Ratio

1. A bag Contains 1,2,5 Rs coins in ratio 4:8:5 and total amount of 90.Rs. how many 5.Rs coins are there in that bag ?
Solution:
Ration is 4:8:5 .
Total amount in 1 Rupee coin is = Rs. 4x*1
Total amount in 2 Rupee coin is = Rs. 8x*2
Total amount in 5 Rupee coin is = Rs. 5x*5

So, 4x+16x+25x = 90; x=2
So no. of 5 rupee coin is = 5x = 10.

(or)

4x*1+8x*2+5x*5=90(coz given ratio is 4:8:5)
45x=90
x=2
So no. of 5 rupee coin is = 5x = 10

2. More than half of the members of a club are ladies.If 4/7 of the ladies and 7/11 of the gents in the club attended the meeting,then what is the smallest number that club could have?
Solution:
it would be 25 in smallest form , as for every 7 women 4 attended the meeting so for this ratio we can assume women to be (2*4/2*7) = (8/14) like there are 8 women attended the meeting from 14 women ,and for every 11 men 7 attended the meeting ,
so overall ( as per condition women are more than half of total members so we can say there are 11 men and 14 women ,as 14 women > 11 men )
 i.e 11 men + 14 women gives 25 members for the club.

3. Three person Anil, Ram and Rishi whose salary together amounts to Rs. 81,000 spends 80%, 85% and 75% of their salaries respectively. If their savings are in the ratio of 8 : 9 : 20, find the salary of Ram.
Solution:
1/5x = 8 ---> x = 40

3/20y = 9 ---> y = 60

1/4z = 20 ---> z= 80

Ratio of their salary's = 40 : 60 : 80 = 2 : 3 : 4
Ram's salary = 3/9 × 81000 = 27000

4. steward assign 1/8th of his monthly salary for food.stewards total food bill for the month is rs.6500.what is stewards yearly salary?
Solution:
let steward monthly sal=x

1/8*x=6500=>x=6500*8=52000(total 1 month sal),
but given there his yearly sal=12*52000=624000

(or)

stewards total one month salary is
8*6500 = 52000
so his one year salary is 52000*12
which is 624000

5. The least whole number which when subtracted from both the terms of the ratio 6 : 7 to give a ratio less than 16 : 21, is?
Solution:
(6 - x) / (7 - x) < 16/21

=> 21 (6 - x) < 16 (7 - x)
=> 5x > 14
=> x > 2.8
So, Answer is 3.

6. A Product is supported each week by the same three customer service Representatives (CSR's). Last month the first CSR took 450 calls, the second took 350 calls, and the third took 300 calls. This month the job will consists of 1500 calls. If the three CSR's each increase their work proportionately, how many more calls will the first CSR take this month than last month? 
Solution:
Last month ratio C1 : C2 : C3 = 45 : 35 : 30 => 9 : 7 : 6
This month total calls = 1500
9x + 7x + 6x = 1500
x= 1500/22 = 750/11 

C1 ratio = 9 * 750/11 = 613.6 => 614 approximately
C2 ratio = 7 * 750/11 = 477.2
C3 ratio = 6 * 750/11 = 409
Calls first CSR ll take more than last month = 614 - 450 = 164


7. A started a business investing Rs 10,000. B joined him after six months with an amount of Rs 8,000 and C joined them with Rs 12,000 after another six months. The amount of profit earned should be distributed in what ratio among A, B and C respectively, four years after A started the business?
Solution:
A started the bussiness with investment 10000 for 4 years ie 48 months
B joined A after 6 months with investnemt 8000 for 3 years and 6 months i.e 42 months
c joined them after another six months i.e 1 year with an investment 12000for 3 years i.e for 36 months

By taking the ratio of 3 of the m we get
10000*48:8000*42:12000*36
=10:7:9

the profit is distributed in the ratio 10:7:9

8. Three partners Rahul, Nayan and Pooja together started a business. Twice the investment of Rahul is equal to thrice the capital of Nayan and the capital of Nayan is four times the capital of Pooja. Find the share of Nayan out of a profit Rs. 22275.
Solution:
let R , N,P be an investment of Rahul Nayan and pooja respectively
Therefore , 2R=3N ( given twice the investment of Rahul is equal to thrice the capital of Nayan) and
N=4P , now ratio of investment of N:R:P = 3N/2 : N :N/4
which is 6:4:1

therefore share of Mayan out of the profit of Rs 22275 = 4/11*22275 = 8100

9. Ramesh, xyz and Rajeev put a partnership. profit is 36000, if Ramesh and xyz ratio is 5 : 4 and xyz and Rajeev 8:9. Find Rajeev's share.
Solution:
ramesh:xyz=5:4
xyz:rajeev=8:9
ramesh:xyz:rajeev=10:8:9

rajeev=(36000*9)/27=12000

10. If 7:13::301:x then the value of ‘x’ is:
Solution:
Given the question;
7:13::301:x
7/13 = 301/x
7x = 301*13
x = 301*13/7

x = 559

11. a man buys a spirit at rs60/litre,adds water to it and then sells it at RS75/litre.what is the ratio of spirit to water if his profit in the deal is 75%
Solution:
c.p of spirit is 60/lit
s.p of 1 lit of mixture is 75
gain 75%
c.p of 1 lit of mixtre=(100/175)*75=300/7

by rule of aligation (60-300/7)/(300/7)=5:2

12. Rs 50000 is divided into two parts One part is given to a person with 10% interest and another part is given to a person with 20 % interest. At the end of first year he gets profit 7000 Find money given by 10%?
Solution:
let x was given to a with 10% interest.
then b got 50,000 - x
interest = p*t*r /100
so
7000 =((50000 - x)*1*20)/100 +(x*1*10)/100
700000 = 1000000 - 10x

x = 30000

13. A rail network consist of 68 routes touching station A, B, C. 5 routes touches each of the three station. 2 routes touches B, C; 3 route touches B, A. 4 Route touches A, C. The ratio of route touching the A:B:C are 11:15:8. Find out how many route touches A, B, C, only. Find out also how many route touches A, B, C station?
Solution:
Given ratio as A : B : C = 11:15:8 
Total routes for A = 68(11/34)= 22 
A only = 22-3-5-4=10 
Similarly total routes for B= 68(15/34)=30
B only = 30-3-5-2=10
Total routes for C= 68(8/34)= 16
C only = 16-4-5-2= 5
Number of routes touches A B C Only = 10+20+5=35

All routes of A B C Is= 22+30+16= 68

14. A milk seller buys milk at Rs 10 per litre and sells it at the same rate and makes a profit of 10% by adding water to it. How much water does he add to a litre of milk?
Solution:
100 ml
just do it by alligation and mixure....... cost price is (100/110)*10=>100/11.
now 0 10
100/11

10-100/11 : 100/11

ratio will be 1:10

15. The ratio of white balls and black balls is 1:2. If 9 gray balls is added it becomes 2:4:3. Then what is number of black balls ?
Solution:
3x=9
=>x=3
now white balls=2*3=6
black balls=4*3=12
so ratio of white and blak balls=1:2(proved)

blk balls=12

16. Find in what ratio will the total wages of the workers of a factory be increased or decreased. If there be a reduction in the ratio 15:11, and a increment in the wages in the ratio 22:25
Solution:
reduction in wage is =15:11
increment in wage is= 22:25
total wages ratio = reduction workers ratio* increment wage ratio
=(15/11)*(22/25)=6/5

total wages ratio =6/5

17. If the ratio of prod of 3 diff comp’s A B & C is 4:7:5 and of overall prod last yr was 4lac tones and if each comp had an increase of 20% in prod level this yr what is the prod of Comp B this yr?
Solution:
If the ratio of prod of 3 diff comp’s A B & C is 4:7:5 and of overall prod last yr was 4lac tones , then 

prod of B last year = 4*7/16 = 7/4 lac tonnes

prod of B this year = 1.2*7/4 = 2.1 lac tonnes

18. Anirudh, Harish and Sahil invested a total of Rs.1,35,000 in the ratio 5:6:4 Anirudh invested has capital for 8 months. Harish invested for 6 months and Sahil invested for 4 months. If they earn a profit of Rs.75,900,then what is the share of Sahil in the profit? 
Solution:
Total units={(5*8)+(6*6)+(4*4)}
=92.
sahil profit={16/92}75,900.

