Quantitative Analysis Part 1

Quantitative Analysis Part 1

Concepts of Ratio & Proportion

• Example 1: The ratio of the ages of A and B is 6 : 5. The difference between the ages of C and A is more than 3 years. The age of D is a prime number between the ages of A and B. The ratio of the ages of B and C is 2 : 3. If the ages of all four are integers, what is the difference between the ages of C and D?
• Solution:

Given ratio of A : B = 6 : 5 and B : C = 2 : 3
So, A : B : C = 12 : 10 : 15
Now let’s use conditions given i.e.
(i) All ages of A, B, C and D should be integers.
(ii) Difference of age between A and C should be more than 3 yrs but from the above ratio difference is only 3 yrs.
So, we have to take A : B : C = 24 : 20 : 30
(iii) age of D lies between A and B and it should be prime number
Numbers between 24 and 20 are = 23, 22, 21 and 23 is the prime number. So, the age of D is 23
Difference of C and D = 30 – 23 = 7yrs

• Example 2: Ten years ago, the ages of Adam and Parker were in the ratio 6 : 7. After six years, Adam's age would be 9.09% less than Parker's age. What would be age of Parker after 9 years?
  Solution:

In percentage topic we have learnt that 9.09% =  1/11, if age of Parker after 6 yrs is X then age of Adam will be X – (1/11)X = (10/11)X hence ratio of Adam and Parker = (10/11)X : X = 10 : 11
                         adam : parker
10yrs ago           6     :     7
after 6yrs          10    :    11
Difference between both these are 16 yrs
So,  4 = 16
1 = 4
hence, age of parker after 6 yrs = 11×4 = 44
age of parker after 9 yrs = 44 + 3 = 47 yrs

The difference between age of father at 5yrs ago and 3 yrs after should be equal to the difference between age of son at 5yrs ago and 3 yrs. To achieve this condition, we manipulate the ratio.
               Father : Son
5yrs ago    21     :    6
3yrs after  25     :   10
now, we can see that 25-21 = 4 and 10-6 = 4
difference between 3yrs after and 5 yrs ago = 8yrs
So, 4 = 8 yrs
      1 = 2 yrs
hence age of father’s 5 yrs ago is 42 yrs and 7yrs ago = 40 yrs.

• Example 3: A stick is broken up into two parts. The ratio of the lengths of the bigger part and the smaller part is equal to the ratio of the lengths of the full stick and the bigger part. What is this ratio?
• Solution: Let the length of a bigger part of the rod  is 1 m and the length of smaller part of the rod is x m, then total length of the rod is (1+x) m

Given, the ratio of bigger to smaller is equal to ratio of the length of full stick to bigger part
So, 1/x = (1+x) /1
x² + x = 1
x² + x – 1 = 0
using quadratics x = (-1 ± √ 5)/2
but x = (-1+√ 5)/2 is the answer, because another will give the negative answer.

• Example 4: The contents of two vessels containing wine and water in the ratio 2 : 3 and 5 : 6 are mixed in the ratio 10 : 7. What would be the ratio of wine and water in the final mixture?
  Solution:

        Wine        Water
vessel 1              2       :        3
vessel 2              5       :        6
to make the total quantity same in both vessel we have to multiply ratio of vessel 1 by (5+6= 11) and vessel 2 by (2+3 = 5)
Now, Vessel 1    22 : 33
          Vessel 2    25 : 30
But, vessel 1 and vessel 2 are  mixed in the ratio of 10 : 7.
So, wine = 22×10+25×7 = 395
  water    = 33×10+30×7 = 540  
It is given that two vessels containing wine and water in the ratio 2 : 3 and 5 : 6 but both vessels have same quantity of total mixture.
So, ratio of wine and water in final mixture = 395 : 540 = 79 : 108
How to express data in a smarter way.
               Wine : water
vessel 1     (2 : 3) × 11 × 10
vessel2      (5 : 6) × 5 × 7
vessel 1     (2 : 3) × 110 = (2 : 3)× 22 = 44 : 66
vessel2      (5 : 6) × 35   = (5 : 6)× 7 = 35 : 42
Hence, ratio will be (44+35) : (66+42) = 79 : 108

