Allegations and Mixtures Practice Questions – Set 5
1. Product
P is produced by mixing chemical C and chemical H in the ratio of 5: 4.
Chemical C is prepared by mixing two raw materials, C1 and C2, in the ratio of
1: 3. Chemical H is prepared by mixing raw materials, C2 and H1, in the ratio
of 2: 1. Then the final mixture is prepared by mixing 864 units of product P
with water. If the concentration of the raw material C2 in the final mixture is
50%, how much water had been added to product P?
(a) 368 units
(b) 328 units
(c) 392 units
(d) 616 units
Answer: A)
Explanation: P is produced by mixing chemical C and
chemical H in the ratio 5: 4
Hence, 5/9th of product P
is chemical C and 4/9th of product P is chemical H
Chemical C has Cs1 and C2
in the ratio 1: 3
so, 3/4th of C is C2
Therefore, fraction of C2
in product P from chemical C = 5/9 * 3/4
Chemical H has C2 and H1
in the ratio 2: 1
so, 2/3rd of H is C2
Therefore, fraction of C2
in product P from chemical H = 4/9 * 2/3
Adding the two, the
fraction of C2 in Product P = 5/9* 3/4 + 4/9* 2/3 = 77/108
The final product is
obtained by mixing 864 units of product P with water
In 864 units of Product
P, amount of C2 = 864 * 77/108 = 616
In the final mixture,
concentration of C2 is 50%.
Therefore, the total
quantity of final mixture = 616 * 2 = 1232
Water added = 1232 – 864
= 368
2. A
caretaker of zoo counted the heads of the animals in a zoo and found it to be
80. When he counted the legs of the animals he found it to be 260. If the zoo
had either Hornbill or jackal, how many jackals were there in the zoo?(In the
zoo, each jackal had four legs and each hornbill had two legs)
(a) 40
(b) 30
(c) 50
(d) 60
(e) Cannot determined
Answer: C)
Explanation: Let the number of jackals be x
Then the number of
hornbills = 80 – x
Each hornbill has two
legs and each jackal has four legs
Therefore total number of
legs = 4x + 2(80 – x) = 260
4x + 160 – 2x =260
2x =100
x = 50
3. An
amount of Rs. 50,000 invested in two parts. The first yields an interest of 9%
p.a. while second 11% p.a. If the total interest at the end of year is 9.75% of
capital, find the amount invested in latter part?
(a) Rs. 17850
(b) Rs. 18500
(c) Rs. 18650
(d) Rs. 18750
(e) None of these
Answer: D)
Explanation:
9 11
9.75
1.25 0.75
Amount invested in latter
scheme = 0.75/2 * 50000 = Rs. 18750
4. There
are two mixtures of wine and water, the quantity of wine in them being 25% and
75% of the mixture. If two gallons of the first are mixed with three gallons of
the second, what will be the ratio of wine to water in the new mixture?
(a) 9 : 11
(b) 11: 9
(c) 9: 11
(d) 2: 11
(e) None of these
Answer: B)
Explanation: Required ratio = [(25/100 × 2) +
(75/100 × 3)]/[(75/100 × 2) + (25/100 × 3)] = 11/9
5. Two
vessels contain mixtures of honey and water in the ratio of 8: 1 and 1: 5
respectively. The contents of both of these are mixed in a specific ratio into
a third vessel. How much mixture must be drawn from the second vessel to fill
the third vessel (capacity 36 gallons) completely in order that the resulting
mixture may be half honey and half water?
(a) 15 gallons
(b) 14 gallons
(c) 13 gallons
(d) 12 gallons
(e) None of these
Answer: B)
Explanation:
8/9 1/6
1/2
1/3 7/18
Required amount = 7/18 ×
36 = 14
6. Bottle B1 contains
milk and water in the ratio 4:5. Bottle B2 contains milk and
water in the proportion 5:1.In what proportion should quantities be taken from
B1& B2 to form a mixture in which milk and water
are in the ratio 5:4?
(a) 2:5
(b) 3:2
(c) 2:3
(d) 5:2
(e) None of these
Answer: D)
Explanation: Let consider the proportion of milk
in each mixture.