=13,200.

(or)

given ratio=5:6:4
Anirudh invest for 8 months,haris 4 6months and sahil for 4 months
so ratio =5*8 : 6*6 : 4*4 
=40:36:16
=10:9:4
sahil's profit=(4/23)*75900

=13200

19. In what proportional should milk and water be mixed to reduce the cost of litre of milk from Rs 18 to Rs 16?
Solution:
Let total milk be x
now,
x milk costs Rs 18
so Rs 16 will have 8x/9 l milk
(unitary method)
means remaining part is water
so x-8x/9=x/9

now ratio is 8:1 

20. Rs 5000 was divided among 5 men, 6 women and 5 boys, such that the ratio of the shares of men, women and boys is 5:3:2 what is the share of the boy?
Solution:
Given that ratio is 5:3:2.
Now the share of all boys will be (2/10)*5000=1000

So share of each boy will be: 1000/5=200

(or)

5x+3x+2x = 5000
x = 500
then 
5 boys share = 2*500

then share of each boy = 2*500/5 = > 200

21. The sides of the triangle are in the ratio 1/2:1/3:1/5 and Its perimeter is 155 cm, what is the difference between the largest sides?
Solution:
Ratio of sides 1/2 : 1/3 : 1/5 =15 : 10 : 6
Given perimeter=155 cm , so if sides are 15x, 10x & 6x, then 15x+10x+6x =155 ,x=5
So side are 15*5=75 ,10*5=50 & 6*5=30 cm and

Difference between largest & smallest side=75 -30=45 cm

22. Kamal started a business investing Rs 9000. After five months, Sameer joined with a capital of Rs 8000. If at the end of the year, they earn a profit of Rs. 6970, then what will be the share of Sameer in the profit ?
Solution:
Ratio of their investment => 9000 : 8000 => 9:8
Kamal's share's calculated for 12 months nd sameer's share calculated for (12-5) months...
Kamal : sameer => 9*12 : 8*7 => 108:56 => 27:14
Kamal's share = 6970*27/(27+14) = 4590

Sameer's share = 6970*14/41 = 2380

(or)

Kamal invested for 12 months and Sameer invested for 7 months.
So
Kamal:Sameer = (9000*12):(8000*7)
= 108:56
= 27:14
Sameer Ratio in profit will be
= (6970*14/41)

= Rs 2380

23. 2 vessels filled with pire contents.v1 contains 20Liter of milk v 2 contains 20 liter of water. 5 liter of milk is taken from v1 and rplaced in v2. then 4 liter of mixture is transferred from v2 to v1. find the ratio of water in v1 amount of milk in v2
Solution:
vessel m w
v1 20 0
v2 0 20
after first transfer
v1 15 0
v2 5 20
ratio of milk to water in v2 is 1:4
for 4 ltrs mixture 4/5 ltrs is milk,16/5ltrs is water
on next trnsfer 
v1 15+4/5 16/5
v2 5-4/5 20-16/5
now ratio of water in v1 to milk in v2 is

(16/5)/(5-4/5)=16/21

24. There r three coins of Re 1, 50 ps, 25 ps having ratio of 13:11:3. the total sum of money is 77, then find out hw mny re. 1 coins is there?
Solution:
There r three coins of Re 1, 50 ps, 25 ps having ratio of 13:11:3.
if There r three coins of Re 1, 50 ps, 25 ps . No of coins are 13x,11x and 3x respectively..
Then 
13x + 11x/2 + 3x/4 =77
x = 77*8/ (104+44+6)
 = 77*8/154 = 4.

so no of 1 rupee coins 
= 13x = 13*4= 52 coins.

25. A person has with him a certain number of weighing stones of 100gm, 500gm and 1 kg in the ratio of 3:5:1. If a maximum of 5 kg can be measured using weighing stones of 500gm alone, then what is the number of 100 gm stones he has?
Solution:
Number of 500gm stones 
= 5kg/500gm = 10
Ratio = 3:5:1 or 6:10:2

Thus no. of 100gm stones 
= 6

26. An alloy of zinc and copper contains the metals in the ratio 5 : 3. The quantity of zinc to be added to 16 kg of the alloy so that the ratio of the metal may be 3 : 1 is:
Solution:
5/8*16=10
3/8*16=6
(10+x)/6=3/1

=8 kg

27. Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats?
Solution:
let 100% seats are there in all then mathematics increased by 40% then 
5*140/100=7
physics is increased by 50% then
7*150/100=21/2
biology increased by 75% then
8*175/100=14
then ratios of increasing seats are 7:21/2:14
14:21:28

2:3:4 is ans

28. A gets Rs.33 when a sum of money was distributed among A, B and C in the ratio 3:2:5 What will be the sum of money?
Solution:
total units=3+2+5=10
A's amount=33
let total = x
so according to d prb
(3/10) * x= 33
x=(33*10)/3

x=110

29. A 20 litre mixture of milk and water contains milk and water in the ratio 3 : 2. 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more. At the end of the two removal and replacement, what is the ratio of milk and water in the resultant mixture?
Solution:
Initially ixture has milk and water in ratio of 12:8.
when 10 ltr of mixture is removed , left mixture has milk and water in qnty/ratio of 6:4.
when 10 ltr milk is added , ratio becomes, 16:4, ratio is now 4:1
now milk is 8 lt and water is 2 lt in 10 lt mixture
again replaced by 10 lt of milk hence milk and water is in 18:2 in 20 lt mix

so ratio becomes 9:1

30. The proportion of milk and water in 3 samples is 2:1, 3:2 and 5:3. A mixture comprising of equal quantities of all 3 samples is made. The proportion of milk and water in the mixture is
Solution:
Proportion of milk in 3 samples is 2/3, 3/5, 5/8.
Proportion of water in 3 samples is 1/3, 2/5, 3/8.

Since equal quantities are taken,
Total proportion of milk is 2/3 + 3/5 + 5/8 = 227/120
Total proportion of water is 1/3 + 2/5 + 3/8 = 133/120

Proportion of milk and water in the solution is 227:133

31. An alloy of zinc and copper contains the metals in the ratio 5 : 3. The quantity of zinc to be added to 6 kg of the alloy so that the ratio of the metal may be 3 : 1 is:
Solution:
in 6kg of alloy 15/4-zinc and 9/4-copper .
so, 
let extra zinc added be x 
(15/4+x)/(9/4)=3/1

x=3

32. It has 20L mixutre conatins milk and water in the ratio 3:5,replace 4 litres of mixture with 4 litres of water what is the final ratio of milk and water.
Solution:
Full mixture has 7.5L milk 12.5L water if we fetch 4L mixture according to ratio it will be 1.5L milk & 2.5L water,replace this mixture from the water then it will be 6L milk and 14L water then ratio will be 
6/14=3:7

(or)

Quantity of water in 16L of mix=16*(5/8)=10L
Quantity of water in 20L of new mix=(10+4)L
Quantity of milk in new mix=20-14=6L

then ratio=6/14=3:7

33. 729 ml of a mixture contains milk and water in ratio 7:2. How much of the water is to be added to get a new mixture containing half milk and half water?
Solution:
milk=(7/9)*729=567
water=(2/9)*729=162
according to question let z ml of water is added of mixture so
(729+z)*(50/100)=162+z
on solving

z=405 ml

34. In a theatre the ratio of car & two wheeler is 1:8. Total Number of tyres are 100. Find how many are cars and how many are two wheelers?
Solution:
let the no of cars be 1x and no of two wheelers be 8x
(1x*4)+(8x*2)=100
20x=100 ...x=5

1x=no of cars=5 
8x=no of two wheelers=40


35. Simran started a software business by investing Rs. 50,000. After six months, Nanda joined her with a capital of Rs. 80,000. After 3 years, they earned a profit of Rs. 24,500. What was Simran's share in the profit?
Solution:
simran=50k*3=150k
nanda=80k*2.5=200k
ratio=150k/200k=3:4

simran=3/7*24500=10500

36. A 20 litre solution conatins oil and kerosene in the ratio 3:5,replace 4 litres of mixture with 4 litres of kerosene what will be the ratio of oil and kerosene?
Solution:
When 4 liters are taken out of 20ml, amount of oil in remaining 
16 lt = 3/8 * 16 = 6
Hence the remaining 10 litres would be kerosene
Now additional 4 litres of kerosene is added to the solution
So total quantity of kerosene = 14 litres

