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# IBPS Clerk Mains Quantitative Aptitude Question Paper 2020

Directions: (Q 1- 6) In the following line graph, Line P shows the per cent by which the marked price of an article is above the cost price, and Line Q shows the discount per cent on 5 different products. Study the Line graph carefully and answer the following questions.

Que 1. If the selling price of article A is Rs.3450, then find the cost price of the article.

A) Rs.  2700

B) Rs. 3100

C) Rs. 2300

D) Rs. 2500

E) Rs. 3000

Explanation :

Selling price (SP) = 3450 ; MP = 150% above CP ; Discount = 40%

CP           MP                         SP

100         250               (250 × 60/100) = 150

150 units(SP) = 3450

=> 100 units (CP) = 3450 × (100/150) = 2300.

∴ The cost price is Rs.2300.

Here the correct option is C.

Que 2. If the discount percent on article C is further increased by 10%, then find the selling price of the article if the cost price of the article is Rs.2500.

A) Rs. 800

B) Rs. 970

C) Rs.1020

D) Rs.1080

E) Rs.1210

Explanation :

Cost price of C= 2500 ; Discount = 80% : MP = 140% above CP

MP = 2500 × 240/100 = 6000

SP = 6000 × 20/100 = 1200

Selling price (SP) after Additional 10% discount = 1200 × 90/100 = Rs.1080

∴ The selling price of the article = Rs.1080.

Here the correct option is D.

Que 3. Find the ratio of the selling price of article E and the marked price of article D, if the cost price of article D is two times the cost price of article E.

A) 9 : 35

B) 12 : 37

C) 15 : 61

D) 7 : 29

E) 12 : 41

Explanation :

Let, CP of E = x

CP of D = 2x

MP of article E = x  × (225/100) = 9x/4.

SP of article E = 9x/4 × (40/100) = 18x/20.

MP of article D = 2x ×(175/100) = 7x/2.

∴ Required ratio = (18x/20) : (7x/2) = 9 : 35

Here the correct option is A.

Que 4. Find the loss percent on article B if a further discount of 9.09% is given on the article.

A) 10%

B) 20%

C) 30%

D)50%

E) 40%

Explanation :

CP of B = 100

MP of B = 220

SP of B = 220 × (25/100) = 55

SP of B after further discount of 9.09%                 = 55 × (90.91/100) = 50

∴ Required loss percentage =(50/100 × 100) = 50%

Here the correct option is D.

Que 5. If the cost price of article D is four times the cost price of article C and the selling price of article C is Rs.1440 then find the difference in the loss on selling article C and article D.

A) Rs. 42

B) Rs. 60

C) Rs. 35

D) Rs. 55

E) Rs. 49

Explanation :

Let the cost price of Article D = 100

The cost price of article C = 25

MP of article D = 175

MP of article C = 25 × 240/100 = 60

Selling price of article D = 175 × (50/100) = 87.5

Selling price of article C = 60 × (20/100) = 12

Now, the selling price of article C = Rs.1440

12 units = 1440

1 unit = 120

So, Cost price of article C = 25 × 120 = Rs.3000

The selling price of article C = Rs.1440

Loss = (3000-1440) = Rs.1560

Cost price of article D = 100 × 120 = Rs.12000

Selling price of article D = 87.5 × 120 = Rs.10500

Loss = (12000-10500) = Rs.1500

∴ The required difference = (1560 – 1500) = Rs.60

Here the correct option is B.

Que 6. Find the ratio of marked price to selling price by selling an article B.

A) 4 : 1

B) 4 : 3

C) 5 : 2

D) 3 : 5

E) 4 : 7

Explanation :

CP           MP                         SP

100         220            220×(25/100)=55

∴ Required ratio of marked price (MP) to selling price (SP) = 220 : 55  = 4 : 1

Here the correct option is A.

Que 7. The area of a square is 1764 m² (approx) and the breadth of a rectangle is 1/3 of the side of a square. If the length is 4m less than breadth, then find the area (in m²) of a rectangle.

