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QUANTITAITVE APTITUDE CLUBBED MCQs
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Question 71. Ahmed and Sahil set out together on bicycle traveling at 15 and 12 kilometers per hour, respectively. After 40 minutes, Ahmed stops to fix a flat tire. If it takes Ahmed one hour to fix the flat tire and Sahil continues to ride during this time, how many hours will it take Ahmed to catch up to Sahil assuming he resumes his ride at 15 kilometers per hour? (consider Ahmed's deceleration/acceleration before/after the flat to be negligible)
Answer: Option B. -> 3.33
:
B
A travels 10 km in 40 min
B travels 8 km in 40 min
After one hour, A would have still traveled only 10 km and B would have traveled 20 km
Representing this on a timeline:
Time in hours123455:20DistanceA101025405560DistanceB82032445660
A gains this 10 km in 103=3.33 hours.
Answer is Option (b).
:
B
A travels 10 km in 40 min
B travels 8 km in 40 min
After one hour, A would have still traveled only 10 km and B would have traveled 20 km
Representing this on a timeline:
Time in hours123455:20DistanceA101025405560DistanceB82032445660
A gains this 10 km in 103=3.33 hours.
Answer is Option (b).
Answer: Option A. -> Row 17
:
A
Perfect squares are the middle term of odd rows. (17)2=289 should be middle term of 17th row. Option (a).
Alternate Solution:
There is 1 odd number in the 1st row.
There are 2 odd numbers in the 2nd row.
There are 3 odd numbers in the 3rd row. ...
Hence this can be considered as a∑n question where, n is the number of odd numbers.
As 289 will be 2892=144145th term in the sequence, find a∑n value which is close to the number. which is equal to 153 and 17 numbers before 153 will be in the 17th row. Hence the answer is 17.
:
A
Perfect squares are the middle term of odd rows. (17)2=289 should be middle term of 17th row. Option (a).
Alternate Solution:
There is 1 odd number in the 1st row.
There are 2 odd numbers in the 2nd row.
There are 3 odd numbers in the 3rd row. ...
Hence this can be considered as a∑n question where, n is the number of odd numbers.
As 289 will be 2892=144145th term in the sequence, find a∑n value which is close to the number. which is equal to 153 and 17 numbers before 153 will be in the 17th row. Hence the answer is 17.
Answer: Option C. -> 103
:
C
After 6 all multiples of 3 would be covered. After 50+6 all 3x+2 tiles would be covered. After 50+50+6 all 3x+1 tiles would be covered. So for 106 and above all tile combinations are possible. 105 again is possible (it is a multiple of 3), 104 is also possible as it is of the form 3x+2. The largest possible value is 103 which can't be obtained.
:
C
After 6 all multiples of 3 would be covered. After 50+6 all 3x+2 tiles would be covered. After 50+50+6 all 3x+1 tiles would be covered. So for 106 and above all tile combinations are possible. 105 again is possible (it is a multiple of 3), 104 is also possible as it is of the form 3x+2. The largest possible value is 103 which can't be obtained.
Answer: Option A. -> 11:21
:
A
Let x (in ml) be the volume of each jug.
Jug1st2nd3rd4thS:W1:33:55:117:9S14x38x516x716xW34x58x1116x916x
(14x+38x+516x+716x)(34x+58x+1116x+916)=(2216)(4216)=2242=1121
:
A
Let x (in ml) be the volume of each jug.
Jug1st2nd3rd4thS:W1:33:55:117:9S14x38x516x716xW34x58x1116x916x
(14x+38x+516x+716x)(34x+58x+1116x+916)=(2216)(4216)=2242=1121
:
Power of x is 1 and that of y is 2 in xy2.
So, we split the term with y into two.
2x+2y+y=12
He know that, AM≥ GM.
2x+2y+y3≥(2x×2y×y)13
Answer: Option D. -> (27×2)13
:
D
AM=(2a1+4a2+8a3)3 and GM =(2a1×4a2×8a3)(13)
Using condition AM≥GM, we get (2a1+4a2+8a3)3≥(2a1×4a2×8a3)(13)
2a1+4a2+8a3≥3(2a1×22a2×23a3)(13)
≥3(2a1+2a2+3a3)(13)
But a1+2a2+3a3=1
Answer =(27×2)13
:
D
AM=(2a1+4a2+8a3)3 and GM =(2a1×4a2×8a3)(13)
Using condition AM≥GM, we get (2a1+4a2+8a3)3≥(2a1×4a2×8a3)(13)
2a1+4a2+8a3≥3(2a1×22a2×23a3)(13)
≥3(2a1+2a2+3a3)(13)
But a1+2a2+3a3=1
Answer =(27×2)13
Question 77. On June 1, 2004 two clubs, S1 and S2, are formed, each with n members. Everyday S1 adds b members while S2 multiplies its current number of members by a constant factor r. Both the societies have the same number of members on June 7, 2004 (end of day). If b =10.5n, what is the value of r?
Answer: Option B. -> 2
:
B
S1 club members increase in an arithmetic progression.
Hence on June 7th,the number of members in club S1 is n+6(10.5n)=64n.
S2 club members increase in an geometric progression.
Hence on June 7th ,the number of members in club S2 is nr6=64n....(As given in the question).
Therefore r=2.
:
B
S1 club members increase in an arithmetic progression.
Hence on June 7th,the number of members in club S1 is n+6(10.5n)=64n.
S2 club members increase in an geometric progression.
Hence on June 7th ,the number of members in club S2 is nr6=64n....(As given in the question).
Therefore r=2.
Answer: Option B. -> nk
:
B
Assumption
Assume n=2
Then
g(2) = g(1) +g(1) = 2g(1)= 2k
substitute n=2 in answer options to eliminate wrong options.
Hence, answer is option (b)
:
B
Assumption
Assume n=2
Then
g(2) = g(1) +g(1) = 2g(1)= 2k
substitute n=2 in answer options to eliminate wrong options.
Hence, answer is option (b)
Answer: Option B. -> n2+2n(n+1)2
:
B
At n=1, we should get 34 (this is the sum to 1 term of the series)
Option (a) gives =24=12
Option (b) gives 34
Option (c) gives38
Option(d) gives 8
Answer= option b
:
B
At n=1, we should get 34 (this is the sum to 1 term of the series)
Option (a) gives =24=12
Option (b) gives 34
Option (c) gives38
Option(d) gives 8
Answer= option b