Friday, 19 April 2013

SBI Clerk Written Exam-2012 Solution

 Quantitative Aptitude


1)   3.6+36.6+3.66+.36+3.0=?
   a)44.22      b)77.22
   c)74.22      d)47.22  e) none of these

Sol>It is just simple addition of given numbers , and answer will be 47.22

2)   23*45 / 15 =?
   a)69           b)65
   c)63           d)71     e)none of these

Sol> 23*45 / 15
       =1035 / 15
       =  69

3)   4(5/6)+7(1/2)-5(8/11)
   
   a)2(10/33)   b)6(20/33)                
   c)2(20/33)   d)6(10/33)      e)none of these
 
Sol> 4(5/6)   +   7(1/2)   -   5(8/11)
       = (29/6)  +  (15/2)   -    (63/11)
       =(319 + 495 - 378) / 66
       =(436) / 66
       =218/33
       =6(20/33)
                        


4)    (210/14) * (17/15) * ? =4046
  a)202         b)218
  c)233         d)227     e)none of these

Sol>  suppose unknown term is x

 so  (210/14) * (17/15) * x = 4046
    => x=(4046 * 14 *15 )  /  (210 * 17)
    => x=238

  So ans will be None Of These
    

5)   83% of 2350= ?
 a)1509.5        b)1950.5
 c)1905.5        d)1590.5    e)none of these

Sol>  83% of 2350
       =2350 * (83/100)
       = 195050/100
       =1950.5

  So ans will be 1950.5


6)   √1089  +3  =  (?)^2
a)5     b)6
c)3     d)8   e)4

Sol>Suppose unknown value is x
   
so  √1089  +3  =  (x)^2

=>33 + 3 = (x)^2
=> 36 = (x)^2
=> √36  = x
=> 6=x

so x= 6 is the answer

7)   96+32*5-31  = ?
a)223    b)225
c)229    d)221   e)none of these

 Sol>
 Suppose unknown value is x
 96+32*5-31  = x
=> 96 + 160 - 31 =x
=>96 + 129  =x
=>x= 225

8)  ? /  36 =(7)^2 - 8
a)1426    b)1449
c)1463    d)1476   e)none of these

Sol>
Suppose unknown value is x

 x / 36 =(7)^2 - 8
=> x / 36  =49 - 8
=> x/36 = 41
=> x=41*36
=>x=1476


9)√8281 =  ?

a)89    b)97
c)93    d)91   e)83

 Sol>
Suppose unknown value is x 

so   √8281 = x
=>  √13*13*7*7  =x
=>  13* 7 =x
=> x= 91


10) (63)^2  -  (12)^2  =?

a) 3528    b)3852
c)3582     d)3825    e)none of these

Sol>
Suppose unknown value is x 
  (63)^2  -  (12)^2  =x
=>  3969  -  144 =x
=>  x= 3825


11) 1(4/5)  +  3(3/5)  =  ?  -  4(3/10)
a) 9 (7/10)    b)7(7/10)
c)9(3/10)      d)7(9/10)  e)none of these


Sol>
Suppose unknown value is x 

1(4/5)  +  3(3/5)  = x - 4(3/10)
=>  (9/5)  +   (18/5)  = x - (43/10)
=>  (9+18)  / 5   = x - (43/10)
=> 27/5   = x - (43/10)
=>  (27/5) + (43/10)  =  x
=> (54 + 43)  /  10  =x
=> 97/10  =x  
=> x=  9 (7/10)


12)17  *  19  *4  /  ?  =161.5 

a)8      b)6
c)7      d)9    e)none of these

sol>  
Suppose unknown value is x 

17  *  19  *4  /  x  =161.5
=> 1292 / x   =161.5 
=>  1292 /  161.5  =x
=>  x=8


13) 1798  /  31  *  ?  =348
a)3     b)6
c)4     d)5   e)none of these

Sol>
Suppose unknown value is x 

so 1798  /  31  *  x  =348

=>  58  *  x=348
=>  x=  348/58
=> x=6


14) (9.8 * 2.3 + 4.46 ) /  3  =(3) ^ ?

a)3    b)9
c)5    d)2   e)none of these

 Sol> 

Suppose unknown value is x 

so  (9.8 * 2.3  + 4.46 ) /  3  =(3) ^ x

=> (27) / 3  =(3) ^ x
=> 9 =(3) ^ x
=>(3)^2  =(3) ^ x
=> so x=2


15) 43% of 600    +  (?)%   of 300  =399

a)45   b)41
c)42   d)47    e)none of these

 Sol> 
Suppose unknown value is x 

 so 43% of 600    +  (x)%   of 300  =399
=>  600* (43 /100 )   +  300 * (x/100)  =399
=>  258  + 3x  =  399
=>  3x = 141
=>  x = 47


 16)What will be the compound interest on a sum of Rs.7500/-  at 4 p.c.p.a  in 2 years?

a) 618/-    b)612/-
c)624/-     d)606/-   e)621/-

Sol>

    A=P+I   where A=Amount, P= Principal Amount , I =Compound Interest

    and A=P(1 + R/100)^N

    =>I= A - P
    =>I=P(1 + R/100)^N  -   P
    =>I=7500(1+ 4/100)^2   -  7500
    =>I=7500(1+0.04)^2   -   7500
    =>I=7500(1.04)^2   -7500
    =>I=8112  -  7500
    =>I=612


So Compound Interest IS Rs. 612/-

17)In how many different ways can the letters of the word 'CREAM' can be arranged?

a)720    b)240
c)360    d)504   e) e)none of these


Sol>

CREAM has 5 letters in word,
So different ways for arranged are = 5P5 =5! =5*4*3*2*1 =120


18)The Circumference of a circle is 792 meters.What will be its radius?
a)120 meters  b)133meters
c)145meters   d) 136meters  e)none of these

Sol>  Circumference of circle is = 2 ∏ r , where r=radius of the circle
                                 =>792 = 2 ∏ r
                                 =>792 = 2*(22/7)*r
                                 => r=(792*7)/(22*2)
                                 =>r  =126 meters.