A) 125 m²

B) 132 m²

C) 140 m²

D) 150 m²

E) 200 m²

Explanation :

Side of the square =√(area) = √1764 = 42 m

Breadth of rectangle = 42 × (1/3) = 14 m

Length of rectangle = (14 – 4) = 10 m

∴ The area of the rectangle = (length × breadth) = 14×10 =140 m²

Here the correct option is C.

Que 8. A person sold an article for Rs.806. If it is sold for Rs. 38 more, then the total selling price will become 20% more than the cost price of the article. Original Selling price is how much percent more than the cost price?

A) 15%

B) 21%

C) 25%

D) 12%

E) 16%

Explanation :

CP                SP

100              120

120 unit = (806+38)

100 unit (CP) = 844 × (100/120) = 704(approximately)

Original Selling price = 806

∴ Profit percent = (806 – 704)/704  × 100 = 15%(approximately)

Here the correct option is A.

Que 9. An article is sold for 12.5% more than the cost price. The selling price is 90% of the marked price. If the article is sold for Rs.31.25 less than the marked price, then find the loss percentage.

A) 7.80%

B) 10.40%

C) 12.75%

D) 15.50%

E) Can’t be determined

Explanation :

12.5% = 1/8

Cost price (CP) = 800

Profit = 100

Selling price = (800 + 100) = 900

Here, selling price > cost price

So, no loss is occurring here.

Here the correct option is E.

Que 10. A boat covers a certain distance in 4 hours in upstream and covers 2 km less distance in downstream in 3 hours. If the speed of the stream is 2 km/hour then find the distance covered by the boat in upstream.

A) 40 km

B) 46 km

C) 56 km

D) 62 km

E) 68 km

Explanation :

Let, Speed of boat = x km/hr

Total distance = D km

Distance = Speed × Time

Now, from the Question

D/(x-2) = 4

=> D = 4x – 8     ———-(1)

(D – 2)/(x + 2) = 3

=> D – 2 = 3x + 6

=> D = 3x + 8   ———–(2)

Equating equation (1) & (2),we get

4x – 8 = 3x + 8

=> x = 16

Speed of boat = 16 km/hr

From equation (1), we get

D = 4 × 16 – 8 = 56 km

∴ Required distance = 56 km

Here the correct option is C.

Directions (Q 11-15): Read the passage carefully and answer the following questions :

In three different colleges, there are three branches Mechanical engineering, Civil engineering and Electrical engineering. The total number of students in college B is 760. The number of students in Mechanical engineering in college B is 300. The number of students in Electrical engineering in college C is twice the number of Mechanical engineering in the same college. The number of students in Mechanical engineering in college B is 60% of the students in the number of students in the same branch in college A. The number of Civil engineering is half the number of Mechanical engineering in college A and 50 more than the number of students in Electrical engineering from college C. The total number of students in Civil engineering in college B is 160 more than the number of students in Electrical engineering in college C. The total number of students in Civil engineering in all the colleges together is 750. The number of students in Electrical engineering in college A is 50% more than the number of students in same branch in college B.

Solution (11-15):

Total number of students in college B = 760

Mechanical engineering students in college B  = 300

Electrical engineering students in college C  = 2 × Mechanical engineering students in  college A

Let Mechanical engineering students in college A = X

Now,

60% of X = 300

=> X = 300 × (100/60)

=> X = 500

Mechanical engineering students in college A  = 500

Civil engineering students in college A = 500/2  = 250

Electrical engineering students in college C + 50 = Civil engineering students in college A

Electrical engineering in college C = 250 – 50 = 200

Mechanical engineering in college C = 200/2 = 100

Civil engineering in college B = Electrical engineering in college C + 160

=> C.E in B = 200 + 160

=> Civil engineering in college B = 360

Civil engineering students in all colleges together = 750

Civil engineering in college C = 750 – (250 + 360) = 140

Electrical engineering in college A = 150% of electrical engineering in college B

Electrical engineering in college B = 760 – (360 + 300) = 100

Electrical engineering students in college A = 100 × (150/100) = 150

Que 11. What is the difference between the number of students in Mechanical engineering in colleges B and C?

