Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

Multiple choice maths introduction to three-dimensional geometry distance between two parallel planes three dimensional geometry - ii distance formula

If $P _1\,,\, P _2\,

,\, P _3$ denotes the perpendicular distances of the plane $2x -3y + 4z + 2 = 0$ from the parallel planes  $2x- 3y + 4z +6 = 0, 4x -6y + 8z + 3 = 0 $ and $2x- 3y + 4z- 6 = 0$ respectively, then

  1. $P _1\, +\, 8P _2\, -\, P _3\, =\, 0$
  2. $P _3\, =\, 16P _2$
  3. $8P _2\, =\, P _1$
  4. $P _1\, +\, 2P _2\, +\, 3P _3\, =\, \sqrt{29}$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

Since the planes are parallel planes

$\displaystyle { P } _{ 1 }=\frac { \left| 2-6 \right|  }{ \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 } }  } =\frac { 4 }{ \sqrt { 4+9+16 }  } =\frac { 4 }{ \sqrt { 29 }  } $
Equation of the plane $4x-6y+8z+3=0$ can be written as $2x-3y+4z+\displaystyle\frac{3}{2}=0$
So, $\displaystyle { P } _{ 2 }=\frac { \left| 2-\frac { 3 }{ 2 }  \right|  }{ \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 } }  } =\frac { 1 }{ 2\sqrt { 29 }  } $
and $\displaystyle { P } _{ 3 }=\frac { \left| 2+6 \right|  }{ \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 } }  } =\frac { 8 }{ \sqrt { 29 }  } $

So, from options:
(A) $\displaystyle { P } _{ 1 }+8{ P } _{ 2 }-{ P } _{ 3 }=\frac { 4 }{ \sqrt { 29 }  } -8\times \frac { 1 }{ 2\sqrt { 29 }  } -\frac { 8 }{ \sqrt { 29 }  } =\frac { 8 }{ \sqrt { 29 }  } -\frac { 8 }{ \sqrt { 29 }  } =0$
(B) $\displaystyle 16{ P } _{ 2 }=\frac { 16 }{ 2\sqrt { 29 }  } =\frac { 8 }{ \sqrt { 29 }  } ={ P } _{ 3 }$
(C) $\displaystyle 8{ P } _{ 2 }=\frac { 8 }{ 2\sqrt { 29 }  } =\frac { 4 }{ \sqrt { 29 }  } ={ P } _{ 1 }$
(D) $\displaystyle { P } _{ 1 }+2{ P } _{ 2 }+3{ P } _{ 3 }=\frac { 4 }{ \sqrt { 29 }  } +\frac { 2 }{ 2\sqrt { 29 }  } +3\times \frac { 8 }{ \sqrt { 29 }  } =\frac { 29 }{ \sqrt { 29 }  } =\sqrt { 29 } $

Multiple choice maths introduction to three-dimensional geometry distance between two parallel planes three dimensional geometry - ii distance formula

A line having direction ratios $3,4,5$ cuts $2$ planes $2x-3y+6z-12=0$ and $2x-3y+6z+2=0$ at point P & Q, then Find length of PQ 

  1. ${{35\sqrt 2 } \over {12}}$
  2. ${{35\sqrt 2 } \over {24}}$
  3. ${{35\sqrt 2 } \over 6}$
  4. ${{35\sqrt 2 } \over 8}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$3r,4r,5r$ be the point that cuts the two plane. 

$\therefore $ for plane $1$, $ 2x+6z-3y-12=0$
$\Rightarrow 6r+30r-12r-12=0$
$\Rightarrow r=\cfrac { 1 }{ 2 } $
for plane $2$, 
$2x-3y+6z+2=0$
$\Rightarrow 6r-12r+30r+2=0$
$\Rightarrow r=\cfrac { -1 }{ 12 } $
point of intersection at plane $1$ $\Rightarrow P(\cfrac { 3 }{ 2 } ,2,\cfrac { 5 }{ 2 } )$
point of intersection at plane $2$ $\Rightarrow Q(\cfrac { -3 }{ 12 } ,\cfrac { -1 }{ 3 } ,\cfrac { -5 }{ 12 } )$
$\therefore $ distance $(PQ)$ $==\sqrt { (\cfrac { 3 }{ 2 } +\cfrac { 3 }{ 12 } )^{ 2 }+(2+\cfrac { 1 }{ 3 } )^{ 2 }+(\cfrac { 5 }{ 2 } +\cfrac { 5 }{ 12 } )^{ 2 } } =\cfrac { 35\sqrt { 2 }  }{ 12 } $
Ans $A$

Multiple choice maths introduction to three-dimensional geometry distance between two parallel planes three dimensional geometry - ii distance formula
 Find the distance between the parallel planes $\vec { r } .(2\hat { i } -3\hat { j } +6\hat { k } )=5$ and$\quad \vec { r } .(6\hat { i } -9\hat { j } +18\hat { k } )\quad +\quad 20\quad =\quad 0. $
  1. $\dfrac {28}{21}$
  2. $\dfrac {5}{3}$
  3. $\dfrac {5}{21}$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The planes are r.(2i - 3j + 6k) = 5 and r.(6i - 9j + 18k) = -20. Dividing the second equation by 3 gives r.(2i - 3j + 6k) = -20/3. The distance is |5 - (-20/3)| / sqrt(2^2 + (-3)^2 + 6^2) = |15/3 + 20/3| / sqrt(4 + 9 + 36) = (35/3) / 7 = 5/3.

