FSB ratio proportion with solutions

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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 Ø Ø Ø I f p q r = 1 t h e nI f p q r = 1 t h e n 11 1 1 pp qq 1 1 1 1 qq rr 1 1 1 1 rr pp 1 1 11 11 + + ++ + + + + ++ + + + + ++ − − − − −− i s e q u i v a l e n t t oi s e q u i v a l e n t t o ((11 )) p p + + q q + + rr ((22)) _p_p 11_q_{q rr} + + ++ ((33)) 11 ((44)) pp qq rr 1 1 11 11 − − − − −− + + ++ Sol.

Sol. From the given condition pqr = 1.From the given condition pqr = 1. Substitute the values of p, q, r

Substitute the values of p, q, r at random such asat random such as pp = = 22 qq = = rr ==

3 3 3 3 2 2 11 ,, ,, . A n s . ( 3 ). A n s . ( 3 ) Ø Ø Ø Ø II f f aa b b cc b b cc aa cc a a bb rr + + = = + + = = + + = = , t h e n r c a n n o t t a k e a n y o t h e r v a l u, t h e n r c a n n o t t a k e a n y o t h e r v a l u e e x c e p te e x c e p t ((11 )) 11// 22 ((22)) ––11 ((33)) 11 // 2 2 o r or –– 11 ((44)) –– 11 // 2 2 oo r r –– 11 Sol. Sol. aa b b cc b b cc aa cc a a bb rr + + = = + + = = + + = = By option, if r = By option, if r = 11 2 2 ⇒ ⇒ 2a – b – c = 02a – b – c = 0 2b – c – a = 0 2b – c – a = 0 2c – a – b = 0 2c – a – b = 0 ⇒ ⇒ 2(a + b + c ) – (a 2(a + b + c ) – (a + b + c ) – + b + c ) – (a + b + c) = 0(a + b + c) = 0

Similarly r = –1 is also satisfied.

Similarly r = –1 is also satisfied. Ans.(3) Ans.(3)

Ø Ø Ø Ø A s t u d e n t g e t s a n a g g r e g a t e o f 6 0A s t u d e n t g e t s a n a g g r e g a t e o f 6 0 % % m a r km a r k s i n f i v e s u b j e c t i n t h e r a ts i n f i v e s u b j e c t i n t h e r a t i o 1 0 : 9 : i o 1 0 : 9 : 8 : 7 : 8 : 7 : 6 . I f t h e6 . I f t h e p a s s i n g m a r k s a r e 5 0 % 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 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 mm 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 , ia r k s , inn h o w m h o w m a n y s u b j e c t s d i d h e p aa n y s u b j e c t s d i d h e p as s t h e e x a m i n a t i o n ?s s t h e e x a m i n a t i o n ? ((11 )) 22 ((22)) 33 ((33)) 44 ((44)) 55 Sol.

Sol. Let his marks be 10, 9, 8, 7 and Let his marks be 10, 9, 8, 7 and 6 in the five subjects. Hence, totally he has 6 in the five subjects. Hence, totally he has scored 40 marks. This constitutesscored 40 marks. This constitutes only 60% of the total marks. Hence, t

only 60% of the total marks. Hence, t otal marks 40/0.6 = 66.7 or 67 approx. , which is the maximum marksotal marks 40/0.6 = 66.7 or 67 approx. , which is the maximum marks in all 5 subjects. Since the

in all 5 subjects. Since the total marks in each subject is the total marks in each subject is the same, hence maximum marks in each subjectsame, 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 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

needs to score 6.5 marks. We can see that only in 1 subject, he scored less than this 1 subject, he scored less than this viz. 6. Hence, heviz. 6. Hence, he passed in 4 subject.

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For PT Faculty use only

Ø Ø Ø Ø T h e c o s t o f T h e c o s t o f d i a m o n d v a r i e s d i r e c t l y a s t h e s q u a r e o f i t s wd i a m o n d v a r i e s d i r e c t l y a s t h e s q u a r e o f i t s w e i g h t . O n c e , e i g h t . O n c e , t h i s d i a m o nt h i s d i a m o n d b r o k e i n t od b r o k e i n t o f o u r p i e c e s w i t h w f o u r p i e c e s w i t h w e i g h t s i n t h e r a t i o 1 : 2 : e i g h t s i n t h e r a t i o 1 : 2 : 3 : 4 . 3 : 4 . W h e n t hW h e n t h e p i e c e s w e r e s o l d , t h e m e r c h a n t g o te p i e c e s w e r e s o l d , t h e m e r c h a n t g o t R s . 7 0 , 0 0 0 l e s s . R s . 7 0 , 0 0 0 l e s s . F i n d t h e o r i g i nF i n d t h e o r i g i n a l p r i c e o f t h e d i a m oa l p r i c e o f t h e d i a m o n d .n d .

