Q.1 Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z are in an arithmetic progression then the common ratio of the geometric progression is
A. 3/6
B. 3/2
C. 5/2
D. 1/6
Detailed Analysis:
Since x, y ,and z are in G.P. and x<y<z, let x = a, y=ar and z=ar2, where a>0 and r>1.
It is also given that, 15x, 16y and 12z are in A.P.
Therefore, 2×16y=5x+12z
Substituting the values of x, y and z we get,
32ar=5a+12ar2
⇒32r=5+12r2
⇒12r2−32r+5=0
On solving the above quadratic equation we get r=1/6 or 5/2.
Since r>1, therefore r=5/2.
Q.2 A tank is fitted with pipes, some filling it and the rest draining it. All filling pipes fill at the same rate, and all draining pipes drain at the same rate. The empty tank gets completely filled in 6 hours when 6 filling and 5 draining pipes are on, but this time becomes 60 hours when 5 filling and 6 draining pipes are on. In how many hours will the empty tank get completely filled when one draining and two filling pipes are on?
Detailed Analysis:
Let the rate of each filling pipes be 'x lts/hr' similarly, the rate of each draining pipes be 'y lts/hr'.
As per the first condition,
Capacity of tank = (6x - 5y)×6..........(i)
Similarly, from the second condition,
Capacity of tank = (5x - 6y) × 60.....(ii)
On equating (i) and (ii), we get
(6x - 5y) × 6 = (5x - 6y)×60
or, 6x - 5y = 50x - 60y
or, 44x = 55y
or, 4x = 5y
or, x = 1.25y
Therefore, the capacity of the tank = (6x - 5y) × 6 = (7.5y - 5y) × 6 = 15y lts
Effective rate of 2 filling pipes and 1 draining pipe = (2x - y) = (2.5y - y) = 1.5y
Hence, the required time = 15y/1.5y=10 hours.
Q.3 If log2(5+log3a)=3 and log5(4a+12+log2b)=3, then a + b is equal to
Detailed Analysis:
5+log3a=23=8⇒a=27
Similarly, 4a+12+log2b=53=125
since a=27,4(27)+12+log2b=125⇒b=32
a + b = 59.
Q.4 If among 200 students, 105 like pizza and 134 like burger, then the number of students who like only burger can possibly be
Detailed Analysis:
Let the number of students who like both pizza and burger be ‘m’ .
The number of students who like neither of them be n
From venn diagram 105 – m + m + 134 – m + n = 200 m – n = 39
∴The possible values of (m, n) are (39, 0) (40, 1)…….(105, 66)
∴ The number of students who like only burger is lies in the range [134 – 105, 134 – 39] = [29, 95]
∴ From options, 93 is a possible answer
Q.5 In an apartment complex, the number of people aged 51 years and above is 30 and there are at most 39 people whose ages are below 51 years. The average age of all the people in the apartment complex is 38 years. What is the largest possible average age, in years, of the people whose ages are below 51 years?
Detailed Analysis:
Let the average age of people aged 51 years and above be x years.
Let the average age of people aged below 51 years be y years.
Let the number of people aged below 51 years be N.
Given, the average age of all the people in the apartment complex is 38 years.
Therefore,
x×30+y×N30+N=38
….(1)
We want to maximize y, which occurs when x is minimum i.e. for x=51.
Substituting the value of x in (1) we get
390=N×(38-y)
Again, when y is maximum, N is also maximum i.e. 39
Therefore maximum value of y = 28.
Q.6 While multiplying three real numbers, Ashok took one of the numbers as 73 instead of 37. As a result, the product went up by 720. Then the minimum possible value of the sum of squares of the other two numbers is
Detailed Analysis:
Let the other two numbers be y and z.
As per the condition
73yz - 37yz = 720
Or 36yz=720
Or yz=20
Minimum possible sum of the squares of the other two numbers would occur when y = z i.e. y=z=√20
Hence the required sum = 40.
