Demo 3
Algebra Quiz
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1. If a - b = 3 and a³ - b³ = 999 then find the value of a² - b².
Detailed Solution
Correct Answer: (d) 63
Given: a - b = 3 and a³ - b³ = 999.
Formula: We use the identity a³ - b³ = (a - b)((a - b)² + 3ab).
Substitute the given values into the formula:
⇒ 999 = 3 × ((3)² + 3ab)
⇒ 999 = 3 × (9 + 3ab)
Divide by 3: ⇒ 333 = 9 + 3ab
⇒ 3ab = 333 - 9 = 324
⇒ ab = 108
Next, we find (a + b) using the identity (a + b)² = (a - b)² + 4ab.
⇒ (a + b)² = (3)² + 4(108)
⇒ (a + b)² = 9 + 432 = 441
⇒ a + b = √441 = 21
Finally, we calculate a² - b², which is (a - b)(a + b).
⇒ a² - b² = 3 × 21 = 63
2. If x, y, z are three integers such that x+y=8, y+z=13 and z+x=17 then the value of x²/yz is:
Detailed Solution
Correct Answer: (b) 18/11
Given equations:
x + y = 8 .....(i)
y + z = 13 .....(ii)
z + x = 17 .....(iii)
Add the three equations together:
⇒ (x + y) + (y + z) + (z + x) = 8 + 13 + 17
⇒ 2x + 2y + 2z = 38
⇒ 2(x + y + z) = 38
⇒ x + y + z = 19 .....(iv)
To find x, subtract equation (ii) from (iv): (x + y + z) - (y + z) = 19 - 13 ⇒ x = 6.
To find y, subtract equation (iii) from (iv): (x + y + z) - (z + x) = 19 - 17 ⇒ y = 2.
To find z, subtract equation (i) from (iv): (x + y + z) - (x + y) = 19 - 8 ⇒ z = 11.
Now, calculate the value of the expression x²/yz:
⇒ (6)² / (2 × 11) = 36 / 22
Simplify the fraction: ⇒ 18/11
3. If 3x+6y+9z=20/3, 6x+9y+3z=17/3 and 18x+27y-z=113/9, then what is the value of 75x+113y?
Detailed Solution
Correct Answer: (a) 163/3
This problem requires solving a system of three linear equations.
Given equations:
1. 3x + 6y + 9z = 20/3 .....(i)
2. 6x + 9y + 3z = 17/3 .....(ii)
3. 18x + 27y - z = 113/9 .....(iii)
Step 1: Eliminate z. Multiply equation (iii) by 3 and add it to equation (ii).
⇒ (54x + 81y - 3z) + (6x + 9y + 3z) = 3(113/9) + 17/3
⇒ 60x + 90y = 113/3 + 17/3 = 130/3
Divide by 10: ⇒ 6x + 9y = 13/3 .....(iv)
Step 2: Eliminate z again. Multiply equation (ii) by -3 and add it to equation (i).
⇒ (3x + 6y + 9z) + (-18x - 27y - 9z) = 20/3 - 3(17/3)
⇒ -15x - 21y = 20/3 - 17 = -31/3
⇒ 15x + 21y = 31/3 .....(v)
Step 3: Solve for x and y. Multiply (iv) by 5 and (v) by -2, then add them.
(30x + 45y = 65/3) + (-30x - 42y = -62/3)
⇒ 3y = 3/3 = 1 ⇒ y = 1/3
Substitute y=1/3 into (iv): 6x + 9(1/3) = 13/3 ⇒ 6x + 3 = 13/3 ⇒ 6x = 4/3 ⇒ x = 2/9
Step 4: Calculate the final expression.
