solved examples of percent, profit loss, numbers ,trigo and series

1. COMBINED FILE OF SSC PROBLEMS
PERCENTAGE
PROFIT AND LOSS
Number system
Decimals
Trigonometry
Series and sequence
SSC CGL PROFILE PROBLEMS ------------------PERCENTAGE
(So –arih –-ssc --solved papers)
PROF -1 ----Successive increase and decrease
1] The salary of A is increased and decreased by 10% successively. Find the
change in the salary --------ans = 1% decr
Use direct formula of succ % change with sign
Or can assume and find
2]the price of an article was first increased by 10% and then again by 20%. If the
last increased price be 33 . Then original price -----ans = 25 rs
3]2 successive price increase of 10% and 10 % of an article are equivalent to a
single price increase of -------ans -----22%
4]The number of seats in cinema hall is increased by 25 %. The cost of each ticket
is also increased by 10% . The effect of these changes on the revenue collection
will be an increase of ----ans = 37.5 %
5] the strength of school increases and decreases in every alternate year by 10%.
It started with increase in 2000. Then the strength of the school in 2003 as

2. compared to that in 2000 was -----ans ---- incr by 8.9%
6]in the expression the values of both x and y are decreased by 20% . by this the
value of the expression is decreased by --------ans ------ 48.8
PROF – 2 ----- Based on fraction and ratio
7] The expenditure and savings of Ram are in ratio 5: 3. If his income increases by
12% and expenditure by 15% , then how much % do his savings increase?---CGL -
08-----ans = 7%
5x/3x is the ratio so for income we need to add so income initial = 8x . now find
the given % values and get the Final income -------its coming as 8.96x and final
exp = 5.75 x -------hence final savings ==== 3.21x ---% saving = 7%
8]The expenses on rice , fish and oil of a family are in the ratio 12:17:3 . The
prices of these articles are increased by 20,30 and 50% respectively. The total
expenses of family on these articles are increased by --------ans = 28(1/8) %
Final initial total --- with x == 32 x -----use % increase in each value ----total final
val = 41 x ----the diff = 9x ------find final % increase.
9] given 10% of A income = 15% of B income = 20% of C income . If the sum of
their income is 7800. Find income of B -----ans ==== 2400(income of B)
Soln-----10a = 15b = 20c ------we need to get terms in ratio so / (divide )by LCM Ie
10/60------15/60------20/60 -------1/6:1/4:1/3 ----------------effectively the ratio we
get --------6:4:3 --- -- as the ratio values -----now sum is given as == 7800
6x + 4x+ 3x = 7800 --------- x= 600 rs
Income of B = 4x -----2400 rs
10] The ratio 5: 4 expressed as a percent equals ----ans = 125%
(5/4) x 100 ===== 125%
PROF -3 ----Based on % increase and decrease in cost of commodity [variation of
2 quantities keeping the third quantity same]
11] If the price of sugar is raised by 25% , find by how much % a householder must
reduce his consumption of sugar so as not to increase his expenditure---ans =20%
Ans -----sugar price === 25% ===fraction ==1/4 –---take initial value = 4
P1 –( 4)---------------------C1 –(Let 1kg)----------------------------------E1(4 x 1 = 4)

3. P2—(4 + 25% of 4)-------C2—(Let be y kg)----------------------------E2(for same E --------
-------------------------------------------------------------------(5y) = 4 ----[y = 4/5]----------------
-------now compare 1 kg with 4/5 --------1/5 ----hence % value ==20%
12] repeat same problem with ------a) 20% price up -----b) 60% price up
13] The price of object rises from Rs 6/kg to Rs 7.50 /kg . If the expenditure is
same the % of reduction in consumption is ------------ans = 20%
Soln------ -- price x consumption = expenditure -for same value of expen just
reverse the two values ----------hence new consumption === 6kg -----------------------
-----hence ---- diff from Rs 7.50 to 6 Rs-----
% value ===== diff /initial x 100 -----------1.50/7.50 x 100 ====20%
14]Different language based problem ----
a) --- production dec by 20% due to retirement of employees. Find increase in
working hours to get same production -----ans (25%)
b) Price of object reduces by 25% . To get the original price ,the new price
must be increased by-------ans( 33(1/3))
c) Duty on article be decr by 40% find incr in consumption for same revenue---
---ans ( 1/6 in fraction ===equal to % value)
d)
15] The price of sugar is increased by 20% . If the expenditure is kept same the
ratio b/w the reduction in consumption and the original consumption is.—ans =
1:6
Soln ----------- let price be 5
p---------------------------------c--------------------------------e
5---------------------------------6-------------------------------30
5 + 20% = 6 --------------------5( to keep same exp)------30
Reduced exp = 6-5 =1 ---ratio ------------------ 1:6
PROF -4 –Based on regaining of the % value
16] The salary of a worker is cut by 20% . He may regain his original salary after
getting a rise of ?----ans -25%
Soln----let sal be 5 -----then after cut sal = 4--------now we need to go from 4 to 5
Hence % increase is 5 -4 = 1 with base as 4 ---25 %

