**Quantitative Aptitude Study Notes**

**Bank & SSC**

Exam

Exam

**PROFIT AND LOSS**

You know that quantitative

aptitude section is most important in

other competitive exams because if you want good score in bank exam then you

have to score good in maths. In competitive exams the most important thing is

time management, if you know how to manage your time then you can do well in

tricks and formula are comes into action. So continuously we are providing

shortcut tricks on different maths topics.

aptitude section is most important in

**bank**

exams in PO andexams in PO and

**Clerk**and forother competitive exams because if you want good score in bank exam then you

have to score good in maths. In competitive exams the most important thing is

time management, if you know how to manage your time then you can do well in

**Bank Exams.**That’s where maths shortcuttricks and formula are comes into action. So continuously we are providing

shortcut tricks on different maths topics.

The one of the most important topic

in maths is

should know how to calculate profit and loss in very short time. For this here

we are providing shortcut tricks and

in maths.

in maths is

**PROFIT AND LOSS**. Youshould know how to calculate profit and loss in very short time. For this here

we are providing shortcut tricks and

**quicker**

methodto calculatemethod

**profit and loss**in maths.

For

understand this rule very well because it is going to be used in almost all the

questions.

**profit and loss**we use rule of fraction is dominant. We shouldunderstand this rule very well because it is going to be used in almost all the

questions.

If our required value is greater

than the supplied value, we should multiply the supplied value with a fraction

which is more than one. And if our required value is less than the supplied

value, we should multiply the supplied value with a fraction which is less than

one.

than the supplied value, we should multiply the supplied value with a fraction

which is more than one. And if our required value is less than the supplied

value, we should multiply the supplied value with a fraction which is less than

one.

If there is a gain of x%, the

calculating figures would be 100 and (100 + x).

calculating figures would be 100 and (100 + x).

If there is a loss of y%, the

calculating figures would be 100 and (100 – y).

calculating figures would be 100 and (100 – y).

If the required value is more

than the supplied value, our multiplying fractions should be 100 + x / 100

than the supplied value, our multiplying fractions should be 100 + x / 100

Or

100 / 100 – y

(both are greater than 1).

If the required value is less than

the supplied value, our multiplying fractions should be

the supplied value, our multiplying fractions should be

Fractions should be 100 / 100 + x

Or

100 – y / 100

(Both are less than 1).

**PROFIT =**

SELLING PRICE (SP) – COST PRICE (CP)

SELLING PRICE (SP) – COST PRICE (CP)

**LOSS = COST**

PRICE (CP) – SELLING PRICE (SP)

PRICE (CP) – SELLING PRICE (SP)

To find the gain or loss per cent

%

%

The profit or loss is generally

reckoned as so much per cent on the cost.

reckoned as so much per cent on the cost.

**GAIN OR LOSS**

PER CENT % = loss or gain × 100 / CP

PER CENT % = loss or gain × 100 / CP

**CASE**

1:

1:

SIMPLE BASIC QUESTIONS FOR PROFIT

AND LOSS:

AND LOSS:

**Example 1:**A woman buys a toy for 25 Rs and

sells it for 30 Rs. Find her gain per cent?

**Solution:**Gain % = gain × 100 / CP

= 5 × 100 / 25

= 20%

**Example 2:**A girl buys a pen for 25 Rs and

sells it for 20 Rs. Find her loss per cent?

**Solution:**Loss % = loss × 100 / CP

= 5 × 100 / 25

= 20%

**Example 3:**If a man purchases 11 oranges

for 10 Rs and sells 10 oranges for 11 Rs. How much profit or loss does he make?

**Solution:**

Suppose that the person bought 11

× 10 = 110 oranges.

× 10 = 110 oranges.

CP of 110 oranges = 10 × 110 / 11

= 100 Rs.

= 100 Rs.

SP of 110 oranges = 11 × 100 / 10

= 121 Rs.

= 121 Rs.

