IBPS SO Prelims Quantitative Aptitude Free Online Mock Test  2
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IBPS SO Prelims Quantitative Aptitude Free Online Mock Test 2, IBPS SO Free Mock Test Series. IBPS Free Mock Test for IT Officer. The below online test on IBPS Specialist Officer Recruitment will help you for preperation of the post of IT Officer, Agricultural Field Officer, Rajbhasha Adhikari, Law Officer, HR/Personnel Officer, Marketing Officer. IBPS Quantitative Aptitude Online Test in English. IBPS SO Quantitative Aptitude Quiz2 2019. The IBPS Full online mock test paper is free for all students. IBPS Question and Answers in Hindi and English. IBPS Mock test for Quantitative Aptitude Subject Via Online Mode. Here we are providing Quantitative Aptitude Question for IBPS SO Exam in English. IBPS SO IT Officer Mock Test Series 2019. Now Test your self for IBPS Exam by using below quiz…
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Question 1 of 50
1. Question
The ratio between the present ages of the Rohit and Rina is 1 : 3. Find the present age of the Rina.
Statement I: Difference between the present ages of the Pooja and Rohit is 22 years.
Statement II:The present age of Pooja is 4 years less than thrice the present age of Rohit.
Statement III:Difference between the present ages of the Rina and Rohit is 26 years.Correct
From statement III: Age of Rina = 26/2 *13 = 39 years
From statement I and II:
Rina = 3Rohit , Pooja – Rohit = 22 and 3Rohit – Pooja = 4
On solving, we get Rina = 39 yearsIncorrect
From statement III: Age of Rina = 26/2 *13 = 39 years
From statement I and II:
Rina = 3Rohit , Pooja – Rohit = 22 and 3Rohit – Pooja = 4
On solving, we get Rina = 39 years 
Question 2 of 50
2. Question
What are the marks obtained by Sushil in Physics?
Statement I: Marks obtained in Biology is as much more than that in Chemistry as the marks obtained in Chemistry is more than that in Physics.
Statement II:The average marks obtained by Sushil in Physics, Biology and Chemistry are 65.
Statement III: Marks obtained by Sushil in Biology is 6 more than that obtained in Physics.Correct
From statement I: Biology – Chemistry = Chemistry – Physics
From statement II: Physics + Chemistry + Biology = 3*65 = 195
From statement III: Biology = Physics + 6
From all the above equations , Physics = 62Incorrect
From statement I: Biology – Chemistry = Chemistry – Physics
From statement II: Physics + Chemistry + Biology = 3*65 = 195
From statement III: Biology = Physics + 6
From all the above equations , Physics = 62 
Question 3 of 50
3. Question
What is the area of the hall?
Statement I: Total cost of flooring the hall is Rs. 14,500.
Statement II: Labour cost of flooring the hall is Rs. 3000.
Statement III: Material cost of flooring per sq. metre is Rs. 150.Correct
Let the area of the hall be x m^2.
Then , total material cost = Rs. 150x
Labour cost = Rs. 3000
Therefore,Total cost = 150x + 3000 = 14500
From this we get the value of x.
Hence , all the three statements are required.Incorrect
Let the area of the hall be x m^2.
Then , total material cost = Rs. 150x
Labour cost = Rs. 3000
Therefore,Total cost = 150x + 3000 = 14500
From this we get the value of x.
Hence , all the three statements are required. 
Question 4 of 50
4. Question
A,B,C,D and E are five friends. Their mean age is 18. What is the age of C ?
Statement I : A’s age is 18
Statement II : B’s age is 2 years less than E and E’s age is 6 years less than D.
Statement III : C’s age is 6 years more than B’s age and 4 years more than E’s age.Correct
A+B+C+D+E = 90
From statement I : B+C+D+E = 72
From statement II: B = E – 2 and E = D – 6
so, D = E + 6
From statement III: D = B + 6 and D = E + 4
Combining all three statements, we get the age of C.Incorrect
A+B+C+D+E = 90
From statement I : B+C+D+E = 72
From statement II: B = E – 2 and E = D – 6
so, D = E + 6
From statement III: D = B + 6 and D = E + 4
Combining all three statements, we get the age of C. 
Question 5 of 50
5. Question
What is the area of the right angled triangle ?
statement I: The perimeter of the triangle is 5 times of the base.
statement II: The one of the angles of the triangle is 60deg.
statement III: The length of hypotenuse is 4 cm.Correct
From statement II and III are sufficient to answer the question.
Incorrect
From statement II and III are sufficient to answer the question.

