Monday 27 March 2017

Maths Capsule : Profit and Loss Short Tricks Part 2

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Important Formulas and Shortcuts to solve Profit and Loss

Today we will be covering a very important topic from the quantitative aptitude section that is - Profit and loss.These formulas and shortcuts will be helpful for your upcoming Exams like SSC CGL & MTS 2017 Exams. If you like it let us know.

Cost Price

The price, at which an article is purchased, is called its cost price, abbreviated as C.P.

Selling Price

The price, at which an article is sold, is called its selling prices, abbreviated as S.P.

Profit/gain = SP – CP
Profit % = Profit/(C P)×100
S P = (100+gain % )/100 ×C P
C P = 100/(100+gain %)×S P

Loss:

If the overall Cost Price exceeds the selling price of the buyer then he is said to have incurred loss.

Loss = C P – S P
Loss % = LOSS/(C P)×100
S P = (100-loss %)/100×C P
C P = 100/(100-loss %)×S P

Profit and Loss Based on Cost Price

To find the percent gain or loss, divide the amount gained or lost by the cost and multiply it by 100.

Example: A toy that cost 80 rupees is sold at a profit of 20 rupees. Find the percent or rate of profit.

Answer:

Gain/cost × 100 = % profit.

20/80 × 100 = 25%. - Answer

To find the loss and the selling price when the cost and the percent loss are given, multiply the cost by the percent and subtract the product from the cost.

Example: A damaged chair that cost Rs.110 was sold at a loss of 10%. Find the loss and the selling price.

Answer:

Cost x percent loss = loss.

110 x 1/10 = 11, loss.

Cost - loss = selling price.

110 - 11 = 99, selling price.

Profit and Loss Based on Selling Price

To find the profit and the cost when the selling price and the percent profit are given, multiply the selling price by the percent profit and subtract the result from the selling price.

Example: A toy is sold for Rs. 6.00 at a profit of 25% of the selling price. Separate this selling price into cost and profit.

Answer :

Selling price x % profit = profit.

Selling price = profit + cost.

6.00 x .25 = 1.50, profit.

6.00 - 1.50 = 4.50, cost.

To find the loss and the cost when the selling price and the percent loss are given, multiply the selling price by the percent loss and subtract the result from the selling price.

Example: At a sale, neckties selling at Rs. 50.00 are sold at a loss of 60% of selling price. What is the loss and the original cost?

Selling price x % loss = loss.

Selling price + loss = cost.

50.00 x .60 = 30.00, loss.

50.00 - 30.00 = 20.00, cost.

To find the selling price when the cost and the percent loss are given, add the percent loss to 100% and divide the cost by this sum.

Example: Socks that cost 7.00 per pair were sold at a loss of 25% of selling price. What was the selling price?

Answer: Cost / (100% + % loss) = selling price.

7.00 / 1.25 = 5.60, selling price.

To find the selling price when the profit and the percent profit are given, or to find the selling price when the loss and the percent loss are given, divide the profit or loss by the percent profit or loss.

Note: This rule should be compared with the one under Profit and Loss Based on Cost. The two rules are exactly similar except that in one case 100% represents cost while in the other case 100% represents selling price.

Example: A kind of tape is selling at a profit of 12% of selling price, equal to 18 per yard. What is the selling price of the tape?

Answer: Profit / % profit = selling price.

18 /.12 = 1.50 selling price.

To find the percent profit or loss, divide the amount gained or lost by the selling price.

Example: A candy bar sells for 1.30 at a profit of 65. What percent of profit on selling price does this represent?

Answer: Gain / selling price = % profit.

65 / 1.30 = .5 or 50% profit.

Mark-up Price

Generally the SP is less than the marked price (MP) the difference MP – SP is known as discount, D.

Discount = M P – S P

Discount %, D% = (Discount) / (M P) ×100

To reduce percent loss on cost to percent loss on selling price, divide percent loss on cost by 100% minus percent loss on cost.

Example: 20% loss on cost is what percent loss on selling price?

Answer:

% loss on cost / (100% - % loss on cost) = % loss on selling price.

0.20 / 80 = .0025 or 25% loss on selling price

To reduce percent loss on selling price to percent loss on cost, divide percent loss on selling price by 100% plus percent loss on selling price.

Example: 20% loss on selling price is what percent loss on cost?

Answer:

% loss on selling price / (100% + % loss on selling price) = % loss on cost.

.20 / 1.20 = .16666 or .16.67% loss on cost.

To reduce percent mark-up (percent profit on cost) to percent profit on selling price, divide percent mark-up by 100% plus percent mark-up.

