For PT Faculty use only Pinnacle Regular Maths - Faculty Support Booklet (2008-09) Ratio, Proportion, Variation, Mixture and Alligation (Chapter 5) Ratio and Proportion Ø
If pqr = 1 then
1 1
+
p
+
(1 ) p + q + r
q
1
−
(2)
1
+
1
+
q +r
1
1
+
−
1
+
r
+
1 p + q +r
is equivalent to
p −1
( 4 ) p −1
(3) 1
+
q −1
+
r −1
Sol. From the given condition pqr = 1. Substitute the values of p, q, r at random such as p
Ø
I f
a b
+
c
=
b c
+
a
=
c a +b
(1 ) 1/ 2
Sol.
a b+c
=
=
c+a
=
c
a+b 1 By option, if r = 2 ⇒ 2a – b – c = 0
=
2 3 , q = ,r 3 2
=
1 . Ans.(3)
r , then r cannot take any other valu e except (2) –1
b
=
(3) 1 / 2 or – 1
(4) – 1 / 2 o r – 1
r
2b – c – a = 0 2c – a – b = 0 ⇒
2(a + b + c ) – (a + b + c ) – (a + b + c) = 0
Similarly r = –1 is also satisfied. Ans.(3) satisfied. Ans.(3) Ø
A student gets an aggregate of 60 % mark s in five subject in the rat io 10 : 9 : 8 : 7 : 6. If the p a s s i n g m a r k s a r e 5 0 % o f t h e m a x i m u m m a r k s a n d e a c h s u b j e c t h a s t h e s a m e m a x i m u m m a r k s , in in h o w m a n y s u b j e c t s d i d h e p as as s t h e e x a m i n a t i o n ? (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 constitutes only 60% of the total marks. Hence, t otal marks 40/0.6 = 66.7 or 67 approx. , which is the maximum marks in all 5 subjects. Since the total marks in each subject is the same, hence maximum marks in each subject will be 67/5 = 13 approx. Out of this 50% is the passing marks . In other words to pass in a subject, he needs to score 6.5 marks. We can see that only in 1 subject, he scored less than this viz. 6. Hence, he passed in 4 subject. Ans.(3) subject. Ans.(3) (1) of (8)
Pinnacle Regular - Maths FSB - 2008-09
For PT Faculty use only Ø
The cost of diamond varies directly as the square of its w eight. Once, this diamon d broke into four pieces with w eights in the ratio 1 : 2 : 3 : 4. When th e pieces were sold, the merchant got Rs.70,000 less. Find the origin al price of the diamo nd. ( 1 ) R s .1 . 4 l a k h
( 2 ) R s .2 l a k h
( 3 ) R s .1 l a k h
( 4 ) R s .2 .5 l a k h
Sol. Let the original weight of the diamond be 10x. Hence, its original price will be k(100x 2 ), where k is a constant. The weights of the pieces after breaking are x, 2x, 3x and 4x. Therefore, their prices will be kx2, 4kx 2 , 9kx 2 and 16kx 2 . So the total pr ice of the pieces = ( 1 + 4 + 9 + 16) kx 2 = 30kx 2. Hence, the difference in the price of the original diamond and its pieces = 100kx2 – 30kx 2 = 70kx 2 = 70000. Hence, kx 2 = 1000 and the original price = 100 kx 2 = 100 × 1000 = 100000 = Rs.1 lakh. lakh. A n s . ( 3 ) Ø
Three friends went for a picnic. First brought five apples and the second brou ght three. The third friend how ever brought only Rs.8. What is the share of the first friend? (1) 8
(2) 7
(3 ) 1
(4 ) N o n e o f t h e s e
Sol. The number of apples = 8, so the amount eaten by each of the three is 8/3 apples therefore first friend should be paid for 5 – (8/3) and second fr iend should be paid for 3–(8/3) apples. They shoul d distribute the sum of Rs.8 in rati o 7/3 : 1/3, i.e., 7 : 1. Ans.(2) 1. Ans.(2) Ø
Total salary of A, B & C is Rs.350. If they spend 75% , 80% & 56% of their salaries respectively their 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 salary C’s saving = 100 – 56 = 44% of his salary 25/100 of A’s salary : 20/100 of B’s salary : 44/100 of C’s salary = 10 : 12 : 33 or 25 × A’s salary : 20 × B’s salary : 44 × C’s s alary = 10 : 12 : 33 or 25 × A’s salary / 20 × B’s salary = 10/12 or A’s salary : B’s salary = 2 : 3, B’s salary : C’s salary = 4 : 5 Thus A : B = 2 : 3, B : C = 4 : 5 No w making B common we h ave A : B = 8 : 12, B : C = 12 : 15, or A : B : C = 8 : 12 : 15 Total salary = 350
⇒
A’s A’s salary = 8 / (8 + 12 + 15) × 350 = 80
B’s salary = 12 / ( 8 + 12 + 15) = 120, and C’ s Salary = 150 Answer. Ø
The ratio of the age of a m an and his w ife is 4 : 3. After 4 years, this ratio w ill be 9 : 7. If at the time of the marriage, the ratio was 5:3, then how many years ago they were marr ied? (1) 1 2 y e a r s
(2) 8 y e a r s
(3 ) 1 0 y e a r s
(4 ) 1 5 y e a r s
Sol. Man’s age = 4k, (say) Wife’s age = 3k, (say) ∴
∴
4k + 4 3k + 4
=
9 7
⇒
k = 8.
