Profit and Loss Questions
Mohit Purchased A watch for 1000rs and then Sold it to Nimesh for 1250rs. Calculate the Profit and Profit %?
Most Simple Question Which You will never get in Any Exam :P( But Basics are Basics we gotta revise it at least)
So What happens here Mohit purchases a Watch( You see word Purchase And You know it's CP) at 1000rs
SO C.P = 1000rs
And then he sells it at 1250Rs( You See the Word SELL Ok that's our SP )
So SP = 1250Rs
Now Profit as I told is nothing but SP - CP So profit = 1250-1000 = 250rs
Now Profit % = [(Profit)/CP*100] So profit % = [(250/1000)*100] = 25%
CASE - 2
Now The Watch That Nimesh purchased for 1250rs Is Sold Again by Her at Rs 625. So what will be Loss and Loss %
Again Usual Stuff
Loss = CP - SP
Loss = 1250 -625 = 625
Loss% = [(loss/CP)*100] = [(625/125)*100] = 50%
So we have a 50% loss here.
Case 3 Inversion case
Profit or loss% is Give and CP or SP is Given and you have to find SP or CP
Steve Sells an article for 1200Rs And he makes a profit of 20% in the Transaction. So What is the Cost price?
I told You once If you don't Know about Something Then assume it as x.
So we take CP = x
Now If i sell an article at 20% profit then what will be our SP in terms of x ?
yeah it's pretty simple 1.2x [ Because is told you percentage to decimal conersion So 20% here is nothing but 0.2x and total SP will be x+ 0.2x = 1.2x remember add in case of profit and subtract in case of loss]
And according to the Question he sold the article at 1200rs
So 1.2x = 1200rs
So x = 1000Rs.
Steve again sells an article for 1200rs but this time he suffers a loss of 20%. What will be the C.P?
Now just Take CP = x
So as i Told S.P will be ? yeah 0.8x ( As i said add in case of profit and subtact in case of loss)
and acoording to the question SP = 1200rs
So 0.8x = 1200
x = 1200/0.8 = 1500rs
So C.P is 1500rs.
If it's given that C.P is 1000rs and profit made is 20%
Then it will be much more simple.
C.P is 1000rs so profit 20% will be 200rs.
So Profit = sp-cp therefore SP = CP + profit = 1000+200=1200rs.
Or If S.P is Given and Also Discount % is given you have to calculate MP
Example S.P = 200rs
Discount % = 80%
Find MP.
Simple Let M.P be X
So S.P after 80% discount will be 0.2x
and according to question S.P = 200
So 0.2x= 200
x = 1000rs = MP
Case 4: Combination Where MP CP SP are Mixed Together.
An article was purchased for Rs. 78,350. Its price was marked up by 30%. It was sold at a discount of 20% on the marked up price. What was the profit percent on the cost price?
Cost price = Rs. 78350
Marked price = 1.3*78350 = 101855( I told in the Start CP is Absolute 100% so If anything is Marked or Sold Above CP by ---% You just have to add that % to 100% and Convert it into it's relative decimal value)
Selling price = 101855*0.8= 81484[ Discount is calculated on MP so here MP is Absolute 100%, 20% discount that Means The product is being Sold at 80% of MP or 0.8MP)
Profit = S.P - CP = 81484 - 78350 = 3134
Profit% = (Profit/CP)*100 = (3134/78350)*100 = 4%
Case 5
A man purchases 11 oranges for 10rs and Sells 10 oranges for 11rs.Find profit or loss%?
1st- Basic method.
Find the C.P of 1 orange that will be 10/11
Find SP of 1 orange that will be 11/10
As it's clear SP>CP hence Profit will be SP-CP = 11/10 - 10/11 = (121 - 100)/110 = 21/110
Profit % = [(profit)/CP]*100 = [(21/110)/(10/11)*100] = [(21*11)/(110*10)*100 = 21%
2nd- LCM method.
