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Answers Paper No2

1C  (A+B)-(B+C)+12
A-C=12
A=C+12   C is 12 years younger than A.
2 D (H+G+S+M)4  - M=12
And  H+G=110
After combining these two equations we can not solve further so data is inadequate    
3D 45%of a+b= 1.35b
9a/20+b=27b/20    
9a/20 =  7b/20
9a =7b    
a/b  = 7 /9     7:9
4D Let x =smaller number and  y = larger number
according to question
y/x+12=xy              and x2 = y/2
Put the value of y in first eq
x3 -x-6=0or ( x-2 ) (x2+2x+3)    it gives the value of x+2 other values are imaginary
y=2x2x2 =8
sun of both numbers
2+8=10
5C M-N = 15 ......(.1)  and  A-N=25 ........(2)
Adding eq 1 and 2
M+A-2N=   40
M+A-N  = 40+N
6B  Ratio between 1st and 2nd row    1:2
 Ratio between 2nd and 3rd row       1:3      Make it  2:6
 Now ratio  1st row :  2nd row   : 3rd row     =  1:2:6                          ( 1+2+6  =13)
Total boys  91  .2nd row boys       91x2/13        7 x2=14
7A Let cost price is x and proft/loss is y
x+yx/100=1920..........1       x-xy/100=1280.............2
adding both eqs
2x =  3200  x= 1600 ( in such type condition take directly the  average of both figures for saving time )
SP of article now  =1600+ 25 %of 1600 = 1600+400 =2000  
8D y>z>x       y-x>0         x-y <0 
Option A is  rejected but options Band C may or may not be  possible .Only  option D is possible 
9D Let  share of PQ and R are 5x, 11x and 23x 
According to question
23 x- (5x+11x) = 23x- 16x=  2800             7x=  2800        x= 400
Now difference  between Qand R share   23x-11x= 12 x = 12 x400 =4800
10D  P, Q and R  are having p,q and r balls respectively  and x balls are replaced from box P to Q
Now according to question's conditions 
4p= q           q=32    p=8 
4p=r+x 
32 =r+x we can not  find the balls   because value of x is not given so data is inadequate
11D  Five odd numbers   a1<  a2 <a3< a4 < a5
 (a3+ a4 + a5 )/3    - (  a1+  a2 +a3)/3=10

   (a4+ a5)- (  a1+  a2 )=30                      (a4+ a5) =42 given
     
  
      (  a1+  a2 )=42-30=12  (D)
                                        
12 E     Neither the speed of trains nor the length of trains  is given  .Can't  calculate speed of train P
13D After combining both  statements ,the age of class teacher is 
14x12 -(14-3)x11 = 168 -121= 47  years
14A CP :SP = 100x :120 x    5:6   statement a alone is sufficient but statement b alone is not sufficient 
15C From statement a
rate of interest =  1800x100/(3x6000)  =  10 %
From statement b
rate of interest =(22/20 -1)100             =   10%
16E (E+M+G+S)/4  = 15     
1   ...  E+S=120           2......    G-M=120 
S can not be determined  with  the help of both statement  so Answer is E 
17B

Total boys in row    16+18 -1 = 34-1 =33  (Get alert to subtract 1  in these type questions because 
                                                                     particular person is counted two times one from left
                                                                     and other from right  .Consider  directions  also  very
                                                                      cautiously)
18E Remember for the given parameter  equilateral triangle  area  is maximum .Here  area of  isosceles triangle  bcm2 is lesser than area of  equilateral triangle   acm2  , so that   b<a
19C Weight of jar = w1g
Weight of liquid = w2-w
Weight of liquid(half filled with liquid) = w2-w1
Weight of liquid filled = 2(w2-w1)
Weight of jar + weight of liquid = w1+2(w2-w1)
= 2w2-w1
Hence [C]
20A Let x be the total value of the property.
x = x/2+x/4+1/5x+10000
= (10+5+4)x/20+10000
= 19x/20+10000
= 20*10000
= 200000
21C  . Total money deposited = Rs.12*150 = Rs.1800
Interest on Rs.150 for 12 months = 150*12/12*6*1/100 = 9
Interest on Rs.150 for 11 months = 150*11/12*6*1/100 = 33/4
Interest on Rs.150 for 1 month = 150*1/12*6*1/100 = 3/4
Total interest = 9/1+33/4+3/4 = Rs.18
Total amount due = 1800+18 = Rs.1818
Hence [C]
22D a+b = 27
5a+11b = 195
solving simultaneously we have,
a = 17 b = 10
23A  6 M = 8W, 2W = 3B, 4B = 5G
1 G = Rs. 50 a day.
Now, 1B = 5/4G 1 W = 3/2 * 5/4G
1 M= 8/6* 3/2* 5/4G = 5/2G
1 M = 5/2 * 50 = 125/-Rs.
24B

