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QUANTITAITVE APTITUDE CLUBBED MCQs

Total Questions : 1394 | Page 8 of 140 pages
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)
  1.    4.5
  2.    3.33
  3.    3.5
  4.    4
 Discuss Question
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).
Question 72. If this pattern continues, where would the number 289 appear?.
If This Pattern Continues, Where Would The Number 289 Appear...
  1.    Row 17
  2.    Row 18
  3.    Row 19
  4.    Row 20
 Discuss Question
Answer: Option A. -> Row 17
:
A
If This Pattern Continues, Where Would The Number 289 Appear...
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 an question where, n is the number of odd numbers.
As 289 will be 2892=144145th term in the sequence, find an 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.
Question 73. Tiles are packed into packets of 6, 9 and 50. What is the largest number of tiles that cannot be purchased with any combination of above packets. Tiles cannot be sold loose.
  1.    449
  2.    451
  3.    103
  4.    49
 Discuss Question
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.
Question 74. Four equal jugs are filled with the mixture of sprit and water. The ratio of sprit to water in four jugs is 1:3, 3:5, 5:11 and 7:9. The mixture of the four jugs is emptied into a single vessel. What is the proportion of sprit to water in the vessel?
  1.    11:21
  2.    11:30 
  3.    12:24
  4.    3:5
 Discuss Question
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
Question 75. If 2x+3y=12; x, y>0, then the maximum possible value of xy2 is ___.
 Discuss Question

:
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
Question 76. If a1+2a2+3a3=1, where a1,a2&a3 are all positive numbers; then what is the minimum value of 2a1+4a2+8a3?
  1.    413
  2.    (3×2)13
  3.    213    
  4.    (27×2)13
 Discuss Question
Answer: Option D. -> (27×2)13
:
D
AM=(2a1+4a2+8a3)3 and GM =(2a1×4a2×8a3)(13)
Using condition AMGM, we get (2a1+4a2+8a3)3(2a1×4a2×8a3)(13)
2a1+4a2+8a33(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?
  1.    3
  2.    2
  3.    1.5
  4.    Cannot be determined
 Discuss Question
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.
Question 78. Let g be a function satisfying g(a+b)= g(a)+g(b) for all a,b which is real. If 
g(1)=k, then find g(n), where n is a natural number
  1.    kn
  2.    nk
  3.    nk
  4.    1n
 Discuss Question
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)
Question 79. Find the sum to n terms of the series 312×22+522×33+732×42...... is
 
  1.    n2+1(n+1)2
  2.    n2+2n(n+1)2
  3.    n2+2(n+1)3
  4.    (n+1)3
 Discuss Question
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
Question 80. The milk and water in two vessels A and B are in the ratio 4:5 and 7:2 respectively. In what ratio the liquids in both the vessels be mixed to obtain a new mixture in vessel C consisting half milk and half water?
  1.    2: 1
  2.    5: 1
  3.    4: 1
  4.    7: 2
 Discuss Question
Answer: Option B. -> 5: 1
:
B
Suppose in the vessels A & B, one litre of liquid is present.
So, Amount of milk in vessel A =49 ; Amount of milk in vessel B = 79
The required mixture in vessel C should be 12 milk.
So, let the ratio be x: 1. Applying alligation:
The Milk And Water In Two Vessels A And B Are In The Ratio 4...
(518)(118)=x:1x=5.

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