⇒ 75x + 113y = 75(2/9) + 113(1/3)
⇒ 50/3 + 113/3 = 163/3
4. If 3x+4y-2z+9=17, 7x+2y+11z+8=23 and 5x+9y+6z-4=18, then what is the value of x+y+z-34?
Detailed Solution
Correct Answer: (c) -31
First, simplify the given equations by moving constants to the right side:
1. 3x + 4y - 2z = 17 - 9 ⇒ 3x + 4y - 2z = 8 .....(i)
2. 7x + 2y + 11z = 23 - 8 ⇒ 7x + 2y + 11z = 15 .....(ii)
3. 5x + 9y + 6z = 18 + 4 ⇒ 5x + 9y + 6z = 22 .....(iii)
The key to this problem is to add all three simplified equations together:
⇒ (3x+7x+5x) + (4y+2y+9y) + (-2z+11z+6z) = 8 + 15 + 22
⇒ 15x + 15y + 15z = 45
Divide the entire equation by 15:
⇒ x + y + z = 3
Now, calculate the value of the required expression: x + y + z - 34
⇒ 3 - 34 = -31
5. If 3x+4y-11=18 and 8x-6y+12=6, then what is the value of 5x-3y-9?
Detailed Solution
Correct Answer: (b) -9
First, simplify the given linear equations:
1. 3x + 4y = 18 + 11 ⇒ 3x + 4y = 29 .....(i)
2. 8x - 6y = 6 - 12 ⇒ 8x - 6y = -6
Divide the second equation by 2 to simplify: ⇒ 4x - 3y = -3 .....(ii)
Now, solve the system of equations. Multiply equation (i) by 3 and equation (ii) by 4 to eliminate y:
⇒ 3(3x + 4y = 29) ⇒ 9x + 12y = 87
⇒ 4(4x - 3y = -3) ⇒ 16x - 12y = -12
Add the two new equations:
⇒ (9x + 16x) + (12y - 12y) = 87 - 12
⇒ 25x = 75 ⇒ x = 3
Substitute x = 3 into equation (i): 3(3) + 4y = 29 ⇒ 9 + 4y = 29 ⇒ 4y = 20 ⇒ y = 5
Finally, calculate the value of the expression 5x - 3y - 9:
⇒ 5(3) - 3(5) - 9 = 15 - 15 - 9 = -9
6. If x+3y-2z/4=6, x+2/3(2y+3z)=33 and 1/7(x+y+z)+2z=9, then what is the value of 46x+131y?
Detailed Solution
Correct Answer: (a) 414
This problem has a trick and does not require solving for x, y, and z individually. First, clear the fractions and simplify each equation:
1. x + 3y - 2z/4 = 6 ⇒ 4x + 12y - 2z = 24 .....(i)
2. x + 2/3(2y + 3z) = 33 ⇒ 3x + 4y + 6z = 99 .....(ii)
3. 1/7(x + y + z) + 2z = 9 ⇒ x + y + z + 14z = 63 ⇒ x + y + 15z = 63 .....(iii)
The goal is to combine these equations to get 46x + 131y. This requires a specific linear combination. Through observation (or trial and error), we find that we need to multiply equation (i) by 21/2 (or 10.5) and then add all three equations.
Multiply (i) by 21/2: (21/2) * (4x + 12y - 2z = 24) ⇒ 42x + 126y - 21z = 252
Now, add this result to (ii) and (iii):
⇒ (42x + 126y - 21z) + (3x + 4y + 6z) + (x + y + 15z) = 252 + 99 + 63
Combine the x terms: 42x + 3x + x = 46x
Combine the y terms: 126y + 4y + y = 131y
Combine the z terms: -21z + 6z + 15z = 0
Combine the constants: 252 + 99 + 63 = 414
The resulting equation is: 46x + 131y = 414
7. If a+b+c=7/12, 3a-4b+5c=3/4 and 7a-11b-13c=-7/12 then what is the value of a+c?
Detailed Solution
Correct Answer: (b) 5/12
Given equations:
1. a + b + c = 7/12 .....(i)
2. 3a - 4b + 5c = 3/4 .....(ii)
3. 7a - 11b - 13c = -7/12 .....(iii)
Step 1: Eliminate b. Multiply (i) by 4 and add to (ii).
