Capgemini Solved Problems On Allegations or Mixtures

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Capgemini Solved Problems On Allegations or Mixtures

Formula to remember :

Quantity of cheaper     C.P of dearer - Mean Price       d - m   
------------------   =  ---------------------------  =   -------
Quantity of dearer    Mean price - C.P of cheaper       m - c

Question 1

Two bag contains different kinds of rice. The first bag contains 40% of worst quality and the rest good quality. The second bag contains 60% of worst quality. How much rice should be mixed from each of the container so as to get 63 kg of rice such that the ratio of worst quality to good quality is 3:4 ?
a)33kg,30kg   b)50kg,13kg   c)54kg,9kg   d)40kg,23kg

Answer : c)54kg,9kg

Solution :

Let the cost of 1 kg rice be Re.1. ...(A)

Good quality in 1 kg mixture in 1st bag = 1 - 40% = 1-(2/5) = 3/5 kg ...(1)
Cost price of 1 kg mixture in 1st bag = (3/5) x 1 = Re.3/5 (Based on our above equation and assumption A )

Good quality in 1 kg mixture in 2nd bag = 1 - 60% = 1-(3/5) = 2/5 kg ...(2)
Cost price of 1 kg mixture in 2nd bag = (2/5) x 1 = Re.2/5 (Based on our above equation and assumption A)

It is given that the final mixture has worst quality to good quality of rice in ration 3:4.
Good quality in 1 kg final mixture = 4/(3+4) x 1 = 4/7kg ...(3)
Then its mean price = Re.4/7 (Based on our above equation and assumption A)

From 1,2 and 3 we get m = 4/7, d = 3/5 and c = 2/5. (Refer formula given in introduction before question 1.)

d - m = (3/5)-(4/7) = 1/35
m - c = (4/7)-(2/5) = 6/35

Applying formula, cheaper quantity:dearer quantity = d-m / m-c = 1/35 : 6/35 = 1:6

Final mixture should contain 63Kg in total as per question.
so, quantity of mixture taken from 2nd bag = 1/(1+6) x 63 = 9 kg
and the quantity of mixture taken from 1st bag = 6/(1+6) x 63 = 54 kg.

hence the answer is 54kg,9kg

Question 2

How many litres of milk worth Rs.12 per litre must be mixed with 60 litres of milk worth Rs.8 per litre so as to gain 20% by selling the mixture at Rs.10.80 per litre?
a)40    b)20    c)25    d)45

Answer : b)20

S.P of 1 litre of mixture = Rs.10.80 and Gain 20%.
Therefore, C.P of 1 litre of mixture = Rs.100/120 x 10.80 = Rs.9 ...(1)

C.P of 1 litre milk of 1st kind = d = Rs.12
C.p of 1 litre milk of 2nd kind = c = Rs.8
Mean Price = m = Rs.9 (from equation 1)

Then d - m = 3 , m - c = 1

Applying above values to our allegations formula (refer introduction before question)

Ratio of quantities of 1st and 2nd kind = 1:3 .

Let X litres of milk of 1st be mixed with 60litres of 2nd kind.

Then, 1 : 3 = X : 60

1/3 = X/60

X = 60/3 = 20.

Thus 20 litres milk of 1st kind to be mixed with 2nd.

Question 3

A bottle labelled X is filled with milk and water is mixed in the ratio 11:4 and another labelled Y has mixture in ratio 3:2. In what ratio these two mixtures be mixed together to get a new one labelled Z containing milk and water in the ratio 2:1?
a)1:1    b)1:2    c)2:1    d)1:3

Answer : a)1:1

Solution:

Let the c.p of milk be Re.1 per litre. ...(A)

Now, milk in 1 litre mixture of X = 11/15 ...(1)
Then c.p of 1 litre mixture of X = (11/15) x 1 = Re.11/15 (Based on value in previous equation and our assumption A)

Milk in 1 litre mixture of Y = 3/5 ...(2)
Then c.p of 1 litre mixture of Y = (3/5) x 1 = Re.3/5 (Based on value in previous equation and our assumption A)

As given in question, milk and water in final mixture are in the ration 2:1. Therefore, Milk in 1 litre mixture of Z = 2/(1 + 2) x 1 = 2/3. ...(3)
Then the c.p of 1 litre mixture of Z = Re.2/3 (Based on value in previous equation and our assumption A)

Here, d = 11/15, c = 3/5 and m = 2/3
Then d - m = 11/15 - 2/3 = (11-10)/15 = 1/15
and m - c = 2/3 - 3/5 = (10-9)/15 = 1/15
Required ratio (as per our formula) = d-m : m-c = 1/15 : 1/15 = 1:1

Hence 1:1 is the required ratio.

