Online Mock Test in Mixture and Allegations For Quantitative Aptitude for Competitive exams like IBPS BANK PO/Clerical,SBI,RRB,SSC,LIC,UPSC-CSAT,SCRA.MAT,CMAT,MBA,SNAP,CAT,NTSE,CLAT
Mixture and Allegations-Test 2
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Subject :- Quantitative Aptitude
Chapter :- Mixture and Allegations-Test 2
Questions :- 25
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A solution of sugar syrup has 15% sugar. Another solution has 5% sugar. How many
liters of the second solution must be added to 20 L of the first solution to make a
solution of 20% sugar.
Correct
Let x L of second solution must be added.
Then, [15*20 + 5*x]/(20 + x) = 10
=> x = 20 L
Incorrect
Let x L of second solution must be added.
Then, [15*20 + 5*x]/(20 + x) = 10
=> x = 20 L
Question 2 of 25
2. Question
1 points
A person has a chemical of Rs. 25 per liter. In what ratio should water be mixed in that
chemical, so that after selling the mixture at Rs. 20 per litre he may get a profit of 25%?
Correct
Selling price of mixture = Rs. 20
Cost price of mixture = (100/125)*20 =
Rs.16
Rule of mixture
25————-0
——–16
16————-9
So, the required ratio = 16 : 9
Incorrect
Selling price of mixture = Rs. 20
Cost price of mixture = (100/125)*20 =
Rs.16
Rule of mixture
25————-0
——–16
16————-9
So, the required ratio = 16 : 9
Question 3 of 25
3. Question
1 points
Three containers X, Y and Z are having mixtures of milk and water in the ratio 1 : 5
, 3 : 5 and 5 : 7 resp. If the capacities of the containers are in the ratio 5 : 4 : 5, then find the ratio of the milk to the water, if the mixtures of all the three containers are
mixed together.
Correct
Ratio of milk and water
= [(1/6)*5 + (3/8)*4 + (5/12)*5] : [(5/6)*5
+ (5/8)*4 + (7/12)*5] = 53 : 115
Incorrect
Ratio of milk and water
= [(1/6)*5 + (3/8)*4 + (5/12)*5] : [(5/6)*5
+ (5/8)*4 + (7/12)*5] = 53 : 115
Question 4 of 25
4. Question
1 points
How many kg of sugar costing Rs. 5.75 per kg should be mixed with 75 kg of cheaper
sugar costing Rs. 4.50 per kg so that the mixture is worth Rs. 5.50 per kg ?
Correct
Sugar I —————- Sugar II
5.75 ————————-4.50
—————–5.50
1——————————0.25
ratio = 4 : 1
The required qty. of sugar I = (75/1)*4 =
300 kg
Incorrect
Sugar I —————- Sugar II
5.75 ————————-4.50
—————–5.50
1——————————0.25
ratio = 4 : 1
The required qty. of sugar I = (75/1)*4 =
300 kg
Question 5 of 25
5. Question
1 points
One test tube contains some acid and another test tube contains an equal quantity
of water. To prepare a solution, 20 g of the acid is poured into the second test tube.
Then, two-thirds of the so formed solution is poured from the second test tube into the
first. If the fluid in the first test tube is four times that in second, what quantity of water
was taken initially.
Correct
Initially, let x g of water and Acid was
taken. Initially 1st process
First test tube = (x – 20) g
Second test tube = (x + 20) g
2nd process
First test tube = (x – 20) + (x + 20)*(2/3)
Second test tube = (x + 20)*(1/3)
Now,
(x -20) + (2/3)(x +20) = 4*(1/3)(x + 20)
=> x = 100 g
Incorrect
Initially, let x g of water and Acid was
taken. Initially 1st process
First test tube = (x – 20) g
Second test tube = (x + 20) g
2nd process
First test tube = (x – 20) + (x + 20)*(2/3)
Second test tube = (x + 20)*(1/3)
Now,
(x -20) + (2/3)(x +20) = 4*(1/3)(x + 20)
=> x = 100 g
Question 6 of 25
6. Question
1 points
A shopkeeper sells two types of books national books and international books .He
sells national books at Rs. 18 / book and incurs at loss of 10% whereas on selling the
international books at Rs. 30 / book ,he gains 20 % .Find the ratio of the national
and international books such that he can
gain a profit of 25% by selling the
combined books at 27.5/ book ?
Correct
Incorrect
Question 7 of 25
7. Question
1 points
One test tube contains some acid and another test tube contains an equal quantity
of water .To prepare a solution , 20 g of the acid is poured into the second test tube
.Then , two –third of the so- formed solution is poured from the second tube
into the first .If the fluid in the first test tube is four times that in the second ,what
quantity of water was taken initially ?
Correct
Let x g of water was taken initially .