 If we assume that the sum = Rs. 100
Then, 1st rate of interest = 100*100/100*5 = 20
2nd rate of interest = 200*100/100*12 = 16 2/3
76n1=5, x1= 42.2 n2 = 4, x2 = 35.5
x12 = n1x1+ n2x2 / n1+n2
= 5 * 42.2 + 4 * 35.5 /9
= 211.0 + 142.0 /9
= 353 /9
= 39 2/9
25A

In one hour, C can complete 1/12th of the work.
In two hours, he can complete 1/12*2 = 2/12th of the work.
A and B start work simultaneously,
together work completed by A and B,
1/6+1/8 = 14/48 work = 7/24 work.
In 2 hours they complete 7/24*2 = 7/12 work
Total work done by A,B,C = 7/12+2/12 = 9/12
Total work to be completed = 12/12 or 1
Remaining work = 3/12
Hour       work done by A
1                           1/6                                         

x                         3/12

3/12*6 = 1.5 hours =x
26A  a)Let my cost price be 100
Selling price = 90
Profit for Mr. X = 10
Profit % = 10/90*100 = 11.11%
b) Let cost price be 100
Selling Price = 900
My loss = 10 = 10/100*100 = 10%
27C  p = Rs.455, n = 4months = 1/3 year, rate = 5%
Amount of Rs.100 = (455*1/3*5)/100 = 107.58
455                   x
107.58             100
x = 423

q = Rs.450, n = 2months = 1/6 year, rate = 5%
Amount of Rs.100 = (450*1/6*5)/100 = 103.75
450                      ?
103.75             100
= 434
Sum to be repaid to p = 434-423 = Rs.11
Hence [C]
28A   B also correct   Both options  A & B  are correct 
29D    Option D is correct 
30B  % = 35 /20 x100 = 175 %  
31E  The answer is A and C both
32B   Average of B and C type vehicles in 1999    30+25 /  2 = 27.50   
33B Rainfall  (rf) in Jan in equatorial  1980 mm 
July in temperate  =   1400 mm 
%  = 1980-1400=580      (580 /1400)x100  41.12%
34E  Steep line angle shows that  TZ has maximum increase rainfall    in  July 
35D  % =( 710 /800)x100  = Approximate 90%
36 B Jan rf  = 1980+2060+690+310+890+1160+730/7= 7820/7=1117.14
July rf   =  2280 +1560+250+710+1400+800+240/7=7240/7=1034.28  
(1117.14 -1034.28)x100/1117.14   =( 82.86 /1117.14  )x100 = 7.00%  less   approximate
37B  From question no 36         1117.14 +1034.28 /2    =2151.42 /2  = 1075.71
38A Since a+b+c = 0
         a+b = -c
      ( a+b)2=(-c)2
     a2+2ab+b2=c2
     a2+b2-c2= -2ab
     (a2+b2+c2) 2=(-2ab) 2

     a4+b4+c4+2a2b2-2b2c2-2c2a2=4a2b2


    a4+b4+c4=2a2b2+2b2c2+2c2a2   =2(a2b2+b2c2+c2a2)

     a4+b4+c4/a2b2+b2c2+c2a2
   = 2
39B Horse 1-   C.P             S.P.
                        x               100
                        ?               120    
S.P. = C.P.(100+profit%/100)
x = 1000/12
Horse 2-          C.P              S.P.
                          y               100    
                          ?                 80
y = 125
Total cost price = cost price of Horse1+cost price of Horse2
= 125 + 1000/12 = 2500/12
Total S.P.= 100+100 = 200
loss = 2500/12- 200    = 100/12  
% of loss = (100/12 ) x100    =  4% 
                    (2500/12)
Short cut Method 
Remember  in these type questions where loss on one unit and same percentage of profit on other unit   there is  always loss  and you can calculate % of loss mentally 
Multiply  profit and loss  percentage  and divide it by 100 , you get required percentage of  loss
Here farmer's gain and loss is 20 %          
20 x20 /100 =  4% our required answer 
First we make matrix of  possible  players  based on first sentence spoken by each player was false,
Raman  Sharven Tejinder Urvinderr  Vinod  Wimpy
Raman      x     x    Yes     Yes     Yes     x
Sharven    Yes     x     x     Yes         Yes     x
Tejinder     x    Yes      x     Yes       x    Yes
Urvinder     x    Yes     x        x      Yes    Yes
Vinod    Yes    Yes      Yes       x      x       x
Wimpy     Yes    x    Yes        x      Yes      x

Table based on second sentence was true.