Man’s age = 32 years
Wife’s Wife’s age age = 24 year years. s. ∴
32 − x 24 − x
=
5 3
⇒
x
=
Suppos Suppose e they they were were marri married ed x year yearss ago. ago. 12. Ans.(1)
Pinnacle Regular - Maths FSB - 2008-09
(2) of (8)
For PT Faculty use only Ø
Two full tank s, one shaped like a cylinder and the other lik e a cone, contain jet fuel. The cylindrical t a n k h o l d s 5 0 0 l i t r e s m o r e t h a n t h e c o n i c a l t a n k . A f t e r 2 0 0 l i t r e s o f fu fu e l h a s b e e n p u m p e d o u t from each tank the cylindrical tank cont ains tw ice the amount of fuel in the conical tank. How many litres of fuel did the cylindrical tank have wh en 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 shall have a capacity of 700 litres. Once 200 litres have been taken out from the same, the remaining holding of each of them shal l be 1000 & 500. A l t e r n a t e : 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. A n s . ( 4 ) Ø
The reduction in the speed of an engine is directly proportion al to the square of the num ber of b o g i e s a t t a c h e d t o i t . Th Th e s p e e d o f t h e t r a i n i s 1 0 0 k m / h r w h e n t h e r e a r e 4 b o g i e s a n d 5 5 k m p h wh en there are 5 bogies. What is the max imum n umber of bogies that can be attached to the train so that it can m ove? (1) 6
(2) 5
(3) 4
(4) N o n e o f t h e s e
Sol. Suppose Reduction in speed is R. Speed of the engine without any bogie = k number of bogies attached = b, proportional ity constant = c, Resultant speed = s We have R = cb 2 and s = k – R = k – cb 2 100 = k – c (4) 2 o r, 1 00 = k – 1 6c and 55 = k –
c(5) 2
.. . ( i )
o r 55 = k – 2 5 c
... (i i )
Solving ( i) and (ii) we ge t k = 180 and c = 5. Now we have S = 180 – 5b 2 . If we put b = 6, S = 0 At most we can attach 5 bogies to the engine. Ans.(2) engine. Ans.(2)
∴
Ø
Arvind Singh pu rchased a 40 seater bus. He started his services on route num ber 2 (from M ahu Naka to Dew as Naka wit h route length of 50 km ). His profit (P) from the bus depends upon the number of passengers over a certain minimum n umber of passengers ‘n’ and upon the distance t r a v e l l e d b y b u s . H i s p r o f i t i s R s .3 .3 6 0 0 w i t h 2 9 p a s s e n g e r s i n t h e b u s f o r a j o u r n e y o f 3 6 k m a n d Rs.6300 with 36 passengers in the bus for a journey of 42 km. What is the min imum n umber of passengers are required so that he w ill 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 = m and let the distance travelled is d, Then P
∝
or
(m – n)d p = k( k( m – n) n) d; d; k is is a co co ns ns ta ta nt nt .
When P = 3600, m = 29 and d = 36, then 3 6 00 = k ( 29 – n ) × 36
.. .( 1)
Again, when p = 6300, m = 36, d = 42, th en 6 3 00 = k ( 36 – n ) × 42
.. .( 2)
Dividing equation (2) by (1) 6300 3600
g ng
k 36 − n =
b
k 29 −
× ×
42 36
g ng
36 − n
⇒
b29
−
=
9 6
⇒
3n = 45
⇒
n = 15
Hence to avoid loss, minimum number of 15 passengers are required. required. A n s . ( 4 ) (3) of (8) Pinnacle Regular - Maths FSB - 2008-09
For PT Faculty use only Allegations Ø
A milkm an mixes 20 litres of w ater with 80 l itres of milk. After selling one–fourth of this mixture, he adds water to replenish th e quantity that he has sold. What is the current propor tion of water to milk? (1) 2 : 3
(2) 1 : 2
(3 ) 1 : 3
(4 ) 3 : 4
W ater Initially
M ilk
20
Sol. After Selling one-fourth
(20 – 5) = 15
After adding water to replenish the quantity
80 (80 – 20) = 60
40
60
Required ratio = 2 : 3. 3. A n s . ( 1 ) Ø
An alloy contains 24% of tin by weight. How mu ch more tin to the nearest kg must be added to 100 kg of the alloy so that th e percentage of tin may be doubled?
