24 questions, 11 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 bag has 26 coins of Rs. 1, Rs. 2 and Rs. 5, worth Rs. 60 in all. There are 7 more Rs. 2 coins than Rs. 1 coins. How many Rs. 5 coins are there?
Let there be x Rs. 1 coins, x + 7 Rs. 2 coins and z Rs. 5 coins.
Count: 2x + z = 19. Value: 3x + 5z = 46.
From the first, x = (19 − z)/2. Substituting: 3(19 − z)/2 + 5z = 46, so 7z/2 = 46 − 572, z = 5.
Check: x = 7, Rs. 2 coins = 14; 7 + 28 + 25 = 60.
For what value of k does the system kx + 6y = 1, x + 3y = 3 have no solution?
No solution needs the lines to be parallel but different: k/1 = 63 ≠ 13.
k = 1 × 2 = 2, and 13 ≠ 2, so the lines are distinct.
A bag has 33 coins of Rs. 1, Rs. 2 and Rs. 5, worth Rs. 90 in all. There are 4 more Rs. 2 coins than Rs. 1 coins. How many Rs. 5 coins are there?
Let there be x Rs. 1 coins, x + 4 Rs. 2 coins and z Rs. 5 coins.
Count: 2x + z = 29. Value: 3x + 5z = 82.
From the first, x = (29 − z)/2. Substituting: 3(29 − z)/2 + 5z = 82, so 7z/2 = 82 − 872, z = 11.
Check: x = 9, Rs. 2 coins = 13; 9 + 26 + 55 = 90.
For what value of k does the system 12x + 4y = k, 6x + 2y = 7 have infinitely many solutions?
The coefficients are in ratio 126 = 42 = 2.
Infinitely many solutions need the same ratio for the constants too: k = 2 × 7 = 14.
3 pens and 3 pencils cost Rs. 114; 2 pens and 4 pencils cost Rs. 102. What is the cost of one pen?
Let a pen cost x and a pencil cost y: 3x + 3y = 114 and 2x + 4y = 102.
Eliminating y: x = (114 × 4 − 102 × 3) / (3 × 4 − 3 × 2) = 1506 = 25.
Then y = 13.
The digits of a two-digit number add up to 8. Reversing the digits gives a number 54 less than the original. Find the number.
Let the number be 10a + b. Reversing changes it by 9(a − b), so |a − b| = 6.
The reversed number is smaller, so the tens digit is the greater one: a − b = 6.
With a + b = 8: a = 7, b = 1. The number is 71.
4 pens and 3 pencils cost Rs. 124; 2 pens and 6 pencils cost Rs. 116. What is the cost of one pen?
Let a pen cost x and a pencil cost y: 4x + 3y = 124 and 2x + 6y = 116.
Eliminating y: x = (124 × 6 − 116 × 3) / (4 × 6 − 3 × 2) = 39618 = 22.
Then y = 12.
How many positive integer pairs (x, y) satisfy 2x + 3y = 67?
One solution is x = 2, y = 21.
Every other solution comes from adding 3 to x and taking 2 from y.
x = 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32 all keep y positive: 11 pairs.
If x + y = 24, y + z = 26 and z + x = 26, find the value of xyz.
Adding all three: 2(x + y + z) = 76, so x + y + z = 38.
z = 38 − 24 = 14, x = 38 − 26 = 12, y = 38 − 26 = 12.
xyz = 12 × 12 × 14 = 2016.
How many non-negative integer pairs (x, y) satisfy 3x + 4y = 108?
One solution is x = 0, y = 27.
Every other solution comes from adding 4 to x and taking 3 from y.
x = 0, 4, 8, 12, 16, 20, 24, 28, 32, 36 all keep y non-negative: 10 pairs.
The digits of a two-digit number add up to 7. Reversing the digits gives a number 45 more than the original. Find the number.
Let the number be 10a + b. Reversing changes it by 9(a − b), so |a − b| = 5.
The reversed number is larger, so the tens digit is the smaller one: a − b = −5.
With a + b = 7: a = 1, b = 6. The number is 16.
For what value of k does the system 16x + 8y = k, 4x + 2y = 8 have infinitely many solutions?
The coefficients are in ratio 164 = 82 = 4.
Infinitely many solutions need the same ratio for the constants too: k = 4 × 8 = 32.
