Complete solution of IBPS PO Prelims Paper 2016 - 22nd October Shift (Based on memory)

Directions (1-5): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer

(1) if x > y
(2) if x >=y
(3) if x < y
(4) if x <=y
(5) if x = y or relationship between x and y cannot be established.

1. 1) 8x^2 -10x +3 = 0
    2) 7y^2 -17y +6 = 0

Sol. For 1st eqn values are,
    x = 0.5, 0.75
     For 2nd equation values are,
    y = 2, 3/7

Thus, no relation between the two.

2. 1) 4x^2 -13x +3 = 0 
    2) 2y^2 -13y +21 = 0

Sol. For 1st eqn values are,
    x = 3, 1/4
     For 2nd equation values are,
    y = 3, 7/2

Thus, y >= x.

3. 1) 5x^2 -32x +12 = 0
    2) y^2 -15y +56 = 0

Sol. For 1st eqn values are,
    x = 6, 2/5
     For 2nd equation values are,
    y = 8, 7

Thus, y > x.

4. 1) 2x^2 +9x +10 = 0
    2) 5y^2 +19y +18 = 0

Sol. For 1st eqn values are,
    x = -2, -5/2
     For 2nd equation values are,
    y = -2, -9/5

Thus, y >= x.

5. 1) x^2 +11x +28 = 0
    2) y^2 +16y +64 = 0

Sol. For 1st eqn values are,
    x = -7, -4
     For 2nd equation values are,
    y = -8

Thus, x > y.

6. An item sold at 10% discount gives 17% profit, what would be the percentage profit if discount increased to 15% ?        

A. 12%
B. 11%
C. 10%
D. 10.5%
E. None of these

Sol. Lets assume C.P. to be 100
Then S.P. will be 117 (as profit 17%)

Now, 117 = (90/100)*M.P.   (As discount is 10%)
     M.P.= 130

When discount 15%, S.P. = (85/100)*130 = 110.5
Thus, profit % = 10.5 %.

7. What is the area of the square whose side is equal to the side of an equilateral triangle with area 9*(3)^0.5 ?

A. 16
B. 36
C. 18
D. 24
E. None of these

Sol. Lets say the side of the square  = side of the equilateral triangle be 'a'.
Now, area of the triangle is given 9*(3)^0.5,thus,
[a*a*(3)^0.5]/2 = 9*(3)^0.5
a = 6

Thus area of the square,
= a^2
= (6)^2
= 36

8. 6 yrs ago, ratio of age of Annu and Nalini was 7:4. The age of Annu 4 yrs hence is equal to age of Nalini 10 yrs hence. Then find the sum of ages of both 2 yrs from now ?

A. 38
B. 36
C. 34
D. 32
E. None of these

Sol. Lets assume Annu's age to be 'x'
     And Nalini's age to be 'y'

The 1st eqn. will be,,
(x-6)/(y-6) = 7/4
The 2nd eqn. will be,
x + 4 = y + 10

Solving both, we get,

x = 20
y = 14

Thus, 20 + 14 + 4 = 38.

9. 15 women can complete 10% of the work in 7 days, 25 women worked for 31.5 days then what fraction of the work is remaining ?

A. 1/5
B. 1/6
C. 1/4
D. 1/3
E. None of these

Sol. 25 women complete the work in = (15*70)/25 = 42 days

25 women per day work = 1/42
25 women 31.5 day work = 31.5/42

Remaining work = (1 - 31.5/42) = 1/4.

10. There are 32 cards in a bag numbered from 1 to 32. 2 cards are drawn one by one without replacement. Find the probability that both are even numbered cards.

A. 1/16
B. 15/62
C. 16/62
D. 15/31
E. None of these

Sol. Required Probability = (16/32)*(15/31) = 15/62

11. A and B enetered into a partnership with the investments of 25500 and 15000. At the end of 1 year B invested 9000 rs more. The total profit earned at the end of 2 yrs is 45,000 then find the difference between the shares of A and B.

A. 5000 RS
B. 6000 RS
C. 7000 RS
D. 8000 RS
E. None of these

Sol. (25500*2):(15000 + 24000)
= 17:13
Required Difference = (4/30)*45000 = Rs 6000.

12. A certain amount of money is invested in a scheme offering 12% simple per annum. After 2 yrs the total amount obtained from this scheme is invested for 5 yrs in another scheme offering 20% simple per annum. If the total amount obtained is 29,760. Then find the sum of money invested initially.

A. 10000
B. 12000
C. 13000
D. 15000
E. None of these

Sol. Let initial sum = 100
100*(124/100)*[(20*5)+100]/100 = 248 rs
For 248 rs total amount = Rs 29760
Thus, for 100 rs total amount will be = 12000 rs.

13. Two trains of length 200m each running in the same direction cross each other in 80 sec. If the speed of faster traiin is 72 km/hr, then find the speed of the slower train.

A. 36 km/hr
B. 45 km/hr
C. 54 km/hr
D. 57 km/hr
E. None of these

Sol. Relative speed = 400/9 = 5m/s
     Speed of faster train = 72*(5/18) = 20m/s
     Speed of slower train = 20 - 5 = 15m/s
     or 15*(18/5) = 54 km/h.

