Instructions

For the following questions answer them individually

Question 1

The average weight of a class of students is 24 kgs. 5 students who together weigh 140 kgs leave the class and 2 students who together weigh 68 kgs join the class. What is the new average of the class?

Solution

Let the number of students be k.

Total weight = 24k.

5 students left and 2 students joined. So, new total number of students = k-3

New total weight = 24k - 140 + 68 = 24(k-3).

New average = 24(k-3)/(k-3) = 24.

Question 2

There are 11 participants in an under-19 competition ( Participants can be up to 19 years old ) . The average age of the group is 13 years. Leaving out the youngest and the oldest, the others are divided into 3 groups of 3 each, such that the average ages of the 3 groups are 11, 13 and 14 years respectively. What is the maximum difference between the ages of the youngest and oldest participants? (All the ages are integers)

Solution

Total age of group = 143.

Total age of the three groups together = 33+39+42 = 114.

Therefore, youngest + oldest = 29 years.

The age of the oldest can be 19.

So, the youngest is 10 years old.

Hence, the difference in ages is 9 years.

Question 3

In what ratio should three salt solutions of concentrations 15%, 20% and 30% be mixed, so that the resultant solution is of concentration 25% ?

Solution

Going from the options, we can see that if the ratio is 1:1:3, the resultant mixture will be of concentration: (1*15+1*20+3*30)/(1+1+3) = 125/5 = 25%. Option b) is the correct answer.

Question 4

Three students A, B, C stole some candies. As it was late in the evening, they decided to divide the candies equally among themselves in the morning and went to sleep. During the night, A woke up, ate one candy, then took one third of the remaining candies and went back to sleep. After some time B woke up, ate one candy, then took one third of the remaining candies and went back to sleep. Lastly, C woke up and did the same as other two, that is, ate one candy and then took one third of the remaining candies. When all three woke up in the morning, they found that the total number of candies were in the form 3k+1. Find the minimum number of candies that were present initially.

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Solution

Let the initial number of candies be x.

After A woke up, number of candies left = 2/3(x-1)

After B woke up, number of candies left = 2/3(2/3(x-1)-1)

Similarly for C it is = 2/3[2/3(2/3(x-1)-1)-1]

2/3[2/3(2/3(x-1)-1)-1] = 3k+1

[2/3(2/3(x-1)-1)-1] = 9k/2+3/2

2/3(2/3(x-1)-1) = 9k/2+5/2

(2/3(x-1)-1) = 27k/4+15/4

2/3(x-1) = 27k/4+19/4

x-1 = 81k/8 + 57/8

x = 81k/8 + 65/8

To minimize x, we need to find the minimum value of k such that 81k+65 is a multiple of 8.

81k+65 can be expressed as (80k+64) + (k+1). The first part is divisible by 8. For the second part to be divisible by 8, minimum value of k is 7.

For k = 7, x = 79.

Question 5

In a class consisting of 150 students, one more student is added and as a result the average weight of class increases by 0.05. From this class one student is taken out and the average weight of the class is reduced by 0.02. If the sum of the weights of the two students mentioned is 2x+b, where x is the original average weight of the class of 150 students, then find the value of b.

Solution

Let the initial average weight of 150 students be x.
$$\frac{Sum\ of\ the\ weights\ of\ 150\ students}{150} = x$$
Sum of the weights of 150 students = 150x
Let the new student, who is added, have weight m. The average weight of the new class becomes x+0.05
Therefore, [150x+m]/151 = x+0.05
150x+m = 151x+7.55
m = x+7.55
Let the weight of the student who is removed be k. Then the new average becomes x+0.05-0.02 = x+0.03
[150x+m-k]/150 = x+0.03
m-k = 150*0.03 = 4.5
k = m-4.5 = x+7.55-4.5 = x+3.05
Now sum of the weights = m+k = (x+7.55)+(x+3.05) = 2x+10.6
Hence b = 10.6

Alternative Solution:

The initial average is x,

The weight of 1st students can be calculated by subtracting the sum of initial weights from the final weights = 151(x+0.05) - 150x = x+7.55
Similarly, weight of 2nd students = 151{x+0.05} - 150{x+0.03} = x+3.05
Weight of 1st student+ weight of 2nd student = x+7.55+x+3.05 = 2x + 10.6 = 2x + b,
=> b=10.6.

