FSB ratio proportion with solutions

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For PT Faculty use only

Pinnacle Regular - Maths FSB - 2008-09

Pinnacle RegularMaths - Faculty Support Booklet (2008-09)

Ratio, Proportion, Variation,Mixture and Alligation (Chapter 5)

Ratio and Proportion

ØØ If pqr = 1 then 1

1 p q1

1 q r1

1 r p1 1 1+ ++

+ ++

+ +− − − is equivalent to

(1) p + q + r (2)1

p q r+ + (3) 1 (4) p q r1 1 1− − −+ +

Sol.From the given condition pqr = 1.

Substitute the values of p, q, r at random such as p q r= = =23

32

1, , . Ans.(3)

ØØ If ab c

bc a

ca b

r+

=+

=+

= , then r cannot take any other value except

(1) 1/2 (2) –1 (3) 1/2 or –1 (4) –1/2 or –1

Sol.a

b cb

c ac

a br

+=

+=

+=

By option, if r =12

⇒ 2a – b – c = 02b – c – a = 02c – a – b = 0

⇒ 2(a + b + c ) – (a + b + c ) – (a + b + c) = 0Similarly r = –1 is also satisfied. Ans.(3)

ØØ A student gets an aggregate of 60% marks in five subject in the ratio 10 : 9 : 8 : 7 : 6. If thepassing marks are 50% of the maximum marks and each subject has the same maximum marks, inhow many subjects did he pass the examination?(1) 2 (2) 3 (3) 4 (4) 5

Sol.Let his marks be 10, 9, 8, 7 and 6 in the five subjects. Hence, totally he has scored 40 marks. This constitutesonly 60% of the total marks. Hence, total marks 40/0.6 = 66.7 or 67 approx. , which is the maximum marksin all 5 subjects. Since the total marks in each subject is the same, hence maximum marks in each subjectwill be 67/5 = 13 approx. Out of this 50% is the passing marks . In other words to pass in a subject, heneeds to score 6.5 marks. We can see that only in 1 subject, he scored less than this viz. 6. Hence, hepassed in 4 subject. Ans.(3)

Pinnacle Regular - Maths FSB - 2008-09 (2) of (8)

For PT Faculty use onlyØØ The cost of diamond varies directly as the square of its weight. Once, this diamond broke into

four pieces with weights in the ratio 1 : 2 : 3 : 4. When the pieces were sold, the merchant gotRs.70,000 less. Find the original price of the diamond.(1) Rs.1.4 lakh (2) Rs.2 lakh (3) Rs.1 lakh (4) Rs.2.5 lakh

Sol.Let the original weight of the diamond be 10x. Hence, its original price will be k(100x2), where k is aconstant. The weights of the pieces after breaking are x, 2x, 3x and 4x. Therefore, their prices will be kx2,4kx2, 9kx2 and 16kx2. So the total price of the pieces = (1 + 4 + 9 + 16) kx2 = 30kx2. Hence, the differencein the price of the original diamond and its pieces = 100kx2 – 30kx2 = 70kx2 = 70000. Hence, kx2 = 1000 andthe original price = 100 kx2 = 100 × 1000 = 100000 = Rs.1 lakh. Ans.(3)

ØØ Three friends went for a picnic. First brought five apples and the second brought three. Thethird friend however brought only Rs.8. What is the share of the first friend?(1) 8 (2) 7 (3) 1 (4) None of these

Sol.The number of apples = 8, so the amount eaten by each of the three is 8/3 apples therefore first friendshould be paid for 5 – (8/3) and second friend should be paid for 3–(8/3) apples. They should distribute thesum of Rs.8 in ratio 7/3 : 1/3, i.e., 7 : 1. Ans.(2)

ØØ Total salary of A, B & C is Rs.350. If they spend 75%, 80% & 56% of their salaries respectivelytheir savings are as 10 : 12 : 33. Find their salaries.

Sol.A’s saving = 100 – 75 = 25% of his salary. B’s saving = 100 – 80 = 20% of his salaryC’s saving = 100 – 56 = 44% of his salary25/100 of A’s salary : 20/100 of B’s salary : 44/100 of C’s salary = 10 : 12 : 33or 25 × A’s salary : 20 × B’s salary : 44 × C’s salary = 10 : 12 : 33or 25 × A’s salary / 20 × B’s salary = 10/12or A’s salary : B’s salary = 2 : 3,B’s salary : C’s salary = 4 : 5Thus A : B = 2 : 3, B : C = 4 : 5 Now making B common we haveA : B = 8 : 12, B : C = 12 : 15, or A : B : C = 8 : 12 : 15Total salary = 350 ⇒ A’s salary = 8 / (8 + 12 + 15) × 350 = 80B’s salary = 12 / (8 + 12 + 15) = 120, and C’s Salary = 150 Answer.