Profit & Loss

• Profit, Loss and Discount are yet another crucial topics of Quantitative Aptitude. To help you all prepare for this topic better, we will discuss this topic today. Before going to detail on this topic, we recommend you to go through the following topics -Percentage and Ratio

Example

• Raunak was travelling to Jaipur from New Delhi by train. At the platform, he purchased a novel “Half Girlfriend”. The printed price of the novel was 250. He negotiated with the vendor and asked for the discount of 30%. The vendor didn’t agree with 30% and finally deal ended at 20% discount. He studied the novel in train and after reaching Jaipur, Raunak sold it to Manish at the MRP of the novel. After studying the novel Manish sold it to Navneet after giving some discount Rs.150.

Now, let us have a look at some common questions or terms that often confuse students. 

1. What is the MRP of the novel?
MRP: Marked Retail Price is the price which is printed on an object. So MRP of the novel is 250.

2. What is the discount?
Discount is calculated on MRP, Raunak and vendor both agreed at 20% discount.
So, Discount = 20% of MRP = 20% of 250
   = (20×250)/100 = 50

3. What is the discount% given to Navneet by Manish?
discount% = (Discount/ CP for Manish)× 100
= (100/250)× 100
= 40%

4. What is the Loss for Manish?
Loss = CP for Manish – SP for Manish
Loss = 250 – 150 = 100 Rs.

Now, we will discuss concepts by solving the questions based on this topic. 

• Example 1: A person sells an article for ₹890 at a loss of 11%. What will be the price of the article when sold at a profit of 10%

Solution:

SP₁ = 890 , loss 11%

We know that loss% = [(CP-SP)/CP]×100

11CP =(CP-890)×100

11CP = 100CP – 890×100

89CP = 890×100

CP = 1000

If he had sold it at 10% profit,

then new SP = CP + 10%CP

New SP = 1000+100 = 1100

Approach: Loss is 11% and it is calculated on CP, So we can say loss = 11% CP

New SP = [(100±new profit or loss %)/100]× CP….(1)

Old SP = [(100±old profit or loss%)/100]×CP….(2)

CP = Old SP/ [(100±old profit or loss%)/100]….(3)

Now, putting eqn 3 it in eqn 1.

New SP = Old SP [(100±new profit or loss %)/(100±new profit or loss %)]

Important: Put + sign for profit and – sign for negative.

New SP = 890 [(100+10)/(100-11)] = 1100

• Example 2: A shopkeeper sold an item for Rs.6080 after giving 20% discount on the labelled price and made 18% profit on the cost price. What would have been the percentage profit if he had not given the discount?

Solution:

SP = MP (100-discount)%

6080 = MP(80%)

MP = 7600

Also, SP = CP [(100+profit%)/100]

SP = CP(118/100)

CP = (6080×100)/118

CP = 5125.54

He sold it at MP. So New SP = MP

profit% = [(MP-CP)/CP]×100

= [(7600-5125.24)/5125.24]×100

=47.5%

Another approach:

Reqd. profit% = [(Discount±Profit or loss)/(100-discount%)]× 100

Reqd. profit% = [(20+18)/(100-20)]× 100

Reqd.profit% = 380/8 = 47.5%

• Example 3: A dishonest dealer professes to sell his goods at cost price, but he uses a weight of 900gm of weight for the kg. Find his gain percent.

Solution:

Let price of 1kg is Rs.100 then, price for 900gm will be Rs.90

Hence, he sells 900gm instead of 1kg for Rs.100 but cost price of it is only Rs.90.

So he earns a profit of Rs.10 on Rs.90 not on Rs.100 

So, profit% = (10/90)×100 = 11(1/9)%

You can also use formula:

gain% = [Error/(true value-error)]×100

gain% = [100/(1000-100)]×100 = 100/9 = 11(1/9)%

• Example 4: A dishonest dealer professes to sell his goods at cost price, but he earns the profit of 25%.Find the weight he has used instead of 800gm?

