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In the summer of $2012$, in New Delhi, the mean temperature of Monday to Wednesday was $41°C$ and of Tuesday to Thursday was $43°C$. If the temperature on Thursday was $15\%$ higher than that of Monday, then the temperature in $°C$ on Thursday was

  1. $40$
  2. $43$
  3. $46$
  4. $49$
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3 Answers

Best answer
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10 votes

Let the temperatures on Monday, Tuesday, Wednesday, and Thursday be $x^∘,  y^∘,  z^∘,$ and $w^∘C$ respectively.

Given that,

  • $x+y+z = 3 \times 41$
  • $y+z+w = 3\times 43$

Thursday temperature is $15\%$ higher than Monday temperature. So, 

  • $w = 1.15x$

After solving above equations, we get $w=46$

Therefore, correct answer is (C).

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$\dfrac{M+T+W}{3}=41$

$\dfrac{T+W+Th}{3}=43$

$M+T+W=123\ \ ....(1)$

$T+W+Th=129\ \ ....(2)$

$Subtract\ (1)\ from\ (2) $

$Th-M=6\ \ ....(3)$

If the temperature on Thursday was 15% higher than that of Monday.

$Th=\dfrac{115}{100}M$

$Substitute\ the\ value\ of\ 'Th'\ in\ (3)$

$\dfrac{115}{100}M-M=6$

$15M=600$

$M=40$

$\therefore Th=46$

 

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MON+TUES+WED=41*3=123-----------(1)

TUE+WED +THURS=43*3=129--------(2)

Substracting (1) from (2) we get THURS - MON= 6

Also it is said that Temperature of THURS is 15 % higher than Monday.

hence 15%=6

hENCE 100%=6/15*100= 40 WHICH IS THE TEMPERATURE of Monday.

The temperature of Thursday is 46  and the option  C is correct.
Answer:

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