To fill a tank, 25 buckets of water is required. How many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to two-fifth of its present?

**Answer: **C

Let required number of buckets \(=x\)

If capacity of buckets are reduced, more number of buckets are required (indirect proportion).

Therefore,

\(25:x=\dfrac{2}{5}:1\\ \Rightarrow 25=\dfrac{2x}{5}\\ \Rightarrow x=62.5\)

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A man has Rs. 480 in the denominations of one-rupee notes, five-rupee notes and ten-rupee notes. The number of notes of each denomination is equal. What is the total number of notes that he has ?

**Answer: **D

Let number of notes of each denomination be \(x\)

Then \(x + 5x + 10x = 480\)

\(\Rightarrow 16x = 480\)

\(\therefore x = 30\)

Hence, total number of notes \(= 3x = 90\)

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Simplfy: \(b - [b -(a+b) - {b - (b - a+b)} + 2a]\)

**Answer: **D

\(b-[b-(a+b)-{b-(b-a+b)}+2a]\)

\(=b-[b-a-b-{b-(2b-a)}+2a]\)

\(=b-[-a-{b-2b+a}+2a]\)

\(=b-[-a-{-b+a}+2a]\)

\(=b-[-a+b-a+2a]\)

\(=b-[-2a+b+2a]\)

\(=b-b\)

\(=0\)

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Simplify: \(16-[5-(6+2(7-\overline{8-5}))]\)

**Answer: **B

\(16-\left [ 5-\left ( 6+2 \left ( 7-\overline{8-5} \right )\right ) \right ]\\ =16-\left [ 5-\left ( 6+2 \left ( 7-8+5 \right )\right ) \right ]\\ =16-\left [ 5-\left ( 6+2 \times 4 \right ) \right ]\\ =16-\left [-9 \right ]\\ =16+9\\ =25\)

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David gets on the elevator at the 11th floor of a building and rides up at the rate of 57 floors per minute. At the same time, Albert gets on an elevator at the 51st floor of the same building and rides down at the rate of 63 floors per minute. If they continue travelling at these rates, then at which floor will their paths cross ?

**Answer: **C

Suppose their paths cross after x minutes.

Then, \(11 + 57x = 51 - 63x \Leftrightarrow 120x = 40\)

\(x = \dfrac 13\)

Number of floors covered by David in \(\left( \dfrac 13 \right )\) min.

\(= \left( \dfrac {1}{3} \times57 \right )=19\)

So, their paths cross at \((11 +19)\) i.e., 30th floor.

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To fill a tank, 25 buckets of water is required. How many buckets of water will be required to fill the same tank if the capacity of the bucket is reduced to two-fifth of its present ?

**Answer: **C

Let the capacity of 1 bucket \(= x\)

Then, the capacity of tank\( = 25x\)

New capacity of bucket \(=\dfrac {2}{5}x\)

\(\therefore\)Required number of buckets

\(=\dfrac {25x}{\left( \dfrac {2x}{5} \right)}\\
=\left( 25x \times \dfrac {5}{2x} \right)\\
=\dfrac {125}{2}\\
=62.5\)

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\(\dfrac{(723 + 1992)^2-(723-1992)^2}{723×1992}=\text{?}\)

**Answer: **A

\(\dfrac{(723+1992)^2-(723-1992)^2}{723×1992}\\=\dfrac{(a+b)^2-(a-b)^2}{ab}\quad \text{(where } a=723 \text{ and }b=1992)\\=\dfrac{a^2+2ab+b^2-\left(a^2-2ab+b^2\right)}{ab}\\=\dfrac{4ab}{4}=4\)

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The price of 80 apples is equal to that of 120 oranges. The price of 60 apples and 75 oranges together is Rs. 1320. The total price of 25 apples and 40 oranges is:

**Answer: **B

Let price of one apple \(=a\)

price of one orange \(=b\)

Price of 80 apples = Price of 120 oranges

\(\Rightarrow 80a=120b\\ \Rightarrow 2a=3b\\ \Rightarrow b = \dfrac{2a}{3}~\cdots(1)\)

Price of 60 apples + Price of 75 oranges =Rs. 1320

\(\Rightarrow 60a+75b=1320\\ \Rightarrow 4a+5b=88\\ \Rightarrow 4a+\dfrac{5(2a)}{3}=88 \quad \text{[ Because substituted value of } b \text{ from }(1)]\\ \Rightarrow 12a+10a=88 \times 3\\ \Rightarrow 6a+5a=44 \times 3\\ \Rightarrow 11a=44 \times 3\\ \Rightarrow a=4 \times 3=12\)

\(b=\dfrac{2a}{3}=\dfrac{2 \times 12}{3}=8\)

Total price of 25 apples and 40 oranges

\(=25a+40b\\=25\times12+40\times8\\=300+320\\=620\)

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There are 6 working days in a regular week and for each day, the working hours are 10. A man earns Rs. 2.10 per hour for regular work and Rs. 4.20 per hour for overtime. If he earns Rs. 525 in 4 weeks, how many hours did he work?

**Answer: **A

Regular working hours in 4 weeks

\(=4 \times 6 \times 10=240\) hours

Amount earned by working in these regular working hours

\(=240 \times 2.10= \text {Rs. } 504\)

Additional amount earned

\(=525-504=\text{Rs. }21\)

Hours he worked overtime

\(=\dfrac{21}{4.2}=\dfrac{210}{42}=5\) hours

Total hours he worked

\(=240+5=245\)

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