This EduRev document offers 20 Multiple Choice Questions (MCQs) from the topic Quadratic Equations & Linear Equations (Level - 2). These questions are of Level - 2 difficulty and will assist you in the preparation of CAT & other MBA exams. You can practice/attempt these CAT Multiple Choice Questions (MCQs) and check the explanations for a better understanding of the topic.
Question for Practice Questions Level 2: Quadratic Equations & Linear Equations - 2
Try yourself:
then which of the following is true?
Explanation
⇒ (xy - x - y + 1)(z - 1) = (xy + x + y + 1)(z + 1)
xyz - xy - xz + x - yz + y + z - 1 = xyz + xy + xz + x + yz + y + z + 1
⇒ -2(xy + yz + zx ) = 2
∴ xy + yz + zx = -1
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Question for Practice Questions Level 2: Quadratic Equations & Linear Equations - 2
Try yourself:The sum of ages of a group of teachers, doctors and lawyers is 2160 years and their average age is 36 years. If each teacher had been older by 1 year, each doctor by 6 years and each lawyer by 7 years, their average age would have increased by 5 years. Find the minimum possible number of doctors in the group.
Explanation
If there are t teachers, d doctors and l lawyers, then
t + d + l = 2160/36 = 60 … (1)
If the average ages of teachers, doctors and lawyers respectively are x, y and z years, then
xt + yd + lz = 2160 ... (2)
Given: (x + 1)t + (y + 6)d + (z + 7) l = 41 60 = 2460
∴ t + 6d + 7l = 300, use the equation 2
i.e. 5d + 6l = 240, use the equation 1
Since d and l are always positive integer values, minimum value of d is possible when l is maximum.
l(maximum) = 35
Hence, from equation (3),
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Question for Practice Questions Level 2: Quadratic Equations & Linear Equations - 2
Try yourself:What is the value of the expression when a = 333, b = 444 and c = 555?
Explanation
Let a - b = x,
b - c = y,
c - a = z.
Then, x + y + z = 0 and the expression becomes
Now, if x + y + z = 0, then x3 + y3 + z3 = 3xyz.
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Question for Practice Questions Level 2: Quadratic Equations & Linear Equations - 2
Try yourself:The number of real solutions of |x|2 + 3 |x| + 2 = 0 is
Explanation
|x|2 is always positive. 3|x| is also positive.
As L.H.S. is always greater than zero, no real solution can exist.
Thus, total number of real solutions = 0
Thus, answer option 1 is correct.
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Question for Practice Questions Level 2: Quadratic Equations & Linear Equations - 2
Try yourself:The condition that the roots of the equation lx2 + mx + n = 0 are in the ratio 3 : 4, is
Explanation
lx2 + mx+ n = 0
Let the roots be α and β.
12m2 = 49nl
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Question for Practice Questions Level 2: Quadratic Equations & Linear Equations - 2
Try yourself:How many possible integral values are there for m if it is known that m2 + 12m + 20 ≤ 0?
Explanation
m2 + 12m + 20 ≤ 0
(m + 2)(m + 10) ≤ 0
Hence, -10 ≤ m ≤ -2
And the set of possible values are {-10, -9, -8, -7, -6, -5, -4, -3, -2}
So, the total possible values for m are 9.
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Question for Practice Questions Level 2: Quadratic Equations & Linear Equations - 2
Try yourself:A and B had certain number of stamps. A said to B, ''If you give me one of your stamps, we shall have equal number of stamps''. B replies, ''If you give me one of your stamps, I shall have twice as many as you will be left with''. Find the total number of stamps with A and B.
Explanation
Let the number of stamps that A had be x.
And, let the number of stamps that B had be y.
Then, x + 1 = y - 1
x - y = - 2 … (i)
And, y + 1 = 2(x - 1)
-2x + y = - 3 … (ii)
Equations (i) + (ii) gives,
x = 5
Substitute the value of x in equation (i), we get
y = 7
Therefore, A had 5 stamps and B had 7 stamps. Thus, the total number of stamps with A and B is 12.