14. Out of three numbers, the first is twice the second and is half of the third. If the avg. of 3 numbers is 56, the three numbers in order are:

A. 48,96,24
B. 48,24,96
C. 96,24,48
D. 96,48,24
E. None of these

Sol. Sum of three numbers = 56*3 = 168
     Let 2nd no. = x
     1st no. = 2x
     3rd no. = 4x
 x + 2x + 4x = 168
The no.s are 48 , 24, 96.

15. A sells a horse to B for Rs 4860, thereby losing 19%, B sells it to C at a price which would have given A 17% profit. Find B's gain.

A. Rs 2160
B. Rs 2610
C. Rs 1260
D. Rs 2260
E. None of these

Sol. C.P. for A = 4860*[100/(100-19)] = 6000
     S.P. with 17% profit for A = 6000*[(100+17)/100] = 7020

     B's gain = 7020 - 4860 = 2160.

Directions(16-20):Find the missing numbers in the given series

16. 11, 14, 23, 50, _

A. 130
B. 132
C. 131
D. 133
E. None of these

Sol. 11 + 3 = 14
     14 + 3^2=23
     23 + 3^3=50

Thus, 50 + 3^4 = 131 

17. 19, 25, 42, 71, 113, _

A. 149
B. 156
C. 169
D. 175
E. None of these

Sol. The difference of the difference is in the A.P.
 = 113 + 56
 = 169.

18. 35, 30, 44, 39, _

A. 60
B. 55
C. 54
D. 53
E. None of these

Sol. 21 + 14 = 35
     35 - 5  = 30
     30 + 14 = 44
     44 - 5  = 39

Thus, 39 + 14 = 53    

19. 23, 39, 32, 48, 41, _

A. 56
B. 57
C. 55
D. 54
E. 52

Sol. 23 + 16 = 39
     39 - 7  = 32
     32 + 16 = 48
     48 - 7  = 41

Thus, 41 + 16 = 57.

20. 13, 17, 33, 97, _

A. 243
B. 341
C. 353
D. 281
E. None of these

Sol. 13 + 4 = 17
     17 + 4^2=33
     33 + 4^3=97



Thus,  97 + 4^4 = 353.  

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Complete solution of IBPS PO Prelims Paper 2016 - 16th October Shift (Based on memory)

Directions (1-5): Find the missing numbers in the given series :

1. 7, 16, 45, 184, 915, _

A. 4560
B. 5640
C. 5496
D. 5696
E. None of these

Sol. 7*2 + 2 = 16
    16*3 - 3 = 45
    45*4 + 4 = 184
   184*5 - 5 = 915

   915*6 + 6 = 5496

2. 11, 20, 38, 74, _

A. 156
B. 164
C. 146
D. 142
E. None of these

Sol. 20 - 11 = 9*1
     38 - 20 = 9*2
     74 - 38 = 9*4

     X - 74 = 9*8
     X = 72 + 74 = 146  

3. 15, 21, 38, 65, 101, _

A. 146
B. 150
C. 145
D. 164
E. None of these

Sol. 21 - 15 = 6
     38 - 21 = 17
     65 - 38 = 27
     101 - 65 = 36

 Now subtracting these differences,
     17 - 6 = 11
     27 - 17 = 10
     36 - 27 = 9

     x - 36 = 8
     x = 44

Now adding this to 101 we will get te next term,

= 101 + 44 = 145

4. 24, 28, 19, 35, 10, _

A. 45
B. 59
C. 46
D. 64
E. None of these

Sol. 24 + 2^2 = 28
     28 - 3^2 = 19
     19 + 4^2 = 35
     35 - 5^2 = 10

     10 + 6^2 = 46

5. 12, 19, 35, 59, 90, _

A. 117
B. 127
C. 147
D. 107
E. None of these

Sol. 19 - 12 = 7
     35 - 19 = 16
     59 - 35 = 24
     90 - 59 = 31

Now subtracting the differences,
     16 - 7 = 9
     24 - 16 = 8
     31 - 24 = 7

     x - 31 = 6
     x = 37

Adding this to 90, we get the next term,

= 90 + 37 = 127

6. Find the probability that a number from 1 to 300 is divisible by 3 or 7 ?

A.37/75
B.32/75
C.36/75
D.28/75
E.26/75

Sol. Multiples of 3 = 100
       Multiples of 7 = 42
       Multiples of both 3 and 7 or the multiple of 21 = 14

Total no. of favorable cases = (100 + 42 - 14) = 128

Total no. of cases = 300

Thus, Probability = 128/300 = 32/75.

7. 14 men can do a work in 18 days ,15 women can do a work in 24 days. If 14 men work for first three days and 10 women work after that for three days find the part of work left after that?

A.3/4
B.1/4
C.1/2
D.1/6
E.1/5

Sol. In 1 day 14 men will do = (1/18)th work
       In 3 days 14 men will do = (1/18)*3 = (1/6)th work.