Sol. Let X kg of tin be added to t he alloy (24 + X) / (100 + X) = 2 (24/100)
⇒
X = 46.
Hence 46 kg of tin must be added to the all oy. oy. Ø
T w o c o n t a i n e r s c o n t a i n e q u a l q u a n t i t i e s o f m i l k a n d w a t e r r e s p e c t i ve ve l y . H a l f t h e c o n t e n t s o f t h e first are poured in the second and then the same quantity is transferred back into the first c o n t a i n e r . Th Th i s i s d o n e t h r e e t i m e s . W h a t i s t h e r a t i o s o f m i l k t o w a t e r i n t h e t w o c o n t a i n e r s a t the end of the pro cess? (1) 5 : 2 , 2 : 5
(2) 5 : 4 , 4 : 5
(3 ) 1 4 : 1 3 , 1 3 : 1 4
(4 ) N o n e o f t h e s e
S o l . Start with a litre of milk in 1st container and a litre of water in 2nd proceed. proceed. A n s . ( 3 ) Ø
A total of ‘a’ litres of pure acid were draw n from a tank con taining 729 l itres of pure acid and w a s r e p l a c e d b y w a t e r . Th Th e r e s u l t w a s t h o r o u g h l y m i x e d t o o b t a i n a h o m o g e n o u s s o l u t i o n a n d then another ‘a’ litres of solution was drawn off, and again replaced by water, and again t h o r o u g h l y m i x e d . Th Th i s p r o c e d u r e w a s p e r f o r m e d s i x t i m e s a n d t h u s t h e t a n k c o n t a i n e d 6 4 l i t r e s of pure acid. Determine ‘a’. (1) 1/ 3
(2 ) 243
(3 ) 81
(4 ) 3
S o l . Here 729 × {(729 – a)/729} 6 = 64 or, or, (729 – a)/ 729 = (64/729) 1/6 or a = 243. A n s . ( 2 ) Ø
T h r e e q u a l i t i e s o f m i l k c o s t i n g R s . 3 , R s . 3 .2 .2 5 a n d R s . 2 . 6 0 p e r l i t r e a r e m i x e d a n d t h e m i x t u r e i s t h e n s o l d a t R s . 3 .5 .5 4 p e r l i t r e t o e a r n a p r o f i t o f 2 0 % . I n w h a t p r o p o r t i o n s h o u l d t h e t h r e e qualities of milk be mixed? (1) 1 : 2 : 3
(2) 2 : 1 : 3
Sol. The mixture is sold at a profit of 20% at Rs.3.54
(3 ) 1 : 1 : 1 ∴
(4 ) 3 : 2 : 1
the actual value of the mixtu re is 3.54/1.2 = 2.95
Now, ratio in which they are to be mixed can be calculated by 3x + 3.25y + 2.60z = 2.95. Putt ing 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) 1. Ans.(3) Pinnacle Regular - Maths FSB - 2008-09 ∴
(4) of (8)
For PT Faculty use only Ø
In a laboratory experiment, a sample of Air, which is a m ixture of only oxygen and w ater vapour is taken. Water vapour cont ains hydrogen and ox ygen gases. If Air contai ns a total of 70% o x y g e n ( i n c l u d i n g t h a t c o n t a i n ed ed i n t h e w a t e r v a p o u r ) b y w e i g h t w h i l e w a t e r v a p o u r c o n t a i n s 2 % o f o x y g e n b y w e i g h t , h o w m a n y k i l o g r a m s o f w a t e r v a p o u r i s p r e s en en t i n 1 k i l o g r a m o f 3 air? 16
( 1 ) 0 .3
( 2 ) 0 .3 6
( 3 ) 0 .3 4
( 4 ) 0 .2 5
Sol. Since 70% of air is oxygen, rema ining 30% is hydrogen . In water vapour, vapour, if one unit is oxygen, i. e., fivesixth is hydrogen.
