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 ?

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:

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:

Explanation:

Solution:
$\begin{array}{rl}& \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{cost}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{a}\phantom{\rule{thinmathspace}{0ex}}\text{chair}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\text{that}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{table}\\ & \phantom{\rule{thinmathspace}{0ex}}\text{be}\phantom{\rule{thinmathspace}{0ex}}Rs.\phantom{\rule{thinmathspace}{0ex}}x\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}Rs.\phantom{\rule{thinmathspace}{0ex}}y\phantom{\rule{thinmathspace}{0ex}}\text{respectively}.\\ & \text{Then},\phantom{\rule{thinmathspace}{0ex}}10x=4y\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}or\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}y=\frac{5}{2}x\\ & \therefore 15x+2y=4000\\ & ⇒15x+2×\frac{5}{2}x=4000\\ & ⇒20x=4000\\ & \therefore x=200\\ & \text{So},\phantom{\rule{thinmathspace}{0ex}}y=\left(\frac{5}{2}×200\right)=500\\ & \text{Hence,}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{cost}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{12}\phantom{\rule{thinmathspace}{0ex}}\text{chairs}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\text{3}\phantom{\rule{thinmathspace}{0ex}}\text{tables}\\ & =12x+3y\\ & =Rs.\phantom{\rule{thinmathspace}{0ex}}\left(2400+1500\right)\\ & =Rs.\phantom{\rule{thinmathspace}{0ex}}3900\end{array}$

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 ?

Explanation:

Solution:
$\begin{array}{rl}& \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{price}\phantom{\rule{thinmathspace}{0ex}}\text{of a}\phantom{\rule{thinmathspace}{0ex}}\text{saree}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\text{a}\phantom{\rule{thinmathspace}{0ex}}\text{shirt}\\ & \phantom{\rule{thinmathspace}{0ex}}\text{be}\phantom{\rule{thinmathspace}{0ex}}Rs.\phantom{\rule{thinmathspace}{0ex}}x\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}Rs.\phantom{\rule{thinmathspace}{0ex}}y\phantom{\rule{thinmathspace}{0ex}}\text{respectively}.\\ & \text{Then,}\phantom{\rule{thinmathspace}{0ex}}2x+4y=1600\phantom{\rule{thinmathspace}{0ex}}....\left(i\right)\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}x+6y=1600\phantom{\rule{thinmathspace}{0ex}}....\left(ii\right)\\ & \text{Divide}\phantom{\rule{thinmathspace}{0ex}}\text{equation}\phantom{\rule{thinmathspace}{0ex}}\left(i\right)\phantom{\rule{thinmathspace}{0ex}}by\phantom{\rule{thinmathspace}{0ex}}2,\\ & \text{we}\phantom{\rule{thinmathspace}{0ex}}\text{get}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{below}\phantom{\rule{thinmathspace}{0ex}}\text{equation}.\\ & =>\phantom{\rule{thinmathspace}{0ex}}x+2y=800\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}---\left(iii\right)\\ & \text{Now}\phantom{\rule{thinmathspace}{0ex}}\text{subtract}\phantom{\rule{thinmathspace}{0ex}}\left(iii\right)\phantom{\rule{thinmathspace}{0ex}}\text{from}\phantom{\rule{thinmathspace}{0ex}}\left(ii\right)\\ & x\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}6y\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1600\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(-\right)\\ & x\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2y\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}800\\ & ----------\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}4y\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}800\\ & ----------\\ & \text{Therefore},\phantom{\rule{thinmathspace}{0ex}}y=200\\ & \text{Nowapply}\phantom{\rule{thinmathspace}{0ex}}\text{value}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}y\phantom{\rule{thinmathspace}{0ex}}\text{in}\phantom{\rule{thinmathspace}{0ex}}\left(iii\right)\\ & =>x+2×200=800\\ & =>x+400=800\\ & \text{Therefore}\phantom{\rule{thinmathspace}{0ex}}x=400\\ & \text{Solving}\phantom{\rule{thinmathspace}{0ex}}\left(i\right)\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\left(ii\right)\phantom{\rule{thinmathspace}{0ex}}\text{we}\phantom{\rule{thinmathspace}{0ex}}\text{get}\phantom{\rule{thinmathspace}{0ex}}\\ & x=400,\phantom{\rule{thinmathspace}{0ex}}y=200\\ & \therefore \text{Cost}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{12}\phantom{\rule{thinmathspace}{0ex}}\text{shirts}\\ & =Rs.\phantom{\rule{thinmathspace}{0ex}}\left(12×200\right)\\ & =Rs.\phantom{\rule{thinmathspace}{0ex}}2400\end{array}$

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:

Explanation:

Solution:
$\begin{array}{rl}& \text{Let}\phantom{\rule{thinmathspace}{0ex}}{C}^{\prime }s\phantom{\rule{thinmathspace}{0ex}}\text{share}=Rs.\phantom{\rule{thinmathspace}{0ex}}x\\ & Then,\phantom{\rule{thinmathspace}{0ex}}{B}^{\prime }s\phantom{\rule{thinmathspace}{0ex}}\text{share}=\phantom{\rule{thinmathspace}{0ex}}Rs.\phantom{\rule{thinmathspace}{0ex}}\frac{x}{4},\\ & {A}^{\prime }s\phantom{\rule{thinmathspace}{0ex}}\text{share}=Rs.\phantom{\rule{thinmathspace}{0ex}}\left(\frac{2}{3}×\frac{x}{4}\right)=Rs.\phantom{\rule{thinmathspace}{0ex}}\frac{x}{6}\\ & \therefore \frac{x}{6}+\frac{x}{4}+x=1360\\ & ⇒\frac{17x}{12}=1360\\ & ⇒x=\frac{1360×12}{17}=Rs.\phantom{\rule{thinmathspace}{0ex}}960\\ & \text{Hence},\phantom{\rule{thinmathspace}{0ex}}{B}^{\prime }s\phantom{\rule{thinmathspace}{0ex}}\text{share}=Rs.\phantom{\rule{thinmathspace}{0ex}}\left(\frac{960}{4}\right)\\ & =Rs.\phantom{\rule{thinmathspace}{0ex}}240\end{array}$

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 ?

