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UB001
Joined: 26 Dec 2018
Last visit: 10 Apr 2022
Posts: 106
Given Kudos: 91
Location: India
Schools: Mannheim"20 Smurfit "20 NUS '21 NTU '20
30 Dec 2018, 11:15
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
76% (01:42) correct
24%
(02:19)
wrong
based on 206
sessions
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How many integers between 24,700 and 58,600 have tens digit 3 and units digit 7?
A. 339
B. 352
C. 353
D. 630
E. 800
A. 339
B. 352
C. 353
D. 630
E. 800
30 Dec 2018, 13:44
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Approach this question as a Series problem.
Between 24,700 and 58,600, the first integer to have tens digit 3 and units digit 7 is 24737. The next integer satisfying the condition is 24837.
Finally, the last integer between 24,700 and 58,600 satisfying the condition is 58537.
We get an AP series 24737, 24837, 24937, ...........58537
Last term of an AP = a + (n - 1)d, where a = first term, d = common difference, and n = number of terms.
Common difference between consecutive terms is 100
Plug in value in the equation, 58537 = 24737 + (n - 1)100
n = 339
The number of integers between 24,700 and 58,600 have tens digit 3 and units digit 7 is the same as the number of terms in the series, i.e. 339.
Hence, the Correct Answer is A. 339
Between 24,700 and 58,600, the first integer to have tens digit 3 and units digit 7 is 24737. The next integer satisfying the condition is 24837.
Finally, the last integer between 24,700 and 58,600 satisfying the condition is 58537.
We get an AP series 24737, 24837, 24937, ...........58537
Last term of an AP = a + (n - 1)d, where a = first term, d = common difference, and n = number of terms.
Common difference between consecutive terms is 100
Plug in value in the equation, 58537 = 24737 + (n - 1)100
n = 339
The number of integers between 24,700 and 58,600 have tens digit 3 and units digit 7 is the same as the number of terms in the series, i.e. 339.
Hence, the Correct Answer is A. 339
12 Jan 2019, 01:24
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UB001
Don't think too much on this, and use logic..
When can be the tens digit 3 and units digit 7 - when LAST two digits are 37..
so the number is ABC37, where ABC is a 3-digit number between 247 and 586, with 247 inclusive...
Since, only one of the two is inclusive, the number of integers will be straight 586-247=339
A