Ratio of oil:kerosene
 = 6/14 
= 3/7

37. It has 20 mixutre conatins mil and water in the ratio 3:5,replace 4 litres of mixture with 4 litres of water what is the final ratio of milk and water.
Solution:
the total mixture quantity=20 litre.
so,milk = 20*(3/8)=7.5 
water = 12.5 (20-7.5)

when 4 litre removed,water and milk will become less in same ratio

milk = 7.5 - (4*3/8) = 6 litre
water = 12.5 - (4*5/8) = 10 litre

when 4 litre water added ,
water amount = 10+4 = 14.

ratio = 6/14 = 3/7

38. A fires 5 shots to B's 3 but A kills only once in 3 shots while B kills once in 2 shots. When B has missed 27 times, A has killed:
Solution:
Given that B has missed 27 times, means he attempted 54 shots..
but A fires 5 shots to B's 3
so A fires 90 shots to B's 54.
but A kills only once in 3 shots...
so

A has killed 30 Birds

(or)

Let the total no. of shots = x
shots fired by A = 5x/8;
shots fired by B= 3x/8;

killed by A= 1/3 of 5x/8= 5x/24;
missed by B= 1/2 of 3x/8= 3x/16;

A.T.Q :- 3x/16=27;
or x= 144;
so A has kiled= 5x/24= (5*144)/24= 30.

39. If A and B get profits of Rs. 6000 and Rs. 4000 respectively at the end of year the ratio of their investments are
Solution:
profit=Investment * Time
so,
6000=i1 *1=6000
4000=i2 * 1= 4000

i1/i2=6000/4000=3/2
=3:2

40. If ratio of profit of A and B is 4:5 and they together invested Rs.90,000 then money invested by B is
Solution:
profit of A and B is 4:5 respectively
4x+5x=90000
9x=90000
x=10000

so,money invested by B is 5x=5*10000=50000 

41. In a company there are 5 partners A,B,C,D,E. They invested money in the ratio of 1:2:3:4:5. If after one year if there is a profit of Rs.15,000. Then share of E will be ?
Solution:
5 partners A,B,C,D,E are invested money in the ratio of 1:2:3:4:5

Total profit = 15000
share of E = Profit*ratio(E)/(ratio(A)+ratio(B)+ratio(C)+ratio(D)+ratio(E))
= 15000*5/(1+2+3+4+5)=5000

so answer will be 5000

42. A and B enter into a partnership and invest Rs. 12,000 and Rs. 16,000 respectively. After 8 months, C also joins the business with a capital of Rs. 15,000. The share of C in a profit of Rs. 45,600 after two years is 
Solution:
For 2yrs means 24 mnths
A=(24*12)=288
B=(24*16)=384
C=(16*15)=240
Profit=45,600

Share of C is=(240/912)*45600
=12000

43. Anil and Ravi invest Rs.60,000 and Rs.40,000 and they got loss of Rs.5,000 at the end of year. Then what was the loss for Ravi in Percentage?
Solution:
ratio of their investment is 3:2
so the lost of ravi is 2000rs
40000*(x/100)=2000
then x=5%

mixture

1. In a mixture of milk and water the proportion of water by weight was 75%. If in 60 gm of mixture 15 gm water was added, what would be the percentage of water? (Weight in gm)
Solution:
Water in 60 gm mixture=60*75/100=45 gm. and Milk=15 gm.After adding 15 gm. of water in mixture,
total water=45+15=60 gm and weight of a mixture=60+15=75 gm.So % of water=100*60/75=80

2. Potatoes are made up of 99% water and 1% potato matter. Jack bought 100 pounds of potatoes and left them outside in the sun for a while. When he returned, he discovered that the potatoes had dehydrated and were now only made up of 98% water. How much did the potatoes now weigh?
Solution:
100 POUNDS = 99 WATER + 1 POTATO MATTER LETS SAY NOW ITS X POUNDS0.98X+1=X=50 POUNDS

(or)

initially water weigh 99 pounds & patato matter weigh 1 pound.
after dehydration water = 98%=> potato matter = 2% = 1 pound100% = (1/2) * 100 = 50 poundsSo, potato now weigh 50 pounds.

3. There are 7 dozen candles kept in a box. If there are 14 such boxes, how many candles are there in all the boxes together?
Solution:
7 dozen=7*12=84 in 1 boxin 14 box=14*84=1176

(or)
there are 7 dozen candles1 dozen contains= 12 candles
so 7*12 = 8414 such boxes so 14 * 84 = 1176


4. In a mixture of 90 litres, the ratio of milk and water is 4 : 2. How much water should be added to the mixture so that the ratio of milk and water becomes 6 : 4 ?
Solution:
Water in the 90 liter mixture= 90*2/6=30 liters and remaining 60 liters is milkIf 'x' liters of water is added and ratio of milk and water becomes 6:4 , then
60/(30+x) =6/4x=10

5. A mixture of milk and water contains 7% water. What quantity of pure milk should be added to 12 litres of mixture to reduce water to 4%?
Solution:
12 litre of mixture contain 7 % of water i.e quantity of water = 12*7/100 =0.84 litre
Now, 11.16 litre is milk in 12 litre of mix. So, 
from question,let x amount of milk is to be added 
(11.16+x)/0.84 = 96/4 
( It's saying the % water is reduced to 4% i.e new ratio of water:milk = 96/4)
Solving this equation you will get x=9.

6. Sugar worth 40 Rs per kg, 50 per kg and x Rs per kg are mixed in the ratio 2:3:4 to get a mixture worth 50 Rs per kg. Find the value of x.
Solution:
40*2+3*50+4*x=50*9
80+150+4x=450
4x=220x=55

7. Two solutions have milk & water in the ratio 7:5 and 6:11. Find the proportion in which these two solutions should Be mixed so that the resulting solution has 1 part milk and 2 parts water?Solution:
7x+6x/5y+11y=1/2
26x=16y
x/y=16/26
i.e 8/13

8. The amount of water (in ml) that should be added to reduce 9 ml lotion, containing 50% alcohol, to a lotion containing 30% alcohol, is:
Solution:
50% of 9ml = 30% of x ml. so, x= 15ml. now,
 amount of water to be added= 15-9= 6ml, 
as alcohol remains same.

9. A vessel is filled to its capacity with pure milk. Ten litres are withdrawn from it and replaced by water. This procedure is repeated again. The vessel now has 32 litres of milk. Find the capacity of the vessel (in litres).
Solution:
by water. After n operations, the quantity of pure liquid=x(1−y/x)^n
quantity of pure liquid is 32 lit
i,e 32=x(1-10/x)^2
=>(x(x-10)^2)/x^2=32
=>(x-10)^2=32x
x^2+100-20x=32x
=>x^2+100-52x=0
by factorization 
(x-2)(x-50)=0
therefore x=50 or 2....since 2 is not possible practically bcaz present milk level is 32 
x=50

10. Pure Milk contains 89% water.How much water should be added to form a sample of 22 litre containing 90% water?
Solution:
a sample of 22 litre containing 90% water will have 10% milk i.e. 2.2 ltr milk.
2.2 ltr milk in a mixture of 11% milk means total mixture is 100*2.2/11= 20 ltr.,
so 2 ltr water should be added in 20 ltr mixture of 89% water to make it 22 ltr mixture having 90% water.

