2. The number of marbles with A and B are in the ratio of10:11. Which

1.	2. The number of marbles with A and B are in the ratio of10:11. Which of the following cannot be a possiblenumber of marbles with A and B together?(A) 189 (B) 210 (C) 231 (D) 153
Answer» Given: The number of marbles with $A$ and $B$ are in the RATIO $10:11$ To FIND: Among the given options which cannot be a possible number of marbles with $A$ and $B$ TOGETHER. Solution: Let number of marbles with $A$ is $10k$ and number of marbles with $B$ is $11k$ , where $k$ is any natural number. ∴Number of marbles with $A$ and $B$ together is $10k+11k=21k$ . Hence, number of marbles with $A$ and $B$ together should be multiple of $21$ . From the given options, the only option which is not multiple of $21$ is option $(D) 153$ , rest all are multiple of Hence, option $(D) 153$ cannot be a possible number of marbles with $A$ and $B$ together.

2. The number of marbles with A and B are in the ratio of10:11. Which of the following cannot be a possiblenumber of marbles with A and B together?(A) 189 (B) 210 (C) 231 (D) 153

Answer»

Given:

The number of marbles with $A$ and $B$ are in the RATIO $10:11$

To FIND:

Among the given options which cannot be a possible number of marbles with $A$ and $B$ TOGETHER.

Solution:

Let number of marbles with $A$ is $10k$ and number of marbles with $B$ is $11k$ , where $k$ is any natural number.

∴Number of marbles with $A$ and $B$ together is $10k+11k=21k$ .

Hence, number of marbles with $A$ and $B$ together should be multiple of $21$ .

From the given options, the only option which is not multiple of $21$ is option $(D) 153$ , rest all are multiple of

Hence, option $(D) 153$ cannot be a possible number of marbles with $A$ and $B$ together.

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