Monday, 24 October 2016

Elitmus Previous Year Questions on Alligations and Mixtures

Elitmus Previous Year Questions on Alligations and Mixtures

SET 1



Alligations Set 01


01.A mixture of 125 gallons of wine and water contains 20% water. How much water must be added to the mixture in order to increase the percentage of water to 25% of the new mixture ? 
(a) 10 gals(b) 8.5 gals(c) 8 gals(d) 8.33 gals
02.4oo students took a mock exam in Delhi, 60% of the boys and 80% of the girls cleared the cut off in the examination. If the total percentage of students qualifying is 65%, how many girls appeared in the examination?
(a) 100(b) 120(c) 150 (d) 300
03.What will be the ratio of petrol and kerosene in the final solution formed by mixing petrol and kerosene that are present in three vessels of equal capacity in ratios 4 : 1, 5 : 2 and 6 : 1 respectively ?
(a) 166 : 22(b) 83 : 22(c) 83 : 44(d) None of these
04.A dishonest grocer professes to sell pure butter at cost price, but he mixes it with adulterated fat and thereby gains 25%. Find the percentage of adulterated fat in the mixture assuming that adulterated fat is freely available ?
(a) 20%(b) 25%(c) 33.33%(d) 40%
05.In a Singapore zoo, there are deers and there are ducks. If the heads are counted, there are 180, while the legs are 448. What will be the number of deers in the zoo
(a) 136(b) 68(c) 44(d) 22
06.Two vessels contain a mixture of spirit and water. In the first vessel and ratio of spirit to water is 8 : 3 and in the second vessel the ratio is 5 : 1. A 35 litre cask is filled from these vessels so as to contain a mixture of spirit and water in the ratio of 4 : 1. How many litre  are taken from the first vessel ?
(a) 11 litres(b) 22 litres(c) 16.5 litres(d) 17.5 litres
07.In what ratio should water be mixed with soda costing Rs 12 per litre so as to make a profit of 25% by selling the diluted liquid at Rs 13.75 per litre ?
(a) 10:1(b) 11:1(c) 1:11(d) 12:1
08.A 20 percent gain is made by selling the mixture of two types of ghee at Rs 480 per kg. If the type costing 610 per kg was mixed with 126 kg of the other, how many kilograms of the former was mixed ?
(a) 138 kg(b) 34.5 kg(c) 69 kg(d) Cannot be det..
09.A bartender stole champagne from a bottle that contained 50% of spirit and he replaced what he had stolen with champagne having 20% spirit. The bottle then contained only 25% spirit. How much of the bottle did he steal ?
(a) 80%(b) 83.33%(c) 85.71%(d) 88.88%
10.A mixture of 20 litres of brandy and water contains 10% water. How much water should be added to it to increase the percentage of water to 25%?
(a) 2 litres(b) 3 litres(c) 2.5 litres(d) 4 litres
Answers : Find Detailed Solutions at the end of the the page.
1. D2. A
3. B4. A
5. C6. A
7. C8. D
9. B10. D