11. How many kg of rice priced at $3.6 per kg should be added to 24 kg of rice priced at $2.7 per kg in order to make 5% profit on selling the mixture at $ 3.15 per kg?
Solution:
If 'x' kg of rice prices at 3.6$/kg is mixed with 24 kg. of rice priced at 2.7$/kg, then average price=
(x*3.6 + 24*2.7)/(24+x) =3.15*100/105
(3.6+64.8)/(24+x) =3

On solving x=12

12. 20 litres of mixture of acid and water contain 10% water. How much water should be added so that percentage of water becomes 20% in this mixture?
Solution:
total mixture=20 ltr
water=20*(10/100)=2
let x ltr water should be added,
so (2+x)=(20+x)*(20/100)
=>(2+x)=(20+x)*5
=>x=2.5 ltr

13. A milkman mixed 10 liters of water to 50 litres of milk of Rs.16 per liter, then cost price of mixture per liter is 
Solution:
Let price of water per liter be re. 1((10*1)+(50*16))/60 =13.33

(or)

Total mixture = 10w + 50m = 60
Total S.P= 16*60= Rs.960 @ Rs.16/lit
Now one can easily see in the statement that the milkman will get profit of 10 litres.
So, Gain = 10*60= Rs.160
Also, Gain = SP - CP 
160=960-CP
CP = 800 for 60litres
so CP for 1 litre = 800/60 = 13.33

14. The concentration of spirit in three different vessels A, B and C are 45%, 30% and 25% respectively. If 4 litres from vessel A, 5 litres from vessel B and 6 litres from vessel C are mixed, find the concentration of spirit in the resultant solution.
Solution:
=(4*45/100)+(5*30/100)+(5*25/100)
=1.8+1.5+1.5=4.8 so 100*4.8(4+5+6)
=32

(or)

(4*45/100)=1.8(5*30/100)=1.5(6*25/100)
=1.51.8+1.5+1.5=4.8(4.8*100)/(4+5+6)=32% 

15. An oil of Rs.42 per kg is mixed with 10 kg oil of Rs.69 per kg. If price of mixture oil is Rs.60 per kg then what is the quantity of first type of oil in the mixture?
 Solution:
Rule of Mixture
(Quantity of cheaper/Quantity of dearer)=(C.P fo dearer)-(mean price)/(mean price)-( C.P of cheaper)
let quantity of cheaper=x 
Quantity of dearer=10 kg
C.P fo dearer=69
mean price=60
C.P of cheaper=42
x/10 =(69-60)/(60-42)
x/10=9/18
x=90/18
so x=5kg

16. A seller mixed 4 dozen bananas costing Rs.12 per dozen with 6 dozen bananas at Rs.8 per dozen then what is the cost price of mixed bananas per dozen?
Solution:
4 dozens of bananas cost is rs12
then 4*12=486 dozens of bananas cost is rs8
then 6*8=4848+48/4+6=96/10=9.6

17. In what ratio must rice of Rs.16 per kg be mixed with rice of Rs.24 per kg so that cost of mixture is Rs.18 per kg? 
Solution:
(18-24)/(16-18)=6/2=3:1

18. If 10 litres of an oil of Rs.50 per litres be mixed with 5 litres of another oil of Rs.66 per litre then what is the rate of mixed oil per litre?
Solution:
50*10 = 50066*5 = 330830/15 = 55.33

19. In what ratio tea of Rs.80 per kg be mixed with 12kg tea of Rs.64 per kg, so thai cost price of mixture is Rs.74 per kg? 
Solution:
Unit Price of Cheapest Unit Price of dearer
64  80
74
80-74 | 74-64
Ratio = 6:10 =3:5

20. In what ratio must tea of Rs.42 per kg be mixed with tea of Rs.50 per kg so that cost of mixture is Rs.45 per kg?
Solution:
50-45:45-42=5:3

21. In what ratio mental A at Rs.68 per kg be mixed with another metal at Rs.96 per kg so that cost of alloy (mixture) is Rs.78 per kg?
Solution:96-78:78-68=9:5

22. A man buys eggs at 2 for Re 1 and an equal number at 3 for Rs 2 and sells the whole at 5 for Rs 3. His gain or loss percent is :
Solution:
Total cp of 2 eggs=1/2+2/3=7/6
Cp for 1 egg=7/12
Now Sp of 1 egg=3/5
Profit=3/5-7/12=1/60
gain %=(1/60)/(7/12)*100=20/7
So ans will be 2 6/7

23. In what ratio must rice at Rs 9.30 per Kg be mixed with rice at Rs 10.80 per Kg so that the mixture be worth Rs 10 per Kg?
Solution:
(9.3x+10.8y)/(x+y)=10
9.3x+10.8y=10x+10y
7x=8y
x/y=8/7
8:7

24. Pure ghee costs Rs 100 per kg. After adulterating it with vegetable oil costing Rs 50 per kg, a shopkeeper sells the mixture at the rate of Rs 96 per kg, thereby making a profit of 20 %. In what ratio does he mix the two ?
Solution:
SP of mixture is 96 making profit 20%
therefor CP is 80

Now using alligation

100 50
/
/
80
/ 

(80-50)=30 (100-80)=20
so the ratio is= 30/20
3:2

25. 4 litres of a 20% solution of alcohol in water is mixed with 6 litres of an 80% solution of alcohol in water. What is the strength of alcohol in the resulting mixture?
Solution:
amount of alcohol in both solution = 4*20/100 + 6*80/100 = 4/5 + 24/5 = 28/5
total solution = 4+6 = 10 litre
percentage of alcohol in mixture = (28/5)/10 * 100 = 56%

26. 1 litre of water is added to 5 litres of a 20% solution of alcohol in water. The strength of alcohol is now?
Solution:
amount of alcohol in solution = 5*20/100 = 1litre
strength of alcohol in solution = 1/6 *100 = 16.67%

27. Two liquids are mixed in the proportion of 3:2 and the mixture is sold at $11 per litre at a 10% profit. If the first liquid costs $2 more per litre than the second, what does it cost per litre?Solution:
Given mixture ratio = 3:2
Lets assume second liquid = x,
So,first liquid = (x+2).
x-10/10-x-2 = 3/2
2x-20 = 24-3x
5x = 44
x=8.8
so first liquid cost is x+2 = 8.80+2 = 10.80

28. Two equal glasses filled with mixture of milk and water in the proportions of 2:1 and 1:1 respectively are emptied into a third glass. What is the proportion of milk and water in the third glass?
Solution:
first glass = 2:1
milk in first glass = 2/3 water in first glass = 1/3

second glass = 1:1
milk in second glass = 1/2 water in second glass = 1/2

third glass = (milk in first glass + milk in second glass)/(water in first glass + water in second glass)


29. 729ml of a mixture contains milk and water in the ratio 7:2. how much water is to be added to get a new mixture containing milk and water in the ratio 2:1?
Solution:
In 729ml milk amount is (7/9)*729=567
Water amount is (2/9)*729=162
our aim is water in the mixture half of milk.
Half of milk is 283.5
water to be added to get 283.5 is 162+121.5
So ans is 121.5ml

30. A mixture of milk and water measures 60 gallons. It contains 20% water. How many gallons of water should be added to it so that water may be 25%?
Solution:
milk+water=60 gallons
60 gallons=20% water+80%milk
20 % of 60=12 gallons
25 % of (60+x)=12+x;
x=4 gallons

31. Concentrations of three wines A, B and C are 10, 20 and 30 percent respectively. They are mixed in the ratio 2: 3 : x resulting in a 23% concentrated solution. Find x.
Solution:
Assume that we have 2a,3a,xa liters of A,B,C wines mixed respectively then

2a(0.10)+3a(0.20)+xa(o.3)=(2a+3a+xa)(0.23)

=0.8+(0.3x)=1.15+(0.23x)
then x=0.35/0.07=5

32. A vessel is full of a mixture of spirit and water in which there is found to be 17% of spirit by measure. Ten litres are drawn off and the vessel is filled up with water. The proportion of spirit is now found to be 15 1/9%. How much does the vessel hold?
Solution:
In 10L of water drawn amount of spirit present is = (17/100)*10 =1.7 L .
x is the capacity of the vessel
from 17% amount of spirit comes down to 15 1/9% .the difference between this percentage is (17% - 15 1/9%) = 17/9 % x 
17/9 % x =1.7 litres . from this equation we get the value of x as 90 L

33. A man buys oranges at Rs 5 a dozen and an equal number at Rs 4 a dozen. He sells them at Rs 5.50 a dozen and makes a profit of Rs 50. How many oranges does he buy ?
Solution:
50(avg of both is 4.5 so he gain 1 rupee and for profit of 50 is 50)

34. A producer of tea blends two varieties of tea from two tea gardens one costing Rs 18 per kg and another Rs 20 per kg in the ratio 5 : 3. If he sells the blended variety at Rs 21 per kg, then his gain percent is
Solution:
(20-x)/(x-18)=5:3
from here x=75/4
now the sell price is 21 and the cost price is 75/4 so profit is 12%

