IBPS PO MAINS 2016 - Quant Quiz-11 (Mixed)

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

1. 4, 16, 26, 34, 40, _

A. 44
B. 42
C. 40
D. 46
E. None of these

Sol. 4 + 12 = 16
    16 + 10 = 26
    26 + 8  = 34
    34 + 6  = 40
    40 + 4  = 44

2.  825, 582, 501, 474, 465, _

A. 475
B. 465
C. 462
D. 483
E. None of these

Sol. 825 - 582 = 243
     582 - 501 = 81
     501 - 474 = 27
     474 - 465 = 9
     465 - 462 = 3

     243/3 = 81
     81/3  = 27
     27/3  = 9
      9/3  = 3

3. 8, 12, 24, 60, 180, _

A. 620
B. 600
C. 610
D. 630
E. None of these

Sol. 8 + 8*0.5 = 12
    12 + 12*1  = 24
    24 + 24*1.5= 60
    60 + 60*2  = 180
   180 + 180*2.5=630

4. 3, 4, 10, 33, 136, _

A. 675
B. 665
C. 685
D. 655
E. None of these

Sol. 3*1 + 1 = 4
     4*2 + 2 = 10
    10*3 + 3 = 33
    33*4 + 4 = 136
   136*5 + 5 = 685

5. 16, 17, 13, 22, 6, _

A. 25
B. 15
C. 31
D. 17
E. None of these

Sol. 17 - 16 = 1
     13 - 17 = -4
     22 - 13 = 9
      6 - 22 = -16
     31 - 6  = 25

6. The front and rear wheels of an off-track vehicle have radii of 2 feet and 5 feet respectively. What is the distance travelled, if the front wheel completes 15 revolutions more than the rear wheel ?

A. 50π feet
B. 64π feet
C. 72π feet
D. 100π feet
E. None of these

Sol. Suppose the front wheel completes n revolutions
Then, the rear wheel completes (n – 15) revolutions.
The distance travelled by the front wheel = 4nπ
And by rear wheel (n – 15) × 10π = 10nπ – 150π.
Since these distances are equal, we get 4nπ = 10nπ – 150π, which on solving yields n = 25.
Thus the distance travelled is 4nπ = 100π feet.

7. The 5000 litres tank of a two-storied apartment complex is 1/5th full. The apartment on the 1st floor consumes 120 litres of water every hour while the apartment on the 2nd floor consumes 180 litres of water every hour. The caretaker turns on the pump that lets in 45 litres of water every minute into the tank. How long will it take to fill the tank ?

A. 115 min
B. 100 min
C. 84 min
D. 68 min
E. None of these

Sol. Suppose the tank is filled in h hours.
In these h hours, the apartment on the 1st floor would have consumed 120h litres of water.
And the apartment on the 2nd floor would have consumed 180h litres of water.
The pump can pour in 45 litres of water every minute, i.e., 2700 litres of water every hour.
So, the total amount of water that has to be let into the tank is 4000 + 120h + 180h = 4000 + 300h litres.
Equating this to 2700h, we get,
4000 + 300h = 2700h, which yields h = 5/3 hours or 100 minutes.

8. Population of a town increases by 15.9% every decade. The urban population increases by 18%, whereas the rural population increases by 4%. What is the ratio of the urban population to the rural population ?

A. 17:3
B. 11:25
C. 17:11
D. 7:11
E. None of these

Sol. Suppose the urban population is U and the rural population is R.
Then, 18U + 4R = 15.9(U + R).
So, (18 – 15.9)U = (15.9 – 4)R, which yields,
U : R = 11.9 : 2.1 = 17 : 3.

9. A Math question paper has 10 questions, each with sub-questions (a) and (b). If a student can solve any sub-question from any five questions, in how many ways can she do so ?

A. 8064
B. 1024
C. 15,504
D. 59,049
E. None of these

Sol. The student can choose 5 out of the 10 questions in 10C5 = 252 ways.
Each of these 5 questions can be answered in 2 ways.
Thus the total number of ways is 10C5 × 2^5 = 252 × 32 = 8064 ways.

10. The height of a right circular cone is decreased by 20%. By what percent should the diameter of the base be decreased so that the volume of the right circular cone is decreased by 35.2%?

A. 6.66%
B. 9%
C. 10%
D. 11.11%
E. None of these

Sol. Suppose the original radius and height are 10 each.
Since (1/3)π is a constant, we can simple work with (r^2)*h and say that the original volume is 1000.
The new height is 8 and the new volume is 648.
If the new radius is R, then R^2 = 648/8 = 81 ⇒ R = 9.
Thus the radius and therefore the diameter is reduced by 10%.

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IBPS PO MAINS 2016 - Quant Quiz-10 (Mixed)

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

1. 2, 3, 7, 16, 32, _

A. 82
B. 67
C. 72
D. 69
E. None of these

Sol. 2 + 1^2 = 3
     3 + 2^2 = 7
     7 + 3^2 = 16
    16 + 4^2 = 32
    32 + 5^2 = 57

2. 6, 4, 5, 11, 39, _

A. 170
B. 185
C. 165
D. 189
E. None of these

Sol. 6*1 - 2 = 4
     4*2 - 3 = 5
     5*3 - 4 = 11
    11*4 - 5 = 39
    39*5 - 6 = 189

3. 2, 4, 10, 22, 42, _

A. 65
B. 64
C. 70
D. 72
E. 76

Sol. 2 + 2 = 4
     4 + 6 = 10
    10 + 12= 22
    22 + 20= 42
    42 + 30= 72

     6 - 2 = 4
    12 - 6 = 6
    20 - 12= 8
    30 - 20= 10

4. 18, 9 , 9, 18, 72, _

A. 624
B. 556
C. 486
D. 576
E. None of these 

Sol. 18*0.5 = 9
     9*1 = 9
     9*2 = 18
    18*4 = 72
    72*8 = 576

5. 25, 35, 49, 67, 89, _

A. 115
B. 120
C. 105
D. 125
E. None of these

Sol. 35 - 25 = 10
     49 - 35 = 14
     67 - 49 = 18
     89 - 67 = 22

     14 - 10 = 4
     18 - 14 = 4
     22 - 18 = 4  

6. A bag contains 4 black balls, 3 red balls and 5 green balls. 2 balls are drawn from the box at random. What is the probability that both the balls are of the same colour?

