IBPS PO Prelims Quantitative Aptitude Question Paper 2019

Directions (1 – 5): The table given below shows the total number of patients admitted to a hospital in five different months and the ratio of child patients to aged patients. Read the data carefully and answer the questions 

Month	Total patients	Ratio of Child : Aged
January	640	5 : 3
February	720	11 : 7
March	450	5 : 4
April	840	7 : 5
May	480	7 : 9

1. Question

Find the ratio of total aged patients admitted to the hospital in January & February together to the total child admitted in March?

a) 52: 25
b) 27: 23
c) 39: 31
d) 11: 19
e) None of these

Answer:-A

Explanation:-

Total aged patients admitted to the hospital in January & February
= ( 640 × 3/8 ) + ( 720 × 7/18 )
= 240 + 280
= 520

Total child admitted in hospital in March 
= 450 ×5/9 
= 250

Required ratio = 520 : 250 = 52 : 25

Thus the required ratio = 52 : 25

Short trick:

From the question, we can see that it is asking for the ratio of the total children admitted in March. So, our answer will be a multiple of 5. from the option only option a has the multiple of 5.

2. Question

The total aged patients admitted in April is what per cent more than that of in March?
a)  50 %
b)  75 %
c). 125 %
d). 65 %
e). 80 %

Answer :-B

Explanation :-

Total aged patient admitted in April 
= 840 ×5/12
= 350
Total aged patient admitted in March 
= 450 ×4/9
= 200

Required percentage = ( 350−200 )/ 200 × 100
= 150 /200× 100 
= 75%

Thus, the required percentage = 75%

3. Question

Find the difference between the total Child patients admitted to the hospital in January and April?

a) 55
b) 60
c) 90
d) 70
e) 95

Answer :- C
Explanation :-

Total Child patients admitted in hospital in January
= 640 ×5/8
= 400

Total Child patients admitted in hospital in April
= 840 ×7/12
= 490
Required difference = (490 – 400) = 90

Thus,the required difference = 90

4. Question

Find the average of child patients admitted to the hospital in February & May?
a) 245
b) 375
c) 325
d) 440
e) 500

Answer:- C
Explanation:-

Total child patients admitted in February
= 720 ×11/18
= 440
Total Child patients admitted in hospital in May
= 480 ×7/16
= 210
Required average 
= ( 440 + 210 )/2
= 325

Thus,the required average = 325

5. Question

If the total aged patients admitted in the hospital in June is 20% more than that of in April and the ratio of Child to aged patients admitted in June is 7 : 9, then find the total children admitted in the hospital in June?

a) 350
b) 470
c) 510
d) 670
e) 540

Answer:- E
Explanation:-

Total aged patient admitted in the hospital in June
= 840 ×5/12×120/100
= 420

Total Child patients admitted to hospital in June 
= 420 ×9/7
= 540

Directions (6-10): 

Find out the approximate value of x in the given questions: (You are not expected to calculate the exact value.

6.Question

130.09 % of 49.98 + x² + 10.03 = (13.98)²

A) 15
B) 12
C) 11
D)19
E) 9

Answer :- C
Explanation :-
130.09 % of 49.98 + x² + 10.03 = (13.98)² 
=> 130/100 × 50 + x² + 10 = (14)²  
=> 65 + 10 + x² = 196
=> x² = 196 – 75 = 121
=> x = 11

7. Question

14.01 ÷ 1/x +140.03 = 419.97

A) 20
B) 25
C) 27
D) 16
E)  35

Answer:- A
Explanation:-
 14.01 ÷ 1/x +140.03 = 419.97
=> 14 * x +140 = 420
=> 14 * x = 420 − 140
=> x =280/14
=> x = 20

8.Question

(16.99)² + x = (21.01)2

A) 175
B) 180
C) 152
D) 130
E) 100

Answer :- C
Explanation :-
 (16.99)² + x = (21.01)²
=> (17)² + x = (21)²
=> 289 + x = 441
=> x = 441 – 289
  x = 152

9.Question

x/20.01 + 99.98 = 24.99% of 840.01

A) 1700
B) 2200
C) 3000
D) 4500
E) 7500

Answer :- B
Explanation :-
  x/20.01 + 99.98 = 24.99% of 840.01
=> x/20 + 100 = 25/100 × 840
=> x/20 = 210 – 100
=> x = 110 × 20
=> x = 2200.

