Tag: maths

Questions Related to maths

A man deposited a certain amount in a bank that would repay double the amount after a year. At the
beginning of the second year, the man took out Rs 8000 and deposited the rest in the same bank. Again at the beginning of the third year, he took out Rs 8000 and deposited the rest in the same bank. At the beginning  of the fourth year, he took out Rs 8000 as before but was not left with any balance in the bank. What was  his initial deposit?

maturity value of recurring deposits banking maths

  1. Rs 6000

  2. Rs 9000

  3. Rs 8000

  4. Rs 7000

Show answer
Correct Option: D
Explanation:

$Let\quad the\quad initial\quad deposit\quad be\quad Rs.\quad x,\ amount\quad after\quad 1st\quad year=Rs.\quad 2x\ amount\quad left\quad after\quad withdrawl=Rs.\quad 2x-8000,\ amount\quad after\quad 2nd\quad year=Rs.\quad 4x-16000\ amonut\quad left\quad after\quad withdrawl=Rs.\quad 4x-24000,\ amount\quad after\quad 3rd\quad year=Rs.\quad 8x-48000=Rs.\quad 8000,\ 8x=56000,\ x=Rs.\quad 7000$

The fixed period of recurring deposit account can be 

maturity value of recurring deposits banking maths

  1. $2$ months

  2. $3$ months

  3. $6$ months

  4. All of the above

Show answer
Correct Option: C
Explanation:

The fixed period of recurring deposit  account can be $6$ months or more.

Therefore, options C is the correct answer.

Kiran deposited Rs.$200$ per month for $36$ months in a bank's recurring deposit account. If the bank pays interest at the rate of $11$ $\%$ per annum, find the amount she gets on maturity.

maturity value of recurring deposits banking maths

  1. Rs.$8412$

  2. Rs.$8421$

  3. Rs.$2481$

  4. Rs.$1234$

Show answer
Correct Option: B
Explanation:

Let the monthly installment be $P$.

Given, $P=200$, $n=36$, $r=11\%$
Interest $=\dfrac {Pn(n+1)r}{2400}$$=\dfrac {200\times 36 \times 37\times 11}{2400}$$=1221$
We know, maturity amount $=(Pn+1221)$$=(200\times 36+1221)$$=$ Rs. $8421$

If $\tan 4x+\tan 5x-\tan 9x=k\tan 4x\tan 5x\tan 9x$ then $k=$

trigonometric equations trigonometric functions trigonometry maths

  1. $1$

  2. $-1$

  3. $ \pm 1$

  4. $2$

Show answer
Correct Option: B
Explanation:

$\begin{array}{l}\tan 9x = \tan (4x + 5x) = \dfrac{{\tan 4x + \tan 5x}}{{1 - \tan 4x\tan 5x}}\ \Rightarrow \tan 9x - \tan 4x\tan 5x\tan 9x = \tan 4x + \tan 5x\ \Rightarrow \tan 4x + \tan 5x - \tan 9x =  - \tan 4x\tan 5x\tan 9x\\therefore k =  - 1\end{array}$

State true or false $\tan(\dfrac{\pi}{4} + \theta) - \tan(\dfrac{\pi}{4} -\theta) = 2\tan\theta$

trigonometric equations trigonometric functions trigonometry maths

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

Using $\tan (A+B)=\cfrac {\tan A+\tan B}{1-\tan A \tan B}$

$\Rightarrow \tan (A-B)=\cfrac {\tan A-\tan B}{1+\tan A+\tan B}$
$\tan \left( \cfrac {\pi}{4}+\theta\right)-\tan \left(\cfrac {\pi}{4}-\theta\right)$
$\Rightarrow A=\cfrac {\pi}{4}, B=\theta$
$\Rightarrow \cfrac {\tan \cfrac {\pi}{4}+\tan\theta}{1-\tan \cfrac {\pi}{4}\tan \theta}-\cfrac {\tan \cfrac {\pi}{4}-\tan \theta}{1+\tan \cfrac {\pi}{4}\tan \theta}$
$\because \tan \cfrac {\pi}{4}=1$
$\Rightarrow \cfrac {1+\tan\theta}{1-\tan \theta}-\cfrac {1-\tan \theta}{1+\tan \theta}$
$\Rightarrow \cfrac {(1+\tan\theta)^2-(1-\tan\theta)^2}{(1-\tan\theta)(1+\tan\theta)}$
$\therefore a^2-b^2=(a+b)(a-b)$
$\Rightarrow \cfrac {1+\tan^2\theta+2\tan\theta-1-\tan^2\theta+2\tan\theta}{1-\tan^2\theta}$
$\Rightarrow \cfrac {4\tan\theta}{1-\tan^2\theta}$
$\because \tan2\theta=\cfrac {2\tan\theta}{1-\tan^2\theta}$
$=2\tan2\theta$.