((11)) RR ss..11 ..4 4 llaa kk hh ((22)) RR ss..2 2 llaakk hh ((33 )) RR ss..1 1 llaa kk hh ((44 )) RR ss..22 ..5 5 llaakk hh

Sol.

Sol. Let the original weight of the diamond be 10x. Hence, its original price will be k(100xLet the original weight of the diamond be 10x. Hence, its original price will be k(100x22_{), where k is a}_{), where k is a}

constant. The weights of the pieces after breaking are x, 2x, 3x and 4x. Therefore, their prices will

constant. The weights of the pieces after breaking are x, 2x, 3x and 4x. Therefore, their prices will be kxbe kx22_,,

4kx

4kx22_{, 9kx}_{, 9kx}22 _{and 16kx}_{and 16kx}22_{. So the total pr}_{. So the total pr ice of the pieces = (}_{ice of the pieces = ( 1 + 4 + 9 + 16) kx}_{1 + 4 + 9 + 16) kx}22 _{= 30kx}_{= 30kx}22_{. Hence, the difference}_{. Hence, the difference}

in the price of the original diamond and its pieces = 100kx

in the price of the original diamond and its pieces = 100kx22 _{– 30kx}_{– 30kx}22_{= 70kx}_{= 70kx}22_{= 70000. Hence, kx}_{= 70000. Hence, kx}22 _{= 1000 and}_{= 1000 and}

the original price = 100 kx

the original price = 100 kx22 _{= 100 × 1000 = 100000 = Rs.1 lakh.}_{= 100 × 1000 = 100000 = Rs.1 lakh. A n s . ( 3 )}_{A n s . ( 3 )}

Ø Ø Ø Ø T h r e e f r i e n d s w e n t f o r a p i c n i c . F i r s t b r o u g h t f i v e a p p l e s a n d t h e s e c o n d b r o uT h r e e f r i e n d s w e n t f o r a p i c n i c . F i r s t b r o u g h t f i v e a p p l e s a n d t h e s e c o n d b r o u g h t t h r e e . T h eg h t t h r e e . T h e t h i r d f r i e n d h o w t h i r d f r i e n d h o w e v e r b r o u g h t o n l y R s . 8 . e v e r b r o u g h t o n l y R s . 8 . W h a t i s t h e s h a r e o f t h e f i r s t f r i e n d ?W h a t i s t h e s h a r e o f t h e f i r s t f r i e n d ? ((11)) 88 ((22)) 77 ((33 )) 11 ((44 )) NN oo nn ee oo ff tt hh ee ss ee Sol.

Sol. The number of apples = 8, so the amount eaten by each of the three is The number of apples = 8, so the amount eaten by each of the three is 8/3 apples therefore first friend8/3 apples therefore first friend should be paid for 5 – (8/3) and second fr

should be paid for 5 – (8/3) and second fr iend should be paid for 3–(8/3) apples. They shouliend should be paid for 3–(8/3) apples. They shoul d distribute thed distribute the sum of Rs.8 in rati

sum of Rs.8 in rati o 7/3 : 1/3, i.e., 7 : 1.o 7/3 : 1/3, i.e., 7 : 1. Ans.(2) Ans.(2)

Ø Ø Ø

Ø T o t a l s a l a r y o f A , T o t a l s a l a r y o f A , B & C i s R s . 3 5 0 . I f t h e y s p e n d 7 5 %B & C i s R s . 3 5 0 . I f t h e y s p e n d 7 5 % , 8 0 % , 8 0 % & 5 6 % & 5 6 % o f t h e i r s a l a r i e s r e s p e c t i v e l yo f t h e i r s a l a r i e s r e s p e c t i v e l y t h e i r s a v i n g s a r e a s 1 0

t h e i r s a v i n g s a r e a s 1 0 : 1 2 : : 1 2 : 3 3 . F i n d t h e i r s a l a r i e s .3 3 . F i n d t h e i r s a l a r i e s .

Sol.

