TISSNET 2017

Let T(n) represent the $$n^{th}$$ term of the series

T(2) - T(1) =10-7 = 3

Similarily T(3) - T(2)  = 9

T(4) - T(3) = 27

T(5) -T(4) = 81

We can see that the difference of the terms are in GP

Hence T(6)-T(5) = 243

T(6) = 243+127 = 370

We can identify the pattern as:

1$$^3$$ +1 =2

2$$^3$$ +2 =10

3$$^3$$ +3 =30

4$$^3$$ +4 = 68

5$$^3$$ +5 = 130

Option B

The above pattern can be seen as following

A BC DE FGH IJK

The next term will have 4 terms after skipping 4 terms from K

PQRS is correct option

3 $$\times 5 \times$$ 4 - 10 = 60 - 10 = 50

The above figure contains 7 vertical and 6 horizontal lines. Selecting any 2 of the vertical and 2 of the vertical lines will give a unique enclosed rectangle. 

This can be done in $$^7C_2\times\ ^6C_{2\ }=\ 21\times\ 15\ =\ 315$$

Sum of first 7 prime numbers = 2 + 3 + 5 + 7 + 11 + 13 + 17 = 58

=> Average = $$\frac{58}{7}=8.28$$

=> Ans - (C)

Let the angles of the triangle be $$\theta,2\theta,90^\circ$$

=> Sum of angles = $$\theta+2\theta+90^\circ=180^\circ$$

=> $$\theta=\frac{90^\circ}{3}=30^\circ$$

=> Ans - (A)

Marked price = 40000

Discount = 8%

Selling price = 40000 $$\times$$ 92/100 = 36800

Profit = 25%

Cost price = 36800 $$\times$$ 100/125 = Rs.29440

Let scores of each candidate respectively be $$a,b,c,d,e,f$$

Average marks scored by Amit, Bimal and Candy = 84

=> $$a+b+c=84\times3=252$$ -----------------(i)

Also, after adding Dorothy marks, average marks of the 4 students = 80

=> $$a+b+c+d=80\times4=320$$ -----------(ii)

Subtracting (i) from (ii), we get : $$d=320-252=68$$

Average marks scored by Bimal, Candy, Dorothy and Ellora = 79

=> $$b+c+d+e=79\times4=316$$ -----------(iii)

Also, $$e=d+3=68+3=71$$

Substituting value of $$d$$ and $$e$$ in equation (iii), => $$b+c=316-68-71=177$$ ------------(iv)

Subtracting (iv) from (i), we get : $$a=252-177=75$$

=> Ans - (B)

Given : ABC is an isosceles triangle, with AB=AC, let $$\angle$$ ABC $$=\angle$$ ACB = $$x$$

=> $$\angle$$ BAC = $$8x$$

Also, AD is angle bisector, => $$\angle$$ BAD $$=\angle$$ CAD = $$4x$$

The angle bisector of an isosceles triangle is perpendicular bisector to the base.

=> $$\angle$$ BDA $$=\angle$$ CDA = $$90^\circ$$

Thus, in right triangle ABD, => $$\angle$$ ABD + $$\angle$$ ADB + $$\angle$$ BAD = $$180^\circ$$

=> $$x+4x=180-90=90^\circ$$

=> $$x=\frac{90}{5}=18^\circ$$

$$\therefore$$ $$\angle$$ BAD = $$4\times18=72^\circ$$

=> Ans - (C)