Simplification Questions and Answers

Answer: C

The value is approximately 350.

Round each part to a convenient number:

\[ \begin{aligned} 13.08 \times 24.98 &\approx 13 \times 25 = 325 \\ \sqrt{624} &\approx \sqrt{625} = 25 \\ 325 + 25 &= 350 \end{aligned} \]

The exact value is about 351.7, so 350 is the nearest option. 375 and 380 overestimate the product, and 320 drops the square root term.

In approximation questions the options are usually far enough apart that rounding each term to the nearest convenient value is enough; here the gap between 350 and its neighbours is 25 or more.

Answer: C

Let required number of buckets \(=x\)
If capacity of buckets are reduced, more number of buckets are required (indirect proportion).
Therefore,
\(25:x=\dfrac{2}{5}:1\\ \Rightarrow 25=\dfrac{2x}{5}\\ \Rightarrow x=62.5\)

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 \(= 3x = 90\)

Answer: A

Multiplication and division have equal precedence and are worked left to right.

\[ 5 \times 5 = 25, \qquad \frac{25}{5} = 5 \]

The 5s cancel, leaving 5.

Answer: D

\(b-[b-(a+b)-{b-(b-a+b)}+2a]\)
\(=b-[b-a-b-{b-(2b-a)}+2a]\)
\(=b-[-a-{b-2b+a}+2a]\)
\(=b-[-a-{-b+a}+2a]\)
\(=b-[-a+b-a+2a]\)
\(=b-[-2a+b+2a]\)
\(=b-b\)
\(=0\)

Answer: B

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

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.

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

Answer: A

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

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