10.Question

x³ – 49.99% of 112  = (13.02)² − (26.98)⅔

A) 9
B) 6
C) 11
D) 17
E) 3

Answer :- B
Explanation :-
  x³ – 49.99% of 112 = (13.02)² – (26.98)⅔
=> x³ – 50/100 × 112 = (13)² – (3³)⅔
=> x³ – 56 = 169 – 9
=> x³ = 160 + 56 = 216
=> x = 6

11. Question

P, Q and R three girls hired a  car for Rs. 720 and used it for 2, 4 and 3 hours respectively. Find the hire charges paid by R.

A)  Rs.180 
B)  Rs.210
C)  Rs.320
D)  Rs.160
E)  Rs.240

Answer :-E
Tricks :-

P : Q : R = 2 : 4 : 3
R = 720 × 3/9 = 240
Thus , the hire charges paid by R is Rs. 240.

12. Question

X, Y and Z three partners started a business jointly. X and Y invested money in the ratio 2 : 1 whereas Y and Z invested capital in the ratio 5 : 3. If their total annual profit is Rs. 16200, then find the Y’s share in the total profit.

A) 6200
B) 5800
C) 7500
D) 4500
E) None of these

Answer :- D
Tricks :-
X : Y = 2 : 1  …..(×5)    = 10 : 5
Y : Z = 5 : 3  …..(×1)    =  5 : 3
A : B : C = 10 : 5 : 3
Y’s share in profit = 16200 × 5/18  = 4500
Here, the correct answer is D.

13. Question

The ratio between the present ages of Tanmay and Vivan is 3:7, respectively. After 4 yr, Vivan’s age will be 39 yr. What was Tanmay’s age 4 yr later?

(a) 11 yr

(b) 13 yr

(c) 19 yr

(d) 18 yr

(e) None of the above

Answer :- C
Explanation :-

Let Tanmay and Vivan’s ages are 3x yr and 7x yr, respectively.
7x + 4 = 39 ⇒ x = 5
=3×5-4=11 yr
Hence, Tanmay’s age 4 yr later
= 3×5 + 4 
= 19 years.

Tricks :-

7 unit ——-> (39-4) = 35

1 unit = 5

Thus, Tanmay’s age after 4 yr = (3 × 5)+4

                                                      = 19

14.Question

The length and radius of a cylindrical rod are 14 m and 2 cm respectively. How many such cylindrical rods can be made out of 0.44 metre³ of the rod?

A) 25
B) 30
C) 45
D) 50
E) 15

Answer :-A
Explanation :-

r = 2 cm = 2/100 m ; h = 14 m
Volume of one rod = πr²h
                                 = 22/7 × 2/100× 2/100 × 14
                                 = 11/625 m³
Volume of  the iron rod = 0.77 m³
:. Number of cylindrical rods = 0.44 × 625/11
                                                   = 25

15.Question

In an IT farm, the ratio of male employees to female employees is 7: 5. If there are 480 employees in the IT farm, then how many female employees are there?
A) 80
B) 70
C) 100
D) 200
E) 90

Correct Option: D
Explanation:-
Let the number of male and female employees are 7x and 5x, respectively.

Given, the total number of employees= 480

⇒ 7x + 5x = 480 
⇒  12x = 480

∴   x = 40

∴    Required number of female employees 
= 5x = 5 × 40 = 200.

Hence, option D is correct.

