Numerical Ability
 From a circular sheet of paper of radius 30 cm, a sector of 10% area is removed. If the remaining part is used to make a conical surface, then the ratio of the radius and height of the cone is __

 13.48
 14.08
 13.08
 13

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90% of area of sheet = Cross sectional area of cone
⇒ 0.9 × π × 30 × 30 = π × r_{1} × 30
⇒ 27 cm = r_{1}
∴ height of the cone =√30^{2}  27^{2}= 13.08 cmCorrect Option: C
90% of area of sheet = Cross sectional area of cone
⇒ 0.9 × π × 30 × 30 = π × r_{1} × 30
⇒ 27 cm = r_{1}
∴ height of the cone =√30^{2}  27^{2}= 13.08 cm
 log tan 1° + log tan 2° + _____ + log tan 89° is 1

 1
 1/√2
 0
 – 1

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log tan 1° + log tan 89° = log(tan 1° × tan 89°)
= log(tan 1° × cot 1°)
= log 1 = 0
Using the same logic total sum is ‘0’.Correct Option: C
log tan 1° + log tan 89° = log(tan 1° × tan 89°)
= log(tan 1° × cot 1°)
= log 1 = 0
Using the same logic total sum is ‘0’.
 If a^{2} + b^{2} + c^{2} = 1, then ab + bc+ ac lies in the interval


1, 2 3 
 1 1, 2 
 1,  1 2  [2, – 4]


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(a + b + c)^{2} >= 0
⇒ a^{2} + b^{2} + c^{2} + 2(ab + bc + ca) >= 0
⇒ 1 + 2(ab + bc + ca) >= 0
Thus, (ab + bc + ca) >= – 1/2
Similarly,using:
(a – b)^{2} + (b – c)^{2} + (c – a)^{2} >=1
We will get ab + bc + ca <= 1.
Hence, [–1/2, 1] will be the right answer.Correct Option: B
(a + b + c)^{2} >= 0
⇒ a^{2} + b^{2} + c^{2} + 2(ab + bc + ca) >= 0
⇒ 1 + 2(ab + bc + ca) >= 0
Thus, (ab + bc + ca) >= – 1/2
Similarly,using:
(a – b)^{2} + (b – c)^{2} + (c – a)^{2} >=1
We will get ab + bc + ca <= 1.
Hence, [–1/2, 1] will be the right answer.
 A tiger is 50 leap of its own behind a deer. The tiger takes 5 leaps per minute to the deer's 4. If the tiger and the deer cover 8 metre and 5 metre per leap respectively, what distance in meters will the tiger have a run before it catches the deer?

 500 ms
 1200 ms
 800 ms
 600 ms

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Tiger – 1 leap ⇒ 8 meter
Speed = 5 leap/hr = 40m/min
Deer → 1 leap = 5 meter
speed = 4hr = 20m/min
Let at time ‘t’ the tiger catches the deer.
∴ Distance travelled by deer + initial distance between them
50 × 8 ⇒ 400m = distance covered by tiger.
⇒ 40 × t = 400 + 20t⇒ t = 400 = 20 min 200
⇒ total distance ⇒ 400 + 20 × t = 800 msCorrect Option: C
Tiger – 1 leap ⇒ 8 meter
Speed = 5 leap/hr = 40m/min
Deer → 1 leap = 5 meter
speed = 4hr = 20m/min
Let at time ‘t’ the tiger catches the deer.
∴ Distance travelled by deer + initial distance between them
50 × 8 ⇒ 400m = distance covered by tiger.
⇒ 40 × t = 400 + 20t⇒ t = 400 = 20 min 200
⇒ total distance ⇒ 400 + 20 × t = 800 ms
 Five teams have to compete in a league, with every team playing every other team exactly once, before going to the next round. How may matches will have to be held to complete the league round of matches?

 20
 10
 8
 5

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For a match to be played, we need 2 teams No. of matches = No. of ways of selections 2 teams out of 5
= 5_{c2} = 10Correct Option: B
For a match to be played, we need 2 teams No. of matches = No. of ways of selections 2 teams out of 5
= 5_{c2} = 10