A) 40

B) 90

C) 150

D) 200

E) 220

Explanation :

Required difference = (300-100) = 200

Here the correct option is D.

Que 12. Find the total number of students in college C.

A) 440

B) 350

C) 380

D) 410

E) 500

Explanation :

Required total number of students in college C  = (100 + 140 + 200) = 440

Here the correct option is A.

Que 13.  Find the average number of students in Mechanical engineering in all the colleges together.

A) 200

B) 250

C) 300

D) 350

E) 400

Explanation :

Total number of Mechanical engineering students in all colleges together = (500+300+100) = 900

Required average = 900/3 = 300

Here the correct option is C.

Que 14.  What is the ratio between the number students in Civil engineering in college A to the number of students in Electrical engineering in college C?

A) 2 : 3

B) 3 : 4

C) 5 : 4

D) 2 : 5

E) 5 : 7

Explanation :

Required ratio = 250 : 200 = 5 : 4

Here the correct option is C.

Que 15. The total number of students in Civil engineering in colleges B and C together is what percentage of the total number of students in Mechanical engineering in the same colleges together?

A) 50%

B) 75%

C) 100%

D) 125%

E) 150%

Explanation :

Total number of students in Civil engineering in colleges B and C together = (360 + 140) = 500

Total number of students in Mechanical engineering in colleges B and C together = (300 + 100) = 400

Required percentage = (500/400) × 100 =125%

Here the correct option is D.

Que 16. If 140% of one-third of a number is equal to 46.2, then the number 132 is how much more than the original number?

A) 18.60%

B) 21.75%

C) 27.64%

D) 33.34%

E) 40.50%

Explanation :

Let the number = P

According to the question,

P ×(1/3) × (140/100) = 46.2

=> P = 99

Required percentage = (132 – 99)/99 × 100 =33.34%

Here the correct option is D.

Que 17. Train A whose length is 180 m crosses a pole in 9 seconds. Another train B whose length is 120 m running at a speed of 63 km/hr crosses train A while moving in the same direction. Find the time required for train A to cross train B, when both trains start at the same time.

A) 120 seconds

B) 130 seconds

C) 150 seconds

D) 165 seconds

E) 140 seconds

Explanation :

Speed of train A = 180/9 = 20 m/s

Speed of train B = 63 km/hr = 63 × 5/18 m/s = 17.5 m/s

Relative speed = (20 – 17.5) = 2.5 m/s

Total distance = (120 + 180) = 300 m

Required time = 300/2.5 = 120 seconds.

Here the correct option is A.

Que 18. Amardeep travels 180 km with a constant speed and again travels a distance which is 180 km less than three times the previous distance covered by him. If the total time taken by him to cover the whole distance is 36 hours, then find the speed of Amardeep.

A) 15 km/hr

B) 20 km/hr

C) 22 km/hr

D) 25 km/hr

E) 30 km/hr

Explanation :

Distance travelled by Amardeep in 1st case = 180 km

Distance travelled by Amardeep in the 2nd case = (3 × 180) – 180 = 360 km

Total time taken by Amardeep to cover the whole distance= 36 hours

Total distance travelled by Amardeep = (360 + 180) = 540 km

∴ Required speed = 540/36 = 15 km/hr

Here the correct option is A.

Directions (Q 19 – 24): Read the graph carefully and answer the following questions.

There are six companies A, B, C, D, E, and F. The pie chart shows the percentage of female employees in different companies. The table shows the percentage of male employees more than female employees in the given company.

Que 19. If the number of females is 1000 then what is the ratio of Females in company A to the males in Company D?

A) 9 : 11

B) 8 : 11

C) 7 : 11

D) 11 : 5

E) 6 : 13

Explanation :

Total number of females = 1000

Females in Company A = 1000 × (20/100) = 200

Females in Company D = 1000 × (22/100) = 220

Males in Company D = 220 × (125/100) = 275

Required ratio = 200 : 275 = 8 : 11

The ratio of Females in company A to the males in company D is = 8: 11

Here the correct option is B.