35. There are two vessels A and B in which the ratio of milk and water are as 5:2 and 8:7 respectively. Two gallons are drawn from vessel A and 3 gallons from vessel B, and are mixed in another empty vessel. What is the ratio of milk and water in it?
Solution:
If 2 gallons are taken from A, the quantity of milk & water=2*5/7:2*2/7=10/7:4/7
If 3 gallons are taken from B,the quantity of milk & water=3*8/15:3*7/15=8/5:7/5
Now in the new mixture qtantity of milk 10/7+8/5=106/35 & water 4/7+7/5=69/35
Therefore ratio of milk:water=106/35:69/35=106:69

36. Gopal purchased 35 kg of rice at the rate of Rs 9.50 per kg and 30 kg at the rate of Rs 10.50 per kg. He mixed the two. Approximately, at what price per kg should he sell the mixture to make 35 % profit in the transaction ?
Solution:
35*9.50+30.10.5=647.5
647.5*1.30=874.125
874.125/65=13.44

37. Ajay bought 15 kg of dal at the rate of Rs 14.50 per kg and 10 kg at the rate of Rs 13 per kg. He mixed the two and sold the mixture at the rate of Rs 15 per kg. What was his total gain in this transaction ?
Solution:
15*14.50=217.5
10*13 =130
(217.5+130)=347.50
15+10=25 so,25*15=375
375-347.50=27.50

38. How much water must be added to 60 liters of milk at 11/2 liters for Rs 20 so as to have a mixture worth Rs 10 2/3 a liter?
Solution:
C.P. of 1 litre of milk = Rs. 20×2/3 = Rs. 40/3.
C.P. of 1 litre of water =Rs 0.
Mean price = Rs. 32/3.
By the rule of alligation, we have :
C.P of 1 litre        C.P of 1 litre
of water             of milk
   (0)                  (Rs. 40/3)
          \                /          Mean Price
          (Rs. 32/3)
         /               \
  40/3−32/3    32/(3−0)
      8/3              32/3
The ratio of water and milk =8/3:32/3.=>8:32=1:4.
Thus, Quantity of water to be added to 60 litres of milk:=>(1/4)×60 litres.=>15 litres.

39. By mixing two qualities of pulses in the ratio 2: 3 and selling the mixture at the rate of Rs 22 per kilogram, a shopkeeper makes a profit of 10 %. If the cost of the smaller quantity be Rs 14 per kg, the cost per kg of the larger quantity is 
Solution:
Cost Price of 5 kg = Rs.(14*2 + x*3) = (28 + 3x). 
Sell price of 5 kg = Rs. (22x5) = Rs. 110. 
(110 - (28 + 3x)/(28 + 3x)) * 100 = 82-3x/28 + 3x = 1 / 10 
820 - 30x = 28 +3x ; 33x = 792 ; 
x = 24

40. A certain amount of solution contains 18% alcohol. 8 litres of the solution is taken out and replaced with water. The resultant solution contains 15% alcohol. Find the volume(in litres) of the solution.
Solution:
8 liters=3%
? =remaing 15%
(15*8)/3=40

41. 6 kgs of tea at $6 per kg and 4 kgs of tea at $7 per kg are mixed together and the mixture is sold at a 10% profit. What is the selling price per kg of mixture?
Solution: 
6kg->$36,4kg->$28, then the mixture will be 10kg->$36+$28=$64.
if 10% profit->$6.4 for 10kg. now 10kg->$64+$6.4=$70.4. Hence per kg=$7.04

42. A 240 gallons solution contains milk and water in the ratio 5:4. How many gallons of water must be added so that the ratio of milk and water in the resulting solution is 4:5?
Solution:
Let X be the number of gallons of water added so that the ratio of milk and water changes to 4:5 from 5:4.
Amount of milk in the initial solution= 5/9 × 240 gallons
Amount of milk in the final solution= 4/9 x (240 + X)gallons
Since there is no change in the content of the milk we have
5/9 × 240 = 4/9 x (240 + X)
i.e. X = 60 gallons

43. A milk vendor sells milk at Cost Price but still gains 20%. Find the ratio of milk and water in every gallon that he sells
Solution:
Let Rs 100 be cost price and selling price
To get 20% profit by selling at Rs 100, quantity to be actual sold =100*100/120=250/3, 
In case of milk to compensate for the quantity water to be mixed=100-250/3=50/3
Therefore the ratio of milk:water=250/3:50/3=5:1

Percentage

1. A mixture of 40 liters of milk and water contains 10% water. How much water should be added to this so that water may be 20% in the new mixture? 
Solution:
Mixture of 40 liters of milk=36M and 4W(Ratio is 90:10)
Now the New mixture shall be in the ratio of 80:20
Now water shall be added but milk shall not be touched
therefore 80% is equivalent to 36
100% is (36/80)*100=45
Thus water shall be added=(45-36-4)=5 liters of water

(or)

In a new mixture.. the milk should be 80%
so 36 ltr= 80%
so total will be 36*100/80= 45 ltr
hence added water is 45-40 = 5 ltr

2. The true discount on Rs. 2562 due 4 months hence is Rs. 122. The rate percent is:
Solution:
si of 4 month ie of 1/3 years=122
so p=2562-122=2440
now R=(si*100)/p*time

so R=122*100*3/2440*1=15

3. The price of petrol is increased by 10%. By how much percent the consumption be reduced so that the expenditure remains the same?
Solution:
9%
 using formulae [r/(100+r%)]*100

(or)

9% consumption has to be reduced.
because (91/100)*110 = 100

i.e 91 percent of (100 + 10) percent = 100 percent.

4. If the sales tax reduced from 3 1/2 % to 3 1/3%, then what difference does it make to a person who purchases an article with market price of Rs. 8400 ?
Solution:
7*8400/200=294
10*8400/300=280

hence person has to pay 14 rs less as tax

5. Consider three brothers Ram, Ravi and Rahul. Consider Ram to be taller than Ravi by 10% and Rahul is taller than Ravi by 30%. Now, by how much percentage Rahul is taller than Ram.
Solution:
Let heights of Ram, Ravi and Rahul be p,q and r respectively. Then as per the conditions given in the question :

p = 110/100r and q = 130/100r
Ratios of heights of q and r will be (130/100r)/(110/100r) = 13/11.

Hence Rahul will be 13/11% taller than that of Ram.

6. Ravi had got twice as much as marks as Ramu. His teacher had made him a promise that, for every mark he scores above Ramu, he would be awarded 50% of those marks as bonus. Find the ratio of his bonus marks to the total marks of Ravi and Ramu.
Solution:
let ramu marks is 100 then ravi marks will be 200
bonus=200-100=100*50/100=50
then ratio: bonus/total marks

50/350=1:7

7. If the price of gold increases by 30%, find by how much the quantity of ornaments must be reduced so that the expenditure may remain the same as before?
Solution:
the formula for this situaion is..
100* (R)/(100+R)

100*30/130= 23.08 %

8. A student gets an aggregate of 60% marks in five subjects in the ratio 10 : 9 : 8 : 7 : 6. If the passing marks are 50% of the maximum marks and each subjects has the same maximum marks, in how many subjects did he pass the exam?
Solution
take the maximum mark in each sub as 100. qualify mark is 50.On a total dt fellow gets 60 x 5= 100.
We can write (10+9+8+7+6)a=300 => a=15/2
In 5 subjects he gets 75,67.5,60,52.5,45.

He passes in 4 subjects

9. The market value of a 10.5% stock, in which an income of Rs. 756 is derived by investing Rs. 9000, brokerage being %, is:
Solution:
an income of rs 756 investment=rs 9000
for an income of rs 21/2 investiment is (9000/756*21/2)=125
for a rs 100 stoks,investment=rs 125
market value of rs 100 stocks=rs(125-1/4)
=rs 124.75

10. The banker's gain on a bill due 2 year hence at 10% per annum is Rs. 8. The true discount is:
Solution:
B.G=8
RATE=10
TIME=2
T.D=B.G*100/RATE*TIME
=8*100/2*10

=40

11. Find the number of ways in which 10 players out of 14 players can be selected such that 3 particular player are always included and 2 particular players are always excluded?
Solution:
Out of 14 players, let us element 2 particular player
which are excluded. Now, there are 12 players for selection of these 12, three have to be included in team always. Thus remaining players are (12 – 3) = 9 and the required players for team (10 – 3) = 7. Now,
selection cannot be done in
9C7 ways.

9C7 = 9!/7!2! = 36 ways.