A. 47/68
B. 1/6
C. 19/66
D. 2/11
E. None of these

Sol. Total no. of balls = 4+3+5 = 12
     Ways of selecting two balls from 12 = 12c2 = 12!/10!2! = 66
     Ways of selecting two balls of same colour = 4c2 + 3c2 + 5c2 = 19

Thus, the required probability = 19/66.

7. A bag contains 5 green, 4 yellow and 3 white marbles. 3 marbles are drawn at random. What is the probability that they are not of the same colour?

A. 13/44
B. 41/44
C. 13/55
D. 152/55
E. None of these

Sol. No. of ways of selecting 3 balls out of 12 = 12c3 = 220
     No. of ways of selecting 3 balls of same colour = 5c3 + 4c3 + 3c3 = 15 ways.

Thus, no. of ways of selecting 3 balls of not same colour = 220 - 15 = 205 ways.
Required probability = 205/220 = 41/44.

8. A committee of 4 is to be formed from among 4 girls and 5 boys. What is the probability that the committee will have number of boys less than number of girls?

A. 1/4
B. 1/5
C. 1/6
D. 1/7
E. None of these

Sol. Selection of 1 boy and 3 girls = 5c1*4c3 = 20
     Selection of 0 boy and 4 girls = 5c0*4c4 = 1*1 = 1

Total ways = 20 + 1 = 21 ways.
     Selecting 4 members from 9 people = 9c4 = 126 ways
Thus, required probability = 21/126 = 1/6.

9. A bag contains 3 red, 6 blue,2 green and 4 yellow balls. If two balls are picked randomly then the probability that either both are red or both are green is 

A. 3/5   
B. 4/105   
C. 2/7
D. 5/91   
E. None of these

Sol. Total ways of selecting 2 balls from 15 = 15c2 = 105
     Ways of selecting 2 red balls = 3c2 = 3
     Ways of selecting 2 green balls = 2c2 = 1
Thus, required probability = (3+1)/105 = 4/105.

10. Out of 15 students studying in a class, 7 are from M.P., 5 from Kerala and 3 from Gujarat. Four students are to be selected at random. What are the chances that at least one is from Kerala ?

A. 12/13
B. 11/13
C. 100/15
D. 51/15
E. None of these

Sol. P(atleast one from kerala) = 1 - P(no one is from kerala)
                                = 1 - 10c4/15c4 = 1 - 2/13 = 11/13. 

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IBPS RRB 2016-17 (Questions asked in 4th Nov. shift)

Directions(1-5): In each question two equations are given, find x and y and give the answer:

A. x > y
B. x < y
C. x >= y
D. x =< y
E. x = y or relation can not be established.

1. (1) 8x^2 - 106x + 323 = 0
   (2) 12y^2 - 109y + 247 = 0

Answer. x = 17/2 , 19/4
        y = 13/3 , 19/4
Thus, x >= y.

2. (1) 4x + 7y = 42
   (2) 3x - 11y = -1

Answer. x = 7
        y = 2
Thus, x > y.

3. (1) 9x^2 - 29x + 22 = 0
   (2) y^2 - 7y + 12 = 0 

Answer. x = 2 , 11/9
        y = 3 , 4
Thus, x < y.

4. (1) 3x^2 - 4x - 32 = 0
   (2) 2y^2 - 17y + 36 = 0

Answer. x = 4 , -8/3
        y = 4 , 9/2
Thus, x =< y

5. (1) 3x^2 - 19x - 14 = 0
   (2) 2y^2 + 5y + 13 = 0

Answer. x = 7 , -2/3
        y = -3/2 , -1
Thus, x > y.

6. A merchant purchases a wrist watch for Rs 450 and fixes its list price in such a way that after allowing a discount of 10%, he earns a profit of 20%. Find the list price ?

A. 550
B. 600
C. 500
D. 650
E. None of these

Sol. C.P. of wrist watch = Rs 450
     Discount = 10%
     Profit earned = 20%
Lets asssume list price to be = 'x'
Then,
450*(120/100) = x*(90/100)
Solving, x = Rs 600.

7. An article was sold at profit of 12%. If the C.P. would be 10% less and selling price is Rs 5.75 more then there would be 30% profit.Then at what price it should be sold to make a profit of 20% ?

A. Rs 136         
B. Rs 138
C. Rs 140
D. Rs 135
E. None of these

Sol. Lets assume C.P. be 'x'
Then, S.P. will be = 1.12x ( as 12% profit)

Now, C.P. = 0.9X (as C.P. is reduuced by 10%)
     S.P. = 1.12x + 5.75
Now as profit is 30%,
(130/100)*0.9x = 1.12x + 5.75
Solving, x = Rs 115

Now at 20% profit S.P. will be = (120/100)*115 = Rs 138.

8. Th simple interest on a sum of money is 8/25 of sum. If the number of years numerically is half the rate of interest then what is the rate of interest ?

A. 5
B. 4
C. 6
D. 8
E. None of these

Sol. Given, S.I. = (8/25)P
            Time = t = R/2

Now, S.I.= PRT/100
     (8/25)P = [P*R*(R/2)]/100
Solving, R = 6%.

9. Three glasses of equal volume contain acid mixed with water. The ratio of acid and water are 2:3, 3:4 and 4:5 respectively. Contents of these glasses are poured in a large vessel. The ratio of acid and water in the large vessel is :

A. 417:564
B. 401:544
C. 407:560
D. 411:540
E. None of these

Sol. Glass 1, acid = 2x ; water = 3x
     Glass 2, acid = 3y ; water = 4y
     Glass 3, acid = 4z ; water = 5z

New Glass composition, acid:water = (2x+3y+4z):(3x+4y+5z)
Also as the volumes of glasses are equal thus,
2x+3x = 3y+4y
5x = 7y
y = 5x/7
Also, 2x+3x = 4z+5z
      5x = 9z
      z = 5x/9

Putting these values above, 401:544.

