If $\frac{X}{3} = \frac{Y}{4} = \frac{Z}{7}, t h e n \frac{X + Y + Z}{Z}$ is equal to [#1692]

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Q1. If $\frac{X}{3} = \frac{Y}{4} = \frac{Z}{7}, t h e n \frac{X + Y + Z}{Z}$ is equal to
Q1. If $\frac{X}{3} = \frac{Y}{4} = \frac{Z}{7}, t h e n \frac{X + Y + Z}{Z}$ is equal to

(A) 2
(A) 2

(B) 3
(B) 3

(D) 5
(D) 5

Answer: (A) 2
Answer: (A) 2

\frac{X}{3} = \frac{Y}{4} = \frac{Z}{7}

Hence, X:Y:Z = 3:4:7

\frac{X + Y + Z}{Z}

\frac{3 + 4 + 7}{7}

\frac{14}{7}

= 2

\frac{X}{3} = \frac{Y}{4} = \frac{Z}{7}

Hence, X:Y:Z = 3:4:7

\frac{X + Y + Z}{Z}

\frac{3 + 4 + 7}{7}

\frac{14}{7}

= 2

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Q1. The conversion of $0.0 \bar{37}$ in the fractional form is
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Q2. What is the name of the mathematical concept that describes a value that never changes, like the ratio of a circle's circumference to its diameter?
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Q5. p, q, r are three numbers such that the LCM of p and q is q and the LCM of q and r is r. The LCM of p, q and r will be
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Q6. 100% of 100 when added to 200% of 200 would result
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Q8. The sum of the four consecutive alternate numbers is 64. The smallest number amongst them is
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Q10. A lady deposits money in her savings bank in such a way that every next day her deposit amount is ₹ 12 more than her previous day deposit. If she starts her deposit with ₹ 12 on the first day, the total amount deposited by Liza at the end of 30 days will be :
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The formula for the sum of the first n terms of an arithmetic progression is

S_{n} = \frac{n}{2} (a + a n)

\frac{30}{2} (12 + 12 * 30)

= 15*(12+360) = 15 * 372 = 5580

The formula for the sum of the first n terms of an arithmetic progression is

S_{n} = \frac{n}{2} (a + a n)

\frac{30}{2} (12 + 12 * 30)

= 15*(12+360) = 15 * 372 = 5580

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If $\frac{X}{3} = \frac{Y}{4} = \frac{Z}{7}, t h e n \frac{X + Y + Z}{Z}$ is equal to [#1692]

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