12. When the price of a pair of shoes is decreased by 10%,the number of pairs sold increased by 20%.what is net effect on sales?
Solution:
Let the S.P be Rs 100 / shoe
Let 100 shoes are sold out
cost of shoe 10% decreased = ( 100- 10)= 90 Rs
20% Increase in sales therefore 120 shoes sold out
90*120=10800
100*100=10000
therefore (10800-10000)/100=800 

=> 8% increase in sales

13. The population of a village is 5500. If the number of males increases by 11 % and the number of females increases by 20 %, then the population becomes 6330. Find the population of females in the town?
Solution:
Let X is the initial population of Male and Y be the initial population of Female 
X + Y = 5500 .........(1)
If the number of males increases by 11 % and the number of females increases by 20 %, then 
Percentage problems for SSC RRB IGNOU Hotel Management exms
1.11 X + 1.2 Y = 6330 , On putting the value of X from equeation (1)
1.11 ( 5500 - Y ) + 1.2 Y = 6330
Y = 2500, 
Population of females in town

14. A shopkeeper sells milk which contains 5% water. What quantity of pure milk should be added to 2 liters of milk (containing 5% water) so that proportion of water becomes 4%?
Solution:
Present proportion of milk and water(in 2 litres of milk) is : 1900:100( in ml)
Now, let the milk should be added x ml
So, the new ratios is (1900+x):100
Hence, the equation becomes 100/(2000+x)=4/100
x comes to 500ml

Therefore, 500 ml of milk should be added so that the proportion of water becomes 4%


15. There are 600 tennis players 4% wear wrist band on one wrist Of the remaining, 25% wear wrist bands on both hands How many players don't wear a wrist band?
Solution:
here 4%=(4/100)*600 = 24
600-24=576
25%(576)=(25/100)*576 =144
no.of persons wear=144+24=168

no. of person not wearing wrist band
=600-168=432

16. In an examination of quantitative aptitude and logical reasoning. 65% examinees cleared quantitative aptitude test while 70% cleared logical reasoning test. If 50% examinees passed both the tests. then how many failed in both tests? 
Solution:
Total %age of student passed=65+(70-50)=85%

So,%age of student failed in both subject=15%

17. In a class the ratio of boys n girls is 5:6.if 25% of boys n 20% of girls r scholarship holders,find the %of students who r not scholarship holders?
Solution:
Let total students be x
No. of boys = 5/11*x
No. of girls = 6/11*x

Total no. of scholarship holders = 25/100*5/11*x + 20/100*6/11*x
=5/44*x + 6/55*x

Total no of students = No. of scholarship holders + No. of students who are not scholarship holders
x = 5/44*x + 6/55*x + No. of students who are not scholarship holders

No. of students who are not scholarship holders = 171/220*x=0.777x
% of students who are not scholarship holders = 0.777x/x*100

=77.7%

18. Shopkeeper rise price by 35% and gives successive discount of 10% and 15%. What is overall % gain or loss?
Solution:
Let d initial price be 100
35 % rise
now price = 135/100*100 = 135

10% discount 
Then price = 135 * 90/100 = 121.5

15 % discount
Then price = 121.5 * 85/100 = 103.275

So Gain = 103.275 - 100 = 3.275

Gain % = Gain * 100 /CP

==> 3.275 * 100 /100 = 3.275%


19. A and B invested Rs.4000 and Rs.5000 and they get 20% profit at the end of year. Then share of B is
Solution:
Total invest is 9000
20% gain of 9000 is 1800
A:B=4:5

B's share will be (5/9)*1800=1000

20. In a certain college, 20% of the boys and 40% of the girls attended the annual college outing. If 35% of all the students are boys, what percent of all the employees went to the outing?
Solution:
assume total students =100 in that 35 students are boys ,remaining 65 are girls........
20% of boys(35) =7
40% of girls(65) =26 
totally (26+7)=33 mems went to outing...... 33 is 33% in 100

so, the answer is 33%

21. Rani's weight is 25% of Meena's weight and 40% of Tara's weight. What percentage of Tara's weight is Meena's weight?
Solution:
Rani's weight=25% meera's weight = 40% tara's weight
meera's weight = 100/25*40% tara's weight

= 160% tara's weight

22. In a party , where there were 200 quests, 60% had veg food, 50% had non veg, 30% of guest had both veg and non veg, how many % did not dine at the party?
Solution:
Total guests =200
Dine members=(veg+nonveg - both)%
(60+50- 30)%
=80%
Not Dine members=100%- 80%
=20% of 200 

=40 members

23. A reduction of 20% in the price of sugar enables a consumer to obtain 2.5kg more for Rs.160. Find the reduced price per kg.of sugar.
Solution:
The extra 2.5 kg is obtained because of 20 reduction in price i.e. 20% of 160 = 32 rs.
Reduced price 1 kg = 32/2.5 = 12.80 rs.
So answer is 12.80 rs.
And original price :-
20% reduction price means you pay 80% of the original price
80% of original price = 12.80 rs/kg

100% of o.p. =16 rs/kg.

24. The total population of a village is 5000. The number of males and females increases by 10% and 15% respectively and consequently the population of the village becomes 5600. what was the number of males in the village ?
Solution:
let x be male, y be female

x + y = 5000 ---> (1)
x+10%x + y+15%y =5600
x + (10/100)x + y + (15/100)y=5600
(110/100)x + (115/100)y=5600
110x+115y=560000 --> (2)

solve (1) & (2)

110x+115y= 560000 --> (2)
-110x-110y=-550000 --> (1) * -110
-------------------
5y= 10000 
y= 2000
x+y =5000
y=2000 then x=3000 

No. of male in the village is 3000

25. In an examination, A got 10% marks less than B, B got 25% marks more than C and C got 20% less than D. If A got 360 marks out of 500. The percentage of marks obtained by D was :
Solution:
A got 369/500 * 100 = 72% 
This was 90% of B. So B = 72 * 100 / 90 = 80%. 

This was 125% of C. So C = 80 * 100 / 125 = 64%. 

This was 80% of D. So D = 64 * 100 / 80 = 80%. 
So D got 80% marks. 

26. The price of sugar is increased by 20%. As a result, a family decreases its consumption by 25%. The expenditure of the family on sugar will be decreased by :
Solution:
Let price of sugar be 100 and consumption be 100
original expenditure=price*quantity
=100*100
=10000
New expenditure=120*75
=9000
decrease in expenditure=[(10000-9000)/10000]*100

=10%

27. In an examination, 35% of the students passed and 455 failed. How many students appeared for the examination?
Solution:
if 35% of students passed then definitely 65% would have failed. rite!
let the total no of students be x.
so 65% of x=455 (i.e no of students failed)
65/100 x X = 455

SO X=455x(100/65)=700 students.

28. A's income is 25% more than B's income. B's income in terms of A's income is
Solution:
let us take b income=100
then A's income=125 
25 rs exceeding the B's
25/125*100=20

B's income is 80

(or)

[r/(100+r)] * 100 =80%


29. The price of oil is increased by 25%. If the expenditure is not allowed to increase, the ratio between the reduction in consumption and the original consumption is :
Solution:
Let the price be Rs 10 per litre and the consumption be 10 Kg
Given, increased price is 25% of 10 = 2.5 + 10 = Rs 12.5 per litre
So, according to the question if the price is not allowed to increase then how much can be consumed on the original price that is Rs 10
So, On 12.5 rupees the consumption is = 1 litre

On 1 rupee the consumption is = 1/12.5
On 10 rupees the consumption is = 1/12.5 X 10 = 4/5

So, for Rs 10 we consume only 4/5 litre instead of 1 litre
So, the reduction in consumption is 1 - 4/5 = 1/5
Therefore reduction in consumption : original consumption

That is, 1/5:1 = 1:5 

30. If the price of petrol increases by 25%, by how much must a user cut down his consumption so that his expenditure on petrol remains constant?
Solution:
let the petrol 1 unit initial price be rs 100.
25% increase on rs 100= rs 125
since expenditure to remain constant
125*x= 100*1
x=.80 unit

therefore consumer reduces =.20 unit i.e 20%

31. Fresh fruit contains 68% water and dry fruit contains 20% water.How much dry fruit can be obtained from 100 kg of fresh fruits ?
Solution:
The fruit content in both the fresh fruit and dry fruit is the same.
given,fresh fruit has 68% water.so remaining 32%(100-68) is fruit content.weight of fresh fruits is 100kg
dry fruit has 20% water.so remaining 80% is fruit content.let weight if dry fruit be y kg.
fruit % in freshfruit = fruit% in dryfruit
32/100 * 100 = 80/100 * y
we get y = 40 kg

