A tin of oil was frac{4}{5} full when 6 bottl | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the total capacity of the tin be $x$ bottles.Initially,the tin was $\frac{4}{5}$ full,so the amount of oil was $\frac{4}{5}x$.When $6$ bottles

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(D) Let the total capacity of the tin be $x$ bottles.Initially,the tin was $\frac{4}{5}$ full,so the amount of oil was $\frac{4}{5}x$.When $6$ bottles were removed,the amount became $\frac{4}{5}x - 6$.Then,$4$ bottles were added,making the amount $\frac{4}{5}x - 6 + 4 = \frac{4}{5}x - 2$.According to the problem,this final amount is $\frac{3}{4}$ of the total capacity,so $\frac{4}{5}x - 2 = \frac{3}{4}x$.Rearranging the equation: $\frac{4}{5}x - \frac{3}{4}x = 2$.Finding a common denominator $(20)$: $\frac{16x - 15x}{20} = 2$.$\frac{x}{20} = 2$.$x = 40$.Therefore,the tin can contain $40$ bottles of oil.

$A$ tin of oil was $\frac{4}{5}$ full when $6$ bottles of oil were taken out. Again,$4$ bottles of oil were poured into it,and it became $\frac{3}{4}$ full. How many bottles of oil can the tin contain?

Similar Questions

What is the smallest value that must be added to $708$ so that the resultant is a perfect square?

What is the value of $5 \frac{1}{2} + 7 \frac{1}{2} + 11 \frac{1}{2}$?

Calculate the total number of prime factors in the expression $(4)^{11} \times (5)^{5} \times (3)^{2} \times (13)^{2}$.

The value of $999 \frac{995}{999} \times 999$ is

What value will come in place of the question mark (?) in the following question? (You are not expected to calculate the exact value)
$10^{3} \times 100^{3} + 999999999 = 10^{?} + 10^{?}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module