Tag: maths

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Questions Related to maths

Multiple choice maturity value of recurring deposits banking 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?

  1. Rs 6000

  2. Rs 9000

  3. Rs 8000

  4. Rs 7000
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

Multiple choice maturity value of recurring deposits banking maths

The fixed period of recurring deposit account can be 

  1. $2$ months
  2. $3$ months
  3. $6$ months

  4. All of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

Therefore, options C is the correct answer.

Multiple choice maturity value of recurring deposits banking maths

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.

  1. Rs.$8412$
  2. Rs.$8421$
  3. Rs.$2481$
  4. Rs.$1234$
Reveal answer Fill a bubble to check yourself
B Correct answer
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. $8421$

Multiple choice trigonometric equations trigonometric functions trigonometry maths

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

  1. $1$
  2. $-1$
  3. $ \pm 1$
  4. $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

Multiple choice trigonometric equations trigonometric functions trigonometry maths

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

Multiple choice trigonometric equations trigonometric functions trigonometry maths

State true or false

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

 

Multiple choice trigonometric equations trigonometric functions trigonometry maths

$tan  5x-tan  3x-tan  2x=$

  1. $\tan 5x \tan 3x \tan 2x$
  2. $\sin 5x \sin 3x \sin 2x$
  3. $\cos 5x \cos 3x \cos 2x$
  4. $\sec 5x \sec 3x \sec 2x$
Reveal answer Fill a bubble to check yourself
A Correct answer
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, $\tan 5x-\tan 3x-\tan 2x=\tan 2x.\tan 3x.\tan 5x$.

Multiple choice trigonometric equations trigonometric functions trigonometry maths

$A, B, C$ 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 ?

  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 $
Reveal answer Fill a bubble to check yourself
A Correct answer
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

Multiple choice trigonometric equations trigonometric functions trigonometry maths

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

  1. $0$
  2. $\pi$
  3. $\pi/2$
  4. $\pi/4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using trigonometric identities for inverse functions and the given range, the sum simplifies to 0. Specifically, the terms cancel out based on the properties of inverse tangents of cotangent functions.

Multiple choice trigonometric equations trigonometric functions trigonometry maths

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

  1. $\frac { 1 }{ 2 } $
  2. $2$
  3. $1$
  4. $-\frac { 1 }{ 2 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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