Take The LCM The two numbers present in the Question i.e LCM of 10 and 11 = 110.
Now this Is the Amount Of oranges you will buy and sell and calculate the profit % on that.
So CP of 110 Oranges = 100rs
S.P of 110 Oranges = 121rs
Profit = 21Rs
Profit % = 21%[ calculates on the CP of 110 Oranges]
3rd - fastest Method but Confusing
Write The Purchase line First --- 11 oranges for 10
-------------------------------------------------cross multiply-----
Selling Line 2nd 10 oranges for 11
11*11 will be SP and 10*10 will Be CP
Profit % = (11*11 - 10*10)/(10*10)*100 = 21%
Lets See one More Example
A man buys 8 pens for rs 9 and sells 9 pens for rs 8, Find profit or loss%?
By LCM methdo
take lcm of 8 and 9 that will be 72
CP of 72 pens will be 81rs
SP of 72 Pens will be 64Rs
Clearly there is loss which will be equal to 81 - 64 = 17rs
Loss % = (17/81)*100 = 20.98 or 21%
BY Fastest Method.
Write the purchase line 8 pen for 9rs
---------------------------------cross multiply
write sale line 9 pen for 8rs
C.P will be 8*8 = 64
S.P will be 9*9 = 81
Loss% will be (17/81)*100 = 20.98%
Case 6:- Dishonest dealer using false Weight and selling at Cost price.
A Dishonest dealer Professes to Sell the product at cost price but Instead of Selling 1000gm He sells only 900gm for 1Kg Wt
You don't need any Formula to Solve This Type of Question. You just have to use your own Mind here.
Now Look what the dealer is actually Doing here.
Dealer says He sell at Cost price Means He say He Sells at the price he purchases.
Now What amount He says He Sells = 1000gm
What Actually He Sells = 900gm
Now you can See here he is only selling 900gm and he is getting the oney for 1000gm
So this money from 100 Grams is His Profit OK.
Now how we calculate profit % ? We take CP as Base and Divide the Profit by CP.
Now look in this Question he is Selling 900 Gram and getting 100gram As profit.
So profit % will be (100/900)*100 = 11.11%
One More Question
A dishonest dealer Professes to sell the goods at cost price but instead of selling 1000 gms he sells only 800 gms for 1KG WT. Find his Gain% ?
Now Just Remember what He says He is Selling or what he gets paid for, he says he sell 1000gram
But What Actually He sells ? yeah exactly 800grams.
So how much he gets Extra or how much he cheats = 200grams
So profit will be (200/800)*100 = 25%
Case 7: Dishonest Dealer and also Selling Above Cost price.
A dishonest dealer Sells his Good 20% above the cost Price and Also cheats the Customer By giving them only 800gm for 1kg wt. What's his Profit % in the whole transaction.
We have to do the same stuff here Just Imagine. If he sells 1kg then how much will get paid for but also remember that he sells his good above 20% of CP Which means that if he sells 1000gm he gets paid for 1200gm. [ 20% above CP ka matlab yahi hua na ?]
So he gets paid for 1200gms and What he actually Sells here is ? Yes 800gms Only
So Profit will be SP - CP = 1200 - 800 = 400gm
Profit % will be (Profit)/CP*100 = [400/800]*100 = 50%
Case 8: When two Articles are Sold at Same Price but one at profit and one at loss and % profit = % loss.
In Such Cases there will always be a loss (%)which will be equal to [(Common Gain or Loss)]/10]^2
Example - A man Sell two Wrist Watches One at a profit of 20% and one at loss of 20%. The selling price of each watch is 200rs.
i) Find the Percentage of profit or loss.
ii) Net Amount of profit or loss.
i) As i told there will always be a loss in this case And % loss = [(common gain or loss)/100]^100
Now just put the value % loss = [(20/10)]^2 = 4%
Hence Loss % = 4%
ii) Net Amount of Loss
So His Total SP was 200 + 200 = 400rs
He Suffers a loss = 4% Which means he sells his watch at 96% of their value i.e CP
So acording to Question 96% of CP = 400rs
or 0.96CP = 400rs
CP = 400/0.96
CP = 416.6667
So Net Amount of Loss = CP - SP = 416.667 - 400 = 16.667Rs
Case 9 - Goods passing through Successive hands.
