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Find the number of natural numbers which lie between 101 and 304 and are divisible by 3 or 5. Also, find their sum.

OR

The gate receipts at the show of a film amounted to Rs. 6500 on the first night and showed a drop of Rs. 110 every succeeding night. If the operational expenses of the show are Rs. 1000 a day, then find on which night the show ceases to be profitable.

Asked by Topperlearning User 4th June 2014, 1:23 PM
Answered by Expert
Answer:

Numbers between 101 and 304, which are divisible by 3 are 102,105,……303.

Clearly, it is an AP with first term = a = 102 and common difference = d = 3

Let there be n terms in this AP

Then, nth term = 303

a + (n-1)d=303

102+ (n-1)3=303

n-1=67

n=68 … (1)

Numbers between 101 and 304, which are divisible by 5 are 105,110,……300.

Clearly, it is an AP with first term = a = 105 and common difference = d = 5

Let there be n terms in this AP

Then, nth term = 300

a + (n-1)d=300

105+ (n-1)5=300

n-1=39

n = 40 … (2)

Numbers between 101 and 304, which are divisible by 15 are 105,120,……300.

Clearly, it is an AP with first term = a = 105 and common difference = d = 15

Let there be n terms in this AP

Then, nth term = 300

a + (n-1)d=300

105+ (n-1)15=300

n-1=13

n = 14 … (3)

From (1), (2) and (3), we find that, there are (68+40-14) = 94 terms between 101 and 304 which are divisible by 3 or 5.

Their sum = S68+S40-S14

= 34[102x2+67x3] +20[105x2+39x5] - 7[105x2+13x15]

=34[204+201]+20[210+195]-7[210+195]

= 13770+8100-2835

= 19035

OR

Here, AP (in rupees) is 6500, 6390, 6280, ……

The show will cease to be profitable on the night when the receipts are just Rs 1000. Let it happen on the nth night. It means an1000.

Here, a=6500, d=-110

Now, an=a+(n-1)d =6500-110(n-1)

Equating it with 1000, we get,

6500-110(n-1)=1000

110(n-1)=5500

n-1=50

n=51

Hence, on 51st night the show will cease to be profitable.

Answered by Expert 4th June 2014, 3:23 PM
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