1st process
First test tube (x- 20)
second test tube (x +20)
2nd process
First test tube = [(x-20) +2/3 (x+20)]
Second test tube = 1/3(x+20)
Now ,
(x -20 ) + 2/3(x+20) = 4* (1/3)(x+20) => x = 100 g
Incorrect
Let x g of water was taken initially .
1st process
First test tube (x- 20)
second test tube (x +20)
2nd process
First test tube = [(x-20) +2/3 (x+20)]
Second test tube = 1/3(x+20)
Now ,
(x -20 ) + 2/3(x+20) = 4* (1/3)(x+20) => x = 100 g
Question 8 of 25
8. Question
1 points
Two brands of detergents are to be combined . Detergent A contains 40 %
bleach and 60 % soap . While detergent B contains 25 % bleach and 75% soap . If
the combined mixture is to be 35 % bleach .What % of the final mixture should be
detergent A?
Correct
Incorrect
Question 9 of 25
9. Question
1 points
A thief has stolen 15 L of beer from a container and replaced with the same
quantity of water .He again repeated this process 3 times .Thus the ratio of the beer
become 343 :169 .Find the initial amount of beer in the container .
Correct
The initial amount of beer in the
container was =343 + 169 = 512 L
Initial amt. of beer : After mixed with water
512 : 343
Taking cube roots on both the sides,
8 :7
For 1 unit of beer -> 15 L
For 8 units of beer -> 120 L
Incorrect
The initial amount of beer in the
container was =343 + 169 = 512 L
Initial amt. of beer : After mixed with water
512 : 343
Taking cube roots on both the sides,
8 :7
For 1 unit of beer -> 15 L
For 8 units of beer -> 120 L
Question 10 of 25
10. Question
1 points
A tank which contains a mixture of syrup and water in ratio 15:6. 25.5 liters of
mixture is taken out from the tank and 2.5 liters of pure water and 5 liters of syrup is
added to the mixture. If resultant mixture contains 25% water, what was the initial
quantity of mixture in the tank before the replacement in liters?
Correct
Quantity of Syrup = 15x
Quantity of water =6x
Total = 21x
Resultant Mixture = 21x – 25.5 + 2.5 + 5 =
21x – 18
Resultant water = 6x – 25.5 * (6/21) + 2.5
= 6x – 7.28
Resultant mixture contains 25% water
(21x – 18)*25/100 = 6x – 7.28
x = 3.7
Initial quantity = 21*3.7 = 77.7
Incorrect
Quantity of Syrup = 15x
Quantity of water =6x
Total = 21x
Resultant Mixture = 21x – 25.5 + 2.5 + 5 =
21x – 18
Resultant water = 6x – 25.5 * (6/21) + 2.5
= 6x – 7.28
Resultant mixture contains 25% water
(21x – 18)*25/100 = 6x – 7.28
x = 3.7
Initial quantity = 21*3.7 = 77.7
Question 11 of 25
11. Question
1 points
Ram covered a distance of 200 km in 10 hrs . The first part of his journey is covered
by auto ,then he hired a car .The speed of the auto and car is 15 km/hr and 30 km /hr
resp. Find the ratio of distance covered by auto and car .
Correct
Incorrect
Question 12 of 25
12. Question
1 points
9 L are drawn from a cask full of water and it is then filled with milk , 9 L of
mixture are drawn and the cask is again filled with milk .The quantity of water now
left in the cask to that of the milk in it is 16: 9 .How much does the cask hold ?
Correct
16 – > water
25 – > milk
=> √(16/25) = 4/5
If 1 -> 9
then 5 = 45 liters
Incorrect
16 – > water
25 – > milk
=> √(16/25) = 4/5
If 1 -> 9
then 5 = 45 liters
Question 13 of 25
13. Question
1 points
If 2 kg metal , of which (1/3) is zinc and the rest is copper , be mixed with 3 kg of
metal , of which (1/4) is zinc and the rest is copper . What is the ratio of zinc to copper
in the mixture ?
Correct
Quantity of zinc in the mixture = 2 (1/3) +
3 (1/4) = ( 2/3) + (3/4)
=17/12
Quantity of copper in the metal = (3+2) –
(17/12) = 43/12
Therefore , 17 / 12 : 43 /12 = 17 : 43
Incorrect
Quantity of zinc in the mixture = 2 (1/3) +
3 (1/4) = ( 2/3) + (3/4)
=17/12
Quantity of copper in the metal = (3+2) –
(17/12) = 43/12
Therefore , 17 / 12 : 43 /12 = 17 : 43
Question 14 of 25
14. Question
1 points
Vessels A and B contain mixtures of milk and water in the ratios 4 : 5 and 5 : 1 resp
.In what ratio should quantities of mixture be taken from A and B to form a mixture in which milk to water is in the ratio 5 : 4 ?
