Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for **LIC AAO Mains 2019** so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.

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**1) Ruma borrowed Rs.___________ from Neeraj at 8% per annum for three years on simple interest. Kavita borrowed the same amount as Ruma from Neeraj at same rate of interest and for the same time but on compound interest. Difference between the interest earned by Neeraj from Kavita and Ruma is Rs._____________.**

**Which of the following option/options satisfy the given condition?**

**I. 60000, 1162.72**

**II. 40000, 788.48**

**III. 80000, 1576.96**

a) Only I and II

b) Only II and III

c) Only I and III

d) Only II

e) All I, II and III

**2) A boat goes 40 km downstream and returns to the initial point in __________ hours. The boat can go 120 Km downstream in ____________ hours. Time taken by the boat to go 60 Km downstream and 56 Km upstream is _____________ hours.**

**Which of the following option/options satisfy the given condition?**

a) 6, 6, 8.6

b) 25, 15, 35.5

c) 12, 6, 17

d) All of the above

e) None of the above

**3) Respective ratio of the incomes of Kamal and Saurav is 16: 15 and respective ratio of their expenditures is 8: 7. Kamal and Saurav saves Rs._____________ each. Income of Ganga is 80% of the income of Kamal and he saves Rs.25000. Expenditure of Ganga is Rs.__________.**

**Which of the following option/options satisfy the given condition?**

**I. 40000, 39000**

**II. 32000, 26200**

**III. 24000, 14000**

a) Only I

b) Only III

c) Only I and III

d) Only I and II

e) All of the above

**4) If numerator of a fraction is increased by 25% and denominator of the fraction is increased by 30%, the fraction becomes ____________. The value of 25% of the sum of the fraction and its reciprocal is ____________.**

**Which of the following option/options satisfy the given condition?**

a) 75/104, 25/48

b) 25/39, 17/24

c) 5/6, 199/390

d) Only a) and c)

e) All of the above

**5) Raghav spent 20% of his monthly income on household, 10% of the rest on education, 25% of the rest on entertainment and 20% of the rest on other expenditures. He saves Rs.____________. Amount spent by Raghav on entertainment is Rs._____________.**

**Which of the following option/options satisfy the given condition?**

a) 10800, 4500

b) 13500, 5625

c) 18900, 7875

d) All of the above

e) None of the above

**Directions (6 â€“ 10): Study the following information carefully and answer the questions given below:**

In five locations P, Q, R, S and T there are only three type of vehicles used: Bike, Car and Jeep. All the three types of vehicles are available in two colours Red and Black only.

Bar graph given below shows the total number of total vehicles (Bikes + Cars + Jeeps), total number of Bikes and total number of Cars in those five locations respectively.

**Total vehicles = Total Bikes + Total Cars + Total Jeeps**

Table given below shows the ratio of Red colour to Black colour vehicles in all the three types in all the five locations.

**6) What is the ratio of total number of red colour Jeeps in the locations P, Q and R together to the to the total number of Black colour Cars in the locations R, S and T together?**

a) 5: 6

b) 3: 4

c) 9: 10

d) 2: 3

e) None of these

**7) Total number of back colour bikes in all the five locations together is approximately what per cent of the total number of all the three types of vehicles in the locations P and R together?**

a) 54(1/3)%

b) 85(2/3)%

c) 46(1/6)%

d) 32(2/9)%

e) None of these

**8) What is the average of difference between the total number of red colour bikes in the locations P and Q, difference between total number of black colour Jeeps in the locations Q and R and difference between the total number of black colour cars in the locations S and T?**

a) 145

b) 160

c) 120

d) 135

e) None of these

**9) Total number of black colours cars in the locations P, Q and R together is approximately what per cent of total number of Jeeps in all the five locations together?**

a) 60%

b) 85%

c) 80%

d) 75%

e) None of these

**10) If 35 red colour bikes from location S are excluded and 25 more red colour bikes are added to location Q and at the same time 15 black colours bikes are excluded from location R and 15 more black colours bikes are added to location Q, then what is the new ratio of total red colour bikes to the total black colour bikes in the locations Q, R and S together?**

a) 89: 115

b) 24: 37

c) 1: 2

d) 46: 83

e) None of these

**Answers:**

**1) Answer: b)**

We know that, for three years

CI â€“ SI = P x (r/100)^{2} x ((300 + r)/100)

**(I)**

CI â€“ SI = 60000 x (8/100)^{2} x 308/100

= 60000 x 64/10000 x 308/100

= Rs.1182.72 â‰ 1162.72

=> Doesnâ€™t satisfy the given condition.

**(II)**

CI â€“ SI = 40000 x (8/100)^{2} x 308/100

= 40000 x 64/10000 x 308/100

= Rs. 788.48 = 788.48

=> Satisfies the given condition.

**(III)**

CI â€“ SI = 80000 x (8/100)^{2} x 308/100

= 80000 x 64/10000 x 308/100

= Rs. 1576.96 = 1576.96

=> Satisfies the given condition.

