Simplification

1.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 ?

1. 45
2. 60
3. 75
4. 90

Answer:Option 1

Explanation:

Solution:

Let number of notes of each denomination be x.
Then x + 5x + 10x = 480
⇒ 16x = 480
∴ x = 30.
Hence, total number of notes = 3x = 90.

2.There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is:

1. 20
2. 80
3. 100
4. 200

Answer:Option 1

Explanation:

Solution:

Let the number of students in rooms A and B be x and y respectively.
Then, x - 10 = y + 10 ⇒ x - y = 20 .... (i)
and x + 20 = 2(y - 20) ⇒ x - 2y = -60 .... (ii)
Solving (i) and (ii) we get: x = 100 , y = 80.
∴ The required answer A = 100.

3.The price of 10 chairs is equal to that of 4 tables. The price of 15 chairs and 2 tables together is Rs. 4000. The total price of 12 chairs and 3 tables is:

1. Rs. 3500
2. Rs. 3750
3. Rs. 3840
4. Rs. 3900

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Let the cost of a chair and that of table \\ be R s . x and R s . y respectively . \\ Then, 10 x = 4 y o r y = \frac{5}{2} x \\ ∴ 15 x + 2 y = 4000 \\ \Rightarrow 15 x + 2 \times \frac{5}{2} x = 4000 \\ \Rightarrow 20 x = 4000 \\ ∴ x = 200 \\ So, y = (\frac{5}{2} \times 200) = 500 \\ Hence, the cost of 12 chairs and 3 tables \\ = 12 x + 3 y \\ = R s . (2400 + 1500) \\ = R s . 3900 \end{aligned}

$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{cost}}\,{\text{of}}\,{\text{a}}\,{\text{chair}}\,{\text{and}}\,{\text{that}}\,{\text{of}}\,\,{\text{table}} \cr & \,{\text{be}}\,Rs.\,x\,{\text{and}}\,Rs.\,y\,{\text{respectively}}. \cr & {\text{Then}},\,10x = 4y\,\,\,or\,\,\,y = \frac{5}{2}x \cr & \therefore 15x + 2y = 4000 \cr & \Rightarrow 15x + 2 \times \frac{5}{2}x = 4000 \cr & \Rightarrow 20x = 4000 \cr & \therefore x = 200 \cr & {\text{So}},\,y = \left( {\frac{5}{2} \times 200} \right) = 500 \cr & {\text{Hence,}}\,{\text{the}}\,{\text{cost}}\,{\text{of}}\,{\text{12}}\,{\text{chairs}}\,{\text{and}}\,{\text{3}}\,{\text{tables}} \cr & = 12x + 3y \cr & = Rs.\,\left( {2400 + 1500} \right) \cr & = Rs.\,3900 \cr}$

4.The price of 2 sarees and 4 shirts is Rs. 1600. With the same money one can buy 1 saree and 6 shirts. If one wants to buy 12 shirts, how much shall he have to pay ?

1. Rs. 1200
2. Rs. 2400
3. Rs. 4800
4. Cannot be determined

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Let the price of a saree and a shirt \\ be R s . x and R s . y respectively . \\ Then, 2 x + 4 y = 1600 . . . . (i) \\ and x + 6 y = 1600 . . . . (i i) \\ Divide equation (i) b y 2, \\ we get the below equation . \\ => x + 2 y = 800 - - - (i i i) \\ Now subtract (i i i) from (i i) \\ x + 6 y = 1600 (-) \\ x + 2 y = 800 \\ - - - - - - - - - - \\ 4 y = 800 \\ - - - - - - - - - - \\ Therefore, y = 200 \\ Nowapply value of y in (i i i) \\ => x + 2 \times 200 = 800 \\ => x + 400 = 800 \\ Therefore x = 400 \\ Solving (i) and (i i) we get \\ x = 400, y = 200 \\ ∴ Cost of 12 shirts \\ = R s . (12 \times 200) \\ = R s . 2400 \end{aligned}

