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14 (a) (i) Round 356 to the nearest ten - OCR - GCSE Maths - Question 14 - 2019 - Paper 1

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14 (a) (i) Round 356 to the nearest ten. (ii) Round 356.052 to 1 decimal place. (b) Find the value of y in each of the following. (i) 3 × 3 × 3 × 3 = 3^y (ii... show full transcript

Worked Solution & Example Answer:14 (a) (i) Round 356 to the nearest ten - OCR - GCSE Maths - Question 14 - 2019 - Paper 1

Step 1

(a) (i) Round 356 to the nearest ten.

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Answer

To round 356 to the nearest ten, we look at the digit in the ones place, which is 6. Since 6 is 5 or greater, we round up. Therefore, 356 rounds to 360.

Step 2

(a) (ii) Round 356.052 to 1 decimal place.

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Answer

To round 356.052 to one decimal place, we look at the digit in the second decimal place. The first decimal place is 0 and the second decimal place is 5. Since 5 is 5 or greater, we round up the first decimal place. Thus, 356.052 rounds to 356.1.

Step 3

(b) (i) Find the value of y in 3 × 3 × 3 × 3 = 3^y.

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Answer

The expression 3 × 3 × 3 × 3 can be rewritten as 3^4, since there are four 3's multiplied together. Therefore, 3^4 = 3^y implies that y = 4.

Step 4

(b) (ii) Find the value of y in 6^3 × 6^5 = 6^y.

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Answer

To combine the powers of 6, we use the exponent addition rule: a^m × a^n = a^(m+n). Therefore, 6^3 × 6^5 = 6^{3+5} = 6^8. This implies y = 8.

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