ØØ The ratio of the age of a man and his wife is 4 : 3. After 4 years, this ratio will be 9 : 7. If at thetime of the marriage, the ratio was 5:3, then how many years ago they were married?(1) 12 years (2) 8 years (3) 10 years (4) 15 years

Sol.Man’s age = 4k, (say)Wife’s age = 3k, (say)

∴ 4 43 4

97

8kk

k++

= ⇒ = .

∴ Man’s age = 32 yearsWife’s age = 24 years. Suppose they were married x years ago.

∴ 3224

53

12−−

= ⇒ =xx

x . Ans.(1)

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ØØ Two full tanks, one shaped like a cylinder and the other like a cone, contain jet fuel. The cylindricaltank holds 500 litres more than the conical tank. After 200 litres of fuel has been pumped outfrom each tank the cylindrical tank contains twice the amount of fuel in the conical tank. Howmany litres of fuel did the cylindrical tank have when it was full?(1) 700 (2) 1000 (3) 1100 (4) 1200

Sol.Work backwards from the options. If the cylinder has a capacity of 1200 litre, then the conical vessel shallhave a capacity of 700 litres. Once 200 litres have been taken out from the same, the remaining holding ofeach of them shall be 1000 & 500.

Alternate: Let the cylinder has a capacity of X litre, then the conical vessel shall have a capacity of (x – 500) litres. (x – 200) = 2 (x – 700) = x = 1200. Ans.(4)

ØØ The reduction in the speed of an engine is directly proportional to the square of the number ofbogies attached to it. The speed of the train is 100 km/hr when there are 4 bogies and 55 kmphwhen there are 5 bogies. What is the maximum number of bogies that can be attached to thetrain so that it can move?(1) 6 (2) 5 (3) 4 (4) None of these

Sol.Suppose Reduction in speed is R. Speed of the engine without any bogie = knumber of bogies attached = b, proportionality constant = c, Resultant speed = sWe have R = cb2 and s = k – R = k – cb2

100 = k – c (4)2 or, 100 = k – 16c ...(i)and 55 = k – c(5)2 or 55 = k – 25 c ...(ii)Solving (i) and (ii) we get k = 180 and c = 5. Now we have S = 180 – 5b2. If we put b = 6, S = 0∴ At most we can attach 5 bogies to the engine. Ans.(2)

ØØ Arvind Singh purchased a 40 seater bus. He started his services on route number 2 (from MahuNaka to Dewas Naka with route length of 50 km). His profit (P) from the bus depends upon thenumber of passengers over a certain minimum number of passengers ‘n’ and upon the distancetravelled by bus. His profit is Rs.3600 with 29 passengers in the bus for a journey of 36 km andRs.6300 with 36 passengers in the bus for a journey of 42 km. What is the minimum number ofpassengers are required so that he will not suffer any loss.(1) 12 (2) 20 (3) 18 (4) 15

Sol.The minimum number of passengers n, at which there is no loss and number of passengers travelling = mand let the distance travelled is d, ThenP ∝(m – n)dor p = k(m – n)d; k is a constant.When P = 3600, m = 29 and d = 36, then

3600 = k(29 – n) × 36 ...(1)Again, when p = 6300, m = 36, d = 42, then

6300 = k(36 – n) × 42 ...(2)Dividing equation (2) by (1)

63003600

36 42

29 36=

− ×

− ×

k n

k nb gb g ⇒

36

2996

−−

=n

nb gb g ⇒ 3n = 45 ⇒ n = 15

Hence to avoid loss, minimum number of 15 passengers are required. Ans.(4)


For PT Faculty use onlyAllegations

ØØ A milkman mixes 20 litres of water with 80 litres of milk. After selling one–fourth of this mixture,he adds water to replenish the quantity that he has sold. What is the current proportion ofwater to milk?(1) 2 : 3 (2) 1 : 2 (3) 1 : 3 (4) 3 : 4

Sol.

Water Milk

Initially 20 80

After Selling one-fourth (20 – 5) = 15 (80 – 20) = 60

After adding water to replenish the quantity 40 60

Required ratio = 2 : 3. Ans.(1)

ØØ An alloy contains 24% of tin by weight. How much more tin to the nearest kg must be added to100 kg of the alloy so that the percentage of tin may be doubled?

Sol.Let X kg of tin be added to the alloy (24 + X) / (100 + X) = 2 (24/100) ⇒ X = 46.Hence 46 kg of tin must be added to the alloy.