$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{price}}\,{\text{of a}}\,{\text{saree}}\,{\text{and}}\,{\text{a}}\,{\text{shirt}} \cr & \,{\text{be}}\,Rs.\,x\,{\text{and}}\,Rs.\,y\,{\text{respectively}}. \cr & {\text{Then,}}\,2x + 4y = 1600\,....(i) \cr & \,\,\,\,\,\,\,\,\,\,\,{\text{and}}\,x + 6y = 1600\,....(ii) \cr & {\text{Divide}}\,{\text{equation}}\,(i)\,by\,2, \cr & {\text{we}}\,{\text{get}}\,{\text{the}}\,{\text{below}}\,{\text{equation}}. \cr & = > \,x + 2y = 800\,\, - - - (iii) \cr & {\text{Now}}\,{\text{subtract}}\,(iii)\,{\text{from}}\,(ii) \cr & x\,\, + \,\,\,\,6y\,\, = \,\,1600\,\,\,( - ) \cr & x\,\, + \,\,\,\,2y\,\, = \,\,\,\,\,\,\,800 \cr & - - - - - - - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4y\,\, = \,\,\,\,\,\,800 \cr & - - - - - - - - - - \cr & {\text{Therefore}},\,y = 200 \cr & {\text{Nowapply}}\,{\text{value}}\,{\text{of}}\,y\,{\text{in}}\,(iii) \cr & = > x + 2 \times 200 = 800 \cr & = > x + 400 = 800 \cr & {\text{Therefore}}\,x = 400 \cr & {\text{Solving}}\,(i)\,{\text{and}}\,(ii)\,{\text{we}}\,{\text{get}}\, \cr & x = 400,\,y = 200 \cr & \therefore {\text{Cost}}\,{\text{of}}\,{\text{12}}\,{\text{shirts}} \cr & = Rs.\,\left( {12 \times 200} \right) \cr & = Rs.\,2400 \cr}$

5.A sum of Rs. 1360 has been divided among A, B and C such that A gets ²/₃ of what B gets and B gets ¹/₄ of what C gets. B's share is:

1. Rs. 120
2. Rs. 160
3. Rs. 240
4. Rs. 300

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Let C^{'} s share = R s . x \\ T h e n, B^{'} s share = R s . \frac{x}{4}, \\ A^{'} s share = R s . (\frac{2}{3} \times \frac{x}{4}) = R s . \frac{x}{6} \\ ∴ \frac{x}{6} + \frac{x}{4} + x = 1360 \\ \Rightarrow \frac{17 x}{12} = 1360 \\ \Rightarrow x = \frac{1360 \times 12}{17} = R s . 960 \\ Hence, B^{'} s share = R s . (\frac{960}{4}) \\ = R s . 240 \end{aligned}

$\eqalign{ & {\text{Let}}\,C's\,{\text{share}} = Rs.\,x \cr & Then,\,B's\,{\text{share}} = \,Rs.\,\frac{x}{4}, \cr & A's\,{\text{share}} = Rs.\,\left( {\frac{2}{3} \times \frac{x}{4}} \right) = Rs.\,\frac{x}{6} \cr & \therefore \frac{x}{6} + \frac{x}{4} + x = 1360 \cr & \Rightarrow \frac{{17x}}{{12}} = 1360 \cr & \Rightarrow x = \frac{{1360 \times 12}}{{17}} = Rs.\,960 \cr & {\text{Hence}},\,B's\,{\text{share}} = Rs.\,\left( {\frac{{960}}{4}} \right) \cr & = Rs.\,240 \cr}$

6.One-third of Rahul's savings in National Savings Certificate is equal to one-half of his savings in Public Provident Fund. If he has Rs. 1,50,000 as total savings, how much has he saved in Public Provident Fund ?

