# LIC AAO Mains Quantitative Aptitude Questions 2019 (Day-06)

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Directions (Q. 1 – 5): Each question contains a statement followed by Quantity I and Quantity II. Read the contents clearly and give answer as,

a) Quantity I > Quantity II

b) Quantity I â‰¥ Quantity II

c) Quantity II > Quantity I

d) Quantity II â‰¥ Quantity I

e) Quantity I = Quantity II or Relation cannot be established

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1) Quantity I: A boat can cover 42 km upstream in 63 minutes. If the speed of the current is 3/7 of the boat in still water, then how much distance (in km) can the boat cover downstream in 42 minutes?

Quantity II: Speed of a man in still water is 16 km/hr and the river is running at 6 km/hr. The total time taken to go to a place and come back is 8 hours. Find the distance travelled by the man?

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2) A bag contains, 4 pink, 7 yellow and 5 black balls.

Quantity I: If 3 balls are drawn randomly, then find the probability of getting at least one yellow ball?

Quantity II: If 2 balls are drawn randomly, then find the probability of getting both the balls are either pink or black?

3) Quantity I: A bangle was sold for Rs. 250 with a profit of 25 %. If it were sold for Rs. 230, then what would have been the percentage of profit?

Quantity II: A person bought two books for Rs. 600 each. He sold one at a profit of 20 % and the other at a loss of 10 %. What would be his overall profit or loss percentage in the whole transaction?

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4) Quantity I: The total area of a Square and rectangle is 496 Sq cm. The side of the square is 14 cm. Find the perimeter of the rectangle, if length of the rectangle is 25 cm?

Quantity II: The side of the equilateral triangle is equal to the diameter of the circle. The area of the equilateral triangle is 196âˆš3 Sq cm. Find the circumference of the circle

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5) Quantity I: A bucket contains some quantity of milk and water in the ratio of 5: 3. If 40 litres of mixture is drawn out and replaced with water and the ratio of milk and water becomes 5: 11, then find the initial quantity of milk

Quantity II: How many litres of water should be added to a 60 litres mixture containing milk and water in the ratio of 2: 1 such that the resultant mixture has 50 % milk in it?

Directions (6 – 10): Study the following information carefully and answer the given questions.

The following table shows the number of pens (In lakhs) produced and sold by three different companies in 5 different years.

6) If the total number of pens sold by company M and N in the year 2013 is 400 and the number of pens sold by company N in 2014 is 200 more than pens sold by the same company in 2013, then find the number of pens produced by company N in 2014?

a) 730 lakhs

b) 400 lakhs

c) 855 lakhs

d) 615 lakhs

e) 550 lakhs

7) If the ratio of the number of pens sold by company M in 2015 to that in 2016 is 8: 17, then find the percentage of pens were sold by company M in 2016?

a) 18 %

b) 22.5 %

c) 25 %

d) 30.5 %

e) 35 %

8) If company O produced â€˜xâ€™ number of pens in 2013 and â€˜x + 200â€™ number of pens in 2015 and the total number of pens sold by company O in these years is 630 lakhs, then find the number of pens were sold by O in 2015?

a) 200 lakhs

b) 250 lakhs

c) 480 lakhs

d) 300 lakhs

e) 350 lakhs

9) Company N has produced 900 lakhs pens in 2016 and the number of pens sold by the same company in 2013 is 10 % more than number of pens sold by the company in 2016, then find number of pens produced by the same company in 2014 if the number of pens sold by company N in 2013 and 2014 is in the ratio of 9 : 8?

a) 200 lakhs

b) 400 lakhs

c) 480 lakhs

d) 660 lakhs

e) 350 lakhs

10) If the average number of pens sold by company M in 2012, 2013 and 2014 are 200 lakhs, then the production of company M in 2014 is what percentage of the total production of company N in 2015?

a) 90 %

b) 80 %

c) 75 %

d) 65 %

e) 55 %

Direction (1-5) :

Quantity I:

Speed of upstream = (42*60)/63 = 40 km/hr

Speed of current : Speed of still water = 3 : 7

Speed of upstream = Speed of boat in still water – speed of Current

40 = 4x

= > x = 10

Speed of current = 30 km/hr

Speed of boat in still water = 70 km/hr

Speed of downstream= Speed of boat in still water + speed of Current

= > 70 + 30 = 100 km/hr

Speed = 100 km/hr, Time = 42 minutes

Distance = 100*(42/60) = 70 km

Quantity II:

Speed of downstream = 16 + 6 = 22 km/hr
Speed of upstream = 16 â€“ 6 = 10 km/hr
Let the distance travelled be x,
(x/22) + (x/10) = 8
x= 55 km

(Or)