Explanation:

Solution:
$\begin{array}{rl}& \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{saving}\phantom{\rule{thinmathspace}{0ex}}\text{in}\phantom{\rule{thinmathspace}{0ex}}\text{N}\text{.S}\text{.C}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\text{P}\text{.P}\text{.F}\text{.}\phantom{\rule{thinmathspace}{0ex}}\text{be}\phantom{\rule{thinmathspace}{0ex}}\\ & Rs.\phantom{\rule{thinmathspace}{0ex}}x\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}Rs.\phantom{\rule{thinmathspace}{0ex}}\left(150000-x\right)\text{respectively}.\\ & \text{Then},\phantom{\rule{thinmathspace}{0ex}}\frac{1}{3}x=\frac{1}{2}\left(150000-x\right)\\ & ⇒\frac{x}{3}+\frac{x}{2}=75000\\ & ⇒\frac{5x}{6}=75000\\ & ⇒x=\frac{75000×6}{5}=90000\\ & \therefore \text{Savings}\phantom{\rule{thinmathspace}{0ex}}\text{in}\phantom{\rule{thinmathspace}{0ex}}\text{Public}\phantom{\rule{thinmathspace}{0ex}}\text{Provident}\phantom{\rule{thinmathspace}{0ex}}\text{Fund}\\ & =Rs.\phantom{\rule{thinmathspace}{0ex}}\left(150000-90000\right)\\ & =Rs.\phantom{\rule{thinmathspace}{0ex}}60000\end{array}$

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:

Explanation:

Solution:
$\begin{array}{rl}& \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{total}\phantom{\rule{thinmathspace}{0ex}}\text{number}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{shots}\phantom{\rule{thinmathspace}{0ex}}\text{be}\phantom{\rule{thinmathspace}{0ex}}x.\\ & \text{Then,}\phantom{\rule{thinmathspace}{0ex}}\text{Shots}\phantom{\rule{thinmathspace}{0ex}}\text{fired}\phantom{\rule{thinmathspace}{0ex}}\text{by}\phantom{\rule{thinmathspace}{0ex}}\text{A}=\frac{5}{8}x\\ & \text{Shots}\phantom{\rule{thinmathspace}{0ex}}\text{fired}\phantom{\rule{thinmathspace}{0ex}}\text{by}\phantom{\rule{thinmathspace}{0ex}}\text{B}=\frac{3}{8}x\\ & \text{Killing}\phantom{\rule{thinmathspace}{0ex}}\text{shots}\phantom{\rule{thinmathspace}{0ex}}\text{by}\phantom{\rule{thinmathspace}{0ex}}\text{A}\\ & =\frac{1}{3}\text{of}\frac{5}{8}x=\frac{5}{24}x\\ & \text{Shots}\phantom{\rule{thinmathspace}{0ex}}\text{missed}\phantom{\rule{thinmathspace}{0ex}}\text{by}\phantom{\rule{thinmathspace}{0ex}}\text{B}\\ & =\frac{1}{2}\text{of}\phantom{\rule{thinmathspace}{0ex}}\frac{3}{8}x=\frac{3}{16}x\\ & \therefore \frac{3x}{16}=27\phantom{\rule{thinmathspace}{0ex}}\text{or}\phantom{\rule{thinmathspace}{0ex}}x=\left(\frac{27×16}{3}\right)=144\\ & \text{Birds}\phantom{\rule{thinmathspace}{0ex}}\text{killed}\phantom{\rule{thinmathspace}{0ex}}\text{by}\phantom{\rule{thinmathspace}{0ex}}\text{A}\\ & =\frac{5x}{24}=\left(\frac{5}{24}×144\right)=30\end{array}$

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 ?

Explanation:

Solution:
$\begin{array}{rl}& \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{capacity}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{1}\phantom{\rule{thinmathspace}{0ex}}\text{bucket}=x\\ & \text{Then,}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{capacity}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{tank}=25k\\ & \text{New}\phantom{\rule{thinmathspace}{0ex}}\text{capacity}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{bucket}=\frac{2}{5}x\\ & \therefore \text{Required}\phantom{\rule{thinmathspace}{0ex}}\text{number}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{buckets}\\ & =\frac{25x}{\left(2x/5\right)}\\ & =\left(25x×\frac{5}{2x}\right)\\ & =\frac{125}{2}\\ & =62.5\end{array}$

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 ?

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 ?

Explanation:

Solution:
Explaination 1

Explaination 2
If number of children was half, each child would have got 16 books.
Therefore, actually each child got
$\frac{16}{2}$
= 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
$=n$

Then, number of books each child will get
$=\frac{n}{8}$

If the number children
$=\frac{n}{2}$
,
number of books each child will get
$=16$

More children, less books (indirect proportion). Therefore,
$\begin{array}{rl}& n:\frac{n}{2}=16:\frac{n}{8}\\ & ⇒\frac{{n}^{2}}{8}=8n\\ & ⇒\frac{n}{8}=8\\ & ⇒n=8\end{array}$

Therefore, total number of books distributed
=8n = 8×64 = 512

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