Question 6

There are three taps A, B, C in a tank. Initially A and B are opened. A fills milk at the rate of 4 L/min and B fills water at the rate of 2 L/min. When the tank is 50% full, tap C is opened. Tap C fills milk at the rate of x L/min. If the concentration of milk is 80% when the tank is full, find the value of x (in L/min)?

Solution

Let the volume of the tank be 300L.
Initially, A and B are opened. Since the flow of A is double that of B, the volume of milk would be double that of water when the tank is 150L full.
Volume of milk = 100L
Volume of water = 50L
Now the tap C is opened.
Final volume of milk = 80% of 300 = 240L
Remaining volume of milk to be filled = 240-100 = 140L
Water to be filled = 150-140 = 10L
$$\frac{140}{4+x} = \frac{10}{2}$$
4+x = 28
x = 24

Question 7

Three friends Lohit, Mohit and Nohit have Rs. 120 together. Lohit gives Mohit an amount which Mohit already has; Mohit then gives Nohit an amount which Nohit already has; Nohit then gives Lohit an amount which Lohit already has. In the end all of them have equal amounts with them. What is the amount which Lohit originally had?

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Solution

Let the initial amount with Lohit, Mohit and Nohit be L, M and N respectively.

After Lohit gives money to Mohit the amount with each of them will be,
Lohit = L-M , Mohit = 2M , Nohit = N.

After Mohit gives money to Nohit the amount with each of them will be,
Lohit = L-M , Mohit = 2M-N , Nohit = 2N.

After Nohit gives money to Lohit the amount with each of them will be,
Lohit = 2(L-M) , Mohit = 2M-N , Nohit = 2N-(L-M).

Now all of them have equal amount with each other,

2(L-M) = 2M-N = 2N-(L-M)

Eqating first and second term,
2N-L+M = 2(L-M)
=> N = 3(L-M)/2....eqn1

Equating second and third term,
2M-N = 2N-(L-M)
L=3N-M....eqn2

Equating first and second term,
2M-N = 2(L-M)
4M=2L+N....eqn3

Substituting eqn 1 in eqn3
=> 4M = 2L + 3(L-M)/2 => 8M = 4L +3L - 3M
=> L/M = 11/7

Substituting eqn 2 in eqn3
4M=2(3N-M)+N => 4M = 6N-2M+N => N/M = 6/7

=> L: M: N = 11: 7: 6
=> L = 55, M = 35, N = 30.

Hence the amount which Lohit originally had is Rs. 55

Alternatively,
Working backwards
Stage Lohit Mohit    Nohit
After 3rd transaction 40 40    40
After 2nd transaction 20   40 60
After 1st transaction 20 70 30
Initially 55   35   30

Question 8

The ratio of incomes of Anil and Ajit is 5:7. The ratio of their expenditures is 3:4, then which of the following is a possible ratio of the savings? (Savings = Income - Expenditure)

Solution

Let the incomes be 5k and 7k and expenditures be 3t and 4t. Ratio of savings is (5k-3t)/(7k-4t).

Let us assume that (5k-3t)/(7k-4t) = 4/5 (Option A).
25k - 15t = 28k -16t
=> t = 3k.
Now, substituting this relationship in the equation (5k-3t)/(7k-4t), we get the numerator as -4k/-5k. Though the overall ratio is 4/5, the numerator is -4k. => 3t>5k in this case. However, expenditure cannot be greater than income and therefore, we can eliminate this case. Checking similarly, we can eliminate options B and C.

Option D:
(5k-3t)/(7k-4t) = 12/17
85k-51t = 84k - 48t
k = 3t.

Substituting this relationship in (5k-3t)/(7k-4t), we get the ratio as 12t/17t. Both the numerator and denominator are positive and hence, option D is a possible solution.

Question 9

Ironite is a mixture of iron and zinc only. It is extracted from an ore that contains iron and zinc in a ratio of 3:2 by weight. In the process of extraction, the entire iron is retained, but some amount of zinc is lost. If ironite contains 10% zinc by weight, find the amount of ore needed to produce 5 tonnes of Ironite.