Q.7 A right circular cone, of height 12 ft, stands on its base which has diameter 8 ft. The tip of the cone is cut off with a plane which is parallel to the base and 9 ft from the base. With π = 22/7, the volume, in cubic ft, of the remaining part of the cone is
A. 158
B. 178
C. 198
D. 218
Detailed Analysis:
We are given that diameter of base = 8 ft. Therefore, the radius of circular base = 8/2 = 4 ft
In triangle OAB and OCD
OAAB=OCCD
⇒
AB = 3×412
= 1 ft.
Therefore, the volume of remaining part = Volume of entire cone - Volume of smaller cone
⇒
13×π×42×12−13×π×12×3
⇒
13×π×189
⇒
227×3×189
⇒
198
cubic ft
Q.8 Train T leaves station X for station Y at 3 pm. Train S, traveling at three quarters of the speed of T, leaves Y for X at 4 pm. The two trains pass each other at a station Z, where the distance between X and Z is three-?fths of that between X and Y. How many hours does train T take for its journey from X to Y?
Detailed Analysis:
Train T starts at 3 PM and train S starts at 4 PM.
Let the speed of train T be t.
=> Speed of train S = 0.75t.
When the trains meet, train t would have traveled for one more hour than train S.
Let us assume that the 2 trains meet x hours after 3 PM. Trains S would have traveled for x-1 hours.
Distance traveled by train T = xt
Distance traveled by train S = (x-1)*0.75t = 0.75xt-0.75t
We know that train T has traveled three fifths of the distance. Therefore, train S should have traveled two-fifths the distance between the 2 cities.
=> (xt)/(0.75xt-0.75t) = 3/2
2xt = 2.25xt-2.25t
0.25x = 2.25
x = 9 hours.
Train T takes 9 hours to cover three-fifths the distance. Therefore, to cover the entire distance, train T will take 9*(5/3) = 15 hours.
Therefore, 15 is the correct answer.
Q.9 Each of 74 students in a class studies at least one of the three subjects H, E and P. Ten students study all three subjects, while twenty study H and E, but not P. Every student who studies P also studies H or E or both. If the number of students studying H equals that studying E, then the number of students studying H is
Detailed Analysis:
Let the number of students who studying only H be h, only E be e, only H and P but not E be x, only E and P but not H be y
Given only P = 0 All three = 10; Studying only H and E but not P = 20
Given number of students studying H = Number of students studying E
= h + x + 20 + 10
= e + y + 20 + 10
h + x = e + y total number of students = 74
Therefore, h + x + 20 + 10 + e + y = 74
h + x + e + y = 44
h + x + h + x = 44
h + x = 22
Therefore, the number of students studying H = h + x + 20 + 10 = 22 + 20 + 10 = 52.
Q.10 A wholesaler bought walnuts and peanuts, the price of walnut per kg being thrice that of peanut per kg. He then sold 8 kg of peanuts at a pro?t of 10% and 16 kg of walnuts at a pro?t of 20% to a shopkeeper. However, the shopkeeper lost 5 kg of walnuts and 3 kg of peanuts in transit. He then mixed the remaining nuts and sold the mixture at Rs. 166 per kg, thus making an overall pro?t of 25%. At what price, in Rs. per kg, did the wholesaler buy the walnuts?
Detailed Analysis:
Let the cost price of peanuts for the wholesaler be x per kg.
Cost price of walnuts for the wholesaler is 3x per kg.
The wholesaler sold 8 kg of peanuts at 10% profit and 16 kg of walnuts at 20% profit to a shopkeeper.
Total cost price to the shopkeeper = (8)(x)(1.1) + 16(3x)(1.2) = 66.4x
The shopkeeper lost 5 kg walnuts and 3 kg peanuts.
The shopkeeper sold the mixture of 11 kg walnuts and 5 kg peanuts.
His total selling price=166(16) = 2656
His total cost price =2656(100125)=2124.8
66.4x=2124.8
x=32
Price at which the wholesaler bought walnuts = 3x = 96/- per kg