Ø
⇒
water vapour = 30%
⇒
water vapour = 6/5 × 30% (of air)
∴
In 1 kg of air, water vapour is 0.36 kg. Ans.(2)
I n w h a t r a t i o s h o u l d t w o v a r i e t i e s o f r i c e c os o s t i n g R s . 1 2 p e r k g a n d R s . 1 8 p e r k g r e s p e ct ct i v e l y b e m i x e d s o t h a t t h e r e s u l t i n g m i x t u r e w h e n m i x e d w i t h a n o t h e r v a r i e t y o f r i c e c o s t i n g R s .2 .2 0 p e r k g in the ratio 4 : 3, would yield a mix ture costing R s.16 per kg? (1) 5 : 1
(2) 7 : 2
(3) 9 : 4
(4) N o n e o f t h e s e
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 is given that 16 − x 20
− 16
3 =
4
⇒
x = Rs.13 per kg.
12
18 13
5
1
Ans.(1) Ø
2 0 l i t r e s o f m i l k w h e n a d d e d t o a 6 0 l i t r e m i l k a n d w a t e r s o l u t i o n i n c r e a s e s t h e c o n c e n t r at at i o n b y same percentage points as decreased by addition of 30 litres of water to the same solution . What is the ratio of milk and w ater 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 concentratio n of milk = k/60. If 20 litres of milk is added, concentration of milk = (k + 20) / 80. If 30 litres of water is added, milk concentration = k/90. So we have
(5) of (8)
k
20 80 +
−
k 60
=
k 60
−
k or k = 180/7. 180/7. So milk to water rati o = (180/7) : (240/7) = 3 : 4. Ans.(3) 4. Ans.(3) 90
Pinnacle Regular - Maths FSB - 2008-09
For PT Faculty use only Ø
In a mi lk Shop th ere are three varieties of milk , ‘Pure’, ‘Cure’ and ‘Lure’. The ‘Pure’ milk has 1 0 0 % c o n c e n t r a t i o n o f m i l k . Th Th e r a t i o o f m i l k t o w a t e r i n t h e ‘ C u r e ’ i s 2 :5 :5 a n d i n t h e L u r e i t i s 3 :8 :8 respectively. Sonali purchased 14 litres of Cure and 22 litres of Lure milk and mix ed them. If she w a n t e d t o m a k e t h e c o n c e n t r at a t i o n o f m i l k i n t h e m i x t u r e o f p u r c h a s ed ed m i l k t o 5 0 % . H o w m a n y litres of ‘Pure’ Milk she is needed? (1) 6 l i t r e s
Sol. Pure
(2) 8 l i t r e s
Cu r e
(3 ) 1 6 l i t r e s
(4 ) 1 8 l i t r e s
Lu re
100% 40% 37.5% 5% 1 2 3 1 5 8 Milk Water
Mixture
Cure
4l
10l
14l
Lure
6l
16l
22l
New mixture
Milk Water 10l
26l
16l
+
26l
26l :
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 t he fixed quantit y of water. water. Ans.(3) Ø
In a m ixture of petrol and kerosene petrol is only 99 litres. If this same quantity of petrol w ould be presented in another mixture of petrol and Kerosene where total volume would be 198 litres less than the actual mixture then th e concentration of petrol in the present mixture w ould have b e e n 1 3 .3 . 3 3 % p o i n t l e s s t h a n t h a t . W h a t i s t h e co co n c e n t r a t i o n o f p e t r o l i n a c t u a l m i x t u r e ? ( 1 ) 20 20%
Sol. Pet r ol
( 2 ) 16 1 6 .6 6 %
( 3 ) 26 2 6 .6 6 %
Ke r o s e n e
T o t a l Mi xt u r e
99
x
99
or ⇒
99 + x
( x – 1 98 )
Again
b
99 x − 99
9900
g
GH x
+
×
100 −
b
99 x + 99
99 − x + 99 x2
−
992
x 2 – 99 2 = 99 2 × 15
g
JK
=
⇒
×
(x – 99)
100 = 13.33 33
x 2 = (99) 2 × (16)
9900 9900(198 198 ) 2 x − 992
or
1333 . ⇒
Therefore the actual concentration of petrol =
Pinnacle Regular - Maths FSB - 2008-09
( 4 ) 8. 8 .3 3 %
b
=
40 3
x = 99 × 4 = 396 litres 99 99 + 396
g
=
20% . Ans.(1)
(6) of (8)
For PT Faculty use only A vessel of capacity 2 litre has 25% alcohol and another vessel of capacity 6 litre had 40% a l c o h o l . Th Th i s t o t a l l i q u i d o f 8 l i t r e w a s p o u r e d o u t i n a v e s s e l o f c a p a c i t y 1 0 l i t r e a n d t h u s t h e r e s t p a r t o f t h e v e s s e l w a s f i l l e d w i t h t h e w a t e r . W h a t i s t h e n e w c o n c e n t r at at i o n o f m i x t u r e ?
