[Download PDF] SBI Associate 2011 PO Paper with Solutions

Instructions

What will come in place of question mark (?) in the following questions:

SBI Associate 2011 PO - Question 101

$$32\cdot05$$% of 259.99=?

Solution

The problem can be solved quickly by rounding off,

32.05% ~32 and 259.99~ 260

Now, 32% of 260 = 83.2

SBI Associate 2011 PO - Question 102

$$\frac{1}{8} of \frac{2}{3} of \frac{3}{5}$$ of 1715=?

Solution

Now, $$\frac{1}{8} of \frac{2}{3} $$= (1/8)*(2/3)=1/12

Also, ( $$\frac{1}{8} of \frac{2}{3}$$ ) of $$\frac{3}{5}$$ = (1/12) * (3/5) = 1/20
Now, (1/20) 0f 1715= (1/20)*1715 = 85

SBI Associate 2011 PO - Question 103

$$25\cdot05\times123\cdot95+388\cdot999\times15\cdot001=?$$

Solution

Here we are approximating 25.05~25, 123.95~124, 388.99~389, 15.001~15.

The question can be rewritten as,

25*123+388*15=8935
Hence the correct option is 8935.

SBI Associate 2011 PO - Question 104

$$561\div35\cdot05\times19\cdot99$$=?

Solution

By approximating 561~560, 19.99~20 and 35.05~ 35 for easier calculation,

the question can be written as,

560*(20/35)=320

SBI Associate 2011 PO - Question 105

$$(15\cdot01)^{2}\times\sqrt{730}=?$$

Solution

The question can be reframed as,

$$15^{2}$$*$$\sqrt(729)$$=$$225\times27$$=607

Instructions

In each of these question a number series is given.In each series only one number is wrong.Find out the wrong numbers

SBI Associate 2011 PO - Question 106

3601, 3602, 1803, 604, 154, 36, 12

Solution

In the given series, nth term is obtained by - (Previous term/(n-1) ) +(n-1).

Hence the 5th term should be (604/4)+4 = 155.

However 5th term is given as 154.
Hence, 154 is wrong term in the given series.

SBI Associate 2011 PO - Question 107

4, 12, 42, 196, 1005, 6066, 42511

Solution

In this question, the nth term is obtained by (previous term*n + n^2).

Hence, after 12, the number would be (12*3+9)= 45

Therefore, the correct option is 45.

SBI Associate 2011 PO - Question 108

2, 8, 12, 20, 30, 42, 56

Solution

Here the consecutive difference between the terms is incremented by 2 for the series.

Hence, after 2, the number should be 6.

SBI Associate 2011 PO - Question 109

32, 16, 24, 65, 210, 945, 5197.5

Solution

Here the sucessive numbers are multiplied by .5,1.5.2.5....

Hence, in place of 65 the number should be 24*2.5=60.

Therefore, the incorrect number for series is 65.

SBI Associate 2011 PO - Question 110

7, 13, 25, 49, 97, 194, 385

Solution

Here, in this series, the nth term is (previous term + 6*2n-2 )

Hence, in place of 194 the term should be (97+6*24) = 193

CAT Content

SBI Associate 2011 PO

SBI Associate 2011 PO

SBI Associate 2011 PO - Question 101

SBI Associate 2011 PO - Question 102

SBI Associate 2011 PO - Question 103

SBI Associate 2011 PO - Question 104

SBI Associate 2011 PO - Question 105

SBI Associate 2011 PO - Question 106

SBI Associate 2011 PO - Question 107

SBI Associate 2011 PO - Question 108

SBI Associate 2011 PO - Question 109

SBI Associate 2011 PO - Question 110

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