Solution

In 5 tonnes of Ironite, zinc present = 0.5 tonnes and iron present = 4.5 tonnes.

Since the ratio given is 3:2, the weight of iron =3x and the weight of zinc =2x

Zinc is lost but iron will remain same in both, hence, 4.5=3x => x=1.5

The total weight = 3x+2x = 5x = 5*1.5 =7.5

Question 10

Lokesh bought Grapes, Milk and Sugar in the ratio of 8:5:3. Further, he knows that grapes are 60% water and milk is 80% water. He put everything together and cooked the mixture for 2 hours during which 25% of the water in the dish got evaporated. What is the ratio of dry grapes to water in the cooked dish? (All ratios and percentages are to be taken by weight)

Solution

Let the quantities of grapes, milk, and sugar he bought be 8k, 5k, and 3k respectively.
In 8k grapes- 60% is water and 40% is dry grapes i.e. 0.6 X 8k water and 0.4 X 8k dry grapes
In 5k milk, 80% is water i.e. 4k is water.
Total water in the mixture= 4.8k+4k=8.8k
Total dry grapes= 3.2k
Now after cooking only 75% of water remained.
=> water remaining= 3/4 X 8.8k= 6.6k
So, the required ratio = 3.2k/6.6k=16/33.

Question 11

A shopkeeper buys 3 varieties of rice for Rs 20, Rs 30 and Rs 40 per kg respectively. He now mixes the 3 varieties in the ratio 5:6:7 by weight and sells the mixture at a profit of 10%. What price per kg does he sell the mixture for?

Solution

CP of rice = 20*5x + 30*6x + 40*7x = 560x.
SP of rice = 560x*110/100 = 616x.
Therefore, price of the mixture = 616x/(5x+6x+7x) = $$34 \frac{2}{9}$$

Question 12

‘n’ years later the sum of ages of Messi and Maradona will be 80. ‘n’ years ago the age of Maradona was 3 times that of Messi. ‘2n’ years ago the age of Maradona was 4 times that of Messi. What will be the difference in their ages after ‘2n’ years from the present?

Solution

Let the current ages of Messi and Maradona be x and y.
(x+n) + (y+n) = 80 = > x+y+2n = 80
$$ 3\times(x-n)=(y-n)$$
$$ 4\times(x-2n)=(y-2n)$$

Solving the 3 equations we get,
n = 5, x = 20 and y = 50.

The difference between their ages will always be the same and is 30 years.

Question 13

A, B, C, D, E and F are six different natural numbers in ascending order. There were 15 numbers - 10, 15, 17, 13, 19, 18, 16, 20, 15, 13, 17, 11, 24, 22, and 20 - on a paper. Each of those 15 numbers were sum of 2 numbers taken at a time among A, B, C, D, E and F. Which of the following could be the value of C?

Solution

Each of the 15 numbers is a sum of 2 numbers taken at a time among A, B, C, D, E and F. Therefore in all possible cases A will be paired with B,C,D,E and F once hence A will be counted 5 times. This is true for each number. Therefore,

5A + 5B + 5C+ 5D + 5E + 5F = 10+15+17+13+19+18+16+20+15+13+17+11+24+22+20

A + B + C + D + E + F = $$\frac{10+15+17+13+19+18+16+20+15+13+17+11+24+22+20}{5}$$ = 50
If the 15 numbers are written in the order of their magnitude, then the sequence is 10, 11, 13, 13, 15, 15, 16, 17, 17, 18, 19, 20, 20, 22, 24.
The least number is obtained by adding the least 2 numbers among A, B, C, D, E ad F.
Hence A + B = 10 => B must be at least 6 => C must be at least 7.
=> C + D + E + F = 40
Highest number is obtained by adding E and F. Second highest number is obtained by adding D and F.
=> D + F = 22 and E + F = 24
=> C + D = 16 => D must be at least 9 and C can be at most 7.

=> The value of C is 7.

