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Delhi University MCA Previous Year Questions (PYQs)
Delhi University MCA Time And Work PYQ
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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2019 PYQ
Solution
Y alone can finish the work in 9 days.
⇒ Y’s 1-day work $= \dfrac{1}{9}$
X works twice as fast as Y.
⇒ X’s 1-day work $= 2 \times \dfrac{1}{9} = \dfrac{2}{9}$
Together, their 1-day work $= \dfrac{2}{9} + \dfrac{1}{9} = \dfrac{3}{9} = \dfrac{1}{3}$
Time taken to complete the whole work $= \dfrac{1}{(1/3)} = 3 \text{ days}$
Answer: $\boxed{3\text{ days}}$ ✅
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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2021 PYQ
Solution
Rates: $A=\frac{1}{15},\; B=\frac{1}{10},\; C=\frac{1}{12}$ (work/day)
Work done by $B+C$ in 4 days: $4\!\left(\frac{1}{10}+\frac{1}{12}\right)=4\cdot\frac{11}{60}=\frac{11}{15}$
Remaining work: $1-\frac{11}{15}=\frac{4}{15}$
Time for $A$: $\dfrac{\frac{4}{15}}{\frac{1}{15}}=4$ days
Answer: $\boxed{4\text{ days}}$ ✅
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Delhi University MCA Previous Year PYQ Delhi University MCA DU MCA 2020 PYQ
Solution
Let rates be $a,b,c$. Given: $a+b=\tfrac{1}{15}$, $b+c=\tfrac{1}{20}$, $a+b+c=\tfrac{1}{10}$.
Using $(a+b)+(a+c)+(b+c)=2(a+b+c)$:
$\tfrac{1}{15}+(a+c)+\tfrac{1}{20}=\tfrac{1}{5}\ \Rightarrow\ a+c=\tfrac{1}{5}-\left(\tfrac{1}{15}+\tfrac{1}{20}\right)=\tfrac{1}{12}$
Time by $A+C = \dfrac{1}{1/12}=12$ days.
Answer: $\boxed{12\text{ days}}$ ✅
Delhi University MCA
Online Test Series,
Information About Examination,
Syllabus, Notification
and More.
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Delhi University MCA
Online Test Series,
Information About Examination,
Syllabus, Notification
and More.
View More