10. The average weight of 36 students is 50 kg. If was found later that the weight of one of the student was misread as 73 kg, whereas his actual weight was 37 kg. Find the correct average weight(in kg).

A. 50
B. 49
C. 51
D. 47
E. 48

Sol. Sum of the weights/36 = 50
     Sum of the weights = 50*36 = 1800

     Correction in sum of the weights = 37 - 73 = -36
     Correct sum of the weights = 1800 - 36 = 1764

Thus, correct average is = 1764/36 = 49.

11. Train takes 4 seconds to pass a man. Another train travelling in opp. direction of same length takes 5 seconds to pass the man. Find the time taken by both trains to cross each other ?

A. 51/9 sec
B. 40/9 sec
C. 42/9 sec
D. 41/9 sec
E. None of these

Sol. Length of trains = x meter
     Speed of 1st train = x/4 m/sec
     Speed of 2nd train = x/5 m/sec

Relative speed = x/4 + x/5 = 9x/20 m/sec
Required Time = 2x/(9x/20) = 40/9 seconds.

12. One third of a certain journey is covered at the rate of 25 km/hr, one fourth at the rate of 30 km/hr and the rest at 50 km/hr. Find the average speed of the whole journey ?

A. 100/3 
B. 35
C. 30
D. 37
E. None of these

Sol. Lets assume the total distance to be 'd'.
The, average speed = d/[(d/3*25)+(d/4*30)+(5d/12*50)] = 100/3 km/hr.

13. 8 men and 6 women can complete a work in 6 days. 1 man works twice as much as 1 woman in a day. 8 men and 4 women started working, after 2 days 4 men left and 4 new women joined. In how many more days will the work be completed ?

A. 6
B. 7
C. 5
D. 4
E. None of these

Sol. (8M + 4W)6 = T
Also, M = 2W
Thus, (16W + 4W)*6 = T
      120W = T
and   60M = T
Now, (8M + 4W)*2 + (4M + 8W)*t = T
     (8M + 2M)*2 + (4M + 4M)*t = 60M
     20 + 8t = 60
     t = 5 days.

14. A motor boat travelling at the same speed can cover 25 km upstream and 39 km downstream in 8 hrs. At the same speed, it can cover 35 km upstream and 52 km downstream in 11 hrs. Find the speed of the stream ?

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

Sol. Lets assume upstream speed to be 'x' km/hr
     And downstream speed to be 'y' km/hr
Then, 25/x + 39/y = 8
And, 35/x + 52/y = 11
Solving these two,we get,
x = 5km/hr
y = 13 km/hr

Thus, speed of stream = (y - x)/2 = (13 - 5)/2 = 4 km/hr.

15. In a lab, 2 bottles contain mixture of acid and water in the ratio of 2:5  in 1st and 7:3 in 2nd. Find the ratio in which these two should be mixed so that the new ratio will become 2:3 ?

A. 21:8
B. 15:8
C. 9:8
D. 19:8
E. None of these

Sol. Using the allegation method, (3/10):(4/35) = 21:8.  

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IBPS RRB 2016-17 (Questions asked in 4th Nov. shift) - Number Series

Directions(1-10): Find the missing numbers in the given series:

1. 2, 4, 11, 37, 153, _

A. 671
B. 771
C. 541
D. 877
E. None of these

Sol. 2*1 + 2 = 4
     4*2 + 3 = 11
    11*3 + 4 = 37
    37*4 + 5 = 153
   153*5 + 6 = 771  

2. 3, 10, 22, 39, 61, _

A. 86
B. 84
C. 83
D. 88
E. None of these

Sol. 3 + 7 = 10
    10 + 12= 22
    22 + 17= 39
    39 + 22= 61
    61 + 27= 88

3. 6, 6, 8, 14, 26, _

A. 42
B. 44
C. 46
D. 40
E. None of these

Sol. 6 + 1^2 - 1 = 6
     6 + 2^2 - 2 = 8
     8 + 3^2 - 3 = 14
    14 + 4^2 - 4 = 26
    26 + 5^2 - 5 = 46

4. 1, 3, 9, 31, 129, _

A. 560
B. 651
C. 543
D. 622
E. None of these

Sol. 1*1 + 2 = 3
     3*2 + 3 = 9
     9*3 + 4 = 31
    31*4 + 5 = 129
   129*5 + 6 = 651

5. 5, 11, 24, 44, 71, _

A. 102
B. 105
C. 112
D. 101
E. None of these

Sol. 5 + 6 = 11
    11 + 13= 24
    24 + 20= 71
    71 + 27= 105

    13 - 6 = 7
    20 - 13= 7
    27 - 20= 7

6. 3, 8, 15, 26, 39, _

A. 64
B. 56
C. 42
D. 48
E. None of these

Sol. 8 - 3 = 5
    15 - 8 = 7
    26 - 15= 11
    39 - 26= 13
    39 + (13+4) = 56

7. 9, 14, 21, 32, 45, _

A. 65
B. 60
C. 55
D. 70
E. None of these

Sol. 14 - 9 = 5
     21 - 14= 7
     32 - 21= 11
     45 - 32= 13
     45 + 15 = 60.

8. 10, 5, 5, 7.5, 15, _

A. 40
B. 37.5
C. 38
D. 45
E. None of these

Sol. 10*0.5 = 5
      5*1   = 5
      5*1.5 = 7.5
     7.5*2  = 15
     15*2.5 = 37.5

9. 6, 11, 18, 29, 46, _

A. 66
B. 71
C. 80
D. 95
E. None of these

10. 7, 15, 28, 59, 114, _

A. 225
B. 228
C. 233
D. 240
E. None of these

Sol. 7*2 + 1 = 15
    15*2 - 2 = 28
    28*2 + 3 = 59
    59*2 - 4 = 114
   114*2 + 5 = 233  

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IBPS RRB 2016-17 - Quant Quiz-15

1. In a family, father’s age is twice that of son's age. Father is 10 years older than mother. Daughter is 20 years younger than his mother and 5 years younger than his brother. What is the age of the father?  

