Question 1
[(74)/(493/2)](1/6) x [(73 /(3434/3)](1/2) = ? / [(74 / 49)1/6]
a) 1 b) 724 c) 76 d) 74
Answer : a) 1
Solution :
Given expression is, [(74)/(493/2)](1/6) x [(73 /(3434/3)](1/2) = ? / [(74 / 49)1/6]
= [(74)/(493/2)](1/6) x [(73 /(3434/3)](1/2) x [(74 / 49)1/6] = ?
= [(74)/(72)(3/2)](1/6) x [(73/(73)(4/3)](1/2) x [(74/72)1/6]
= (74)/(73)](1/6) x [(73)/(74)](1/2) x [(74/72)1/6]
= (74-3)(1/6) x [(73-4)](1/2) x [(74-2)1/6
= (71)(1/6) x [(7-1)](1/2) x [(72)1/6
= (7)(1/6) x (7)(-1/2) x (7)1/3
= (7)(1/6 -1/2 + 1/3)
= (7)(2-6+4/12)
= (70 )(1/12) = 1
Hence the answer is 1.
Question 2
sqrt(104) x sqrt(3x1016 + 1016) = ? x sqrt(108) x 2
a) 104 b) 106 c) 108 d) 1016
Answer : b) 106
Solution :
Given expression is, sqrt(104) x sqrt(3 x 1016 + 1016) = ? x sqrt(108) x 2
= (104)(1/2) x (3 x 1016 + 1016)(1/2) = (? x 2) x (108)(1/2)
= (10)(4/2) x(4 x 1016)(1/2) = (? x 2) x (108/2)
= 102 x 2 x 108 = (? x 2) x (104)
= 2 x 102x 10(4+4) = (? x 2) x (104)
= 2 x 102 x 104 x 104 = (? x 2) x (104)
= 2 x 10(2+4) x 104 = (? x 2) x (104)
= 2 x 106 x 104 = (? x 2) x (104)
Therefore, ? = 10(6).
Hence the answer is 106.
Question 3
366 x 122514 x 2258 /(1966 x 441(11/2)) = ? x 317 x 544 / 75
a) 1 b) 3-7 c) 5-2 d) 73
Answer : a) 1
Solution :
Given expression is, 366 x 122514 x 2258 /(1966 x 441(11/2)) = ? x 317 x 544 / 75
The terms in RHS are the powers of 3, 5 and 7.
So let us simplify the LHS as follows:
366 x 122514 x 2258 /(1966 x 441(11/2)) = (62)6 x (352)14 x (152)8 / (142)6 x (212)(11/2))
= (6)12 x (35)28 x (15)16 / (14)12 x (21)11)
= (2x3)12 x (5x7)28 x (3x5)16 / (7x2)12 x (7x3)11)
= 212 x 312 x 528 x 728 x 316 x 516 / 712 x 212 x 711 x 311
= 212-12 x 312+16-11 x 528+16 x 728-12-11
= 20 x 317 x 544 x 7-5
= 317x 544 / 75
Now, RHS = ? x 317x 544 / 75 = LHS = 317x 544 / 75
Therefore, ? x 317x 544 / 75 = 317x 544 / 75
Hence the answer is 1.
Question 1
In a factory, there are two machines A and B. A machine can produce 1000 watts power in 5 hours and
by running both the machines together can produce same watts power in 3 hours. Find the time taken
by B machine alone to produce same watts of power.
a) 5 hrs b) 7 hrs & 30 mins c) 6 hrs d) 6 hrs & 30 mins
Answer : b) 7 hrs & 30 mins
Solution :
A type machine can produce 1000 watts power in 5 hours.
Then A's 1 hour work = 1/5.
Both A and B can produce in 3 hours.
And 1 hour work of (A + B)'s = 1/3.
B's 1 hour work = 1/3 = 1/5 = 2/15.
Therefore, machine B can produce 1000 watts power in 15/2 hours.
i.e., 15/2 hours = 7.5 hours = 7 hours and 30 minutes.