Que 20. If the number of females is 1000, find the average number of males in Company B, D and E.

A) 210

B) 225

C) 242

D) 249

E) 275

Explanation :

Total number of females = 1000

Females in Company B = 1000 × (15/100) = 150

Males in Company B =150 × (130/100) = 195

Females in Company D = 1000 × (22/100) = 220

Males in Company D = 220 × (125/100) = 275

Females in Company E =1000 × (15/100) = 150

Males in Company E = 150 × (170/100) = 255

Total number of males in Company B,D and E together = (195 + 275 + 255) = 725

Required Average = 725/3 = 242(approx)

The average number of males in Company B, D and E is = 242

Here the correct option is C.

Que 21. If the number of females is 1000, what is the difference between the number of females in Company E and the number of Males in Company C?

A) 5

B) 12

C) 13

D) 25

E) 3

Explanation :

Females in Company E = 150

Females in Company C =1000 × (10/100) =100

Males in Company C= 100 × (145/100) = 145

Required difference = (150 – 145) = 5

The difference between the number of females in Company E and the number of Males in Company C is = 5

Here the correct option is A.

Que 22. The number of males in Company F is how much percentage more/less than the number of males in Company A?

A) 25% more

B) 25% less

C) 30% more

D) 31% less

E) 35% more

Explanation :

Let the total number of females in all companies together = 1000

Now Females in Company F = 180

Males in Company F = 180 × (115/100) = 207

Females in Company A = 200

Males in Company A = 200 × (150/100) = 300

Required less percentage = (300 – 207)/300 × 100% = 31%

The number of males in Company F is 31% percent less than the number of males in Company A.

Here the correct option is D.

Que 23. If the total number of females is 1500, then find the difference between the number of females in Company B and the number of males in Company A.

A) 225

B) 255

C) 300

D) 315

E) 550

Explanation :

Total number of females = 1500

Females in Company B = 1500 × (15/100) = 225

Females in Company A = 1500 × (20/100) = 300

Males in Company A = 300 × (150/100) = 450

Required difference = (450 – 225) = 225.

Here the correct option is A.

Que 24. What is the central angle corresponding to the number of females in Company C and F together?

A) 98.5°

B) 100.8°

C) 110.5°

D) 75°

E) 130°

Explanation:

Angles corresponding to the number of females in Company C and F together =(10% + 18%) = 28%

Required central angle = 360 × (28/100) =100.8°

The central angle corresponding to the number of females in Company C and F together is 100.8°.

Here the correct option is B.

Que 25.  Pipe A can fill 3/4th of a cistern in 15 hours and pipe B can fill 2/3 rd of the same cistern in 20 hours. If both the pipes open simultaneously then find the time taken by them to fill 80% of the cistern.

A) 6 hour 30 minutes

B) 8 hours

C) 9 hours 36 minutes

D) 10 hours 45 minutes

E) None of these

Explanation :

Pipe A can fill the tank in = 15 × 4/3 = 20 hours

Pipe B can fill the tank in =20 × 3/2 = 30 hours

Let total work = LCM of 20 and 30 = 60 units.

The efficiency of A and B becomes 60/20 and 60/30 i.e., 3 units and 2 units respectively.

80% of 60 = 48 units

Time is taken by A and B to fill 48 units of the tank = 48 / 5 = 9 hours 36 minutes.

Here the correct option is C.

Que 26. A can do a piece of work in 20 days while B can do the same piece of work in 30 days. They started working together and after 6 days A left work. Two days later B is joined by C. If the work is completed in the same time as before then how much percentage efficient is C more than A?

