IBPS 2012 CWE for PO/MT

Quantitative Aptitude

1)4003 * 77 - 21015 = ? * 116

a)2477

b)2478

c)2467

d)2476

e)none of these

Sol> Suppose unknown value is x

then  

4003 * 77 - 21015 = x * 116

=> (4003 * 77 -21015 )/116  = x

=>(308231 -21015 )/116  = x

=> (287216)/116 =x

=>x=2476

2)[( 5 √7 + √7 ) * (4√7 + 8√7)] - (19)^2  = ?

a)143

b)72√7

c)134

d)70√7

e)none of these

Sol>

 [ (5 √7 + √7 ) * (4√7 + 8√7)] - (19)^2

=[6√7  * 12 √7 ] -  361

= 72 √7 √7  - 361

=72 * 7 - 361

=504 - 361

=143

3)(4444 / 40) + (645 / 25) + (3991 / 26)  = ?

a)280.4

b)290.4

c)295.4

d)285.4

e)none of these

Sol> 

(4444 / 40) + (645 / 25) + (3991 / 26)

=>(1111/10) + ( 129/5) + (307 / 2 )

=>(1111 + 258 + 1535) /10

=>2904/10

=>209.4

4)√33124 * √2601 - (83)^2  =  (?)^2   + (37) ^ 2

a)37

b)33

c)34

d)28

e)none of these

Sol> Suppose unknown value is x

 √33124 * √2601 - (83)^2  =  (x)^2   + (37) ^ 2

=>182  * 51 - 6889  =    (x)^2   + (37) ^ 2

=> 9282 - 6889 = (x)^2   + (37) ^ 2

=> 2393 = (x)^2   + (37) ^ 2

=> 2393 =(x)^2  +1369

=>1024 =(x)^2 

=> x=√1024

=>x=32

5) 5  17/37  * 4  51/52     * 11 1/7  + 2  3/4

a)303.75

b)305.75

c)303  3/4

d)305  1/4

e)none of these

Sol>

 5  17/37  * 4  51/52     * 11 1/7  + 2  3/4

= 202/37  *  259 /52  *    78 /7   +  11/4

=303  + 11/4

=305.75

 6 -10 What approximate value come in place of ? in the following questions

(Not need of exact value)

6) 8737 / 343  * √50 = ?

a)250

b)140

c)180

d)100

d)280

Sol>

8737 / 343  * √50

=>

8737 / 343 = 25.4  (nearly)

√50  =7.0 (nearly)

  

so 8737 / 343  * √50 =177.8 (nearly)

so ans from above options is 180

7) 54821^ 1/3  * (303/8) =?^2

a)48

b)38

c)28

d)18

e)58

Sol>

54821^ 1/3 = 37.  (nearly)

303/8 = 37.8 (nearly)

Suppose unknown value is x

54821^ 1/3  * (303/8) =x^2

=> 37. * 37.8  =x^2

so ans is 37. nearly 38

so ans is 38

8) 5/8 of 4011.33 + 7/10 of 3411.22 = ?

a)4810

b)4980

c)4890

d)4930

e)4850

Sol> 

 2507.0  + 2387.8

=4894

So ans nearly 4890

9)23% of 6783 + 57 % 8431 = ?

a)6460

b)6420

c)6320

d)6630

e)6360

Sol>

1560.0 + 4805.6  (nearly)

= 6365  (nearly)

so nearest answer from above options is

6360

10) 335.01 * 244.99 / 55 = ?

a)1490

b)1550

c)1420

d)1590

e)1400

Sol>

 335.01 * 244.99 / 55

=335.01  * 4.4 (nearly)

 =1492  (nearly)

so nearest answer from above options is

 1490

 in questions a number series is given.In each series only one number is wrong.Find out wrong number.

11)5531  5506  5425  5304  5135  4910  4621

a)5531

b)5435

c)4621

d)5135

e)5506

Sol>

5531  - 5506 = 25     square root is 5

5506  - 5425  =81     square root is  9

5425  -  5304  =121     square root is 11

5304   - 5135  =169     square root is 13

5135   - 4910  =225     square root is 15

4910   -4621 =289     square root is 17

difference of square roots are same is 2 except first two terms

so the first term is wrong in series

that is 5531

12)A certain amount was to be distributed among A,B, and C in the ratio 2:3:4 respectively, but was erroneously distributed in the ratio 7:2:5 respectively. As a result of this, B got Rs.40 less, What is the amount?

a)Rs 210/-

b)Rs.270/-

c)Rs.230/-

d)Rs.280/-

e) None of these

Sol>

Suppose certain amount is x

then True distribution is (Respectively A,B,C)

(2/9)x , (3/9)x  , (4/9)x 

and  erroneously distribution is (Respectively A,B,C)

(7/ 14)x , (2/ 14)x, (5/ 14)x

 Now, B got Rs.40 less (given)

so  

(3/9)x -  (2/ 14)x = 40

=>(42x - 18x)/126   = 40

=>24x  = 40 * 126

=>x=210

So certain amount is Rs. 210/-

13 )Rachita enters a shop to buy ice-creams ,cookies and pastries. She has to buy at least 9 unit of each. She buys more cookies than ice-creams and more pastries than cookies. She picks up a total of 32 items.

