Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

Multiple choice maths gst (goods and service tax) more about gst commission, brokerage and vat taxation, calculation of income tax

A dealer pays following VAT @ of 14% on 20th April Rs. 2500, 20th May Rs. 1200. Find his sales for March and April.

  1. $16428.55$
  2. $26428.55$
  3. $36428.55$
  4. $46428.55$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Sales for April:
Total VAT = VAT% of sales
$2500=\cfrac{14}{100}\times sales$
Sales of April $= Rs. 17857.14$
Sales for May:
Total VAT = VAT% of sales
$1200=\cfrac{14}{100}\times sales$
Sales of April $= Rs. 8571.42$
Total sales for March and April $= 17857.14 + 8571.41 = Rs. 26428.55$

Multiple choice maths gst (goods and service tax) more about gst commission, brokerage and vat taxation, calculation of income tax

Shri. Batliwala sold shares of Rs $30350$ and purchased shares of Rs $69650$ in a day. He paid $0.1\%$ brokerage on both sale and purchase of shares. $18\%$ GST was charged on the brokerage. Find his total expenditure on brokerage and tax.

  1. $120$
  2. $118$
  3. $139.5$
  4. $105$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Value of shares sold = $30350$
Brokerage =$0.1%$
$\therefore$ brokerage  value =$\cfrac{0.1}{100}\times30350$
= $303.5/10$=$30.35$
Value of shares purchased =$69650$
Brokerage =$0.1%$
$ \text {brokerage value} =\cfrac{0.1}{100}\times69650$
=$696.5/10$
=$69.65$
$\therefore$ total  brokerage  value =$(30.35+69.65)$
=$Rs 100$
$G. S. T$= $18%$ of brokerage  value 
$\therefore$ amount of gst = $\cfrac{18}{100}\times100$
=$18$
$\therefore$  total amount of brokerage +gst =$100$+$18$
=$118$.
Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

The range of values of $\lambda$ for which the circles $ { x }^{ 2 }+{ y }^{ 2 }=4$ and ${ x }^{ 2 }+{ y }^{ 2 }-2\lambda y+5=0$ have two common tangents only is-

  1. $\lambda \epsilon \left( -\sqrt { 5 } ,\sqrt { 5 } \right) $
  2. $\lambda <-\sqrt { 5 } or\quad \lambda >\sqrt { 5 }$
  3. $-\sqrt { 5 } <\lambda <1$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Two circles have two common tangents if the distance between their centers is less than the sum of their radii and greater than the difference of their radii. Here, centers are (0,0) and (0, lambda), radii are 2 and sqrt(lambda^2 - 5). The condition leads to the specified range.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

Intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre.

  1. True

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

For any two parallel tangents to a circle, the segment of a third tangent intercepted between them subtends a 90-degree angle at the center because the radii to the points of tangency are perpendicular to the tangents.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

$\overline { M N }$ and $\overline { M Q }$ are two tangents from a point $M$ to a circle with centre $0$ If $m \angle N O Q = 120 ^ { \circ } ,$ then ?

  1. $N Q = M N = M Q$
  2. $N Q = O M$
  3. $O Q = O M$
  4. $O N = M N$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If angle NOQ = 120 degrees, then in the quadrilateral MONQ, the angles at N and Q are 90 degrees. Thus, angle M = 180 - 120 = 60 degrees. Triangle MNQ is isosceles with angle M = 60, so it is equilateral, meaning NQ = MN = MQ.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

The chord of contact of the pair of tangents to the circle $x^2+y^2=1$ drawn from any point on the line $2x+y=4$ passes through a fixed point. 

  1. True

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

If chords are drawn to the circle from a fixed point $(x _1,y _1)$ and then tangents are drawn at point of contact, the point of intersection of all tangents lie on a fixed point.


The fixed point is called pole and fixed line is called polar.


Equation of polar is $T=0$.

$C:x^2+y^2-1=0$

Equation of polar is $T=0$.

$xx _1+yy _1-1=0$

The line is identical to given line $2x+y-4=0$.

By comparing coefficients, we get,
$\dfrac{x _1}{2}=\dfrac{y _1}{1}=\dfrac{-1}{-4}$

$x _1=\dfrac{1}{2},y _1=\dfrac{1}{4}$

Hence, the fixed point is $(\dfrac{1}{2}, \dfrac{1}{4})$.

Multiple choice maths tangents and intersecting chords other theorems related to circles touching circles angle made by a chord and a tangent

From a point $P$ which is at a distance of $13$ cm from the centre $O$ of a circle of radius $5$ cm, the pair of tangents $PQ$ and $PR$ to the circle are drawn. Then the area of the quadrilateral $PQOR$ is:

  1. $60$ cm$^{2}$
  2. $65$ cm$^{2}$
  3. $30$ cm$^{2}$
  4. $32.5$ cm$^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The radius perpendicular tangent at the pt. of contact, therefore, $OQ\perp PQ$ and $OR\perp PR$
In rt. $\triangle OPQ$, we have
$PQ=\sqrt{OP^{2}-OQ^{2}}$
   $=\sqrt{169-25}=\sqrt{144}=12$ cm
$\Rightarrow $ $PR=12$ cm (Two tangents from the same external pt. to a circle are equal)
Now area of quad. $PQOR=2\times $Area of $\triangle POQ$
   $\displaystyle =\left ( 2\times \frac{1}{2}\times 12\times 5 \right )$ cm$^{2}=60$ cm$^{2}$