Sol. A’s saving = 100 – 75 = 25% of his salary. B’s saving = 100 – 80 = 20% of his salary 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

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 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

or 25 × A’s salary : 20 × B’s salary : 44 × C’s s alary = 10 : 12 : 33alary = 10 : 12 : 33 or 25 × A’s salary / 20 × B’s salary = 10/12

or 25 × A’s salary / 20 × B’s salary = 10/12 or A’s salary : B’s salary = 2 : 3,

or A’s salary : B’s salary = 2 : 3, B’s salary : C’s salary = 4 : 5 B’s salary : C’s salary = 4 : 5

Thus A : B = 2 : 3, B : C = 4 : 5 No

Thus A : B = 2 : 3, B : C = 4 : 5 No w making B common we hw making B common we h aveave A : B = 8 : 12, B : C = 12 : 15,

A : B = 8 : 12, B : C = 12 : 15, or A : B : C = or A : B : C = 8 : 12 : 158 : 12 : 15 Total salary = 350

Total salary = 350 ⇒⇒ A’A’s salary = 8 / s salary = 8 / (8 + 12 + 15) × 350 = 80(8 + 12 + 15) × 350 = 80

B’s salary = 12 / (

B’s salary = 12 / ( 8 + 12 + 15) = 120, and C’8 + 12 + 15) = 120, and C’ s Salary = 150 Answer.s Salary = 150 Answer.

Ø Ø Ø Ø T h e r a t i o o f t h e a g e o f a mT h e r a t i o o f t h e a g e o f a m a n a n d h i s wa n a n d h i s w i f e i s 4 : 3 . i f e i s 4 : 3 . A f t e r 4 y e a r s , t h i s r a t i o wA f t e r 4 y e a r s , t h i s r a t i o w i l l b e 9 : 7 . I f a t t h ei l l b e 9 : 7 . I f a t t h e t i m e o f t h e m a r r i a g e , t h e r a t i o w a s 5 : 3 , t i m e o f t h e m a r r i a g e , t h e r a t i o w a s 5 : 3 , t h e n h o w t h e n h o w m a n y y e a r s a g o t h e y w e r e m a r rm a n y y e a r s a g o t h e y w e r e m a r r i e d ?i e d ? ((11)) 11 2 2 yy ee aa rr ss ((22)) 8 8 yy ee aa rr ss ((33 )) 11 0 0 yy ee aa rr ss ((44 )) 11 5 5 yy ee aa rr ss Sol.

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

∴ ∴ 44 44 3 3 44 9 9 7 7 88 k k k k k k + + + + = = ⇒ ⇒ == .. ∴

∴ Man’s age = 32 yearsMan’s age = 32 years

Wif

Wife’s ae’s age = 24 yge = 24 yearears.s. SupSuppospose thee they wey were mare marrirried x yed x yearears agos ago..

∴ ∴ 3232 24 24 5 5 3 3 1212 − − − − = = ⇒ ⇒ == xx xx xx .. Ans.(1) Ans.(1)

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For PT Faculty use only

Ø

Ø Ø

Ø Two full tankTwo full tank s, one shaped like a cylinder and the other liks, one shaped like a cylinder and the other lik e a cone, e a cone, contain jet fuel. The cylindricalcontain jet fuel. The cylindrical t a n k h o l d s 5 0 0 l i t r 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 fe 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 e l h a s b e e n p u m p e d o u tu e l h a s b e e n p u m p e d o u t f r o m e a c h t a n k t h e c y l i n d r i c a l t a n k c o n t f r o m e a c h t a n k t h e c y l i n d r i c a l t a n k c o n t a i n s t wa i n s t w i c e t h e a m o u n t o f f u e l i n t h e c o n i c a l t a n k . i c e t h e a m o u n t o f f u e l i n t h e c o n i c a l t a n k . H o wH o w m a n y l i t r e s o f f u e l d i d t h e c y l i n d r i c a l t a n k h a v e w h m a n y l i t r e s o f f u e l d i d t h e c y l i n d r i c a l t a n k h a v e w h e n i t w a s f u l l ?e n i t w a s f u l l ? ((11)) 770000 ((22)) 11000000 ((33)) 11110000 ((44)) 11220000 Sol.

Sol. Work backwards from the options. If the cylinder has a capacity of 1200 litre, then Work backwards from the options. If the cylinder has a capacity of 1200 litre, then the conical vessel shallthe conical vessel shall have a capacity of 700 litres. Once 200 litres have been taken out from the same, the remaining holding of have a capacity of 700 litres. Once 200 litres have been taken out from the same, the remaining holding of each of them shal

each of them shal l be 1000 & 500.l be 1000 & 500. A l t e r n a t e :