A) 50%

B) 60%

C) 40%

D) 75%

E) 25%

Explanation :

LCM of 20 and 30 = 60 (total work)

A’s 1 day work = 60/20 = 3

B’s 1 day work = 60/30 = 2

(A+B) together complete the work in = 60/(2+3) =12 days

A’s 6 days work = 3 × 6 = 18

B’s (6+2)=8 days work = 2 × 8 = 16

Remaining work [60 – (18 + 16)] = 26 done by B and C together in (12 – 8) = 4 days.

B’s 4 days work = 4 × 2 = 8

Remaining work done by C in 4 days.

So C’s 4 days work = (26 – 8) = 18

C’s 1 day work = 18/4 = 4.5

Efficiency ratio of A and C = 3 : 4.5

Required percentage = (4.5 – 3)/3 ×100 =50%

∴ C is 50% more efficient than A.

Here the correct option is A.

Que 27.  A square is inscribed in a circle and another circle is inscribed in the square in such a way that it touches the sides of a square. If the radius of the smaller circle is 14√2 cm, then find the circumference of the larger circle.

A) 77 cm

B) 88 cm

C) 144 cm

D) 154 cm

E) 176 cm

Explanation :

Side of the square = 2 × 14√2 = 28√2 cm

Here, the diameter of the bigger circle is equal to the diagonal of the square.

Diagonal of the square = √2 × side = √2 × 28√2 = 56 cm

Diameter of the bigger circle = 56 cm

Radius of the bigger circle = 56/2 = 28 cm

Circumference of the larger circle = 2πr  =2 × 22/7 × 28 = 176 cm

∴ The required circumference of the larger circle = 176 cm.

Here the correct option is E.

Que 28.  The present age of the mother is 42 years. After 12 years the ratio of ages of father and mother is 10 : 9 and the ratio of present ages of their son and daughter is 5 : 6. Father’s present age is four times the present age of daughter. Find the sum of the ages of son and daughter after 12 years.

A) 36 years

B) 46 years

C) 42 years

D) 50 years

E) 55 years

Explanation :

Let, 12 years ago the age of the Father and mother be 10x and  9x respectively.

9x = 42 + 12

x = 6

12 years ago father’s age = 10 × 6 = 60 years.

Present age of Father = (60 – 12) = 48 years.

Present age of daughter = 48 × (1/4) = 12 years

The ratio of Present age of son and daughter = 5 : 6

Let, the present age of the son and daughter be 5y and 6y respectively.

6y = 12

y = 2

Present age of son = 5 × 2 = 10 years

After 12 years,  son’s age =10 + 12 = 22 years

After 12 years, daughter’s age =12 + 12 = 24 years

Sum of ages of son and daughter after 12 years = (22 + 24) = 46 years.

Here the correct option is B.

Que 29. A container contains 96 litres mixture of milk and water with 35% milk. If 16 litres of the mixture is taken out and then 10 litres of water is added. What is the ratio of milk and water in the new mixture?

A) 9 : 14

B) 13 : 27

C) 11 : 17

D) 14 : 31

E) 50 : 51

Explanation :

Remaining mixture after taken out 16 litres mixture = (96 – 16) = 80 L

Milk in the remaining mixture = 80 × (35/100) L = 28 L

Water in the remaining mixture = (80 – 28) = 52 L

Total Quantity of water after 10 L of water added = (52 + 10) = 62 L

The ratio of milk and water in the new mixture = 28 : 62 = 14 : 31

∴ the required ratio of milk and water in the new mixture = 14 : 31

Here the correct option is D.

Que 30. The simple Interest for 2 years is Rs.4200. The difference between the compound interest and simple interest for 2 years is Rs.420. Find the Compound interest for 3 years.

A) 6220

B) 6680

C) 7644

D) 7995

E) 8550

Explanation :

Simple interest for 1 year = 4200/2 = Rs.2100

Rate of interest = (420/2100) × 100% = 20%

Difference between CI and SI for 2 years = P×(R/100)²

=> 420 = P × (20 × 20)/(100 × 100)

=> P = 10500

Rate = 20% = 1/5

Principal            Amount

5                           6

(5)³                      (6)³       (for 3 years)

125                       216

CI for 3 years = (216 – 125) = 91

125 units = 10500

91 units = (10500/125) × 91 =Rs. 7644

∴  The compound interest for 3 years is = Rs.7644

Here the correct option is C.