Question 6 of 50
6. Question
The average monthly salary of employees , consisting of officers and workers of an organization is Rs. 3000. The average salary of an officer is Rs. 10,000 while that of a worker Rs. 2000/ month . If there are total 400 employees in the organization . Calculate the number of workers and officers separately.
Correct
Let the no. of officers be x
and no. of workers be (400 – x)
=> 3000*400 = 10000x + 2000(400 – x)
=>x = 50
Therefore ,
No. of workers = 350
No. of officers = 50Incorrect
Let the no. of officers be x
and no. of workers be (400 – x)
=> 3000*400 = 10000x + 2000(400 – x)
=>x = 50
Therefore ,
No. of workers = 350
No. of officers = 50 
Question 7 of 50
7. Question
Rohit ate in a coffee shop and got a membership discount of 30% on the original bill amount but he had to pay 10% as service tax and 8% service charge on the discounted bill amount. If Rohit paid Rs.4,743 ,which included a tip of Rs. 200, how much money did he give as service charge?
Correct
Let the bill after 30% discount be Rs. x
(x*118)/100 = 4743 – 200
=> (x*118)/100 = 4543
=> x= Rs. 3850
Therefore ,
service charge = (3850*8)/100 = Rs.308Incorrect
Let the bill after 30% discount be Rs. x
(x*118)/100 = 4743 – 200
=> (x*118)/100 = 4543
=> x= Rs. 3850
Therefore ,
service charge = (3850*8)/100 = Rs.308 
Question 8 of 50
8. Question
There is a one type of liquid contains 25% of milk ,the other contains 30% of milk .A milk container is filled with 6 parts of the first liquid and 4 parts of the second liquid .Find what is the percentage of milk in the mixture ?
Correct
Formula : (nx + my) /(n + m) %
Required % = (6*25 + 4*30)/ (6+4) = 27%Incorrect
Formula : (nx + my) /(n + m) %
Required % = (6*25 + 4*30)/ (6+4) = 27% 
Question 9 of 50
9. Question
Ishita invested in 3 schemes P,Q and R the amounts in the ratio 2 : 3 : 4 resp. If the schemes offered interest at 20 p.c.p.a , 16 p.c.p.a and 15 p.c.p.a resp. Calculate respective ratio of amount after one year.
Correct
Required ratio = 2x*(120/100) : 3x*(116/100) : 4x*(115/100)
60 : 87 : 115Incorrect
Required ratio = 2x*(120/100) : 3x*(116/100) : 4x*(115/100)
60 : 87 : 115 
Question 10 of 50
10. Question
A man takes 1(7/9) times as long to row a distance upstream as to row the same distance downstream. What is the speed of the boat in still water if it takes 3 hours to travel 38.4 downstream (in km/hr)?
Correct
Let speed of boat in still water = x kmph
speed of current = y kmph
4/(x+y) = 3
=> x+y = 38.4/3 = 12.8kmph
Again,
38.4/(xy) = (16/9)*3 = 16/3
=> x – y = 7.2 kmph
Therefore ,
speed of boat in still water = (12.8+7.2)/2 = 10 kmphIncorrect
Let speed of boat in still water = x kmph
speed of current = y kmph
4/(x+y) = 3
=> x+y = 38.4/3 = 12.8kmph
Again,
38.4/(xy) = (16/9)*3 = 16/3
=> x – y = 7.2 kmph
Therefore ,
speed of boat in still water = (12.8+7.2)/2 = 10 kmph 
Question 11 of 50
11. Question
The number of copies circulated for a first – rank newspaper is 25, 42, and 745 in the world. If the number of copies circulated for the second rank newspaper is less by 6, 15, 430 of the first – rank newspaper, how many copies of second rank newspaper are circulated in the world?
Correct
Required answer = 25,42,745 – 6,15,430
Incorrect
Required answer = 25,42,745 – 6,15,430