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 Let the cylinder has a capacity of X litre, then the conical vessel shall have a capacity of (x – 500) litres. (x – 500) litres. (x – 200) = 2 (x – 700) = x = (x – 200) = 2 (x – 700) = x = 1200.1200. A n s . ( 4 ) A n s . ( 4 ) Ø Ø Ø Ø T h e r e d u c t i o n i n t h e s p e e d o f a n e n g i n e i s d i r e c t l y p r o p o r t i o nT h e r e d u c t i o n i n t h e s p e e d o f a n e n g i n e i s d i r e c t l y p r o p o r t i o n a l t o t h e s q u a r e o f t h e n u ma l t o t h e s q u a r e o f t h e n u m b e r o f b e r o f b o g i e s a t t a c h e d t o i t . T b o g i e s a t t a c h e d t o i t . Th e s p e e d o f t h e t r a i n i s 1 0 0 k mh 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 hh 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 he 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 w h w h e n t h e r e a r e 5 b o g i e s . e n t h e r e a r e 5 b o g i e s . W h a t i s t h e m a xW h a t i s t h e m a x i m u m ni m u m n u m b e r o f b o g i e s t h a t c a n b e a t t a c h e d t o u m b e r o f b o g i e s t h a t c a n b e a t t a c h e d t o t h et h e t r a i n s o t h a t i t c a n m t r a i n s o t h a t i t c a n m o v e ?o v e ? ((11)) 66 ((22)) 55 ((33)) 44 ((44)) NN oo nn ee oo ff tt hh ee ss ee Sol.

Sol. Suppose Reduction in speed is R. Speed of Suppose Reduction in speed is R. Speed of the engine without any bogie = the engine without any bogie = k k number of bogies attached = b, proportional

number of bogies attached = b, proportional ity constant = c, Resultant speed = sity constant = c, Resultant speed = s We have R = cb

We have R = cb22 _{and s = k – R =}_{and s = k – R = k – cb}_{k – cb}22

100 = k – c (4)

100 = k – c (4)22 _o_orr,_{, 1}₁₀₀₀₀₌_{= k}_{k –}_–₁₁₆_6cc _...((ii))

and 55 = k – c(5)

and 55 = k – c(5)22 _o_or_{r 5}₅₅₅_{= k}₌_k_–_–₂₂₅₅_cc _...((iiii))

Solving (

Solving ( i) and (ii) we gei) and (ii) we ge t k = 180 and c = 5. Now we have S = 180 – 5bt k = 180 and c = 5. Now we have S = 180 – 5b22_{. If we put b = 6, S =}_{. If we put b = 6, S = 0}₀

∴

∴ At most we can attach 5 bogies to the engine.At most we can attach 5 bogies to the engine. Ans.(2) Ans.(2)

Ø Ø Ø Ø A r v i n d S i n g h p u r c h a s e d a 4 0 s e a t e r b u s . H e s t a r t e d h i s s e r v i c e s o n r o u t e n u mA r v i n d S i n g h p ur c h a s e d a 4 0 s e a t e r b u s . H e s t a r t e d h i s s e r v i c e s o n r o u t e n u m b e r 2 ( f r o m Mb e r 2 ( f r o m M a h ua h u N a k a t o D e w N a k a t o D e w a s N a k a w i ta s N a k a w i t h r o u t e l e n g t h o f 5 0 k mh r o u t e l e n g t h o f 5 0 k m ) . ) . H i s p r o f i t ( P ) f r o m t h e b u s d e p e n d s u p o n t h eH i s p r o f i t ( P ) f r o m t h e b u s d e p e n d s u p o n t h e n u m b e r o f p a s s e n g e r s o v e r a c e r t a i n m i n i m u m n n u m b e r o f p a s s e n g e r s o v e r a c e r t a i n m i n i m u m n u m b e r o f p a s s e n g e r s ‘ n ’ a n d u p o n t h e d i s t a n c eu m b e r o f p a s s e n g e r s ‘ n ’ a n d u p o n t h e d i s t a n c e 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 . 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 6 0 0 w3 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 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 da n d R s . 6 3 0 0 w i t h 3 6 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 4 2 k m . W h a t i s t h e m i n R s . 6 3 0 0 w i t h 3 6 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 4 2 k m . W h a t i s t h e m i n i m u m ni m u m n u m b e r o f u m b e r o f p a s s e n g e r s a r e r e q u i r e d s o t h a t h e w p a s s e n g e r s a r e r e q u i r e d s o t h a t h e w i l l n o t s u f f e r a n y l o s s .i l l n o t s u f f e r a n y l o s s . ((11)) 1122 ((22)) 2200 ((33)) 1188 ((44)) 1155 Sol.

Sol. The minimum number of passengers n, at which there is no loss and number of passengers travelling = mThe 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

and let the distance travelled is d, Then P

P ∝∝(m – n)d(m – n)d

o r

o r p = kp = k( m ( m – n– n) d) d; k i; k is a cs a co no ns ts ta na nt .t . When P = 3600, m = 29 and d = 36, then When P = 3600, m = 29 and d = 36, then

3 366000 0 = = kk((229 9 – – nn) ) × × 3366 ...((11)) Again, when p = 6300, m = 36, d = 42, th Again, when p = 6300, m = 36, d = 42, th enen 6 633000 0 = = kk((336 6 – – nn) ) × × 4422 ...((22)) Dividing equation (2) by (1) Dividing equation (2) by (1) 6300 6300 3600 3600 3 366 4422 2 299 3366 = = − − ×× − − ×× k k nn k k nn

gg

b b

gg

⇒⇒ 36 36 29 29 9 9 6 6 − − − − = = n n n n

gg

b b

gg

⇒⇒ 3n = 453n = 45 ⇒⇒ n = 15n = 15

Hence to avoid loss, minimum number of 15 passengers are required.

