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## In a bag, there are a certain number of toy-blocks with alphabets A, B, C and D written on them. The ratio of blocks A:B:C:D is in the ratio 4:7:3:1. If the number of ‘A’ blocks is 50 more than the number of ‘C’ blocks, what is the number of ‘B’ blocks?

In a bag, there are a certain number of toy-blocks with alphabets A, B, C and D written on them. The ratio of blocks A:B:C:D is in the ratio 4:7:3:1. If the number of ‘A’ blocks is 50 more than the number of ‘C’ blocks, what is the number of ‘B’ blocks? A. 650 B. 480 C. 578 D. 350

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Given, the ratio of blocks of A,B,C,D are in the ratio 4:7:3:1

Let us consider the common ratio to be 'x'.

So, toy blocks with alphabet A is 4x and,

toy blocks with alphabet B is 7x and,

toy blocks with alphabet C is 3x and,

toy blocks with alphabet D is x.

Again, the number of 'A' blocks is 50 more than the number of 'C' blocks.

As no. of 'A' and 'C' blocks are 4x and 3x respectively.

So,

4x = 50 + 3x

⇒ x = 50

Thus, the number of 'B' blocks is 7x = 7(50) = 350

350 is the required number.

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