CAT 2018 – Slot 2 – Quantitative Ability – Let a(base1), a(base2), … , a(base52) be positive integers such that a(base1) < a(base2) < … < a(base52)
Q. 1: Let a(base1), a(base2), … , a(base52) be positive integers such that a(base1) < a(base2) < … < a(base52). Suppose, their arithmetic mean is one less than the arithmetic mean of a(base2), a(base3), …, a(base52). If a(base52) = 100, then the largest possible value of a(base1) is?
1. 20
2. 23
3. 48
4. 45
We want to maximize the value of a1, subject to the condition that
(a2+ a3 +⋯….+ a52 )/51 -(a1+ a2 +⋯….+ a52 )/52 = 1
Since a(base52) = 100 and all the numbers are positive integers, maximizing a1 entails maximizing a2, a3 ….a51. The only way to do this is to assume that a2, a3…. A52 are in an AP with a common difference of 1.
Let the average of a2, a3, …, a52 i.e.
(a2+ a3 +⋯….+ a52 )/51= a27 =A ( using the average of an odd number of terms in an Arithmetic Progression is equal to the value of the middle-most term)
So a52 = a27 + 25*1 = a27 + 25 and
given a52 = 100
=> a27 = A = 100 – 25 = 75
a2 + a3 + … + a52 = 75×51 = 3825
Given a1 + a2 +… + a52 = 52(A – 1) = 3848
Hence a1 = 3848 – 3825 = 23
Ans : 23
Q. 2: Points A and B are 150 km apart. Cars 1 and 2 travel from A to B, but car 2 starts from A when car 1 is already 20 km away from A. Each car travels at a speed of 100 kmph for the first 50 km, at 50 kmph for the next 50 km, and at 25 kmph for the last 50 km. The distance, in km, between car 2 and B when car 1 reaches B is
Time taken to cover first 50 km at 100 km/hr = 1/2 hr.
Time taken to cover second 50 km at 50 km/hr = 1 hr. Time taken to cover last 50 km at 25 km/hr = 2 hr.
When car 2 starts, car 1 has already covered 20 km.
So, time taken by car 1 to reach B after car 1 starts = total time – time required to travel first 20 km
= 3 hr 30 min – 12 min = 3 hr 18 min
Distance travelled by car 1 = (50 + 50 + 45) = 145 km Distance from B = (150 – 145) km = 5 km
Hence, 5 is the correct answer.
Q. 3: The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u=(x+y)/2 and v=(y+z)/2. If x ≥ z, then the minimum possible value of x is
Given, (x+y+z)/3=80
x+y+z=240——1)
And (x+y+z+u+v)/4=75
x+y+z+u+v =375——2)
From eq 1) & eq 2) u+v =135
And from question , u+v =(x+2y+z)/2
Or 270=x+2y+z
y = 30 and x+z = 210
as x ≥ z
so x will be minimum if x =z
minimum value of x =210/2 = 105
Q. 4: If the sum of squares of two numbers is 97, then which one of the following cannot be their product?
1. −32
2. 48
3. 64
4. 16
Solution: let numbers are a and b.
Given, a^2+b^2=97
Using A.M.≥G.M
(a^2+b^2)/2 ≥ (a^2 b^2 )^(1/2)=ab
ab ≤ 97/2 = 48.5
Thus product of numbers can not be greater than 48.5
Option 3 64 is correct
Q. 5: For two sets A and B, let AΔB denote the set of elements which belong to A or B but not both. If P = {1,2,3,4}, Q = {2,3,5,6,}, R = {1,3,7,8,9}, S = {2,4,9,10}, then the number of elements in (PΔQ)Δ(RΔS) is
1. 9
2. 7
3. 6
4. 8
P = {1,2,3,4} and Q = {2,3,5,6,} PΔQ = {1, 4, 5, 6}
R = {1,3,7,8,9} and S = {2,4,9,10}
RΔS = {1, 2, 3, 4, 7, 8, 10}
(PΔQ)Δ(RΔS) = {2, 3, 5, 6, 7, 8, 10}
Thus, there are 7 elements in (PΔQ)Δ(RΔS) . hence, 7 is the correct answer.
Checkout Other Questions of CAT 2018 Slot 2 Paper:
Verbal Ability : | Q.01- Q.05 | Q.06- Q.9 | Q.10- Q.14 | Q.15- Q.19 | Q.20- Q.24 | Q.25- Q.29 | Q.30 – Q.34 |
Logical Reasoning : | Q.01- Q.04 | Q.05- Q.08 | Q.09- Q.12 | Q.13- Q.16 | Q.17- Q.20 | Q.21- Q.24 | Q.25 – Q.28 | Q.29 – Q.32 |
Quantitative Aptitude: | Q.01- Q.05 | Q.06- Q.10 | Q.11- Q.15 | Q.16- Q.20 | Q.21- Q.25 | Q.26- Q.30 | Q.31 – Q.34 |
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