A. 52 years 
B. 50 years 
C. 58 years 
D. 55 years 
E. None of these

Sol. Le the age of father be 'x'
Then age of son will be = x/2
Age of mother will be = (x-10)
Age of daughter will be = (x-10-20) = (x-30)
Also given,
x/2 - 5 = x - 30
x = 50 yrs.

2. After replacing an old member by a new member, it was found that the average age of five members of a club is same as it was 3 years ago. The difference between the ages of the replaced and the new members is: 

A. 2 years 
B. 4 years 
C. 8 years 
D. 15 years
E. None of these

Sol. Increase in ages of five members in 3 years = 3*5 = 15 yrs
Since the avg. age remains same, therefore required difference = 15 yrs.

3. The ratio of the present age of X to that of Y is 3 : 11. Y is 12 years younger than Z. Z’s age after 7 years will be 85 years. What is the present age of X’s mother, who is 25 years older than X ?


A. 43 years
B. 67 years 
C. 45 years 
D. 69 years 
E. None of these 

Sol. Present age of Z = 85 - 7 = 78
     Present age of Y = 78 - 12 = 66
     Present age of X = (3/11)*66 = 18
     Present age of X's mother = 18 + 25 = 43 yrs.

4. At present, Tina is eight times her daughter’s age. 8 years from now, the ratio of the ages of Tina and her daughter will be 10 : 3. What is Tina’s present age?  

A. 32 years 
B. 40 years  
C. 36 years 
D. Cannot be determined 
E. None of these

Sol. Given T = 8D
where T is the age of tina
And D is the age of daughter
Also, (8D+8)/(D+8)=10/3
Solving, we get D=4
Tina's present age = 8D = 8*4 = 32 yrs.

5. The age of the father is 30 years more than the son’s age. Ten years hence, the father’s age will become three times the son’s age that time. What is the son’s present age in years?   

A. Eight  
B. Seven 
C. Five 
D. Cannot be determined 
E. None of these 

Sol. Let the son's present age be x yrs.
Then the father's present age is (x+30) yrs.
Father's age after 10 yrs = (x+40) yrs
Son's age after 10 yrs = (x+10) yrs
According to the question:
(x+40)=3(x+10)
x = 5 yrs

Directions(6-10): Find the missing number in the given series:

6. 4, 2, 8, -4, 16, _

A. -10
B. -12
C. -14
D. -8
E. None of these

Sol. 4 - (1*2) = 2
     2 + (2*3) = 8
     8 - (3*4) =-4
    -4 + (4*5) =16
    16 - (5*6) =-14

7. 1, 6, 9, 44, 85, _

A. 146
B. 136
C. 126
D. 116
E. 156

Sol. 1 + (1^2 + 2^2) = 6
     6 + (2^2 + 3^2) = 19
    19 + (3^2 + 4^2) = 44
    44 + (4^2 + 5^2) = 85
    85 + (5^2 + 6^2) = 146

8. 36, 202, 44, 290, _

A. 52
B. 42
C. 48
D. 50
E. 56

Sol. 10^2 - 8^2 = 36
     11^2 + 9^2 = 202
     12^2 - 10^2 = 44
     13^2 + 11^2 = 290
     14^2 - 12^2 = 52      

9. 9 16 44 107 ?

A. 362
B. 288
C. 282
D. 364
E. None of these

Sol. 9 + 2³ – 1 = 16
16 + 3³ + 1 = 44
44 + 4³ – 1 = 107; 107 + 5³ + 1 = 233

10. 13 17 33 97 ? 1377

A. 332
B. 353
C. 388
D. 396
E. 340
E. 233  

Sol. 13 + 2² = 17; 17 + 4² = 33; 33 + 8² = 97….

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IBPS RRB 2016-17 - Quant Quiz-14

1. Find the number of ways of distributing 8 identical balls into 3 boxes so that no box is empty and each box being large enough to accommodate all balls ?

A. 24
B. 28
C. 36
D. 21
E. 18

Sol. (8-1)C(3-1) = 7C2 = 7*6/2*1 = 21.

2. A group consists of 4 couples  in which each of the 4 men have one wife each. In how many ways could they arranged in a straight line so that the men and women occupy alternate position ?

A. 576
B. 982
C. 1152
D. 1024
E. None of these

Sol. 4!*4! + 4!*4! = 576+576 = 1152.

3. A five digit number is formed with the digits 0,1,2,3 and 4 without repetition.Find the chance that the number is divisible by 2 ?

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

Sol. 5 digit number = 5! = 120
Divisible by 2 then the last digit should be 0, 2, 4
Then no. of cases with 0 as last digit = 4! = 24
     no. of cases with 2 as last digit = 18
     no. of cases with 4 as last digit = 18
Total cases = 24 + 18 + 18 = 60
P = 60/120 = 1/2.

4. From a group of 4 men and 3 women , 2 persons are selected at random. Find the probability that at least one woman is selected ?

A. 2/3
B. 3/5
C. 5/7
D. 1/3
E. 1/2

Sol. Total cases = 7c2 = 7!/5!*2! = 21
     Cases when no women is selectd = 4c2 = 4!/2!*2! = 6
     Cases when atleast 1 woman is selected = (21-6) = 15
     Probability = 15/21 = 5/7.

5. 16 persons are participated in a party. In how many different ways can they host the seat in a circular table, if the 2 particular persons are to be seated on either side of the host ?

A. 16! * 2
B. 14! * 2
C. 18! * 2
D. 14!
E. None of these

Sol. (16 – 2)! * 2 = 14! * 2.

6. A and B start a business with investments of Rs. 10000 and Rs. 9000 respectively. After 4 months, A takes out 1/2 of his capital. After 2 more months, B takes out 1/3 of his capital while C joins them with a capital of Rs. 14000. At the end of a year, they earn a profit of Rs. 10160. Find the share of each member in the profit?