1. Rs. 30,000
2. Rs. 50,000
3. Rs. 60,000
4. Rs. 90,000

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Let saving in N .S .C and P .P .F . be \\ R s . x and R s . (150000 - x) respectively . \\ Then, \frac{1}{3} x = \frac{1}{2} (150000 - x) \\ \Rightarrow \frac{x}{3} + \frac{x}{2} = 75000 \\ \Rightarrow \frac{5 x}{6} = 75000 \\ \Rightarrow x = \frac{75000 \times 6}{5} = 90000 \\ ∴ Savings in Public Provident Fund \\ = R s . (150000 - 90000) \\ = R s . 60000 \end{aligned}

$\eqalign{ & {\text{Let}}\,{\text{saving}}\,{\text{in}}\,{\text{N}}{\text{.S}}{\text{.C}}\,{\text{and}}\,{\text{P}}{\text{.P}}{\text{.F}}{\text{.}}\,{\text{be}}\, \cr & Rs.\,x\,{\text{and}}\,Rs.\,\left( {150000 - x} \right){\text{respectively}}. \cr & {\text{Then}},\,\frac{1}{3}x = \frac{1}{2}\left( {150000 - x} \right) \cr & \Rightarrow \frac{x}{3} + \frac{x}{2} = 75000 \cr & \Rightarrow \frac{{5x}}{6} = 75000 \cr & \Rightarrow x = \frac{{75000 \times 6}}{5} = 90000 \cr & \therefore {\text{Savings}}\,{\text{in}}\,{\text{Public}}\,{\text{Provident}}\,{\text{Fund}} \cr & = Rs.\,\left( {150000 - 90000} \right) \cr & = Rs.\,60000 \cr}$

7.A fires 5 shots to B's 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed:

1. 30 birds
2. 60 birds
3. 72 birds
4. 90 birds

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Let the total number of shots be x . \\ Then, Shots fired by A = \frac{5}{8} x \\ Shots fired by B = \frac{3}{8} x \\ Killing shots by A \\ = \frac{1}{3} of \frac{5}{8} x = \frac{5}{24} x \\ Shots missed by B \\ = \frac{1}{2} of \frac{3}{8} x = \frac{3}{16} x \\ ∴ \frac{3 x}{16} = 27 or x = (\frac{27 \times 16}{3}) = 144 \\ Birds killed by A \\ = \frac{5 x}{24} = (\frac{5}{24} \times 144) = 30 \end{aligned}

$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{total}}\,{\text{number}}\,{\text{of}}\,{\text{shots}}\,{\text{be}}\,x. \cr & {\text{Then,}}\,{\text{Shots}}\,{\text{fired}}\,{\text{by}}\,{\text{A}} = \frac{5}{8}x \cr & {\text{Shots}}\,{\text{fired}}\,{\text{by}}\,{\text{B}} = \frac{3}{8}x \cr & {\text{Killing}}\,{\text{shots}}\,{\text{by}}\,{\text{A}} \cr & = \frac{1}{3}{\text{of}}\frac{5}{8}x = \frac{5}{{24}}x \cr & {\text{Shots}}\,{\text{missed}}\,{\text{by}}\,{\text{B}} \cr & = \frac{1}{2}{\text{of}}\,\frac{3}{8}x = \frac{3}{{16}}x \cr & \therefore \frac{{3x}}{{16}} = 27\,{\text{or}}\,x = \left( {\frac{{27 \times 16}}{3}} \right) = 144 \cr & {\text{Birds}}\,{\text{killed}}\,{\text{by}}\,{\text{A}} \cr & = \frac{{5x}}{{24}} = \left( {\frac{5}{{24}} \times 144} \right) = 30 \cr}$

8.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 ?

1. 10
2. 35
3. 62.5
4. Cannot be determined

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Let the capacity of 1 bucket = x \\ Then, the capacity of tank = 25 k \\ New capacity of bucket = \frac{2}{5} x \\ ∴ Required number of buckets \\ = \frac{25 x}{(2 x / 5)} \\ = (25 x \times \frac{5}{2 x}) \\ = \frac{125}{2} \\ = 62.5 \end{aligned}