Question 2
There is a building with 4 floors and each floor has equal area. There are three sweepers P, Q and R. P
can clean 1 floor in 10 hours, Q can clean 8/5 of the floors in 40 hours and R can clean 4/3 of the floors
in 13 hours. Then who will clean all the 4 floors at first?
a) R b) R & P c) Q d) Q & P
Answer : a) R
Solution :
P can sweep 1 floor in 10 hours.
Then 4 floors will be cleaned by P in = 10 x 4 = 40 hours.
Q can sweep 8/5 part in 40 hours.
Then 1 floor will be done by Q in 40/(8/5) = 25 hours.
And 4 floors will be done by Q in 25 x 4 = 100 hours.
R can clean 4/3 part of the floors in 13 hours.
Then 1 floor will be cleaned by R in 13/(4/3) = 13 x 3/4 hours.
And 4 floors will be done by R in 13 x 3/4 x 4 = 13 x 3 = 39 hours.
Thus P, Q and R takes 40, 100 and 39 hours respectively.
Hence the sweeper R will complete the work at first.
Question 3
A man and his son planned to paint their own house. The man alone can do in 20 days. He paints for 4
days and then his son completed in 8 days. Find how many days will they take working together?
a) 12 days b) 11 days c) 7 days d) 8 days
Answer : c) 7 days
Solution :
Man's 1 day work = 1/20
Then 4 days work = 4/20 = 1/5
Remaining work = 1 - 1/5 = 4/5
Now, 4/5 work is done by the son in 8 days.
Then, whole work will be done by him in (8 x 5/4) days = 10 days.
Then his 1 day work = 1/10
The man and his son's 1 day work = 1/10 + 1/10 = 3/10
They will complete the work together in 20/3 days = 6 2/3 days = 7 days(nearly).
Hence the answer is 7 days.
Question 1
A boy drops a bouncing toy from 1st floor to ground floor. Each time, it bounces vertically to a height of
1/2 of the preceding bounce's height. Find the distance the toy will cross if it totally takes 6 bounces and
the distance between each floor is 10m.
a)39 m b)30 m c)40 m d)15 m
Answer : b)30 m
Solution :
The distance between two floors is 10m.
The boy dropped the bouncing toy from a height of 10 m.
As the toy is initially dropped, the toy falls 10 meters. Then, the toy rebounds to 10 x 1/2 = 5 meters and
falls back down 5 meters. This process will repeat for 5 more times.
Then:
Distance that the toy travels down = 10 + 10x1/2 + 10x 1/2 x 1/2 + ...+ 10x(1/2)6
= 10 + 10x(1/2)1 + 10(1/2)2 + 10(1/2)3 + 10x(1/2)4 + 10(1/2)5 + 10(1/2)6 which is a g.p series with 7 terms
and first term = 10 common term = 1/2.
we know that, sum of n terms in a g.p series is
sum = a(1 - rn)/1 - r when r is less than one
Here, distance = 10 x [1 - (1/2)7]/(1-1/2) = 10 x 127/128 / 1/2
= 20x127/128 = 19.84 meters.
Distance that the toy travels up = 10x(1/2)1 + 10(1/2)2 + 10(1/2)3 + 10x(1/2)4 + 10(1/2)5 + 10(1/2)6
= 19.84 - 10 = 9.84 meters.
Thus, the total distance is 19.84 + 9.84 = 29.68 meters = 30 meters (approximately)
Question 2
A spider starts jumping up from the bottom of a ladder with 32 steps. Each minute, it jumps 4 steps and
slips back 2 steps. How much time would the spider take to reach the top of the ladder?
a)15 minutes b)14 minutes c)16 minutes d)18 minutes
Answer : a)15minutes.
Solution :
Think of it like a number line: In the first minute, the spider goes 4 steps up, and slips back 2 steps.
0...1...2...3...4...5...6
|---|---|---|---|---|---|
---------------->
........<------(Imagine that's going up instead of horizontal.)
The spider effectively climbs 4 steps - 2 steps = 2 steps higher after the first minute.
In the second minute, the spider goes up 4 more steps, and slips back 2 steps. So he's at 2 steps + 4
steps - 2 steps = 4 steps.
