Q1.
If two numbers are in the ratio of 10:3 and 2 is subtracted from each,
the resulting numbers are in the ratio 9:2. Find the numbers
Ans1. Since the ratio is 10:3, let the numbers be 10X and 3X
Therefore (10X -2) / (3X-2) = 9/2
7X = 14, X = 2
Therefore numbers are 20 and 6
Q2. A common foodstuff is found to contain 2.5% iron. The serving size
is 90.0 grams. If the recommended daily allowance is 18 gm of iron, how
many servings would a person have to eat to get 100% of the daily
allowance of iron?
Ans2. Iron percentage = 2.5/100 x 90 = 2.25 gms
Total grams required = 18
Servings required = 18/2.25 = 8 servings in a day
Q3. A substance is 99% water. Some water evaporates, leaving a substance
that is 98% water. How much of the water evaporated?
Ans3. The substance has 99% water and 1 % other substance
Let the amount of water be X and the other substance be Y
Now X/(X+Y) = 99/100, Y = X/99
After some evaporation suppose water left = Z
Therefore Z/(Z+Y) = 98/100, Y = 2Z/98
Equating Y, X/99 = 2Z/98
Z = 0.495X
Therefore Current water level is 0.495 for earlier water level, so 50.5 % water has been evaporated.
Q4.
If I clean a 3200 square foot building five nights per week for a sum
of Rs. 575 per month, what is the cost per square foot?
Ans4. Total square foot = 3200
Total sum = Rs. 575
Cost per square foot = 575/3200 = Rs. 0.18 per square foot
Q5. A and B started a business by investing Rs 50000 and Rs 25000. what is the share of each if yearly profit is Rs 2000.
Ans5. Total investment = 50000 + 25000 = Rs.75000
A’s share = 50000/75000 x 2000 = Rs.1333.34
B’s share = 25000/75000 x 2000 = Rs. 666.66
Q6.The
salary of Ravi, Ajay and Bhuvan is Rs 350000. If they spend 70%, 75%,
and 80% of their salaries respectively, their savings are in ratio of
15:10:25. Find their salaries.
Ans 6. Total Salary of the three = 350000
Ravi’s spent = 70%, therefore Ravi’s saving = 30%
Ajay’s spent = 75%, therefore Ajay’s saving = 25%
Bhuvan’s spent = 80%, therefore Bhuvan’s saving = 20%
30 % of Ravi’s Salary: 25 % of Ajay’s Salary: 20 % of Bhuvan’s Salary = 15:10:25
30/100 R : 25/100 A : 20/100 B = 15:10:25
30R:25A:20B = 15:10:25
From here 30R/25A = 15/10, R/A = 375/300 = 17/12 = 34/24
Also 25A/20B = 10/25, A/B = 8/25 = 24/75
Now R:A:B = 34:24:75
Ravi’s Salary = 34/133 x 350000 = 89474
Ajay’s Salary = 24/133 x 350000 = 63158
Bhuvan’s Salary = 75/133 x 350000 = 197368
Q7.
Divide Rs 435 among A, B and C so that if Rs 9, Rs 4 Rs 2 be subtracted
from their respective shares, the shares left may be in the ratio 6:4:5.
Ans7. Here the ratio of shares is given 3:4:5
The total is 435, and 9+4+2 = 15 needs to be subtracted from it
= 435 – 15 = 420
Now diving 420 in ratio of 6:4:5
A’s Share = 6/15 x 420 = 168, 168 + 9 = 177
B’s Share = 4/15 x 420 = 112, 112 + 4 = 116
C’s Share = 5/15 x 420 = 140, 140 + 2 = 142
Q8.
A, B and C partnered in a business. A contributed Rs 12000 , B Rs 10000
and C Rs 8000 and their profit was Rs. 2400. What is the share of each?
Ans8. A’s Contribution = 12000
B’s Contribution = 10000
C’s Contribution = 8000
Total = 30000
A’s share = 12000/30000 x 2400 = Rs 960
B’s share = 10000/30000 x 2400 = Rs 800
C’s share = 8000/30000 x 2400 = Rs 640
Q9.
A, B and C partnered in a business for an year. A contributed Rs 12000
for 6 months, B Rs 10000 for 8 months and C Rs 8000 for the entire year
and their profit was Rs. 2728. What is the share of each?
