Actual IPMAT - Indore questions on this topic from past papers, newest first. Each links back to its paper.
Suppose a, b and c are three real numbers such that Max(a, b, c) + Min(a, b, c) = 15, and Median(a, b, c) − (a, b, c) − Mean(a, b, c) = 2. Then the median of a, b and c is
Answer: B
A written solution for this question isn't available yet.
Let 50 distinct positive integers be chosen such that the highest among them is 100, and the average of the largest 25 integers among them exceeds the average of the remaining integers by 50 . Then the maximum possible value of the sum of all the 50 integers is _________.
Answer: 3150
To calculate the maximum sum of 50 integers, we can assume the largest 25 integers as 100, 99, 98, ..., 76.
Average of these 25 integers = (100 + 76)/2 = 88
The average of remaining 25 integers must be 88 - 50 = 38
∴ Maximum sum of all 50 integers = 25 × 88 + 25 × 38 = 25 × 126 = 3150
Hence, 3150.
In a room, there are n persons whose average height is 160 cm. If m more persons, whose average height is 172 cm, enter the room, then the average height of all persons in the room becomes 164 cm. Then m : n is
Answer: A
Average height of n persons initially = 160
∴ Sum of the heights of these n persons = 160n
Sum of the heighs of m persons joining = 172m
Total height of (n + m) persons = 160n + 172m
⇒ Average = 164 =
⇒ 164n + 164m = 160n + 172m
⇒ 4n = 8m
⇒ m : n = 1 : 2
Hence, option (a).
In a bowl containing 60 ml orange juice, 40 ml of water is poured. Thereafter, 100 ml of apple juice is poured to make a fruit punch. Madhu drinks 50 ml of this fruit punch and comments that the proportion of orange juice needs to be higher for better taste. How much orange juice should be poured into the fruit punch that remained, in order to bring up the level of orange juice to 50 percentage?
Answer: D
Fruit Punch contains 60 ml orange juice, 40 ml of water and 100 ml of apple juice i.e., a total of 200 ml.
Madhu drinks 50ml i.e., 1/4th of the this mixture, hence 3/4th of the mixture is left..
Total solutio left = 150 ml
Orange left = 60 × 3/4 = 45 ml
Water left = 40 × 3/4 = 30 ml
Apple left = 100 × 3/4 = 75 ml
Now let x ml of orange is added to bring orange to 50% i.e., 1/2.
∴ 45 + x = 1/2 × (150 + x)
⇒ 90 + 2x = 150 + x
⇒ x = 60 ml
Hence, option (d).
A class consists of 30 students. Each of them has registered for 5 courses. Each course instructor conducts an exam out of 200 marks. The average percentage marks of all 30 students across all courses they have registered for, is 80%. Two of them apply for revaluation in a course. If none of their marks reduce, and the average of all 30 students across all courses becomes 80.02%, the maximum possible increase in marks for either of the 2 students is
Answer: 6
Initial sum of the marks of all students across all subjects = 30 × 80% × 200 × 5 = 24,000
Sum of the marks of all students across all subjects after revaluation = 30 × 80.02% × 200 × 5 = 24,006
∴ Marks of these students increased by 6.
Least increase in marks of one of the students can be 0, hence maximum increase in marks of the other student must be 6.
Hence, 6.
Fifty litres of a mixture of milk and water contains 30 percent of water. This mixture is added to eighty litres of another mixture of milk and water that contains 20 percent of water. Then, how many litres of water should be added to the resulting mixture to obtain a final mixture that contains 25 percent of water?
Answer: B
Fifty litres of a mixture of milk and water contains 30 percent of water.
⇒ Quantity of milk = 70% of 50 = 35 liters
⇒ Quantity of water = 30% of 50 = 15 liters
Eighty litres of another mixture of milk and water that contains 20 percent of water
⇒ Quantity of milk = 80% of 80 = 64 liters
⇒ Quantity of water = 20% of 80 = 16 liters
Let x liters of water is added along with these two solutions to make water 25% of the whole mixture.
⇒ 15 + 16 + x = 25% of (50 + 80 + x)
⇒ 31 + x = 25% of (130 + x)
⇒ 31 + x = 32.5 + 0.25x
⇒ 0.75x = 1.5
⇒ x = 2 liters.
Hence, option (b).
The average marks of 6 students in a test is 64 . All the students got different marks, one of the students obtained 70 marks and all other students scored 40 or above. The maximum possible difference between the second highest and the second lowest marks is
Answer: B
Total marks of 6 students = 6 × 64 = 384
Marks of all the students are distinct and greater than 40.
We need to maximise the difference of 2nd highest and 2nd lowest marks
For this we need to minimise the 2nd lowest marks and maximise the 2nd highest marks.
Now, least possible marks of bottom 2 students = 40 and 41.
Marks of the students is 70.
Let the marks of remaining three students be A, B and C
∴ 40 + 41 + 70 + A + B + C = 384
⇒ A + B + C = 233.
Now, if C is higest and B is second highest, we can maximise B by minimising A.
Least possible marks of A now is 42.
⇒ B + C = 191
For B to be maximum possible, B and C should be as close as possible.
∴ C = B + 1
⇒ B + B + 1 = 191
⇒ B = 95
∴ The 6 numbers are: 40, 41, 42, 70, 95, 96
⇒ Differnce between 2nd highest and 2nd lowest marks = 95 - 41 = 54.
Hence, option (b).
The average of five distinct integers is 110 and the smallest number among them is 100. The maximum possible value of the largest integer is
Answer: 144
A written solution for this question isn't available yet.
In a given village there are only three sizes of families: families with 2 members, families with 4 members and families with 6 members. The proportion of families with 2,4 and 6 members are roughly equal. A poll is conducted in this village wherein a person is chosen at random and asked about his/her family size. The average family size computed by sampling 1000 such persons from the village would be closest to
Answer: B
A written solution for this question isn't available yet.
An alloy P has copper and zinc in the proportion of 5: 2 (by weight), while another alloy Q has the same metals in the proportion of 3: 4 (by weight). If these two alloys are mixed in the proportion of a : b (by weight), a new alloy R is formed, which has equal contents of copper and zinc. Then, the proportion of copper and zinc in the alloy S, formed by mixing the two alloys P and Q in the proportion of b : a (by weight) is
Answer: C
A written solution for this question isn't available yet.