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There are four consecutive even integers. Sum of the reciprocals of the first two number is (11/60). Find the reciprocal of third largest number of the series.

A) 1/12	B) 2/14
C) 1/14	D) 2/13

Answer & Explanation Answer: C) 1/14

Explanation:

Ans. C.
Let 4 consecutive even numbers are x, x+ 2, x+ 4, x+ 6
According to the question -
1/x + 1/x+2 = 11/60
x+2+x/ x(x+2) = 11/60
2(x+1)/x^2 + 2x = 11/60
120x + 120 = 11x 2 + 22x
11x 2 - 98x – 120 = 0
So ,
X = -24/22 , 10
X = -12/11 , 10
So, the numbers will be 10,12,14,16.
Hence,
Third largest number = 14
Reciprocal of third number = 1/14

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Exam Prep: AIEEE , Bank Exams , GATE

Q:

Find the quadratic equations whose roots are the reciprocals of the roots of $2 x^{2} + 5 x + 3$ ?

A) $3 x^{2} + 5 x + 2 = 0$

B) $5 x^{2} + 3 x + 2 = 0$

C) $3 x^{2} - 5 x - 2 = 0$

D) None

A) A	B) B
C) C	D) D

Answer & Explanation Answer: A) A

Explanation:

The quadratic equation whose roots are reciprocal of $2 x^{2} + 5 x + 3 = 0$ can be obtained by replacing x by 1/x.

Hence, 2(1/x)(1/x)+ 5(1/x) + 3 = 0

=> $3 x^{2} + 5 x + 2 = 0$

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Filed Under: Mathematical Operations
Exam Prep: GATE , CAT , Bank Exams , AIEEE
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Q:

If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to ?

A) 13/125	B) 14/57
C) 11/120	D) 16/41

Answer & Explanation Answer: C) 11/120

Explanation:

Let the numbers be a and b.
We know that product of two numbers = Product of their HCF and LCM
Then, a + b = 55 and ab = 5 x 120 = 600.
=> The required sum = (1/a) + (1/b) = (a+b)/ab
=55/600 = 11/120

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Filed Under: HCF and LCM
Exam Prep: AIEEE , Bank Exams , CAT
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