Algebra
- x varies inversely as square of y. Given that y = 2 for x = 1, the value of x for y = 6 will be equal to
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x ∝ 1 y2 ⇒ x = k y2
where k is a constant of proportionality.
When, x = 1, y = 2⇒ 1 = k ⇒ k = 4 4 ∴ x = 4 y2 x = 4 = 1 6 × 6 9 Correct Option: D
x ∝ 1 y2 ⇒ x = k y2
where k is a constant of proportionality.
When, x = 1, y = 2⇒ 1 = k ⇒ k = 4 4 ∴ x = 4 y2 x = 4 = 1 6 × 6 9
- If (a – 3)2 + (b – 4)2 + (c – 9)2 = 0, then the value of √a + b + c is :
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(a – 3)2 + (b – 4)2 + (c – 9)2 = 0
⇒ a – 3 = 0 ⇒ a = 3
b – 4 = 0 ⇒ b = 4
and c – 9 = 0 ⇒ c = 9
∴ √a + b + c = √3 + 4 + 9
= √16 = ± 4Correct Option: C
(a – 3)2 + (b – 4)2 + (c – 9)2 = 0
⇒ a – 3 = 0 ⇒ a = 3
b – 4 = 0 ⇒ b = 4
and c – 9 = 0 ⇒ c = 9
∴ √a + b + c = √3 + 4 + 9
= √16 = ± 4
- If a2 + b2 + c2 + 3 = 2 (a – b – c), then the value of 2 a – b + c is :
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a2 + b2 + c2 + 3 = 2a – 2b – 2c
⇒ a2 – 2a + 1 + b2 + 2b + 1 + c2 + 2c + 1 = 0
⇒ (a – 1)2 + (b + 1)2 + (c + 1)2 = 0
∴ a – 1 = 0 ⇒ a = 1
b + 1 = 0 ⇒ b = –1
c + 1 = 0 ⇒ c = –1
∴ 2a – b + c = 2 + 1 – 1 = 2Correct Option: D
a2 + b2 + c2 + 3 = 2a – 2b – 2c
⇒ a2 – 2a + 1 + b2 + 2b + 1 + c2 + 2c + 1 = 0
⇒ (a – 1)2 + (b + 1)2 + (c + 1)2 = 0
∴ a – 1 = 0 ⇒ a = 1
b + 1 = 0 ⇒ b = –1
c + 1 = 0 ⇒ c = –1
∴ 2a – b + c = 2 + 1 – 1 = 2
- If x2 − y2 = 80 and x – y = 8, then the average of x and y is
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x2 − y2 = 80
x – y = 8∴ x + y = x2 − y2 = 80 = 10 x – y 8 ∴ Required average = x + y = 10 = 5 2 2 Correct Option: D
x2 − y2 = 80
x – y = 8∴ x + y = x2 − y2 = 80 = 10 x – y 8 ∴ Required average = x + y = 10 = 5 2 2
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The third proportional to 
x + y 
and √x2 + y2 is y x
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Third proportional of a and b = b2 a = (√x2 + y2)2 y + y y x = x2 + y2 x2 + y2 xy
= xyCorrect Option: A
Third proportional of a and b = b2 a = (√x2 + y2)2 y + y y x = x2 + y2 x2 + y2 xy
= xy
