Q.1 The value of the sum 7 x 11 + 11 x 15 + 15 x 19 + ...+ 95 x 99 is
A. 80707
B. 80773
C. 80730
D. 80751
Detailed Analysis:
Nth term of the series can be written as
tn=(4n+3)(4n+7)
=16n2+40n+21
Σtn=16Σn2+40Σn+21Σ1
=16n(n+1)(2n+1)6+40n(n+1)2+21n
here n = 23 (7, 11, 15….. 95 is an AP with common different 4 with 23 terms)
∑tn=16×23×24×476+20×23×24+21×23
= 80707
Q.2 How many two-digit numbers, with a non-zero digit in the units place, are there which are more than thrice the number formed by interchanging the positions of its digits?
Detailed Analysis:
Let 'ab' be the two digit number. Where b ≠
0.
On interchanging the digits, the new number will be ‘ba’
As per the condition 10a+b > 3×(10b + a)
7a > 29b
For b = 1, a = {5, 6, 7, 8, 9}
For b = 2, a = {9}
For b = 3, no value of 'a' is possible.
Hence, there are a total of 6 such numbers
Q.3 Gopal borrows Rs. X from Ankit at 8% annual interest. He then adds Rs. Y of his own money and lends Rs. X+Y to Ishan at 10% annual interest. At the end of the year, after returning Ankit’s dues, the net interest retained by Gopal is the same as that accrued to Ankit. On the other hand, had Gopal lent Rs. X+2Y to Ishan at 10%, then the net interest retained by him would have increased by Rs. 150. If all interests are compounded annually, then find the value of X + Y.
A. 3000
B. 4000
C. 5000
D. 6000
Detailed Analysis:
Interest to be repaid to Ankit at the end of the year = 0.08X
Interest that Gopal would receive from Ishan in two cases are as given.
Case I: if he lends X + Y
Interest received = (X + Y) × 0.1 = 0.1X + 0.1Y
Interest retained by Gopal after paying to Ankit
= (0.1X + 0.1Y) – (0.08X) = 0.02X + 0.1Y
Given that Interest retained by Gopal is same as that accrued by Ankit
=> (0.02X + 0.1Y) = 0.08X
=> Y = 0.6X
Case II: if he lends X + 2Y
Interest received = (X + 2Y) × 0.1 = 0.1X + 0.2Y
Interest retained by Gopal after paying to Ankit
= (0.1X + 0.2Y) – (0.08X) = 0.02X + 0.2Y
Given that interest retained by Gopal would increase by 150
=> (0.02X + 0.2Y) – (0.02X + 0.1Y) = 150
0.1Y = 150
=> Y = 1500 and X = 1500×0.6= 2500
Hence X + Y = 2500 + 1500 = 4000
Q.4 On a long stretch of east-west road, A and B are two points such that B is 350 km west of A. One car starts from A and another from B at the same time. If they move towards each other, then they meet after 1 hour. If they both move towards east, then they meet in 7 hrs. The difference between their speeds, in km per hour, is
Detailed Analysis:
Let 'x' and 'y' be the speed (in km/hr) of cars starting from both A and B respectively.
If they both move in east direction, then B will overtake A only if y > x.
Also, relative speed of both the cars when they move in east direction = (y - x) km/hr
It is mentioned that they take 7 hours to meet. i.e. they travel 350 km in 7 hours with a relative speed of (y-x) km/hr.
Hence, (y - x) = 350/7 = 50 km/hr
Q.5 A 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1:3 by volume. This mixture is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution, then the unknown concentration of S is
Detailed Analysis:
Let the volume of the first and the second Solution be 100 and 300.
When they are mixed, quantity of ethanol in the mixture
= (20 + 300S)
Let this Solution be mixed with equal volume i.e. 400 of third Solution in which the strength of ethanol is 20%.
So, the quantity of ethanol in the final Solution
= (20 + 300S + 80) = (300S + 100)
It is given that, 31.25% of 800 = (300S + 100)
or, 300S + 100 = 250
or S = 12
= 50%
Hence, 50 is the correct answer.
Q.6 A water tank has inlets of two types A and B. All inlets of type A when open, bring in water at the same rate. All inlets of type B, when open, bring in water at the same rate. The empty tank is completely filled in 30 minutes if 10 inlets of type A and 45 inlets of type B are open, and in 1 hour if 8 inlets of type A and 18 inlets of type B are open. In how many minutes will the empty tank get completely filled if 7 inlets of type A and 27 inlets of type B are open?
Detailed Analysis:
Let the rate of filling of Type A and Type B pipes be a and b respectively.
Given 30×(10a + 45b) = 1 and 60×(8a + 18b) = 1
=> 30 × (10a + 45b) = 60 × (8a + 18b)
=> 10a + 45b = 16a + 36b
=> 3b = 2a or a = 1.5b
The total work = 30 × (10a + 45b) = 30 × (15b + 45b)
= 1800b
Required answer =1800b7a+27b=1800b10.5b+27b=48
Q.7 A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal takes out 10% of the mixture and substitutes it by water of the same amount. The process is repeated once again. The percentage of water in the mixture is now
A. 35.2
B. 36.3
C. 34.3
D. 32.1
Detailed Analysis:
Final quantity of alcohol in the mixture = 700700+175×(90100)2×[700+175]
= 567 ml
Therefore, final quantity of water in the mixture = 875 - 567 = 308 ml
Hence, the percentage of water in the mixture = 308875×100
= 35.2 %
Q.8 In a tournament, there are 43 junior level and 51 senior level participants. Each pair of juniors play one match. Each pair of seniors play one match. There is no junior versus senior match. The number of girl versus girl matches in junior level is 153, while the number of boy versus boy matches in senior level is 276. The number of matches a boy plays against a girl is
A. 1098
B. 998
C. 1198
D. 898
Detailed Analysis:
Among a group of n persons, number of matches played = n(n – 1)/2
Among the Junior participants, let the number of girls be n.
The number of matches played among girls
= n(n – 1)/2 = 153
=> n(n – 1) = 306 = 18 × 17 => n = 18
Number of boys = 43 – 18 = 25
The number of matches played between a boy and a girl = 25×18 = 450
Among the Senior level participants, let the number of boys be n.
The number of matches played between two boys
= n(n – 1)/2 = 276
=> n(n – 1) = 552 = 24 × 23 => n = 24
The number of girls = 51 – 24 = 27
The number of matches played between a boy and a girl = 27 × 24 = 648
Required answer = 450 + 648 = 1098
Q.9 Let a1, a2, ... , a52 be positive integers such that a1 < a2 < ... < a52. Suppose, their arithmetic mean is one less than the arithmetic mean of a2, a3, ..., a52. If a52 = 100, then the largest possible value of a1 is
Detailed Analysis:
We want to maximize the value of a1, subject to the condition that a1 is the least of the 52 numbers and that the average of 51 numbers (excluding a1) is 1 less than the average of all the 52 numbers. Since a52 is 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. a27 be A.
(Note: The average of an odd number of terms in an Arithmetic Progression is equal to the value of the middle-most term)
Since a52 = a27 + 25 and a52 = 100
=> A = 100 – 25 = 75
a2 + a3 + … + a52 = 75×51 = 3825
Given a1 + a2 +… + a52 = 52(A – 1) = 3848
Hence a1 = 3848 – 3825 = 23
Q.10 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
Detailed Analysis:
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.
