Quantitative Aptitude – Arithmetic – Pipes and Cisterns – A water tank has inlets of two types A

Quantitative Aptitude – Arithmetic - Pipes and Cisterns – A water tank has inlets of two types A

Slot -2 – Quantitative Aptitude – Arithmetic – Pipes and Cisterns – A water tank has inlets of two types A

A water tank has inlets of two types A and B. All inlets of type A when open, bring in water at the same rate. All inlets of type B, when open, bring in water at the same rate. The empty tank is completely filled in 30 minutes if 10 inlets of type A and 45 inlets of type B are open, and in 1 hour if 8 inlets of type A and 18 inlets of type B are open. In how many minutes will the empty tank get completely filled if 7 inlets of type A and 27 inlets of type B are open?

Answer: 48

Suppose inlets of type A fill A liters per minute and type B fills B liters per minute.

So, capacity of tank = 30(10A + 45B) = 60(8A + 18B)

⇒10A + 45B = 16A + 36B

⇒6A = 9B

⇒A = 1.5B

⇒Capacity of Tank = 30(15B + 45B) = 30*60B = 1800 B

Time taken to fill the tank with 7A and 27B, which is 10.5B and 27B, which is 37.5B = 1800 B / 37.5 B = 48 minutes

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Quantitative Aptitude – Arithmetic – Time Speed and Distance – Points A, P, Q and B lie on

Quantitative Aptitude – Arithmetic - Time Speed and Distance – Points A, P, Q and B lie on

Slot -2 – Quantitative Aptitude – Arithmetic – Time Speed and Distance – Points A, P, Q and B lie on the same line

Points A, P, Q and B lie on the same line such that P, Q and B are, respectively, 100 km, 200 km and 300 km away from A. Cars 1 and 2 leave A at the same time and move towards B. Simultaneously, car 3 leaves B and moves towards A. Car 3 meets car 1 at Q, and car 2 at P. If each car is moving in uniform speed then the ratio of the speed of car 2 to that of car 1 is?

a) 2 : 7

b) 2 : 9

c) 1 : 4

d) 1 : 2

Answer: c) 1 : 4

Solution:
Car 2 and car 3 meets at Q, so ratio of their speed = AQ : BQ = 200 : 100 = 2:1

Car 1 and car 3 meets at P, so ratio of their speed = AP : BP = 100 : 200 = 1 : 2

So required ratio of speed of car 1 and car 2 = 1 : 4

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Quantitative Aptitude – Arithmetic – Simple Interest and Compound Interest – Gopal borrows Rs. X

Quantitative Aptitude – Arithmetic - Simple Interest and Compound Interest – Gopal borrows Rs. X

Slot -2 – Quantitative Aptitude – Arithmetic – Simple Interest and Compound Interest – Gopal borrows Rs. X from Ankit at 8%

Gopal borrows Rs. X from Ankit at 8% annual interest. He then adds Rs. Y of his own money and lends Rs. X+Y to Ishan at 10% annual interest. At the end of the year, after returning Ankit’s dues, the net interest retained by Gopal is the same as that accrued to Ankit. On the other hand, had Gopal lent Rs. X+2Y to Ishan at 10%, then the net interest retained by him would have increased by Rs. 150. If all interests are compounded annually, then find the value of X + Y?

Answer: 4000

Solution:
Interest to be repaid to Ankit at the end of the year = 0.08X

Interest that Gopal would receive from Ishan in two cases are as given.

Case I: if he lends X + Y Interest received = (X + Y) × 0.1 = 0.1X + 0.1Y
Interest retained by Gopal after paying to Ankit = (0.1X + 0.1Y) – (0.08X) = 0.02X + 0.1Y
Given that Interest retained by Gopal is same as that accrued by Ankit
So (0.02X + 0.1Y) = 0.08X
=> Y = 0.6X

Case II: if he lends X + 2Y Interest received = (X + 2Y) × 0.1 = 0.1X + 0.2Y
Interest retained by Gopal after paying to Ankit = (0.1X + 0.2Y) – (0.08X) = 0.02X + 0.2Y

Given that interest retained by Gopal would increase by 150

=> (0.02X + 0.2Y) – (0.02X + 0.1Y) = 150
0.1Y = 150

> Y = 1500 and X = 1500×0.6= 2500
Hence X + Y = 2500 + 1500 = 4000

Ans : 4000

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Quantitative Aptitude – Arithmetic – Ratios – The scores of Amal and Bimal in

Quantitative Aptitude – Arithmetic - Ratios – The scores of Amal and Bimal in

Slot -2 – Quantitative Aptitude – Arithmetic – Ratios – The scores of Amal and Bimal in

The scores of Amal and Bimal in an examination are in the ratio 11 : 14. After an appeal, their scores increase by the same amount and their new scores are in the ratio 47 : 56. The ratio of Bimal’s new score to that of his original score is?
a) 8 : 5

b) 3 : 2

c) 4 : 3

d) 5 : 4

Answer: c) 4 : 3

Solution: let their scores were 11x and 14x and it increase by n then

(11x+n)/(14x+n)=47/56

616x+56n=658x+47n

42x=9n

n = 42x/9

So Bimal’s new score = 14x + 42x/9 = 168x/9

So required ratio = (168x/9)/14x = 4∶ 3

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Quantitative Aptitude – Arithmetic – Mixtures – The strength of a salt solution is p%

Quantitative Aptitude – Arithmetic - Mixtures – The strength of a salt solution is p%

Slot -2 – Quantitative Aptitude – Arithmetic – Mixtures – The strength of a salt solution is p%

The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. If three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%. If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%. A fourth solution, D, is produced by mixing B and C in the ratio 2 : 7. The ratio of the strength of D to that of A is?

a) 1 : 3

b) 2 : 5

c) 1 : 4

d) 3 : 10

Answer: a) 1 :3

Solution:

let the concentration of A, B & C are a% , b% & c% respectively .