19)Cost of 36 pens and 42 pencils is Rs 460/- ,what is the cost of 18 pens and 21 pencils?

a)Rs 230/-   b)Rs 203/-
c)Rs 302/-   d)Rs320/-    e)none of these

Sol>   Suppose x is for pen  and y is for pencils

           then  Cost of 36 pens and 42 pencils is Rs 460/-
                  => 36x + 42y =460
                  =>2(18x + 21y) =460
                  =>  18x + 21 y =460/2  =230               - - - -  - - (1)
               
            Now  the cost of 18 pens and 21 pencils?
                 =>18x+21y=?

          From (1)  Ans is Rs 230/-

20)The ratio of ages of A and B seven years ago was 3:4 respectively.The ratio of their ages nine years from now will be 7:8 respectively. What is the B's age at present?

a)16 years   b)19  years
c)28 years   d)23 years    e)none of these

Sol>

Suppose Today age of A is x years and age of B is y years

before 7 years (From today)
  =>A's Age is (x-7)years     and    B's Age is (y-7)years and ratio of their ages 3:4

=>  (x-7)/(y-7)  =  3/4
=>  4x - 28 =3y - 21
=> 4x-3y=7                         - - - - - (1)


Now after 9 years (From today) 
=> A's Age is (x+9)years     and    B's Age is (y+9)years and ratio of their ages 7:8

=>(x+9) / (y+9)  = 7/8
=> 8x + 72 = 7y +63
=> 8x - 7y = -9                  - - - -    (2)


Now from equation (1)  and  equation (2)  solving

    4x - 3y =7   - - - - -    x2
    8x-7y= -9


    8x-6y=14
    8x-7y= -9
 ------------------
 -      +     +
 ------------------

 y=23 years


where y is B's Age at present so it is the answer


21) In how many years will Rs.4600 amount to Rs.5428 at 3 p.c.p.a simple interest?

a)3    b)5
c)6    d)4     e)none of these

Sol>


Formula for interest is
I=PRN /100 ,where P=Principal Amount , R=Rate Of Simple Interest, N=years

                           I =simple interest


and  A=P+I   , where A=Amount


here, A= Rs.5428  , P=Rs.4600

So, I=A-P=5428-4600=828

Now   I=PRN /100
     =>828 = (4600 * 3 * N ) /100
     =>828*100 / 4600 *3  =  N
     => N=6 years


22)What will be the average of the following set of scores?
     59,84,44,98,30,40,58

  a)62   b)66
  c)75    d)52  e)59

Sol> Average =Sum of all n numbers/ n , here n=7

     =>Average=(59+84+44+98+30+40+58)/7
     =>Average=413/7
     => Average=59

23) The sum of three consecutive odd numbers is 1383.what is the largest number?

 a)463   b)459
 c)457    d)461  e)none of these

Sol>

Suppose three consecutive odd numbers are x-2,x,x+2

Their sums are 1383
so  x-2+x+x+2=1383

  =>3x=1383
  =>x=461

Numbers are x-2=459 , x=461 , x+2=463

So the largest number is 463



24-26 Study the information given below and answer the questions that follow:

An article was bought for Rs.5600. Its price was marked up by 12%. Thereafter it was sold at a discount of 5% on the marked price.


24)what was the marked price of the article?
a)Rs 6207  b)Rs.6242
c)Rs.6292  d)Rs.6192  e)Rs.6272

Sol> article marking up by 12%,
    so the marked price is = 5600 + 12% of 5600
                                   = 5600 + ( 5600*12/100 )
                                   =5600 +672
                                   =6272
So the answer is Rs.6272


25)What was the percent profit on the transaction?
a)6.8%    b)6.3%
c)6.4%    d)6.6%  e)6.2%

Sol>

here the marked price is Rs.6272 from above solution

Now 5% discount given on marked price

so new price is=marked price - (5% of marked price)
                    =6272 - (5% of 6272)
                    =6272 - (6272 * 5/100)
                    =6272 - (313.6)
                    =5958.4

So profit (in %)=(gain / cost price )*100
                     =(358.4/5600 )* 100    , because gain=5958.4 - 5600=258.4
                     =6.4%


26)What was the amount of discount given?
a)Rs.319.6     b)Rs.303.6
c)RS.306.3     d)RS.313.6   e)Rs.316.9


Sol>  discount = 5% of marked price
                     = 5% of 6272
                     = 313.6

  discount=Rs 313.6

27)The area of the rectangle is 1209 square meters. Its length measures 39 meters.How much is its perimeter?

a)122 meter   b)134  meter  
c)148 meter   d)144 meter   e)none of these


Sol> Area of rectangle=length * width

here length=39 meter , area=1209 square meters

so              1209 =39 * width
             =>width= 1209/39
             =>width=31 meter

Now Perimeter of rectangle= 2 (length + width)
                                     =2(39+31)
                                     =2(70)
                                     =140meter

So ans is  none of these

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