      T T     Bad.   L.T
        R     yes        no       yes
         S     yes       yes        no
         T     yes       no        yes
        U     no       yes        yes
        V     yes       yes        no
       W     no       yes        yes

On the basis of these two tables we can answer  all the  questions very easily 
40B  With whom did Raman play a game?          
(A) Wimpy                                    
(B) Vinod                      
(C) Tejinder                             
(D) Urvinder
Table  1 indicate Raman can play with Sharven ,Vinod And Wimpy ,but table 1again indicates only Vinod  can play with Raman  answer is B   
41A Who played Table Tennis?
(A) Raman and Vinod                                                               
(B) Sharvan and Urvinder          
(C) Tejinder and Wimpy                                                  
(D) Wimpy and Sharvan 
2nd table indicates that   only option A is valid  Table one also support it 
42A Who played Badminton? 
(A) Sharvan and Urvinder                                                                       
(B) Wimpy and Sharvan
(C) Raman and Vinod                                                   
(D) Tejinder and Wimpy 
Options A and B are valid according to table 2 but table one only support option A 
43B

Who played Lawn Tennis?
(A) Sharvan and Urvinder                                                                       
(B) Tejinder and Wimpy 
(C) Raman and Vinod                                                   
(D) Wimpy and Sharvan
Option B is only valid according to table 2. Table one also support it
44A  Let X= 3 ,Y =5 and Z = 2
which of the following is not true

(A) (X-Z)2 Y is even     (3-2)2 5 = 5 false      (B) (X-Z) Y is odd      (3-2)5  =5  odd = true
(C) (X-Z) Y2 is odd      (3--2) 52  =25 true     (D) (X-Y)2 Z is even    (3-5)2 2  = 8  = true
Only A  is false 
45C The red one flashes 3 times every minute and the green one flashes 5 times every two minutes
Time of red   60/3  = 20 seconds    
Time of green 120/5= 24seconds
Take LCM of both  20 and 24        = 120 seconds = 2 Min 
both lights blink in one hour  =    60 / 2 = 30 times
46C  N = 1421 x 1423 x 1425 what is the remainder when N is divided by 12
Divide only 1421 by 12 gives remainder 5  
1423 gives remainder 7 
1425 gives remainder  9
Multiplying all these remainders 
5x7x9  and divide by 12 again         315 /12    Here we get remainder of 3 
It is our required answer 
47D a1 = 1 and an+1 = 2an + 5, for n being a natural number. The value of a100 is
a2 = 2a1 +5  = 2x1+5=7  
a3 =2a2+5 =    2x7+5= 19
a4=2a3 +5  =   2x19+5=38+5 = 43
now  a1 =1 and  options  give a rough  idea that option D is most appropriate answer, convert it into nth term  and check a2 a3 a4  after putting n=1,2,3,4
nth term = 6x2n-1 - 5               Put  n=2      a2 =   6x22-1 - 5    =6x2-5   =7                                                  
                                              Put n=3       a3=   6x23-1 - 5     =24-5 = 19  it is value of a3so that Option D  
                                                                                                              is correct answer you can again
                                                                                                           confirm to it by  a4 if time permits
                                                                                                            to you .

We use this method  to save time only  .It takes few more min. if we solve step by   step                                                                       
48B  x      x       x
x-4  x-4 x-4 
3x -12  =  x   
2x=12 
x=6
total bullets  6x3=18 
49C It is very interesting question .Make three groups of 9 balls in each group of these balls and weigh any two group of them If both are equal ,heavy ball is in the third group . otherwise balance shows  the  heavy one Take heavy  one group and again make three groups of  3  balls in each group   and weigh any two groups of balls  If both are equal third one is heavy otherwise balance shows  the  heavy one .Take heavy group  and weigh  any two balls If both  are equal third one is heavy otherwise balance shows  the  heavy ball By this way we find heavy balls by  weighing only three times.
50C Suppose length  of square's side is x   .Area  =x2 
                                                  Increased area  69%  =1.69 x2    
                                                  side of increased square     = 1.3x 
                                                    % increase                     .3x100/x  = 30%

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