Question
Easy
If $B_{1}$, $B_{2}$ and $B_{3}$ are three kinds of birds on a tree in the ratio 3 : 7 : 5 and the number of $B_{2}$ birds are more than $B_{1}$ birds by a multiple of 9 and 7 both, then the minimum number of birds on the tree are
1
630
2
238
3
942
4
945
Question Details
Time to Solve: 12
Exam: UPTET
Level/Paper: Paper 2
Chapter: Fractions & Decimals
Topic: Ratio & Proportion
Correct Answer
Option
Explanation
Here's a detailed explanation for why Option 4 is the correct answer and why the other options are incorrect: Step-by-step Derivation of the Correct Answer (Option 4): 1. Represent the number of birds using the given ratio: Let the common multiple for the ratio be $x$. * Number of $B_1$ birds = $3x$ * Number of $B_2$ birds = $7x$ * Number of $B_3$ birds = $5x$ Since the number…Read More
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