Online Quiz 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 1
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Subject :- Quantitative Aptitude
Chapter :- Mixture and Allegations-Test 1
Questions :- 25
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A vessel is filled with liquid, which is 3 parts water and 5 parts milk. How much of the liquid should be drawn of and replaced by water to make it half water and half milk?
Correct
Suppose the vessel initially contains 8 litres
of liquid.
Let x litres of this liquid be replaced with
water.
water in new mixture =(3-3x/8+x)
syrup in new mixture =(5-5x/8)
Then (3-3x/8+x) = (5-5x/8)
5x + 24 = 40 – 5x
10x=16==>x=8/5
So part of mixture replaced is 8/5*1/8=1/5
Incorrect
Suppose the vessel initially contains 8 litres
of liquid.
Let x litres of this liquid be replaced with
water.
water in new mixture =(3-3x/8+x)
syrup in new mixture =(5-5x/8)
Then (3-3x/8+x) = (5-5x/8)
5x + 24 = 40 – 5x
10x=16==>x=8/5
So part of mixture replaced is 8/5*1/8=1/5
Question 2 of 25
2. Question
1 points
Milk and water are in a Can A as 4:1 and in Can B as 3:2. For Can C, if one takes equal
quantities from A and B, find the ratio of milk to water in C.
Correct
Ratio of only milk in vessel A = 4 : 5
Ratio of only milk in vessel B = 3 : 5
Let ‘x’ be the quantity of milk in vessel C
4/5………………….3/5
……………x
3/5-x………………x – 4/5
(3/5-x)/(x-4/5)=1/1
X=7/10
Therefore, quantity of milk in vessel C = 7
=> Water quantity = 10 – 7 = 3
Hence the ratio of milk & water in vessel 3 is 7 : 3
Incorrect
Ratio of only milk in vessel A = 4 : 5
Ratio of only milk in vessel B = 3 : 5
Let ‘x’ be the quantity of milk in vessel C
4/5………………….3/5
……………x
3/5-x………………x – 4/5
(3/5-x)/(x-4/5)=1/1
X=7/10
Therefore, quantity of milk in vessel C = 7
=> Water quantity = 10 – 7 = 3
Hence the ratio of milk & water in vessel 3 is 7 : 3
Question 3 of 25
3. Question
1 points
A mixture contains alcohol and water in the ratio 3:2. If it contains 3 liters more alcohol
than water, the quantity of alcohol in the mixture
Correct
If quantity of water as x and alcohol as
x+3.
(x+3)/x=3/2
Water x=6 and alcohol = x+3 = 9 liters
Incorrect
If quantity of water as x and alcohol as
x+3.
(x+3)/x=3/2
Water x=6 and alcohol = x+3 = 9 liters
Question 4 of 25
4. Question
1 points
Three types of Rice of Rs. 1.27, Rs. 1.29 and Rs. 1.32 per kg are mixed together to
be sold at Rs. 1.30 per kg. In what ratio should this rice be mixed?
Correct
127………………………..132
………………130
2…………………………….3
Then
129………………132
………….130
1…………………….2
Hence final ratio is 2 : 2: 3+1 ==>1:1:2
Incorrect
127………………………..132
………………130
2…………………………….3
Then
129………………132
………….130
1…………………….2
Hence final ratio is 2 : 2: 3+1 ==>1:1:2
Question 5 of 25
5. Question
1 points
A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains 25%. The percentage of water in the mixture is:
Correct
Let CP of 1 litre milk be Rs 1
Then SP of 1 litre of mixture =Rs 1 Gain
25%
CP of 1 litre mixture =Rs (100/125*1)=4/5
Milk………………..Water
1………………………0
…………..4/5
4/5……………………1/5
Ratio 4:1
Hence %ge of water in the mixture
=(1/5*100)=20%
Incorrect
Let CP of 1 litre milk be Rs 1
Then SP of 1 litre of mixture =Rs 1 Gain
25%
CP of 1 litre mixture =Rs (100/125*1)=4/5
Milk………………..Water
1………………………0
…………..4/5
4/5……………………1/5
Ratio 4:1
Hence %ge of water in the mixture
=(1/5*100)=20%
Question 6 of 25
6. Question
1 points
A container contains 50 liters of milk. From this container 5 liters of milk was
taken out and replaced by water. This process was repeated further two times.
How much milk is now contained by the container?
Correct
Amount of milk left after 3 operations ={50
(1-5/50)3}
=50 * 9/10 * 9/10 * 9/10
=36.45
Incorrect
Amount of milk left after 3 operations ={50
(1-5/50)3}
=50 * 9/10 * 9/10 * 9/10
=36.45
Question 7 of 25
7. Question
1 points
An alloy of gold and copper weights 50 g. It contains 80% gold. How much gold
should be added to the alloy so that percentage of gold is increased to 90?