$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{capacity}}\,{\text{of}}\,{\text{1}}\,{\text{bucket}} = x \cr & {\text{Then,}}\,{\text{the}}\,{\text{capacity}}\,{\text{of}}\,{\text{tank}} = 25k \cr & {\text{New}}\,{\text{capacity}}\,{\text{of}}\,{\text{bucket}} = \frac{2}{5}x \cr & \therefore {\text{Required}}\,{\text{number}}\,{\text{of}}\,{\text{buckets}} \cr & = \frac{{25x}}{{\left( {2x/5} \right)}} \cr & = \left( {25x \times \frac{5}{{2x}}} \right) \cr & = \frac{{125}}{2} \cr & = 62.5 \cr}$

9.In a regular week, there are 5 working days and for each day, the working hours are 8. A man gets Rs. 2.40 per hour for regular work and Rs. 3.20 per hours for overtime. If he earns Rs. 432 in 4 weeks, then how many hours does he work for ?

1. 160
2. 175
3. 180
4. 195

Answer:Option 1

Explanation:

Solution:

Suppose the man works overtime for x hours.
Now, working hours in 4 weeks = (5 * 8 * 4) = 160.
∴ 160 * 2.40 + x * 3.20 = 432
⇒ 3.20x = 432 - 384 = 48
⇒ x = 15.
Hence, total hours of work = (160 + 15) = 175.

10.Free notebooks were distributed equally among children of a class. The number of notebooks each child got was one-eighth of the number of children. Had the number of children been half, each child would have got 16 notebooks. Total how many notebooks were distributed ?

1. 256
2. 432
3. 512
4. 640

Answer:Option 1

Explanation:

Solution:

Explaination 1

\begin{aligned} Let number of children = n \\ Then, number of books each child will get = \frac{n}{8} \\ Total books distributed = n \times \frac{n}{8} = \frac{n^{2}}{8} \\ If the number children = \frac{n}{2}, \\ number of books each child will get = 16 \\ Total books distributed = \frac{n}{2} \times 16 = 8 n \\ ∴ \frac{n^{2}}{8} = 8 n \\ \Rightarrow \frac{n}{8} = 8 \\ \Rightarrow n = 64 \\ Total number of books distributed \\ = 8 n = 8 \times 64 = 512 \end{aligned}

$\eqalign{ & {\text{Let number of children}} = n \cr & {\text{Then, number of books each child will get }} = \frac{n}{8} \cr & {\text{Total books distributed}} = n \times \frac{n}{8} = \frac{{{n^2}}}{8} \cr & {\text{If the number children}} = \frac{n}{2}, \cr & {\text{number of books each child will get}} = 16 \cr & {\text{Total books distributed }} = \frac{n}{2} \times 16 = 8n{\text{ }} \cr & \therefore \frac{{{n^2}}}{8} = 8n \cr & \Rightarrow \frac{n}{8} = 8 \cr & \Rightarrow n = 64 \cr & {\text{Total number of books distributed}} \cr & = 8n = 8 \times 64 = 512 \cr}$

Explaination 2
If number of children was half, each child would have got 16 books.
Therefore, actually each child got

\frac{16}{2}

$\frac{{16}}{2}$ = 8 Books And the number of children is 8 × 8 = 64
Hence, total number of books distributed = 64 × 8 = 512

Explaination 3
Let number of children

= n

$= n$
Then, number of books each child will get

= \frac{n}{8}

$= \frac{n}{8}$
If the number children

= \frac{n}{2}

$= \frac{n}{2}$ ,
number of books each child will get

= 16

$= 16$
More children, less books (indirect proportion). Therefore,

\begin{aligned} n : \frac{n}{2} = 16 : \frac{n}{8} \\ \Rightarrow \frac{n^{2}}{8} = 8 n \\ \Rightarrow \frac{n}{8} = 8 \\ \Rightarrow n = 8 \end{aligned}

$\eqalign{ & n : \frac{n}{2} = 16 : \frac{n}{8} \cr & \Rightarrow \frac{{{n^2}}}{8} = 8n \cr & \Rightarrow \frac{n}{8} = 8 \cr & \Rightarrow n = 8 \cr}$
Therefore, total number of books distributed
=8n = 8×64 = 512