Quant Mix Quiz-2 for IBPS PO 2016-17

1.A jar contains a mixture of two liquids A and B in the ratio 3:1 . When 15 ltrs of this mixture is drawn off and replaced with 9 ltrs of B,then ratio between A and B becomes 3:4. How many ltrs of liq. A was initially present in the jar ?

Sol. Let assume A = 3x ltrs and B = x ltrs.

Now 15 ltrs is drawn and 9 ltrs of B is added,

A = 3x - (3/4)*15 = (12x-45)/4

B = x - (1/4)*15 + 9 = (4x+21)/4

Now this ratio is equal to 3:4 , thus,

3/4 = (12x-45)/(4x+21)

x = 6.75

Thus A = 3x = 3*6.75 = 20.25 ltrs


2. Two equal glasses filled with mixture of alcohol and water in the ratio 2:1 and 1:1 respectively emptied in third glass. Find the ratio of alcohol and water in the third glass ?

Sol. For 1st glass,

Lets assume, alcohol = 2x and water = x

For 2nd glass,

Alcohol = y and water = y

Now as both glasses are equal, thus,

3x =2y
x = 2y/3

Now for 3rd glass,

Alcohol = 2x + y  and water = x + y

Ratio of this will be,

(2x+y)/(x+y)

Put x=2y/3 and solve,

=7/5

Thus alcohol:water is 7:5.

Directions( 3-7):Find the missing numbers in the given series

3.12, 19, 35, 59, 90, _

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

Now,
     16-7 = 9
     24-16= 8
     31-24= 7

     x-31 = 6
     x = 37

NOW, 90 + 37 = 127


4. 4, 10, 27, 112, 555, _

Sol. 4*2 + 2 = 10
     10*3 - 3 = 27
     27*4 + 4 = 112
     112*5 - 5 = 555

   = 555*6 + 6 = 3336  


5. 19, 23, 14, 30, 5, _

Sol. 19 + 2^2 = 23
     23 - 3^2 = 14
     14 + 4^2 = 30
     30 - 5^2 = 5

     5 + 6^2 = 41


6. 5, 12, 33, 136, 675, _

Sol. 5*2 + 2 = 12
     12*3 - 3 = 33
     33*4 + 4 = 136
     136*5 - 5 = 675

     675*6 + 6 = 4056


7. 14, 6, 4, 4, 8, _

A. 17
B. 12
C. 16
D. 15

Sol. 14*0.5 - 1 = 6
         6*0.5 + 1 = 4
         4*1.5 - 2 = 4
         4*1.5 + 2 = 8
   
     8*2.5 - 3 = 17

8. A seller mark the price 50% above the cost price and give 10% discount on an item.While selling he cheats customer by giving 20% less in weight .Find his overall profit percent(approximate)? 

A. 70%
B. 65%
C. 68.75%
D. 64.8%
E. None of these

Sol. Let cost price be Rs 100
Then marked price will be Rs 150

He gives 10% discount, so,
= 0.9*150 = Rs 135

Now this will be our selling price = Rs 135

As he cheats and gives 20% less in weight,

Cost Price becomes = 0.8*100 = Rs 80

Profit % = [(135-80)/80]*100 = 68.75 %.

9. There are 81 liters pure milk in a container. One-third of milk is replaced by water in the container. Again one-third of mixture is extracted and equal amount of water is added. What is the ratio of milk to water in the new mixture?

A. 4:5
B. 3:5
C. 3:4
D. 1:2
E. None of these

Sol. Amount of milk left after two withdrawls,

= 81*[(81-27)/81]^2
= 36 ltrs

Amount of water left = (81-36) = 45 ltrs

Ratio milk:water ,

= 36/45

= 4:5.

10. A rectangle is circumscribed by a circle. Ratio of length and breadth of the rectangle is 4:3, then the ratio of circumference of the circle to perimeter of the rectangle is ?

A. 127/140
B. 137/140
C. 147/140
D. 157/140
E. None of these

Sol. Given,

L/B = 4/3

B= 3L/4

Also,

L^2 + B^2 = (2R)^2           (where R is the radius of the circle)

L^2 + (3L/4)^2 = (2R)^2

2R = 5L/4

Now ratio of circumference:perimeter wil be,

= (2R*3.14)/2(L + 3L/4)

= (5L*3.14/4)/2(7L/4)

= 157/140

= 157:140

11. Find the probability of getting a multiple of 2 or 6 from numbers [1,2,3,.......,200] ?

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

Sol. As 6 is a multiple of 2 thus all the numbers multiple of 6 are already included in the set of multiples of 2.

No. of multples of 2 = 100
Total numbers = 200

Probability = 100/200 = 1/2.

12. The average marks in Science subject of a class of 20 students is 68. If the marks of two students were misread as 48 and 65 of the actual marks 72 and 61 respectively, then what would be the correct average?

(a) 68.5
(b) 69
(c) 69.5
(d) 70
(e) 66

Sol. Total marks in science/20 = 68

Total marks in science = 68*20

Now net correction in tota marks is  = (72+61)-(48+65) = 20

Correct total marks = 68*20 + 20

Thus correct avg will be,

= (68*20 + 20)/20

= 69 marks.

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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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