4. PROF -5 ---Based on comparisons of magnitude in terms of percentage
17]If income of A is 50% less than that of B’s , then B’s income is what percent
more than that of A-----ans 100%
Soln ---let income of b = 2 (as per % value reference) -----a = 2 minus 50% = 1
Compare 1 with 2 ---diff- =1 and base = 1(since compared with a)
18] The salary of geeta is 25% more than that of sita . FIND what % of sita’s
income is less than that of geeta -----ans ----20%
Let sita sal = 4 and solve
19] use the above profile with ---------40%, 12.5%, 33(1/3)% ,16(2/3)%
PROF -6 ---Based on word problems of percent
20] when 60% of a number is substracted from other number, the second number
reduces to its 52%, then the ratio of the first number to the second number is
-------ans = 4:5
48y = 60x equation will come
PROF -7 ---Based on mixtures
21]—A vessel has 60 L of solution of acid and water having 80% acid. How much
water is to be added to make it solution in which acid forms 60%---Ans = 20L
--we need to equate either the water or the acid value to get the final percentage
value----- 80% acid = 48 l acid -------------now let x l water be added----eqn is
48 /( 60 + x) = 60% = ie 60/100 --------- so x = 20 L
22] Similar ques-----if new amount of water is added then find the strength of acid
in the new solution
PROF -8 ---Based on percent expenses and savings
23] A man spends 12.5% of his salary on items of daily use and 30% of the
remainder on house rent. After that he is left with 2940 Rs . find his original
salary----ans = 4800 rs
12.5/100 * x = x/8 ------now remaining = x –x/8 === 7x/8 -----now find 30% of
7x/8---------= 21x/80 -----remaing = 49x/80 ====== 2940 rs
X = 4800 rs
PROF -9 ---Based on percent drop /increase in price of quantity and related
consumption
24] A reduction of 21% in the price of an item enables a person to buy 3 kg more

5. for Rs 100 . The reduced price of item per kg is? --------------ans = 7 Rs
SOLN --number of articles = total cost /price of 1 article
3 more articles -----for 100 rs -------hence ---------now reduction = 21% = x minus
(21% of x) == 0.79x
(100/x) minus (100/0.79x) = 3 ---X = 8.86 SO ans = 0.79 x
25] reduction of 10% in price of sugar ---5 kg more can be got by a person for rs
270 ---find original price----ans === 6 rs
26] The price of rice is reduced by 2% . how many kg of rice can be bought for the
money which was sufficient to buy 49 kg of rice.---ans = 50 kg
Soln ----here we will use the concept ---- rate* quantity = total price -----------------
----------------total price will remain same
(r)(49) = T ----Eqn 1----------now ( r – 2% of r) ( Q) = T ------Eqn 2 ----solve 1 and 2
we get --------- Q = 50 Kg
PROF -10 ---Based on the successive percent change
27]Present population of village is 67600 rs . It has been increasing annually at the
rate of 4%. Find the popu of village 2 year ago ans ---62500
P( 1+r/100)(1+r/100) = 67600 ---------(67600 is the present value)------- take r = 4%
and solve
We get ans --------------we need to divide the present value by the successive
product term --- --------62500
28] value of a car decreases every year at the rate of 5% . if its present value is
411540 ---what was the value 3 year ago----------------------------ans = 480x100
Same concept as above ques but we need to use ( 1- r/100) ie minus the % value
and solve
29]---calc based -----The popu of a village was 9800. In a year with the increase in
popu of males by 8% and that of females by 5% , the popu of the village became
10458. Find the no of males in the village before increase. --------------------5600
Let males =x and then females = 9800 –x and solve
PROF -6(B) –Based on word problems of percent
30] A candidate got 55% of the valid votes. If 2% of the total votes were invalid
and the total voter were 104000, then no of valid votes in the favour of candidate
--------56056

6. Soln ---- 100 % - 2% = 98% votes received by the person
32]The allowances of person make 165% of his basic pay, if he gets 11925 as
gross salary, then his basic pay is--------------------------------ans = 4500rs
Basic pay =x -----now x + allowances % of x =====gross sal
33]calc ----method –imp-----The cost of an apple is twice that of a banana and the
cost of a banana is 25% less than that of guava. If the cost of each type of fruit
increases by 10%, then the % incr in the cost of 4 bananas, 2 apples and 3 guavas
is ---------- ans = 10%
Soln ---- take price of banana = Rs 1 and solve ----these type of qs get solved easily
by taking values rather than taking x
34]----approach based --------------A man spends 75% of his income . His income
increased by 20% and he increased his expenditure by 15% . His savings will then
be increased by ans = 35%
Let initial income be 100 rs and solve
35] confidence check ------ In a class avg score of girls in an exam is 73 and that of
boys is 71. The avg score for the whole class is 71.8 . find the % of girls ----
ans(40%)
Soln -------- solve as per avg concept but in this ques we need to solve the entire
problem and go ahead till the end –we get finally b/g = 3/2 so % of g = (2/5)*
100 ==== 40%
36] language -------- A 24 carat gold is taken to be 100% pure , then the % of pure
gold in 22 carat gold is ----------------------------------------------------------ans –91(2/3)
Soln -----24 carat ----100% ---so 1 carat --we will get and so 22 carat =
(100/24)*22
37] 5 STAR --calc + concept ------- If the income tax is increased by 19 % the net
income is reduced by 1% . The rate of income tax is ? ---------------5%
Soln ----- let the income be 100 rs and let rate of tax = r%
Case 1 ----- tax = r% of income ==(r) Rs-----value of income tax == r
Net income in case 1 ==== 100 –r
Case 2 -----now we are increasing the tax value ( don’t confuse it with the tax
rate the tax rate is same but the tax value is increasing)
So now new tax ==== r RS + 19% increase in rRs
So ---- r + ( 19/100)* r ======= 1.19r
Now given that the net income gets reduced by 1 %
So net income case 1 - net income case 2 / income case 1 * 100 ------ = 1%
Because net income is reducing finally
(100-r) –(100 –(1.19r)/(100 –r) * 100 ======= 1