Tricks :-
(7 + 5) = 12 unit ——-> 480
                 1 unit ——–> 40
Thus, the number of female employee= 5 × 40 = 200

Directions ( 16 -20 ):- In the following questions contain two equations I and II. You have to solve both the equations and determine the relationship between them and give answers as,

a)  x > y

b)  x ≥ y

c)  x < y

d)  x ≤ y

e)  x = y or the relation cannot establish.

16. Question

I)   x² – 44x – 141 = 0
II) y² + 62y + 177 = 0

Answer :- B
Explanation :-
 x² – 44x – 141 = 0
=> x² – 47x + 3x – 141 = 0
=> (x – 47)( x + 3) = 0
x = 47 , – 3

 y² + 62y + 177 = 0
=> y² + 59y + 3y + 177 = 0
=> (y + 59) ( y + 3) = 0
y = – 59 , – 3

So,  x ≥ y .

17.Question

I) 19x² – 23x + 4 = 0
II) 7y² + 52y + 21 = 0

Answer :- A
Explanation :-

    19x² – 23x + 4 = 0
=> 19x² – 19x – 4x + 4 = 0
=> ( 19x – 4) ( x – 1) = 0
=> x = 1 , 4/19

    7y² + 52y + 21 = 0
=> 7y² + 49y + 3y + 21 = 0
=> ( 7y + 3) ( y + 7) = 0
=> y = – 3/7 , – 7

So, Here  x > y .

18.Question

I) 12x² + 67x + 21 = 0
II) 11y² – 52y – 15 = 0

Answer :- C
Explanation :-

   12x² + 67x + 21 = 0
=> 12x² + 4x + 63x + 21 = 0
=> ( 4x + 21) ( 3x + 1) = 0
=> x = – 21/4 , – 1/3

     11y² – 52y – 15 = 0
=> 11y² – 55y + 3y – 15 = 0
=> ( 11y + 3) ( y – 5) = 0
=> y = 5 , – 3/11

Here , x < y

19.Question

I)  x² – 30x + 176 = 0
II) 20y² + 20y – 15 = 0

Answer :- A
Explanation :-

     x² – 30x + 176 = 0
=> x² – 8x – 22x + 176 = 0
=> ( x – 8) ( x – 22) = 0
=> x = 8 , 22

    20y² + 20y – 15 = 0
=> 20y² + 30y – 10y – 15 = 0
=> ( 10y – 5 ) ( 2y + 3) = 0
=> y = 5/10 , -3/2

So, x > y .

20.Question

I)  15x² + 40x + 20 = 0
II) 4y² – 23y + 19 = 0

Answer :- C
Explanation :-

    15x² + 40x + 20 = 0
=> 15x² + 30x + 10x + 20 = 0
=> ( 15x + 10) ( x + 2) = 0
=> x = – 10/15 , – 2

      4y² – 23y + 19 = 0
=> 4y² – 4y – 19y + 19 = 0
=> ( 4y – 19) ( y – 1) = 0
=> y = 1 , 19/4

So ,here  x < y .

21. Question

What is the probability of getting a 3 or a 4 when a single 6 – sided dies is rolled?

A) 1/8

B) 1/18

C) 1/2

D) 1/3

E) 1/6

Answer :- D
Explanation :-
Total outcomes = 6
Probability of getting a number in a single rolling =1/6
So, P(3) = 1/6 and P(4) = 1/6
Now,the probability getting either 3 or 4
= P ( 3 or 4)
= P(3) + P(4)
= 1/6 + 1/6
= 1/3
Thus , the required probability = 1/3

Tricks:-

P(3 or 4) = 1/6 + 1/6   = 1/3

22. Question

8 cm and 12 cm are the sides of a triangle. The angle included between the two sides is 45°. Find out the area of the triangle.

a) 80√2  cm²

b) 70√3   cm²

c) 30√2    cm²

d) 24√2    cm²

e) 150√3  cm²

Answer :- D

Explanation :-

Let  a = 8 cm

      b= 12 cm

Area = 1/2 × ab Sinθ

       =1/2 × 8 ×12 × Sin45°

       =1/2 ×8 × 12 ×1/√2

       = (48 × √2) / (√2 × √2)   

       = 24√2  cm²

Thus, the area of the triangle = 24√2  cm².