Question 12 of 50
12. Question
A bag contains 4 green and 6 red balls. Another bag B contains 3 green and 4 red balls. If one ball is drawn from each bag. Find the probability that both are green.
Correct
Total balls in bag A = 4 + 6 = 10
Total balls in bag B = 7
Probability of green balls in bag A = 4/10
Probability of green balls in bag B = 3/7
Therefore ,
Total probability = (4/10)*(3/7) = 6/35Incorrect
Total balls in bag A = 4 + 6 = 10
Total balls in bag B = 7
Probability of green balls in bag A = 4/10
Probability of green balls in bag B = 3/7
Therefore ,
Total probability = (4/10)*(3/7) = 6/35 
Question 13 of 50
13. Question
In a rectangular tank there are two garbage tanks , X and Y with lengths 12m and 15m resp. in a square field . If the total area of the square field excluding the rectangular tanks is 360 sq. m and the breadth of both the rectangular tanks is (1/3) of the side of the square field , what is the perimeter of the square field (in m)?
Correct
Let length of square be x m.
Breadth of each tank = (x/3) m
Area of square field – Area of both tanks = 360
x^2 – {[(12*x)/3] +[(15*x)/3]} = 360
=> x =24
Therefore ,
Perimeter of the square field = 4*24 = 96mIncorrect
Let length of square be x m.
Breadth of each tank = (x/3) m
Area of square field – Area of both tanks = 360
x^2 – {[(12*x)/3] +[(15*x)/3]} = 360
=> x =24
Therefore ,
Perimeter of the square field = 4*24 = 96m 
Question 14 of 50
14. Question
The ratio between the rates of travelling of P and Q is 3 : 4 and then P takes 10 min. more than the time taken by Q to reach a destination . If P had walked at double the speed , at what time he would have covered the distance ?
Correct
Time taken by P = (x+10) min.
Time taken by Q = x min.
Therefore ,
=> 3 / 4 = x/ (x+10)
=> x = 30 min.
Time taken by P = 40min.
If P doubles his speed ,then time taken by P = 20min.Incorrect
Time taken by P = (x+10) min.
Time taken by Q = x min.
Therefore ,
=> 3 / 4 = x/ (x+10)
=> x = 30 min.
Time taken by P = 40min.
If P doubles his speed ,then time taken by P = 20min. 
Question 15 of 50
15. Question
In how many different ways can the letters of the word “CANDIDATE” be arranged in such a way that the vowels always come together?
Correct
Vowels = A ,I ,A , E
Consonants = C,N,D,D,T
Here ,D and A comes twice.
No. of arrangements = (6! * 4!)/ (2! *2!) = 4320Incorrect
Vowels = A ,I ,A , E
Consonants = C,N,D,D,T
Here ,D and A comes twice.
No. of arrangements = (6! * 4!)/ (2! *2!) = 4320 
Question 16 of 50
16. Question
Shravan and Ronny decided to go on a holiday to B on a particular day from A. Shravan leaves for B at 10 :00 am , at speed of 72km/hr. Ronny leaves for B at 10:30 am same day as Shravan left. At what speed should Ronny travel to catch up with Shravan in 4 hours?[in kmph]
Correct
Distance covered by Shravan in half an hour = 72*(1/2)= 36km
Relative speed of Ronnny = (x72) kmph
Therefore , 36 /(x72) =4
X = 81 kmphIncorrect
Distance covered by Shravan in half an hour = 72*(1/2)= 36km
Relative speed of Ronnny = (x72) kmph
Therefore , 36 /(x72) =4
X = 81 kmph 
Question 17 of 50
17. Question
A started a business. After 4 months from the start of the business, B and C joined. The respective ratio between the amounts invested by A, B and C was 6:11:12. If the C’s share in annual profit was Rs. 720 more than A’s share, what was the total annual profit earned?
Correct
Ratio of the equivalent capitals of A,B and C for 1 month
= 6x * 12 : 11x * 8 : 12x *8
= 9 : 11 : 12
sum of the ratios = 9 + 11+ 12 = 32
If total amount profit be Rs. y
then,
12y/32 – 9y/32 = 720
=> 720 *32/3 = Rs. 7680Incorrect
Ratio of the equivalent capitals of A,B and C for 1 month
= 6x * 12 : 11x * 8 : 12x *8
= 9 : 11 : 12
sum of the ratios = 9 + 11+ 12 = 32
If total amount profit be Rs. y
then,
12y/32 – 9y/32 = 720
=> 720 *32/3 = Rs. 7680 
Question 18 of 50
18. Question
In a school the number of boys and that of the girls are in the respective ratio of 2:3. If the number of boys is increased by 20% and that of girls is increased by 10%, what will be the new ratio of number of boys to that of the girls?