7. r = 5%
Soln –method 2 ---the reduction in net income ie 1% will be same as the reduction
in the tax values -----so we can also use the tax values and their change and
equate them to 1%
38] calc approach ---The number of employees in a firm is increased by 25% and
the wages per reduced by 25%. If it results in x% decrease in total wages , value of
x =?----ans(25/4)
Soln ----in this type assuming values is good
Let e1 = 4 then e 2 = 5 and wages per head w1 = 100 then w2 == 75
Find the diff
39]concept ----- fresh fruit contains 68% water and dry fruit contains 20% water.
How much dry fruit can be obtained from 100 kg fresh fruits? Ans -----52 kg
Best approach ------- The fruit % in both the different weights of 2 types of fruits is
same ie ------(32/100)* 100 kg ===== (80/100)* x ------ --- x =40 kg
Soln------let we have 1000 kg ice creams of both types
1 type has 680 kg water and 2 type has 200 kg water as per 68% and 20% water
value. Now extra water available is 680 kg – 200 kg = 480 kg ------------- so after
removing 200 kg water now we have 480 kg more water
The bigger ice cream has total weight = 1000kg and 480 kg water is extra so if we
remove this 480 kg water we have 520 kg ice cream left ------which is of type 2
So if 1000------520 then if 100kg type 1 ice cream then ----type 2 is ----52 kg
Soln -----method 2 –diff in % ==== 68 -20 ====48 % water extra hence –20%
water of type 1 is absorbed or evaporated ----------then along with fruit
content(cream) we can make 100 – 48 ===52 kg type 2 ice cream( or fruit)
PROF – 7 ---- Based on student passing and failing
40] approach--- In an exam a student had to obtain 33% of the max marks to pass.
He got 125 marks and failed by 40 marks. The max marks were ans ----500
Soln ---- 33% of x –( 125) = 40
41] A student gets 30% marks fails by 5 marks , while another student who
scores 40% marks gets 10 more than minimum pass marks . The max marks are
Ans ---50
Method –1 ---using 1 variable only ---- let pass marks be p and max marks be x
So ---- p – ( 30%x) = 5 eqn -----1]
40%x = 10 + p
Method -2 ---we can also do above ques by taking x and y ie x as obtained marks
and y as the max marks and solve 2 equations

8. 42] calc method ---- 8% of the voters did not cast their votes. In the election there
were only 2 candidates . The winner by obtaining 48% of the total votes defeated
his opponent by 1100 votes. The total no of voters in the election was -----
ans(27500)
Soln ----- let total votes = 100 –votes not casted = 8 ---- votes got by winner = 48
By loser = 44 ( since 44+ 48 = 92) --------hence diff of votes = 4
When diff = 4 total were = 100
When diff = 1100 total find by unitary
Quantitative aptitude problems ------------so ----arh papers ssc
Profit and loss -----------past ssc papers
Profile 1 -----basic concept of profit and loss
1] A retailer buys a sewing machine at a discount of 15% and sells it for 1955 rs.
Hence he makes a profit of 15%. The discount is -------------------ans = 300
Soln –let mrp = x and get cp ---now use profit = 15% and get sp and , now use the
given sp and make it equal to 1955
2] A man sold 20 apples for 100 and gained 20% . H ow many apples did he buy rs
100. Ans = 24
Soln ---- let cp be x ---use profit % and get no of apples by dividing 100 by cp of 1
apple
3]The diff b/w SP and CP of an article is 210 rs. If the profit % is 25 then the SP is---
-----ans = 1050 rs
Soln --- sp –cp = 210 and then use profit % relation
4]if the total cost of 73 articles having equal cost is 5110 and the total selling price
of 89 such articles is 5607,then in the transaction there will be ----ans = loss of
10%
Soln ---we need to make the articles same to compare the sp and cp
So --- cp of 73 articles is given and so we can find cp of 89 articles ---- sp of 89 is
given use it and find loss %
5] A fruit seller buys lemons at 2 for Rs 1 and sells them at 5 for rs 3. The profit %
is ---------ans = 20%
Soln – get cp and sp of 1 lemon and have same no of lemons for easy calculation

9. 6]A dealer sold ¾ of his articles at gain of 20% and the remaining at cost price.
The gain percent earned by him in the whole transaction is - -------ans = 15%
Soln ----let no of articles be 4 so that ¾ x 4 === 3 and let cp of 1 article be rs 1
7]A man gains 20% by selling an article for a certain price . If he sells it at double
the price , the percentage of profit will be -----ans = 140
Soln ----let cp = 100 and solve
PROFILE 2 ---- When CP of x articles is equal to SP of y articles
8] if the CP of 16 tables be equal to SP of 12 tables , then the gain percent is --------
Ans = 33(1/3)%
Soln ---let cp of 1 table =1 rs ---- get the sp of 1 table by comparing the data as per
the equation 16(1) = 12( SP1) ------ and find the Profit %
9]A man buys pencils at the rate of 6 for rs 5 and sells them at 5 for rs 6 . his profit
% is ------- ans = 44%
Soln ---- find sp and cp of 1 pencil and get the combined value by using lcm of the
denominator values -----and then find the profit %
PROFILE 3 ---When the sp and cp are having some values
10. A fruit seller buys 700 oranges at the rate of rs 500 for 100 oranges and other
variety of 500 oranges at the rate of rs 700 for 100 oranges and sells them at 84
per dozen. The profit percent is -----------------------------ans === 20%
Soln -----get total cp and total sp and compare
11. If I purchased 11 books for rs 100 and sold 10 books for 110 the profit % per
book sold is ans = 21%
Soln ---1 book cp = 100/11 ---------1 book sp = 110/10 –hence the deno has 11 and
10 so let total books be lcm of 10 and 11 ie 110 and get the values of the sp and
cp
12. Some toffees are got at the rate of 11 for Rs 10 and the same number at the
rate of 9 for rs 10. If the whole toffees are sold at rs 1 per toffee find the gain or
loss percent ans = loss = 1%
Soln ---- for each category find the no of toffees for rs 1 and then use the same
concept of using the denominator and the lcm and getting the combined values
13. A man purchased some eggs at 3 for Rs 5 and sold them at 5 for Rs 12, thus
he gained Rs 143 in all. The number of eggs he bought is
Soln – method 1 ---let the no of eggs be x and then form the eqn to equate it to
143 and find the value of x
Method 2 ---find the cp and sp for 1 egg and then use the lcm method of