It's a Lot like the method i told you yesterday about consecutive increase or decrease.
But Let's just Check it again.
A sells a good a profit of 20% to B and B Sells That Good to C at a profit of 25% If C pays 225 For It. What was it's Cpst Price for A.
So Just Asumme that CP for A was x
So he sells it at 20% profit That means he sells it at 1.2x
Now S.P of A = C.P of B
So Now B sells it to C at 25% profit
That means B sells it at 1.2*1.25*x
Now C pays 225rs
That mean 1.2*1.25*x = 225
so x = 150Rs
Answer CP for A = 150Rs.
Or you can Also use the formula which i told yesterday [x + y + (xy)/100]
Same way you can solve for 3 persons also.
Case 10 - CP of X articles = SP of Y Articles.
Very Simple If you know the trick behind it.
Eg - CP of 25 Articles is Equal to the SP of 20 Articles. Find the Profit or loss %.
Just write it This was 25CP = 20SP
Now Cross multiply So that Variables gets on One side of the equation and Constant gets on the Other Side.
So SP/CP = 25/20
Now you just have to take that Elements on the opposite sides of Equation represents their corresponding value.
So in Equation SP/CP = 25/20. In front of SP the value is 25( So our SP will be 25)
And in front of CP the value is 20( So out CP is 20)
Now You know CP and SP calculating profit or loss is a child's play now but still we have to play it[ Personal Advice Always Believe in complete solution of the question, never leave the question in Mid Way ]
So as SP>CP there is profit
And profit will be [(SP-CP)/CP]*100 = (5/20)*100 = 25%
Another Example
CP of 10 articles is equal to the SP of 12 Articles Find the profit or Loss %?
Do the same stuff again
10CP=12SP
Cross multiply now.
SP/CP = 10/12
So SP = 10
and CP = 12
So clearly there is a loss And loss % = (Loss/CP)*100 = (2/12)*100 = 16.66%
Case 1: Marking Above x% and giving discount of y%, Total profit or loss.
Eg A person Marks his good above 50% of CP and Gives Discount of 20% Find his Profit %.
The easiest way to solve this type of question is to assume the CP as 100
So CP is 100
M.P will be 50% above CP that will be 150
Now he gives discount of 20%
As discount is caculated on MP so SP after deducting the discount will be 150*0.8 = 120
Now SP = 120, CP = 100 So profit % will be 20%.
Case 12 - Decrease in Price of Commodity allows A person to Buy X quantity more of an item.
EG - When the price of sugar decreases by 10%, a man could buy 1 kg more for Rs 270. Then, the
original price and final price of the sugar are ?
Now remember i told you a formula yesterday Which Goes something like this[ (How Much It is decreased)/( What It Becomes after decrease)*100].
So this Question is Implementation of that Formula only.
Price is decreased by 10%.
So Man can purchase how much extra now ? Apply the formula ( How much decrease/ What It becomes )*100 = (10/90)*100 = 100/9 %
So man can buy 100/9 % more sugar now.
Lets Assume that originally He used to buy x kg of sugar
And as it's given in the question He can Buy 1 KG more. So that means that 100/9 % of x = 1kg
(100/900)*x = 1
x = 9kg.
Now Original Quantity = 9kg
So Original Cost = 270/9 = 30Rs/KG
Increased Quantity = 9+1 = 10kg
So Final Price = 270/10 = 27RS/kg
Case 13
A trader allows a discount of 25% on his articles but wants to gain 50% gain. How many times the CP should be marked on the items?
Simple way to solve this Question is By Assuming MP as X and CP as Y.
