Algebra
- If 999x + 888y = 1332 888x + 999y = 555, then the value of x + y is
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999x + 888y = 1332
888x + 999y = 555
On adding,
1887x + 1887y = 1887
⇒ 1887 (x + y) = 1887⇒ x + y = 1887 = 1 1887 Correct Option: C
999x + 888y = 1332
888x + 999y = 555
On adding,
1887x + 1887y = 1887
⇒ 1887 (x + y) = 1887⇒ x + y = 1887 = 1 1887
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If bc + ab + ca = abc, then the value of b + c + a + c + a + b = 1, then the value of a2 + b2 + c2 is a2 + bc b2 + ca c2 + ab a2 + bc b2 + ac c2 + ab
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a2 − bc + b2 − ca + c2 − ab = 1 a2 + bc b2 + ca c2 + ab ⇒ 
a2 − bc + 1 
+ b2 − ca + 1 + c2 − ab + 1 = 4 a2 + bc b2 + ca c2 + ab ⇒ a2 − bc + a2 + bc + b2 − ca + b2 + ca + c2 − ab + c2 + ab = 4 a2 + bc b2 + ca c2 + ab ⇒ 2a2 + 2b2 + 2c2 = 4 a2 + bc b2 + ca c2 + ab ⇒ a2 + b2 + c2 = 4 = 2 a2 + bc b2 + ca c2 + ab 2 Correct Option: D
a2 − bc + b2 − ca + c2 − ab = 1 a2 + bc b2 + ca c2 + ab ⇒ a2 − bc + 1 + b2 − ca + 1 + c2 − ab + 1 = 4 a2 + bc b2 + ca c2 + ab ⇒ a2 − bc + a2 + bc + b2 − ca + b2 + ca + c2 − ab + c2 + ab = 4 a2 + bc b2 + ca c2 + ab ⇒ 2a2 + 2b2 + 2c2 = 4 a2 + bc b2 + ca c2 + ab ⇒ a2 + b2 + c2 = 4 = 2 a2 + bc b2 + ca c2 + ab 2
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If bc + ab + ca = abc, then the value of b + c + a + c + a + b is bc(a − 1) ac(b − 1) ab(c − 1)
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Given, bc + ab + ca = abc
∴ bc + ab = abc – ac
ab + ca = abc – bc
bc + ca = abc – ab∴ Expression = b + c + a + c + a + b abc − bc abc − ac abc − ab = b + c + a + c + a + b ab + ac bc + ab bc + ca = b + c + a + c + a + b a(b + c) b(c + a) c(a + b) = 1 + 1 + 1 a b c = bc + ac + ab abc = abc = 1 abc Correct Option: D
Given, bc + ab + ca = abc
∴ bc + ab = abc – ac
ab + ca = abc – bc
bc + ca = abc – ab∴ Expression = b + c + a + c + a + b abc − bc abc − ac abc − ab = b + c + a + c + a + b ab + ac bc + ab bc + ca = b + c + a + c + a + b a(b + c) b(c + a) c(a + b) = 1 + 1 + 1 a b c = bc + ac + ab abc = abc = 1 abc
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If x = a − b , y = b − c , z = c − a , then (1 − x)(1 − y)(1 − z) is equal to a + b b + c c + a (1 + x)(1 + y)(1 + z)
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x = a − b 1 a + b
By componendo and dividendo,
= a + b − a + b = b a + b + a − b a
Similarly,1 − y = c ; 1 − z = a 1 + y b 1 + z c ∴ Expression = (1 − x)(1 − y)(1 − z) (1 + x)(1 + y)(1 + z) = b × c × a = 1 a b c Correct Option: A
x = a − b 1 a + b
By componendo and dividendo,= a + b − a + b = b a + b + a − b a
Similarly,1 − y = c ; 1 − z = a 1 + y b 1 + z c ∴ Expression = (1 − x)(1 − y)(1 − z) (1 + x)(1 + y)(1 + z) = b × c × a = 1 a b c
- If a, b, c are real numbers and a2 + b2 + c2 = 2 (a – b – c) – 3, then the value of a + b + c is
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a2 + b2 + c2 = 2a – 2b – 2c – 3
⇒ a2 – 2a + 1 + b2 + 2b + 1 + c2 + 2c + 1 = 0
⇒ (a – 1)2 + (b + 1)2 + (c + 1)2 = 0
⇒ a – 1 = 0, b + 1 = 0, c + 1 = 0
⇒ a = 1, b = –1, c = –1
∴ a + b + c = 1 – 1 – 1 = –1Correct Option: A
a2 + b2 + c2 = 2a – 2b – 2c – 3
⇒ a2 – 2a + 1 + b2 + 2b + 1 + c2 + 2c + 1 = 0
⇒ (a – 1)2 + (b + 1)2 + (c + 1)2 = 0
⇒ a – 1 = 0, b + 1 = 0, c + 1 = 0
⇒ a = 1, b = –1, c = –1
∴ a + b + c = 1 – 1 – 1 = –1