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For PT Faculty use only

Allegations

Ø Ø Ø Ø A m i l k mA m i l k m a n m i x e s 2 0 l i t r e s o f wa n m i x e s 2 0 l i t r e s o f w a t e r w i t h 8 0 la t e r w i t h 8 0 l i t r e s o f m i l k . A f t e r s e l l i n g o n e – f o u r t h o f t h i s m i x t u r e ,i t r e s o f m i l k . A f t e r s e l l i n g o n e – f o u r t h o f t h i s m i x t u r e , h e a d d s w a t e r t o r e p l e n i s h t h h e a d d s w a t e r t o r e p l e n i s h t h e q u a n t i t y t h a t h e h a s s o l d . W h a t i s t h e c u r r e n t p r o p o re q u a n t i t y t h a t h e h a s s o l d . W h a t i s t h e c u r r e n t p r o p o r t i o n o f t i o n o f w a t e r t o m i l k ? w a t e r t o m i l k ? ((11)) 22 :: 33 ((22)) 11 :: 22 ((33 )) 11 :: 33 ((44 )) 33 :: 44 Sol. Sol. W

W aatteerr MM iillk k IInniittiiaallllyy 2200 8800 A

Afftteer r SSeelllliinng g oonnee--ffoouurrtthh ((220 0 – – 55) ) = = 1155 ((880 0 – – 2200) ) = = 6600 A

Afftteer r aaddddiinng g wwaatteer r tto o rreepplleenniissh h tthhe e qquuaannttiitty y 4400 6600 Required ratio = 2 : 3. Required ratio = 2 : 3. A n s . ( 1 ) A n s . ( 1 ) Ø Ø Ø Ø A n a l l o y c o n t a i n s 2 4 % A n a l l o y c o n t a i n s 2 4 % o f t i n b y w e i g h t . H o w m uo f t i n b y w e i g h t . H o w m u c h m o r e t i n t o t h e n e a r e s t k g m u s t b e a d d e d t oc h m o r e t i n t o t h e n e a r e s t k g m u s t b e a d d e d t o 1 0 0 k g o f t h e a l l o y s o t h a t t h 1 0 0 k g o f t h e a l l o y s o t h a t t h e p e r c e n t a g e o f t i n m a y b e d o u b l e d ?e p e r c e n t a g e o f t i n m a y b e d o u b l e d ? Sol.

Sol. Let X kg of tin be added to tLet X kg of tin be added to t he alloy (24 + X) / (100 + X) = 2 (24/100)he alloy (24 + X) / (100 + X) = 2 (24/100) ⇒⇒ X = 46.X = 46.

Hence 46 kg of tin must

Hence 46 kg of tin must be added to the allbe added to the all oyoy..

Ø Ø Ø 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 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 wa n d w a t e r r e s p e c t i va t e r r e s p e c t i ve l y . H a l f t h e c o n t e n t s o f t h ee l y . H a l f t h e c o n t e n t s o f t h e f i r s t a r e p o u r e d i n t h e s e c o n d a n d t h e n t h e s a m e q u a n t i t y i s t r a n s f e r r e d b a c k i n t o t h e f i r s t f i r s t a r e p o u r e d i n t h e s e c o n d a n d t h e n t h e s a m e q u a n t i t y i s t r a n s f e r r e d b a c k i n t o t h e f i r s t c o n t a i n e r . T c o n t a i n e r . 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 wh 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 wa t e r i n t h e t w o c o n t a i n e r s a to c o n t a i n e r s a t t h e e n d o f t h e p r o t h e e n d o f t h e p r o c e s s ?c e s s ? ((11)) 5 5 : : 22 , , 2 2 : : 55 ((22)) 5 5 : : 44 , , 4 4 : : 55 ((33 )) 11 4 4 : : 11 33 , , 11 3 3 : : 11 44 ((44 )) NN oo nn e e oo f f tt hh ee ss ee S o l .