Directions (Q 31-33): Two friends Aasu and Chandan started a business in which Aasu invests Rs.12600 and Chandan invested 1/3 rd of the amount invested by Aasu. Aasu left the business after 5 months and at the same time Sandhu joins with the investment of Rs.6000. At the end of one year, Chandan receives Rs.3000 as a share of the profit.

Solution (31-33):

Investment of Aasu = 12600

Investment of Chandan = 12600 × 1/3 = 4200

Investment of Sandhu = 6000

Profit share of Aasu,Chandan and Sandhu = (12600 × 5) : (4200 × 12) : (6000 × 7) = 105 : 84 : 70

According to the question,

84 units = 3000

1 unit = 3000/84

Profit share of Aasu = 105 × (3000/84) = 3750

The profit share of Chandan = 3000

Profit of Sandhu = 70×(3000/84) = 2500

Que 31. The share of Aasu is how much percent more than the share of Sandhu at the end of the year?

A) 50%

B) 25%

C) 30%

D) 60%

E) 40%

Explanation :

Required percentage  = (3750 – 2500)/2500 × 100% = 50%

Here the correct option is A .

Que 32. Find the total profit earned in the business.

A) 8550

B) 9250

C) 9500

D) 10000

E) 12000

Explanation :

Total profit earned = (3750 + 3000 + 2500)

= Rs.9250

Here the correct option is B.

Que 33. Find the share of Aasu from the profit after one year.

A) 2500

B) 3250

C) 3750

D) 4000

E) 4200

Explanation :

The share of Aasu from the profit after one year = Rs.3750

Here the correct option is C.

Directions (Q 34-40): What approximate value should come in place of question mark (?) in the following questions?

Que 34.  (11.91)² ÷ (3.98)² × (12.94)² – (18.89)² = ?

A) 1160

B) 1220

C) 1140

D) 1250

E) 1300

Explanation :

(11.91)² ÷ (3.98)² × (12.94)² – (18.89)² = ?

=> (12²/4²) × 13² – 19² = ?

=> 9 × 169 – 361 = ?

=> ? = (1521 – 361)

? = 1160

Here the correct option is A.

Que 35.  26.91 ÷ 2.97 × (13.84)² – 169.89 = ?

A) 1356

B) 1436

C) 1594

D) 1280

E) 1680

Explanation :

26.91 ÷ 2.97 × (13.84)² – 169.89 = ?

=> (27/3) × 14² – 170 = ?

=> 9 × 196 – 170 = ?

=> ? = (1764 – 170)

? = 1594

Here the correct option is C.

Que 36.  22.87% of 1199.79 + 29.95% of 229.92 = ?

A) 345

B) 397

C) 381

D) 359

E) 435

Explanation:

22.87% of 1199.79 + 29.95% of 229.92 = ?

=> (23/100) × 1200 + (30/100) × 230 = ?

=> 276 + 69 = ?

? = 345

Here the correct option is A.

Que 37. 124.97% of 300 + 24.99% of 100 – 9.99% of 29.79 = (?)

A) 392

B) 397

C) 380

D) 364

E) 350

Explanation :

124.97% of 300 + 24.99% of 100 – 9.99% of 29.79 = (?)

=> (125/100) × 300 + (25/100) × 100 – (10/100) × 30 = ?

=> 375 + 25 – 3 = ?

? = 397

Here the correct option is B.

Que 38. (69.98% of 99.99) – (19.98% of 224.99) = (?)

A) 22

B) 25

C) 28

D) 33

E) 41

Explanation :

(69.98% of 99.99) – (19.98% of 224.99) = (?)

=> (70/100)×100 – (20/100)×225 = ?

=> 70 – 45 = ?

? = 25

Here the correct option is B.