Multiple choice maths introduction to three-dimensional geometry distance between two parallel planes three dimensional geometry - ii distance formula

The distance between the parallel planes $2x+y+2z-8=0 $ and $4x+2y+4z+5=0$ is

  1. $\displaystyle \frac{7}{2}$
  2. $\displaystyle \frac{5}{2}$
  3. $\displaystyle \frac{3}{2}$
  4. $\displaystyle \frac{9}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Distance between parallel planes $ax+by+cz+d _1$ and $ax+by+cz+d _2$ is given by $\dfrac{|d _1-d _2|}{\sqrt{a^2+b^2+c^2}}$.

The equation of first plane we can taken as $4x+2y+4z-16=0$.
Hence the distance is $\dfrac{|-16-5|}{\sqrt{4^2+2^2+4^2}}=\dfrac{7}{2}$

Multiple choice banks introduction, recurring deposit accounts and calculation of interest on a fixed deposit account banking and taxation banking maths

Which type of bank account is operated by businessman?

  1. fixed deposit

  2. save deposit

  3. recurring deposit

  4. current deposit
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Current deposit type of bank account is operated by businessman.

Current accounts are basically meant for businessman and are never used for the purpose of investments or savings.

Multiple choice banks introduction, recurring deposit accounts and calculation of interest on a fixed deposit account banking and taxation banking maths

In which account interest is calculated half yearly on the minimum balance between $11^{th}$ and the last day of the month?

  1. fixed deposit

  2. recurring deposit

  3. saving deposit

  4. current deposit
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In saving deposit account interest is calculated half yearly on the minimum balance between $11^{th}$ and the last day of the month.

Multiple choice banks introduction, recurring deposit accounts and calculation of interest on a fixed deposit account banking and taxation banking maths

You have Rs. 1500 in your savings account at the beginning of the month.the record below shows all of your transactions during the month.How much money is in your account after these transactions?

Date Withdrawal Deposit
4/9/14 Rs 1200 Rs 2000
22/9/14 Rs. 2100 Rs.2500
  1. $Rs.\ 2000$
  2. $Rs.\ 3100$
  3. $Rs.\ 2500$
  4. $Rs.\ 2700$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have, Amount of money in savings account at the beginning of the month =Rs.1500
Total amount of money withdrawal Rs 1200+2100=Rs.3300
Total amount of money deposited =Rs(2000+2500)=Rs.4500
So, total amount of money after transactions=Rs (1500-3300+4500)=Rs.2700 

Multiple choice banks introduction, recurring deposit accounts and calculation of interest on a fixed deposit account banking and taxation banking maths

The rates of simple interest in two banks A and B are in the ratio 5 : 4. A person wants to deposit his total savings in two tanks in such a way that he received equal half yearly interest form both. He should deposit the savings in banks A and B in the ratio of

  1. 2 : 5

  2. 4 : 5

  3. 5 : 2

  4. 5 : 4
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that $ SI = \dfrac {PNR}{100} $
We clearly see that when Simple Interest and the time $ N $ is constant or same, then Principal $ P $ varies inversely with $ R $
So, if $ R $ is in the ratio $ 5:4 $ then $ P $ deposited should be in the inverse $ 4: 5 $

Multiple choice banks introduction, recurring deposit accounts and calculation of interest on a fixed deposit account banking and taxation banking maths

John had a Savings Bank Account in a bank. In the months of April, $'97$ and May, $'97$ he had the following entries in his passbook.
Find the amounts on which John will get interest for the months of April, $2011$ and May, $2011$. 

Date Particular Withdrawals (In Rs.) Deposits (In Rs.) Balance (In Rs.)
April $1$ By Balance $4,600.00$
April $7$ By Cash $1,200.00$ $5,800.00$
April $24$ To Cheque $800.00$ $5,000.00$
May $16$ By Cheque $2,000.00$ $7,000.00$
May $29$ To Cash $1,500.00$ $5,500.00$



  1. Rs. $5000$, Rs. $5500$
  2. Rs. $5000$, Rs. $2500$
  3. Rs. $2500$, Rs. $5000$
  4. Rs. $2500$, Rs.$ 2500$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

According to the entries in the passbook:

The minimum balance after $10^{th}$ April $2011$ and up to last of April $2011$ is Rs.$5000$.
$\therefore $ the amount on which John will earn interest for the month of April $2011=$ Rs.$5000$.
Similarly the minimum balance after $10^{th}$ May $2011$ and up to last of May $2011$ is Rs.$5500$
$\therefore $ the amount on which John will earn interest for the month of May $2011=$ Rs.$5500$.