23. Question

In a classroom, a girl multiplied a number by 3/5 instead of 5/3. What is the percentage error in this calculation?

a) 64 %

b) 35 %

c) 26 %

d) 58 %

e) none of these

Answer :- A

Explanation :-

3/5  ,  5/3

LCM = 15

3/5 × 15 = 9

5/3 × 15 = 25

Required percentage =(25 – 9) / 25 × 100 %

                                   = 64 %

Thus, the error is = 64 %

24. Question

A young lady does a piece of work in 3 days. That piece of work was done by an old woman in 4 days. If the young girl and old woman worked together, they got total wages of Rs. 3500. How much did the old woman get?

a) 1500

b) 2000

c) 1000

d) 1200

e) None of these

Answer:-A

Tricks:-

        Young girl      :   old woman

Time   =          3     :     4

Efficiency=     4     :     3

(Time and efficiency are inversely proportional)

Old Woman get = 3500 ×  3/7

                   = 1500

Thus, the old woman gets the wages of Rs 1500.

25. Question

Priya covered a certain distance between her house and the cinema hall by walking. With an average speed of 15 km/hr, she is late by 10 mins. If she increases her speed to 20 km/hr, she reaches the cinema hall 5 min earlier. What is the distance between her house and the cinema hall?

a) 15 km

b) 10 km

c) 12 km

d) 18 km

e) 35 km

Answer: A

Explanation:-

Let the distance = x

time difference= (10 + 5 ) min = 15/60 hrs

According to question,

x / 15 – x /20 = 1/4

4x – 3x = 15

x = 15

Thus the distance is = 15 km

Short Trick:

=15×25×15/5×60

=15 km

Directions:- (26-30): Read the data carefully and answer the questions :

There are four government Schools A, B, C and D . Total of 2600 students studying in the three schools A, B, & C and the ratio of total students studying in these three (A : B : C) is 21: 16: 15 respectively. The number of boys in A school is 250 more than that of girls in C school and the ratio of the number of girls in A school to that of boys in C school is 9: 8. The number of girls in B school is 70% of the number of boys in A school. Total 1000 students studying in D school and the number of boys in D school is 200 more than that of in C school.

Explanation (26 – 30)

Total students in A school = 2600 ×21/52
= 1050
Total students in B School= 2600 ×16/52
= 800
Total students in C School= 2600 ×21/52
= 750

Let the number of girls in A school and number of boys in C school be 9x and 8x respectively
The number of boys in A school= (1050−9????)
Number of girls in C school= (750-8x)
According to the question,
(1050−9????)−(750 − 8x) = 250
⇒???? = 50
Number of girls in A school = 450
Number of boys in A school= 1050 – 450= 600
Number of boys in C school= 400
Number of girls in C school= 7500-400= 350
Number of girls in C school= 70/100× 600 = 420
Number of boys in B School= 800 – 420= 380
Total Boys in D school = 400 + 200 = 600
So, total girls in D school= 1000 – 600 =400

SCHOOLS	BOYS	GIRLS	TOTAL
A	600	450	1050
B	380	420	800
C	400	350	750
D	600	400	1000

Note:- Above explanation covers questions 26 to 30.

26. Question

Find the average number of boys in A & D School?

A) 400
B) 500
C) 600
D) 700
E) 350

Answer :- C
Explanation :-
Required average =( 600+600)/2
= 600

27. Question

Total girls in A School are what per cent more than that of in D School?

A) 10 %
B) 12.5 %
C) 14.5 %
D) 15 %
E) 20%

Answer :- B
Explanation :-
Required percentage =( 450−400)/400× 100
= 50/400× 100 = 12.5%

28. Question

Find the ratio of total students in C school to total girls in A & C together?