Distance = Time *[(Speed of Still water2 â€“ speed of Stream2)/(2*Speed of Still water)]

= > Distance = 8*[(162 â€“ 62)/(2*16)]

= > Distance = 8*[(256 – 36)/32]

= > Distance = 55 km

Quantity I > Quantity II

Total number of balls = 4 + 7 + 5 = 16 balls

Quantity I:

n(S) = 16C3 = (16*15*14)/(1*2*3)

The probability of getting at least one yellow ball = 1- P (None is yellow ball)

P (None is yellow ball)

n(E) = 9C3 = (9*8*7)/(1*2*3)

P(E) = n(E)/n(S) = 9C3 /16C3

= > [(9*8*7)/(1*2*3)] / [(16*15*14)/(1*2*3)]

= > 3/20

Required probability = 1 – (3/20) = 17/20

Quantity II:

n(S) = 16C2 = (16*15)/(1*2)

n(E) = Probability that both the balls are either pink or black

n(E) = 4C2 or 5C2

P(E) = n(E)/n(S)

= > [4C2 or 5C2] / 16C2

= > [6 + 10] / [(16*15)/(1*2)]

= > (16*2)/(16*15) = 2/15

Quantity I > Quantity II

Quantity I:

According to the question,

(125/100)*CP = 250

CP = 250*(4/5) = Rs. 200

SP = 230

Profit % = (Profit/CP)*100

= > (30/200)*100

= > 15 %

Quantity II:

According to the question,

CP1= 600, Profit = 20 %

SP1 = 600*(120/100) = 720

CP2 = 600, Loss = 10 %

SP2= 600 Ã— (90/100) = 540

Total selling price = S.P1 + S.P2 = 720 + 540 = 1260

Total cost price = 600 + 600 = 1200

Profit % = (60/1200)*100 = 5 %

Quantity I > Quantity II

Quantity I:

a2 + lb = 496

142 + 25b = 496

196 + 25b = 496

25b = 496 â€“ 196

25b = 300

B = 12 cm

Perimeter of rectangle = 2*(l + b) = 2*(25 + 12)

= > 2 * 37 = 74 cm

Quantity II:

The area of the equilateral triangle = 196âˆš3 Sq cm

The area of the equilateral triangle = (âˆš3/4)*a2

(âˆš3/4)*a2 = 196âˆš3

a2 = 196*4

Side (a) = 14*2 = 28 cm

The diameter of the circle = 28 cm

Circumference of the circle = 2Ï€r = 2*(22/7)*14 = 88 cm

Quantity I < Quantity II

Quantity I:

Total quantity of mixture = 40 litres

Water = 40*(3/8) = 15 lit, Milk = 40*(5/8) = 25 lit

(5x – 25)/(3x â€“ 15 + 40) = 5/11

55x – 275 = 15x + 125

= > x = 10 litres

Initial quantity of milk = 10*5 = 50 litres

Quantity II:

Total mixture = 60 litres

Milk = 40 lit, water = 20 lit

According to the question,

40/(20 + x) = 1/1

40 = 20 + x

X = 20 litres

Quantity I > Quantity II

Direction (6-10) :

Number of pen sold by company N in 2013

= > 400 â€“ 600*(43/100)

= > 400 â€“ 258 = 142 lakhs

Number of pen sold company N in 2014 = 142 + 200 = 342 lakhs

Number of pen produced by company N in 2014

= > 342*100/40 = 855 lakhs

Number of pens sold by company M in 2015

= > 480*15/100= 72 lakhs

Number of pens sold by company M in 2016

= > 72*17/8 = 153 lakhs

Required percentage = (153/680)*100 = 22.5 %

Number of pens sold in 2013

= > x*15/100 + (x+200)*40/100 = 630

= > 15x/100 + 40x/100 + 80 = 630

= > (15x+ 40x)/100 = 630 – 80

= > 55x = 550*100

= > x = 1000 lakhs

Number of pens sold in 2015 = (1000+200)*(40/100) = 480 lakhs

Number of pens sold by company N in 2016 = 900*(30/100) = 270 lakhs

Number of pens sold by company N in 2013 = 270*(110/100) = 297 lakhs

Number of pens sold by company N in 2014 = 297*8/9 = 264 lakhs

Number of pens produced by company N in 2014

= > 264*(100/40) = 660 lakhs

Total pens sold by company M in 2012, 2013 and 2014 together

= > 200*3= 600 lakhs

Number of pens sold by company M in 2014

= > 600 â€“ [(60/100)*450 + (43/100)*600]

= > 600 â€“ 270 â€“ 258 = 72 lakhs

Number of pens produced by company M in 2014 = 72*(100/20) = 360 lakhs

Required percentage = (360/450)*100 = 80%