A. Rs A – Rs. 3300, B – Rs. 3500, C – Rs. 3360
B. Rs A – Rs. 3200, B – Rs. 3600, C – Rs. 3360
C. Rs A – Rs. 3200, B – Rs. 3700, C – Rs. 3260
D. Rs A – Rs. 3200, B – Rs. 3500, C – Rs. 3460
E. None of these 

Sol. A : B : C = (10,000 x 4 + 5000 x 8) : (9000 x 6 + 6000 x 6) : (14000 x 6)
= 80000 : 90000 : 84000 = 40 : 45 : 42
A’s share = Rs. 10160 x 40/127 = Rs. 3200;
B’s share = Rs. 10160 x 45/127 = Rs. 3600;
C’s share = Rs. 10160 x 42/127 = Rs. 3360.

7. 1, 4, 19, 54, 117, _

A. 194
B. 256
C. 216
D. 200
E. None of these

Sol. 1 + 1*3 = 4
     4 + 3*5 = 19
    19 + 5*7 = 54
    54 + 7*9 = 117
   117 + 911 = 216

8. Prachi started a business investing Rs. 50,000 in 2015, In 2016, she invested an additional amount of Rs. 20,000 and Arsh joined him with an amount of Rs. 70,000. In 2017, Prachi invested another additional amount of Rs. 20,000 and Heena joined them with an amount of Rs. 70,000. What will be Arsh’s share in the profit of Rs. 300,000 earned at the end of 3 years from the start of the business in 2015 ?

A)Rs Rs. 250,000.
B)Rs Rs. 120,000.
C)Rs Rs. 100,000.
D)Rs Rs. 150,000.
E)None of these

Sol. Prachi : Arsh : Heena
= (50000 x 12 + 70000 x 12 + 90000 x 12) : (70000 x 24) : (70000 x 12)
= 2520000 : 1680000 : 840000 = 3 : 2 : 1
Arsh’s share = Rs.300,000 x 2/6 = Rs. 100,000.

9. The ratio of the monthly salaries of A and B is in the ratio 15 : 16 and that of B and C is in the ratio 17 : 18. Find the monthly income of C if the total of their monthly salary is Rs 1,87,450. 

A. Rs 66,240
B. Rs 72,100
C. Rs 62,200
D. Rs 65,800
E. Rs 60,300

Sol. A/B = 15/16 and B/C = 17*18
So A : B : C = 15*17 : 16*17 : 16*18
= 255 : 272 : 288
So C’s salary = [288/(255+272+288)] * 1,87,450

10. X takes 6 days less than Y to finish the work individually. If X and Y working together complete the work in 4 days, then how many days are required by Y to complete the work alone ?

A. 7 days
B. 10days
C. 5 days
D. 12days
E.None of these

Sol. Lets assume Y takes = a days
Then X will take = (a-6) days

Now given, 1/a + 1/(a-6) = 1/4
Solving, we get,
a = 12 days.

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IBPS PO MAINS 2016 - Quant Quiz-9 (Mixed)

1. In a vessel containiing, 60 litres of mixture of milk and water, water is only 30%. 30 litres of mixture is taken out and replaced with x litres of pure milk. As a result the percentage of water in the mixture becomes 20%. What is the value of x ?

A. 5
B. 10
C. 15
D. 8
E. 12  

Sol. Amount of water initially = (30/100)*60 = 18 litres
     Amount of milk initially = (60 - 18) = 42 litres

Now, 30 litres of mixture is taken out and replaced with x litres of milk.
     Amount of water = 18 - (18/60)*30 = 9 litres
     Amount of milk = 42 - (42/60)*30 + x = (21 + x) litres
     Total quantity = 9 + 21 + x = (30 + x) litres.
Now its given that percentage of water is 20%. so,
     (20/100)*(30+x) = 9
     x = 15 litres.

2. Ramesh won a competition and got some prize money. He gave Rs. 2000 less than the half of prize money to Suresh and Rs. 1000 more than the two third of the remaining to his Mahesh. If both of them got the same amount, what is the prize money Ramesh got ?

A. 20000  
B. 25000
C. 26000  
D. 24000
E. None of these

Sol. Assume Ramesh got x rupees.
     He gave x/2 - 2000 to Suresh.
     And (2/3)*( x/2 + 2000 ) + 1000 to Mahesh.
     x/2 - 2000 = x/3 + 7000/3
     x = 26000.

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

3. 3, 4, 12, _, 576, 27648

A. 48
B. 64
C. 36
D. 72
E. None of these

Sol. 4*3 = 12
    12*4 = 48
    48*12= 576
   576*48= 27648

4. 4, 12, 60, 420, _, 41580

A. 4620
B. 3780
C. 4200
D. 5040
E. None of these

Sol. 4*3 = 12
    12*5 = 60
    60*7 = 420
   420*9 = 3780
  3780*11= 41580

5. 686, 438, 314, 252, 221, _

A. 180
B. 187.5
C. 201
D. 205.5
E. 175.5

Sol. 686 – 248 = 438
438 – (248/2) = 438 – 124 = 314
314 – (124/2) = 314 – 62 = 252
252 – (62/2) = 252 – 31 = 221

6. 4, 3, 12, 6, 36, 9, 108, 12, _ 

A. 324
B. 252
C. 156
D. 182
E. 236

Sol. It is a double series.
*3 :– 4, 12, 36, 108
+3 :– 3, 6, 9, 12

7. 120, 170, 224, 280, 360, _ 

A. 392
B. 442
C. 458
D. 528
E. 540

Sol. 11^2 – 1 = 120
13^2 + 1 = 170
15^2 – 1 = 224
17^2 + 1 = 280
19^2 – 1 = 360
21^2 + 1 = 442

8. A rectangular floor is fully covered with square tiles of identical size. The tiles on the edges are white and the tiles in the interior are red. The number of while tiles is the same as the number of red tiles. A possible value of the number of titles along one edge of the floor is 

A. 10
B. 12
C. 14
D. 16
E. None of these

Sol. Let the rectangle has m and n tiles along its length and breadth respectively.
The number of while titles
W = 2 m + 2 (n - 2) = 2(m + n - 2)
And the number of red tiles
R = mn - 2(m + n - 2)
Given
W = R ⇒ 4 (m + n - 2) = mn
⇒ mn - 4m - 4n = -8
⇒ (m - 4) (n - 4) = 8
As m and n are integers so (m - 4) and (n - 4) are both integers. The possibilities are (m - 4, n - 4) = (1 , 8) or (2, 4) giving, (m, n) as (5, 12) or (6, 8) so the edges can have 5, 12, 6 or 8 tiles. Answer is (b) only.

