4. none of these

Show answer
Correct Option: A
Explanation:

We have 

$AB=A\Rightarrow A\left( BA \right) =A\quad$, subsitute $BA=B$ 
${$ $A(BA) =(AB)A$ $}$
$\Rightarrow \left( AB \right) A=A$
Subsitute $AB = A$
$\Rightarrow AA=A\quad \quad \left[ \therefore AB=A \right] \ \Rightarrow { A }^{ 2 }=A$
Again $BA=B$

$\Rightarrow B\left( AB \right) =B\quad \quad \left[ \because AB=A \right] $
$\Rightarrow \left( BA \right) B=B$
$\Rightarrow BB=B$
$\Rightarrow { B }^{ 2 }=B$

If for suitable matrices $A, B$; $AB=A$ and $BA=B$; then ${A}^{2}$ equals-

maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. $I$

  2. $A$

  3. $B$

  4. $0$

Show answer
Correct Option: B,C
Explanation:

Taking $AB=A$


Multiplying by a matrix $A$

$ABA=A^2$

$\Rightarrow A(BA)=A^2$      (Associative property)

$AB=A^2$      ($\because BA=B$)

$\Rightarrow A=A^2$    ($\because AB=A$)

If A is invertible, then which of the following is not true?

maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

  1. $\displaystyle { A }^{ -1 }={ \left| A \right| }^{ -1 }$

  2. $\displaystyle { \left( { A }^{ 2 } \right) }^{ -1 }={ \left( { A }^{ -1 } \right) }^{ 2 }$

  3. $\displaystyle { \left( { A }^{ ' } \right) }^{ -1 }={ \left( { A }^{ -1 } \right) }^{ ' }$

  4. None of these

Show answer
Correct Option: A
Explanation:
A is invertible
$\Rightarrow { A }^{ -1 }$ exists
Option A: ${ A }^{ -1 }={ \left| A \right|  }^{ -1 }$
But we cannot write that a matrix and its determinant are both equal
$\therefore $ option A is not true
Option B: ${ \left( { A }^{ 2 } \right)  }^{ -1 }={ \left( { A }^{ -1 } \right)  }^{ 2 }$
This option is true from the property
${ \left( { A }^{ n } \right)  }^{ -1 }={ \left( { A }^{ -1 } \right)  }^{ 2 }$
Option C: ${ \left( { A }^{ -1 } \right)  }^{ 1 }={ \left( { A }^{ 1 } \right)  }^{ -1 }$
Consider $\left( { A }^{ T } \right) { \left( { A }^{ -1 } \right)  }^{ T }={ \left( { A }^{ -1 }A \right)  }^{ T }={ I }^{ T }=I$
Similarly
${ \left( { A }^{ -1 } \right)  }^{ T }{ \left( { A }^{ T } \right)  }={ \left( A{ A }^{ -1 } \right)  }^{ 1 }{ I }^{ 1 }=1$
From $1$ and $2$
${ A }^{ T }{ \left( { A }^{ -1 } \right)  }^{ T }={ \left( { A }^{ -1 } \right)  }^{ T }{ \left( { A }^{ T } \right)  }=I$
$\Rightarrow { A }^{ 1 }$ is multiplicative inverse of ${ \left( { A }^{ -1 } \right)  }^{ 1 }$
$\Rightarrow { \left( { A }^{ T } \right)  }^{ -1 }={ \left( { A }^{ -1 } \right)  }^{ T }$