Question 14

A rice trader selling basmati rice makes a profit on the trade by adulterating the rice with an inferior substitute surti kolam. When he mixes the basmati and kolam in the ratio 3:2 he makes a profit of 33.33%. When he mixes them in the ratio 4:1 he makes a profit of Rs 25 per kg. In what ratio must he mix the two rices to make a profit of 100%? (Assume he sells at the stated price of basmati per kg)

Solution

Let the price of basmati and kolam be x and y respectively. Assuming that the quantity is 5 kg, the shopkeeper will mix 3 kg of basmati with 2 kg of surti kolam. So, the selling price will be 5x since he sells all quantity as basmati. But the actual cost price is less which is equal to 3kg of basmati and 2kg of surti kolam = 3x+2y. Profit = 5x- (3x+2y) = 2x-2y Hence, profit %= 33.33% = (2x-2y)/(3x+2y) => x=8y/3.

From, mixture 2, profit per kg = Rs 25. Hence profit for 5kg = 25*5 = (5x-4x-y)=x-y. Hence x=200 and y=75.

Let's assume for 100% profit, he is mixing them in the ratio of k:1.

In 100% profit case, the selling price will be double of the cost price.

S.P= 200(k+1) and C.P= (200 X k) + (75 X 1)

=> 200k+200= 2 X [200k + 75]

=> k=1/4

So the ratio is 1:4

Question 15

The price of a watch is marked up by 40% on its previous selling price and a discount of 50% is offered on the marked up price. If the profit made after the discount is 20%, find the profit made on the watch (in percentage) if the discount offered was 25%.
(If the answer is a%, enter a as your answer).

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Solution

Let SP = 100 => MP = 140.
New SP = 70.
Profit = 20% => CP = 70*100/120.
If the discount is 25%, the new SP is 105.
Therefore, profit % = $$\frac{105 - 70*100/120}{70*100/120} * 100$$% = 80%

Question 16

In a class consisting of 150 students, one more student is added and as a result the average weight of class increases by 0.05. From this class one student is taken out and the average weight of the class is reduced by 0.02. If the sum of the weights of the two students mentioned is 2x+b, where x is the original average weight of the class of 150 students, then find the value of b.

Solution

Let the initial average weight of 150 students be x.
$$\frac{Sum\ of\ the\ weights\ of\ 150\ students}{150} = x$$
Sum of the weights of 150 students = 150x
Let the new student, who is added, have weight m. The average weight of the new class becomes x+0.05
Therefore, [150x+m]/151 = x+0.05
150x+m = 151x+7.55
m = x+7.55
Let the weight of the student who is removed be k. Then the new average becomes x+0.05-0.02 = x+0.03
[150x+m-k]/150 = x+0.03
m-k = 150*0.03 = 4.5
k = m-4.5 = x+7.55-4.5 = x+3.05
Now sum of the weights = m+k = (x+7.55)+(x+3.05) = 2x+10.6
Hence b = 10.6

Alternative Solution:

The initial average is x,

The weight of 1st students can be calculated by subtracting the sum of initial weights from the final weights = 151(x+0.05) - 150x = x+7.55
Similarly, weight of 2nd students = 151{x+0.05} - 150{x+0.03} = x+3.05
Weight of 1st student+ weight of 2nd student = x+7.55+x+3.05 = 2x + 10.6 = 2x + b,
=> b=10.6.

Question 17

Two alloys are such that the ratio of iron to copper in the first alloy is 7:13 and that in the second alloy is 5:12. In what ratio should the two alloys be mixed such that resulting alloy has copper and iron in the ratio 7:3.

Solution

Let the ratio of alloy A : alloy B = 1:k.
Copper to Iron in the final alloy = [(7/20)*1+(5/17)*k]/[(13/20)*1+(12/17)*k] = 3:7

Solving this, we get k = 17/2 => the required ratio = 2:17.

Question 18

A, B and C are 3 varieties of beer which contains starch, hops and water in varying proportions. A contains x% starch and y% hops. B contains y% starch and y% water. C contains x% hops and y% water. When A, B and C are mixed in the ratio 2:3:5, then the final mixture has 38% starch and 28% hops. Find the sum of x and y?