9. A telecom service provider engages male and female operators for answering 1000 calls per day. A male operator can handle 40 calls per day whereas a female operator can handle 50 calls per day. The male and the female operators get a fixed wage of Rs 250 and Rs 300 per day respectively. In addition, a male operator gets Rs 15 per call he answers and female operator gets Rs 10 per call she answers. To minimize the total cost, how many male operators should the service provider employ assuming he has to employ more than 7 and maximum 12 number of the females?   

A. 15
B. 14
C. 12
D. 10
E. None of these

Sol. Let us form both equations first:
40m + 50f = 1000
250 m + 300 f + 40 × 15 m + 50 × 10 × f = A
850m + 8000 f = A
When M and F are the number of Males and Females and A is the amount paid by the service provider.
Then the possible values of F are 8, 9, 10, 11, 12
If F = 8, then, M = 15
If F = 9, 10, 11 then M will not be an integer while F = 12 then M will be 10.
By putting F = 8 and M = 15, A = 18800. When F = 12 and M = 10, then A = 18100.


Hence the number of males will be 10.

10. The cost price of goods with a bankrupt is Rs. 25500 and if the goods had realised in their full value, his creditiors would have received 85 paise in the rupee. But 2/5 of the goods were sold at 17% and the remainder at 22% below their cost price. How many paise in a rupee was received by the creditors? 

A. 72 paise 
B. 68 paise  
C. 55 paise 
D. 52 paise  
E. None of these

Sol. Total debt=25500× 100/85=Rs.30000
Money received by selling the goods=25500(2/5×83/100+3/5×78/100)
=25500/500 (166+234)
=51×400=Rs.20400
Therefore, money received by the creditors for a rupee=Rs.(20400/30000)=Rs.0.68=68 paise
Hence, the creditor received 68 paise in a rupee.

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IBPS PO MAINS 2016 - Quant Quiz-8 (Mixed)

1. X borrows a sum of Rs.100,000 from a bank @ 10% p.a. compounded annually for 2 years. He then lends Rs.20,000, compounded every 8 months @ 10%, to each of his five friends for a period of two years. At the end of two years, X collects all the money from his friends and clears his debt with the bank. How much money did X make at the end of the two-year period ?

A. 12,100
B. 14,100
C. 15,100
D. 4,971
E. 13,200

Sol. The amount to be paid to bank after 2 years will be 100000 × (1.1)^2 = Rs.121,000.
        A 2-year period will be made up of three 8-month periods, each of which will cost each of the friends 10%.
        The amount that each of the friends will pay back after 2 years will be 20000 × (1.1)^3 =Rs.26620.
        So, the total sum of money collected from the five friends will be 5 × 26620 = Rs.133,100.
        After clearing his debt with the bank, X will have made 133100 – 121000 = Rs.12,100.


2. X lent Rs 2500 to Y for 4 years and Rs 4000 to Z for 3 years on simple interest at the same rate of interest and received Rs 4400 in all from both of them as interest. The rate of interest per annum is:

A. 25%
B. 17%
C. 22%
D. 20%
E. None of these

Sol. Let the rate of interest be R%. Then,
       (2500*R*4)/100 + (4000*R*3)/100 = 4400
       R = 4400/220 = 20%.

Directions(3-6): Find the missing numbers in the given series:

3. 7, 9, 23, 121, 839, _

A. 7650
B. 7561
C. 7542
D. 7560
E. 7652

Sol. 7*1 + 2 = 9
     9*3 - 4 = 23
    23*5 + 6 = 121
   121*7 - 8 = 839
   839*9 + 10 =7561

4. 6, 8, 16, 34, 66, _

A. 120
B. 122
C. 115
D. 116
E. 124

Sol. 6 + 2*1^2 = 8
     8 + 2*2^2 = 16
    16 + 2*3^2 = 34
    34 + 2*4^2 = 66
    66 + 2*5^2 = 116

5. 90, 89, 58, 53, 34, _

A. 22
B. 24
C. 25
D. 26
E. 21

Sol. 10^2 - 10 = 90
      9^2 + 8  = 89
      8^2 - 6  = 58
      7^2 + 4  = 53
      6^2 - 2  = 34
      5^2 + 0  = 25

6. 50, 26, 14, 8, 5, _

A. 4
B. 3
C. 3.5
D. 2.5
E. 4.5

Sol. 50/2 + 1 = 26
     26/2 + 1 = 14
     14/2 + 1 = 8
      8/2 + 1 = 5
      5/2 + 1 = 3.5

          
7. 3 years ago, the average age of a family of 5 members was 17 years. With the birth of a new baby, the average age of six members remains the same even today. Find the age of the new baby.

A. 1 year
B. 2 year
C. 3/2 year
D. 4 year
E. None of these

Sol. Sum of present age of the six members = (17 × 6) = 102 years.
     Average of the present age of the 5 members (excluding baby) = 17 + 3 = 20 years.
     Sum of present ages of the 5 members = 5 × 20 = 100 years.
     Age of the baby = 102 – 100 = 2 years.

8. X gets on the elevator at the 11th floor of a building and rides up at the rate of 57 floors per minute. At the same time, Y gets on an elevator at the 51st floor of the same building and rides down at the rate of 63 floors per minute. If they continue travelling at these rates, then at which floor will their paths cross?

A. 19 
B. 28
C. 30 
D. 37
E. None of these

Sol. Suppose their paths cross after x minutes.
     Then, 11 + 57x = 51 - 63x
     120x = 40
     x = 3
   
     Number of floors covered by X in (1/3) min. =  1 x 57  = 19.
     
     So, their paths cross at (11 +19) i.e., 30th floor.

9. X, Y and Z started a business investing Rs 9000, 12000, 10000 respectively. X kept the money for whole year. Y took his entire amount after 3 months and Z with drew 3000 Rs after 3 months. If after a year, the total profit was Rs 7,900, what was X’s share in it?