Que 39. 14.98% of 100.11 ÷ 4.99 × (?) = 29.98

A) 5

B) 7

C) 10

D) 16

E) 21

Explanation :

14.98% of 100.11 ÷ 4.99 × (?) = 29.98

=> (15/100)×100 ×(1/5)×(?) = 30

=> 3 × (?) = 30

? = 10

Here the correct option is C.

Que 40. √624.98 ÷ (29.99 ÷ √36.12) – (?) = √3.99

A) 3

B) 7

C) 9

D) 12

C) 17

Explanation :

√624.98 ÷ (29.99 ÷ √36.12) – (?) = √3.99

=> √625 ÷ (30÷√36) – (?) = √4

=> 25 ÷ 5 – (?) = 2

=> 5 – (?) = 2

? = 3

Here the correct option is A.

Directions (Q 41-45): The table shows the number of students, the percentage of boys who participated, the percentage of girls who participated, and the percentage of students who did not participate in the final examination in 5 years. Some values are missing in the table. Study the following table carefully and answer the questions given below.

Que 41.  If the number of total students present in 2015 is 20% more than that in 2016, then find the number of girls who participated in the final examination of 2015.

A) 151

B) 168

C) 175

D) 192

E) 201

Explanation :

Number of total students present in 2015 = (120/100) × 500 = 600

Participated students = (80/100) × 600 = 480

Participated girls = (35/100) × 480 = 168

The number of girls who participated in the final examination in 2015 = 168.

Here the correct option is B.

Que 42.  Find the ratio of the number of students not participated in 2016 to the number of students not participated in 2017.

A) 25 : 28

B) 28 : 25

C) 25 : 32

D) 23 : 31

E) 20 : 27

Explanation :

The number of students who have not participated in 2016 = (15/100) × 500 = 75

Number of students who have not participated in 2017 = (12/100)×700 = 84

Required ratio = 75 : 84 = 25 : 28

The ratio of the number of students not participated in 2016 to the number of students not participated in 2017 is = 25 : 28

Here the correct option is A.

Que 43.  If the number of total students present in 2019 is 250 more than the number of total students present in 2018, then find the number of girls who participated in the examination of 2019.

A) 385

B) 345

C) 250

D) 275

E) 295

Explanation :

Total students present in 2019 = (750+250) =1000

Total Students who participated in examination = (70/100) × 1000 = 700

Total girls who participated in the examination   = (55/100) × 700 = 385

Here the correct option is A.

Que 44.  Find the average number of total students in 2016, 2017 and 2018 together.

A) 590

B) 620

C) 690

D) 660

E) 650

Explanation :

Required average = (500 + 700 + 750)/3 = 650

Here the correct option is E.

Que 45.  Find the ratio of the number of girls who participated in 2017 to the number of girls who participated in 2018.

A) 78 : 151

B) 77 : 90

C) 77 : 150

D) 78 : 149

E) 77 : 153

Explanation :

Number of students who participated in 2017 = (88/100) × 700 = 616.

Number of girls students who participated in 2017 = (75/100) × 616 = 462.

Number of students who participated in 2018 = (90/100) × 750 = 675.

Number of girls students who participated in 2018 = (80/100) × 675 = 540.

Required ratio = 462 : 540 = 77 : 90

Here the correct option is B.

Directions (Q 46-50): Given below are quantities named A and B. Based on the given information, you have to determine the relationship between the two quantities. You should use the given data and your knowledge of mathematics to choose among the possible answer.

Que.46

Quantity A: The average weight of 4 students is 40 kg. If one more person joins the team then the average weight becomes 42 kg, then find the weight of the new person.

Quantity B: The price of sugar increases by 20% and the person can buy 9 kg less sugar. Find the final quantity of sugar he has.

A) Quantity A > Quantity B

B) Quantity A < Quantity B

C) Quantity A ≥ Quantity B

D) Quantity A ≤ Quantity B

E) Quantity A = Quantity B or No relation.