Correct
Quantity of milk in vessel A = 4 / (4+5) =
4/9
Quantity of milk in vessel B = 5/(5+1) =
5/6
Quantity of milk in resultant mixture = 5 /(
5+4) = 5/9
A B
4/9 5/6
. 5/9
5/18 1/9
Required ratio= 5:2
Incorrect
Quantity of milk in vessel A = 4 / (4+5) =
4/9
Quantity of milk in vessel B = 5/(5+1) =
5/6
Quantity of milk in resultant mixture = 5 /(
5+4) = 5/9
A B
4/9 5/6
. 5/9
5/18 1/9
Required ratio= 5:2
Question 15 of 25
15. Question
1 points
Two barrels contain a mixture of ethanol and gasoline is 60% in the first barrel and
30% in the second barrel .In what ratio must the mixtures from the first and the
second barrels be taken to form a mixture containing 50% alcohol ?
Correct
Mix. I Mix. II
3/5 3/10
. 1/2
1/5 1/10
Required ratio = 2 : 1
Incorrect
Mix. I Mix. II
3/5 3/10
. 1/2
1/5 1/10
Required ratio = 2 : 1
Question 16 of 25
16. Question
1 points
A mixture of a certain quantity of milk with 15 ltr of water is sold at 100 paisa per ltr. If
pure milk be worth Rs 1.15 ltr, then how
much milk is there in the mixture?
Correct
Incorrect
Question 17 of 25
17. Question
1 points
In a mixture of 75ltr the ratio of milk to water is 2 : 1. The amount of water to be
further added to the mixture so as to make the ratio of milk to water 1 : 2 will be?
Correct
M : W
BEFORE 2 : 1…………..(1)
AFTER 1*2: 2*2
. 2 : 4………….(2)
multiply 2 in equation (2) to make milk
same……so…..1 =25
4 = 100
100 – 25 = 75ltr
Incorrect
M : W
BEFORE 2 : 1…………..(1)
AFTER 1*2: 2*2
. 2 : 4………….(2)
multiply 2 in equation (2) to make milk
same……so…..1 =25
4 = 100
100 – 25 = 75ltr
Question 18 of 25
18. Question
1 points
A container contained 60ltr milk. Out of this 6ltr of milk was taken out and replaced
with water. This process was further repeated two times. How much milk is now
in container?
In an alloy zinc & copper are in the ratio of 1 :1. In the second alloy the same element
are in the ratio 3 : 5. If these two alloys be mixed to form a new alloy in which two
elements are in the ratio 2 : 3, find the ratio of these two alloys in the new alloy?
Correct
½…………………..3/8
. 2/5
. 1/40…………………1/10
. 1 : 4
Incorrect
½…………………..3/8
. 2/5
. 1/40…………………1/10
. 1 : 4
Question 20 of 25
20. Question
1 points
In a class of 20 students the average of their marks is 59. If one student left the class then average become 60. Find the marks of that student?
Correct
Average increases by 1 when 1 leaves, so
for 19 students::
. 59 – 19 = 40
Incorrect
Average increases by 1 when 1 leaves, so
for 19 students::
. 59 – 19 = 40
Question 21 of 25
21. Question
1 points
If the sum of 5 consecutive odd number is 265. Then the largest number would be?
Correct
Average = 265/5 = 53
(average)53…..55……57…….ans is 57
Incorrect
Average = 265/5 = 53
(average)53…..55……57…….ans is 57
Question 22 of 25
22. Question
1 points
In a bag there are three types of coins, 1rupee, 50paisa, 25paisa in the ratio of
5:10:16. The total value is Rs 700. The total number of coins is?
A can contain a mixture of two liquids P & Q in proportion 3 :5. When 8ltr of mixture are drawn off and the can is filled with Q, the proportion of P & Q becomes 3:7. How
many ltr of liquid P was contained in the can initially?
Correct
. P…………..Q
initial 3…………….5
after 3……………..7
7-5 = 2
2 =8
1 =4
after (7+3) =10= 40ltr
so initial = 3/3+5*40 = 15ltr
Incorrect
. P…………..Q
initial 3…………….5
after 3……………..7
7-5 = 2
2 =8
1 =4
after (7+3) =10= 40ltr
so initial = 3/3+5*40 = 15ltr
Question 24 of 25
24. Question
1 points
300 ltr of mixture contains 20% water in it and rest is milk. The amount of milk that
must be added so that the resulting mixture contains 90% milk is?
Correct
20% of 300 = 60
now we have to make milk 90% then water
will become 10%
10% = 60
100% = 600
so 600 – 300 = 300ltr
Incorrect
20% of 300 = 60
now we have to make milk 90% then water
will become 10%
10% = 60
100% = 600
so 600 – 300 = 300ltr
Question 25 of 25
25. Question
1 points
8kg of tea consisting Rs240 per kg is mixed with 9kg of tea costing Rs25o per kg. The
average price per kg of the mixed tea is ?