A. Rs 4,000
B. Rs 3,600
C. Rs 3900
D. Rs 3500
E. None of these

Sol. In partnership problems profit earned is directly proportional to the product of Investment and Time period for which investment was made in the partnership business.

     Required ratio of profit distribution among X, Y, and Z respectively will be as followed,
     = 9 × 12: 12 × 3: 10 × 3 + 7 × 9 = 36: 12:31,
     hence X will get 36/(36 +  12 +  31) × 7900 = Rs 3600.

10. X sells a tube to B at a profit of 25% and Y sells it to Z at profit of 20%. If Z pays Rs 450 for it, what did X pay for it?

A. Rs 240
B. Rs 247.5
C. Rs 300
D. Rs 500
E. None of these

Sol. Let X paid = Rs x
     120% of 125 % of x = 450
     (120/100) × (125/100) × x = 450
     x = Rs 300.

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IBPS RRB 2016-17 - Quant Quiz-13

1. (989/34) ÷ (65/869) * (515/207) = ?

A. 845
B. 870
C. 945
D. 745
E. 890

2. 67% of 801 – 231.17 = ? – 23% of 789

A. 400
B. 490
C. 550
D. 600
E. 750

3. (32.13)2 + (23.96)2 – (17.11)2= ?


A. 1410
B. 1310
C. 1550
D. 1650
E. 1810

4. 5907 – 1296 ÷ 144 = ? * 8


A. 700.25
B. 658.25
C. 628.25
D. 737.25
E. 630.5

5. √7378 * √1330 ÷ √660 = ?


A. 150
B. 160
C. 120
D. 170
E. 140

6. 22240 ÷ √? = 34 * 12


A. 3065
B. 3085
C. 3025
D. 3075
E. None of these

7. 8451 + 793 + 620 – ? = 6065 + 713


A. 3486
B. 3586
C. 3286
D. 3186
E. None of these

8. (12.25)2 – √625 = ?


A. 145.1625
B. 125.0625
C. 155.1625
D. 165.0625
E. None of these

9. 156 + 16 * 1.5 – 21 = ?


A. 126
B. 149
C. 141
D. 159
E. None of these

10. (√7921 – √2070.25) * (1/4) = ?


A. 15
B. 16
C. 17
D. 19
E. 11




                                     ANSWERS

1) C. 945
2) B. 490
3) B. 1310
4) D. 737.25
5) C. 120
6) C. 3025
7) E. None of these
8) B. 125.0625
9) D. 159
10) E. 11

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IBPS PO MAINS 2016 - Quant Quiz-7 (Time and Work)

1. A can complete a piece of work in 4 days. B takes double the time taken by A, C takes double that of B, and D takes double that of C to complete the same task. They are paired in groups of two each. One pair takes two-thirds the time needed by the second pair to complete the work. Which is the first pair?

(a) A and B

(b) A and C

(c) B and C

(d) A and D

(e) C and D

Sol. Work done in one day by A, B and C are 1/4,1/8,1/16 and 1/32 respectively.

Using answer choices, we note that the pair of B and C does 3/16 of work in one day; the pair of A and D does 1/4+1/32=9/32 of work in one day

Hence, A and D take 32/9 days

B and C take 16/3=32/6 days

Hence, the first pair must comprise of A and D.

2. R finishes a work in 7 days. P finishes the same job in 8 days and Q in 6 days. They take turns to finish the work. R worked on the first day, P on the second day and Q on the third and then again R and so on. Who was working on the last day when work got finished?

(a) R

(b) P

(c) Q

(d) P and Q

(e) Cannot be determined

Sol. Three day’s work = 1/7+1/8+1/6=73/168

Six day’s work = 73/84

Seventh day work = 1/7, done by R

Since 73/84+1/7=85/84 > 1, therefore, R was working on the last day.

3. Construction of a road was entrusted to a civil engineer. He has to finish the work in 124 days for which he employed 120 workers. Two-third of the work was completed in 64 days. How many workers can be reduced now without affecting the completion of the work on time?

(a) 56

(b) 64

(c) 80

(d) 24

(e) None of these

Sol. 2/3rd of the work was completed in 64 days by 120 workers.

1/3rd of the work was completed in 32 days by 120 workers.

Also 1/3rd of the work is to be completed in 60 days by (120 – x) workers, where x is the number of men reduced in order to finish the work on schedule.

So, (120 – x) × 60 = 120 ⇒ x = 56.

4. Two workers earned Rs. 225 first worked for 10 days and the second for 9 days. How much did each of them get daily if the first worker got Rs. 15 more for working 5 days than the second worker got for working 3 days?

(a) Rs. 11.70; Rs. 12.00

(b) Rs. 10.80; Rs. 13.00

(c) Rs. 11.25; Rs. 12.50

(d) Rs. 12.60; Rs. 11.00

(e) None of these

Sol. Let A got Rs. x per day and B got Rs. y per day. So, 10x + 9y = 225 and 5x = 3y + 15 ⇒ x = 10.80, y = 13.

5. Two pipes A and B can fill a tank in 20 and 30 h respectively. Both the pipes are opened to fill the tank but when the tank is 1/3 full, a leak develops in the tank. Due to this leakage one-third of the water supplied by pipes A and B goes waste. What is the total time to fill the tank if the leak if not closed ?

(a) 12 h

(b) 16 h

(c) 18 h

(d) 20 h

(e) None of these

Sol. Let us assume total work = 180 (we are not assuming it to be LCM of 20 and 30 = 60 because in that case 1/3rd of A + B will be fractional)

Time taken to fill 1/3rd of the tank = 180/ (9 + 6) = 4 h

Due to leakage, net inflow = 2/3 (9 + 6) = 10 units

Time taken to fill remaining 120 units = 12 h

So total time taken = 12 + 4 = 16 h

6. P, Q and R can do a piece of work in 16, 24 and 30 days respectively. They started the work simultaneously but P stops the work after 4 days and Q called off the work 2 days before the completion. In what time the work is finished ?

a) 100/9 days
b) 100/11 days
c) 100/7 days
d) 100/13 days
e) None of these

Sol. 4/16 + (T -2)/24 + T/30 = 1 where T is the time taken to complete the job.

       T = 100/9 days.