Example: A coat marked up 60% carries what percent of profit on selling price?

Answer : % profit on cost / ( 100% + % profit on cost ) = % profit on selling price.

.60 / 1.60 = .375 or 37.5% on selling price.

Here we are providing you all the types of questions that have been asked in SSC Exams and How to solve it in an Easy way with Grade Stack methods:-

Type 1:

The cost price of 40 articles is the same as the selling price of 25 articles. Find the gain per cent. (CGL-2012)

(a) 65%

(b) 60%

(d) 75%

Answer: (b) Gain per cent

=(40-25)/25×100

=15/25×100=60%

Grade Stack methods

In Above question We take x = 40 , y = 25

Then Gain % = (x –y) x 100/ y

Type2:

Bananas are bought at the rate of 6 for Rs. 5 and sold at the rate of 5 for Rs. 6. Profit per cent is: (CGL-2004)

(a) 36%

(b) 42%

(d) 48%

Answer : (c) To avoid fraction, let the number of bananas bought

LCM of 5 and 6 = 30

CP of 30 bananas

= 5 x 5 = Rs. 25

SP of 30 Bananas = 6 x 6

= Rs. 36

Profit = Rs. (36-25) = Rs. 11

Profit %

= 11/25×100=44%

Grade Stack Method

[(6 x 6 -5x 5)/ (5 x 5)] x 100 = 44%

Type 3:

A man bought oranges at the rate of 8 for Rs 34 and sold them at the rate of 12 for Rs. 27. How many oranges should be sold to earn a net profit of Rs 45? (CGL-2011)

(a) 90

(b) 100

(d) 150

Answers: (a) Let the man buy 24 (LCM of 8 and 12) oranges.

C.P. of 24 oranges = 34/8 ×24 = Rs. 102

S.P. of 24 oranges = 27/12×24= Rs. 114

Gain = 114 – 102 = Rs. 12

Rs. 12 = 24 oranges

Rs. 45 = 24/12×45= 90 oranges

Type 4:

A shopkeeper earns a profit of 12% on selling a book at 10% discount on printed price. The ratio of the cost price to printed price of the book is ?(CGL-2013)

(a) 45 : 56

(b) 50 : 61

(d) 99 : 125

Answer: (a) C.P. of the book = Rs. x

Printed price = Rs. y

(y×90)/100=x × 112/100

x/y=90/112=45/56

Type 5:

A dealer sold two types of goods for Rs 10,000 each. On one of them, he lost 20% and on the other he gained 20%. His gain or loss per cent in the entire transaction was (CGL-2012)

(a) 2% loss

(b) 2% gain

(d) 4% loss

Answers: (d) Here, S.P. is same, Hence there is always a loss. Loss per cent =(20×20)/100=4%

Gradestack Trick

Loss % = (n^2)/100= (20)^2/100= 4%

Where n= 20

Type 6:

On selling an article for Rs170, a shopkeeper loses 15%. In order to gain 20%, he must sell that article at rupees: (CGL-2013)

(a) 215.50

(b) 212.50

(d) 210

Answer ; (c) C.P. of article = (200×120)/100 = Rs. 240

Type 7:

An article is sold at a loss of 10%. Had it been sold for Rs. 9 more, there would have been a gain of 12 1/2% on it. The cost price of the article is (CGL – 2002)

(a) Rs. 40

(b) Rs. 45

(d) Rs. 35

Answers: (a) Let the cost price of the article = Rs. x

S.P. at 10% loss

= x×90/100= Rs. 9x/10

P. at 12 1/2 % gain

x × (100+12 1/2)/100 = Rs. 225x/200

According to the question

9x/10 + 9 = 225x/200

180x + 1800 = 225x

x = Rs. 40

Type 8:

A sells a suitcase to B at 10% profit. B sells it to C at 30% profit. If C pays Rs 2860 for it, then the price at which a bought it is (CGL-2013)

(a) 1000

(b) 1600

(d) 2500

Answer: (c) If the C.P. of the suitcase for A be Rs. x, then

x ×110/100×130/100=2860

x=(2860×100×100)/(110×130) = Rs. 2000

Type 9:

Arun marks up the computer he is selling by 20% profit and sells them at a discount of 15%. Arun’s net gain percent is

(CGL-2013)

(a) 4

(b) 2

(d) 2.5

Answer (b)

Gradestack method:

r1 = 20 , r2 = 15

Formula = r1 – r2 – (r1 x r2)/100

(20-15-(20×15)/100)

= 20 -18 = 2%

Type10:

A tradesman sold an article at a loss of 20%. If the selling price had been increased by Rs. 100, there would have been a gain of 5%. The cost price of the article was: (CGL-2004)

(a) Rs. 200

(b) Rs. 25

(d) Rs. 250

Answer (c) Let the C.P. of article be Rs. x.