S o l . Start with a litre of milk Start with a litre of milk in 1st container and a litre of in 1st container and a litre of water in 2nd proceed.water in 2nd proceed. A n s . ( 3 ) A n s . ( 3 )

Ø Ø Ø Ø A t o t a l o f ‘ a ’ l i t r e s o f p u r e a c i d w e r e d r a wA t o t a l o f ‘ a ’ l i t r e s o f p u r e a c i d w e r e d r a w n f r o m a t a n k c o nn f r o m a t a n k c o n t a i n i n g 7 2 9 lt a i n i n g 7 2 9 l i t r e s o f p u r e a c i d a n di t r e s o f p u r e a c i d a n d w a s r e p l a c e d b y w a t e r . T w a s r e p l a c e d b y w a t e r . Th e r e s u l t w a s t h o r o u g h lh 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 dy 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 t h e n a n o t h e r ‘ a ’ l i t r e s o f s o l u t i o n w a s d r a w n o f f , a n d a g a i n r e p l a c e d b y w a t e r , a n d a g a i n t h e n a n o t h e r ‘ a ’ l i t r e s o f s o l u t i o n w a s d r a w n o f f , a n d a g a i n r e p l a c e d b y w a t e r , a n d a g a i n t h o r o u g h l y m t h o r o u g h l y m i x e d . Ti x e d . 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 mh 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 se 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 o f p u r e a c i d . D e t e r m i n e ‘ a ’ . o f p u r e a c i d . D e t e r m i n e ‘ a ’ . ((11)) 11// 33 ((22)) 224433 ((33 )) 8811 ((44 )) 33 S o l .

S o l . Here 729 × {(729 – a)/729}Here 729 × {(729 – a)/729}66 _{= 64 or}_{= 64 or, (729 – a)/}_{, (729 – a)/ 729 = (64/729)}_{729 = (64/729)}1/61/6 _{or a = 243.}_{or a = 243. A n s . ( 2 )}_{A n s . ( 2 )}

Ø Ø Ø 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 nT 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 , g R s . 3 , R s . 3 .R s . 3 .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 t2 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 xh e m i x t u r e i st u r e i s t h e n s o l d a t R s . 3 . t h e n s o l d a t R s . 3 .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 %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 wI n w h a t p r o p o r t i o n s h o uh 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 el d t h e t h r e e q u a l i t i e s o f m i l k b e m i x e d ? q u a l i t i e s o f m i l k b e m i x e d ? ((11) 1 ) 1 : : 2 2 : : 33 ((22) 2 ) 2 : : 1 1 : : 33 ((33 ) 1 ) 1 : : 1 1 : : 11 ((44 ) 3 ) 3 : : 2 2 : : 11 Sol.

Sol. The mixture is sold at a The mixture is sold at a profit of 20% at Rs.3.54profit of 20% at Rs.3.54∴∴ the actual value of the mixtuthe actual value of the mixtu re is 3.54/1.2 = 2.95re is 3.54/1.2 = 2.95

Now, ratio in which they are to be mixed can be calculated by Now, ratio in which they are to be mixed can be calculated by 3x + 3.25y + 2.60z = 2.95. Putt

3x + 3.25y + 2.60z = 2.95. Putt ing x = y = z ing x = y = z = 1/3 we get, 1 + 1.0833 + = 1/3 we get, 1 + 1.0833 + .8667 = 2.95.8667 = 2.95

∴

(5)

For PT Faculty use only

Ø Ø Ø Ø I n a I n a l a b o r a t o r y e x p e r i m e n t , l a b o r a t o r y e x p e r i m e n t , a s a m p l e o f A i r , w h i c h i s a ma s a m p l e o f A i r , w h i c h i s a m i x t u r e o f o n l y o x y g e n a n d wi x t u r e o f o n l y o x y g e n a n d w a t e r v a p o u ra t e r v a p o u r i s t a k e n . W a t e r v a p o u r c o n t i s t a k e n . W a t e r v a p o u r c o n t a i n s h y d r o g e n a n d o xa i n s h y d r o g e n a n d o x y g e n g a s e s . y g e n g a s e s . I f A i r c o n t a iI f A i r c o n t a i n s a t o t a l o f 7 0 %n s a t o t a l o f 7 0 % 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 e 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 i n t h e wd i n t h e w a t e r v a p o u r ) b y w e i g h t wa 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 sh i l e w a t e r v a p o u r c o n t a i n s 1 16622 3 3 % % 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 eo 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 t i n 1 k i l o g r a m o f n t i n 1 k i l o g r a m o f air? air? ((11)) 00..33 ((22)) 00..3366 ((33)) 00 ..3344 ((44)) 00..2255 Sol.

Sol. Since 70% of air is oxygen, remaSince 70% of air is oxygen, rema ining 30% is hydrogenining 30% is hydrogen . In water vapour. In water vapour, if one unit is oxygen, i., if one unit is oxygen, i. e., five-e., five-sixth is hydrogen.

sixth is hydrogen.

