Simplification Questions and Answers

Q: A basket had 250 mangoes. \(\frac{1}{10}\) were rotten. Rakesh and his friends had \(\frac{1}{9}\) of the remaining mangoes. What fraction of the total number of mangoes was left ?

The fraction left is \(\frac{4}{5}\).Take the rotten mangoes out first, then the share eaten from what remained:\[ \begin{aligned} \text{rotten} &= \tfrac{1}{10} \times 250 = 25 \\ \text{eaten} &= \tfrac{1}{9} \times 225 = 25 \\ \text{left} &= 250 - 25 - 25 = 200 \end{aligned} \]So \(\frac{200}{250} = \frac{4}{5}\) of the mangoes were left. \(\frac{5}{4}\) and \(\frac{6}{4}\) are greater than 1, which is impossible for a part of the basket, and \(\frac{3}{4}\) would mean 62.5 mangoes had gone.

Q: Rounak had Rs.300. He spent \(\frac{1}{3}\)rd of his money on notebooks and \(\frac{1}{4}\)th of the remainder on stationery items. How much money is left with him ?

Rounak is left with Rs. 150.\[ \begin{aligned} \text{notebooks} &= \tfrac{1}{3} \times 300 = 100 \\ \text{remainder} &= 300 - 100 = 200 \\ \text{stationery} &= \tfrac{1}{4} \times 200 = 50 \\ \text{left} &= 200 - 50 = 150 \end{aligned} \]The trap is taking one quarter of the original Rs. 300 (Rs. 75) instead of the remainder, which gives Rs. 125, not an option. Rs. 200 is what is left after the notebooks alone, Rs. 100 is the amount spent on notebooks, and Rs. 250 matches no step of the working.

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61.

A basket had 250 mangoes. \(\frac{1}{10}\) were rotten. Rakesh and his friends had \(\frac{1}{9}\) of the remaining mangoes. What fraction of the total number of mangoes was left ?

A.\(\frac{4}{5}\)
B.\(\frac{5}{4}\)
C.\(\frac{3}{4}\)
D.\(\frac{6}{4}\)

Exam Year(s)

Tripura Public Service Commission - 2021,

Answer: A

The fraction left is \(\frac{4}{5}\).

Take the rotten mangoes out first, then the share eaten from what remained:

\[ \begin{aligned} \text{rotten} &= \tfrac{1}{10} \times 250 = 25 \\ \text{eaten} &= \tfrac{1}{9} \times 225 = 25 \\ \text{left} &= 250 - 25 - 25 = 200 \end{aligned} \]

So \(\frac{200}{250} = \frac{4}{5}\) of the mangoes were left. \(\frac{5}{4}\) and \(\frac{6}{4}\) are greater than 1, which is impossible for a part of the basket, and \(\frac{3}{4}\) would mean 62.5 mangoes had gone.

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62.

Rounak had Rs.300. He spent \(\frac{1}{3}\)rd of his money on notebooks and \(\frac{1}{4}\)th of the remainder on stationery items. How much money is left with him ?

A.Rs. 250
B.Rs. 200
C.Rs. 150
D.Rs. 100

Exam Year(s)

Tripura Public Service Commission - 2019,

Answer: C

Rounak is left with Rs. 150.

\[ \begin{aligned} \text{notebooks} &= \tfrac{1}{3} \times 300 = 100 \\ \text{remainder} &= 300 - 100 = 200 \\ \text{stationery} &= \tfrac{1}{4} \times 200 = 50 \\ \text{left} &= 200 - 50 = 150 \end{aligned} \]

The trap is taking one quarter of the original Rs. 300 (Rs. 75) instead of the remainder, which gives Rs. 125, not an option. Rs. 200 is what is left after the notebooks alone, Rs. 100 is the amount spent on notebooks, and Rs. 250 matches no step of the working.

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63.

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?

A.63
B.64.5
C.62.5
D.60.5

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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64.

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 ?

A.45
B.60
C.75
D.90

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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65.

5×5/5 = ?

A.5
B.1
C.10
D.25

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.

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66.

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

A.a
B.2a
C.4a
D.0

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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67.

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

A.23
B.25
C.28
D.30

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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68.

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 ?

A.19
B.28
C.30
D.37

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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69.

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 ?

A.10
B.35
C.62.5
D.Can not be determined
E.None of these

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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70.

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

A.4
B.33
C.6
D.1

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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