Then (a+2b+3c )/600=20/100=1/5

a+2b+3c=120——1)

And (3a+2b+c )/600=30/100=3/10

3a+2b+c=180——-2)

By eq 1)/ eq 2) , (a+2b+3c )/(3a+2b+c )=2/3
3a+6b+9c=6a+4b+2c

(2b+7c)/a=3

So correct answer : a) 1 :3

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Quantitative Aptitude – Arithmetic – Mixtures – A 20% ethanol solution is mixed

Quantitative Aptitude – Arithmetic - Mixtures – A 20% ethanol solution is mixed

Slot -2 – Quantitative Aptitude – Arithmetic – Mixtures – A 20% ethanol solution is mixed

A 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1:3 by volume. This mixture is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution, then the unknown concentration of S is?

a) 55%

b) 52%

c) 48%

d) 50%

Answer: d) 50 %

Solution:
Let the volume of 20% ethanol solution in first mixtures = x so
Volume of S = 3x

Total volume of first solution = 4x

Total volume of final solution =2×4x=8x

Volume of ethanol solution in final mixture = x + 4x = 5x ,

So volume of S in final solution = 3x

Thus (31.25-20)/(s-31.25) = 3x/5x = 3/5
56.25=3s-93.75
3s=150
s=50%

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Quantitative Aptitude – Arithmetic – Time Speed and Distance – On a long stretch of east-west

Quantitative Aptitude – Arithmetic - Time Speed and Distance – On a long stretch of east-west

Slot -2 – Quantitative Aptitude – Arithmetic – Time Speed and Distance – On a long stretch of east-west road

On a long stretch of east-west road, A and B are two points such that B is 350 km west of A. One car starts from A and another from B at the same time. If they move towards each other, then they meet after 1 hour. If they both move towards east, then they meet in 7 hrs. The difference between their speeds, in km per hour, is?

Answer: 50

Solution:

Let the speeds of Cars are v and u . then, v+u=350———–1)
and v-u=350/7=50———-2)

So required difference in speeds=50 km/hr

Correct answer∶50

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Quantitative Aptitude – Arithmetic – Mixtures – A jar contains a mixture of 175 ml

Quantitative Aptitude – Arithmetic - Mixtures – A jar contains a mixture of 175 ml

Slot -2 – Quantitative Aptitude – Arithmetic – Mixtures – A jar contains a mixture of 175 ml

A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal takes out 10% of the mixture and substitutes it by water of the same amount. The process is repeated once again. The percentage of water in the mixture is now?

a) 20.5

b) 30.3

c) 25.4

d) 35.2

Answer: d) 35.2

Solution:
initial volume of Alcohol = 700, total volume = 700+ 175 = 875 , replaced quantity = 87.5 & n =2

Using, final volume =initial volume ( 1-(replaced quantity)/(total volume))^n = 700(1-87.5/875)^2 = 700×81/100 = 567
So water in final mixture = 875 – 567 = 308
Required percentage = 308/875×100 = 35.2

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Quantitative Aptitude – Arithmetic – Time Speed and Distance – Points A and B are 150 km apart

Quantitative Aptitude – Arithmetic - Time Speed and Distance – Points A and B are 150 km apart

Slot -2 – Quantitative Aptitude – Arithmetic – Time Speed and Distance – Points A and B are 150 km apart

Points A and B are 150 km apart. Cars 1 and 2 travel from A to B, but car 2 starts from A when car 1 is already 20 km away from A. Each car travels at a speed of 100 kmph for the first 50 km, at 50 kmph for the next 50 km, and at 25 kmph for the last 50 km. The distance, in km, between car 2 and B when car 1 reaches B is (TITA )?

Answer: 5

Solution:
Time taken by Car 1 to travel first 20 km =20/100 hr =1/5 hr

Thus Car 2 will reach B 12 minutes after Car 1 will reach.

Thus Required distance = time×speed=1/5×25=5 km

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Quantitative Aptitude – Arithmetic – Pipes and Cisterns – A tank is emptied every day

Quantitative Aptitude – Arithmetic - Pipes and Cisterns – A tank is emptied every day

Slot -2 – Quantitative Aptitude – Arithmetic – Pipes and Cisterns – A tank is emptied every day at a fixed time

A tank is emptied every day at a fixed time point. Immediately thereafter, either pump A or pump B or both start working until the tank is full. On Monday, A alone completed filling the tank at 8 pm. On Tuesday, B alone completed filling the tank at 6 pm. On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. At what time was the tank filled on Thursday if both pumps were used simultaneously all along?

a) 4:24 pm

b) 4:12 pm

c) 4:48 pm

d) 4:36 pm

Answer: a) 4 : 24 pm

Solution:

Let pump A alone can fill the tank in t hours so time taken by pump B alone = t-2 hours

As per question , (t-3)/t+2/(t-2)=1

t^2-3t+6=t^2-2t
Or t=6

So the time at which pumps tank is emptied = 8 pm – 6 hours = 2 pm

Time taken by both pumps together to fill the tank =(6×4)/(6+4)=2.4 hours=2 hour 24 minutes

Thus tank will be filled by 2 pm + 2 hour 24 minutes=4:24 pm

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