7. If P can do 1/3 of the work in 5 days and Q can do 1/4 of the work in 6 days, then how much money will Q get if they were paid a total of 390 rupee?

a) 120
b) 150
c) 170
d) 190
e) None of these

Sol. P can alone complete the whole  work in 15 days and Q can complete the same work alone in 24 days. So ratio of work done by them 1/15: 1/24 i.e. 8: 5
Q get = (5/13)*390 = 150.

8. A does half as much work as B in one third of the time taken by B. If together they take 20 days to finish the work then what will be the share of A if 1000 rupees is given for the whole work?

a) 400
b) 500
c) 600
d) 700
e) None of these

Sol. Let B take x days to complete the work, then A will take  = x/3 + x/3 = 2x/3 days (as half work is completed in one third of the time)
3/2x + 1/x = 1/20
X = 50. So A will complete the work in 100/3 days and B will complete the work in 50 days.
Ratio of work done by A and B – 3/100: 1/50 = 3:2
So A share = 3/5*1000 = 600.

9. A does half as much work as B does in one sixth of the time. If together they take 20 days to complete the work, then what is the time taken by A to complete the work independently.

a) 80/3 days
b) 100/3 days
c) 60/3 days
d) 140/3 days
e) None of these

Sol. Let B complete the work in X days so in one day work done by B is 1/x
as A do half work in one-sixth of the time so A will complete work in 2*x/6 = x/3 days
One day work of A and B i.e. 3/x + 1/x = 1/20. So we get x = 80
So time taken by A alone = 80/3 days.

10. A and B can do a piece of work in 20 and 25 days respectively. They began to work together but A leaves after some days and B completed the remaining work in 12 days. Number of days after which A left the job-

a) 5.7/9 days
b) 6.7/9 days
c) 7.7/9 days
d) 11.7/9 days
e) None of these

Sol. (1/20 + 1/25)*T + 12/25 = 1
We will get T = 52/9 i.e. 5.7/9 days.

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IBPS RRB 2016-17 - Quant Quiz-12 (Based on previous year papers)

1. The average age of some males and 15 females is 18 years. The sum of the ages of 15 females is 240 yrs and average age of males is 20 yrs. Find the number of males ?

A. 10
B. 12
C. 15
D. 18
E. 20

Sol. Lets assume no. of males to be 'x'.

Thus, sum of ages of males/x = 20
      sum of ages of males = 20x

Therefore, (240 + 20x)/(15 + x) = 18
Solving, we get,
x = 15yrs

2. If the sum of the smaller number x and two times the other number is equal to sum of two times the smaller number and 16. The difference between the numbers is 6. Find the smaller number ?

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

Sol. Lets assume the larger number be 'y'
Thus, x + 2y = 2x + 16........1st equation
Also, y - x = 6...............2nd eqution

Solving these two, we get,
y = 10
x = 4

3. 16 men and 12 women can complete a work in 8 days, if 20 men can complete the same work in 16 days, in how many days 16 women can complete the same piece of work ?

A. 8
B. 12
C. 10
D. 15
E. 9

Sol. Lets assume it take 'w' days for 1 women to complete the job.
Then, 16/(16*20) + 12/w = 1/8
      w = 160 days
Now for 16 women it will be, = 160/16 = 10 days.

4. If A's salary is 10,000 less than B's salary and B's salary is 15,000 less than C's salary and sum of their salaries is 65,000. Find the salary of A.

A. 15,000
B. 10,000
C. 12,000
D. 20,000
E. 5,000

Sol. Lets assume the salary of C be 'x'
Then, salary of B = x - 15,000
      salary of A = x - 25,000
Now sum of these salaries is,
x + x - 15,000 + x - 25,000 = 65,000
Solving,
x = 35,000
A's salary = 35,000 - 25,000 = 10,000.

5. The sum of the ages of P and Q is 25 yrs more than the age of R.The present age of Q is 5 yrs more than the age of R. Find the present age of P ?

A. 15
B. 20
C. 12
D. 10
E. 16

Sol. Given, P + Q = 25 + R
Also, given, Q = 5 + R
Putting the value of Q in the 1st eqn, we get,
P + 5 + R = 25 + R
P = 20 yrs.

6. The perimeter of the rectangular field is 240 m. The ratio of length and breadth is 8:7. Find the area of the rectangle ?

A. 3854
B. 3584
C. 3485
D. 3845
E. None of these

Sol. Perimeter of the rectangle = 2*(l + b) = 240
Also given, l/b = 8/7, using this in 1st eqn,
8b/7 + b = 120
b = 56
l = 64
Area of the rectangle = 56*64 = 3584 sq.m.

7. One fourth of two-fifth of 30% of a number x is equal to 15. Find 20% of the same number ?

A. 100
B. 120
C. 105
D. 80
E. None of these

Sol. (1/4)*(2/5)*(30/100)*x = 15
     x = 100.

8. The difference between compound interest and simple interest on a sum for 2 yrs at 20% per annum, when the interest is compounded annually is Rs 240. If the interest were compounded half yearly, the difference in two interests over same period would be ?

A. Rs 384.6
B. Rs 344.2
C. Rs 324.8
D. Rs 316.5
E. None of these

Sol. Difference for 2 yrs = P*20*20/100*100 = 240
Solving, we get,

P = 6000

So SI for 2 yrs = 6000*20*2/100 = 2400
Amount at compounded half yearly = 6000 [1 + 10/100]^4

Solving,we get,

Amount at compounded half yearly = 8784.6
So CI at compounded half yearly = 8784.6 – 6000 = 2784.6
So difference = 2784.6 – 2400 = Rs 384.6


Directions(9-10): What should come in place of the question mark(?) in the following number series:

9. 7, 16, 34, 61, 97, ?

A. 142
B. 154
C. 148
D. 164
E. None of these

Sol. 7 + 9*1, 16 + 9*2, 34 + 9*3, 61 + 9*4, ..

10. 8, 2, 2, 4.5, 18, ?

A. 112.5
B. 122.5
C. 134.2
D. 142.5
E. None of these

Sol. *0.5^2, *1^2, *1.5^2, *2^2, *2.5^2,  

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