So Let MP be X, So SP after 25% discount = 0.75x
And He also Wants to Again 50% on CP, So SP in Terms of y will be = 1.5y
Now Both SP are Equal So
0.75x = 1.5y[ Now we have to find MP with respect to CP So express the equation in terms of Y]
x = 2y
or x = 200% of Y
So he Should marks his Goods 100% above the CP.
Case 14:- Successive Discount.
We all used to get Amazed when we heard deals like 50% + 49% discount, I always used to wonder how can they sell their product at 1% price LOL. Then i studied % in class 7th and it became clear to me that's it's another way of looting commom man.
So what actually is Successive Discount.
Successive simply means anything which is applied in succession ( ek ke baad ek apply karna )
So When Pantaloons Say 50% + 30% off It doesn't mean you will get the discount of 80%. If they wanted to give you 80% discount( which they would never do) then they would simply have written 80% instead of 50% + 30%.
For Example You wen to Pantaloons or levis whatever And You Like a jeans Whose MP is 1000rs, and there is a discount of 50% + 30%. So Now You have to apply the 50% discount, By applying that New MP will become 500rs and Now On this 500Rs you will apply the next 30% discount to get the final SP which will be 350Rs.
So Lets See some Examples.
There are 2 Successive discount on Watch Whose MP is 2000rs. the first Discount is of 40% and other is of 20%.
The Good thing with successive discount is that you can apply The discount in any way you want, that means you can apply 20% discount first and if you want you can apply 40% discount first. The answer will remain the same.
So now Lets Apply 40% discount on 2000rs. After applying 40% discount the MP will become 1200rs and On that 1200 we apply another 20% discount So the final SP will become 960Rs.
Now Do the Other Way. First Apply 20% discount on 2000rs So new MP will be 1600 Rs Now apply 40% discount again. And the Final SP will be 960 Rs
You can see the answer is same in both the cases.
But I will tell you simple method Just Multiply It.
I means MP is 2000 You want to apply 40% and 20% Discount Just do it like this was 2000*0.6*0.8 = 960
Sometimes It's Also Asked two successive discount of 30% and 40% is Equal to what Single Discount.
No need to worry Just do the regular Stuff. If MP was 1 after 30% discount it will become 0.7 and after 40% it will be 0.6
So multiply the values.
0.7*0.6 = 0.42.
Now This 0.42 is The Final SP
So total Discount will be equal to 1- 0.42 = 0.58 or 58%
Lets see 1 more example.
What will the Single Equivalent discount for two Successive Discount of 40% and 50%?
Let MP = 1
Now apply discount 0.6*0.5 = 0.3 = SP
So Discount = 1- 0.3 = 0.7 or 70%.
Case-15: Equation Based Question,
it's not a single case many Question can be Made From This case But basic idea is you have to make a mathematical Equation To Solve Such type of Questions.
A trader gets a profit of 25% on an article. If he buys the article at 10% lesser price and sells it for Rs. 2 less, he still gets 25% profit. Find the actual CP of the article.
Let Assume the CP of the article was x. So according to Question The SP must have been 1.25x
Now He buys the article at 10% lesser price that means he buys it at 0.9x
And he sells it 2rs less which means at 1.25x - 2
He will still get 25% profit But This 25% will be calculated on 0.9x because it's the new CP
So 1.25x - 2 - 0.9x = 25% of 0.9x
0.35x -2 = 0.225x
0.35x - 0.225x = 2
0.125x = 2
x = 2/0.125 = 2000/125 = 16
So CP = 16.
Another One
A trader Sells an Article at 25% profit If he had Sold the item at 10 Rs. more the profit would have been 30%. Find The CP?
It's very simple question In this type of question just assume CP as x.
And Convert the % value of Profit into decimal and Then Solve the question Accordingly.