10. denominator to get the value of cp and sp for some fixed no of eggs .
Now use the unitary method using 11 as profit and entire gain as 143 to get the
number of eggs
PROFILE 4 ----When 2 values of selling price are given and some certain value of
profit or loss % is also given
14. By selling an article for rs 102, there is a loss of 15% when the article is sold for
Rs 134.40 , the net result in the transaction ans --- 12% gain
Soln ---- for sp = 102 and loss % get the cp . Now use the cp for the new given sp
15. A man sold his watch at a loss of 5% . had he sold it for Rs 56.25 more , he
would have gained 10%. What is the cost price of the watch? Ans = 375
Soln ---let cp = 100 sp for case 1 = 95 and sp for case 2 = 110 ---the difference in
the sp’s is 110 – 95 === 15
Now use the unitary method ---------------15 is diff ---original cp = 100
56.25 is diff of the sp then find
16. When an article is sold at a gain of 20% it yields rs 60 more than when it is
sold at a loss of 20%. The cost price of the article is ans = 150 rs
Soln --- let the cp = 100 then sp in 1 case = 120 and sp in case 2 = 80 . The diff in
the profits = 40 rs . Now use the unitary method for the diff to be 40 then cp was
100 ,if diff is 60(given ) then cp will be ===== ans
17. A businessman sells a commodity at 10% profit . If he had bought it at 10%
less and sold it for rs 2 less then he would have gained 16(2/3)% . Find the cost
price ans = 40 rs
Soln –let cp =100-- sp in case 1 = 110 -----now cp for case 2 = 100- 10% = 90 rs
Now sp in case 2 = 100+ 16(2/3%) of 90 === 105 rs
Diff of sp = 110 -105 = 5 ---------use unitary when diff is 2 and get cp
PROFILE 5 ---When two articles are sold and loss on 1st
and profit on 2nd
18. A cloth merchant sold half of his cloth at 40% profit , half of remaining at 40%
loss and the rest was sold at cost price . In the total transaction his gain or loss will
be ----- ---10% gain
Soln let the cp be 100 rs----- find the sp as per the given condition ie 40% profit
and 40 % loss and rest ie 25 m cloth
19. A dealer sold 2 tv sets for 7400 rs each. On one he gained 10% and on the
other he lost 10%. The dealer’s loss or gain in the transaction is -------1% loss
Soln ---consecutive gain and loss always gives loss ---by formula ---(r^2/100)%

11. PROFILE 6 ----based on mixtures ,fractions and ratios etc:
20. A milkman got 70 L of milk for 630 rs and added 5 L of water . if he sells it at 9
rs per litre , his profit % is ----7(1/7)%
Soln ---he got 70 L from market and also added 5 L in it ---so his net cp for 75 L is
630 rs and net sp is 75x9 = 675 ---now we can get prof%
21. A man buys a certain number of oranges at 20 for Rs 60 and an equal number
at 30 for rs 60. He mixes them and sells them at 25 for rs 60. Find the gain or loss
percent -----loss of 4%
Let oranges be 60 (lcm of 20, 30 ) ---get the cp of 120 oranges ----now get the sp
of 120 pieces and find the profit and loss
22.A shopkeeper got 80 kg of sugar at the rate of rs 13.50 per kg . he mixed it with
120 kg of sugar costing rs 16 per kg. Inorder to make a profit of 20% he must sell
the mixture at ----rs 18/kg
Soln ---cp of 200 kg sugar ==== 80 x13.50 + 120x16 = 3000
Now get the cp of 1 kg sugar === 3000/200 = 15 rs -----------now get the sp as per
20 % profit ----
23. Two third of a consignment was sold at a profit of 5% and the remainder at a
loss of 2% . If the total profit was 400 then the value of the consignment was
---ans = 15000
Soln – let cp = x -----2/3(x) *(5/100) - 1/3(2/100) = 400 get cp ie x
24. Nita blends 2 varieties of tea one costing rs 180 per kg and other costing rs
200 per kg in the ratio 5:3. If she sells the blended variety at rs 210 per kg find the
gain% ----ans = 12%
Soln ---cp initial = 180 x5 + 200 x 3 --------sp of the blend = 210x( 5+3) ---get the
gain %
25. A person got some articles at the rate of 5 per rupee and the same number at
the rate of 4 per rupee. He mixed both the types and sold at the rate of 9 for rs
2.In this business he suffered a loss of rs 3.FIND THE total articles ----ans = 1080
Soln ---- let articles be 10 -------- total cp = 5x (1/5) + 5x(1/4) and also sp =
2/9(10) -------- now get the loss as per the condition ==== loss = 1/36
Use the unitary method if the loss is 1/36 then the articles = 10 so if the loss is rs
3 articles = 1080
PROFILE 7 ---When SP is given and condition of the loss and the gain are given
26. geeta buys a plot of land for rs 96000. She sells 2/5 of it at a loss of 6% .she
wants to make a profit of 10% on the whole transaction by selling the remaining
land . the gain% on the remaining land is ----------------------ans = 20(2/3)

