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HCF and LCM

HCF means highest common factor or greatest common divisor.

LCM means the least common multiple.

HCF of two or more than two number can be find by prime factorization method.

Whereas the LCM is the multiplication of common factors as well as the which are not 
common.  

     1) Find HCF of 12, 18 and 27

Solution:- by prime factorization method
12=3*2*2
18=3*3*2
27=3*3*3
HCF= factors which are common in all the three number
HCF=3

But the method is not applicable in case of having large numbers.
So there is a method known as ‘long division method’ to find the HCF of large numbers.

 Method:-

1: find the smallest and biggest number from the given number.

2: consider the smallest number as the divisor and the biggest number as dividend.

3: divide them and get the remainder.

4: check whether the remainder is zero or not.

Case 1: if the remainder is zero then stop the procedure. The divisor of the step having remainder zero is the required HCF.

Case 2: if the remainder is not zero then consider this remainder as the divisor and the divisor of the previous step as the dividend and follow the step 3 & 4.

     2)Find the HCF of 3442 and 4267

Solution:- consider the smallest number as the divisor and the large number as the dividend.

3442) 4268 (1
         -3442
         ͞   ͞   ͞   ͞   ͞   ͞   ͞   ͞   ͞   ͞   ͞
             826) 3442 (4
                    -3304
                     ͞   ͞   ͞   ͞   ͞   ͞   ͞   ͞   ͞   ͞   ͞   ͞   ͞ 
                      138) 826 (5
                             -690
                          ͞   ͞   ͞   ͞   ͞   ͞   ͞   ͞   ͞   ͞   ͞   ͞   ͞   ͞
                               136 ) 138 (1
                                       -136
                                            ͞   ͞   ͞   ͞   ͞   ͞   ͞   ͞   ͞   ͞   ͞   ͞
                                           2 ) 136 (68
                                                -136
                                                ͞   ͞   ͞   ͞   ͞
                                                   000

      Hence the HCF=2 

      

     3) Find the LCM of 25 and 60.

Solution:- by prim factorization method

25=5*5
60=5*4*3
LCM=5*5*4*3
LCM=300

But the method is limited. If we have given big or more than 2 number then it would be very difficult to find the LCM.

There is one method which can be used to find the LCM of two or more (big) numbers.

Method:-

1: write down the given numbers in a tabular form.

2: either select a prim number which divides at least two of them or a composite 
 number which divides all the numbers.

3: if you have selected a prime number which divides two of them then simply do the division and write the quotient in the next row and if the numbers which are not divisible by the number ‘which has been selected as the divisor’ then write those numbers as it is.

4: follow the step 2 & 3 again and again till you reach to the step where you don’t find any divisor.

5: now the LCM=product of numbers in the first column and the last row

     

     4)    Find the LCM of 20, 25, 30 & 40

Solution:- write down the given numbers in a tabular form.
_____|__20__|__25__|__30__|__40__|
__5__|__4___|__5___|__6___|__8___|
__2__|__2___|__5___|__3___|__4___|
__2__|__1___|__5___|__3___|__2___|

LCM=5*2*2*1*5*3*2=600

     5)    A farmer has 20 apples plants, 35 orange plants and 75 strawberry plants. He wanted to arrange them in rows such that each row has the same number of trees and all are of same type. So find the minimum number of rows that can be formed?

Solution:-  here we first find the number of plants can be planted in a row
That means we will have to find the HCF of smallest and the biggest number from 20,35 & 75

So we find HCF of 20 & 75 by long division method.

20)75(3
    -60
   ͞   ͞   ͞   ͞
      15) 20 (1
           -15
         ͞   ͞   ͞   ͞   ͞
             5)15 (3
               -15
                ͞   ͞   ͞
                00

Hence the HCF=5

=> 5 plants can be planted in one row

Total number of rows=total number of plants / number of plants in one row
Total number of rows=(20+35+75)/5=26 

6)     There are 5 bells which tolls after each 10 sec, 12 sec, 20sec, 25sec and 40 sec. At what time will they all toll together?