25% of x = 0.25x
and 30% of x= 0.3x
Now in the Question it is said The dealer would get 10rs more if the profit is 30% Or the difference between 25% profit and 30% profit is 10Rs
So 0.3x - 0.25x = 10
0.05x = 10
x = 10/0.05 = 1000/5 = 200
One More Question, A dealer Sells an Article at 20% profit If he had sold the article at 500rs less he would have suffered a loss of 30%. Find CP
Just Take CP as x
so 20% profit will be = 0.2x
30% loss = -0.3x[ remember loss is assigned as negative]
So according to Question the Difference between 20% profit and 30% loss is 500rs
So 0.2x - ( -0.3x) = 500
0.2x + 0.3x = 500
0.5x = 500
x = 500/0.5 = 5000/5 = 1000
Quiz :
Time: (5-6) minutes.
1. Aadesh bought a combined total of 25 monitors and printers. He marked up the monitors by 20% on CP while each printer was marked up by Rs. 2000. He was able to sell 75% of the monitors and 2 printers and make a profit of Rs. 49,000. The remaining monitors and 3 printers could not be sold by him. Find his overall profit or loss if he gets no return on unsold items and it is known that a printer costs 50% of a monitor.
(a) Loss of Rs. 48,500
(b) Loss of Rs. 21,000
(c) Loss of Rs. 41,000
(d) Data Inadequate
(e) None of these
2. A rickshaw dealer buys 30 rickshaws for Rs.4725. Of these, 8 are four seaters and rest are two seaters. At what price must he sell the four seaters so that if he sells the two seaters at 3/4th of this price, he makes a profit of 40% on his outlay.
(a) Rs. 180
(b) Rs. 270
(c) Rs. 360
(d) Rs. 450
(e) None of these
3. Ritesh bought 25 washing machines and microwave ovens for Rs. 2,05,000. He sold 80% of the washing machines and 12 microwaves ovens for a profit of Rs 40,000. Each washing machine was marked up by 20% over cost and each microwave oven was sold at a profit of Rs. 2,000. The remaining washing machines and 3 microwave ovens could not be sold. What is Raghav’s overall profit/loss?
(a) Rs. 1000 profit
(b) Rs. 2500 loss
(c) Rs. 1000 loss
(d) Cannot be determined
(e) None of these.
4. A flat and a piece of land were bought by two friends Tarun and Varun respectively at prices of Rs. 2Lakh and Rs. 2.2 Lakh. The price of the flat rises by 20% every year and that of land by 10% every year. After two years, they decided to exchange their possessions. What is approx. percentage gain of the gainer?
(a) 7.56%
(b) 6.36%
(c) 4.39%
(d) 3.36%
(e) None of these
5. Sunil calculates his profit percentage on the selling price whereas Sujeet calculates his profit on the cost price. They find that the difference of their profits is Rs. 900. If the selling price of both of them are the same, and Sunil gets 50% profit and Sujeet gets 40% profit, then find their selling price.
(a) Rs 4200
(b) Rs 4500
(c) Rs 4000
(d) Rs 4800
(e) None of these
6. A reduction of 10% in the price of salt enables a person to buy 2 kg more for Rs.180. Find the reduced and the original price per kg of salt respectively.
(a) Rs 10, Rs 9
(b) Rs 9, Rs 10
(c) Rs 18, Rs 20
(d) Rs 20, Rs 18
(e) Rs 18, Rs 16.2
7. A person sold his watch for Rs. 24. If the percentage of his loss was equal to the cost price , then the watch would have cost him
(a) Rs. 40
(b) Rs. 60
(c) Rs. 50
(d) Rs. 80
(e) None of these
8. A man buys two horses for Rs. 1550. He sells one so as to lose 23% and other so as to gain 27%. On the whole transaction he neither gains nor loses. What does each horse costs?
(a) 807,743
(b) 817,733
(c) 827,723
(d) 837,713
(e) None of these
9. An orange vendor makes a profit of 20% by selling oranges at a certain price. If he charges Rs. 1.2 higher per orange he would gain 40%. Find the original price at which he sold an orange.