Every minute, it gains 4 steps and loses 2 steps, for a total gain of 2 steps. So after the 14th minute, it is
at 28 steps height.
Now, here's the twist: in the 15th minute, it jumps up 4 steps. Since 28 steps + 4 steps = 32 steps, it is at
the top of the ladder.
Hence the answer is 15 minutes.
Question 3
A man throws a ball from certain height to the ground and it covers a distance of X meters on the
ground. Each time it's bounce covers 2/3 of previous distance. The ball stops bouncing after 4 bounces.
Then the total distance it covers on the ground is:
a)191X/43 meters b)250X/21 meters c)172X/75 meters d)211X/81 meters
Answer : d)211X/81 meters
Solution :
This problem is slightly different from the 1st one.
Here, we have to find the distance which covered by the ball on the ground.
Given that, it covers X meters by the through.
And it takes four bounce, each bounce covers 2/3 of previous distance.
Then the distance covered in 1st bounce = (2/3)X
The distance covered in 2nd bounce = (2/3) of (2/3)X = X(2/3)2
The distance covered in 3rd bounce = (2/3) of X(2/3)2 = X(2/3)3
And the distance covered in 4th bounce = (2/3) of X(2/3)3 = X(2/3)4
Required distance = X + (2/3)X + X(2/3)2 + X(2/3)3 + X(2/3)4
The above series is a G.P series of 5 terms with a = X and r = 2/3
Then sum = a(1 - rn)/(1 - r)
= X[1 - (2/3)5]/(1 - 2/3)
= X[1 - 32/243]/(1/3)
= 3X(211/243) = 211X/81
Hence the required answer is 211X/81 meters.
Question 1
A man spent 5/16 of his age plus two as student, 1/40 plus 1 as a husband, 1/4 as a good politician and
3/40 as a father. And the remaining 6 years as a good grand father. Then the living days of the man is:
a) 72 b) 89 c) 80 d) 69
Answer : c)80
Solution :
Lets say his age is X.
So years spent as student = 5X / 16 + 2,
as Husband = X/4 + 1,
as Politician = X/4,
and as Father = 3X/40.
Remaining 6 years (as grandfather) = X - { 5X/16 + 2 + X/4 + 1 + X/4 + 3X/40}
i.e., 6 = X - { 5X / 16 + 2 + X / 4 + 1 + X / 4 + 3X / 40}
6 = X - [13X / 16 + 3 + 3X / 40]
6 = X - [71X / 80 + 3]
9X/80 = 9
X = 80.
Hence the age of the man is 80 years.
Question 2
In a village, every weekend, 1/4 th of men and 1/6 th of women participate in the social activity. If the
total number of participants is 75, out of them 25 are men then the total number of men and women in
the village is:
a) 100 b) 200 c) 300 d) 400
Answer : d)400
Solution :
Number of men who participated in social activities = 25
Therefore, 1/4 th of the men = 25
Or, total number of men in the village = 100
Number of women who participated in social activities = 75 - 25 = 50
Therefore, 1/6 th of the women = 50
Or, total number of women in the village = 300
Hence the required answer is 300 + 100 = 400.
Question 3
A father has divided his properties in such a way that one-half of total properties goes to A, two-third of
the remaining shared equally to B, C and D and the rest to E, F, G and H such they equally gets Rs.30,000
by their sharing property. Then the amount will D get is:
a) Rs.80,000 b) Rs.68,000 c) Rs.24,000 d) Rs.72,000
Answer : a) Rs.80,000
Solution :
A's share = 1/2
Then remaining = 1 - 1/2 = 1/2 ...(A)
2/3 of this remaining 1/2 (as in eq A) goes to B,C and D.
Total share of B, C and D = 2/3 of 1/2 = 1/3
Therefore, now remaining property = Total share of E,F,G and H
= value of (A) - total share of B,C and D
= 1/2 - 1/3 = 1/6
Then, E, F, G and H 's individual shares = (1/6) / 4 = 1/6 x 1/4 = 1/24
Given 1/24 = Rs.30,000
Then, total property amount = Rs.24 x 30,000 = Rs.7,20,000
Total share of B, C and D = 1/3 x 7,20,000
Then their individual share amount = (1/3 x 7,20,000) / 3 = Rs.80,000.