ØØ Two containers contain equal quantities of milk and water respectively. Half the contents of thefirst are poured in the second and then the same quantity is transferred back into the firstcontainer. This is done three times. What is the ratios of milk to water in the two containers atthe end of the process?(1) 5 : 2, 2 : 5 (2) 5 : 4, 4 : 5 (3) 14 : 13, 13 : 14 (4) None of these

Sol. Start with a litre of milk in 1st container and a litre of water in 2nd proceed. Ans.(3)

ØØ A total of ‘a’ litres of pure acid were drawn from a tank containing 729 litres of pure acid andwas replaced by water. The result was thoroughly mixed to obtain a homogenous solution andthen another ‘a’ litres of solution was drawn off, and again replaced by water, and againthoroughly mixed. This procedure was performed six times and thus the tank contained 64 litresof pure acid. Determine ‘a’.(1) 1/3 (2) 243 (3) 81 (4) 3

Sol. Here 729 × {(729 – a)/729}6 = 64 or, (729 – a)/729 = (64/729)1/6 or a = 243. Ans.(2)

ØØ Three qualities of milk costing Rs.3, Rs.3.25 and Rs.2.60 per litre are mixed and the mixture isthen sold at Rs.3.54 per litre to earn a profit of 20%. In what proportion should the threequalities of milk be mixed?(1) 1 : 2 : 3 (2) 2 : 1 : 3 (3) 1 : 1 : 1 (4) 3 : 2 : 1

Sol.The mixture is sold at a profit of 20% at Rs.3.54 ∴ the actual value of the mixture is 3.54/1.2 = 2.95Now, ratio in which they are to be mixed can be calculated by3x + 3.25y + 2.60z = 2.95. Putting x = y = z = 1/3 we get, 1 + 1.0833 + .8667 = 2.95∴ the three qualities should be mixed in the same ratio i.e. 1 : 1 : 1. Ans.(3)

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ØØ In a laboratory experiment, a sample of Air, which is a mixture of only oxygen and water vapouris taken. Water vapour contains hydrogen and oxygen gases. If Air contains a total of 70%oxygen (including that contained in the water vapour) by weight while water vapour contains

1623

% of oxygen by weight, how many kilograms of water vapour is present in 1 kilogram of

air?(1) 0.3 (2) 0.36 (3) 0.34 (4) 0.25

Sol.Since 70% of air is oxygen, remaining 30% is hydrogen. In water vapour, if one unit is oxygen, i.e., five-sixth is hydrogen.⇒ water vapour = 30%⇒ water vapour = 6/5 × 30% (of air)∴ In 1 kg of air, water vapour is 0.36 kg. Ans.(2)

ØØ In what ratio should two varieties of rice costing Rs.12 per kg and Rs.18 per kg respectively bemixed so that the resulting mixture when mixed with another variety of rice costing Rs.20 per kgin the ratio 4 : 3, would yield a mixture costing Rs.16 per kg?(1) 5 : 1 (2) 7 : 2 (3) 9 : 4 (4) None of these

Sol.Let, the mixture of varieties of rice costing Rs.12 per kg and Rs.18 kg per kg is costing Rs.x per kg. It isgiven that

1620 16

34

−−

=x

⇒ x = Rs.13 per kg.

5

12 18

113

Ans.(1)

ØØ 20 litres of milk when added to a 60 litre milk and water solution increases the concentration bysame percentage points as decreased by addition of 30 litres of water to the same solution.What is the ratio of milk and water initially?(1) 1 : 2 (2) 2 : 1 (3) 3 : 4 (4) 4 : 5

Sol.Let k litres be the amount of milk in 60 litres of solution. So the concentration of milk = k/60. If 20 litres ofmilk is added, concentration of milk = (k + 20) / 80. If 30 litres of water is added, milk concentration = k/90.

So we have k k k k+

− = −20

80 60 60 90 or k = 180/7. So milk to water ratio = (180/7) : (240/7) = 3 : 4. Ans.(3)


For PT Faculty use onlyØØ In a milk Shop there are three varieties of milk, ‘Pure’, ‘Cure’ and ‘Lure’. The ‘Pure’ milk has

100% concentration of milk. The ratio of milk to water in the ‘Cure’ is 2:5 and in the Lure it is 3:8respectively. Sonali purchased 14 litres of Cure and 22 litres of Lure milk and mixed them. If shewanted to make the concentration of milk in the mixture of purchased milk to 50%. How manylitres of ‘Pure’ Milk she is needed?(1) 6 litres (2) 8 litres (3) 16 litres (4) 18 litres

Sol. Pure Cure Lure100% 40% 37 5%

11

25

38

.