⇒

⇒ water vapour = 30%water vapour = 30% ⇒

⇒ water vapour = 6/5 × 30% (of water vapour = 6/5 × 30% (of air)air) ∴

∴ In 1 kg of air, water vapour is 0.36 kg.In 1 kg of air, water vapour is 0.36 kg. Ans.(2) Ans.(2)

Ø Ø Ø Ø I n wI n w h a t r a t i o s h o u l d t wh 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 oo v a r i e t i e s o f r i c e c os 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 cs 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 i v e l y b et 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 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 ht u r e w h e n m i x e d we 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 .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 0 p e r k g2 0 p e r k g i n t h e r a t i o 4 : 3 , w o u l d y i e l d a m i x i n t h e r a t i o 4 : 3 , w o u l d y i e l d a m i x t u r e c o s t i n g Rt u r e c o s t i n g R s . 1 6 p e r k g ?s . 1 6 p e r k g ? ((11)) 5 5 : : 11 ((22)) 7 7 : : 22 ((33)) 9 9 : : 44 ((44)) NN oo nn e e oo f f tt hh ee ss ee Sol.

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 isLet, 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 given that 16 16 2 20 0 1166 3 3 4 4 − − − − = = x x ⇒ ⇒ x = Rs.13 per kg.x = Rs.13 per kg. 5 5 1 122 1188 1 1 13 13 Ans.(1) Ans.(1) Ø Ø Ø Ø 2 0 l i t r e s o f m i l k 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 w h e n a d d e d t o a 6 0 l i t r e m i l k 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 aa 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 i o n b yt i o n b y s a m e p e r c e n t a g e p o i n t s a s d e c r e a s e d b y a d d i t i o n o f 3 0 l i t r e s o f w a t e r t o t h e s a m e s o l u t i o n s a m e p e r c e n t a g e p o i n t s a s d e c r e a s e d b y a d d i t i o n o f 3 0 l i t r e s o f w a t e r t o t h e s a m e s o l u t i o n .. W h a t i s t h e r a t i o o f m i l k a n d w W h a t i s t h e r a t i o o f m i l k a n d w a t e r i n i t i a l l y ?a t e r i n i t i a l l y ? ((11)) 11 :: 22 ((22)) 22 :: 11 ((33)) 33 :: 44 ((44)) 44 :: 55 Sol.

Sol. Let k litres be the amount of milk in 60 litres of solution. So the concentratioLet k litres be the amount of milk in 60 litres of solution. So the concentratio n of milk = k/60. If 20 n of milk = k/60. If 20 litres of litres of milk is added, concentration of milk =

milk is added, concentration of milk = (k + 20) / (k + 20) / 80. If 30 litres of 80. If 30 litres of water is added, milk concentration = k/90.water is added, milk concentration = k/90. So we have

So we have kk ++2020 − − kk = = kk −− kk

8

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For PT Faculty use only

Ø Ø Ø Ø I n a m iI n a m i l k S h o p t hl k S h o p t h e r e a r e t h r e e v a r i e t i e s o f m i l ke r e a r e t h r e e v a r i e t i e s o f m i l k , , ‘ P u r e ’ , ‘ C u r e ’ a n d ‘ L u r e ’ . ‘ P u r e ’ , ‘ C u r e ’ a n d ‘ L u r e ’ . T h e ‘ P u r e ’ m i l k h a sT h e ‘ P u r e ’ m i l k h a s 1 0 0 % 1 0 0 % c o n c e n t r a t i o n o f m i l kc o n c e n t r a t i o n o f m i l k . T. Th e r a t i o o f m i l k t o wh 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 :a t e r i n t h e ‘ C u r e ’ i s 2 :5 a n d i n t h e L u r e i t i s 3 :5 a n d i n t h e L u r e i t i s 3 :88 r e s p e c t i v e l y . S o n a l i p u r c h a s e d 1 4 l i t r e s o f C u r e a n d 2 2 l i t r e s o f L u r e m i l k r e s p e c t i v e l y . S o n a l i p u r c h a s e d 1 4 l i t r e s o f C u r e a n d 2 2 l i t r e s o f L u r e m i l k a n d m i xa n d m i x e d t h e m . I f s h ee d t h e m . I f s h e w a n t e d t o m a k e t h e c o n c e n t r a w a n t e d t o m a k e t h e c o n c e n t r at 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 et 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 m i l k t o 5 0 %d m i l k t o 5 0 % . H o w m a n y. H o w m a n y l i t r e s o f ‘ P u r e ’ M i l k s h e i s n e e d e d ? l i t r e s o f ‘ P u r e ’ M i l k s h e i s n e e d e d ? ((11)) 6 6 ll ii tt rr ee ss ((22)) 8 8 ll ii tt rr ee ss ((33 )) 11 6 6 ll ii tt rr ee ss ((44 )) 11 8 8 ll ii tt rr ee ss Sol.