(a) Rs. 3
(b) Rs. 12
(c) Rs. 4.8
(d) Rs. 6.0
(e) None of these
10. After selling a watch, shyam found that he had made a loss of 10%. He also found that had he sold it for Rs.27 more, he would have made a profit of 5%. the actual initial loss was what percentage of the profit earned,had he sold the watch for a 5% profit?
(a) 23%
(b) 150%
(c) 200%
(d) 180%
(e) None of these.
Answers & Explanation;
1. a
2. b
3. c
4. e (8.189 approx. )
5. a
6. b
7. e (Either Rs 40 or Rs 60)
8. d
9. e (Rs. 3.60)
10. c
Explanation:
1.Total Number of printers = 5 (2 sold , 3 unsold)
Monitors = 20.
Profit made on Printers sold = 2000*2 = 4000.
Monitors sold = 20*75% = 15
Profit made on Monitors sold = 49000-4000 = Rs.45000.
Profit made per monitor = 45000/15 = 3000.
20% of CP of Monitor = 3000
CP of Monitor = 15000.
CP of Printer = 7500
Total CP = 15000*20 + 7500*5 = 3,37,500
Total SP = 18000*15 + 9500*2 = 2,89,000
Loss = 48,500
2. Total investement = Rs. 4725
Total SP = 1.4*4725 = 6615
Now, Let the price of 4 seater be x then price of two seater will be .75x.
8x + 22*0.75x = 6615
24.5x= 6615 or x = 270
3. Total number of Microwave ovens = 15 (12 sold +3 unsold)
Hence, Washing machines = 10
He sold 12 ovens and 8 washing machines
Hence, In total he sold 80% of both
Thus, He sells 80% of both at a profit of Rs. 40,000.
Cost of 80% of the goods = 0.8*2,05,000 = 1,64,000
Hence, Total SP = 1,64,000+40,000 = 2,04,000
CP = 2,05,000
Loss = Rs.1000
4. After 2 years :-
Flat would be worth = 2Lakh* 1.2*1.2 = Rs. 288000
Land would be worth= 2.2Lakh*1.1*1.1 = Rs. 266200
Profit of the Gainer = Rs. 21800
Profit % of the gainer = 21800*100/266200= 8.189(approx)
Also if loss% woudd have been asked of the loser
loss% = 21800*100/288000 = 7.56 (approx. )
5. Let SP be Rs. 100
CP for Sunil = => (SP-CP)*100/SP = 50
CP for Sunil ==> (100-CP)*100/100 = 50 or CP = Rs. 50
(Divided by SP as Profit calculated on SP)
Profit for Sunil = 100-50 = Rs 50
Now, CP for Sujeet = (SP-CP)*100/CP = 40
(100-CP)*100= 40CP or CP for Sujeet= Rs. 1000/14
Profit for Sujeet = 100-100/14 = 400/14
Now, Difference in profit when SP is 100 = 50-400/14 = 300/14.
Now, Equating difference and SP, we have
300/14 : 100 : : 900 : SP
SP = 900*100*14/300 = Rs. 4200
6 . Let originally he buy X kg for Rs. 180
Now, he will buy X+2 kg for Rs. 180.
Reduction in original price =10%
(180/X)/kg*90/100 = [180/(X+2)]/kg
90(X+2) = 100X
X = 18
Therefore, Originally he bought 18kg.