Correct
let the ratio of the no. of boys and girls be 2x and 3x respectively .
No. of boys after 20% increase = 1.20 * 2x = 2.4x
No. of girls after 10% increase = 1.10 * 3x = 3.3x
Required ratio
= 2.4x/3.3x = 8 /11 = 8:11Incorrect
let the ratio of the no. of boys and girls be 2x and 3x respectively .
No. of boys after 20% increase = 1.20 * 2x = 2.4x
No. of girls after 10% increase = 1.10 * 3x = 3.3x
Required ratio
= 2.4x/3.3x = 8 /11 = 8:11 
Question 19 of 50
19. Question
The area of a square is 1024 sq. cm . What is the respective ratio between the length and the breadth of a rectangle whose length is twice the side of the square and breadth is 12 cm less than the side of the square ?
Correct
side of the square = root(1024) = 32 cm
Length of the rectangle = 2* 32 = 64
Breadth of the rectangle = 32 – 12 = 20 cm
Required ratio = 64 : 20 = 16 : 5Incorrect
side of the square = root(1024) = 32 cm
Length of the rectangle = 2* 32 = 64
Breadth of the rectangle = 32 – 12 = 20 cm
Required ratio = 64 : 20 = 16 : 5 
Question 20 of 50
20. Question
In a container there are 25 eggs out of which 5 eggs are rotten . If two eggs are chosen at random, what will be the probability that at least one egg is rotten ?
Correct
Total outcomes= 25c2= 300
Favourable outcomes = 20c1*5c1 +5c2 = 100 +10 = 110
Now, Probability of selecting at least one rotten egg = 110/300
= 11/30Incorrect
Total outcomes= 25c2= 300
Favourable outcomes = 20c1*5c1 +5c2 = 100 +10 = 110
Now, Probability of selecting at least one rotten egg = 110/300
= 11/30 
Question 21 of 50
21. Question
In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly
I. 5×2 + 11x – 12 = 0
II. 4y2 – 20y + 21 = 0
A) If x > y
B) If x < y C) If x ≥ y D) If x ≤ y E) If x = y or relation cannot be establishedCorrect
x = 3, 4/5
y = 1.5, 3.5
Put all values on number line and analyze the relationship
3………4/5……..1.5…….3.5Incorrect
x = 3, 4/5
y = 1.5, 3.5
Put all values on number line and analyze the relationship
3………4/5……..1.5…….3.5 
Question 22 of 50
22. Question
In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly
A) If x > y
B) If x < y C) If x ≥ y D) If x ≤ y E) If x = y or relation cannot be established I. 2x2 + 11x + 14 = 0, II. 3y2 – 10y – 8 = 0Correct
x = 3.5, 2
y = 2/3, 4
Put all values on number line and analyze the relationship
3.5……….2………2/3……….4Incorrect
x = 3.5, 2
y = 2/3, 4
Put all values on number line and analyze the relationship
3.5……….2………2/3……….4 
Question 23 of 50
23. Question
In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly
A) If x > y
B) If x < y C) If x ≥ y D) If x ≤ y E) If x = y or relation cannot be established I. 2x2 + 17x + 30 = 0, II. 4y2 – 13y – 12 = 0Correct
x = 6, 5/2
y = 3/4, 4
Put all values on number line and analyze the relationship
6……5/2…….. 3/4…….4Incorrect
x = 6, 5/2
y = 3/4, 4
Put all values on number line and analyze the relationship
6……5/2…….. 3/4…….4 
Question 24 of 50
24. Question
In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly
A) If x > y
B) If x < y C) If x ≥ y D) If x ≤ y E) If x = y or relation cannot be established I. 3x2 – 10x + 8 = 0, II. 3y2 + 8y – 16 = 0Correct
x = 2, 4/3
y = 4, 4/3
Put all values on number line and analyze the relationship
4…….4/3…….. 2Incorrect
x = 2, 4/3
y = 4, 4/3
Put all values on number line and analyze the relationship
4…….4/3…….. 2 
Question 25 of 50
25. Question
In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly
A) If x > y
B) If x < y C) If x ≥ y D) If x ≤ y E) If x = y or relation cannot be established I. 3x2 – 4x – 4 = 0, II. 4y2 + 23y + 15 = 0Correct
x = 2/3, 2
y = 5, 3/4
Put all values on number line and analyze the relationship
5…….. 3/4…….2/3…….2Incorrect
x = 2/3, 2
y = 5, 3/4
Put all values on number line and analyze the relationship
5…….. 3/4…….2/3…….2 
Question 26 of 50
26. Question
22% of 150 + 29 × 36 + 24% of 750 = x What value should come in place of the x in the following questions?
Correct
33 + 1044 + 180 = 1257
Incorrect
33 + 1044 + 180 = 1257

