A right pyramid stands on a square base of side10 cm. If the height of the pyramid is 12 cm, the area (in cm2) of its slant surface is
520
420
360
260
Answer : Option D. As the side of Square base = 10
Then GE = 1/2 * 10 = 5
As Triangle OGE is a right angle triangle , so
OE2 = OG2 + GE2 = 144+25 = 169
hence, OE = 13
In triangle OCD, height of the triangle = OE = 13
DC = 10
Hence, the Area of Slant surface = 4* Area of triangle OCD ( as the pyramid is regular )
= 4 * (1/ 2 * 10 * 13) = 260cm 2
Hence, the answer is option D.
If each interior angle of a regular polygon is 150°, the number of sides of the polygon is
8
10
15
None of these
Answer : Option D.
Each interior angle in a regular polygon = (n-2) x 180/n
where, n is the no of sides.
Here, (n-2) x 180/n = 150
=> 180n - 360 = 150n
=> 30n = 360
=> n = 12
So, Ans. is option D.
If the Altitude of a right prism is 10 cm and its base is an equilateral triangle of side 12 cm, then its total surface area ( in cm2 ) is
5 + 3√3
36√3
360
72( 5 + √3 )
Answer : Option D.
Total Surface area = Perimeter of base x height
+ 2 x area of base
= 36 x 10 - 2 x √3/4 x 12 2
= 360 +72 √ 3
= 72(5+ √ 3)
so, as option D is correct.
The value of tan 10° tan 15° tan 75° tan 80° is
0
1
-1
2
Answer : Option B.
tan10 tan15 tan75 tan80
we know that tan10 = tan(90 - 80) = cot 80o
and tan15 = tan( 90 - 75) = cot 75o
Therefore,tan10 tan15 tan75 tan80
= cot80 cot75 tan75 tan80
=1/tan80 x 1/tan75 x tan80 =1 =so,Answer is option B
The minimum value of 4 tan2θ + 9 cot2θ is equal to
=>sin 9x.cos 2x + sin 9x.sin 2x =cos9x.cos2x +cos9x.sin2x
=>sin 9x.(cos 2x + sin 2x) = cos 9x.(cos 2x +sin 2x)
=>sin 9x/cos 9x = 1
=>tan 9x = 1
Now,tan 9x+ cot 9x
= tan 9x + 1 /tan 9x
= 1 + 1/1 = 2
A person sets to cover a distance of 12 km in 45 minutes. If he covers of the distance in of time, then what is the speed in the remaining time?
16 km/hr
8 km/hr
12 km/hr
55 km/hr
10 km/hr
Answer : Option C.
Distance already covered = 3/2*12 = 9 km, Time spent = 2/3*45 min = 30 min
Distance left = (12 – 9) km = 3 km, Time left = (45 – 30) min = 15 min
Therfore, Required speed = 3/ 15/60 km/hr = 12 km/hr.
If a train 110 metres in length passes a man walking at the rate of 6 km/hr against it in 6 seconds, it will pass another man walking at the same speed in the same direction in time of
91/3sec
101/3sec
8 sec
6 sec
71/3sec
Answer : Option E.
Let the speed of the train = x km/hr. Relative speed = (x + 6) km/hr = (x + 6) X 15/8 m/sec
(x + 6)x 5/18x6 = 110, x = 60. Speed of train = 60 km/hr for 2nd person,
Relative speed = (60 – 6) km/hr = 54x5/18 m/sec = 15 m/hr. Time taken to cross 2nd person =110/15=22/3= 7 1/3 sec.
(91.66 % of 24 +3)1/2 = ? +2
9
5
7
8
3
Answer : Option E.
(11/12*24+3)1/2=?+2
(22+3)1/2-2=?
?=5-2=3
45% of 401 ÷ 3 - ? = 6.022
38
52
6
24
12
Answer : Option D.
45% of 400 ÷ 3 -? = (6)2
? = 60 – 36 = 24
As the side of Square base = 10