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Question for Practice Questions Level 2: Quadratic Equations & Linear Equations - 2
Try yourself:If p = x2 - yz, q = y2 - zx and r = z2 - xy, then the value of
Explanation
px + qy + rz = x3 - xyz + y3 - xyz + z3 - xyz
= x3 + y3 + z3 - 3xyz
= (x + y + z)(x2 + y2 + z2 - xy - yz - zx)
= (x + y + z)(p + q + r)
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Question for Practice Questions Level 2: Quadratic Equations & Linear Equations - 2
Try yourself:If α, β and γ are the roots of the equation 2x3 - 3x2 - 5x + 6 = 0, then α2 + β2 + γ2 is equal to
Explanation
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Question for Practice Questions Level 2: Quadratic Equations & Linear Equations - 2
Try yourself:If a + b = 2c, then what is the value of
Explanation
a + b = 2c
∴ a - c = c - b
= 2
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Question for Practice Questions Level 2: Quadratic Equations & Linear Equations - 2
Try yourself:The set of value of λ for which both the roots of the equation x2 - (λ + 1)x + λ + 4 = 0 are negative, is
Explanation
The given equation is:
x2 - (λ + 1)x + λ + 4 = 0
Both roots are negative, if
(i) discriminant ≥ 0
(ii) sum of the roots is negative, and
(iii) product of the roots is positive
i.e. if (λ + 1)2 - 4(λ + 4) ≥ 0
and λ + 1 < 0 because (-(b/a) = -(-(λ + 1)) => (λ + 1)
and λ + 4 > 0
i.e λ2 -2λ -15 ≥ 0 and λ < -1 and λ > -4
i.e. (λ - 5) (λ + 3) ≥ 0 and -4 < λ < -1
i.e. (λ ≤ -3 or λ ≥ 5) and -4 < λ < -1
i.e. -4 < λ ≤ -3 ( -4 is not included and -3 is included in the range)
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Question for Practice Questions Level 2: Quadratic Equations & Linear Equations - 2
Try yourself:If x2 = y + z, y2 = z + x and z2 = x + y, then the value of
Explanation
x2 = y + z
i.e. x + x2 = x + y + z ……… (1)
y2 = z + x
y + y2 = z + x + y ……… (2)
z2 = x + y
z + z2 = x + y + z ……… (3)
Eqn. (1) = Eqn. (2) = Eqn. (3)
⇒ x(1 + x) = y(1 + y) = z(1 + z) = x + y + z = k (let)
Adding L.H.S. and R.H.S., we get
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Question for Practice Questions Level 2: Quadratic Equations & Linear Equations - 2
Try yourself:If and then is equal to
Explanation
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Question for Practice Questions Level 2: Quadratic Equations & Linear Equations - 2
Try yourself:A famous cricketer Mr. Michael Swrfieai went to purchase a certain number of balls. There were two types of balls in the outlet: tennis balls and leather balls. A leather ball was 70 cents costlier than a tennis ball. If it is known that he purchased a total of 30 balls and spent $32(1$ = 100 cents) on purchasing the balls, what could be the possible cost of a tennis ball?
Explanation
Let the cost of each tennis ball be t cents. So, cost of each leather ball will be (t + 70) cents.
Let us suppose Michael bought k tennis balls and (30 – k) leather balls. So, total cost = kt + (30 – k)(t + 70)
Or 3200 = kt + 30t + 2100 – kt – 70k
1100 = 30t – 70k
When t = 60, we get feasible value of k.
So, cost of each tennis ball = 60 cents.
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Question for Practice Questions Level 2: Quadratic Equations & Linear Equations - 2
Try yourself:If p, q are roots (non-zero) of the equation x2 + px + q = 0, then the least value of x2 + px + q is
Explanation
p + q = -p
And, pq = q
⇒ p = 1 ( q ≠ 0)
Hence,
1 + q = -1
⇒ q = -2
x2 + px + q = x2 + x - 2
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Question for Practice Questions Level 2: Quadratic Equations & Linear Equations - 2
Try yourself:If x, y and z are real numbers, then x2 + 4y2 + 9z2 - 6yz - 3zx - 2xy is always
Explanation
Given expression:
x2 + (2y)2 + (3z)2 - (2y)(3z) -(3z)x - x(2y)
= a2 + b2 + c2 - bc - ca - ab, where a = x, b = 2y, c = 3z
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Question for Practice Questions Level 2: Quadratic Equations & Linear Equations - 2
Try yourself:If y = x - 1/x, the expression 5x4 + 3x3 - 7x2 - 3x + 5 can be expressed as
Explanation
5x4 + 3x3 - 7x2 - 3x + 5
= x2(5x2 + 3x - 7 - 3/x + 5/x2)
= x2[5(x2 + 1/x2 - 2) + 3(x - 1/x) + 3]
= x2[5(x - 1/x)2 + 3(x - 1/x) + 3]
= x2(5y2 + 3y + 3)
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Question for Practice Questions Level 2: Quadratic Equations & Linear Equations - 2
Try yourself:If then find the value of (x5 - 2x4 - x3 - x2 - 2x)
Explanation
⇒ x2 + 1 = 3x ........(1)
Multiply by x , we get
⇒ x3 + x = 3x2 ........(2)
Multiply by x, we get
⇒ x4 + x2 = 3x3 ........(3)
Again, Multiply by x, we get
⇒ x5 + x3 = 3x4 ........(5)
Now , Add the eq. (1), (2), (3) and (4) , we get
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Question for Practice Questions Level 2: Quadratic Equations & Linear Equations - 2
Try yourself:There is a father, a mother and 2 sons in a family and their total age is 60 years. The difference between the sons' age is 3 years, mother's age exceeds the sum of the sons' age by 17 years and the difference of age of father and mother is equal to the age of the elder son. How old is the father?
Explanation
F + M + E + Y = 60 (F = Father's age, M = Mother's age, E = Elder son's age and Y = Younger son's age)
E = Y + 3,
M = E + Y + 17 = 2Y + 20,
F - M = E
F = M + E = (2Y + 20) + (Y + 3) = 3Y + 23
Now, F + M + E + Y = (3Y + 23) + (2Y + 20) + (Y + 3) + Y = 60
⇒ Y = 2 and F = 3Y + 23 = 29 years
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Question for Practice Questions Level 2: Quadratic Equations & Linear Equations - 2
Try yourself:The values of a and b for which 3x3 - ax2 - 74x + b is a multiple of x2 + 2x - 24 are
Explanation
3x3 - ax2 - 74x + b is a multiple of x2 + 2x - 24.
x2 + 2x - 24 = 0
Or, x2 + 6x - 4x - 24 = 0
Or, x(x + 6) - 4(x + 6) = 0
Or, (x - 4)(x + 6) = 0
Or, x = 4, x = -6 satisfies 3x3 - ax2 - 74x + b.
When x = 4:
192 - 16a - 296 + b = 0 … (1)
When x = -6:
-648 - 36a + 444 + b = 0 … (2)
Solving the above two equations, we get a = -5, b = 24.
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