Correct
Gold in alloy =50*80% =40gm
Copper in alloy =50*20%=10gm
Now,
(40+x)/10=90/10
X=50gm
Incorrect
Gold in alloy =50*80% =40gm
Copper in alloy =50*20%=10gm
Now,
(40+x)/10=90/10
X=50gm
Question 8 of 25
8. Question
1 points
A trader sells total 315 TV sets. He sells black and white TV sets at a loss of 6% and
color TV sets at a profit of 15% thus he gains 9% on the whole. What are the no. of
black and white sets which he has sold?
4 kg of a metal contains 1/5 copper and rest in Zinc. Another 5 kg of metal contains
1/6 copper and rest in Zinc.The ratio of Copper and Zinc into the mixture of these
two metals:
Correct
Copper in 4 kg = 4/5 and Zinc in 4 kg =
4*4/5=16/5
Copper in 5 kg = 5/6 and Zinc in 5 kg =
5*5/6=25/6
Therefore, Copper in mixture = 4/5 +
5/6=49/30
and Zinc in the mixture = 16/5 +
25/6=221/30
Therefore the required ratio = 49 : 221
Incorrect
Copper in 4 kg = 4/5 and Zinc in 4 kg =
4*4/5=16/5
Copper in 5 kg = 5/6 and Zinc in 5 kg =
5*5/6=25/6
Therefore, Copper in mixture = 4/5 +
5/6=49/30
and Zinc in the mixture = 16/5 +
25/6=221/30
Therefore the required ratio = 49 : 221
Question 10 of 25
10. Question
1 points
Rs. 69 were divided among 115 students so that each girl gets 50 paise less than a boy. Thus each boy received twice the paise as each girl received. The no. of girls in the class is:
Correct
Here each girl receives 50 paise and each
boy receives 100 paise and the average
receiving of each student.
=6900/115=60paise
50…………………….100
…………..60
40……………………..10
4:1
5 == 115
4 ? == 92
Incorrect
Here each girl receives 50 paise and each
boy receives 100 paise and the average
receiving of each student.
=6900/115=60paise
50…………………….100
…………..60
40……………………..10
4:1
5 == 115
4 ? == 92
Question 11 of 25
11. Question
1 points
A shopkeeper purchase two quantities of rice at the rate of Rs. 280/kg and Rs.
260/kg . In 52 kg of the second quantity, how much rice of the first quantity should
be mixed so that by selling the resulting mixture at Rs.300/ kg , he gains a profit of
25%.
Correct
profit % = 25/100 = ¼
CP = 4 and profit = 1
SP = 5
Now , SP = 300
5———300
1———60
CP = 4 * 60 = R.240/kg
Rice 1————— Rice 2
200——————–260
————–240
20———————40
1:2
52 kg of the second quantity .
so Rice 1 : Rice 2 = 1*26 : 2 * 26= 26 : 52
Hence , 26 kg of Rice 1 is added in the
mixture.
Incorrect
profit % = 25/100 = ¼
CP = 4 and profit = 1
SP = 5
Now , SP = 300
5———300
1———60
CP = 4 * 60 = R.240/kg
Rice 1————— Rice 2
200——————–260
————–240
20———————40
1:2
52 kg of the second quantity .
so Rice 1 : Rice 2 = 1*26 : 2 * 26= 26 : 52
Hence , 26 kg of Rice 1 is added in the
mixture.
Question 12 of 25
12. Question
1 points
If the average weight of the whole class is 50 kg. And the average weight of boys in
the class is 30 kg and the average weight of girls in the same class is 22 kg. What could
be the possible strength of boys and girls in the class respectively?
Correct
No. of boys : No. of girls
22 ————— 30
———- 25
5 —————– 3
3 : 5
Hence , the possible strength of the boys
and girls in the whole class = 5 : 3
Incorrect
No. of boys : No. of girls
22 ————— 30
———- 25
5 —————– 3
3 : 5
Hence , the possible strength of the boys
and girls in the whole class = 5 : 3
Question 13 of 25
13. Question
1 points
A woman travels 200 km in 5 hours in two parts. In the first part of the journey, she
travels by car at the speed of 50 km/hr . In the second part of the journey , she travels
by bus at the speed of 30 km/hr . How much distance did she travel by bus?
Correct
speed of car —————- speed of bus
50 ———————————- 30
———————200/5
10 ———————————– 10
= 1 : 1
Time taken by both the vehicles = 5/2 = 2.5
hrs.