Hence D's share is Rs.80000.
Question 1
A boy drops a bouncing toy from 1st floor to ground floor. Each time, it bounces vertically to a height of
1/2 of the preceding bounce's height. Find the distance the toy will cross if it totally takes 6 bounces and
the distance between each floor is 10m.
a)39 m b)30 m c)40 m d)15 m
Answer : b)30 m
Solution :
The distance between two floors is 10m.
The boy dropped the bouncing toy from a height of 10 m.
As the toy is initially dropped, the toy falls 10 meters. Then, the toy rebounds to 10 x 1/2 = 5 meters and
falls back down 5 meters. This process will repeat for 5 more times.
Then:
Distance that the toy travels down = 10 + 10x1/2 + 10x 1/2 x 1/2 + ...+ 10x(1/2)6
= 10 + 10x(1/2)1 + 10(1/2)2 + 10(1/2)3 + 10x(1/2)4 + 10(1/2)5 + 10(1/2)6 which is a g.p series with 7 terms
and first term = 10 common term = 1/2.
we know that, sum of n terms in a g.p series is
sum = a(1 - rn)/1 - r when r is less than one
Here, distance = 10 x [1 - (1/2)7]/(1-1/2) = 10 x 127/128 / 1/2
= 20x127/128 = 19.84 meters.
Distance that the toy travels up = 10x(1/2)1 + 10(1/2)2 + 10(1/2)3 + 10x(1/2)4 + 10(1/2)5 + 10(1/2)6
= 19.84 - 10 = 9.84 meters.
Thus, the total distance is 19.84 + 9.84 = 29.68 meters = 30 meters (approximately)
Question 2
A spider starts jumping up from the bottom of a ladder with 32 steps. Each minute, it jumps 4 steps and
slips back 2 steps. How much time would the spider take to reach the top of the ladder?
a)15 minutes b)14 minutes c)16 minutes d)18 minutes
Answer : a)15minutes.
Solution :
Think of it like a number line: In the first minute, the spider goes 4 steps up, and slips back 2 steps.
0...1...2...3...4...5...6
|---|---|---|---|---|---|
---------------->
........<------(Imagine that's going up instead of horizontal.)
The spider effectively climbs 4 steps - 2 steps = 2 steps higher after the first minute.
In the second minute, the spider goes up 4 more steps, and slips back 2 steps. So he's at 2 steps + 4
steps - 2 steps = 4 steps.
Every minute, it gains 4 steps and loses 2 steps, for a total gain of 2 steps. So after the 14th minute, it is
at 28 steps height.
Now, here's the twist: in the 15th minute, it jumps up 4 steps. Since 28 steps + 4 steps = 32 steps, it is at
the top of the ladder.
Hence the answer is 15 minutes.
Question 3
A man throws a ball from certain height to the ground and it covers a distance of X meters on the
ground. Each time it's bounce covers 2/3 of previous distance. The ball stops bouncing after 4 bounces.
Then the total distance it covers on the ground is:
a)191X/43 meters b)250X/21 meters c)172X/75 meters d)211X/81 meters
Answer : d)211X/81 meters
Solution :
This problem is slightly different from the 1st one.
Here, we have to find the distance which covered by the ball on the ground.
Given that, it covers X meters by the through.
And it takes four bounce, each bounce covers 2/3 of previous distance.
Then the distance covered in 1st bounce = (2/3)X
The distance covered in 2nd bounce = (2/3) of (2/3)X = X(2/3)2
The distance covered in 3rd bounce = (2/3) of X(2/3)2 = X(2/3)3
And the distance covered in 4th bounce = (2/3) of X(2/3)3 = X(2/3)4
Required distance = X + (2/3)X + X(2/3)2 + X(2/3)3 + X(2/3)4
The above series is a G.P series of 5 terms with a = X and r = 2/3
Then sum = a(1 - rn)/(1 - r)
= X[1 - (2/3)5]/(1 - 2/3)
= X[1 - 32/243]/(1/3)
= 3X(211/243) = 211X/81
Hence the required answer is 211X/81 meters.