105% of x - 80% of x = Rx. 100

25% of x = Rx. 100

x = Rs. (100×100)/25

= Rs. 400

That’s how Profit and loss questions are solved easily. Here is a short cut quiz for profit and loss to test your practice.

Maths Capsule : Fraction Problem

Quant Short Tricks for Fraction Problems

Hello Readers,

In every competitive exam, you have to compare the given fractions in a number of problems in Data Interpretation and Quantitative Ability. Let us study some of the common methods of identifying out the largest or smallest of a given set of fractions which are useful for Mental Maths.

Type – 1

When the numerators are same and the denominators are different, the fraction with the largest denominator is the smallest.

Have a look at the following example.

Example: Which of the following fractions is the smallest?

(3/5) , (3/7) , (3/13), (3/8)

Here, 13 is the largest denominator, so, (3/13) is the smallest fraction. 5 is the smallest denominator, hence (3/5) is the largest fraction.

Here logic is very simple,

Situation: (i) Assume that you are 5 Children in your family. Your Dad brought an Apple and mom cut it into 5 pieces and distributed among all the children including you. (1/5)

Situation (ii) : Assume that you are 8 Children in your family. Your Dad brought an Apple and mom cut it into 8 pieces and distributed among all the children including you. (1/8)

Type - 2

When the numerators are different and the denominators are same, the fraction with the largest numerator is the largest. Have a look at the following example.

Example: Which of the following fractions is the smallest?

(7/5) , (9/5), (4/5), (11/5)

As 4 is the smallest numerator, the fraction 4/5 is the smallest.

As 11 is the largest numerator, the fraction 11/5 is the largest.

Here too logic is very simple,

Situation 1 : Assume that you are 4 Children in your family. Your Dad brought 8 Apples and mom distributed them among all the children including you. (8/4)

Situation 2 : Assume that you are 4 Children in your family. Your Dad brought 12 Apples and mom distributed them among all the children including you. (12/4)

Type - 3

The fraction with the largest numerator and the smallest denominator is the largest.

Example: Which of the following fractions is the largest?

(19/16), (24/11), (17/13), (21/14), (23/15)

Solution : As 24 is the largest numerator and 11 is the smallest denominator, 24/11 is the largest fraction.

Type - 4

When the numerators of two fractions are unequal, we try and equate them by suitably cancelling factors or by suitably multiplying the numerators. Thereafter we compare the denominators as in TYPE 1. Have a look at the following examples.

Example: Which of the following fractions is the largest?

(64/328), (28/152), (36/176), (49/196)

Solution : 64/328 = 32/164 = 16/82 = 8/41 this is approximately equal to 1/5

Note : In these type of problems, approximate values will be enough. No need to get EXACT values.

25/152 = 14/76 = 7/38 this is approximately equal to 1/5.5

36/176 = 18/88 = 9/44 this is approximately equal to 1/5

49/196 = 7/28 = ¼

As all the numerators are 1 and the least denominator is 4, the fraction 49/196 is the largest

Example: Which of the following fractions is the largest?

(71/181), (214/519), (429/1141)

Solution : (71/181) = (71 X 6) / (181 X 6) = 426/1086

(214/519) = (214 X 2) / (519 X 2) = 428/1038

The numerators are now all ALMOST equal (426, 428 and 429). The smallest denominator is 1038.

So, the largest fraction must be 428/1038 that is 214/519 :)

Type - 5

For a fraction Less than 1 :

If the difference between the numerator and the denominator is same then the fraction with the larger values of numerator and denominator will be the largest. Have a look at the following example.

Example: Which of the following fractions is the largest?

(31/37), (23/29), (17/23), (35/41), (13/19)

Solutions: difference between the numerator and the denominator of each fraction is 6.... So the fraction with the larger numerals i.e., 35/41 is the greatest and the fraction with smaller numerals i.e., 13/19 is the smallest.

Type - 6

For a fraction Greater than 1

If the difference between the numerator and denominator is same, then the fraction with the smaller values will be the largest.

Example: Which of the following fraction is largest ?

(31/27), (43/39), (57/53), (27/23), (29/25)

Solution : As the difference between the numerator and the denominator is same, the fraction with the smaller values i.e., 27/23 is the largest.

Maths Capsule : HCF And LCM.