How many cookies does she buy?

a)Either 12 or 13

b) Either 11 or 12

c)Either10 or 11

d)Either 9 or 11

e)Either 9 or10

Sol>

For solving this question, I recommend you to go through options

here,

 No of Pastries >  No of Cookies  >  No of Ice creams

and has to  buy at least 9 unit of each. 

So suppose Ice creams for 9 units, then No of cookies must  >9 so option d) and e) is not possible

Now Suppose    No of Cookies   is         Either10 or 11  then     No of Pastries must >11 , so its smallest value becomes 13

then sum of three = 9 +11+ 12= 32 , which is correct

(If we take cookies for Either 11 or 12 then Pastries>12 then sum is=9+12+13=34 not correct

If we take cookies for Either 12 or 13 then Pastries>13 then sum is=9+13+14=36 not correct)

14)The product of three consecutive even numbers is 4032. The product of the first and the third number is 252.What is the five times the second number?

a)80

b)100

c)60

d)70

e)90

Sol>

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

The product of the first and the third number is 252  so

(x-2)(x+2)=252

=>x^2 - 4 =252

=>x^2 = 256

=> x=√256

=>x=16

which is second number,

five times of it is =16*5 = 80

15)The sum of the ages of 4 members of a family 5 years ago was 94 years . Today, when the daughter has been married off and replaced by a daughter in law , the sum of their ages is 92 .Assuming that there has been no other chance in the family structure and all the people are alive, what is the difference in the age of the daughter and the daughter in law?

a)22 years

b)11 years

c)25years

d)19years

e)15years

Sol>

The sum of the ages of 4 members of a family 5 years ago was 94 years 

so their sum of recent ages = 94 + 4(5) 

                                         =114

Now the daughter has been married off and replaced by a daughter in law,the sum of their ages is 92

so 

sum of 4 members ages -  age of daughter + daughter in law 's age = 92

=>114 -  age of daughter + daughter in law 's age = 92

=>114-92 = age of daughter - daughter in law 's age 

=> 22= age of daughter - daughter in law 's age 

So,the difference in the age of the daughter and the daughter in law = 22 years

16) Akash scored 73 marks in subject A. He scored 56% marks in subject B and X marks in subject C. Maximum marks in each subject were 150.The overall percentage marks obtained by Akash in all the three subjects together were 54%. How many mark did he score in subject C?

a)84

b)86

c)79

d)73

e)none of these

Sol>

Mark for subject A=73

Mark for subject B=56% of 150 =150*56 / 100 =84

Mark for subject C=X

Now  overall percentage marks = 54%

so 

(73 + 84 +X ) /450  *  100  =54

=>(73 + 84 +X )=243

=>73 + 84 +X=243

=> 157+X=243

=>X=243-157

=>X=86

Marks scored in C=86

17) The area of a square is 1444 square meters. The breadth of a rectangle is 1/4 th the side of the square and the length of the rectangle is thrice the breadth. What is the difference between the area of the square and the area of the rectangle?

a)1552.38 sq.meter

b)1169.33 sq.meter

c)1181.21 sq.meter

d)1173.25 sq.meter

e)None of these

Sol>

The area of a square is 1444 square meters

Now area of square =(side)^2

=>1444= (side)^2

=> √1444=side

=>side=38

Now breadth of a rectangle is 1/4 th the side of the square 

So breadth of a rectangle = 1/4 *38

 length of the rectangle is thrice the breadth = 3 (1/4 *38)

So Area of the rectangle = length * breadth 

                                     = 3 (1/4 *38)  * ( 1/4 *38)

                                     = 270.75 square meters

Now,the difference between the area of the square and the area of the rectangle 

=1444  -  270. 75

=1173.25 square meters

.

Concept of Inequality

Readers, Our Today's topic is   Concept of Inequality. It is important as Inequality related questions asked  in bank exams repeatedly.

First of all , when two values are equal ,

                  like  a=b

            then it is the concept of Equality .

When two values are not equal 

                 like  a ≠b

          then it is the concept of inequality

 ->  In inequality, there is two possibilities ,

     either a > b  or a < b

How to Solve Inequality Equation - Rules

A) we can add / subtract same number on both sides

like a-10>5

then adding 10 both side , give  result as a>15

B) we can switched both sides , for that we change the orientation of  inequality sign.

like a-10 > b-5 then b-5 < a-10

C) we can multiply/divide  by same ,  positive numbers on both sides

like 2a > 4 , divide by 2 both sides give result as  a > 2

D) we can multiply/divide  by same ,  negative  numbers on both sides , that time orientation of  inequality sign.must change.

like  -5a > 10 , divide both sides by -5 give result as a < -2

Important Rules(Summary)

1)  a > b and a>0 and b>0  then 1/a  <  1/b

2)  a>b and a<0 and b<0   then 1/a < 1/b   (same result as above as both are <0)

3) a>b  then a-b>0 and a=b +p   where p>0

4) a<b then a-b<0  and a=b - p where p>0

5) a>b  and k>0  then a+k>b+k

6) a>b and k>0 then ka > kb and a/k > b/k

7)a>b and k<0 then ka < kb and a/k < b/k

8) a>b then -a < -b

9) a<b and (x-a)(x-b) < 0 then a<x<b

10) a<b and (x-a)(x-b) <= 0 then a<= x <= b

11)  a<b and (x-a)(x-b) > 0 then x<a or x>b

12) a<b and (x-a)(x-b) >= 0 then x<= a or x>= b

Thanks .

  

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