In Vessel A, the
proportion of milk in 4/((4+5) )=4/9 and
in vessel B, the
proportion of milk is 5/(5+1) = 5/6
The amount of milk in the
mixture= 5/((5+4)) = 5/9
(5/6)-(5/9)=5/18
(5/9)-(4/9)=1/9
4/9 5/6
5/9
5/18 1/9
The ratio is (5/18) :
(1/9) = 5 : 2
7. Digvijay
has Rs.5000. He invests a part of it at 3% per annum and the remainder at 8%
per annum simple interest. His total income in 3 years is Rs.750. Find the sum
invested at different rates of interest?
(a) 3000, 2000
(b) 3000, 4000
(c) 3500,2000
(d) 1900,3300
(e) Cannot be determined
Answer: A)
Explanation: He is investing part of it at 3% at
annum and remaining at 8% per annum since he is getting Rs 750 as profit for 3
years, rate percent of the whole amount is 5%
PNR/100 = 750
5000*3*R/100= 750
R= 5%
3% 8%
5%
3 2
From above The ratio of
investment will be 3:2
8. Quantity1:
How many kgs of best quality wheat costing Rs.42/kg should a shopkeeper mix
with 25 kgs of ordinary wheat costing Rs.24 per kg so that he makes a profit of
20% on selling the mixture at Rs.40/kg?
Quantity2:
25 kg
(a) Quantity1 > Quantity2
(b) Quantity1 ≤ Quantity2
(c) Quantity1 < Quantity2
(d) Quantity1 ≥ Quantity2
(e) Quantity1 = Quantity2 or No relation
Answer: C)
Explanation: As the trader makes 20% profit by
selling the mixture at Rs.40/kg, his cost per kg of the mixture = Rs.32/kg.
C.P of 1 kg of best
quality wheat = Rs. 42
C.P of 1 kg of wheat
of 2nd kind = Rs. 24
Mean price = Rs. 40
Rs.42 Rs.24
Rs.32
8 10
Let the amount of best
quality wheat being mixed be x kg
8: 10 = x: 25
x = 20 kg
Quantity1 (20 kg) <
Quantity2 (25 kg)
9. In what
ratio must a person mix three kinds of rice costing him Rs 1.20, Rs 1.44 and Rs
1.74 per kg so that the mixture may be worth Rs 1.41 per kg?
(a) 11: 7: 77
(b) 26: 45: 8
(c) 27: 25: 6
(d) 11: 77: 7
(e) None of these
Answer: D)
Explanation: Mix rice of first and third kind to
get a mixture worth Rs 1.41 per Kg.
C.P of 1 Kg rice of 1st
kind 120paise
C.P of 1 Kg rice of 3rd
kind 174paise
Mean Price 141paise
120p 174p
141p
33 21
Resultant ratio (1st
kind of rice : 3rd kind of rice) = 33 : 21 = 11 : 7
Mix rice of 1st
kind and 2nd kind to obtain a mixture worth of Rs.1.41per kg
120p 144p
141p
3 21
Resultant ratio (1st
kind of rice : 2nd kind of rice) = 1 : 7
Required Ratio rice (1st
: 2nd : 3rd) = 11 : 77 : 7
10. A
sample of xlitres from a can having a 60 litre mixture of beer and water
containing beer and water in the ratio of 2: 3 is replaced with beer so that
the can will have beer and water in equal proportions. What is the value
of x?
(a) Cannot determined
(b) 30 litres
(c) 10 litres
(d) 6 litres
(e) None of these
Answer: C)
Explanation: The easy way to solve this question
is go through options
the mixture of 60 litres
has in it 24 litres of beer and 36 litres of water. (2: 3: : beer: water)
When you remove x litres
from it, you will remove 0.4x litres of beer and 0.6x litres of water from it
Take option c) According
to this option, x=10
So, when one removes, 10
litres of the mixture, one is removing 4 litres of beer and 6 litres of water
Therefore, there will be
20 litres of beer and 30 litres of water in the can
Now, when you add 10
litres of beer, you will have 30 litres of beer and 30 litres of water