Milk

Milk

Water

Water

Mixture

Cure

Lure

New mixture

4l

10l

6l

16l

26l

10l

26l

16l

26l

14l

22l

+

Required mixutre 1 : 1

Since in the required mixture the ratio of milk and water is 1 : 1 so she has to add up 16 litre of more milk(pure) to get it, for the fixed quantity of water. Ans.(3)

ØØ In a mixture of petrol and kerosene petrol is only 99 litres. If this same quantity of petrol wouldbe presented in another mixture of petrol and Kerosene where total volume would be 198 litresless than the actual mixture then the concentration of petrol in the present mixture would havebeen 13.33% point less than that. What is the concentration of petrol in actual mixture?(1) 20% (2) 16.66% (3) 26.66% (4) 8.33%

Sol.Petrol Kerosene Total Mixture99 x 99 + x99 (x – 198) (x – 99)

Again 99

99100

9999

100 13 33x x−

× −+

× =b g b g .

or 990099 99

9913 332 2

x xx

+ − +−

FHG

IKJ = . or 9900 198

994032 2

( )x −

=

⇒ x2 – 992 = 992 × 15 ⇒ x2 = (99)2 × (16) ⇒ x = 99 × 4 = 396 litres

Therefore the actual concentration of petrol =99

99 39620%

+=b g . Ans.(1)

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ØØ A vessel of capacity 2 litre has 25% alcohol and another vessel of capacity 6 litre had 40%alcohol. This total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus therest part of the vessel was filled with the water. What is the new concentration of mixture?(1) 31% (2) 71% (3) 49% (4) 29%

Sol.Amount of alcohol in first vessel = 0.25 × 2 = 0.5 litreamount of alcohol in second vessel = 0.4 × 6 = 2.4 litreTotal amount of alcohol out of 10 litres of mixture is 0.5 + 2.4 = 2.9 litre

Hence, the concentration of the mixture is 29% = ×FHG

IKJ

2 910

100. . Ans.(4)

ØØ Alloy A contains 40% gold and 60% silver. Alloy B contains 35% gold and 40% silver and 25%copper. Alloys A and B are mixed in the ratio 1:4. What is the ratio of gold and silver in thenewly formed alloy is?(1) 20% and 30% (2) 36% and 44% (3) 25% and 35% (4) 49% and 36%

Sol.Assume the weight of Alloy A is 100 kg.∴ The weight of alloy B is 400 kg.∴ Gold Silver Copper

A 40 kg 60 kg 0 kgB 140 kg 160 kg 100 kgtotal → 180 kg 220 kg 100 kg

∴ Ratio of Gold and Silver in new alloy = 180500

200500

: = 36%:44%. Ans.(2)

ØØ Dia and Urea are two chemical fertilizers. Dia is consists of N, P and K and Urea consists of onlyN and P. A mixture of Dia and Urea is prepared in which the ratio of N, P and K is 26%, 68% and6% respectively. The ratio of N, P and K in Dia is 20%, 70% and 10% respectively. What is theratio of N and P in the Urea?(1) 27% and 63% (2) 33% and 67% (3) 35% and 65% (4) 70% and 30%

Sol. Urea

N N

N

P P

P

K K

K

x 20%

26%

y 70%

68%

0 10%

6%

Dia

Mixture

This 6% of K is obtained only from Dia.

N N

N

P P

P

K K

K

x 120

260

y

680

0 60

60

Dia

Mixture

Urea∴


For PT Faculty use onlyNU + ND = NM ⇒ NU + 120 = 260 ⇒ NU = 140.

and PU + PD = PM ⇒ PU + 420 = 680 ⇒ PU = 260U, D, M → Urea, Dia and mixture∴ Amount of N in Urea = 140 and amount of P in Urea = 260∴ Ratio of N : P = 7:13 ⇒ 35 : 65. Ans.(3)

ØØ Last year in CAT, each section of the question paper had a different weightage. The weightageof QA, DI and VA/RC sections was 8, 9 and 10 respectively. The maximum marks in all the threesections together were 810. Wrong answer did not carry negative marks as a penalty. If Padmahad gotten 20% more marks in QA and 8% more marks in DI and 7.14% more marks in VA/RC,then she must had gotten 100% marks in all the three sections. The total marks that Padma hadscored(1) 730 (2) 700 (3) 750 (4) 775

Sol.8x + 9x + 10x = 810⇒ x = 30Total marks in QA → 240

DI → 270

VARC

→ 300

Now her score in QA → 24012

200.

= Her score in DI = 2701 08

250.

=

Her score in VARC

= 300

1 0714280

.= Her total score = 200 + 250 + 280 = 730. Ans.(1)

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FSB ratio proportion with solutions