Question
Find first four terms of an A.P., whose first term is 3 and common difference is 4.
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Answer
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A trader from Surat, Gujarat sold cotton clothes to a trader in Rajkot, Gujarat. The taxable value of cotton clothes is Rs. 2.5 lacs. What is the amount of GST at 5% paid by the trader in Rajkot?
Sint. Aruna Thakkar purchased 100 shares of FV 100 when the MV is Rs. 1200. She paid brokerage at the rate of 0.3% and 18% GST on brokerage.
Find the following -
(1) Net amount paid for 100 shares.
(2) Brokerage paid on sum invested.
(3) GST paid on brokerage.
(4) Total amount paid for 100 shares.
Find the following -
(1) Net amount paid for 100 shares.
(2) Brokerage paid on sum invested.
(3) GST paid on brokerage.
(4) Total amount paid for 100 shares.
Which of the following are quadratic equations?
$\text{x}^2+\frac{1}{\text{x}^2}=5$
→$\text{x}^2+\frac{1}{\text{x}^2}=5$
$\triangle\text{ABC}$ and $\triangle\text{DBC}$ lie on the same side of BC, as shown in the figure. From a point P on BC, PQ || AB and PR || BD are drawn, meeting AC at Q and CD at R respectively. Prove that QR || AD.
→The sides of certain triangles are given below. Determine which of them are right triangles.
$a = 16\ cm, b = 3.8\ cm$ and $c = 4\ cm.$
→$a = 16\ cm, b = 3.8\ cm$ and $c = 4\ cm.$
In the given pairs of triangles, find which pair of triangles are similar. State the similarity criterior and write the similarity relation in symbolic from.
Very-Short and Short-Answer Questions:
Show that the system $2x + 3y - 1 = 0$, $4x + 6y - 4 = 0$ has no solution.
→Show that the system $2x + 3y - 1 = 0$, $4x + 6y - 4 = 0$ has no solution.
The maximum bowling speeds (in km/h) of 33 players at a cricket coaching centre are given below:
Calculate the median bowling speed.
|
Speed in km/h
|
85-100
|
100-115
|
115-130
|
130-145
|
|
No. of players
|
10
|
4
|
7
|
9
|
Write the discriminant of the following quadratic equation.
$(x - 1)(2x - 1) = 0$
→$(x - 1)(2x - 1) = 0$
Evaluate:
$4(\sin^430^\circ+\cos^460^\circ)-\frac{2}{3}(\sin^260^\circ-\cos^245^\circ)+\frac{1}{2} \ \tan^260^\circ$
→$4(\sin^430^\circ+\cos^460^\circ)-\frac{2}{3}(\sin^260^\circ-\cos^245^\circ)+\frac{1}{2} \ \tan^260^\circ$