A) 15 : 16
B) 20 : 21
C) 10 : 11
D) 5 : 6
E) None of these

Answer :- A
Explanation :-
Required ratio = 750
(450+350)
= 15 : 16

29. Question

Find the difference between total boys in B & C School together and total girls in A School?
A) 300
B) 330
C) 370
D) 450
E) 510

Answer :-B
Explanation :-
Required difference = (380 + 400) – 450 = 330

30. Question

Total girls in A school is what per cent of total boys in D School?

A) 25 %
B) 50%
C) 60 %
D) 75 %
E) 80 %

Answer:- D
Explanation:-
Required percentage = 450/600× 100 = 75%

Directions ( 31 – 32):-

Study the following information carefully and answer the questions given below. 
In a football tournament, the probability that A club will win this match is 1/3 and the probability that B club will win this match is 1/2.

31. Question
What is the probability that both A and B club will win this match?

a) 1
b) 1/3
c) 2/3
d) 1/6
e) None of these

Answer:- D
Explanation:-

The probability that A club win the match = 1/3
The probability that B club will win the match = 1/2
The probability that both clubs will win the match = Probability that A club will win and the probability that B club will win
= 1/3 x 1/2     
= 1/6   
Thus, the probability that both A and B club will win the match =1/6  

32. Question

What is the probability that only 1 club [either A club or B club] will win the match?

1. 1/2

2. 1/3

3. 1

4. 2/3

Answer:- A
Explanation:-

The probability that the A club will win the match = 1/3
So, the Probability that A the club will lose the match = 2/3.
The probability that the B club will win the match = 1/2
So, the Probability that the B club will lose the match = 1/2.
The probability that only one club will win is when A club wins and B club loses or A club loses and B club wins.
= (1/3) x (1/2) + (1/2) x (2/3)    
= 1/2

33. Question

Three lady workers A, B, and C can do work in 20 days, if they work together. But B lady can work only half time due to her pregnancy. The work can be finished in how many days? 
(a) 12 days 
(b) 17 days 
(c) 18 days 
(d) 24 days 
(e) None of these

Answer:- D
Explanation:-

Let, the efficiency of each worker is 1 unit/day
For three workers, it is 3 units/day
Hence in 20 days, the Total work to be done 
= (20×3) = 60 units
If B lady worker works for half a day only  
Then total work done in a day by three worker = 1+ 1 + 0.5 = 2.5  
Hence, No. of days required to finish = 60/2.5 days = 24 days

34.Question
A toy seller marks his toys at a certain price. He allows a discount on the marked price is 15% and still he gains 25 %. At what per cent above the cost price does he marks in his toys? 

(a) 288/3 %
(b) 35% 
(c) 520/7 %
(d) 800/17 %
(e) None of these

Answer :- D
Explanation :-
Let, Marked price (MP) = 100
Selling price ( SP) = (100 – 15) = 85
Cost price ( CP) = 85 × 100/125 = 68
MP – CP = (100 – 68) = 32
Required % = 32/68 × 100 % 
                   = 800/17%
Thus, the required percentage above the cost price, he marks his toy= 800/17 %

35. Question

Two jars P and Q contain a mixture of Alcohol and water. The ratio of alcohol and water in jar P and Q are in ratios 2 : 1 and 2 : 3 respectively. In what ratio these two mixtures should be mixed so that the new mixture contains 50% alcohol and 50% water. 

(a) 3 : 10
(b) 2 : 7
(c) 5 : 12
(d) 1 : 2
(e) 3 : 7

Answer:-A
Explanation:-

         P                     Q
       2/3                  2/5
                 \      /
                   \ /
                   1/2
                /         \
              /             \
   (1/2 – 2/5)       ( 2/3 – 1/2)
    = 1/10                = 1/3

P : Q = 1/10 : 1/3
         = 3 : 10
Thus, the required ratio = 3 : 10