Profit = 121 Rs – 100 Rs = 21 Rs

And % profit = profit × 100 / CP

= 21 × 100 / 100

=21%

**We can also**

use shortcut method or quicker method for this type of questions:

use shortcut method or quicker method for this type of questions:

**Quicker**

method:rewrite the

method:

statements as follows:

Purchase 11 oranges for 10 Rs.

Sell 10 oranges for 11 Rs.

Now, percentage profit and loss

is given by:

is given by:

11 * 11 – 10 * 10 × 100

10 * 10

= 21%

Since the sign is +ve, there is a

gain of 21%.

gain of 21%.

The above form of structural

adjustment should be remembered. The first line deals with purchase whereas the

second line deals with sales. Once you get familiar with the form, you need to

write only the figures and not the letters.

adjustment should be remembered. The first line deals with purchase whereas the

second line deals with sales. Once you get familiar with the form, you need to

write only the figures and not the letters.

**Example 4:**A man purchases 8 pens for 9 Rs

and sells 9 pens for rupees 8. How much profit or loss does he make?

**Solution:**we will solve this questions

quicker trick:

Purchases 8 pens for 9 Rs

Sells 9 pens for 8 Rs

% profit or loss = 8 × 8 – 9 × 9 ×

100

100

9

× 9

× 9

= -1700/81

= -20.98%

Since the sign is –ve, there is a

loss of 20.98%.

loss of 20.98%.

**CASE**

2:

2:

**Example 5:**A dishonest dealer professes to

sell his goods at cost price, but he uses a weight of 960 gm for the kg weight.

Find his gain per cent.

Solution: suppose goods cost the

dealer Re 1 per kg. He sells for Re 1 what cost him Re 0.96.

dealer Re 1 per kg. He sells for Re 1 what cost him Re 0.96.

Gain on Re 0.96 = Re 1 – Re 0.96

= Re 0.04

= Re 0.04

Gain on Rs 100 = 0.04 × 100 /

0.96

0.96

= Rs 25 / 6

Gain % = 25/6%

We can also solve this question

by using direct formula, it will save your time in the exam.

by using direct formula, it will save your time in the exam.

**Direct**

formula:

formula:

**% Gain = ERROR × 100**

True value – Error

True value – Error

**Or**

**% Gain**

= true weight – false weight × 100

= true weight – false weight × 100

**False**

weight

weight

= 40 × 100 / 1000 – 40

=25/6%

**CASE**

3:

3:

In the profit and loss topic the

next example on;

next example on;

**TO FIND THE**

SELLING PRICE:

SELLING PRICE:

**Example 6:**A man bought a cycle for Rs 250.

For how much should he sell it so as to gain 10%?

**Solution:**

If CP is Rs 100, the SP is Rs

110.

110.

If CP is Rs 1, the SP is Rs 110 /

100.

100.

If CP is Rs 250, the SP is Rs

110×250 / 100

110×250 / 100

= Rs 275.

Another suggested method (by rule

of fraction)

of fraction)

If he wanted to sell the bicycle

at a gain of 10%, the selling price (required value) must be greater than the

cost price (supplied value), so we should multiply Rs 250 with a more than one

value fraction. Since there is a gain, our calculating figures should be 100

and (100+10) and the fraction should be 110/100.

at a gain of 10%, the selling price (required value) must be greater than the

cost price (supplied value), so we should multiply Rs 250 with a more than one

value fraction. Since there is a gain, our calculating figures should be 100

and (100+10) and the fraction should be 110/100.

Thus, selling price = 250 ×

110/100

110/100

= Rs 275.

**CASE**

4:

4:

**TO FIND THE**

COST PRICE:

COST PRICE:

**Example 7:**By selling goods for Rs 352.88, I lost 12%. Find

the cost price?

**Solution:**CP should be more than SP; so we

multiply SP by

100 / 100 – 12 = 100 / 88

(a fraction which is more than

one)

one)

CP = 352.88 × 100 / 88 = Rs 401.