**2) Answer: d)**

Let, speed of the boat in still water = x Km/h

And speed of the stream = y Km/h

**(a)**

40/(x + y) + 40/(x â€“ y) = 6 ———— (i)

120/(x + y) = 6

=> x + y = 120/6

=> x + y = 20

From (i)

40/20 + 40/(x â€“ y) = 6

=> 40/(x â€“ y) = 6 â€“ 2

=> 40/(x â€“ y) = 4

=> x â€“ y = 40/4

=> x â€“ y = 10

Let, required time = t minutes

60/20 + 56/10

= 3 + 5.6

= 8.6 = 8.6

=> Satisfies the given condition.

**(b)**

40/(x + y) + 40/(x â€“ y) = 25 ———— (i)

120/(x + y) = 15

=> x + y = 120/15

=> x + y = 8

From (i)

40/8 + 40/(x â€“ y) = 25

=> 40/(x â€“ y) = 25 â€“ 5

=> 40/(x â€“ y) = 20

=> x â€“ y = 40/20

=> x â€“ y = 2

Let, required time = t minutes

60/8 + 56/2

= 7.5 + 28

= 35.5 = 35.5

=> Satisfies the given condition.

**(c)**

40/(x + y) + 40/(x â€“ y) = 12 ———— (i)

120/(x + y) = 6

=> x + y = 120/6

=> x + y = 20

From (i)

40/20 + 40/(x â€“ y) = 12

=> 40/(x â€“ y) = 12 â€“ 2

=> 40/(x â€“ y) = 10

=> x â€“ y = 40/10

=> x â€“ y = 4

Let, required time = t minutes

60/20 + 56/4

= 3 + 14

= 17 = 17

=> Satisfies the given condition.

**3) Answer: d)**

Let, incomes of Kamal and Saurav be Rs.16k and Rs.15k respectively.

**(I)**

(16k â€“ 40000)/(15k â€“ 40000) = 8/7

=> 112k â€“ 280000 = 120k â€“ 320000

=> 120k â€“ 112k = 320000 â€“ 280000

=> 8k = 40000

=> k = 40000/8

=> k = 5000

Income of Kamal = 16k = 16 x 5000 = Rs.80000

Income of Ganga = 80/100 x 80000 = Rs.64000

Expenditure of Ganga = 64000 â€“ 25000 = Rs.39000 = 39000

=> Satisfies the given condition

**(II)**

(16k â€“ 32000)/(15k â€“ 32000) = 8/7

=> 112k â€“ 224000 = 120k â€“ 256000

=> 120k â€“ 112k = 256000 â€“ 224000

=> 8k = 32000

=> k = 32000/8

=> k = 4000

Income of Kamal = 16k = 16 x 4000 = Rs.64000

Income of Ganga = 80/100 x 64000 = Rs.51200

Expenditure of Ganga = 51200 â€“ 25000 = Rs.26200 = 26200

=> Satisfies the given condition

**(III)**

(16k â€“ 24000)/(15k â€“ 24000) = 8/7

=> 112k â€“ 168000 = 120k â€“ 192000

=> 120k â€“ 112k = 192000 â€“ 168000

=> 8k = 24000

=> k = 24000/8

=> k = 3000

Income of Kamal = 16k = 16 x 3000 = Rs.48000

Income of Ganga = 80/100 x 48000 = Rs.38400

Expenditure of Ganga = 38400 â€“ 25000 = Rs.13400 â‰ 14000

=> Doesnâ€™t satisfy the given condition.

**4) Answer: a)**

Let, the fraction be a/b.

(a x 125/100)/(b x 130/100) = 25a/26b

**(a)**

25a/26b = 75/104

=>a/b = 75/104 x 26/25

=>a/b = Â¾

Now,

25/100 x (3/4 + 4/3) = Â¼ x (9 + 16)/12 = 25/48 = 25/48

=> Satisfies the given condition

**(b)**

25a/26b =25/39

=>a/b = 25/39 x 26/25

=>a/b = 2/3

Now,

25/100 x (2/3 + 3/2) = Â¼ x (4 + 9)/6 = 13/24 â‰ 17/24

=> Doesnâ€™t satisfy the given condition.

**Â (c)**

25a/26b = 5/6

=>a/b = 5/6 x 26/25

=>a/b = 13/15

Now,

25/100 x (13/15 + 15/13) = Â¼ x (169 + 225)/195 = Â¼ x 394/195 = 197/390 â‰ 199/390

=> Doesnâ€™t satisfy the given condition.

**5) Answer: d)**

Let, total income of Raghav be Rs. P

**(a)**

(100 â€“ 20)/100 x (100 â€“ 10)/100 x (100 â€“ 25)/100 x (100 â€“ 20)/100 x P = 10800

=> 80/100 x 90/100 x 75/100 x 80/100 x P = 10800

=> P = 10800 x 100/80 x 100/90 x 100/75 x 100/80

=> P = Rs.25000

Amount spent by Raghav on entertainment

= 80/100 x 90/100 x 25/100 x 25000

= Rs.4500 = 4500

=> Satisfies the given condition.