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Solution

Given
Each of three varieties contain starch, hops and water in varying proportions
A contains x% starch and y% hops. So 100-(x+y) % of water.
B contains y% starch and y% water. So 100 - (2y) % of hops.
C contains x% hops and y% water. So 100- (x+y) % of starch

Concentration of starch in final mixture is 38%,then

(2x/10) + (3y/10) + (5(100 -(x+y))/10) = 38

3x + 2y = 120 - (1)

Also, given concentration of hops in final mixture is 28%, then

(2y/10) + (3(100-2y)/10) + (5x/10) = 28

-5x + 4y = 20 - (2)

Solving (1) and (2) we get x = 20 and y = 30

∴ x + y = 50

Question 19

There are 25 students in a class and the number of students who scored 5, 7 and 3 marks in a test are 8, 10 and 7 respectively.

The teacher can decide how many students should go to the remedial class, but can't pick the students for it. Once the number is decided, the students are selected at random. The teacher wants every student in the class to have atleast three others who scored the same marks in the examination. What is the minimum number of students who should be sent to the remedial class for this to happen?

Solution

Assume that the teacher divided the groups in such a way that the conditions given are not satisfied. In such a case let us find the maximum number of students in group B.
In the worst possible scenario we can assume that in group B there are 10 students with 7 marks each, 8 students with 5 marks each and 3 students with 3 marks each. There will be 21 students in group B but the condition given is still not satisfied.
So, if there are 21 students, in group B, there is a possibility that the condition given is not satisfied. But, if there are 22 students in group B, then there will be at least five students with n marks each.
So, the minimum number of students to be in group B such that the criteria is fulfilled is 22.
Hence, option (2) is the correct choice.

Question 20

Three students A, B, C stole some candies. As it was late in the evening, they decided to divide the candies equally among themselves in the morning and went to sleep. During the night, A woke up, ate one candy, then took one third of the remaining candies and went back to sleep. After some time B woke up, ate one candy, then took one third of the remaining candies and went back to sleep. Lastly, C woke up and did the same as other two, that is, ate one candy and then took one third of the remaining candies. When all three woke up in the morning, they found that the total number of candies were in the form 3k+1. Find the minimum number of candies that were present initially.

Solution

Let the initial number of candies be x.

After A woke up, number of candies left = 2/3(x-1)

After B woke up, number of candies left = 2/3(2/3(x-1)-1)

Similarly for C it is = 2/3[2/3(2/3(x-1)-1)-1]

2/3[2/3(2/3(x-1)-1)-1] = 3k+1

[2/3(2/3(x-1)-1)-1] = 9k/2+3/2

2/3(2/3(x-1)-1) = 9k/2+5/2

(2/3(x-1)-1) = 27k/4+15/4

2/3(x-1) = 27k/4+19/4

x-1 = 81k/8 + 57/8

x = 81k/8 + 65/8

To minimize x, we need to find the minimum value of k such that 81k+65 is a multiple of 8.

81k+65 can be expressed as (80k+64) + (k+1). The first part is divisible by 8. For the second part to be divisible by 8, minimum value of k is 7.

For k = 7, x = 79.

Question 21

A rice trader selling basmati rice makes a profit on the trade by adulterating the rice with an inferior substitute surti kolam. When he mixes the basmati and kolam in the ratio 3:2 he makes a profit of 33.33%. When he mixes them in the ratio 4:1 he makes a profit of Rs 25 per kg. In what ratio must he mix the two rices to make a profit of 100%? (Assume he sells at the stated price of basmati per kg)

Solution

Let the price of basmati and kolam be x and y respectively. Assuming that the quantity is 5 kg, the shopkeeper will mix 3 kg of basmati with 2 kg of surti kolam. So, the selling price will be 5x since he sells all quantity as basmati. But the actual cost price is less which is equal to 3kg of basmati and 2kg of surti kolam = 3x+2y. Profit = 5x- (3x+2y) = 2x-2y Hence, profit %= 33.33% = (2x-2y)/(3x+2y) => x=8y/3.

From, mixture 2, profit per kg = Rs 25. Hence profit for 5kg = 25*5 = (5x-4x-y)=x-y. Hence x=200 and y=75.

Let's assume for 100% profit, he is mixing them in the ratio of k:1.

In 100% profit case, the selling price will be double of the cost price.