Original Price = Rs. 10/kg
Reduced Price = Rs. 9/kg
7. SP = Rs. 24
Let CP be X hence, Loss% = X
(X-SP)*100/X = X or (X-24)*100/X = X
X^2-100X+2400 = 0
(X-60)(X-40) = 0
X= 60 or 40
8. Let CP of one be X and other be Y
X+Y = 1550….(i)
ATQ:-
0.77X + 1.27Y = 1550…(ii) (as no profit and no loss is there)
Solving both, we get
50Y = 35650 or Y = 713
Hence, X = 1550-713 = 837
Therefore, CP of each horse = 837,713
9. Let the CP be Rs. x/ orange
Profit = 20%
SP = Rs. 1.20x
Now, If case :-
SP =x+ Rs.1.2
Profit = 40%
Therefore, we can say
1.40x = x+1.2 or x = Rs. 3
Hence, Original SP = Rs 1.2x = Rs. 3.60/-
10. Profit= 5% (If case )
5% of CP ------> Rs. 27
So, CP = Rs. 540
Now, Loss% = 10
Loss =Rs. 54
Required % = 54*100/27 = 200%
Unit Digits of Squares
First we need to remember unit digits of all squares from 1 to 10. The figure below shows the unit digits of the squares.
Now from the picture we can say that, whenever the unit digit of a number is 9, unit digit of the square root of that number will be definitely 3 or 7. Similarly, this can be applied to other numbers with different unit digits.
How to Find Square Root of a Four Digit Number
Let’s learn how to find square root by taking different examples.
Example: Find the square root of 4489.
We group the last pair of digits, and the rest of the digits together.
Now, since the unit digit of 4489 is 9. So we can say that unit digit of its square root will be either 3 or 7.
Now consider first two digits i.e. 44. Since 44 comes in between the squares of 6 and 7 (i.e. 62 < 44 < 72), so we can definitely say that the ten’s digit of the square root of 4489 will be 6. So far, we can say that the square root will be either 63 or 67.
Now we will find the exact unit digit.
To find the exact unit digit, we consider the ten’s digit i.e. 6 and the next term i.e. 7.
Multiply these two terms
Since, 44 is greater than 42. So square root of 4489 will be the bigger of the two options i.e. 67.
Let us take another example.
Example: What is the square root of 7056?
Unit digit will be 4 or 6.
Since, 82 < 70 < 92
So the square root will be either 84 or 86.
Now consider 8 and 9
Since, 70 is less than 72. So square root will be the lesser of the two values i.e. 84.
Let’s try it out with five digit numbers now!
How to Find Square Root of a Five Digit Number
We pair the digits up starting from the right side. Since there is one extra left over after two pairs are formed, we club it with the pair closest to it.
Example: √(16641) = ?
Unit digit will be 1 or 9.
Since, 122 < 166 < 132
So, the square root will definitely be 121 or 129.
Now, consider 12 and 13
Since, 166 is greater the 156, we pick the larger of the options i.e. 129.
Let’s take another example, so that this trick will be clear to you.
Example: √(33489) = ?
Unit digit will be 3 or 7.
Since, 182 < 334 < 192
So, square root will be 183 or 187.
Now consider 18 and 19.
Now, 334 is less than 342. So, the square root will be lesser of the two numbers i.e. 183.
How to Prepare IBPS PO Quant – Tips & Tricks for Prelims 2017
IBPS PO 2017 is scheduled to take place from 18th September to 22nd September of this year. IBPS POis one of the most awaited banking exams for the aspirants in the banking sector. Changes in the difficulty level of Quant have been reflected in exams such as SBI PO 2017, BOB PO 2017 and more. In all these exams, the pattern of Quant has remained the same but the difficulty level of questions has increased. Therefore, it becomes necessary for you to not just study more but also study smartly. In this article, we will provide you with how to prepare tips for IBPS PO Quant.
Quantitative Aptitude involves a lot of calculations. This can become frustrating sometimes. However, this occasional dullness of Quant can be removed by studying it with quizzes and reading. Make sure that your practice is a blend of reading important tips for the section as well as practising sums. The best way to prepare is following a strategy prioritising topics that are easy and have high weightage.
Go through the exam pattern before you begin this to ensure complete familiarity what you are going to face in exams.
IBPS PO Exam Pattern – 2017
IBPS PO Quant – Divide Different Topics
Knowing which topic has what weightage will allow you to categorise questions into those which need intense practice and which don’t. Moreover, you can also systematically decide which topics are hard for you need more attention.