A bag has 37 coins of Rs. 1, Rs. 2 and Rs. 5, worth Rs. 132 in all. There are 6 more Rs. 2 coins than Rs. 1 coins. How many Rs. 5 coins are there?
Let there be x Rs. 1 coins, x + 6 Rs. 2 coins and z Rs. 5 coins.
Count: 2x + z = 31. Value: 3x + 5z = 120.
From the first, x = (31 − z)/2. Substituting: 3(31 − z)/2 + 5z = 120, so 7z/2 = 120 − 932, z = 21.
Check: x = 5, Rs. 2 coins = 11; 5 + 22 + 105 = 132.
For what value of k does the system kx + 2y = 17, 3x + y = 9 have no solution?
No solution needs the lines to be parallel but different: k/3 = 21 ≠ 179.
k = 3 × 2 = 6, and 179 ≠ 2, so the lines are distinct.
If x + y = 26, y + z = 22 and z + x = 26, find the value of xyz.
Adding all three: 2(x + y + z) = 74, so x + y + z = 37.
z = 37 − 26 = 11, x = 37 − 22 = 15, y = 37 − 26 = 11.
xyz = 15 × 11 × 11 = 1815.
How many positive integer pairs (x, y) satisfy 7x + 3y = 49?
One solution is x = 1, y = 14.
Every other solution comes from adding 3 to x and taking 7 from y.
x = 1, 4 all keep y positive: 2 pairs.
How many positive integer pairs (x, y) satisfy 2x + 7y = 94?
One solution is x = 5, y = 12.
Every other solution comes from adding 7 to x and taking 2 from y.
x = 5, 12, 19, 26, 33, 40 all keep y positive: 6 pairs.
A bag has 35 coins of Rs. 1, Rs. 2 and Rs. 5, worth Rs. 85 in all. There are 2 more Rs. 2 coins than Rs. 1 coins. How many Rs. 5 coins are there?
Let there be x Rs. 1 coins, x + 2 Rs. 2 coins and z Rs. 5 coins.
Count: 2x + z = 33. Value: 3x + 5z = 81.
From the first, x = (33 − z)/2. Substituting: 3(33 − z)/2 + 5z = 81, so 7z/2 = 81 − 992, z = 9.
Check: x = 12, Rs. 2 coins = 14; 12 + 28 + 45 = 85.
The digits of a two-digit number add up to 6. Reversing the digits gives a number 36 less than the original. Find the number.
Let the number be 10a + b. Reversing changes it by 9(a − b), so |a − b| = 4.
The reversed number is smaller, so the tens digit is the greater one: a − b = 4.
With a + b = 6: a = 5, b = 1. The number is 51.
If x + y = 12, y + z = 7 and z + x = 11, find the value of xyz.
Adding all three: 2(x + y + z) = 30, so x + y + z = 15.
z = 15 − 12 = 3, x = 15 − 7 = 8, y = 15 − 11 = 4.
xyz = 8 × 4 × 3 = 96.
If x + y = 24, y + z = 23 and z + x = 19, find the value of xyz.
Adding all three: 2(x + y + z) = 66, so x + y + z = 33.
z = 33 − 24 = 9, x = 33 − 23 = 10, y = 33 − 19 = 14.
xyz = 10 × 14 × 9 = 1260.
6 books and 2 notebooks cost Rs. 64; 3 books and 5 notebooks cost Rs. 64. What is the cost of one book and one notebook together?
Let a book cost x and a notebook cost y: 6x + 2y = 64 and 3x + 5y = 64.
Eliminating y: x = (64 × 5 − 64 × 2) / (6 × 5 − 2 × 3) = 19224 = 8.
Then y = 8. One of each costs Rs. 16.
6 books and 4 notebooks cost Rs. 196; 5 books and 2 notebooks cost Rs. 150. What is the cost of one book and one notebook together?
Let a book cost x and a notebook cost y: 6x + 4y = 196 and 5x + 2y = 150.
Eliminating y: x = (196 × 2 − 150 × 4) / (6 × 2 − 4 × 5) = −208/-8 = 26.
Then y = 10. One of each costs Rs. 36.
The digits of a two-digit number add up to 13. Reversing the digits gives a number 27 less than the original. Find the number.
Let the number be 10a + b. Reversing changes it by 9(a − b), so |a − b| = 3.
The reversed number is smaller, so the tens digit is the greater one: a − b = 3.
With a + b = 13: a = 8, b = 5. The number is 85.
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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