Solution:- here we have to find the LCM

_____|__10__|__12__|__20__|__25__|__40__|
__2__|__5___|__6___|__10__|__25__|__20__|
__2__|__5___|__3___|__5___|__25__|__10__|

LCM=2*2*5*3*5*25*10=75000 sec 

So far we have seen how to find the LCM & HCF of integer number. Now we 

will see how to find the LCM & HCF of fractions.

HCF of fraction= HCF of numbers in numerator / LCM of numbers in denominator

LCM of fraction= LCM of numbers in numerator / HCF of numbers in denominator

1)    Find the LCM of 0.25 and 1.55

Solution:- we can write 0.25 and 1.55 as follows.

25/100=1/4 and 155/100=31/20

LCM of fraction= LCM of numbers in numerator / HCF of numbers in denominator

LCM of fraction= LCM ( 1,31)/ HCF (4, 20)=31/4

2)    Find the HCF of 0.45 and 2.15

Solution:- 0.45=45/100=9/20 and 2.15=215/100=43/20

HCF of fraction= HCF (9,43) / LCM (20,20)=1/20

Note:- it is necessary to write down the given fractions into the simplest form.

                  

Profit and Loss

Profit and Loss  Aptitude basics, practice questions, answers and explanations

Profit=SP-CP
Loss=CP-SP
Profit %= (Profit/CP) *100
Loss %= (Loss/ CP) *100
Discount= MP-SP
Discount %= (discount/ MP) * 100
where SP= Selling Price, CP= Cost Price, MP= Marked Price

Exercise questions

1. A trader makes a profit equal to the selling price of 75 articles when he sold 100 of the articles. What % profit did he make in the transaction?

A) 33.33%     

B) 75%      

C) 300%     

D)150%

Solution:-profit=SP of 75 articals CP=SP of 25 articals and SP=SP of 100 articals

Profit%=(profit/CP) *100=((10-25)/25)*100=300%

2. A merchant buys two articles for Rs.600. He sells one of them at a profit of 20% and the other at a loss of 10% and makes no profit or loss in the end. What is the selling price of the article that he sold at a loss?

A) Rs. 400    

B) Rs.440     

C) Rs. 536.80                

D) Rs. 160

Solution:- Assume that the cost prices be x and y

Total cost price=600=x+y……(1)

Total selling price=120% of x+90% of y

But he get no profit/loss

600=120% of x+90% of y

4x+3y=2000……(2)

Now solve equations (1) and (2) for y

Y=400 Rs.

3. A trader professes to sell his goods at a loss of 10% but weights 900 grams in place of a kg weight. Find his real loss or gain per cent.

A) 2% loss  

B) 0.00% gain

C) 2% gain  

D)None of these

Solution:- Use the following formula

(100+g)*Faulty measure=(100+x)*True measure

Where,

g-overall gain/loss

x-profit or loss (as per CP & SP)

here we have x=-10 (loss) ,g=? ,true measure=1000gm, faulty measure=900gm

now put the given values in to the formula

(100+g)*900=(100-10)*1000

100+g=900/9

G=0.00%

4. If apples are bought at the rate of 30 for Rs.100.  How many apples must be sold for Rs.100 so as to gain 20%?

A) 28    

B)25         

C) 20        

D) 22

Solution:- To gain 20% profit he will have to sell the 30 apples for 120 Rs. Assume that he sells x number of apples for Rs. 100

i.e:- 120% of x=30

120/100*x=30  è x=30*100/120=25

That means to have 20% profit he will sell 25 apples for 100 Rs.

5.Rajiv sold an article for Rs.56 which cost him Rs.x. If he had gained x% on his outlay, what was his cost?