State true or false

$\tan{ 18 }^{ 0 }+\tan27^{ 0 }+\tan{ 18 }^{ 0 }\cdot \tan27^{ 0 }=1$

trigonometric equations trigonometric functions trigonometry maths

  1. True

  2. False

Show answer
Correct Option: B
Explanation:

$tan180+tan270+tan180+tan270=1$

$0.32491969 + 0.50952544+0.32491969 + 0.50952544 = 1$

$1.66889026 = 1$

The left side $1.66889026 $ not equal to rightside$ 1$.

Which means the given statement is false.

 

$tan  5x-tan  3x-tan  2x=$

trigonometric equations trigonometric functions trigonometry maths

  1. $\tan 5x \tan 3x \tan 2x$

  2. $\sin 5x \sin 3x \sin 2x$

  3. $\cos 5x \cos 3x \cos 2x$

  4. $\sec 5x \sec 3x \sec 2x$

Show answer
Correct Option: A
Explanation:

We've,

$\tan (3x+2x)=\tan 5x$

or, $\dfrac{\tan 3x+\tan 2x}{1-\tan 3x.\tan 2x}=\tan 5x$

or, $\tan 3x+\tan 2x=\tan 5x-\tan 2x.\tan 3x.\tan 5x$

or, $\tan 5x-\tan 3x-\tan 2x=\tan 2x.\tan 3x.\tan 5x$.

$A, B, C$ are three angles such that $\tan  A+\tan  B+\tan  C=\tan  A  \tan  B  \tan  C.$ Which of the following statements is always correct ?

trigonometric equations trigonometric functions trigonometry maths

  1. $ABC$ is a triangle, i.e. $A+B+C=\pi $

  2. $A=B=C. i.e., $ $ABC$ is an equilateral triangle

  3. $A+B=C, $ i.e., $ABC$ is a right- angled triangle

  4. $A+B=\pi $

Show answer
Correct Option: A
Explanation:

(A) $tan\left [ (A+B)+C \right ]$
$=\frac{tan (A+B)+tan C}{1-tan (A+B)  tan  C}=\frac{\frac{tan  A+tan   B}{1-tan  A   tan  B}+tan  C}{1-\frac{tan  A+tan  B}{1-tan   A   tan  B}.  tan  C}$
$=\frac{tan  A+tan  B+tan  C-tan  A   tan  B   tan  C}{Denominator}$
$=0$
$\left [ since,  tan  A+tan  B+tan  C =tan  A   tan  B   tan  C \right ]$
$\therefore A+B+C=\pi $ i.e.,  A, B, C is a triangle

If $\dfrac{\pi}{4}<A<\dfrac{\pi}{2}$ then $\tan^{-1}\left(\dfrac{1}{2}\tan 2A\right)+\tan^{-1}(\cot A)+\tan^{-1}(\cot^{3}A)$=

trigonometric equations trigonometric functions trigonometry maths

  1. $0$

  2. $\pi$

  3. $\pi/2$

  4. $\pi/4$

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Correct Option: A

If $A+B+C=\pi $ and cosA=cosB cosC, then tanB tanC is equal to 

trigonometric equations trigonometric functions trigonometry maths

  1. $\frac { 1 }{ 2 } $

  2. $2$

  3. $1$

  4. $-\frac { 1 }{ 2 } $

Show answer
Correct Option: B
Explanation:

Given $A+B+C=\pi\implies A=\pi-(B+C)$

And also given $\cos A=\cos B\cos C$
$\implies \cos (\pi-(B+C))=\cos B\cos C$
$\implies -\cos B\cos C+\sin B\sin C=\cos B\cos C$
$\implies \sin B\sin C=2\cos B\cos C$
$\implies \tan B\tan C=2$