Study Notes of HCF and LCM and Aptitude Quiz

TIPS FOR SOLVING QUESTIONS RELATED TO HCF and LCM:

Prime number : A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

For example, 2, 3, 5, 7, 11, 13, etc. are prime numbers.

Co-Prime Number: Two numbers are said to be relatively prime, mutually prime, or co-prime to each other when they have no common factor or the only common positive factor of the two numbers is 1.

In other words, two numbers are said to be co-primes if their H.C.F. is 1.

Factors: The numbers are said to be factors of a given number when they exactly divide that number.

Thus, factors of 18 are 1, 2, 3, 6, 9 and 18.

Common Factors: A common factor of two or more numbers is a number which divides each of them exactly.

Thus, each of the numbers - 2, 4 and 8 is a common factor of 8 and 24.

Multiple: When a number is exactly divisible by another number, then the former number is called the multiple of the latter number.

Thus, 45 is a multiple of 1, 3, 5, 9, 15 and 45.

Common Multiple: A common multiple of two or more numbers is a number which is exactly divisible by each of them.

For example, 12, 24 and 36 is a common multiple of 3, 4, 6 and 12.

Prime Factorisation: If a natural number is expressed as the product of prime numbers, then the factorisation of the number is called its prime factorisation.

A prime factorisation of a natural number can be expressed in the exponential form.

For example:

(1) 24 = 2 x 2 x 2 x 3 = 2^3 x 3.

(2) 420 = 2 x 2 x 3 x 5 x 7 = 2^2 x 3 x 5 x 7

Highest Common Factor (H.C.F.) or Greatest Common Divisor (G.C.D.) or Greatest Common Measure (G.C.M.) are synonymous terms:

The H.C.F of two or more than two numbers is the greatest numbers which divides each of them without any remainder.

Methods of finding the H.C.F. of a given set of numbers:

Method I: Prime Factorisation method :

Express each one of the given numbers as the product of prime factors. The product of least powers/index of common prime factors gives H.C.F.

Example I:

Find the H.C.F. of 8 and 14 by Prime Factorisation method?

Solution:

8 = 2 x 2 x 2

14 = 2 x 7

Common factor of 8 and 14 = 2.

Thus, Highest Common Factor (H.C.F.) of 8 and 14 = 2.

Example II:

Find the H.C.F. of 24, 36 and 72 by Prime Factorisation method?

24 = 2 x 2 x 2 x 3

36 = 2 x 2 x 3 x 3

72 = 2 x 2 x 2 x 3 x 3

H.C.F. of 24, 36 and 72 = Product of common factors with least powers/index = 2^2 x 3

Thus, Highest Common Factor (H.C.F.) of 24, 36 and 72 = 12

Method II: Successive Division method :

Divide the larger number by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is the required H.C.F.

Example I:

Find the H.C.F. of 8 and 14 by Successive Division method?

8 | 14 | 1

6 | 8 | 1

2 | 6 | 3

Least Common Multiple (L.C.M.):

L.C.M. of two or more given numbers is the smallest number which is divisible by all the given numbers.

Methods of finding the L.C.M. of a given set of numbers:

Method I: Prime Factorisation method :

Express each one of the given numbers as the product of prime factors. The product of greatest powers/index of common prime factors gives L.C.M.

Example I:

Find the L.C.M. of 8 and 14 by Prime Factorisation method?

Solution:

8 = 2 x 2 x 2

14 = 2 x 7

L.C.M. of 8 and 14 = Product of all the prime factors of each of the given number with greatest index of common prime factors

= 2^3 x 7 = 56.

Thus, L.C.M. of 8 and 14 = 56.

Method II: Division method :

Find the L.C.M. of 8 and 14 by using Division method?

2 | 8, 14

|4, 7

L.C.M. of the given numbers = product of divisors and the remaining numbers = 2 x 4 x 7 = 56.

Other important formula related to H.C.F. and L.C.M.:

(1) H.C.F. of given fractions = H.C.F. of numerator / L.C.M. of denominator

2) L.C.M. of given fractions = L.C.M. of numerator / H.C.F. of denominator

(3) Product of two numbers (First number x Second Number) = H.C.F. X L.C.M.

(4) H.C.F. of a given number always divides its L.C.M.

(5) Largest number which divides x, y, z to leave remainder R in each case = H.C.F. of (x-R), (y-R), (z-R).

(6) Largest number which divides x, y, z to leave same remainder = H.C.F. of (y-x), (z-y), (z-x).

