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 __

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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

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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

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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?

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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?

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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