12. Soln ----for 10% profit === sp = 110/100( 96k) = 105600
Cp pf 2/5 land = 96k(2/5) = 38400 and its sp = 94% of 38400 = 36096
Cp of remaining part ie 3/5 part of land = 96k( 3/5) = 57600
And its sp = 105600 – 36096 ====== 69504-------Because 36096 has already been
invested by the person
Gain % = 69504- 57600/57600 = 20(2/3)%
PROFILE 8 ---Successive passing through profit and loss
27. A sold a horse to B for rs 4800 by losing 20%. B sells it to C at a price which
would have given A a profit of 15% . The gain of B is ans = 2100
Soln --- let cp for A = x ------for loss –20% -----80%(x) = 4800 ---x= 6000
Sp for B = 15% profit on 6000 ------= 6900 ----now cp for B = 4800 ---profit = 6900-
4800
28. An article passing through 2 hands is sold at a profit of 38% at the original cost
price . If the first dealer makes a profit of 20% then the profit made by the second
is - ----ans =15
Soln –let cp = 100
100 ---------1-------2--------138
20------18 --------------------------38 -20 = 18
Profit % for second ===== (18/120)*100 ==== 15%
29. By selling 60 articles a vendor gains the selling price of 15 articles . find his
gain % -----ans = 33(1/3)%
Soln ----- profit = sp of 15 articles
Sp of 15 = sp of 60 - cp of 60 -------= sp fof 45
Cp of 60 = sp of 45 ----------sp/cp =4/3 ----------------prof = 33(1/3)%
30. By selling 1 dozen ball pens a shopkeeper earned the profit equal to the
selling price of 4 ball pens . his prof % ans = 50%
Soln ----profit = sp of 4 ball pens -------------------
Sp of 4 = sp of 12 – cp of 12
31 . A seller got 200 books for 12000. He wanted to sell them at a profit so that he
got 20 books free. Find the prof % ans = 10%
Soln ---cp of 1 book = 12k/200 = 60 rs
Prof = 60 x 20 =1200--------prof % = (1200/12k)x100 = 10%
32. 12 copies of a book were sold for 1800 thereby gaining cost price of 3 copies .
the cost price of a copy is ans = 120 rs
Soln ----let cp of 1 book = x --- ---cp of 12 books =12x-----sp of 12 books = 1800 -----
-given profit ==== cp of 3 books = 3x-----------net sp = cp + prof --------------------------

13. ---------------- 1800 = 12x + 3x ======= 15x = 1800 ----x = 120
PROFILE 9 ------ Miscellaneous questions
33. A dishonest trader claims that his selling price is equal to his cost price but
weighs 900g instead of 1 kg, then his prof % ans = 11(1/9)%
Soln ----let 1gm has cost = 1 rs ------------------hence he gives 900 gm instead of
1000gm ---------------- hence = gain = 1000- 900 = 100gm
Prof % ==== 100/900 = 1/9 =========11(1/9)%
34. If books bought at prices ranging from 150 to 300 are sold at prices ranging
from 250 to 350,what is the greatest possible profit that might be made in selling
15 books. Ans = 3000
Soln ----for greatest profit cp = lowest and sp = highest –hence
We have ========== 350 -150 === profit value ----------- 15 books x 200 = 3000
35. cost of a packet of coffee powder and a litre of milk are 20 and 30 respectively
.10 cups of coffee is made with one packet coffee powder and for each cup
200ml of milk is used . if the coffee is sold at 25% profit the selling price of each
cup of coffee is ans = 10 rs
Soln --- take 1 cup at initial level ------for 10 cups ---1 packet is used so ----in 1 cup
the cost of powder used is 20/10 ===rs 2
Now for milk ----------------1000ml milk cost ---rs 30
200ml ======= rs 6 -------------cost of 1 cup = 6 + 2 = 8 --------now the profit = 25%
on 1 cup ==== 125/100(8) ===== 10 rs
Profile 10 ----double cheat problem
36. A dealer cheats to the extent of 10% while buying as well as selling by using
false weight . What is his increase in the profit % ans = 21%
Soln ---let the initial wt = 100 gm ------------due to 10% cheating during buying he
gets 110 gm -----now again 10% cheating is done by him to sell quantity –so ----11
gm more and so ----121 gm –now comparing 121 gm with 100 gm we get the net
profit % = 21%
Other types -----profile 11 ---use of SP as the base
37. A man purchases 2 fans for Rs 2160 . By selling one fan at a profit of 15% and
other at a loss of 9%, he neither gains nor loses in the whole transaction . find the
cost price of each fan ans = 810,1350
Soln --- let the cp of each be x and y -----the profit amount has to be equal to the
loss amount for no loss or gain in the transaction and so

