Excluded number = (27 x 5) - (25 x 4) = 135 - 100 = 35.
We Know that sum of the n natural numbers =
Then sum of 50 natural numbers= = 1250
Average of the 50 natural numbers = = 25.5
Marks in English = 85×7 ? 83×6
= 595 ? 498
= 97
Assume Hebah has Rs. M
Since 25% more money at Poonam
=> Money at Poonam = M + (25% of M)
=> Money at Poonam = M + 0.25M = Rs. 1.25M
Money with Navaneet is thrice the money with Poonam,
=> Money at Navaneet = 3(1.25M) = Rs. 3.75M
Now, sum of all Money = (M + 1.25M + 3.75M) = Rs. 6M
But given the average of the money is Rs. 350
=> 6M/3 = 350
=> 2M = 350
=> M = 350/2 = Rs. 175
=> Amount of money Navaneet has = Rs. 3.75M = 3.75 x 175 = Rs. 656.25.
The third proportional of two numbers p and q is defined to be that number r such that
p : q = q : r.
Here, required third proportional of 10 & 20, and let it be 'a'
=> 10 : 20 = 20 : a
10a = 20 x 20
=> a = 40
Hence, third proportional of 10 & 20 is 40.
Numbers of students in section A = x
? Numbers of students in section B and C = (100 ? x)
? x 70 + (100 ? x) 87.5 = 84 100
=> 70x + 87.5 100 ? 87.5x = 8400
=> 8750 ? 17.5x = 8400
=> 17.5x = 8750 ? 8400 => x = 20.
Let x, x+2, x+4, x+6, x+8 and x+10 are six consecutive odd numbers.
Given that their average is 52
Then, x + x+2 + x+4 + x+6 + x+8 + x+10 = 52×6
6x + 30 = 312
x = 47
So Product = 47 × 57 = 2679
You could solve this by drawing a Venn diagram. A simpler way is to realize that you can subtract the number of students taking both languages from the numbers taking French to find the number taking only French. Likewise find those taking only German. Then we have:Total = only French + only German + both + neither
78 = (41-9) + (22-9) + 9 + neither.
Not enrolled students = 24
Let their prices be 3x, 5x and 7x.
Then, 3x + 5x + 7x = (15000 * 3) or x = 3000.
Cost of cheapest item = 3x = Rs. 9000.
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