Solution:

Let cost price of 800gm goods is Rs. 100

He sells well at cost price i.e. Rs.100 but earns 25% profit.

So, CP of goods he sold = [SP/(100+profit)]× 100

CP of goods = (100/125)× 100 = 80

100Rs. costs for 800gm

80Rs. costs for (800/100)×80 = 640 gm.

He used 640gm instead of 800gm. 

Simple & Compound Interest

• Let’s discuss the basic difference between Simple Interest and Compound Interest for Principal = 1000, rate of interest (r) = 10%, time = 3yrs

Simple Interest

• SI for 1^st yr = (1000×10×1)/100 = 100,
• SI for 2^nd yr = 100 (In SI it will be the same as 1^st yr)
• SI for 3^rd yr = 100

Compound Interest

• CI for 1^st yr = 100
• CI for 2^nd yr will not be same as 1^st yr because principal for 2^nd yr is the amount of 1^st yr.
• So, CI (2^nd yr) = (1100×10×1)/100 = 110
• CI for 3^rd yr will also not be the same as 1^st yr and 2^nd yr because principal for 3^rd yr is the amount of 2^nd yr.
• principal (3^rd yr) = Amount (2^nd yr) = Principal(2^nd yr)+Interest(2^nd yr) = 1100+110 = 1210
• CI (3^rd yr) = (1210×10×1)/100 = 121

Hence total CI for 3yrs = 100+110+121 = 331

Amount after 3 yrs = 1331 

Interest is always calculated on the Principal. But in case of CI, Principal is get changed every year.

If we calculate it by net rate concept then the Principal will remain same.

Concept 1: How to calculate net CI rate for 2 years?

• Let rate is r% per annum for 2 years
• Net CI rate for 2yrs can be calculated by = 2r+(r²/100)
• If rate is 1%, net CI rate for 2yrs = 2×1+(1²/100) = 2.01%
• If rate is 3%, net CI rate for 2yrs = 2×3+(3²/100) = 6.09%
• If rate is 14%, net CI rate for 2yrs = 2×14+(14²/100) = 29.96%
• We suggest you to learn the table given below:

Concept 2: How to calculate net CI rate for 3 years?

• Let rate is r% per annum for 3 years
• Net CI rate for 3yrs can be calculated = 3r+3(r²/100)+1(r³/10000)
• If rate is 3% p.a., net CI rate for 3 yrs
  = 3×3+3(9/100)+1(27/10000)
  = 9+.27+.0027 = 9.2727
• If rate is 12% p.a., net CI rate for 3 yrs
  = 3×12+3(144/100)+1(1728/10000)
  = 36+4.32+.1728
  = 40.4928
• Representation while calculating net rate %.
• Let’s calculate it for the rate 3% p.a.
  write, r/r²/r³ = 3/9/27
  then,3r/3r²/1r³ = 9/27/27
   = 9.2727
• We suggest you learn the table given below:

Concept 3: If the r% p.a. is a fraction

• Example: if rate is 16(2/3) % and principal is 216, then calculate CI for 2yrs and 3yrs.

Solution: We can write 16(2/3)% = 1/6 (Discussed in percentage study notes)
 For 2 years
216×(1/6)= 36, Now multiply 36 by 2 = 72
36× (1/6) = 6 , multiply 6 by 1 = 6
Add both the above value = 72+6 = 78
CI for 2yrs = 78
For 3 years
216×(1/6) = 36, Now multiply 36 by 3 = 108
36× (1/6) = 6, multiply 6 by 3 = 18
6× (1/6) = 1, multiply 1 by 1 = 1
Add all the above values = (108+18+1)= 127
CI for 3yrs = 127

Concept 4: When r% is given p.a. and CI has to be calculated half-yearly or quarterly basis.

• Example: If r% = 10% per annum. Find the CI on 5000 for 2 years if it is compounded half-yearly.

