Read the following and answer the questions that follow.
If 5 people are transferred from E to R and another independent set of 5 people are transferred back from R to E, then after this operation (Assume that the set transferred from R to E contains none from the set of students that came to R from E)
What will happen to R’s average?
Will always increase since the net value transferred from R to E will be higher than the net value transferred from E to R.
Read the following and answer the questions that follow.
If 5 people are transferred from E to R and another independent set of 5 people are transferred back from R to E, then after this operation (Assume that the set transferred from R to E contains none from the set of students that came to R from E)
What can be said about E’s average?
Since the lowest score in Class R is 23 which is more than the highest score of any student in Class E. Hence, E’s average will always increase.
Read the following and answer the questions that follow.
If 5 people are transferred from E to R and another independent set of 5 people are transferred back from R to E, then after this operation (Assume that the set transferred from R to E contains none from the set of students that came to R from E)
At the end of the 2 steps mentioned above (in the direction) what could be the maximum value of the average of class R?
The maximum possible value for R will happen when the E to R transfer has the maximum possible value and the reverse transfer has the minimum possible value.
Read the following and answer the questions that follow.
If 5 people are transferred from E to R and another independent set of 5 people are transferred back from R to E, then after this operation (Assume that the set transferred from R to E contains none from the set of students that came to R from E)
For question 3, what could be the minimum value of the average of class R?
For the minimum possible value of R we will need the E to R transfer to be the lowest possible value while the R to E transfer must have the highest possible value.
Thus, E to R transfer > 18* 5 while R to E transfer will be 31*5. Hence the answer is 22.4.
Read the following and answer the questions that follow.
If 5 people are transferred from E to R and another independent set of 5 people are transferred back from R to E, then after this operation (Assume that the set transferred from R to E contains none from the set of students that came to R from E)
What could be the maximum possible average achieved by class E at the end of the operation?
The maximum value for E will happen in the case of Q4. Then the increment for group E is:
31* 5 – 18 * 5 = 5 * (31 – 18) = 65.
Thus the maximum possible value is 465 / 20 = 23.25.
Read the following and answer the questions that follow.
If 5 people are transferred from E to R and another independent set of 5 people are transferred back from R to E, then after this operation (Assume that the set transferred from R to E contains none from the set of students that came to R from E)
What could be the minimum possible average of class E at the end of the operation?
The minimum possible average will happen for the transfer we saw in Q. 3. Thus the answer will be 405 / 20 = 20.25.
The average of 7 consecutive numbers is E. If the next three numbers are also added, the average increases by
(IIFT 2013)
If the numbers are 1,2,3,4,5,6,7 and we add 8, 9,10.
Initial average = (1 + 2 + 3 + 4 + 5 + 6 + 7) / 7 = 4 Final average = (1 + 2 + 3 + 4 + 10) / 10 = 5.5
Therefore, the average increases by 1.5.
In 2001 there were 6 members in Barney’s family and their average age was 28 years. He got married between 2001 and 2004 and in 2004 there was an addition of a child in his family. In 2006, the average age of his family was 32 years. What is the present age (in 2006) of Barney’s wife (in years) is:
If the present age of Barney’s wife is x years. Then according to the question:
(33 X 6 + x + 2) / 8 = 32
x + 2 + 198 = 256 x + 2 = 58
x = 56
x = 56 years
The weight of a metal piece as calculated by the average of 7 different experiments is 53.735 gm. The average of the first three experiments is 54.005 gm, of the fourth is 0.004 gm greater than the fifth, while the average of the sixth and seventh experiment was 0.010 gm less than the average of the first three. Find the weight of the body obtained by the fourth experiment.
You can take 53 as the base to reduce your calculations. Otherwise, the question will become highly calculationintensive. Let the fifth experiment’s measurement be ‘x’ above 53.
Then you get:
0.735 X 7 = 1.005 X 3 + (x + 0.004) + x + 0.995 X 2
> 5.145 = 3.015 + 2x + 0.004 + 1.99.
On solving this you get x = 0.068. Hence, the weight of the fifth body is 53.068 and the weight of the fourth body is 53.072. Hence, the option (d) is correct.
One collective farm got an average harvest of 21 tons of wheat and another collective farm that had 12 acres of land less given to wheat, got 25 tons from a hectare. As a result, the second farm harvested 300 tons of wheat more than the first. How many tons of wheat did each farm harvest?
If the dates of birth, of four of them are prime numbers, then find the maximum average of the sum of their dates of birth.
The prime dates must be 29th, 23rd, 19th and 5th.
This represents a reduction in the totals from 30 + 25 + 20 + 5 to 29 + 23 + 19 + 5 – a drop of 4.
Hence, the maximum possible average will reduce by 4/5 = 0.8. Hence, the answer will be 27.2.
The average age of a group of persons going for a movie is 20 years. 10 new persons with an average age of 10 years join the group on the spot due to which the average of the group becomes 18 years. Find the number of persons initially going for the movie.
Read the following and answer the questions that follow.
If 5 people are transferred from C to B, further, 5 more people are transferred from B to A, then 5 are transferred from A to B and finally, 5 more are transferred from B to C.
What is the maximum possible average achieved by class C?
The maximum possible value for C will be achieved when the transfer from C is of five 26’s and the transfer back from B is of five 31’s.
Hence, difference is totals will be +25. Hence, max. average = (900 + 25)/30 = 30.833.
An alloy is prepared by mixing three metals A, B and C in the proportion 3: 4: 7 by volume. Weights of the same volume of the metals A, B and C are in the ratio 5: 2: 6. In 130 kg of the alloy, the weight, in kg, of the metal C is
Required weight of C=(7×63×5+4×2+7×6)×130=84C=(7×63×5+4×2+7×6)×130=84 kg
Read the following and answer the questions that follow.
If 5 people are transferred from C to B, further, 5 more people are transferred from B to A, then 5 are transferred from A to B and finally, 5 more are transferred from B to C.
What is the maximum possible average value attained by class A?
A will attain maximum value if five 33’s come to A from C through B and five 18’s leave A. In such a case the net result is going to be a change of +75. Thus the average will go up by 75/20 = 3.75 to 23.75.
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