Ans9. A’s Contribution = 12000 for 6 months = 12000 x 6 = 72000
B’s Contribution = 10000 for 8 months = 10000 x 8 = 80000
C’s Contribution = 8000 for 12 months = 8000 x 12 = 96000
Total = 248000
A’s share = 72000/248000 x 2728 = Rs 792
B’s share = 80000/248000 x 2728 = Rs 880
C’s share = 96000/248000 x 2728 = Rs 1056
Q10.
Ravi and Mayank enter into a partnership by investing Rs. 7000 and Rs.
3000 respectively. At the end of one year, they divided their profits
such that 1/3 of the profit is divided equally for the efforts they
have put into the business and the remaining amount of profit is
divided in the ratio of the investments they made in the business. If
Ravi received Rs. 8000 more than Mayank did, what was the profit made
by their business in that year?
Ans10. Let the profit made during the year be X
The Profit to be divided equally = X/3
The profit to be divided as per contribution = 2X/3
The diving ratio is 7000:3000 = 7:3
Ravi shall get more profit in 2X/3, by 70-30 = 40% more
Therefore 40/100 x 2X/3 = 8000
The profit, X = Rs. 30000
Q11. A, B and C enter into a partnership by investing Rs.3000, Rs.4000 and Rs.2000. A is a working partner and gets 1/4th
of the profit for his services and the remaining profit is divided
amongst the three in the rate of their investments. What is the amount
of profit that B gets if A gets a total of Rs. 8000?
Ans11. Let the profit be X
The investment is in the ratio 3000:4000:2000 = 3:4:2
Profit to be taken by A exclusively = 1X/4
Profit to be divided equally =3X/4
A's profit = X/4 + 1/3 x 3X/4 = 8000
Form here, X = 16000
Therefore B's profit = 4/9 (3/4 x 16000) = Rs. 5333.33
Q12.
A, B and C, each of them working alone can complete a job in 6, 8 and
12 days respectively. If all three of them work together to complete a
job and earn Rs.2340, what will be B’s share of the earnings?
Ans12. A does 1/6th of job in a day (total job in 6 days)
B does 1/8th of job in a day (total job in 8 days)
C does 1/12th of job in a day (total job in 12 days)
Ratio of Jobs done = 1/6:1/8:1/12 = 4:3:2
B’s Share = 3/9 x 2340 = Rs. 780
Q13
A petroleum distributor has two gasohol storage tanks, the first
containing 9 percent alcohol and the second containing 12 percent
alcohol. They receive an order for 300,000 gallons of 10 percent
alcohol. How can they mix alcohol from the two tanks to fulfill it?
Ans13. Let total volume to be delivered is 100 and X is the amount of 9% alcohol, therefore the amount of 10% alcohol is 100-X
Therefore 0.09X + 0.12(100-X) = 10 (10% of 100)
X = 66.67, 9% alcohol, therefore 12 % alcohol = 33.33 %
In 300000 gallons, 9% alcohol = 66.67/100 x 300000 = 200010
In 300000 gallons, 12% alcohol = 33.33/100 x 300000 = 99990
Q14.
Milk that has 5% fat is mixed with milk that has 2% fat. How much of
each is needed to obtain 60 litres of milk that has 3% fat?
Ans14. Let total volume to be delivered is 100 and X is the amount of 5% fat, therefore the amount of 2% fat is 100-X
Therefore 0.05X + 0.02(100-X) = 3 (3% of 100)
X = 33.33
X = 33.33, 5% fat, therefore 2 % fat = 66.67 %
In 60 litres, 2% fat = 66.67/100 x 60 = 40.002
In 60 litres, 5% fat = 33.33/100 x 60 = 19.998
Q15. A chemist has 6 liters of a 25% alcohol solution. To make a solution containing 50% alcohol how much alcohol should he add?
Ans15. Solution volume = 6 litres, with 25% alcohol
Therefore alcohol = 25/100 x 6 = 1.5 L
Suppose X litres of alcohol is added to make it 50% solutions
Therefore (1.5 + X )/(6+X) = 0.5
X = 3 litres
Q16.Two
vessels contain mixtures of spirit and water. In the first vessel the
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 litres are
taken from the first vessel?
Ans16. Let X litres be the volume taken from the first vessel;
Therefore (35-X) litres are taken from the second.