Therefore, distance travelled by bus =30 *
2.5 = 75 km
Incorrect
speed of car —————- speed of bus
50 ———————————- 30
———————200/5
10 ———————————– 10
= 1 : 1
Time taken by both the vehicles = 5/2 = 2.5
hrs.
Therefore, distance travelled by bus =30 *
2.5 = 75 km
Question 14 of 25
14. Question
1 points
Somnath bought two different kinds of oil, one is soya oil and another is olive
oil.There are two mixtures of these two oils . In the first mixture the ratio of the soya and olive oil is in the ratio of 3 : 4 and in the second mixture the ratio of the soya
and olive oil is 5 : 6 . If he mixes these two mixtures and makes a third mixture of 36
liters in which the ratio of the soya oil and olive oil is 4 : 5. Find the quantity of the
second mixture that is needed to make 36 liters of third type of mixture.
Correct
MixI —————– MixII
(3/7) —————- (5/11)
————–(4/9)
(1/99)——————(1/63)
Ratio = 7 : 11
Required quantity of the second mixture to
make the third mixture
= (11/18)*36 = 22 litres
Incorrect
MixI —————– MixII
(3/7) —————- (5/11)
————–(4/9)
(1/99)——————(1/63)
Ratio = 7 : 11
Required quantity of the second mixture to
make the third mixture
= (11/18)*36 = 22 litres
Question 15 of 25
15. Question
1 points
A vessel which contains 100 liters of salt and sugar solution in the ratio of 22 : 3 .
From the vessel 40 liters of mixture is taken out and 4.8 liters of pure salt solution and pure sugar solution , both are added to the mixture . What is the percentage of the
quantity of sugar solution in the final mixture less than the quantity of salt solution?
Correct
40 L is taken out remaining 60 L
salt solution = (22/25)*60 = 52.8 L
sugar solution = (3/25)*60 = 7.2 L
On adding salt and sugar solution
salt solution = 52.8 + 4.8 = 57.6 L
sugar solution = 7.2 + 4.8 = 12 L
Require % = (57.6 – 12)/57.6 = 79(1/6)%
Incorrect
40 L is taken out remaining 60 L
salt solution = (22/25)*60 = 52.8 L
sugar solution = (3/25)*60 = 7.2 L
On adding salt and sugar solution
salt solution = 52.8 + 4.8 = 57.6 L
sugar solution = 7.2 + 4.8 = 12 L
Require % = (57.6 – 12)/57.6 = 79(1/6)%
Question 16 of 25
16. Question
1 points
The average marks of the students in four sections P, Q ,R and S together is 60% .
The average marks of the students of P, Q, R and S separartely are 45% , 50%, 72%
and 80% respectively. If the average marks of the students of P and Q together is 48%
and that of the students of Q and R is 60%. What is the ratio of number of students in
sections A and D ?
Correct
A —————- D
45—————80
——–60
20———–15
4 : 3
Hence , the required ratio = 4 : 3
Incorrect
A —————- D
45—————80
——–60
20———–15
4 : 3
Hence , the required ratio = 4 : 3
Question 17 of 25
17. Question
1 points
A shopkeeper has two types of wheat . The percentage of first type of wheat is 80%
and the percentage of second type of wheat is 60%. If he mixes 28 kg of first type of
wheat to the 32 kg of second type of wheat , then find the percentage of resultant wheat
in the mixture.
Correct
Type I ————- Type II
60———————–80
————-x
32———————-28
8 : 7
7 : 8 (reverse ratio)
Now,(80-60)*7/(7+8) = 20 *(7/15) = 9.33
Required % = 60+9.33 = 69.33
Incorrect
Type I ————- Type II
60———————–80
————-x
32———————-28
8 : 7
7 : 8 (reverse ratio)
Now,(80-60)*7/(7+8) = 20 *(7/15) = 9.33
Required % = 60+9.33 = 69.33
Question 18 of 25
18. Question
1 points
From a container of wine , 8 liters of wine is drawn and replace the same quantity
with water. This is performed three more times, now the ratio of the quantity of wine
to that of water in the container becomes 16 : 65. What is the initial quantity of wine in
the container?
Correct
Let x be the initial quantity of the wine .
After 4 operations the quantity of wine left
= [x{1-(8/x)^4}]L
=> [x{1-(8/x)}^4] = 16 / 81
=>{1 – (8/x)}^4 = 16/81
=> (x -8)/x = 2/3
=> x = 24 L
Incorrect
Let x be the initial quantity of the wine .