SET 2



Alligations Set 02


01.Two containers of equal capacity are full of a mixture of oil and water. In the first, the ratio of oil to water is 4 : 7 and in the second it is 7 : 11. Now both the mixtures are mixed in a bigger container. What is the resulting ratio of oil to water ? 
(a) 149 : 247(b) 247 : 149(c) 143 : 241(d) 241 : 143
02.Two vessels contain spirit and water mixed respectively in the ratio of 1 : 3 and 3 : 5. Find the ratio in which these are to be mixed to get a new mixture in which the ratio of spirit to water is   1 : 2 ?
(a) 2 : 1(b) 3 : 1(c) 1 : 2(d) 1 : 3
03.The price of a pen and pencil is Rs 35. The pen was sold at a 20% profit and the pencil at a 10% loss. If in the transaction a man gains Rs 4, how much is cost price of the pen ?
(a) Rs 10(b) Rs 25(c) Rs 20(d) None of these
04.A person purchased a cupboard and a cot for Rs 18,000. He sold the cupboard at a profit of 20% and the cot at a profit of 30%. If his total profit was 25.83%, find the cost price of the cupboard ?
(a) Rs 10,500(b) Rs 12,000(c) Rs 7500(d)  Rs 10,000
05.A vessel is full of a mixture of kerosene and petrol in which there is 18% kerosene. Eight litre are drawn off and then the vessel is filled with petrol. If the kerosene is now 15%, how much does the vessel hold ?
(a) 40 litres(b) 32 litres(c) 36 litres(d) 48 litres
06.Two solutions of 90% and 97% purity are mixed resulting in 21 litres of mixture of 94% purity. How much is the quantity of the first solution in the resulting mixture ?
(a) 15 litres(b) 12 litres(c) 9 litres(d) 6 litres
07.In what ratio should water be mixed with soda costing Rs 12 per litre so as to make a profit of 25% by selling the diluted liquid at Rs 13.75 per litre ?
(a) 10 : 1(b) 11 : 1(c) 1 : 11(d) 12 : 1
08.A sum of Rs 36.90 is made up of 90 coins that are either 20 paise coins or 50 paise coins. Find out how many 20 paise coins are there in the total amount ?
(a) 47(b) 43(c) 27(d) 63
09.A bonus of Rs 9,85,000 was divided among 300 workers of a factory. Each male workers gets 5000 rupees and each female worker gets 2500 gets. Find the number of male workers in the factory ?
(a)  253(b) 47(c) 94(d) 206
10.In what proportion must water be mixed with milk so as to gain 20% by selling the mixture at the cost price of the milk ? ( Assume that water is freely available )
(a) 1 : 4(b) 1 : 5(c) 1 : 6(d) 1 : 12
Answers : Find Detailed Solutions at the end of the the page.
1. A2. C
3. B4. C
5. D6. C
7. C8. C
9. C10. B