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Question 1

If the average of 4 consecutive odd numbers is 92, then the least number of those numbers is:
a) 89    b) 71    c) 81    d) 91

Answer : a) 89

Solution :

Let the four consecutive odd numbers be x, x+2, x+4 and x+6

We have to find the least number i.e., x

Given that the average is 92. That is, x + x+2 + x+4 + x+6 / 4 = 92
4x + 12 = 92(4)
4x + 12 = 368
4x = 368 - 12 = 356
x = 356/4 = 89

Hence the required number is 89.

Question 2

If the sum of five consecutive even numbers is 1580, then find the average of the next five consecutive even numbers
a) 326    b) 312    c) 325    d) 318

Answer : b) 312

Solution:

Let the five consecutive even numbers be x, x+2, x+4, x+6 and x+8

The sum of the above five numbers is 1580.
i.e., x + x+2 + x+4 + x+6 + x+8 = 1580
5x + 20 = 1580
5x = 1580 - 20
x = 312

Therefore, the 5 even numbers are 312, 312+2, 312+4, 312+6 and 312+8
Or 312, 314, 316, 318 and 320

Then the next 5 consecutive even numbers are 322, 324, 326, 328 and 330

Now, the required average = 322+324+326+328+330 / 5 = 1630/5 = 326.

Hence the answer is 326.

Question 3

If the sum of the 4 consecutive even numbers is 4 more than three times the largest number, then the average of those numbers is:
a) 32    b) 12    c) 25    d) 13

Answer : d) 13

Solution :

Even numbers can be represented to have the form 2n where n = 0,1,2,3....

Based on our above understanding, any four consecutive even numbers can be represented as,

2k, 2k + 2, 2k + 4, 2k + 6

Now, adding these we get,

2k + (2k + 2) + (2k + 4) + (2k + 6) = 8k + 12

But the above sum is 4 more than 3 times the largest number(which is 2k + 6) .

i.e., 8k + 12 = 4 + 6k + 18

2k = 10
k = 5

Therefore, the sum of the numbers = 8k + 12 = 52 (by substituting the value of k)

Then the required average = 52/4 = 13

[vedha.v]

Below are three problems of boats and streams.

Question 1

Find the speed of boat in still water if it takes 5 hours to cover a certain distance upstream and takes just 2 and half hours to cover the same distance downstream. The rate of stream is 10 kmph.
a)20km/hr   b)25km/hr   c)30km/hr   d)cannot be determined

Answer : c)30km/hr

Solution:

If the speed downstream is a km/hr and the speed upstream is b km/hr, then:

Speed in still water = (a + b)/2 km/hr.
Rate of stream = (a - b)/2 km/hr.

Let boat's upstream speed be X km/hr and downstream be Y km/hr
Distance covered in upstream for 5 hours = Distance covered in downstream for 2 1/2 or 5/2 hours.
By using distance = speed x time, we get
X x 5 = Y x 5/2
2(5X) = 5Y
2X = Y
Y - 2X = 0 ...eqn 1

Given that, the speed of stream is 10 km/hr
i.e., (Y-X)/2 = 10
Y - X = 20 ...eqn 2

Solving eqn 1 and eqn 2, we get
X = 20 and Y = 40

Speed of Boat in still water = (X + Y)/2
(20 + 40)/2 = 30.
Hence the answer is 30km/hr.

Question 2

The rate of current is 5km/hr and the speed of boat in still water is 19km/hr. The ratio between the distance travelled downstream in 15 minutes to the distance travelled upstream in 30 minutes is:
a)2:3    b)3:5   c)3:2.5   d)3:3.5

Answer : d)3:3.5

Solution :

If the speed of boat in still water is u km/hr and the speed of the stream is v km/hr, then:

Speed downstream = (u + v) km/hr.

Speed upstream = (u - v) km/hr.

Given that the speed of boat in still water is 19 km/hr and the rate of current is 5 km/hr.

