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30 Jun 2019, 22:41
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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:
95%
(hard)
Question Stats:
30% (02:35) correct
70%
(02:42)
wrong
based on 86
sessions
History
Date
Time
Result
Not Attempted Yet
The value of the sum 7 x 11 + 11 x 15 + 15 x 19 + ...+ 95 x 99 is
A. 80707
B. 80773
C. 80730
D. 80751
E. 80734
A. 80707
B. 80773
C. 80730
D. 80751
E. 80734
27 Feb 2021, 23:02
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raghavrf
There is always a smarter and quicker way to solve questions. The more you keep your eyes open for such opportunities thrown at you, the easier quant will become.
Look at the options. All have different units digit
Can we use this to our advantage?
\(7*11+11*15+15*19+19*23+23*27+27*31+........95*99.\)
Let us check on the units digit for each term
\(7*1+1*5+5*9+9*3+3*7+7*1....\)
So the units digit of each term in the series repeats after every 5 terms.
7,5,5,7,1,7,5,5,7,1.....
The Sum of each set of five terms is 7+5+5+7+1=5
Five-term means 5*4 =20 more
So number of complete set of five terms =\(\frac{87-7}{20}=4\). Thus, the sum of units digits of 7*11+11*15+....83*87 becomes 4*5=20 or units digit is 0.
Now what we are left with is 87*91+91*95+95*99, that is 7+5+5=17 or 7.
Only A has units digit as 7.
The test makers may not have designed the questions for you to take advantage of such situations but finding one and using it to your advantage is what can make a difference between Q42 and Q50.
General Discussion
WiziusCareers1
Joined: 27 Apr 2009
Last visit: 23 Apr 2026
Posts: 176
Given Kudos: 35
Status:Not Applying
Location: India
Schools: HBS '14 (A)
GMAT 1: 730 Q51 V36
30 Jun 2019, 23:04
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The nth term of this sequence is given by (4n + 3)(4n + 7) = 16n^2 + 40n + 21n
Here the minimum value of n is 1 and the maximum is 23.
Sum of squares is given by n(n + 1)(2n + 1)/6. Here, the sum is 23(24)(47)/6 = 4324
Sum of first n natural numbers is given by n(n + 1)/2. Here, the sum is 23(24)/2 = 276
Hence, the sum of the sequence is (16*4324) + (40*276) + 21*23 = 69184 + 11040 + 483 = 80707
Note: This is definitely not a GMAT Question
Here the minimum value of n is 1 and the maximum is 23.
Sum of squares is given by n(n + 1)(2n + 1)/6. Here, the sum is 23(24)(47)/6 = 4324
Sum of first n natural numbers is given by n(n + 1)/2. Here, the sum is 23(24)/2 = 276
Hence, the sum of the sequence is (16*4324) + (40*276) + 21*23 = 69184 + 11040 + 483 = 80707
Note: This is definitely not a GMAT Question
