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happypuppy
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14 Jan 2022, 10:21
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C
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62% (01:56) correct
38%
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We are given a decimal number P = 12.abc where a, b, and c are the tenths, hundredths, and the thousandths digit respectively. Is c + 5, a two-digit number?
(1) b is not an odd number.
(2) If 10P were rounded to the nearest tenths, the result would be 124.7.
(1) b is not an odd number.
(2) If 10P were rounded to the nearest tenths, the result would be 124.7.
14 Jan 2022, 12:12
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I can answer this..
(1) just tells us about b, nothing about c, so not sufficient.
(2) tells us that 10P = 12a.bc = 124.7, so we can infer that a = 4. But, b can be 6 and c can be > 5 and rounding to the nearest tenth would give us b as 7; or we could have b = 7 and c < 1 s.t rounding gives us b = 7. So (2) by itself is also not sufficient.
Looking at both of the statements together, we know that b can't be odd, and b has to be < 9 (digit) so b can be 2, 4, 6, 8. 2 and 4 are eliminated because it can't give us 7 when rounding. If b = 6, and c > 5, rounding will give us b = 7. If b = 8, no value of c can give us b = 7 (because if c <5, b= 8 and if c > 5, b = 9). Therefore, b = 6 and c>5. The smallest value of c is 6, and 6+5 > 9, hence both statements together are sufficient.
(1) just tells us about b, nothing about c, so not sufficient.
(2) tells us that 10P = 12a.bc = 124.7, so we can infer that a = 4. But, b can be 6 and c can be > 5 and rounding to the nearest tenth would give us b as 7; or we could have b = 7 and c < 1 s.t rounding gives us b = 7. So (2) by itself is also not sufficient.
Looking at both of the statements together, we know that b can't be odd, and b has to be < 9 (digit) so b can be 2, 4, 6, 8. 2 and 4 are eliminated because it can't give us 7 when rounding. If b = 6, and c > 5, rounding will give us b = 7. If b = 8, no value of c can give us b = 7 (because if c <5, b= 8 and if c > 5, b = 9). Therefore, b = 6 and c>5. The smallest value of c is 6, and 6+5 > 9, hence both statements together are sufficient.
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14 Jan 2022, 12:20
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1) B is not an odd number (not giving any information) (Insufficient)
2) 10P = 124.7
P = 12.47(round off value)
Here, A = 4
B can be 6 or 7 as the value is rounded off.
So, the value of BC can range from 65 to 74.
Now,
If B is 7, then adding 5 to C's values = 0+5, 1+5, 2+5, 3+5, 4+5 won't give two-digit number.
But if B = 6 then adding 5 to C's values = 5+5, 6+5, 7+5, 8+5, 9+5 will a give two-digit number. (Not sufficient)
Combining both,
We know B = 6
So adding 5 in C's values will always give 2 digit number.
Answer is C
Posted from my mobile device
2) 10P = 124.7
P = 12.47(round off value)
Here, A = 4
B can be 6 or 7 as the value is rounded off.
So, the value of BC can range from 65 to 74.
Now,
If B is 7, then adding 5 to C's values = 0+5, 1+5, 2+5, 3+5, 4+5 won't give two-digit number.
But if B = 6 then adding 5 to C's values = 5+5, 6+5, 7+5, 8+5, 9+5 will a give two-digit number. (Not sufficient)
Combining both,
We know B = 6
So adding 5 in C's values will always give 2 digit number.
Answer is C
Posted from my mobile device