(7) Largest number which divides x, y, z to leave remainder a,b,c = H.C.F. of (x-a), (y-b), (z-c).

(8) Least number which when divided by x, y, z and leaves a remainder R in each case = (L.C.M. of x, y, z) + R

1.The product of two 2 digit numbers is 2028 and their HCF is 13. What are the numbers ?

A. 26, 78

B. 39, 52

C. 13, 156

D. 36, 68

2.Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?

A. 4

B. 10

C. 15

D. 16

3.N is the greatest number which divides 1305, 4665 and 6905 and gives the same remainder in each case. What is the sum of the digits in N?

A. 4

B. 3

C. 6

D. 5

4.Which greatest possible length can be used to measure exactly 15 meter 75 cm, 11 meter 25 cm and 7 meter 65 cm

A. 45cm

B. 255cm

C. 244cm

D. 55cm

5.Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:

A. 4

B. 5

C. 6

D. 8

6.A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?

A. 36 minutes 22 seconds

B. 46 minutes 22 seconds

C. 36 minutes 12 seconds

D. 46 minutes 12 seconds

7.An electronic device makes a beep after every 60 sec. Another device makes a beep after every 62 sec. They beeped together at 10 a.m. The time when they will next make a beep together at the earliest, is

A. 10:28 am

B. 10:30 am

C. 10:31 am

D. None of above

8.What is the LCM of 2/3, 5/6 and 4/9 ?

A. 3/10

B. 3/20

C. 10/3

D. 20/3

9.The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:

A. 1008

B. 1015

C. 1022

D. 1032

10.What is the HCF of 9/10,12/15, 18/35 and 21/40 is :

A. 3/5

B. 252/5

C. 1/280

D. 63/700

ANSWER AND SOLUTION :
1(B)Explanation :
Let the numbers be 13x and 13y (? HCF of the numbers = 13)
13x × 13y = 2028
=> xy = 12
co-primes with product 12 are (1, 12) and (3, 4) (? we need to take only co-primes with product 12. If we take two numbers with product 12, but not co-prime, the HCF will not remain as 13)
Hence the numbers with HCF 13 and product 2028
= (13 × 1, 13 × 12) and (13 × 3, 13 × 4)
= (13, 156) and (39, 52)
Given that the numbers are 2 digit numbers
Hence numbers are 39 and 52

2(D)Explanation:
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds(2 minutes).
In 30 minutes, they will toll together 30/2 + 1 = 16 times.

3(A)Explanation :
If the remainder is same in each case and remainder is not given, HCF of the differences of the numbers is the required greatest number
6905 - 1305 = 5600
6905 - 4665 = 2240
4665 - 1305 = 3360
Hence, the greatest number which divides 1305, 4665 and 6905 and gives the same remainder, N
= HCF of 5600, 2240, 3360
= 1120
Sum of digits in N
= Sum of digits in 1120
= 1 + 1 + 2 + 0
= 4

4(A)Explanation:
Convert first all terms into cm.
i.e. 1575 cm, 1125cm, 765cm.
Now whenever we need to calculate this type of question, we need to find the HCF. HCF of above terms is 255.

5(A)Explanation:
N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

6(D)Explanation :
LCM of 252, 308 and 198 = 2772
Hence they all will be again at the starting point after 2772 seconds
or 46 minutes 12 seconds

7(C)Explanation:
L.C.M. of 60 and 62 seconds is 1860 seconds
1860/60 = 31 minutes
They will beep together at 10:31 a.m.

8(D)Explanation :
LCM for fractions = LCM of Numerators / HCF of Denominators
LCM of 2/3, 5/6 and 4/9=LCM (2, 5, 4) / HCF (3, 6, 9) = 20/3

9(B)Explanation:
Required number = (L.C.M. of 12,16, 18, 21, 28) + 7
= 1008 + 7
= 1015

10(C)Explanation :
HCF for fractions = HCF of Numerators / LCM of Denominators
HCF of 9/10,12/15, 18/35 and 21/40=HCF (9,12,18,21) / LCM (10,15,35,40) = 3/840=1/280

Shiksha

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Monday 27 March 2017

Maths Capsule : Profit and Loss Short Tricks Part 2

Important Formulas and Shortcuts to solve Profit and Loss

Here we are providing you all the types of questions that have been asked in SSC Exams and How to solve it in an Easy way with Grade Stack methods:-

Maths Capsule : Fraction Problem

Quant Short Tricks for Fraction Problems

Maths Capsule : HCF And LCM.

Study Notes of HCF and LCM and Aptitude Quiz

Daily English Capsule Day 22

Other Important Posts

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