After 4 operations the quantity of wine left
= [x{1-(8/x)^4}]L
=> [x{1-(8/x)}^4] = 16 / 81
=>{1 – (8/x)}^4 = 16/81
=> (x -8)/x = 2/3
=> x = 24 L
Question 19 of 25
19. Question
1 points
The price of the diesel is Rs. 70 per litre and the price of the petrol is Rs. 40 per
litre. If the profit after selling the mixture at Rs. 75 per litre be 25 %. Find the ratio of
the diesel and petrol in the mixture.
There are two factories, one in India and another in US. Mr. Anish purchased these
two factories for total 80 crores. Later on, he sold the Indian factory at the rate
of 16% profit and the US factory at 32% profit, thereby he gained 20%. What is the
selling price of the factory?
Correct
Indian Factory ————– US factory
16———————32
————-20
12———————4
= 3 : 1
The CP of Indian Factory = (80/4)*3 = 60
crores
SP = 69.6 crores
Incorrect
Indian Factory ————– US factory
16———————32
————-20
12———————4
= 3 : 1
The CP of Indian Factory = (80/4)*3 = 60
crores
SP = 69.6 crores
Question 21 of 25
21. Question
1 points
A chemist has 10L of a solution that is 10% nitric acid by volume. He wants to dilute
the solution to 4% strength by adding water. How many litres of water must be add?
Correct
Quantity of nitric acid = 10 *(1/10) = 1 L
Water = 10 – 1 = 9 L
Let x litre of water be added,
(10 + x ) * (4/100) = 1
=> x = 15 L
Incorrect
Quantity of nitric acid = 10 *(1/10) = 1 L
Water = 10 – 1 = 9 L
Let x litre of water be added,
(10 + x ) * (4/100) = 1
=> x = 15 L
Question 22 of 25
22. Question
1 points
A bottle contains (3/4) of milk and the rest water. How much of the mixture must be
taken away and replaced by an equal quantity of water so that the nixtude has half milk and half water?
Correct
Ratio of milk : water = 3 : 1
water = (1/4)*100 = 25
Let x L is taken out , then
qty. of milk left= (3 – 3x/4)
water left = (1 – x/4) + x
Now , 3 – 3x/4 = (1 – x/4) + x
=> x = 4/3
Required % = 4/(3*4)*100 = 33(1/3)%
Incorrect
Ratio of milk : water = 3 : 1
water = (1/4)*100 = 25
Let x L is taken out , then
qty. of milk left= (3 – 3x/4)
water left = (1 – x/4) + x
Now , 3 – 3x/4 = (1 – x/4) + x
=> x = 4/3
Required % = 4/(3*4)*100 = 33(1/3)%
Question 23 of 25
23. Question
1 points
P and Q are two alloys of gold and copper prepared by mixing metals in the ratio 7 : 2
and 7 : 11 resp. If equal quantities of the alloys are melted to form a third alloy R, Find the ratio of gold and copper.
Correct
In 1 kg of alloy P, Gold = 7/9
Copper = 2/9
In 1 kg of alloy Q, Gold = 7/18
Copper = 11/18
Therefore, Ratio of Gold and Copper in
alloy R
= 7/9 + 7/18 : 2/9 + 11/18
= 21 : 15 = 7 : 5
Incorrect
In 1 kg of alloy P, Gold = 7/9
Copper = 2/9
In 1 kg of alloy Q, Gold = 7/18
Copper = 11/18
Therefore, Ratio of Gold and Copper in
alloy R
= 7/9 + 7/18 : 2/9 + 11/18
= 21 : 15 = 7 : 5
Question 24 of 25
24. Question
1 points
A container has 30 L of water. If 3 L of water is replaced by 3 L of spirit and this
operation is repeated twice , what will be the quantity of water in the new mixture?
Correct
Suppose a container contains x units of
liquid from which y units are taken out and
replaced by water. After n operations, the
quantity of pure liquid.
= x(1 – y/x)^n units
= Remaining water = 30(1 – 3/30)^2 = 24.3 L
Incorrect
Suppose a container contains x units of
liquid from which y units are taken out and
replaced by water. After n operations, the
quantity of pure liquid.
= x(1 – y/x)^n units
= Remaining water = 30(1 – 3/30)^2 = 24.3 L
Question 25 of 25
25. Question
1 points
Two barrels contain a mixture of ethanol and gasoline. The content of the ethanol 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% ethanol?
Correct
Mixture I ————– Mixture II
Ethanol – (3/5)———Ethanol- (3/10)
———————(1/2)
(1/5)—————————(1/10)
= 2 : 1
Incorrect
Mixture I ————– Mixture II
Ethanol – (3/5)———Ethanol- (3/10)
———————(1/2)
(1/5)—————————(1/10)
= 2 : 1