While preparing, as will be understood by the tips below, there are too many topics in IBPS PO Quant section. Therefore, what you have to do is practice the questions in a particular order. This order will be in such a way:
- More Weightage + Easy Questions
- Time Consuming + Easy Questions
- More Weightage + Difficult Questions
- Time Consuming + Difficult Questions
Topic
|
Weightage
|
Simplification
|
5
|
Average
|
1
|
Percentage
|
1 – 2
|
Ratio &
Proportion
|
1 – 2
|
Data
Interpretation
|
5 – 10
|
Mensuration
|
1 – 3
|
Quadratic
Equations
|
0 – 5
|
Interest
|
1 – 2
|
Problem
on Ages
|
0 – 1
|
Profit &
Loss
|
1 – 3
|
Number
Series
|
5
|
Speed, Time,
Distance
|
1 – 3
|
Time &
Work
|
1 – 2
|
Number
System
|
0 – 1
|
Data
Sufficiency
|
0 – 5
|
Linear
Equations
|
0 – 1
|
Permutations &
Combination/
Probability
|
0 – 1
|
Mixtures &
Alligitions
|
1
|
Before you are go through the preparation strategies we give here, make sure you are completely familiar with the syllabus. This will help you to know what exactly to expect. You can read the detailed syllabus for IBPS PO here:
Detailed Syllabus for IBPS PO Prelims
IBPS PO Quant – How to Prepare Easy Topics
- Topics: Simplification, Average, Percentage, Ratio & Proportion, Interest, Profit & Loss.
- These topics are very scoring in nature.
- In terms of solving, they are less time consuming.
- Therefore, more accuracy is possible in the questions devoted to these topics.
Strategies:
- Simplification –
- Start targeting for these questions early in the preparations.
- Concepts are easy to understand and so you should keep yourself familiar with them.
- Practice quick mental calculations to save time while solving.
- Try using short tricks and techniques for faster calculations.
- Practice a mix of problems on BODMAS Rule, Approximations, Square & Cubes, Surds & Indices, Decimals and Percentages.
IBPS PO Quant – How to Prepare Moderate Topics
- Topics: Number Series, Data Interpretation, Number System, Linear and Quadratic Equations, Data Sufficiency, Problem on Ages.
- These topics are easy to moderate and are easier if proper steps are followed carefully.
Strategies:
- Number Series –
- These questions are easy and quick to solve.
- However, learning different approaches to solve this topics needs constant practice.
- Solve around 100 questions of Number Series on different patterns.
- Your confidence increases after your familiarity increases.
- Data Interpretation –
- Daily practice is a must for DI questions.
- Do 1 set each on Tables, Pie Chart, Line Graph, Bar Graph and Miscellaneous questions involving word problems everyday.
- These topics can be solved accurately due to knowledge of concepts.
- So practice around 100 questions on this chapter.
- With increase in practice, you can increase your conceptual understanding.
- Don’t ignore these top
IBPS PO Quant – How to Prepare Difficult Topics
- Topics: Permutations and Combinations/Probability, Speed, Time and Distance, Time and Work, Mensuration, Mixtures and Alligations.
- There are relatively less chances of solving them correctly.
- They also take more time in solving.
- However, practice is the key in these topics most importantly.
Strategies:
- Do 2-3 sub-topics every week.
- Moreover, in Permutations & Combinations, properly do arrangements of letters of a word.
- In Time, Speed & Distance, do problems on trains, Boat & Stream, Average Speed.
- In Time & Work, questions on Work Efficiency, Wages and Pipes & Cisterns should be stressed.
- Try at least 50 problems of each type.
- In Mensuration, do topics involving Cubes, Cuboids, Square, Rectangle, Cylinder, Sphere, Circle, Semicircle, Cone etc.
To find the number of factors of a given number, express the number as a product of powers of prime numbers.
Take the example of 576. 576 = 64*9 = 26 x 32
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