⇒ (4a + 4b + 4c) + (3a - 4b + 5c) = 4(7/12) + 3/4
⇒ 7a + 9c = 7/3 + 3/4 = 28/12 + 9/12 = 37/12 .....(iv)
Step 2: Eliminate b again. Multiply (i) by 11 and add to (iii).
⇒ (11a + 11b + 11c) + (7a - 11b - 13c) = 11(7/12) - 7/12
⇒ 18a - 2c = 77/12 - 7/12 = 70/12 = 35/6
Divide by 2: 9a - c = 35/12 .....(v)
Step 3: Solve for a and c. From (v), c = 9a - 35/12. Substitute this into (iv).
⇒ 7a + 9(9a - 35/12) = 37/12
⇒ 7a + 81a - 315/12 = 37/12
⇒ 88a = (315 + 37)/12 = 352/12
⇒ a = (352/12) / 88 = 4/12 = 1/3
Substitute a=1/3 into (v): 9(1/3) - c = 35/12 ⇒ 3 - c = 35/12 ⇒ c = 3 - 35/12 = 36/12 - 35/12 = 1/12
Step 4: Calculate a+c.
⇒ a + c = 1/3 + 1/12 = 4/12 + 1/12 = 5/12
8. If x-4y=0 and x+2y=24, then what is the value of (2x+3y)/(2x-3y)?
Detailed Solution
Correct Answer: (b) 11/5
Given equations:
1. x - 4y = 0 .....(i)
2. x + 2y = 24 .....(ii)
From equation (i), we can express x in terms of y: x = 4y.
Substitute this into equation (ii):
⇒ (4y) + 2y = 24
⇒ 6y = 24 ⇒ y = 4
Now find the value of x:
⇒ x = 4y = 4(4) = 16
Now, substitute the values of x and y into the expression (2x+3y)/(2x-3y):
⇒ (2(16) + 3(4)) / (2(16) - 3(4))
⇒ (32 + 12) / (32 - 12)
⇒ 44 / 20
Simplify the fraction: ⇒ 11/5
9. If 3x+5y+7z=49 and 9x+8y+21z=126, then what is the value of y?
Detailed Solution
Correct Answer: (c) 3
Given equations:
1. 3x + 5y + 7z = 49 .....(i)
2. 9x + 8y + 21z = 126 .....(ii)
The key here is to notice that the coefficients of x and z in equation (ii) are exactly 3 times the coefficients in equation (i). We can use this to eliminate both x and z in one step.
Multiply equation (i) by 3:
⇒ 3(3x + 5y + 7z = 49) ⇒ 9x + 15y + 21z = 147 .....(iii)
Now, subtract equation (ii) from our new equation (iii):
⇒ (9x + 15y + 21z) - (9x + 8y + 21z) = 147 - 126
The x and z terms cancel out:
⇒ 15y - 8y = 21
⇒ 7y = 21
⇒ y = 3
10. A man buys 2 apples and 3 kiwi fruits for ₹37. If he buys 4 apples and 5 kiwi fruits for ₹67 then what will be the total cost of 1 apple and 1 kiwi fruit?
Detailed Solution
Correct Answer: (c) ₹15
Let the cost of one apple be 'x' and the cost of one kiwi be 'y'.
From the problem statement, we can form two linear equations:
1. 2x + 3y = 37 .....(i)
2. 4x + 5y = 67 .....(ii)
To solve this system, multiply equation (i) by 2 to make the coefficient of x the same as in equation (ii):
⇒ 2(2x + 3y = 37) ⇒ 4x + 6y = 74 .....(iii)
Now, subtract equation (ii) from equation (iii):
⇒ (4x + 6y) - (4x + 5y) = 74 - 67
⇒ y = 7. The cost of one kiwi is ₹7.
Substitute y = 7 back into equation (i):
⇒ 2x + 3(7) = 37
⇒ 2x + 21 = 37
⇒ 2x = 16 ⇒ x = 8. The cost of one apple is ₹8.
The total cost of 1 apple and 1 kiwi is x + y.
⇒ 8 + 7 = ₹15
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