There are 12 (4 x 3) possible outcomes as listed below
(1,2), (1,4), (1,6), (3, 2), (3,4), (3, 6), (5, 2), (5,4), (5,6), (7,2). (7, 4) (7, 6) (ii) (a) only (1, 6), (3,.4), (5, 2) given sum as 7
\(\therefore\)Required probability = \(\frac {2} {13}=\frac {1} {4}\) 
(b) There is no even sum
\(\therefore\)Required probability= \(\frac {0} {12}=0\)
(c) All the sums are odd
\(\therefore\)Required probability =\(\frac {12} {12} =1\)
(d) only (3, 6), (5, 4), (5, 6), (7, 2). (7, 4) and (7, 6) gives sum as more than 7
\(\therefore\)Required probability = \(\frac {6} {12}\)
=\(\frac {1} {2}\)

Let ABCD be a trapezium with AB 75 m,
DC = 50 m, ㄥABC = 60⁰ and ㄥDAB = 90⁰.
Draw CE ⊥AB.
∴ AE = 50m and EB = 75-50 = 25m
We know that sum of angles of a trapezium is 360⁰.
∴ Area of four sectors = πr²
\(={22\over 7}\times17.5\times17.5\)
=962.5m²
In rt .ㄥed ΔCEB,
\({EC\over EB}=tan60^0\)
\(⇒ {EC\over 25}=\sqrt{3}\)
⇒ EC=25√3m
Now, area of trapezium\(={1\over 2}(AB+DC)\times EC\)
\(={1\over 2}(75+50)\times25\sqrt3\)
\(={1\over 2}\times125\times25\sqrt3\)
=2706.25
Area of remaining portion = 2706.25 - 962.5
=1743.75m²

17.29 cm²

5.7 cm

Steps of Construction : 
1. Draw any line l and take any point X on it. 
2. At X, construct a right angle.
3. With X as centre and of radius 4 cm, draw an arc intersecting XY in A.
4. At A and initial line AX, draw angle of 30^o  to each sides and let these lines intersect line I in B and C.
5. Join AB and AC to get the given triangle.
6. At B, construct an acute angle ㄥCBZ (<90^o).
7. Mark three points on BZ such that BB₁= B₁B₂ = B₂B₃.
8. Join B₃C.
9. Through B₂, draw B₂C' || B₃C, intersecting line I in C'.
10. Through C', draw C'A' || CA, intersecting BA in A'. Thus, ΔA'BC' is the required triangle.

\(x={ 58 }^{ \circ }\)

21.96m/s

17.32m, 40m

Since the sum of the first n terms of an A.P. is given as 3n² + 5n
\(\therefore\)  S_n = 3n² + 5n                ....(i)
\(\therefore\)   S_n-1 = 3( n - 1)2 + 5 ( n - 1 ) 
= 3( n² + 1 - 2n ) + 5( n - 1 )
= 3n² + 3 - 6n + 5n - 5 
= 3n² - n - 2                  ...(ii)
Now,  an = S_n - S_n-1
= 3n² + 5n - 3n² + n + 2
an =  6n + 2
Put n = 1 and n = 2, we have
a₁ = 8 and a₂ = 14
Hence, the common difference is 6.

Given, height of the cone=8 cm and radius of the cone = 5 cm

Volume of cone=\(\frac{1}{3}\)πr²h
=\(\frac{1}{3}\)π(5)²8 cm³=\(\frac{200}{3}\)π cm³
Radius of one spherical lead shot=0.5 cm
Volume of one spherical lead shot=\(\frac{4}{3}\)πr ³=\(\frac{4}{3}\)π(0.5)³ cm³
=\(\frac{4\times{0.125}}{3}\)π cm³=\(\frac{0.5}{3}\)π cm³
When spherical lead are dropped in the vessel, one fourth of water flows out
Let number of lead shots be n
Volume of n spherical shots = \(\frac{1}{4}\) volume of conical vessel
n\(\left(\frac{0.5}{3}\pi\right)\)=\(\frac{1}{4}\left(\frac{200}{3}\pi\right)\)⇒n(0.5)=50⇒ n=\(\frac{50\times{10}}{5}\)=100