Question 27 of 50
27. Question
65% of 540 + 868 ÷ 14 – 12 × 39 = x What value should come in place of the x in the following questions?
Correct
351 + 62 – 468 = 55
Incorrect
351 + 62 – 468 = 55

Question 28 of 50
28. Question
√1849 × 1326 ÷ 51 + x% of 350 – 12% of 950 = 1165 What value should come in place of the x in the following questions?
Correct
43 × 26 + x% of 350 – 114 = 1165
x = 46
Incorrect
43 × 26 + x% of 350 – 114 = 1165
x = 46

Question 29 of 50
29. Question
26 × 56 + √1369 – 76% of 150 = x What value should come in place of the x in the following questions?
Correct
1456 + 37 – 114 = 1379
Incorrect
1456 + 37 – 114 = 1379

Question 30 of 50
30. Question
√7921 + 39 × 36 – √2704 ÷ 13 × 23 = x What value should come in place of the x in the following questions?
Correct
89 + 1404 – 4 × 23 = 1401
Incorrect
89 + 1404 – 4 × 23 = 1401

Question 31 of 50
31. Question
The number of clerical cadre employees recruited by bank E in 1999 was approximately what percentage of number of officer’s cadre employees recruited by bank A in 1997 ?
Correct
Required % = (1650/415)*100= 400%
Incorrect
Required % = (1650/415)*100= 400%

Question 32 of 50
32. Question
In the year 1998, which two banks together recruited the highest number of officer cadre employees?
Correct
Bank –B =725
Bank –D =678
Incorrect
Bank –B =725
Bank –D =678

Question 33 of 50
33. Question
In which year was the total number of employees both clerical and officers together recruited by Bank D the maximum?
Correct
The maximum number of employees (both clerical and officers)
= Year 1997 = 2400 + 786 = 3186
Incorrect
The maximum number of employees (both clerical and officers)
= Year 1997 = 2400 + 786 = 3186

Question 34 of 50
34. Question
Which bank shows the continuous decrease in the recruitment of employees (in officer’s cadre) over the years?
Correct
From the table its clear : Bank C
Incorrect
From the table its clear : Bank C

Question 35 of 50
35. Question
What is the average number of employees recruited in Bank C in clerical cadre?
Correct
Required average = 4392/6 = 732
Incorrect
Required average = 4392/6 = 732

Question 36 of 50
36. Question
3, 4, 9, 28, 113, 565, 3397 The following series are based on certain pattern. One number in the series is wrong. Find the odd one out
Correct
*1 + 1, *2 + 1, *3 + 1, *4 + 1, *5 + 1
565 > 566
Incorrect
*1 + 1, *2 + 1, *3 + 1, *4 + 1, *5 + 1
565 > 566

Question 37 of 50
37. Question
4, 6, 12, 30, 60, 315, 1260 The following series are based on certain pattern. One number in the series is wrong. Find the odd one out
Correct
*1.5, *2, *2.5, *3, *3.5, *4
So 60 > 90
Incorrect
*1.5, *2, *2.5, *3, *3.5, *4
So 60 > 90

