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Estimate the given numbers to the nearest 1000’s place and find the estimated sum and actual sum. The numbers are 751724 and 1971843.

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Last updated date: 20th Jun 2024
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Answer
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Hint:
Complete step-by-step solution
Now, the numbers in question are 751724 and 1971843.
So, for number 751724, we have two possible 1000’s digit numbers which are 751000 or 752000.
Now, 751000 – 751724 = -724 and 752000 – 751724 = 276
As, $\left| -724 \right|>\left| 276 \right|$
So, the estimate of 751724 to the nearest 1000’s place is equal to 752000.
Similarly for, 1971843, we have two possible 1000’s digit numbers which are 1971000 or 1972000.
Now, 1971000 – 1971843 = -843 and 1972000 – 1971843 = 157
As, $\left| -843 \right|>\left| 157 \right|$
So, the estimation of 1971843 to the nearest 1000’s place is equal to 1972000.
Then, the estimated sum of numbers 751724 and 1971843 will be equal to 752000 + 1972000, which is equal to 2724000.
And the actual sum of numbers 751724 and 1971843 will be equal to 751724 + 1971843, which is equal to 2723567.

Note: Always remember that $\left| -a \right|=\left| a \right|=a$. While estimation, try not to do any calculation mistakes as the numbers are so big, the calculation error may occur while subtraction or addition. Also, two possible nearest 1000’s digit numbers must be calculated carefully. Don’t forget to check the modulus values of the difference obtained in the last step of the estimation of the nearest 1000’s digit number.