S.P= 200(k+1) and C.P= (200 X k) + (75 X 1)

=> 200k+200= 2 X [200k + 75]

=> k=1/4

So the ratio is 1:4

Question 22

The price of a iPhone is increased by 15% and then decreased by 10%. The price of an iPod is increased by 10% and then decreased by 15%. If the final prices of both the items are the same, find the ratio of the initial price of iPhone to the initial price of iPod.

Solution

Let the price of iPhone be 100x => the final price is 103.5x.

Let the price of iPod be 100y.

The final price is 93.5y. => 103.5x = 93.5y => x/y = 935/1035 = 187/207.

Question 23

Ram added some water to the solution of juice and water such that the initial ratio a/b of juice to water gets inverted. Shyam sees Ram adding the water and he adds the juice to the solution such that the ratio of juice to water is back to its initial value. What is the possible value of ratio of amount of juice in the original solution to the final solution if both a and b are natural numbers?

Solution

Let the initial volume of juice and water be a ml and b ml respectively.

Initially
Juice = a, Water = b

After first mixing,
Juice = a, Water = x

As the ratio gets inverted,
a/x = b/a
$$x = a^2/b$$
Juice = a , Water =$$a^2/b$$

After second mixing,
Juice = y, Water =$$a^2/b$$
As the ratio comes back to its original value,
y/[$$a^2/b$$] = a/b
y =$$a^3/b^2$$
Juice = $$a^3/b^2$$, Water = $$a^2/b$$

Ratio of juice in the original mixture to that in final mixture = a/($$a^3/b^2$$) = $$b^2/a^2$$
As a and b are natural numbers, the only plausible value is in option d where both numerator and denominator are perfect squares.

Alternate Solution:

Let the initial volume of the solution be $$O_1$$. After Ram mixes water, let the volume of the solution be $$O_2$$ and let the volume of the final solution (after Shyam mixes juice) be $$O_3$$

The amount of juice is the same in $$O$$ and $$O_1$$ as Ram only added water. Hence, $$\frac{a}{a+b} \times O = \frac{b}{a+b} \times O_1$$
This would imply that $$O_1 = \frac{a}{b} \times O$$
Similarly, The amount of water is the same in $$O_1$$ and $$O_2$$ as Shyam only added juice. Hence, $$\frac{a}{a+b} \times O_1 = \frac{b}{a+b} \times O_2$$
This would imply that $$O_2 = \frac{a}{b} \times O_1$$
Therefore, $$O_2 = \frac{a}{b} \times \frac{a}{b} \times O = \frac{a^2}{b^2} \times O$$
Therefore, $$\frac{O}{O_2} = \frac{b^2}{a^2}$$

Hence, the required ratio is the ratio of the squares of two natural numbers. Of the given options, only 4:9 is of that form and hence, that is the answer.

CAT 2023 Averages, Ratio and Proportion Questions with Answers

Averages, Ratio and Proportion

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The average weight of a class of students is 24 kgs. 5 students who together weigh 140 kgs leave the class and 2 students who together weigh 68 kgs join the class. What is the new average of the class?

In what ratio should three salt solutions of concentrations 15%, 20% and 30% be mixed, so that the resultant solution is of concentration 25% ?

The ratio of incomes of Anil and Ajit is 5:7. The ratio of their expenditures is 3:4, then which of the following is a possible ratio of the savings? (Savings = Income - Expenditure)

A shopkeeper buys 3 varieties of rice for Rs 20, Rs 30 and Rs 40 per kg respectively. He now mixes the 3 varieties in the ratio 5:6:7 by weight and sells the mixture at a profit of 10%. What price per kg does he sell the mixture for?

‘n’ years later the sum of ages of Messi and Maradona will be 80. ‘n’ years ago the age of Maradona was 3 times that of Messi. ‘2n’ years ago the age of Maradona was 4 times that of Messi. What will be the difference in their ages after ‘2n’ years from the present?

Two alloys are such that the ratio of iron to copper in the first alloy is 7:13 and that in the second alloy is 5:12. In what ratio should the two alloys be mixed such that resulting alloy has copper and iron in the ratio 7:3.

The price of a iPhone is increased by 15% and then decreased by 10%. The price of an iPod is increased by 10% and then decreased by 15%. If the final prices of both the items are the same, find the ratio of the initial price of iPhone to the initial price of iPod.

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