Explanation :

Quantity A :

Average weight of 4 students = 40 kg

Total weight of 4 students = 40 × 4 = 160 kg

Average weight of 5 students = 42 kg

Total weight of 5 students = 42 × 5 = 210 kg

The weight of new person = (210 – 160) = 50 kg

Quantity B :

20% = 1/5

Original        Increased

5                    6                 (price)

6                    5                (Consumption)

As we know price is inversely proportional to consumption.

(6 – 5)=1 unit => 9 kg

Consumption of sugar after increment of price     = 5 × 9 = 45 kg

So, Quantity A > Quantity B.

Here the correct option is A.

Que 47.  Determine the total cost to paint all the walls of the cuboid.

Statement I. The perimeter of the cuboid is 30 m.

Statement  II. The cost of painting is Rs.25 per meter square.

A) Statement I alone is sufficient to answer the question.

B) Statement II alone is sufficient to answer the question.

C) Either statement I alone or statement II alone is sufficient to answer the question.

D) Statement I and statement II together are not sufficient to answer the question.

E) Both the Statement I and II are needed to answer the question.

Explanation :

To paint all the walls of the cuboid we have to find the total surface area of the cuboid.

Statement I :

Perimeter of cuboid = 4 × (length + breadth + height)

So, 4(length + breadth + height) = 30

From here we can not find the total surface area. Because length, breadth, and height are not given.

Statement II :

Total surface area of the cuboid = 2(lb + bh + lh)

Total cost of paint of the walls of cuboid =2(lb+bh+lh) × 25

l,b and h is not given, so we can not find the total cost.

Statement I and statement II together are not sufficient to answer the question.

Here the correct option is D.

Que 48.  Find the weight of the new person.

Statement I. The average of the 5 person’s weight is 65 kg.

Statement II. One more person joins the group, then the new average weight becomes 68 kg.

A) Statement I alone is sufficient to answer the question.

B) Statement II alone is sufficient to answer the question.

C) Either statement I alone or statement II alone is sufficient to answer the question.

D) Statement I and statement II together are not sufficient to answer the question.

E) Both the Statement I and II are needed to answer the question.

Explanation :

Statement I :

Total age of 5 person = 5 × 65 = 325 kg.

Statement I alone is not sufficient to answer the question.

Statement II:

Total age of 6 person = 6 × 68 = 408 kg.

Statement II alone is not sufficient to answer the question.

From statements I and II, we get

Weight of new person = (408 – 325) = 83 kg.

Both Statements I and II are needed to answer the question.

Here the correct option is E.

Que 49.  The compound interest after 5 years will be?

Statement I. The sum of the amount Rs.5000 is received at a rate of 5% per annum.

Statement II. The simple interest after one year is Rs.250.

A) Statement I alone is sufficient to answer the question.

B) Statement II alone is sufficient to answer the question.

C) Either statement I alone or statement II alone is sufficient to answer the question.

D) Statement I and statement II together are not sufficient to answer the question.

E) Both the Statement I and II are needed to answer the question.

Explanation :

Statement I :

R = 5% , T = 5 years , Amount = 5000 , CI = ?

Principal × (5/100) = 5000

=> Principal = 100000

It is clear from the statement I that we can calculate compound interest easily.

Statement II :

Simple interest = 250

We can not calculate compound interest from statement II.

Statement I alone is sufficient to answer the question.

Here the correct option is A.

Que.50 Given below are quantities named A and B. Based on the given information you have to determine the relationship between the two quantities. You should use the given data and your knowledge of mathematics to choose among the possible answer.

Quantity A: After selling an article X for Rs.225, a shopkeeper incurred a loss of 10%. Find the cost price of X.

Quantity B: The cost price of Y is Rs.240

A) Quantity A > Quantity B

B) Quantity A < Quantity B

C) Quantity A ≥ Quantity B

D) Quantity A ≤ Quantity B

E) Quantity A = Quantity B or No relation.

Explanation :

Quantity A :

Cost price = 225 × (100/90) = Rs.250

Quantity B :

Cost price = Rs.240

Quantity A > Quantity B.

Here the correct option is A.

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