Let the number of boys = x
From the given data,
=> [21x + 16(18)]/(x+16) = 19
=> 21x - 19x = 19(16) - 16(18)
=> 2x = 16
=> x = 8
Therefore, the number of boys in the class = 8.
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
Age of the teacher = (37 * 15 - 36 * 14) years = 51 years.
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.
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
Average = (76 + 65 + 82 + 67 + 85 )/ 5 = 375/5 = 75.
Let their prices be 3x, 5x and 7x.
Then, 3x + 5x + 7x = (15000 * 3) or x = 3000.
Cost of cheapest item = 3x = Rs. 9000.
Let the expenditure on the 95 students be 'x'
Then,
95x + 500 = 120(x ? 5)
95x + 500 = 120x - 600
120x - 95x = 1100
25x = 1100
=> x = 44
Therefore, total expenditure of the hostel becomes
95 x 44 + 500 = 4680.
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