24 questions, 2 of them short answer (type the number, as in IPMAT Indore). Each is marked with the exam level it suits. Answer, then open the step-by-step solution.
A starts a business with Rs. 6000. After 9 months B joins with Rs. 9000. Out of a year-end profit of Rs. 22000, what is A's share?
Capital × time: A = 6000 × 12 = 72000; B = 9000 × 3 = 27000.
A's share = 22000 × 7200099000 = Rs. 16000.
The average of 12 consecutive odd numbers is 68. What is the largest of them?
For an evenly spaced list the average is the middle value; the numbers run from 57 to 79.
Largest = 68 + 2 × 112 = 79.
The average weight of 9 people decreases by 1 kg when one of them, weighing 63 kg, is replaced by a new person. Find the new person's weight.
The total falls by 9 × 1 = 9 kg.
New person = 63 − 9 = 54 kg.
The average weight of 10 people decreases by 1.5 kg when one of them, weighing 66 kg, is replaced by a new person. Find the new person's weight.
The total falls by 10 × 1.5 = 15 kg.
New person = 66 − 15 = 51 kg.
The average weight of 6 people increases by 3 kg when one of them, weighing 52 kg, is replaced by a new person. Find the new person's weight.
The total rises by 6 × 3 = 18 kg.
New person = 52 + 18 = 70 kg.
A container holds 40 litres of milk. 8 litres are taken out and replaced with water; this is done 2 times in all. How much milk is left?
Each time, the milk is multiplied by (40 − 8)/40 = 45.
Milk left = 40 × (45)2 = 25.6 litres.
A container holds 60 litres of milk. 20 litres are taken out and replaced with water; this is done 3 times in all. How much milk is left?
Each time, the milk is multiplied by (60 − 20)/60 = 23.
Milk left = 60 × (23)3 = 17 79 litres.
Two numbers are in the ratio 8 : 7. If 12 is added to each, the ratio becomes 52 : 47. Find the larger number.
Let the numbers be 8k and 7k: (8k + 12)/(7k + 12) = 5247.
Solving: k = 5.
The numbers are 40 and 35; the larger is 40.
The average weight of 5 people increases by 3 kg when one of them, weighing 63 kg, is replaced by a new person. Find the new person's weight.
The total rises by 5 × 3 = 15 kg.
New person = 63 + 15 = 78 kg.
In what ratio must rice at Rs. 22/kg be mixed with rice at Rs. 101/kg so that the mixture costs Rs. 45/kg?
By alligation: cheaper : dearer = (dearer − mean) : (mean − cheaper).
= (101 − 45) : (45 − 22) = 56 : 23 = 56 : 23.
The average of 15 consecutive even numbers is 28. What is the largest of them?
For an evenly spaced list the average is the middle value; the numbers run from 14 to 42.
Largest = 28 + 2 × 7 = 42.
In what ratio must rice at Rs. 24/kg be mixed with rice at Rs. 93/kg so that the mixture costs Rs. 79/kg?
By alligation: cheaper : dearer = (dearer − mean) : (mean − cheaper).
= (93 − 79) : (79 − 24) = 14 : 55 = 14 : 55.
Two numbers are in the ratio 6 : 5. If 3 is added to each, the ratio becomes 25 : 21. Find the larger number.
Let the numbers be 6k and 5k: (6k + 3)/(5k + 3) = 2521.
Solving: k = 12.
The numbers are 72 and 60; the larger is 72.
A container holds 60 litres of milk. 4 litres are taken out and replaced with water; this is done 2 times in all. How much milk is left?
Each time, the milk is multiplied by (60 − 4)/60 = 1415.
Milk left = 60 × (1415)2 = 78415 litres.
In what ratio must rice at Rs. 59/kg be mixed with rice at Rs. 74/kg so that the mixture costs Rs. 64/kg?
By alligation: cheaper : dearer = (dearer − mean) : (mean − cheaper).
= (74 − 64) : (64 − 59) = 10 : 5 = 2 : 1.
A starts a business with Rs. 5000. After 2 months B joins with Rs. 7000. Out of a year-end profit of Rs. 26000, what is A's share?
Capital × time: A = 5000 × 12 = 60000; B = 7000 × 10 = 70000.
A's share = 26000 × 60000130000 = Rs. 12000.
Two numbers are in the ratio 3 : 2. If 15 is added to each, the ratio becomes 48 : 37. Find the larger number.
Let the numbers be 3k and 2k: (3k + 15)/(2k + 15) = 4837.
Solving: k = 11.
The numbers are 33 and 22; the larger is 33.
In what ratio must rice at Rs. 28/kg be mixed with rice at Rs. 103/kg so that the mixture costs Rs. 60/kg?
By alligation: cheaper : dearer = (dearer − mean) : (mean − cheaper).
= (103 − 60) : (60 − 28) = 43 : 32 = 43 : 32.
A container holds 50 litres of milk. 10 litres are taken out and replaced with water; this is done 3 times in all. How much milk is left?
Each time, the milk is multiplied by (50 − 10)/50 = 45.
Milk left = 50 × (45)3 = 25.6 litres.
The average of 7 consecutive numbers is 48. What is the largest of them?
For an evenly spaced list the average is the middle value; the numbers run from 45 to 51.
Largest = 48 + 1 × 3 = 51.
A starts a business with Rs. 6000. After 8 months B joins with Rs. 2000. Out of a year-end profit of Rs. 32000, what is A's share?
Capital × time: A = 6000 × 12 = 72000; B = 2000 × 4 = 8000.
A's share = 32000 × 7200080000 = Rs. 28800.
A starts a business with Rs. 7000. After 8 months B joins with Rs. 9000. Out of a year-end profit of Rs. 5000, what is A's share?
Capital × time: A = 7000 × 12 = 84000; B = 9000 × 4 = 36000.
A's share = 5000 × 84000120000 = Rs. 3500.
The average of 14 consecutive numbers is 33.5. What is the largest of them?
For an evenly spaced list the average is the middle value; the numbers run from 27 to 40.
Largest = 33.5 + 1 × 132 = 40.
Two numbers are in the ratio 4 : 9. If 10 is added to each, the ratio becomes 29 : 59. Find the larger number.
Let the numbers be 4k and 9k: (4k + 10)/(9k + 10) = 2959.
Solving: k = 12.
The numbers are 48 and 108; the larger is 108.
These are original practice questions written by Target Campus in the pattern, syllabus and difficulty of IPMAT Indore, IPMAT Rohtak, JIPMAT, NPAT and SET. They are not previous-year papers.
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