Exams

The given figure can be labelled as :

Straight lines :

The number of straight lines are 19

i.e. BC, CD, BD, AF, FE, AE, AB, GH, IJ, KL, DE, AG, BH, HI, GJ, IL, JK, KE and DL.

The given figure can be labelled as shown :

The horizontal lines are AK, BJ, CI, DH and EG i.e. 5 in number.
The vertical lines are AE, LF and KG i.e. 3 in number.
The slanting lines are LC, CF, FI, LI, EK and AG i.e. 6 in number.
Thus, there are 5 + 3 + 6 = 14 straight lines in the figure.

The given figure can be labelled as :

The simplest triangles are AJF, FBG, HDI, GCH and JEI i.e 5 in number.
The triangles composed of the three components each are AIC, FCE, ADG, EBH and DJB i.e 5 in number.
Thus, there are 5 + 5 = 10 triangles in the given figure.

The simplest triangles are AKI, AIL, EKD, LFB, DJC, DKJ, KIJ, ILJ, JLB, BJC, DHC and BCG i.e. 12 in number.
The triangles composed of two components each are AKJ, ALJ, AKL, ADJ, AJB and DBC i.e. 6 in number.
The triangles composed of the three components each are ADC and ABC i.e. 2 in number.
There is only one triangle i.e. ADB composed of four components.
Thus, there are 12 + 6 + 2 + 1 = 21 triangles in the figure.

National Youth Day is celebrated in India on 12 January on the birthday of Swami Vivekananda. In 1984 the Government of India declared the day as the National Youth Day and since 1985 the event is celebrated in India every year.

Relative speed = 42 + 36 = 78 km/hr = 653 m/s

Distance = (520 + 520) =1040 mts.

Time = 1040×365= 48 sec

As Trains are moving in same direction,

Relative Speed = 60-20 = 40 kmph

--> 40×518=1009 m/sec

Length of Train= Speed * Time

Length = 1009×6

$$\begin{gathered} ⇒2003=6623 mts \\ \end{gathered}$$ 

Given speed = 63 km/hr = 63×518=352 m/s

Let the length of the bridge = x mts

Given time taken to cover the distance of (170 + x)mts is 30 sec.

We know speed = distancetimem/s

⇒352=170+x30

 -->  x = 355 mts.