Solution:

Rate is calculated half yearly so new r% = (10/2)% = 5%
Given time is 2 yrs, acc.to half yearly, it will be 2×2 = 4
Now we have to calculate CI for 4yrs @ 5%
We know 5% = (1/20)
So, 5000×(1/20) = 250, multiply 250 by 4 = 1000
250× (1/20) = 12.5, multiply 12.5 by 6 = 75
12.5× (1/20) = 0.625, multiply 0.625 by 4 = 2.5
0.625× (1/20)= .03125 multiply .03125 by 1 = .03125
Add all the above values
(1000+75+2.5+0.03125)
= 1077.53125

Concept 5: When different rates are given for 2 years.

• If a% is given for 1^st year and b% is given for 2^nd year.
• Net rate of CI for 2 yrs = (a+b+ab/100) % (discussed in percentage study notes)
• Note: The net CI rate will be the same if b% is given for 1^st year and a% is given for 2^nd year.

Example: If principal is 1000 Rs and r(1^st yr) = 4% and r(2^nd yr) = 6%. Calculate the CI after 2yrs.
Solution:

Net CI rate = 4+6+(4×6)/100
= 10.24%
Now CI = 1000×10.24% = 102.4 Rs

Concept 6: When difference between CI and SI is given.

• We know, net CI for 2yrs = 2r+(r²/100) %,net SI for 2 yrs = 2r%
  So, difference = (r²/100)%
• Example: If the difference between CI and SI is Rs.10 and the principal is Rs.1000.Calculate the rate % per annum.

Solution: difference = 10 Rs.

So difference% = (10/1000)×100 = 1%
We know that, if rate of interest is 10%
then, net CI rate (2yrs) = 21%
net SI rate (2yrs) = 20%
difference  = 1%
Definitely we can say r% per annum is 10%.

• Example: A man borrowed Rs.8,400 at 10% p.a. CI. He pays equal annual repayment of X rs and clear off his debts in 2 yrs. What is the value of X?

For 3 yrs: If r% p.a. is given, convert it into fraction(a/b)

• Example: A man borrowed Rs.1820 at 20% p.a. CI. He pays equal annual repayment of X rs and clear off his debts in 3 yrs. What is the value of X?

Annual Instalments for Simple Interest:

Let's discuss a real example to understand  instalment concepts:

• A person deposit Rs.140 to the bank every year up to 5 years . The bank gives him 5% rate of interest simple annually.
• And at the end of 5 years he get total amount of Rs.770
• So, 140 is the instalment, time is 5 years rate of interest is 5% and the amount or debt is Rs.770
• This Instalment is also known as the annual payment. Debt is total amount, so don’t confuse between these two terms.

Installment =   where A = debt, r = rate of interest and t = time period
Example 4: What annual payment will discharge a debt of Rs.848 in 4yrs at 4% per annum simple interest?

• In case if you forget formula then how to approach this question.
• Let installment is X. There are 4 installments and rate of interest is also 4%
  Debt (A) = four installments + (r%) × installments × (0+1+2+… (t-1))
  So, 848 = 4X + (4%)(X)(0+1+2+3)
        848 = 4X+
        848 = 4X+
        848 = 424X/100
        X = 200

Some Important examples based on Simple Interest

• Example 1: A sum amounts to Rs. 702 in 2 years and Rs. 783 in 3 years. Calculate the sum, rate of interest and the amount after 5 years?

Solution:
Amount for 2 years(A₂) = 702
Amount for 3 years (A₃)= 783
Interest for 1 year (I) = 783-702 = 81
So Sum = A₂ – 2I = 702 – 2×81
              = 702-162 = 540
rate of interest = (81/540)×100
                             = 15%
Amount after 5 years = Sum+5I
                                    = 540+ 5×81
                                    = 945

• Example 2: Rs.4000 is divided into two parts such that if one part be invested at 3% and the other at 5%, the annual interest from both the investments is Rs. 144. Find each part.

Solution: Let the amount lent at 3% rate be Rs.X, then amount lent at 5% rate is 4000-X
So, 3% of X + 5% of (4000-X) = 144
   5% of 4000 – 2% of X = 144
  200 – 2% of X = 144
  2% of X = 56
  X = (56/2)×100
 X = 2800
And 4000 -X = 1200.

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