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For breads to be equal, A+B = 6+7 = 13, which cannot be divided equally among three. Thus C has some bread initially. For, the breads to be equally distributable among the three, the total number of breads should be a multiple of 3, greater than 6+7 (13) = 15, 18, 21, 24…. If Total breads = 15, this means C had 2 breads initially Bread in each’s share = 5 This means A gave 1 and B gave 2 breads to C Now, suppose the total number of breads = 18 Breads initially with C = 18-13 = 5 This means each’s share is 6 bread each. For this to be the case, A should give nothing to C and B should give 1 bread to C to make all the three have equal breads but A has definitely shared some bread with C. Thus this case not possible. Thus, C initially had 2 breads and A gave 1 and B gave 2 breads to C. Now, money will be divided among A and B in the ratio 1:2 Rs. 42/3 = 14, B’s share = 14 x 2 = 28 Rs
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If the mean of four observations is 20 and when a constant C is added to each observation the mean becomes 22 the value of C is –
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Find the simplified value of the expression:
32 + 12 x 108 ÷ 12 + 4 + 32 ÷ 16 x 3
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