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Q:

A is twice efficient as B and together they do the same work in as much time as C and D together. If C and D can complete the work in 20 and 30 daysrespectively, working alone ,then in how many days A can complete the work individually:

A) 12 days	B) 18 days
C) 24 days	D) 30 days

Answer & Explanation Answer: B) 18 days

Explanation:

A + B = C + D

| | | |

Ratio of efficiency 10x + 5x 9x + 6x

|________| |_________|

15x 15x

Therefore , ratio of efficiency of A:C =10:9

Therefore, ratio of days taken by A:C = 9:10

Therefore, number of days taken by A = 18 days

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Filed Under: Time and Work

Q:

The ratio of efficiency of A is to C is 5:3. The ratio of number of days taken by B is to C is 2:3. A takes 6 days less than C, when A and C completes the work individually. B and C started the work and left after 2 days. The number of days taken by A to finish the remaining work is:

A) 4.5	B) 5
C) 6	D) 9 1/3

Answer & Explanation Answer: C) 6

Explanation:

A : C

Efficiency 5 : 3

No of days 3x : 5x

Given that, 5x-6 =3x => x = 3

Number of days taken by A = 9

Number of days taken by C = 15

B : C

Days 2 : 3

Therefore, Number of days taken by B = 10

Work done by B and C in initial 2 days = $2 [\frac{1}{10} + \frac{1}{15}]$ = 1/3

Thus, Rest work =2/3

Number of days required by A to finish 2/3 work = (2/3) x 9 = 6 days

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Q:

From a container, 6 liters milk was drawn out and was replaced by water. Again 6 liters of mixture was drawn out and was replaced by the water. Thus the quantity of milk and water in the container after these two operations is 9:16. The quantity of mixture is:

A) 15	B) 16
C) 25	D) 31

Answer & Explanation Answer: A) 15

Explanation:

Let quantity of mixture be x liters.

Suppose a container contains x units of liquid from which y units are taken out and replaced by Water. After operations , the quantity of pure liquid = $x {(1 - \frac{y}{x})}^{n}$ units, Where n = no of operations .

So, Quantity of Milk = $x {(1 - \frac{6}{x})}^{2}$

Given that, Milk : Water = 9 : 16

=> Milk : (Milk + Water) = 9 : (9+16)

=> Milk : Mixture = 9 : 25

Therefore, $\frac{x {(1 - \frac{6}{x})}^{2}}{x} = \frac{9}{25}$

=> x = 15 liters

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Filed Under: Alligation or Mixture

Q:

The ratio of petrol and kerosene in the container is 3:2 when 10 liters of the mixture is taken out and is replaced by the kerosene, the ratio become 2:3. Then total quantity of the mixture in the container is:

A) 25	B) 30
C) 45	D) cannot be determined

Answer & Explanation Answer: B) 30

Explanation:

pool : kerosene

3 : 2(initially)

2 : 3(after replacement)

$\frac{R e m a i n i n g Q u a n t i t y}{I n i t i a l Q u a n t i t y} = (1 - \frac{R e p l a c e d Q u a n t i t y}{T o t a l Q u a n t i t y})$

(for petrol) $\frac{2}{3} = (1 - \frac{10}{k})$

=> K = 30

Therefore the total quantity of the mixture in the container is 30 liters.

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Filed Under: Alligation or Mixture

Q:

In a mixture of milk and water, there is only 26% water. After replacing the mixture with 7 liters of pure milk , the percentage of milk in the mixture become 76%. The quantity of mixture is:

A) 65 liters	B) 91 liters
C) 38 liters	D) None of these

Answer & Explanation Answer: B) 91 liters

Explanation:

Milk Water

74% 26% (initially)

76% 24% ( after replacement)

Left amount = Initial amount $(1 - \frac{r e p l a c e d a m o u n t}{t o t a l a m o u n t})$

24 = 26 $(1 - \frac{7}{k})$

=> k = 91

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Filed Under: Alligation or Mixture

Q:

A jar was full with honey. A person used to draw out 20% of the honey from the jar and replaced it with sugar solution. He has repeated the same process 4 times and thus there was only 512 gm of honey left in the jar, the rest part of the jar was filled with the sugar solution. The initial amount of honey in the jar was filled with the sugar solution. The initial amount of honey in the jar was:

A) 1.25 kg	B) 1 kg
C) 1.5 kg	D) None of these

Answer & Explanation Answer: A) 1.25 kg

Explanation:

Let the initial amount of honey in the jar was K, then

$512 = K {(1 - \frac{1}{5})}^{4}$ $[∵ 20 % = \frac{20}{100} = \frac{1}{5}]$

or

$512 = K {(\frac{4}{5})}^{4}$

Therefore, K = 1250

Hence initially the honey in the jar= 1.25 kg

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Filed Under: Alligation or Mixture

Q:

From a container of wine, a thief has stolen 15 liters of wine and replaced it with same quantity of water.He again repeated the same process. Thus in three attempts the ratio of wine and water became 343:169. The initial amount of wine in the container was:

A) 75 liters	B) 100 liters
C) 150 liters	D) 120 liters

Answer & Explanation Answer: D) 120 liters

Explanation:

$\frac{W i n e (l e f t)}{W a t e r (a d d e d)} = \frac{343}{169}$

It means $\frac{W i n e (l e f t)}{W i n e (i n i t i a l a m o u n t)} = \frac{343}{512}$ $[∵ 343 + 169 = 512]$

Thus ,

$343 x = 512 x {(1 - \frac{15}{K})}^{3}$

$\Rightarrow \frac{343}{512} = {(\frac{7}{8})}^{3} = {(1 - \frac{15}{k})}^{3}$

=> K = 120

Thus the initial amount of wine was 120 liters.

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Filed Under: Alligation or Mixture

Q:

From the 50 liters of milk, 5 liters of milk is taken out and after it 5 liters of water is added to the rest amount of milk. Again 5 liters of milk and water is drawn out and it was replaced by 5 liters of water. If this process is continued similarly for the third time, the amount of milk left after the third replacement:

A) 45L	B) 36.45L
C) 40.5L	D) 42.5L

Answer & Explanation Answer: B) 36.45L

Explanation:

General Formula:

Final or reduced concentration = initial concentration x ${(1 - \frac{a m o u n t b e i n g r e p l a c e d i n e a c h o p e r a t i o n}{t o t a l a m o u n t})}^{n}$

where n is the number of times the same operation is being repeated. The "amount being replaced" could be pure or mixture as per the case. similarly ,"total amount" could also be either pure or mixture. Here amount being replaced denotes the quantity which is to be withdrawn in each time.

Therefore, $50 \times {(1 - \frac{5}{50})}^{3}$

= 36.45 L

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Filed Under: Alligation or Mixture

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