Question 38 of 50
38. Question
200, 196, 192, 180, 160, 130, 88 The following series are based on certain pattern. One number in the series is wrong. Find the odd one out
Correct
Pattern should be like
………2…………6………….12………..20…………..30…………..42
……………..4…………6………….8…………10…………..12
So 196 > 198
Incorrect
Pattern should be like
………2…………6………….12………..20…………..30…………..42
……………..4…………6………….8…………10…………..12
So 196 > 198

Question 39 of 50
39. Question
9.2, 10.6, 7.6, 12.4, 6, 14, 4.4 The following series are based on certain pattern. One number in the series is wrong. Find the odd one out
Correct
9.2 – 1.6 = 7.6, 7.6 – 1.6 = 6, 6 – 1.6 = 4.4
10.8 + 1.6 = 12.4, 12.4 + 1.6 = 14
So 10.6 > 10.8
Incorrect
9.2 – 1.6 = 7.6, 7.6 – 1.6 = 6, 6 – 1.6 = 4.4
10.8 + 1.6 = 12.4, 12.4 + 1.6 = 14
So 10.6 > 10.8

Question 40 of 50
40. Question
1, 730, 973, 1054, 1081, 1069, 1093 The following series are based on certain pattern. One number in the series is wrong. Find the odd one out
Correct
+3^{6}, +3^{5}, +3^{4}, +3^{3}, +3^{2}, +3^{1}
So 1069 > 1090Incorrect
+3^{6}, +3^{5}, +3^{4}, +3^{3}, +3^{2}, +3^{1}
So 1069 > 1090 
Question 41 of 50
41. Question
Following bargraph shows the percentage profit earned by three companies Wipro, TCS and Infosys in the period of 2011 to 2018.
If the expenditure of TCS in the year 2011 and the income of Infosys in the year 2014 are equal then what is the ratio of the income of TCS in the year 2011 to the expenditure of Infosys in the year 2014?
Correct
Let Expenditure TCS = Income Infosys = x
Income TCS = x × (100 + 60)/100 = 8x/5
Expenditure Infosys = x × 100/100 + 25 = 4x/5
Required ratio = Income TCS/Expenditure Infosys
= 8x/ 5 × 5/4x
= 2/1
Incorrect
Let Expenditure TCS = Income Infosys = x
Income TCS = x × (100 + 60)/100 = 8x/5
Expenditure Infosys = x × 100/100 + 25 = 4x/5
Required ratio = Income TCS/Expenditure Infosys
= 8x/ 5 × 5/4x
= 2/1

Question 42 of 50
42. Question
Following bargraph shows the percentage profit earned by three companies Wipro, TCS and Infosys in the period of 2011 to 2018.
If the total expenditure of Wipro in the year 2011 and Infosys in the year 2015 together is 94 lakh then what is the sum of the total income of Wipro in 2011 and Infosys in 2015?
Correct
% profit of Wipro in 2011 and % profit of Infosys in 2015 are equal and are 40%,
Total income = 94 × (100 + 40)/100
= 94 × 1.4
= 131.6 lakh
Incorrect
% profit of Wipro in 2011 and % profit of Infosys in 2015 are equal and are 40%,
Total income = 94 × (100 + 40)/100
= 94 × 1.4
= 131.6 lakh

Question 43 of 50
43. Question
Following bargraph shows the percentage profit earned by three companies Wipro, TCS and Infosys in the period of 2011 to 2018.
If the expenditure of Wipro in the year 2013 is 55.5 lakh then what is its income in that year?
Correct
% Profit 2013 = 48%,
Expenditure = 55.5 lakh
Income = 55.5 × (100 + 48)/100
= 55.5 × 1.48
= 82.14
Incorrect
% Profit 2013 = 48%,
Expenditure = 55.5 lakh
Income = 55.5 × (100 + 48)/100
= 55.5 × 1.48
= 82.14