32. If the price of sugar rises from Rs. 6 per kg to Rs. 7.50 per kg, a person, to have no increased in his expenditure on sugar, will have to reduce his consumption of sugar by 
Solution:
600/7.50=80%

so 20% reduced

33. The price of sugar increases by 32%. A family reduces its consumption so that the expenditure of the sugar is up by 10% only. If the total consumption of sugar before the price rise was 10 kg per month, then the consumption of sugar per month at present (in kg) is :
Solution:
let initially 10kg sugar= 100rs.
on increased price 10kg SUGAR= 132rs.
consumer uses sugar= 110rs.
therefore 1rs. sugar = 10/132kg

and 110rs. sugar = 10*110/132=8.3333

(or)

10kg-100rs
Now 10kg-132

so for 110rs u will get 1100/132=8.3333

34. Deven Invests Rs. 2,34,558 which is 25% of his annual income, in National Saving schemes. Whar is his monthly income?
Solution:
LET INCOME=X
25% OF X=234558
THEREFORE X=938232

monthly income= 938232/12=78186

35. In an examination, 80% of the students passed in English, 85% in Methematics and 75% in both English and Mathematics. If 40 students failed in both the subjects, the total number of students is :
Solution:
x*20%+x*15%-x*25%=40

x=Total Student=400

36. A reduction of 21% in the price of wheat enables a person to buy 10.5 kg more for Rs. 100. What is the reduced price per kg ?
Solution:
100 rs ------------xkg
79rs---------------x+10.5
21 rs----------10.5kg
x---------------1kg

x=21/10.5=2 

37. The present population of a country estimated to be 10 crores is expected to increase to 13.31 crores during the next three years. The uniform rate of growth is :
Solution:
population after n years = P(1+R/100)^n
13.31=10(1+R/100)^3
R=10% 

38. The population of a town increases 4% annually but is decreased by emigration annually to the extent of (1/2)%. What will be the increase percent in 3 years ?
Solution:
p(1+r/100)^n
r=3.5 
n=3
p=population 

ans 10.8

(or)

let initial population be 100.
after the end of year 1 population=100+3.5=103.5
after the end of year 2 population=103.5+103.5*3.5/100=107.12
after the end of year 3 population=107+107*3.5/100=110.8


so increase percentage =110.8-100=10.8


39. Ravi's salary was reduced by 25%.Percentage increase to be effected to bring the salary  to the original level is
Solution:
If initial salary is Rs 100.
After 25% reduction, salary will become Rs 75.
To make it again Rs 100, the salary should be increased by Rs 25.
So % increase required = 100*25/75= 100/3 % = 33 1/3 %

(or)

Taking 100 Rs as salary
Salary reduced by 25 % so salary is reduced 100 x 0.25 =25
So present salary is 100-25 = 75 Rs.
To take it Original value

75 Rs = 100 (I take 75 as 100%)
100 Rs =?

100 x 100
-----------
75
= 133.33 %
Difference

= 133.33-100 % = 33.33 % = 33 1/3% 

40. The boys and girls in a college are in the ratio 3 : 2. If 20% of the boys and 25% of the girls are adults, the percentage of students who are not adults is :
Solution:
let total students will be 100 divide it in 3:2 =60/40
20 % 60=12 boys adult
25 % 40= 10girls adults
total adults = 22
no of students =100

100-22=78

41. Water tax is increased by 20% but its consumption is decreased by 20%. Then, the increase or decrease in the expenditure of the money is 
Solution:
assume total amount is 100 
increased tax = 100+20 = 120
decrease in consumption = 100-20 =80
total amount 100*100 = 10000
after change 120*80 = 9600


(10000-9600/10000)*100 = 4 % decrease

42. A man’s income was reduced by 10%. How much percent it must now be raised so that it may be equal to the original amount?
Solution:
let income be 100
reduced by 10%
reduced income-90
diff required to cover original-10
10/90=11.111111111

So,answer is 11%

43. In an election contested by two parties, Party D secured12% of the total votes more than Party R. If party R got 132,000 votes, by how many votes did it lose the election?
Solution:
Let total % of votes secured by D = x 
Therefore, by R = ( x - 12)%
As there are only two parties, 
Thus, 
x + (x - 12) = 100
x = 56%
Therefore, R got (56 - 12) = 44%
By question, R got = 132,000
i.e; 44% (T) = 132,000
T = 3000,00 
Now, Party R lost by 12% (T)

= 36,000

44. In an election between two candidates, a candidate who gets 40% of total votes is defeated by 1500 votes. The number of votes polled by the winning candidate is :
Solution:
total votes be x
40% of x =60% of x - 1500
therefore,
20% of x=1500

i.e 60% of x =1500*3=4500

45. A cricket team won 40% of the total number of matches it played during a year. If it lost 50% of the matches played and 20 matches were drawn, the total number of matches played by the team during the year was :
Solution:
consider team played total x matches then,
40% of x+50% of x+20=x
90 5 of x+20=x
20=x-9x/10
20=x/10
x=200

so total matches played by team is 200

46. In a certain office, 72% of the workers prefer tea and 44% prefer coffee. If each of them prefers tea or coffee and 40 like both, the total number of workres in the office is :
Solution:
If the total no.of workers is 100 then 72 prefer tea and 44 prefer coffee.
n(Tea U Coffee) = n(Tea)+ n(Coffee) - n (Tea ^ Coffee)
100 = 72 + 44 - x (x: is workers who take both coffee and tea)
x = 116 - 100 = 16.
Therefore Out of 100 workers, 16 take both coffee and tea.
But as per the problem 40 take both coffee and tea

100 --- 16
? ----- 40

(40/16) * 100 = 250.

47. A student scores 55% marks in 8 papers of 100 marks each. He scores 15% of his total marks in English. How much does he score in English? 
Solution:
the total mark is (55/100)*800=440

15% of 440 is (440/100)*15=66

(or)

Given student scores 55% marks in english in 8 papers of 100 marks each.
so,his total marks are (55*800)/100 =>440
15% of his 440 marks is 440*(15/100)=>66

so,he scored 66 marks in english

48. There are 600 boys in a hostel. Each plays either hockey or football or both. If 75% play hockey and 45% play football, how many play both ?
Solution:
((75*600)/100)+((45*600)/100)-600
=120

(or)

450 members plays only hockey., 270plays only football. So both game players 120


49. Prices register an increase of 10% on food grains and 15% on other items of expenditure. If the ratio of an employee's expenditure on food grains and other items be 2 : 5, by how much should his salary be increased in order that he may maintain the same level of consumption as before, his present salary being Rs. 2590 ?
Solution:
Total expenditure= Rs. 2590
Let common ratio be x
then, expenditure on food grains=2x and expenditure on other items = 5x
therefore, 2x+5x=2590
7x=2590
x=370
so 2x = 2*370=Rs. 740 and 5x = 5*370=Rs. 1850
increase of 10% on food grains so new price= 740+10% of 740
 = 740+74=Rs. 814
increase of 15% on other items so new price= 1850+15% of 1850 
= 1850+277.5=Rs. 2127.50

Sum of increased prices=814+2127.50=Rs. 2941.50
difference between new expenditure and old expenditure
= 2941.50-2590=Rs. 351.50


50. In an examination, 35% candidates failed in one subject and 42% failed in another subject while 15% failed in both subjects. If 2500 candidates appeared at the examination, how many passed in either subject but not in both ? 
Solution:
both fail=15% only fail in sub1=20% and 
only fail in subject2
=(42-15)%=27%
so only pass in subject1=27% and only pass in subject2
= 20% so (20+27)%=47% 
passed in either subject but not both
now 100% =2500

so 47%=(2500/100) * 47
=1175

51. If the price of petrol increases by 25% and Kevin intends to spend only an additional 15% on petrol, by what percentage must he reduce the quantity of petrol purchased?
Solution:
Say cost of petrol is 100. 
New cost = 125. 
Now he spends only 115, so would get only 115/125 = 0.92 of the quantity. 

