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 price of an article is first decreased by 40% and then increased by 20%. The net change in the price will be :
Q1. The price of an article is first decreased by 40% and then increased by 20%. The net change in the price will be :

(A) 20%
(A) 20%

(B) 40%
(B) 40%

(D) 28%
(D) 28%

Answer: (D) 28%
Answer: (D) 28%

Net change (-40 + 20 +

\frac{(- 40) * 20}{100}

)% = (-20 +

\frac{- 800}{100}

)% = (-20 - 8)% = -28% = 28%

Net change (-40 + 20 +

\frac{(- 40) * 20}{100}

)% = (-20 +

\frac{- 800}{100}

)% = (-20 - 8)% = -28% = 28%

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Q2. Shyam stored Rs 35 in the form of 1 rupee coin and 50 paise coins in the ratio 2 : 3. The number of 50 paise coins are
Q2. Shyam stored Rs 35 in the form of 1 rupee coin and 50 paise coins in the ratio 2 : 3. The number of 50 paise coins are

(A) 20
(A) 20

(B) 25
(B) 25

(D) 35
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Answer: (C) 30
Answer: (C) 30

1*20 + 0.5*30 = 20+15 = 35

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2024-04-19

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Q3. The largest of 2⁶, 3⁵, 4⁴ and 5³ is
Q3. The largest of 2⁶, 3⁵, 4⁴ and 5³ is

(A) 2⁶
(A) 2⁶

(B) 3⁵
(B) 3⁵

(D) 5³
(D) 5³

Answer: (C) 4⁴
Answer: (C) 4⁴

2⁶ = 64 3⁵ = 243 4⁴ = 256 5³ = 125

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Q4. 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
Q4. 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

(A) q
(A) q

(B) r
(B) r

(D) pqr
(D) pqr

Answer: (B) r
Answer: (B) r

LCM will be r. px= q and qy = r, hence r = pxy.

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Q5. In a range of consecutive numbers starting with 1, all the even numbers are removed. From the remaining, consider the first 7 numbers. The sum of these 7 numbers is
Q5. In a range of consecutive numbers starting with 1, all the even numbers are removed. From the remaining, consider the first 7 numbers. The sum of these 7 numbers is

(A) 35
(A) 35

(B) 42
(B) 42

(D) 56
(D) 56

Answer: (C) 49
Answer: (C) 49

1+3+5+7+9+11+13 = 49

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Q6. If A : B = 3, then 144A : 36B is
Q6. If A : B = 3, then 144A : 36B is

(A) 4
(A) 4

(B) 12
(B) 12

(D) 16
(D) 16

Answer: (B) 12
Answer: (B) 12

A : B = 3 =>

\frac{A}{B}

= 3

\frac{144 A}{36 B}

\frac{144}{36} \times \frac{A}{B}

\frac{144}{36}

x 3 = 4 x 3 = 12

A : B = 3 =>

\frac{A}{B}

= 3

\frac{144 A}{36 B}

\frac{144}{36} \times \frac{A}{B}

\frac{144}{36}

x 3 = 4 x 3 = 12

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Q7. When three times of a given number is subtracted from the square of the number, the result is the number itself. The number is
Q7. When three times of a given number is subtracted from the square of the number, the result is the number itself. The number is

(A) 4
(A) 4

(B) 2
(B) 2

(D) -4
(D) -4

Answer: (A) 4
Answer: (A) 4

4^{2 - (4*3)
= 16 - 12
= 4
4^{2 - (4*3)
= 16 - 12
= 4}}

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Q8. The count of prime numbers between 80 and 100 is
Q8. The count of prime numbers between 80 and 100 is

(A) 2
(A) 2

(B) 5
(B) 5

(D) 4
(D) 4

Answer: (C) 3
Answer: (C) 3

Prime numbers are numbers that are only divisible by 1 and themselves. Between 80 and 100, the prime numbers are 83, 89, and 97. Therefore, there are 3 prime numbers in this range.

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Q9. Consider the four numbers given numbers. If the digits of each number are arranged in ascending order, which new number would be the smallest?
Q9. Consider the four numbers given numbers. If the digits of each number are arranged in ascending order, which new number would be the smallest?

I. 376 II. 629 III. 921 IV. 397

I. 376 II. 629 III. 921 IV. 397

(A) III
(A) III

(B) II
(B) II

(D) IV
(D) IV

Answer: (A) III
Answer: (A) III

376 -> 367 629 -> 269 921 -> 129 * 397 -> 379

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Q10. Find the least number by which 1250 must be multiplied to make it a perfect square.
Q10. Find the least number by which 1250 must be multiplied to make it a perfect square.

(A) 4
(A) 4

(B) 3
(B) 3

(D) 2
(D) 2

Answer: (D) 2
Answer: (D) 2

1250 = 2*5*5*5*5 1250*2 = 2*2*5*5*5*5 = 2500

\sqrt{2500}

= 2*5*5 = 50

1250 = 2*5*5*5*5 1250*2 = 2*2*5*5*5*5 = 2500

\sqrt{2500}

= 2*5*5 = 50

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If X3=Y4=Z7, then X+Y+ZZ is equal to [#1692]

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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]

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