Question 44 of 50
44. Question
Following bargraph shows the percentage profit earned by three companies Wipro, TCS and Infosys in the period of 2011 to 2018.
If the income of Wipro in year 2011 and expenditure of TCS in year 2012 are equal and 91 lakh each then what is the difference between the income of TCS in 2012 and the expenditure of Wipro in the year 2011?
Correct
% Profit Wipro = 40%
Income Wipro = 91 lakh
Expenditure Wipro = 91 ×100/140
= 65 lakh
% Profit TCS = 50%,
Expenditure TCS = 91 lakh
Income TCS = 91 × 150/100
= 136.6 lakh
Difference = 136.5 – 65
= 71.5 lakh
Incorrect
% Profit Wipro = 40%
Income Wipro = 91 lakh
Expenditure Wipro = 91 ×100/140
= 65 lakh
% Profit TCS = 50%,
Expenditure TCS = 91 lakh
Income TCS = 91 × 150/100
= 136.6 lakh
Difference = 136.5 – 65
= 71.5 lakh

Question 45 of 50
45. Question
Following bargraph shows the percentage profit earned by three companies Wipro, TCS and Infosys in the period of 2011 to 2018.
What is the percentage rise in the percentage profit of TCS from 2013 to 2014?
Correct
% Profit 2013 = 40%
% Profit 2014 = 45%
% Rise = (45 – 40)/40 × 100
= 500/40
= 12.5%
Incorrect
% Profit 2013 = 40%
% Profit 2014 = 45%
% Rise = (45 – 40)/40 × 100
= 500/40
= 12.5%

Question 46 of 50
46. Question
What is the income of Redmi in the year 2015?
Correct
Income of Redmi in 2015
= 25.4 + 25.4 × 21.5/100
= 25.4 + 5.461
= 30.861 crore
Incorrect
Income of Redmi in 2015
= 25.4 + 25.4 × 21.5/100
= 25.4 + 5.461
= 30.861 crore

Question 47 of 50
47. Question
The income of Vivo in the year 2014 is approximately what per cent of the expenditure of Oppo in the year 2013?
Correct
Income of Vivo in 2014
= 30.33crore
Expenditure of Oppo in 2013 = 21 crore
Required % = 30.33 /21 × 100
= 144.42
= 145%
Incorrect
Income of Vivo in 2014
= 30.33crore
Expenditure of Oppo in 2013 = 21 crore
Required % = 30.33 /21 × 100
= 144.42
= 145%

Question 48 of 50
48. Question
The expenditure of Oppo in the year 2011 and 2018 together is approximately what percent of the expenditure of Redmi in the year 2012 and 2014 together?
Correct
Expenditure of Oppo
= 17.8 + 23.2 = 41 crore
Expenditure of Redmi
= 27.5 + 22.5 = 50 crore
Required% = 41/50 × 100
= 82%
Incorrect
Expenditure of Oppo
= 17.8 + 23.2 = 41 crore
Expenditure of Redmi
= 27.5 + 22.5 = 50 crore
Required% = 41/50 × 100
= 82%

Question 49 of 50
49. Question
The percentage profit of Redmi in the year 2013 is approximately what per cent more or less than the percentage profit of Oppo in the year 2011?
Correct
Percentage profit of Redmi in 2009 = 28.4%
Percentage profit of Oppo in 2011 = 16.2%
Required % = (28.4 – 16.2)/16.2 × 100
= 12.2 × 100/16.2
= 75.3%
= 75%
Incorrect
Percentage profit of Redmi in 2009 = 28.4%
Percentage profit of Oppo in 2011 = 16.2%
Required % = (28.4 – 16.2)/16.2 × 100
= 12.2 × 100/16.2
= 75.3%
= 75%

Question 50 of 50
50. Question
What is the difference between the profits of Oppo and Vivo in the year 2018?
Correct
Profit of Oppo in 2018
= 23.2 × 31.5/100
= 7.308
Profit of Vivo in 2018
= 24.8 × 27.5/100
= 6.82
Difference = 7.308 – 6.820
= 0.488 crore
= 48.8 lakh
Incorrect
Profit of Oppo in 2018
= 23.2 × 31.5/100
= 7.308
Profit of Vivo in 2018
= 24.8 × 27.5/100
= 6.82
Difference = 7.308 – 6.820
= 0.488 crore
= 48.8 lakh