Quant Aptitude by Marzan

1. A bucket contains a mixture of two liquids A and B in the proportion 7:5. If 9 litres of the mixture is replaced by 9 litres of liquid B, then the ratio of the two liquids becomes 7:9. How much of the liquid A was there in the bucket?

Solution:

Ratio of A B initially 7:5

9l liquid withdrawn

Ratio still remains 7:5

now 9 l of B is added and ratio becomes 7:9

Since only B is added quantity of A must remain same.

Ratio changes from 7:5 to 7:9 after adding 9 l of B

9–5=4 of B is added in 5+7=12

9 must be added in 12*9/4 = 27

Original mixture 27+9=36 litre.

Volume of A at initial = 7/12*36 = 21 litres

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1. An amount of tk. 10,000 becomes tk. 20,736 in 2 years. If the rate of interest is compounded half yearly, what is the annual rate of interest?
Solution:
When compounded semi-annually:
R%=r/2 and T=2t
Here,
T=2t=2*2=4 years
Given that,
10000 : 20736
=>625 : 1296 [divided by 16]
=>∜625 ∶ ∜1296 [4 years]
=>5 : 6
So, 1/5 = 20%
Now,
R%=r/2
=>20%= r/2
=> r=40% (ans)
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1. A box contains 200 marbles, 25% of which are of black color. Hanna took some marbles from the box and found that 30% of them are black. Of the remaining marbles, 10% were black marbles. How many marbles did Hanna take?

Solution:
Here,
50 black marbles, originally in the box.

Let X = the number of marbles removed from the box.

30% of X are black marbles.

Let Y = the number of marbles remaining in the box.

10% of Y are black marbles.

X + Y = 200

X = 200 - Y

30% of X + 10% of Y = 50

0.3X + 0.1Y = 50

0.3(200 - Y) + 0.1Y = 50

60 - 0.3Y + 0.1Y = 50

60 - 0.2Y = 50

- 0.2Y = - 10

Y = 50

X = 200 - Y = 200 - 50 = 150

Hanna removed 150 marbles from the box.

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1. A and B can finish a work together in 12 days, and B and C together in 16 days. If A alone works for 5 days and then B alone continues for 7 days, then remaining work is done by C in 13 days. In how many days can C alone finish the complete work?

Here LCM of 12 and 16 is taken as total work.

Assume total work =48 units

Then work done by (A+B) in one day = 48/12 = 4 units

Similarly, by (B+C) in one day = 48/16 = 3 units

Now according to question,
A works 5 days , B for 7 days and C for 13 days to complete total work

So, 5A + 7B + 13C = 48 units
5(A+B) + 2(B+C) + 11C = 48 units
5*4 + 2*3 +11C =48 units
11C = 22 units
C = 2 units ( C does 2 units of work daily)

Therefore, 48/2 = 24 days

C requires 24 days to complete the work alone.

1. A alone can do a piece of work in 30 days, while B alone can do it in 15 days, and C alone can do it in 10 days. If in every second day B, and in every third day C helps A in doing the work, how many days will be required to complete the whole work?

Let the total units of work to be done is 30. (LCM of 30, 15 & 10)

A do 1 units of work per day. (30/30)

B do 2 units of work per day. (30/15)

C do 3 units of work per day. (30/10)

Therefore, 1st 6 days (A+B+C) can do= (1×6+2×3+3×1)= 18 units

[1st day A, 2nd day (A+B), 3rd day (A+C), 4th day (A+B), 5th day A, 6th day (A+B+C) work. So, A works 6 days, B works 3 days and C works 2 days.]

Remaining work= (30-18 ) units= 12 units

2nd 3 days (A+B+C) can do= (1×3+2×1+3×1)= 8 units

Remaining work= (12-8 ) units= 4 units

Next 2 days (A+B) can do= (1×2+2×1)= 4 units

Remaining work= (4–4) units = 0 units

Total required days=(6+3+2) days= 11 days

11 days will be required to complete the whole work .

Answer: 11 days.

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1. Profit on selling 10 candles equals selling price of 3 bulbs. While loss on selling 10 bulbs equals selling price of 4 candles. Also profit percentage equals to the loss percentage and cost of a candle is half of the cost of a bulb. What is the ratio of selling price of candle to the selling price of a bulb?