14. 15% of x = 9% of y –-------------------get x/y ratio and then find x and y using the
total quantity as 2160-ie x/(x+y) multiply by 2160
38. A person sold an article at 20% profit on the sp . Then when the cp reduced by
10% then he also reduced the sp by 10%. His % of profit on the cp will be ----ans
=25%
Soln ---let the sp =100 --------prof = 20 ---so cp === 80
Now 10% reduction in both cp and sp we have ------80 - 8 = 72 and for sp 100-10
= 90 ---------------so prof = 90 – 72 = 18 ----------now get %prof using 18 as the
profit on 72 as the cp ------------prof = 25%
39. A man purchased a bed sheet for rs 450 and sold it at a gain of 10% calculated
on the sp . The selling price of the bedsheet ans = 500 rs
Soln ---- let the sp be x rs -----now he is selling the bedsheet at a gain of 10% on
the sp and so -----450+ 10% of x = x ----we get the ans
40. One trader calculates the % of profit on the butying price and another
calculates on the selling price . When the selling prices are the same then the
difference of their actual profits is rs 85 and both claim to have made 20% profit .
What is the sellinf price of each? Ans = 2550 rs
Soln --- let the cp be 100 rs for 1st
trader ---then his sp = 120
For second trader given that sp has to be same ie sp = 120 and also 20% profit
now we need to find the cp ---let the cp be x ---the prof % s found on the sp
So ----(120 –x/120)* 100 = 20 ---- x= 96 ---------------------gain = 24 rs ---diff of profit
= 24 -20 =4 rs
When diff = 4 rs then sp was 120
When diff = 85 then sp = 2550
41. by selling a table for 350 rs instead of 400 rs , loss % incr eases by 5% . The
cost price of the table is ans = 1000
Soln --- diff of sp = 400 – 350 = 50 rs
Ie 5% of cp = 50 rs –let cp =x --------------find the value of x
42. A man had 100 kg of sugar , part of which he sold at 7% profit and rest at 17%
profit . he gained 10% on the whole . H ow much did he sell at 7% profit ---ans =
70 kg
Soln ---- let cp = rs 1/ kg --------- total cp = 100 rs ------------ let 7% quantity = x and
other ie 17 % = 100 –x -------------------- the profit value will be same so we will
equate the profit from both sides ---ie 10% of 100 ===== 7% of xkg( rs 1) + 17% of
(100-x kg)(rs 1) ----here rs 1 is the cp ----find value of x
43. A dealer makes a profit of 20% even after giving a 10% discount on the

15. advertised price of a vehicle . If he makes a profit of 7500 rs on the sale of the
vehicle , the advertised price was ? ans = 50,000 rs
Soln --- let the marked price = 100 ---------------- disc = 10% ---sp = 90 ------and
profit on this sp –20% –we will get cp =75---now profit = 15 rs
Apply the unitary method based on diff of profit ie 15 rs and if initial mrp = 100
---
44. The cost price of 2 dozen bananas is 32 rs . after selling 18 bananas at the rate
of 12rs per dozen , the shopkeeper reduced the rate as 4 per dozen . the loss % is
Ans = 37.5%
Soln ---initial 1 banana cost = 32/24 rs and noe he sells the same quantity ie 24
bananas --------- but 18 and 6 separately -----------rate for 18 is ---- 12/12 = rs 1 ie
total 18 rs ---------and rate for 6 is 1/3 rs ie for 6 pieces ===2 rs
Now compare both the values of cost prices ie --------------case 1 cp -= 32 rs
And case 2 cp = 18 + 2=20 rs -------find prof % now
QUANTITATIVE APTITUDE QUESTIONS ------PROFILE BASED
CHAPTER ---NUMBER SYSTEM So ---Arihant ssc papers
SSC CGL LEVEL -------PREVIOUS QUES OF THE SSC
Profile problems ------
1. If x is a prime number and -1≤2x-7/5≤1 then the number of
values of x is------
Ans ---4 ------ solve and select prime nos
2. From given options one of the factors of x(c) – 3x(s) + 3x + 7
ie x cube -3x(square) + 3x + 7 is
Ans ----put x = -1 then we get f(-1) = 0 so x+1 is factor of above exp and
then divide it to get ans
St ----- put x = 1 in the exp and we get 8 now put x=1 in options and
check
3.rational no b/w 3/4 and 3/8 is
Ans ---find mean

16. 3. If a and b are 2 distinct natural no’s then which condition is true
for their under root addition
Ans --- √a +√b > √(a+b) -----for these type of problems use the hit and
trial rule and also --------------and take those nos whose sq root is easy to
find
4. Find the number of terms b/w 75 and 97
No of terms b/w x and y are ------(y – x) + 1
5. The digit in the units place of 251^ 98 + 21^ 29 - 106 ^100 +
705^35 - 16^4 + 259
Ans ------solve as per the 4 div method ---unit places are
1+1 - 6+ 5- 6 + 9 ===== 4
6. If the sum of the digits of any integer lying b/w 100 and 1000 is
substracted from the number the result always is divisible by
which number?
Ans ------is 9-----take 102 and check ---
7. Both the end digits of a 99 digit number N are 2 . If n is divisible by
11 then all the middle digits are equal to
Ans ------------------- 4
Given the con 2 - - - - - - - - - - 2
The b/w positions are all same and let them be x
Now total digits = 99 so odd positions =50
And even = 49 -------------now we have 97 digits in b/w and also if we
count the positions 1, 2,3 ,4 ------------------till 99
We will have 48 odd positions( because 50 -2 positions = 48) in b/w if
we count from 3 and we get 49 even positions if we count from 2
So sum of odd pos = 2 + 48x+ 2 = 4 + 48x
And sum of even pos = 49x together 49 even and 48 odd will give 97
as sum ------------using condition of 11 divisibility rule we have
4 + 48x – 49 x = 0 and so ans =========4