**(b)**

100 â€“ 20)/100 x (100 â€“ 10)/100 x (100 â€“ 25)/100 x (100 â€“ 20)/100 x P = 13500

=> 80/100 x 90/100 x 75/100 x 80/100 x P = 13500

=> P = 13500 x 100/80 x 100/90 x 100/75 x 100/80

=> P = Rs.31250

Amount spent by Raghav on entertainment

= 80/100 x 90/100 x 25/100 x 31250

= Rs.5625 = 5625

=> Satisfies the given condition.

**(c)**

100 â€“ 20)/100 x (100 â€“ 10)/100 x (100 â€“ 25)/100 x (100 â€“ 20)/100 x P = 18900

=> 80/100 x 90/100 x 75/100 x 80/100 x P = 18900

=> P = 18900 x 100/80 x 100/90 x 100/75 x 100/80

=> P = Rs.43750

Amount spent by Raghav on entertainment

= 80/100 x 90/100 x 25/100 x 43750

= Rs.7875 = 7875

=> Satisfies the given condition.

**Directions (6 – 10): **

Total number of Jeeps is location P = 1600 â€“ (600 + 750) = 250

Total number of Jeeps is location Q = 1500 â€“ (800 + 400) = 300

Total number of Jeeps is location R = 1400 â€“ (750 + 450) = 200

Total number of Jeeps is location S = 1000 â€“ (500 + 350) = 150

Total number of Jeeps is location T = 800 â€“ (450 + 250) = 100

Now, tabulate this data in to the table:

**6) Answer: c)**

Total number of red colour Jeeps in location P = 250 * (7/10) = 175

Total number of red colour Jeeps in location Q = 300 * (5/12) = 125

Total number of red colour Jeeps in location R = 200 * (3/4) = 150

Total number of red colour Jeeps in the locations P, Q and R together

= 175 + 125 + 150 = 450

Total number of Black colour Cars in location R = 450 * (4/9) = 200

Total number of Black colour Cars in location S = 350 * (4/7) = 200

Total number of Black colour Cars in location T = 250 * (2/5) = 100

Total number of Black colour Cars in the locations R, S and T together

= 200 + 200 + 100 = 500

Required ratio = 450: 500 = 9: 10

**7) Answer: a)**

Total number of back colour bikes in location P = 600 * (7/15) = 280

Total number of back colour bikes in location Q = 800 * (3/8) = 300

Total number of back colour bikes in location R = 750 * (3/5) = 450

Total number of back colour bikes in location S = 500 * (4/5) = 400

Total number of back colour bikes in location T = 450 * (4/9) = 200

Total number of back colour bikes in all the five locations together

= 280 + 300 + 450 + 400 + 200 = 1630

Total number of all the three types of vehicles in the locations P and R together = 1600 + 1400 = 3000

Required per cent = (1630/3000) * 100

= 54(1/3)%

**8) Answer: d)**

Total number of red colour bikes in the location P = 600 * (8/15) = 320

Total number of red colour bikes in the location Q = 800 * (5/8) = 500

Difference between the total number of red colour bikes in the locations P and Q= 500 â€“ 320 = 180

Total number of black colour Jeeps in the location Q = 300 * (7/12) = 175

Total number of black colour Jeeps in the location R = 200 * (1/4) = 50

Difference between total number of black colour Jeeps in the locations Q and R

= 175 â€“ 50 = 125

Total number of black colour cars in the location S = 350 * (4/7) = 200

Total number of black colour cars in the location T = 250 * (2/5) = 100

Difference between the total number of red colour cars in the locations S and T = 200 â€“ 100 = 100

Required average = (180 + 125 + 100)/3 = 135

**9) Answer: b)**

Total number of black colours cars in the location P = 750 * (2/3) = 500

Total number of black colours cars in the location Q = 400 * (3/8) = 150

Total number of black colours cars in the location R = 450 * (4/9) = 200

Total number of black colours cars in the locations P, Q and R together

= 500 + 150 + 200 = 850

Total number of Jeeps in all the five locations together

= 250 + 300 + 200 + 150 + 100 = 1000

Required per cent = (850/1000) * 100

= 85%

**10) Answer: a)**

Red colour bikes in location Q = 800 * (5/8) = 500

Black colour bikes in location Q = 800 * (3/8) = 300

Red colour bikes in location R = 750 * (2/5) = 300

Black colour bikes in location R = 750 * (3/5) = 450

Red colour bikes in location S = 500 * (1/5) = 100

Black colour bikes in location S = 500 * (4/5) = 400

After all the operations:

Total number of red colour bikes in location Q, R and S together

= (500 + 25) + (300) + (100 â€“ 35) = 890

Total number of black colour bikes in location Q, R and S together

= (300 + 15) + (450 â€“ 15) + (400) = 1150

Required ratio = 890: 1150 = 89: 115