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Quantitative Aptitude--new TCS pattern - useful for banks also.

Thanks to Maniraj, campus preparation.

  1. Which of the following numbers must be added to 5678 to give a remainder of 35 when divided by 460?
a)      980                b) 797                   c) 955                   d) 618
Solution:
797,
If you need of remainder 35, while dividing a number by 460. The Unit place of that number must be 5. So 5678+797= 6475
2. Raj writes a number of two digits exceeds four times the sum of its digits by 3. If the number increased by 18, the result is the same as the number formed by reversing the digits. Find the number?
a)      35
b)      42
c)       49
d)      57
Solution:
Equation we got from the question is
10X+Y =4(X+Y) +3 è 2X-Y=1
10X+Y+18 = 10Y+X èx-y=-2
Solving above two equations we get, X = 3 and Y = 5.
So the number is 35.
3. 28a + 30b + 31c = 365 .Find a, b, c and a + b + c?
a)      1, 4, 7
b)      7, 2, 3
c)       1, 5, 5
d)      8, 7, 6
Solution:
It’s like sum of the number of days in a leap year.
“a” represents the number of months having 28 days
“b” represents the number of months having 30 days
“c” represents the number of months having 31 days
So
a =1 (February),
 b=4(April, June, September, November)
c=7(January, March, May, July, August, October, December)
 4. How many motor vehicles registration number plates can be formed with the digits 1, 2, 3, 4, and 5(no digits being repeated) if it is given that registration number can have 1 to 5 digits?
a)      205
b)      100
c)       325
d)      105
Solution:
For one digit: 5 = 5 chances
For two digit: 5*4 = 20 chances
For three digit: 5*4*3 = 60 chances
For four digit: 5*4*3*2 = 120 chances
For five digit: 5*4*3*2*1 = 120 chances
Total chances = 5+20+60+120+120=325 chances
 5. A city in the U S has a basketball league with three basketball teams, the Arêtes, the Braves and the Deities. A sports writer notices that the tallest player of the Arêtes is Shortest than the shortest player of the Braves. The shortest of the Deities is shortest of the Arêtes, while the tallest of the braves is taller than the tallest of the Arêtes. Which of the following can be judged with certainty?
X) Paul, a brave is taller than David, an Arêtes.
Y) David, a Deities, is shorter than Edward, an Arêtes.
a) X only
b) Both X and Y
c) Neither X and Y
d) Y only
6.  A rectangle of height 100 squares and width 200 squares. Squares is drawn on a graph paper. It is colored square by square from top left corner and moving across in a spiral turning right. Whenever a side of this rectangle or a colored square is reached. Which square is colored last (give its row and column numbers). The bottom right square is on row 100, column 200?
a)      51,150
b)      51, 50
c)       50, 150
d)      50, 50
Solution:
1,1==>==>==>==>==>==>==>
==>==>==>==>==>==>==>||
||3,2<==<==<==<==<==||
<==<==<==<==<==<==<==<==

For (4, 8) rectangle, ends at (3, 2)
Likewise,
For (2, 4) rectangle, ends at (2, 1)
For (3, 6) rectangle, ends at (2, 5)
For (4, 8) rectangle, ends at (3, 2)
For (5, 10) rectangle, ends at (3, 8)
For (6, 12) rectangle, ends at (4, 3)
For (7, 14) rectangle, ends at (4, 11)
For (8, 16) rectangle, ends at (5, 4)
For (9, 18) rectangle, ends at (5, 14)
For (10, 20) rectangle, ends at (6, 5)
Analyzing the above 10 points:
We can get some idea that,
For Even number of rows, End points column increases 1 by previous column.
For Odd number of rows, End points column difference of the present column and the previous even number column.
For End point row, the value repeats two times from row 2.
In our problem, (100, 200) i.e.
Even number row,
So End point column value must be 50 for 200 columns and End point row value must be 51 for 100 rows.
For even number of rows and columns,
Condition: Column =2*Row
Endpoint (R, C) = {((Row/2) +1), Column/4}
Answer is (51, 50)
7.  George and Mark can paint 720 boxes in 20 days. Mark and Harry in 24 days and Harry and George in 15 days. George works for 4 days, Mark for 8 days and Harry for 8 days. The total number of boxes painted by them is
a)      252
b)      516
c)       348
d)      492
Solution:
One day work:
G+M = (1/20)*720 = 36 boxes;
 M+H= (1/24)*720 = 30 boxes;
 H+G = (1/15)*720 = 48 boxes.
Together:
  2(G+M+H) = 36+30+48 = 114
(G+30) = 114 /2=57 boxes
G= 27 boxes.
G‘s one day work is 27 boxes. So G’s four days work is 108 boxes
M+H‘s one day work is 30 boxes. So M+H‘s eight days work is 240 boxes.
So Total number of boxes painted = 108+240 = 348 boxes painted.
 8. Masha drives along the perimeter of the square field of side 10kms. She drive along the first side at 10 kmph along the second side at 20 kmph, along the third side at 30 kmph and along the fourth side at 40 kmph. Her average speed is?
a)      19.2 kmph
b)      18 kmph
c)       20 kmph
d)      30 kmph
Solution:
Total Average Speed = Total Distance / Total time = 40 / (1+0.5+0.33+0.25) =19.2kmph
9.  A merchant buys 20 kg of wheat of Rs.30 per kg and40 kg wheat at Rs.25 per kg. He mixed them and sells one third of the mixture at Rs.26 per kg. The price at which the merchant should sell the remaining mixture, So that he may earn a profit of 25 % in his whole outlay is
a)      Rs.30
b)      Rs.360
c)       Rs.40
d)      Rs.37
Solution:
Cost price =20*30 + 40*25 = Rs.1600
Profit = 25 % of (1600) = Rs.400
Selling price = Rs.2000
One third of mixture is (1/3)*60 = 20 kg
Remaining mixture is 40 kg
2000 = 26*20 + 40*X;
X = Rs.37
 10) The mean of three number is 10 more than the least of these number and 15 less than the greatest of the Three. If the median of the three number is 5, then sum of the three is
a)      30
b)      20
c)       5
d)      25
Solution:
X is the greatest value, Y be the Middle value and Z is the least value.
(X+Y+Z)/3 = 10+Z = X-15;
If Y=5, X=25, Z=0. So the sum of the three number is 30.
Quantitative Aptitude--new TCS pattern - useful for banks also. Reviewed by sambasivan srinivasan on 12:30:00 AM Rating: 5

2 comments:

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Linda Roverson said...

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