17. 8. The number of odd composite divisors of 1848 are
Express in factor form -----------2^3 x 3 x 7 x 11
Now form combinations of 3,7,11 separately
We get ans =====4 divisors
9. Which of the below numbers will always divide a 6 digit number
of the form xyxyxy ( where x and y lie from 1 to 9)
Ans --------expand the above in expanded form ie the power system
and take common and check from the options
10. (2^ 16) - 1 is divisible by
Expand using a^2 – b^ 2 and check by options
Ans = 17
11. The least number which is to be added to the greatest
number of 4 digits so that the sum may be divisible by 345 is
Ans == 6 ---------9999 + x / 345 ======find the rem by dividing
9999 we get 339 ---now logically if we add 6 to this 339 we get
345 and so ans is 6
PROF -2 ---Divisibility rules and remainder
12. The least no which is to be added to the greatest no of 4
digits so that the sum is divisible by 345? Ans =6
Soln ----divide 9999 by 345 ---remainder = 339 now to get perfect
divisibility we need 345 hence add 6 to 339
13. When k is divided by 3 the remainder is 1 and when k+1 is
divided by 5 the remainder is 0 . From the below the possible
value of k is---------options ---62, 63,64,65 ---------------ans = 64
Soln ---check from options
14. When n is divided by 5 the remainder is 2. Find the
remainder when n^2 is divided by 5? ------------------ans = 4
Soln --- n =5q+ 2 -----square both sides
n^2 = 5(5q^2 +4q) + 4
15. If a and b are 2 odd positive integers by which of the

18. following integers is a^4 – b^4 always divisible.------------ans = 8
Soln ---expand relation by using a^2 – b^2 and use any 2 odd integers ie
1 and 3
16. A 4 digit no is formed by repeating a 2digit no such as
1515,3737 etc: Any no of this form is exactly divisible by –ans -101
Soln ---- xyxy = (xy) x 100 + xy = = = (100+1)xy
17. How many 3 digit numbers in all are divisible by 6 –ans = 150
Use concept of AP ---1ST
3 digit no divisible by 6 = 102---now last 3 digit
div by 6 ====996 ----- use the relation of nth term of ap -------with nth
term = 996 and 1st
term = 102
18. It is given that (2^32+ 1) is exactly divisible by a certain no.
which one of the following is also divisible by the same no?---------
---------------options = 2^96+ 1 ----ans ------------1st
option
Soln ---- convert 2^96 + 1 into a^3 + b^3 formula and we get 2^32 +1
19. The numbers 2272 and 875 are divided by 3 digits number
N,giving the same remainders. The sum of the digits of N is --------
-------------------------------------------------------------------------ans = 10
Soln --- let the remainder in each case be x , then 2272 –x and 875 –x
are exactly divisible by that 3 digits number .
Hence their difference be [(2272 - x ) – ( 875 –x)] = 1397
1397 = 11 x 127 which is divisible by both 11 and 127 ----the 3 digits
divisor = 127 =====1+ 2+7 ===10
20. If 17^200 is divided by 18 the remainder is --ans = 1
Soln --- 17^200/( 17+1) --------------a^n/(a+1) ===== rem = 1
21. Concept-------- How many numbers less than 1000 are
multiples of both 10 and 13? ------------------ans =7
Soln ---- the no of multiples of 10x13 =130 are obtained by dividing
1000 by 130 ---hence quotient of 1000/130 = 7
Eg 2 x 5 = 10 -------hence 2,4,6,8,10 are the numbers divisible by 2 ie
the quotient of 10/2 ie 5 numbers
22. The smallest no to be added to 1000 so that 45 divides the
sum exactly is ----------------------------------------------ans = 35

19. Soln ---- 1000/45 ====22 and (10/45) and so rem = 10 now to be
divisible by 45 ----45-10 =====35 must be added
Ssc quantitative aptitude -------decimals and fractions –ssc questions
So ---arh ssc papers
1] The decimal fraction of 2.3(49 bar) -----ans = -----------2326/990
Soln ---2+ 0.3(49 bar) and solve by removing bar
2]The diff of 5.(76bar) and 2.(3bar) ans = 3.(bar 43)
Solve by removing bar
3] The no 0.121212 ----in the form of p/q is ------4/33
Soln ----0.12(bar)
4]the least among fraction 15/16, 19/20,24/25,34/35 ----------------ans ----15/16
Soln ---convert into decimals and compare
SSC CGL ---QUANTITATIVE APTITUDE ----SERIES PROBLEMS
So –arh ssc papers
Profile 1 ---based on rearrangement of terms
1. The middle terms of the series ----2+4+6+---------198 -------ans = 100
Soln--- use the formula for the nth term of AP and find the number of terms --- n = 99 ----
Mid term = n+1/2 ==== 50th
term
2. The sum of the series 1+ 0.6 + 0.06 + 0.006 + 0.0006------------------ans = 5/3
Soln ---add the series we get 1.66666 ie 1.6 bar ( bar is on 6) --- let this be = X and then solve

