At 4 am- the temperature in Eastbourne was -5- C- By 1 pm- it had increased by 4- C- The temperature in Eastbourne at 1 pm is- A- 9- CB- -9- CC- 1- CD- -1- C

The correct option is D 1^∘ CTemperature at 4 am = nbsp; 5^∘ C Rise in temperature = 4^∘ C Thus, temperature at 1 pm = 5^∘ C + nbsp;4^∘ C = 1^∘ C ...

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Byju's Answer

Standard VII

Mathematics

Application of Addition of Negative Numbers in Everyday Life Situation

At 4 am, the ...

Question

At 4 am, the temperature in Eastbourne was $- 5^{\circ} C$ . By 1 pm, it had increased by $4^{\circ} C$ . The temperature in Eastbourne at 1 pm is .

9^{\circ} C

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- 9^{\circ} C

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1^{\circ} C

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- 1^{\circ} C

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Solution

The correct option is D $- 1^{\circ} C$
Temperature at 4 am = $- 5^{\circ} C$

Rise in temperature = $4^{\circ} C$
Thus, temperature at 1 pm = $- 5^{\circ} C + 4^{\circ} C = - 1^{\circ} C$

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At 4 am, the temperature in Eastbourne was −5∘ C. By 1 pm, it had increased by 4∘ C. The temperature in Eastbourne at 1 pm is .

The correct option is D −1∘ CTemperature at 4 am = −5∘ C Rise in temperature = 4∘ C Thus, temperature at 1 pm = −5∘ C+4∘ C=−1∘ C

At 4 am, the temperature in Eastbourne was $- 5^{\circ} C$ . By 1 pm, it had increased by $4^{\circ} C$ . The temperature in Eastbourne at 1 pm is .

The correct option is D $- 1^{\circ} C$
Temperature at 4 am = $- 5^{\circ} C$

Rise in temperature = $4^{\circ} C$
Thus, temperature at 1 pm = $- 5^{\circ} C + 4^{\circ} C = - 1^{\circ} C$