SO he should reduce his consumption by 8%

(or)

assume petrol price=100rs and he spends 100rs for petrol
25% increase --> 125rs
but he spends additional 15% only --> 115 rs

tat means reduce % = (115/125)*100 = 8%

52. Peter got 30%, of the maximum marks in an examination and failed by 10 marks. However, Paul who took the same examination got 40% of the total marks and got 15 marks more than the passing marks. What was the passing marks in the examination?
Solution:
let the max mark be x n passing mark be y.
peter got 30% of max mark =.3x i.e (x*30/100)
but he failed by 10 mark
therefore passing mark i.e y=.3x+10
now, paul got 40% of max mark=.4x
but he get 15 marks more than passing mark 
therefore passing mark i.e y= .4x-15

now from eq (i) n eq(ii)
.4x-15=.3x+10
x=250 i.e max mark
therefore passsing mark(put value of x)

y=.4x-15=.3x+10= 85

53. In an election between two candidates, one got 55% of total valid votes. 20% of the votes were invalid. If the total number of votes was 7500 the number of valid votes that the other candidate got was :
Solution:
total number of votes : 7500
invalid votes are : 7500*(20/100) =1500
then valid votes are : 7500-1500 = 6000
one member got : 6000*(55/100) = 3300

other candidate got : 6000-3300 = 2700

54. At an election involving two candidates, 68 votes were declared invalid. The wining candidate scores 52% and wins by 98 votes. The total number of votes polled is :
Solution:
let total valid votes be 100 winner gets 52 votes loser gets 48 votes difference 52-48=4 
so if difference is 4 then valid votes be 100 by this if difference is 98 then valid votes be 
(98/4)*100=2450 
total votes =2450+68(invalid votes)=2518

55. On increasing the price of T.V. sets by 30%, their sale decreases by 20%. What is the effect on the revenue receipts of the shop ?
Solution:
Let, price = Rs.100, sale = 100
Then, sale value = Rs.(100×100)=Rs.10000
New sale value = Rs.(130×80)=Rs.10400

INCREASE%=400/10000*1OO=4% increse

56.  5 litres of water is added to a certain quantity of pure milk costing $3 per litre. If, by selling the mixture at the same price as before a profit of 20% is made, what is the amount of pure milk in the mixture?
Solution:
3 rs/lit = 1.2 *[x (rs/lit)]

hence actual price = 2.5 rs /lit
hence 2.5/(3-2.5) = amount/5

amount = 25 lit

57. Salary of Suresh is 25% less than the salary of Ramesh. By how much percent the salary of Ramesh is more than the salary of Suresh?
Solution:
using (r/100-r)100%
=25/75*100

=33 1/3%

58. The value of a machine depreciation at the rate of 10% every year. It was purchased 3 years ago. If its present value is Rs. 8748, its purchase price was :
Solution:
let x be the initial price.....
a/c question x*(1-(1/10))^3=8748

i-e:x=8748*(1000/729)=12000

59. At an election a candidate who gets 84% of the votes is elected by a majority of 476.Votes what is the total number of votes polled?
Solution:
84x-16x=476
68x=476

x=700

60. 5% of (25% of Rs1600) is
Solution:
25% of 1600=(25*1600)/100=400

5% of 400=(5*400)/100=20

61. If the cost of rice decreases by 20%, a man is able to buy 25 kg more for Rs.1000. Find the Cost Price per kg
Solution:
let earlier cost was Rs 100. 20% dec means Rs 80
let earlier he could bought x kg and now x+25
so x*1000=80*(x+25)
x=100
so 1000/100=10 Rs
now 1000/125=Rs 8

so price per kg is rs 8 

62. The price of sugar increases by 20%. By what percent must a house wife reduce the consumption of sugar, so that the expenditure on sugar is the same as before ?
Solution:
20/120*100=16.67

63. Raman's salary was decreased by 50% and susequently increased by 50%. He has a loss of :
Solution:
first take salary is 100%
100-50/100*100 = 50%
50+50/100*50 = 75%
so 100-75=25%

so decrease is 25%

64. 30% of the men are more than 25 years old and 80% of the men are less than or equal to 50 years old. 20% of all men play football. If 20% of the men above the age of 50 play football, what percentage of the football players are less than or equal to 50 years?
Solution:
20% of the men are above the age of 50 years. 20% of these men play football. Therefore, 20% of 20% or 4% of the total men are football players above the age of 50 years. 

20% of the men are football players. Therefore, 16% of the men are football players below the age of 50 years.


Therefore, the % of men who are football players and below the age of 50=(16/20)×100= 80%


65. 5000 votes in an election, 14% invalid. The Winner won by a margin of 15%. Find the number of votes secured by the winner
Solution:
14% are invalid...so 5000*14/100 =4300.
if we consider no of votes secured by winner is y, then the other candidate secures 4300-y votes. given difference between them is 15 % of the valid votes.
hence, y-(4300-y)=15*4300/100....

y=2472.5...approximately 2473

66. if there are 5,000 voters out of which 20% are not eligible to vote and there are two candidates contesting? The winning candidates won by 15% of votes? what is the total number of votes he got ?
Solution:
5000*.8=4000
4000*.15=600
x-(4000-x)=600

x=2300

67. A candidate who gets 20%marks fails by 11 marks but another candidate who gets 42% marks gets 22 marks more than the passing marks. Find the maximum marks.
Solution:
x*20/100+11=x*42/100-22

solve this question x=150

68. 75% of the participants in a graduation ceremony will receive an honorary award If 400 people participated in the ceremony, how many more people will receive awards than those who will not?
Solution:
25%400=100
75%400=300

300-100=200.

69. The Maruti price has risen by 25% and the sales have come down by 4%. What is the total percentage change in revenue?
Solution:
let initial price=100 n sales=100
initial revenue=100*100=10000
final price=125 n sales=96
final revenue=125*96=12000

%age change=(12000-10000/10000)*100=20

70. A man spends 75% of his income. His income increases by 20% and he increased his expenditure by 15%. His saving are then increased by 
Solution:
Let income = Rs. 100 
.-. Expenditure = Rs. 75 Saving = Rs. 25
New income = Rs. 120 
.-. Expenditure = (115/100) X 75 = 345/4
Saving = 120 - (345/4) = 135/4

Increase % in saving = (35/4) X (1/25) X 100 = 35%

71. Mukesh has twice as much money as Sohan and Sohan has 50% more money than what Pankaj has. If the average money with them is Rs. 110, then Mukesh has
Solution:
Let Mukesh ,Sohan and m,s,p resp.
then m=2s ,s=p*3/2
then avg =(2s+s+2/3*s)/3=110
so,s=90

m=2*s=2*90=180.

72. 1100 boys and 700 girls are examined in a test; 42% of the boys and 30% of thegirlsPass.The percentage of the total who failed is:
Solution:
Number of boys passed in the exam=(1100*42)/100=462
Boys failed in the exam=1100-462=638
Number of girls passed in the exam=(700*30)/100=210
Girls failed in the exam are=700-210=490

Total number of students failed in the exam=638+490=1128

Percentage of students who failed in exam 
=(1128*100)/1800=62.66

73. Anil's height is 20% less than Deepak's. How much is Deepak's height more than Anil's ?
Soluton:
{100*20/(100-20)}%=25%

74. If the price of petrol increases by 25 and kevin intends to spend only 15% more on petrol.By how much percent should he reduces the quantity of petrol that he buys?
Solution:
Let C.P = 100
petrol increased by 25 = 125 (new C.P)
Kevin spend only 15% = 115
(115/125)*100 = 92 ltr

100-92 = 8 %

75. Mr.John used to purchase certain number of mangoes for $360. Since the price of mangoes is reduced by 10% he got 12 more mangoes today. Find the original price of 120 mangoes
Solution:
let cost of mangoes be 'm'
10% of 360 = 12m
(10/100) * 360 = 12m
36 = 12m
m=3

cost of 120 mangoes = 120*3 
= $360

76. The retail price of rice decreased from Rs. 16 kg to Rs. 12 per kg. Find the percentage decrease. 
Solution:
decrease:16-12=4
%decrease=4/16*100

ans:25

77. The monthly salary of Mr. Anand is Rs. 8500. He spends 20% on education of his children, 25% as house rent, 30% on food, 5% on travels, 8% on miscellaneous things and rest he saves. Find his annual savings. 
Solution:
his total spendings are :20+25+30+5+8=88%
so total savings=12% of 8500 =1020

but anually he saves =12*1020 =12240