June 25, 1982, fell on a Friday. On which day of the week did June 25

Question

June 25, 1982, fell on a Friday. On which day of the week did June 25, 1987, fall? (Note: 1984 was a leap year.)

A. Sunday
B. Monday
C. Tuesday
D. Wednesday
E. Thursday

PS02525

Mackieman · Accepted Answer

One year = 365 days except for leap years = 366 days
52 weeks in a year 7 days per week = 364 days

1982 25th a Friday, if we move 365 days in the next year we end up on the 25th, but the weekday will be Saturday since we have moved 52 weeks + 1 day into the future.
Hence,

1982 25th Friday
1983 25th Saturday (+1 day)
1984 25th Monday (+1 day and +1 extra day for leap year)
1985 25th Tuesday (+1 day)
1986 25th Wednesday (+1 day)
1987 25th = Thursday

If the system had been 364 days per year, every year we would have end up on the same day every year.

jayantheinstein · Answer

good question :-

remainder theorem can be used.

since 84 is a leap it contains one day extra..so totally 1826 days.

now every 7 days gives next friday..so 1826/7 will give remainder 6 indicating the last set of division alone is not complete to give :thursday !!

MHIKER · Answer

1982 -Friday
1983 -Fd+1=Sat
84+1=Sun
85+1=mon
86+1=tue
87+1=wed
1987 wed+1=Thursday

fluke · Answer

June 25, 1987

The following is a formula/mathematical approach to solving this type of question. It is kind of cumbersome; but once acquainted, you can easily solve these types of questions.

year 1600-> 0 odd days
In every 100 years, there are 5 odd days
+300-> 5*3 = 15 odd days = 15%7= 1 odd day

Upto 1900-> 1 odd day

Upto 1986: 
86/4 = 21 leap years
86-21=65 ordinary years

Every leap year: 2 odd days,
ordinary year: 1 odd day
So; 21*2+65 = 42+65=107 odd days = 107%7 = 2 odd days

So; from 1600-1986 end = 1+2 = 3 odd days

in 1987(non-leap year)
Jan=31; Feb=28; March=31; April: 30 May:31 June: 25
31+28+31+30+31+25 = 176%7 = 1 odd day

Total: 3+1=4 odd days

odd days-day
0-Sunday
1-Monday
2-Tuesday
3-Wednesday
4-Thursday
5-Friday
6-Saturday

Ans: Thursday

*****************************
June 25, 1982, fell on a Friday. On which day of the week did June 25, 1987, fall?

june 25,1983: Saturday
june 25,1984: Monday (Leap year: Feb would be 29 days- so days 2+)
june 25,1985: Tuesday
june 25,1986: Wednesday
june 25,1987: Thursday

Bunuel · Answer

Similar question to practice: the-19th-of-september-1987-was-a-saturday-if-1988-was-a-126468.html

Raffio · Answer

I set this up like a remainder problem. There are 5 years between 6/25/1982 and 6/25/1987...

365 * 5 = 1825
Add one day for leap year = 1825+1 = 1826
Divide 1826 days by 7 = 7/18 = 2, 7/42 = 6, 7/6 0 remainder 6 days.

So since the remainder is 6 days, the answer is one day less than Friday, which is Thursday.

dabaobao · Answer

Total years = 5
Total days = 5 *365 + 1 (for leap year) = 1826
1826/7 gives a remainder of 6.

Therefore, Friday + 6 days = Thursday

ScottTargetTestPrep · Answer

Solution:

Since June 25, 1987, is exactly 5 years after June 25, 1982, including the 1 extra leap day in 1984, we see that 5 x 365 + 1 = 1826 days have elapsed between the two dates. Since 1826/7 = 260 R 6, we see that 260 weeks and 6 days have elapsed between the two dates. Since June 25, 1982, fell on a Friday, then June 25, 1987, would be 6 days after Friday; hence, June 25, 1987, fell on a Thursday.

Answer: E

ktzsikka · Answer

Hi buddy
You did a mistake by counting year '87 twice. You should add the extra 1 in '84 for it's a leap year.

Posted from my mobile device

DanTheGMATMan · Answer

Potentially useful in real life:

https://www.youtube.com/watch?v=xeOYGUrEAOc

egmat · Answer

I can see why this calendar problem might seem tricky at first - we're dealing with multiple years and a leap year thrown in. Let me walk you through the key insight that makes this much simpler than it appears.

Here's how to think about this systematically:

The beautiful thing about day-of-week problems is that days repeat every 7 days. So instead of trying to count 5 years worth of days, you just need to find how many "extra" days beyond complete weeks passed between these dates.

Step 1: Count the total days 
From June 25, 1982 to June 25, 1987 is exactly 5 years.
- Normal years:  5 	imes 365 = 1825  days
- But 1984 was a leap year, so add 1 day:  1825 + 1 = 1826  total days

Step 2: Use the 7-day cycle principle  
Since days repeat every 7 days, you only need the remainder when dividing by 7:
 1826 \div 7 = 260  remainder  6

You can verify:  260 	imes 7 = 1820 , and  1826 - 1820 = 6

Step 3: Count forward from Friday 
Starting from Friday (June 25, 1982), count 6 days forward:
Friday → Saturday → Sunday → Monday → Tuesday → Wednesday →  Thursday

Answer: E. Thursday

Notice how the leap year is crucial here - without that extra day, you'd have gotten Wednesday instead. The modular arithmetic approach saves you from counting over 1,800 days individually!

Want to master the systematic framework for all calendar problems?  You can check out the  comprehensive solution on Neuron  to see the complete pattern recognition approach and common trap avoidance techniques. You can also practice with detailed solutions for  many other official questions on Neuron  to build consistent accuracy across all problem types.

June 25, 1982, fell on a Friday. On which day of the week did June 25

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GMAT Problem Solving (PS) Questions

June 25, 1982, fell on a Friday. On which day of the week did June 25

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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