Permutation And Combinations Solution 01


1.
Greatest five digit number : 43210
Smallest five digit number : 10234
Difference = 43210-10234= 32976
(c)
2. We consider vice-chairman and the chairman as 1 Unit. Now, 9 persons can be arranged along a circular table in 8! ways. And vice-chairman and chairman can be arranged in 2 different ways. Hence required number of ways=2 x 8!
(b)
3.  3 Girls can be selected out of 5 girls in 5C3 ways.
Since number of boys to be invited is not given, hence out of 4 boys, he can invite them (2)4 ways.
Hence required number of ways is = 5C3 x (2)4  = 160
(b)
4.Total number of ways of filling the 5 boxes numbered as (1, 2, 3, 4 and 5) with either blue or red balls=25=32.
Two adjacent boxes with blue can be got in 4 ways.
i.e. (1, 2) (2, 3) (3, 4) (4, 5)
Three adjacent boxes with blue can be got in 3 ways
i.e. (1, 2, 3), (2, 3, 4) and (3, 4, 5)
Four adjacent boxes with blue can be be got in 2 ways.
i.e. (1234) , (2345)
and five boxes with blue can be got in 1 way.
Hence total number of ways of filling the boxes such that adjacent boxes have blue
(4 + 3 + 2 + 1)=10
(d)
5. To construct 2 roads, three towns can be selected out of 4 in 4 x 3 x 2 = 24 ways. Now, if the third road goes from the third town to the first town, a triangle is formed and if it goes to the fourth town, a triangle is not formed. So there are 24 ways to form a triangle and 24 ways of avoiding a triangle.
(d)
6. Total number of 4 digit numbers that can be formed = 4!. If the number is divisible by 25, then the last two digit are 25. So the first two digits can be arranged in 2! ways.
Hence required probability = 2!/4! = 1/12
(a)
7. Keeping one digit in fixed position, other four can be arranged in 4! ways= 24 ways. Thus each of the 5 digits will occur in each of the five place 4! times. Hence the sum of digits in each position is 24(1 + 3 + 5 + 7 + 9) = 600. So, the sum of all numbers
=600(1+10+100+1000+10000)=6666600
(a)
8. Required number of matches played will be (139 – 1)=138
(c)
9. Required probability is = 1/6
(c)
10. Let the number of students in the front row be x and the number of rows be n.
Hence, number of students in the next rows would be (x – 3) , (x – 6), (x – 9),……. and so on.
Now we have to check for each value of n=3, 4, 5, 6
Firstly take n=3
x + (x – 3) + (x – 6) =630
» 3x = 639 i.e. x = 213 ( Thus, n=3 is  possible)
Likewise if n=4
x + (x – 3) + (x – 6)+(x – 9) =630
» 4x – 18 = 630 i.e. x = 162 ( Thus, n=4 is possible)
Likewise if n=5
x + (x – 3) + (x – 6)+(x – 9) +( x -12) =630
» 5x – 30 = 630 i.e. x = 132 ( Thus, n=5 is possible)
Likewise if n=6
x + (x – 3) + (x – 6)+(x – 9)+(x -12)+(x – 15)=630
» 6x – 45 = 630 i.e. x =112.5 (NOT AN INTEGER)  ( Thus, n=6 is not possible)
(d)
1. Let there be m boys and n girls.
Then nC2 = 45 » n(n – 1)=90» n =10
mC2 = 190  » m(m – 1) = 380 » m=20
Number of games played between one boy and one girl
= 10C1 x 20C1= 10 x 20=200
(a)
2. Any of the 4 colours can be chosen for the first stripe. Any of the remaining 3 colours can be used for the second stripe. The stripe can again be colored in 3 ways. ( We can repeat the colour of the first stripe, but not use the colour of the second stripe).
Similarly, there are 3 ways to colour each of the remaining stripes.
» The number of ways the flag can be coloured is
4 x (3)5 = (12)(3)4
(a)
3. The available digits are 0, 1, 2 …………… 9. The first digit can be chosen in 9 ways ( 0 not acceptable ), the second digit can be accepted in 9 ways ( digits repetition not allowed). Thus the code can be made in 9 x 9=81 ways.
Now, there are only 4 digits which can create confusion 1, 6, 8, 9. The same can be given in the following ways.
Total number of ways confusion can arise = 4 x 3=12
Thus, required answer= 81-12=69
(d)
4.
Number of ways for selecting single digit=2
Number of ways for selecting two digit= 2 x 3=6
Number of ways for selecting three digit= 2 x 3 x 3=18
Number of ways for selecting four digit= 2 x 3 x 3 x 3=54
Number of ways for selecting five digit=2 x 3 x 3 x 3 x 3=162
Number of ways for selecting six digit= 2 x 3 x 3 x 3 x 3 x 3=486
Hence, total number of ways = (2 + 6 + 18 + 54 + 162 + 486)=728
(d)
5. There are 32 black and 32 white square on a chess-board then number of ways in choosing one  white and one black square on the chess.
32C1 x 32C1= 32 x 32=1024
Number of ways in which square lies in the same row
White square=4
Black  square=4
Number of rows=8
4C1 x 4C1 x 8 = 128
»Number of ways in which square lies in the same column= 128
Total number in which square lie on the same row or same column= 128 + 128=256.
(d)
6. Ist place of the four letter password can be filled in 11 ways.
IInd place of four letter password can be filled in 10 ways.
IIIrd place of four letter password can be filled in 9 ways.
IVth place of four letter password can be filled in 8 ways.
Hence, required number of ways= 11 x 10 x 9 x 8=7920 ways
(a)
7. Three letter password from 26 letters  can be selected in 26 x 25 x 24 ways. Three letter password from 15 asymmetric letters can be selected in 15 x 14 x 13 ways.
Hence, three letter password with at least one symmetric letter can be made in (26 x 25 x 24)-(15 x 14 x 13)=12870 ways.
(c)
8. At lease one candidate out of (2n + 1) candidates can be selected in (2n+1 - 1) ways.
» 22n+1 - 1 = 63  » 22n+1 = 64 = (2)6» n=2.5
Since n cannot be a fraction. Hence n=3.
(a)
9. Required number of triangles formed
10C2 x 11 + 11C2 x 10= 45 x 11 + 55 x 10 = 1045
(c)
10. The digit in the unit’s place should be greater than that in the ten’s place. Hence, if digit 5 occupies the unit place then remaining four digits need not to follow any order.
Hence required number of ways = 4!
However, if digit 4 occupies the unit place then 5 cannot occupies the ten’s positions. Hence, digits at the ten’s place will be one among 1, 2 or 3. This can happen in 3 ways. The remaining 3 digits can be filled in the remaining three places in 3! ways. Hence in all we have (3 x 3!) numbers ending in 4.
Similarly, if we have 3 in the unit’s place, the ten’s place can be either 1 or 2. This can happen in 2 ways. The remaining 3 digits can be arranged in the remaining 3 places in 3! ways. Hence we will have (2 x 3!) numbers ending in 3. Similarly, we can find that there will be 3! numbers  ending in 2 and no number ending with 1. Hence total number of numbers
= 4! + (3) x 3! + (2)3! + 3!
= 4! + 6 x 3! = 24 + (6 x 6) = 60
(b)


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