A) Rs. 40      

B) Rs. 45 

C) Rs. 36 

D)Rs. 28

Solution:- Given, SP=56 ,CP=X ,Profit%=X%

we know that

Profit%=(profit/CP)*100

X=[(56-X)/X]*100

X^2+100X-5600=0……..solve for X

X=CP=40

6. One year payment to the servant is Rs. 200 plus one shirt.  The servant leaves after 9 months and receives Rs. 120 and a shirt. Then find the price of the shirt.

A) Rs. 80

B) Rs. 100

C) Rs. 120

D) Cannot be determined

Solution:- Let x be the price of the shirt then

payment for 12 months=200+x

payment for 1 month=(200+x)/12………(1)

but, payment for 9 months=120+x

payment for 1 month=(120+x)/9…………..(2)

from (1) & (2)

(200+x)/12=(120+x)/9

Solve for x and we get

X=120

7. Two merchants sell, each an article for Rs.1000. If Merchant A computes his profit on cost pricwhile Merchant B computes his profit on selling price, they end up making profits of 25% respectively. By how much is the profit made by Merchant B greater than that of Merchant A?

A) Rs.66.67

B) Rs. 50         

C) Rs.125         

D) Rs.200

Solution:- SP1=SP2=1000

Merchant A =CP+Profit=1000

if profit%=25%

èprofit=200…….merchant A

Profit=25% of 1000=250…….merchant B

Difference=250-200=50

8. A merchant marks his goods in such a way that the profit on sale of 50 articles is equal to the selling price of 25 articles. What is his profit margin?

A) 25%         

B)50%         

C) 100%         

D)66.67%

Solution:- here CP=SP of 25 articles and SP=SP of 50 articles èprofit=SP of 25 articles

Hence profit%=100%

9. A merchant marks his goods up by 75% above his cost price. What is the maximum % Amount that he can offer so that he ends up selling at no profit or loss?

A)75%         

B) 46.67%         

C) 300%         

D)42.85%

Solution:-Assume that the CP=100 & SP=175 also assume that he reduces x% amount to get no profit  no loss.

X% of 175=100

 X=400/7

That means he will offer=100-400/7=300/7=42.85%

10. A merchant marks his goods up by 75% above his cost price. What is the maximum % Amount that he can offer so that he ends up selling at no profit or loss?
A)75%         
B) 46.67%         
C) 300%         
D)42.85%

Solution:-Assume that the CP=100 & SP=175 also assume that he reduces x% amount to get no profit  no loss.

X% of 175=100

X=400/7

That means he will offer=100-400/7=300/7=42.85%

11. A dealer buys a product at Rs.1920. he sells at a discount of 20% still he gets teh profit of 20%. what is the selling price?   ( 1 Marks )

A)  534

B)  2300

C)  2304

D) 2403

Solution:- cost price is 1920
now discount part is to confuse you because the actual selling price will never
depend on discount
so 20% of 1920 is 384
and adding the profit to calculate selling price 1920+384=2304

12. Mani sells vegetables and he marks up the prices at 5% above his cost price. Also the weighing stones used by him weigh only 90% of the correct weight. Find his effective percentage of mark-up.   (2 Marks )

A)  15%

B)  16^2/3%

C)  14^1/2%

D)  20%

Solution:- let the cost price be 100 of 1 kg....now he will sell 1 kg in 105 but due to error in weighing stones he will sell only 900 gram in 105 but he has paid 900*(100/1000)=90 rs for 900 grams. net profit=105-90=15 Rs...
percentage=100*(15/90)=16^2/3

13. Two merchants sell an article each for Rs.1000.one of them computes profit as a % of cost price, while the second calculates it incorrectly as a % of selling price. If both of them claim to have made a profit of 10%, who made more profit and by what amount? (1 Mark)

A) second and 9 rs

B) second and 10 rs

C)first and 9 rs

D)first and 10 rs

Solution:- first merchant get profit of 10% from cp=> (cp*110/100)=1000 => cp=Rs.909
so first merchant get profit of Rs.91.
second merchant get profit of 10% from sp=> profit=1000*10/100 =Rs.100
so the profit of second merchant is high and it is more than first merchant by 9Rs.