Downstream speed = (19 + 5) km/hr = 24km/hr
Distance travelled downstream in 15 minutes with a speed of 24km/hr = 24 x (15/60) = 6 km

Upstream speed = (19 - 5) km/hr = 14km/hr
Distance travelled upstream in 30 minutes with a speed of 14km/hr = 14 x(30/60) = 7 km

Then the required ratio = 6:7 = 3 : 3.5

Hence the answer is 3 : 3.5

Question 3

A man takes certain time to row 8 km downstream. But when he swims upstream for the same time, he can cover 6 km. If he rows a distance of 96 km upstream and comes back in 28 hours, then find the rate of the stream ?

a)2km/hr   b)1.5km/hr   c)1km/hr   d)2.5km/hr

Answer : c)1km/hr

Solution:

Let X hours be the time taken to move 8 km downstream. Then as per the question data, X will be the time taken to cover 6 km upstream.

Total distance travelled = 96 km
Downstream speed = 8/X km/hr
Time taken to row 96 km downstream = 96 / (8/X) = 12X hrs (time = speed / distance )

Upstream Speed = 6/X km/hr
Time taken to row 96 km upstream = 96 / (6/X) = 16X hrs

Time taken to return back is 28 hrs
Total Time taken = Time taken to row downstream + Time taken to row upstream
28 = 12X + 16X
28 = 28X
X = 1

Substituting X = 1 we get,
Downstream Speed = 8 km/hr and Upstream Speed = 6 Km/hr.
Rate of stream = 8 - 6 / 2 = 1 Km/hr

[vedha.v]

Below are three dimensions based problems using simple calculations.

Question 1

A manufacturer reduces the size of his machine of Material1 from 30" to 28" and that of Material2 from 24" to 23". He normally spends Rs.1440 for Material1 and Material2. 3/4th of the amount spent would be for Material1. How much will he save under the size of new size (Assuming that the costs incurred by the machines are directly proportional to their sizes)?
a)Rs.85   b)Rs.87   c)Rs.86   d)Rs.88

Answer : b)Rs.87

Solution:

The manufacturer totally spends Rs.1440 for both material 1 and 2.
Out of this 3/4th is entirely for Material1.

Amount spent on Material1 = 1440 x 3/4 = Rs.1080
i.e, he spent Rs.1080 for Material1 of 30" and he reduces 2".

   Size      Amount
   30"      Rs.1080
   2"        ?
Amount saved on Material1 = 1080 x 2/30 = Rs.72

Amount spent on Material2 = total amount - amount for Material1 = 1440 - 1080 = Rs.360.
i.e., he spent Rs.360 for Material2 of 24" and he reduces 1".

   Size      Amount
   24"      Rs.360
   1"        ?
Amount saved on Material2 = 360 x 1/24 = Rs.15
Therefore total amount saved after size reduction = 72 + 15 = Rs.87

Question 2

The dimensions of a certain machine are X" x Y" x Z". If the average of its dimensions equals 680" and none of the dimensions is less than 640", what is the greatest possible length of one of the dimensions?
a)760"    b)800"    c)720"    d)745"

Answer : a) 760"

Solution:

Given (X+Y+Z)/3 = 680".
Therefore the total length of the dimensions is X + Y + Z = 3 x 680 = 2040.
Let us assume Z is the dimension with greatest possible length.

From above equation, Z = 2040 - X - Y ...(1)

Based on eq (1), for Z to be of maximum value, X and Y should be as low as possible. Since it is given that none of the dimensions is lesser than 640", the lowermost value that X and Y can get is 640".

Substituting X = Y = 640" in eq (1), we get,
Z = 2040 - 640 -640 = 760".

Question 3

What will be the ratio of the shortest side of new dimensions to the greatest side of the old dimensions if the size of a certain machine is increased proportionally until the sum of its dimensions equals 800" and the dimensions of old size are 130" x 60" x 360" ?.
a)1:2    b)3:5    c)5:18    d)18:3

Answer : c) 5:18

Solution:

From the given data, we have
The dimensions of the machine before size increment = 130" x 70" x 360".
Sum of the dimensions = 130+70+360 = 560"
Note that the greatest side = 360" & smallest side = 70". ...(A)

It is given that the sum of the dimensions of new size = 800"

Since dimensions are increased proportionately, the smallest side of the old dimensions before increment will be the smallest side of new dimension after size increment.

   Sum      Side
   560"       70"
   800"        ?
Smallest side will be increased to 800 x 70 / 560 = 100"
i.e., the smallest side of the new dimensions is 100".
We have already found that the the greatest side of old dimensions is 360".
Required Ratio = Smallest side of the New Dimensions : Greatest side of the Old
Dimensions
= 100 : 360 = 5 : 18
Hence the answer is 5 : 18.