In 1 day 15 women will do = (1/24)th work
In 3 days 15 women will do = (1/24)*3 = (1/8)th work
In 3 days 10 women will do =(1/8)*(10/15) = (1/12)th work

Thus work left will be,

= 1-(1/6 + 1/12)
=3/4.

8. Perimeter of a rectangle is x and circumference of a circle is 8 more than the perimeter of the rectangle. Ratio of radius of circle and length of the rectangle is 1:2 and ratio of length and breadth of rectangle is 7:3. Find the length of the rectangle?

A. 14
B. 21
C. 28
D. 35
E. 7

Sol. Given,

2*(l+b) = x
where l = length
And b = breadth

Also given,

2*3.14*r = x + 8         ( circumference of the circle = 2*3.14*r)
Where r = radius of the circle

2*3.14*r = 2*(l+b) + 8

Also given,
l/b = 7/3
b = 3*l/7
also,
r/l = 1/2
r = l/2

Using both these values,

2*3.14*l/2 = 2*(l + 3l/7) + 8

Solving we get,

Length of the rectangle = l = 28.

9. A invest on some scheme at 5% and B at 3% for two year. If the total sum invested by A and B is 4000 and the simple interest received by both is same then find the amount invested by A ?

A.1300
B.1500
C.2500
D.2700
E.2100

Sol. Lets assume amount invested by A is 'x'
Then amount invested by B will be (4000-x).

S.I. received by A = (p*5*2)/100
S.I. received by B = ((4000-p)*3*2)/100

As given both interests are same, thus,

(p*5*2)/100 = ((4000-p)*3*2)/100

Solving,

p = Rs 1,500.

10. Two trains crosses each other in 14 sec when they are moving in opposite direction, and when they are moving in same direction they crosses each other in 3 minute 2 sec. Find the speed of the faster train by what percent more than the speed of the slower train?

A.16.67%
B.17.33%
C.16.33%
D.17.67%
E.18.33%

Sol. Lets assume speed of faster train be 'a' m/sec.
And assume speed of slower train be 'b' m/sec and its length to be 'x' metres.

Now, keeping slower train stationary, relative speed of faster train will be (when both in opp. direction) (a+b) m/sec and distance to cover will be 'x' metres.

14 = x/(a+b)

Similarly when both in same direction, relative speed of faster train will be (a-b) m/sec and distance to cover will be 'x' metres.

182 = x/(a-b)                 where (3min 2sec = 182 seconds)


Now dividing the two equations,

182/14 = (a+b)/(a-b)

Solving,

a/b = 7/6

Now % by which speed of faster train is more w.r.t. slower train is,

= [(a-b)/b]*100

Simplifying this,

= (a/b - 1)*100

Thus putting the value in the eqn.

= (7/6 - 1)*100

= 16.67 %.

Directions (11-15): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer

(1) if x > y
(2) if x >=y
(3) if x < y
(4) if x <=y
(5) if x = y or relationship between x and y cannot be established.

11 I. 3x^2 - 22x + 7 = 0
     II.y^2 - 15y + 56 = 0

Sol. Solving quadratic 1 and 2,

    x= 1/3, 7
    y= 7,8

Thus, x<=y

12 I. 2x^2 - 17x + 36 = 0
   II. 2y^2 - 19y + 44 = 0

Sol. Solving quadratic 1 and 2,

    x= 9/2, 4
    y= 11/2, 4

Thus, x<=y

13.I. x - 169^0.5 = 0
   II.y^2 - 169 = 0

Sol. Solving quadratic 1 and 2,

    x= 13
    y=-13, 13

Thus, x>=y

14 I. 3x^2 + 20x + 25 = 0
     II. 3y^2 + 14y + 8 = 0

Sol. Solving quadratic 1 and 2,

    x= -5/3, -5
    y= -2/3, -4

Thus, no relation can be established.

15 I. 3x^2 + 5x + 2 = 0
     II. 3y^2 + 18y + 24 = 0

Sol. Solving quadratic 1 and 2,

    x= -2/3, -1
    y= -2,-4

Thus, x>y.

Directions (16-20): What should come in the place of question mark (?) in the questions given below, 

16. 40% of 265 + 35% of 180 = 50% of ?+ ?% of 80

Sol. Solving this,

x = 130
 
(a) 80 
(b) 95.5
(c) 130 
(d) 125.5 
(e) 115

17. (0.25×0.16)^0.5 of 15/7 = ?

Sol. Solving this,

x=0.43

(a) 0.43
(b) 0.76
(c) 0.91
(d) 0.20
(e) 0.62

18.?/529=324/?

Sol. Solving this,

x= 414

(a) 404
(b) 408
(c) 410
(d) 414
(e) 416

19.(682% of 782) ÷ 856 =?

Sol. Solving this,

x = 6.25

(a) 4.50
(b) 10.65
(c) 2.55
(d) 8.75
(e) 6.25

20. 15.5% of 850 + 24.8% of 650 = ?

Sol. Solving this,

x = 295

(a) 295
(b) 330
(c) 270
(d) 375
(e) 220

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