20. 3. 1^2 -2^2 +3^2 -4^2 --------- -10^2 --------------------------------------------ANS = -55
SOLN --- taking pairs ie a^2 – b^2 ==== (a+b)(a-b) and solve 1-2 = -1, 3-4 = -1 so use in this
manner
4. The sum of first 50 even numbers is - ---2550
Soln ---use reln ---n(n+1) =============== 50(50+1)
5. The sum of first 20 terms of the series 1/5x6 + 1/6x7 + 1/7x8 ---- ans = 0.16
Soln ---- 1/5x6 ==== 1/5 – 1/6 and use this for other terms also
6. 1+ 10 +10^2 + ---upto n terms = 10^n -1/9 , then the sum of the series 4+44+444+-----
upto n terms is ans------40( 10^n -1)/81
PROFILE 2 ---BASED ON SPLITTING THE TERMS IN TWO DIFF TERMS
Soln --- 4( 1+11+111-----) now ------- div and multiply by 9 ans then use 9 = 10-1, 99 = 100 -1
Separate the terms and use the gp relation and solve
7. When simplified the sum ( 1/2+ 1/6+ 1/12+ ----------1/n(n+1) -----------ans = n/n+1
Soln ----write ½ as 1-1/2 and 1/6 = 1/2 -1/3 and separate and solve and so we will get finally
1 – 1/(n+1) ===== after cancellation of various terms
8. The largest natural no which exactly divides the product of any 4 consecutive nat
numbers is --------ans = 24
Soln ---1x2x3x4 = 24
9. The value of 3/1^2.2^2 + 5/2^2.3^2 + 7/3^2.4^2 ----------19 / 9^2.10^2 -----ans = 99/100
Soln ----by observing we get 3/1^2.2^2 === 3/1.4 ---in the deno we can see that 4 -1 = 3 and
similarily -----we have 5/ 4.9 --------in the deno we have ----9-4 = 5 so
By splitting we get -------3/1.4 = 1/1 -1/4 ---and same for the other terms
10. (10^12 + 25)^2 – (10^12 -25)^2 = 10^n then the value of n is ------------ans = 14
Soln ----use a^2 – b^2 === 4ab and compare the LHS and RHS
11. WHAT is the 507th
term of the sequence 1,-1,2,-2,1,1,-1,2-2 -------ans = 2
Soln ---here the repeatition takes after 4th
term and so cycle goes at a period of 4---- 507/4 =
remainder = 2 and so ans = 2
QUNTITATIVE APTITUDE PROFILE PROBLEMS --------- SO -- A rh papers ssc
SSC CGL LEVEL -----------SELF assembled
TRIGNOMETRY
1. If θ is positive acute angle and 4cos^θ - 4cosθ + 1 = 0, then the value of

21. tan(θ- 15) =? Ans = 1
Soln --- given expression forms perfect square ---find cos θ –it will be equal to ½
get theta
2. If ( rcosθ - √3)^2 + (rsinθ -1)^2 = 0 , then the value of rtanθ +
secθ/(rsecθ+tanθ) = ans = 4/5
Soln --- both terms of the lhs will be individually equal to 0 so put 1 and 2 term =
to 0 get the value of rcosθ and rsinθ and square both sides to eliminate using
identity and finally get the value of r = 2 --------------now convert the given exp
into sin and cos and solve
3. If 0< θ< 90 and 2sin^2(θ) + 3cosθ = 3 then the value of θ is
Ans = 60 deg
Soln --- convert the sin term into cos term and solve the quadratic we will get
cosθ = 1 ie theta = 0 and cos θ = ½ ------now check the options
4. If sinx + sin^2(x) = 1 then the value of cos^2(x) + cos^4(x) is
Ans =1
Soln --- change sin ^ 2 term to cos ^ term and put in the given exp
5. If asecθ = x and btanθ = y then how x and y are connected with a and b
Ans = b^2x^2 – a^2y^2 = a^2b^2
Use sec^ 2 – tan ^2 = 1 and separate the terms
6. If x =a( sin θ + cosθ) and y =b(sin θ – cos θ) then the value of x^2/a^2 +
y^2/b^2 is
Ans = 2
Soln --- separate x/a and y/b and square both sides and solve and use identiy
7. If sin5θ = cos 20 ----------theta lies b/w 0 and 90 then the value of θ is
Ans = 14
Cos θ= sin ( 90 – θ)
8. If sin 3A = cos ( A -26) where 3A is an acute angle then the value of A is
Ans = 29 deg
Soln ---- sin3a = cos( 90 – 3a) an solve
9. If a+ b= 90 then the value of 2[sin ^2(a) + sin^2(b)]/cosec^2(a+b)] is
Ans = 2
Change sin^2 into cos^2 so that the identity can be used and solve
10. If 7sin^2(θ) + 3cos^2(θ) = 4 then the value of cosθ
Ans = √3/2
Soln --- split 7sin^2 into 4 + 3 so that the identity can be used and then we will
get 4sin^2θ + 3 =4 ------------now get sin theta and solve
11.If cos^4(x) + cos^2(x) = 1 then the value of tan^4(x) + tan^2(x) ----ans = 1

22. Soln --- cos^4(x) = sin^2(x) ----now convert the expression of tan into sin and cos
and solve and sin^2 + cos^2 = 1 will be used as a identity
12. If xsinθ – ysinθ = full under root[(x^2+y^2)] and cos^2(θ)/a^2 +
sin^2(θ)/b^2 = 1/ (x^2+y^2) ans =====1
Soln --- square the first eqn from both sides and take x^2 and y^2 as common
and so we will use (a-b) ^ 2 formula and the expression of [xcosθ + ysinθ]^2 = 0
And so finally we get xcosθ = ysinθ and we get tanθ = x/y
And we can make the triangle with hypo , base and perpendicular from the
expression of tanθ = x/y
Get sinθ and cosθ and then square them and use in the target expression having
a and b and solving it we get ans =====1
13.The min value of sin^2(θ) + cos^2(θ) -------------------ans ===== 3/4
Soln ---- by checking options we get that min value is got at θ = 45 deg
14. The value of sin^2(1deg) + sin^2(3) + sin^2(5) ---------sin^2(89) ----ans =
22(1/2)
Soln ----89 = 90 -1 and use sin ^2 + cos^ 2 = 1 identity and solve
15. The min value of cos ^2(θ) + sec^2(θ) ------------------------ ans = 2
Soln ---both cos and sec are min at -1 and put in square form
16. Cos θ + secθ = 2 theta varies from 0 to 90 deg then the value of cos 10θ
+sec 11θ ans = 2
Soln --- get full eqn in cos and finally we get cosθ = 1 ie theta = 0 deg now put
in the target expression
17.