Quantitative Aptitude for All Bank Exam -SBIPO 2015

1.(8.34%of793) - (12.51%of286)= ?
1)33.3756.      2)36.3657.   3)30.3576. 4)28.3675. 

2.48+8*0.75-5=?
1)22.                  2)36.           3)49.             4)56.     

3.6784+2213+844-?=6734+775
1)2332.             2)2323.         3)2344.        4)2442.    

4.11960÷?=65*23
1)6.                     2)11.            3)3.               4)8.             

5.One-fifth of a number is 62. What will 73% of that number be?
1)198.7.             2)212.5.      3)226.3.       4)234.8        
 .

6.. How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per Kg so that there may be a gain of 10 % by selling the mixture at Rs. 9.24 per Kg ?
1). 60 Kg	2). 63 kg
3). 58 Kg	4). 56 Kg

7.A tank is filled in 10 hours by three pipes A, B and C. The pipe C is twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank?
1). 70 hours	2). 30 hours
3). 35 hours	4). 50 hours

Answer
1.(3)?=8.34*793/100- 12.5*286/100
=66.1362-35.7786=30.3576

2.(3)
?=48+8*0.75-5
     =48+6-5=49
3.(2)6784+2213+844-?=6743 +775
=> 9841-?=7518=2323

4(4); 11960/? = 65*23
=>? =11960/65*23 = 8
   
5.(3) Let the number be x
According to the question,
1/5x=62
=>x=62*5 = 310
73%of310 = 73/100*310 =226.3
6.(2)

Selling Price(SP) of 1 Kg mixture= Rs. 9.24

Profit = 10%

Cost Price(CP) of 1 Kg mixture= 100(100+Profit%)×SP=100110×9.24=92.411=Rs.8.4

By the rule of alligation, we have

CP of 1 kg sugar of 1st kind					CP of 1 kg sugar of 2nd kind
Rs. 9					Rs. 7
Mean Price
Rs.8.4
8.4 - 7 = 1.4					9 - 8.4 = .6

ie, to get a cost price of 8.4, the sugars of kind1 and kind2 should be mixed in the

Ratio  = 14 : 6 = 7 : 3

Let x Kg of kind1 sugar is mixed with 27 kg of kind2 sugar

then x : 27 = 7 : 3


⇒x27=73⇒x=27×73=637.(1)

Let the pipe A can fill the tank in x hours

Then pipe B can fill the tank in x2 hours and pipe C can fill the tank in x4 hours

Part filled by pipe A in 1 hour = 1xPart filled by pipe B in 1 hour = 2xPart filled by pipe C in 1 hour = 4xPart filled by pipe A, pipe B and pipe C in 1 hour = 1x+2x+4x=7xi.e., pipe A, pipe B and pipe C can fill the tank in x7 hours

Given that pipe A, pipe B and pipe C can fill the tank in 10 hours
X/7 = 10 =>X 7* = 70

⇒x=10×7=70 hour