Ø
( 1 ) 31%
( 2 ) 71%
( 3 ) 49%
( 4 ) 29%
Sol. Amount Amount of alcohol in first vessel = 0.25 × 2 = 0.5 litre amount of alcohol in second vessel = 0.4 × 6 = 2.4 litre Total amount of alcoh ol out of 10 litres of mixture is 0.5 + 2.4 = 2.9 litre Hence, the concentration of the mixture is 29%
Ø
HG
=
2.9 10
×
JK
100 . Ans.(4)
A l l o y A c o n t a i n s 4 0 % g o l d a n d 6 0 % s i l v er e r . A ll l l o y B c o n t a i n s 3 5 % g o l d a n d 4 0 % s i l v er er a n d 2 5 % copper. Alloys A and B are mixed in t he ratio 1:4. What is t he ratio of gold and silver in t he newly formed alloy is? (1) 2 0 % an d 3 0 %
(2) 36% and 44%
(3) 2 5 % an d 35 %
(4) 49 % an d 36 %
Si l ve r
Cop pe r
Sol. Assume the weight of Alloy A is 100 kg. ∴
The weight of alloy B is 400 kg. Gold
∴
A
40 kg B total
∴
Ø
→
60 kg
0 kg
1 40 k g
16 0 k g
10 0 k g
1 80 k g
22 0 k g
10 0 k g
Ratio of Gold and Silver in new alloy =
180 200 : = 36%:44%. 36%:44%. A n s . ( 2 ) 500 500
D i a a n d U r e a a r e t w o c h e m i c a l f e r t i l i z e r s . D i a i s c o n si si s t s o f N , P a n d K a n d U r e a c o n s i s t s o f o n l y N and P. A mixture of Dia and Urea is prepared in which th e ratio of N, P and K is 26% , 68% and 6 % r e s p ec e c t i v e ly l y . Th Th e r a t i o o f N , P a n d K i n D i a i s 2 0 % , 7 0 % a n d 1 0 % r e s p ec e c t i v e ly ly . W h a t i s t h e ratio of N and P in the Urea? (1) 2 7 % an d 6 3 %
Sol.
Urea N x
P y
(2) 33% and 67%
(3) 3 5 % an d 65 %
(4) 70 % an d 30 %
Dia K 0
N 20% Mixture N P 68% 26%
P K 70% 10% K 6%
This 6% of K is obtained only from Dia. ∴
Urea N x
(7) of (8)
P y
Dia K 0
N P 120 Mixture N P K 260 680 60
K 60
Pinnacle Regular - Maths FSB - 2008-09
For PT Faculty use only NU + N D = N M
⇒
N U + 120 = 260
and P U + P D = P M ⇒ P U + 420 = 680 U, D, M
Ø
→
⇒
⇒
NU = 140.
P U = 260
Urea, Dia and mixture
∴
Amount of N in Urea Urea = 140 and amount of P in Urea = 260
∴
Ratio of N : P = 7:13
⇒
35 : 65. Ans.(3) 65. Ans.(3)
L a s t y e a r i n C A T , e a c h s e c t i o n o f t h e q u e s t i o n p a p e r h a d a d i f f e r e n t w e i g h t a g e . Th Th e w e i g h t a g e of QA, DI and VA/ RC sections was 8, 9 and 10 respectively. The maximum m arks in all the three sections together were 810. Wrong answer did not carry negative marks as a penalty. If Padma h a d g o t t en e n 2 0 % m o r e m a r k s i n Q A an a n d 8 % m o r e m a r k s i n D I a n d 7 .1 .1 4 % m o r e m a r k s i n V A / R C , t h e n s h e m u s t h a d g o t t e n 1 0 0 % m a r k s i n a l l t h e t h r e e s e c t i o n s. s . Th Th e t o t a l m a r k s t h a t P a d m a h a d scored (1 ) 730
(2 ) 700
(3 ) 750
(4 ) 775
Sol. 8x + 9x + 10x = 810 ⇒
x = 30
Total marks in QA → 240 DI
→
270
VA → 300 RC Now her score in QA →
Her score in
240 1.2
300 VA = 1.0714 0714 RC
=
=
200
280
Pinnacle Regular - Maths FSB - 2008-09
Her score in DI =
270 1.08
=
250
Her total sco re = 200 + 250 + 280 = 730. Ans.(1) 730. Ans.(1)
(8) of (8)