Explanation:
Candle —————— Bulb
CP ….x ——————– y
SP…..a———————b
and y = 2x
Profit = 10(a – x) = 3b
Loss = 10(y – b) = 4a
Profit% = (3b/10x)*100———–(1)
and Loss% = (4a/10y)*100———-(2)
Again,
equating (1) & (2), we get
a/b = 3/2

2. Two persons A and B take a field on rent. A puts on it 21 horses for 3 months and 15 cows for 2 months; B puts 15 cows for 6months and 40 sheep for 7 1/2 months. If one day, 3 horses eat as much as 5 cows and 6 cows as much as 10 sheep, what part of the rent should A pay?

Explanation:

3h = 5c

6c = 10s

A = 21h*3 + 15c*2

= 63h + 30c

= 105c + 30c = 135c

B = 15c*6 + 40s*7 1/2

= 90c + 300s

= 90c + 180c = 270c

A:B = 135:270

27:52

A = 27/79 = 1/3

3. The speed of Boat in Still water is 40 Km/hr and speed of the stream is 20 Km/hr. The distance between Point A and Point B is 480 Km. The boat started traveling downstream from A to B, in the midway, it is powered by an Engine due to which speed of the Boat increased. Now Boat reached Point B and started back to point A with help of the same engine. It took 19 hours for the entire journey. Then with the help of the engine, the speed of the boat increased by how many Km/hr?

Solution:
19 = 240/60 + 240/60+x + 480/20+x
x = 20 Km/hr

4. Arun will be half as old as Lilly in 3 years. Arun will also be one-third as old as James in 5 years. If James is 15 years older than Lilly, how old is Arun?

Explanation:
let age of Arun =x, Lilly =y James = z
(x+3) =1/2 *(y+3) so we have 2x-y =-3 -(1)
(x+5) =1/3 * (z+5) ; => 3x-z=-10 -(2)
From (1)&(2) we get, x+y-z =-7 -(3)
we have z =15+y – (4)
from equation 3 and 4 we get x=8

1. A boat travels upstream at the speed of 20 km/h and downstream at the speed of 30 km/h. What is the speed of the boat in still water?

Explanation:

Let the speed of boat be x km/hr and speed of the stream be y km/hr

According to the question

x + y = 30 ---(1)

x – Y = 20 --- (2)

From equation (1) and equation (2)

⇒ x = 25

Speed of boat in still water is 25 km/hr.

2. A boat takes 5 hours for traveling downstream from point P to point Q and coming back to point P upstream. If the velocity of the stream is 3 kmph and the speed of the boat in still water is 5 kmph, what is the distance between P and Q?

Solution:

Let the time taken going downstream be x hours

Then time taken going upstream will be 5-x hours

Since distance = speed x time and distance each way is same ,

x* (5+3) = (5-x)*(5–3)

8x = 10–2x

10x = 10

x =1 hour

Distance = 1 x (5+3) = 8 km (Answer)

3. A candidate who gets 20% marks fails by 30 marks, but another candidate who gets 32% marks gets 42 marks more than the minimum pass marks. The maximum marks will be.

Explanation:

20% of maximum marks = minimum pass marks - 30

32% of maximum marks = minimum pass marks + 42

12% of maximum marks = 42 + 30

12% of maximum marks = 72

Therefore, Maximum marks = 600

1. Two pipes A & B can fill a cistern in 24 minutes and 36 minutes respectively. If both the pipes are opened together, after how much time B should be closed so that the tank is full in 20 minutes?

Solution:

Time taken by pipe A to fill the cistern = 24 minutes
time taken by pipe B to fill the cistern = 36 minutes
the total time taken to fill the cistern = 20 minutes.

Concept used:
If a pipe can fill a cistern in x minutes, then it will fill 1/x part in 1 minute.

Calculation:
Let the pipe B be closed after x minutes.

then, part filled by both the pipes in x minutes = (1/24 + 1/36)x = 5x/72

now, the part filled by A in ( 20-x ) minutes = (20-x)/24

∴ 5x/72 + 20-x/24 = 1

⇒ (5x + 60 - 3x )/72 = 1

⇒ 2x + 60 = 72

∴ x = 6

∴ pipe B should be closed after 6 minutes.

2. A thief is chased by police and caught. The thief was 40 m ahead of the police and the speeds of the thief and the police are 10 m/s and 15 m/s, respectively. Find the total distance travelled by police.

Solution:

Let Police will catch the thief after thief has run further x m.

Distance travelled by the police = (40 + x) m

Time taken by police to travel (40 + x) m,

⇒ (40 + x)/15 sec

Time taken by thief to travel x m,

⇒ x/10 sec

As, time taken by both them will be same,

(40 + x)/15 = x/10

⇒ 400 + 10x = 15x

⇒ 5x = 400

⇒ x = 80 m

Total distance travelled by police = 40 + 80

⇒ 120 m

∴ Total distance travelled by police is 120 m