Amount of spirit from vessel I = 8X/11 (ratio in vessel I is 8:3)
Amount of spirit from vessel I = 5/6(35-X) (ratio in vessel I is 8:3)
Total spirit = 8X/11 + 5/6(35-X) = 4/5 x 35 (mixture ratio is 4:1)
Solving, X = 11 litres
Q17.
An alloy of aluminum, iron and zinc contains 90% of zinc, 7% of
aluminum and 3% of iron. A second alloy of zinc and iron only is melted
with the first and the mixture contains 85% of zinc, 5% of aluminum and
10% of iron. Find the percentages in the second alloy.
Ans17. Let original alloy had 100 gms, and X gms of zinc and Y gms of iron in the second alloy. Total weight now = 100 + X + Y
Now (90 + X)/(100 + X + Y) x 100 = 85
And 7/(100 + X + Y) x 100 = 5
And (3+Y)/(100 + X + Y) x 100 = 10
From second equation X + Y = 40
Using it in first equation, (90 + X)/(100 + 40) x 100 = 85
Therefore X = 29 and Y = 40 – 29 = 11
X % = 29/40 x 100 = 72.5 % and Y = 100 – 72.5 = 27.5 %
Q18.
How many kilograms of Basmati rice costing Rs.42/kg should a shopkeeper
mix with 25 kgs of ordinary rice costing Rs.24 per kg so that he makes
a profit of 25% on selling the mixture at Rs.40/kg?
Ans18. Let the amount of Basmati rice being mixed be X kgs.
As the trader makes 25% profit by selling the mixture at Rs.40/kg, his cost for one kilogram = 40 x 100/125 = Rs. 32/kg
Therefore (X x 42) + (25 x 24) = 32 (X + 25)
Solving X = 20 kgs.
Q19. A grocer wishes to mix 6 pounds of Ray’s dog food worth $3 per pound
with Bram’s dog food, worth $4 per pound. How much of the Bram’s dog
food should be mixed if the dog food mixture is to sell for $3.20?
Ans19. Quantity of Ray’s dog food = 6 pounds
Cost = $3 per pound
Let Quantity of Ray’s dog food = A pounds
Cost = $4 per pound
Using rule of allegation
6/A = (4 – 3.2)/(3.2 – 3), from Here A = 1.5 pounds
Q20.
6kg of inferior quality tea are mixed with 3kg of high quality tea
which costs Rs.2.00 per kilogram more than the inferior tea. The total
price of the mixture is Rs.24.00. What is the price of the inferior
tea?
Ans20. Let x be the price per kg of the inferior tea
Therefore price per kg of expensive tea = x + 2
Total cost will be 6x + 3(x+2) = 24
6x + 3x + 6 = 24
9x = 18
x = 2
So cheap tea is Rs. 2 per kg.
Q21. Peanuts sell for Rs.3.00 per Kilogram. Cashews sell for Rs.6.00 per Kilogram.
How many Kilograms of cashews should be mixed with 12 Kilograms of peanuts
to obtain a mixture that sells for Rs.4.20 per Kilograms?
Ans21. Let A be the weight of cashews to buy.
Total cost of mixture = (12 x 3) + (A x 6) = (12+A) x 4.2
36 + 6A = 50.4 + 4.2A
1.8A = 14.4
A = 8
So you need to buy 8 kgs of cashews.
Q22.
In what ratio must a person mix three kinds of tea costing Rs.60/kg,
Rs.75/kg and Rs.100 /kg so that the resultant mixture when sold at
Rs.96/kg yields a profit of 20%?
Ans22. The resultant mixture is sold at a profit of 20% at Rs.96/kg
Therefore the cost = 96 x 100/120 = Rs. 80
Using rule of allegation for 75 and 100,
(Quantity of 75)/ (Quantity of 100) = (100 – 80)/(80-75)= 4/1
Using rule of allegation for 60 and 100,
(Quantity of 60)/ (Quantity of 100) = (100 – 80)/(80-60)= 1/1
Therefore (Quantity of 60): (Quantity of 75): (Quantity of 100)
is 1:4:2
Q23.
From a cask of milk containing 30 litres, 6 litres are drawn out and
the cask is filled up with water. If the same process is repeated a
second, then a third time, what will be the number of litres of milk
left in the cask?
Ans23. As per the formula x [1-y/x]n
where x is initial quantity of milk in the cask y is the quantity of
milk withdrawn in each process and n is the number of process.
Therefore Quantity of milk left after the 3rd operation = [1 – 6/30]3 x 30 = 15.36 liters
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