Sol. PPuurree CCuurree LLuurree 1 10000%% 4400%% 337 57 5%% 1 1 1 1 2 2 5 5 3 3 8 8 .. Milk Milk Milk Milk Water Water Water Water Mixture Mixture Cure Cure Lure Lure New mixture New mixture 4 4ll 10 10ll 6 6ll 16 16ll 26 26ll 10 10ll 26 26ll 16 16ll 26 26ll 14 14ll 22 22ll + + Required mixutre Required mixutre 11 :: 11

Since in the required mixture the ratio

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

(pure) to get it, for t he fixed quantithe fixed quantit y of watery of water.. Ans.(3) Ans.(3)

Ø Ø Ø Ø I n a mI n a m i x t u r e o f p e t r o l a n d k e r o s e n e p e t r o l i s o n l y 9 9 l i t r e s . i x t u r e o f p e t r o l a n d k e r o s e n e p e t r o l i s o n l y 9 9 l i t r e s . I f t h i s s a m e q u a n t i t y o f p e t r o l wI f t h i s s a m e q u a n t i t y o f p e t r o l w o u l do u l d b e p r e s e n t e d i n a n o t h e r m i x t u r e o f p e t r o l a n d K e r o s e n e w h e r e t o t a l v o l u m e w o u l d b e 1 9 8 b e p r e s e n t e d i n a n o t h e r m i x t u r e o f p e t r o l a n d K e r o s e n e w h e r e t o t a l v o l u m e w o u l d b e 1 9 8 l i t r e sl i t r e s l e s s t h a n t h e a c t u a l m i x t u r e t h e n t h l e s s t h a n t h e a c t u a l m i x t u r e t h e n t h e c o n c e n t r a t i o n o f p e t r o l i n t h e p r e s e n t m i x t u r e we c o n c e n t r a t i o n o f p e t r o l i n t h e p r e s e n t m i x t u r e w o u l d h a v eo u l d h a v e b e e n 1 3 . b e e n 1 3 .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 cp 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 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 to 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 ?u r e ? ((11) 2) 200%% ((22) 1) 166..6666%% ((33 ) 2) 266..6666%% ((44 ) 8) 8..3333%% Sol.

Sol. PPeettrrooll KKeerroosseennee TToottaal l MMiixxttuurree 9 999 xx 9999 ++xx 9 999 ((xx –– 119988)) ((xx –– 9999)) Again Again 9999 99 99 100100 99 99 99 99 110000 113 33 333 xx −− xx × × −− + + × × ==

b

b g

g

b

b gg

.. or or ₉₉₀₀₉₉₀₀ 9999 9999 99 99 13331333 2 2 22 xx xx xx + + − − ++ − −

HHGG

KKJJ

== .. oror 99990000 191988 99 99 40 40 3 3 2 2 22 (( )) xx −− = = ⇒ ⇒ xx22 – 99– 9922 = 99= 9922 ×× 1515 ⇒⇒ xx22 = (99)= (99)22 ×× (16)(16) ⇒⇒ x = 99 × 4 = 396 litresx = 99 ×4 = 396 litres

Therefore the actual concentration of petrol =

Therefore the actual concentration of petrol = 9999 9

999 3++39966 20%20%

= =

(7)

For PT Faculty use only

Ø Ø Ø Ø A v e s s e l o f c a p a c i t y 2 l i t r e h a s 2 5 % A v e s s e l o f c a p a c i t y 2 l i t r e h a s 2 5 % a l c o h o l a n d a n o t h e r v e s s e l o f c a p a c i t y 6 l i t r e h a d 4 0 %a l c o h o l a n d a n o t h e r v e s s e l o f c a p a c i t y 6 l i t r e h a d 4 0 % a l c o h o l . T a l c o h o l . Th i s t o t a l l i q u ih i s t o t a l l i q u i d o f 8 l i t r e wd 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 ta 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 eh 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 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 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 ac o n c e n t r at i o n o f m i x t u r e ?t i o n o f m i x t u r e ? ((11)) 3311%% ((22)) 7711%% ((33)) 4499%% ((44)) 2299%% Sol.

Sol. Am Amount of alcohol in ount of alcohol in first vessel = 0.25 × 2 first vessel = 0.25 × 2 = 0.5 litre= 0.5 litre amount of alcohol in second vessel = 0.4 × 6 = 2.4 litre amount of alcohol in second vessel = 0.4 × 6 = 2.4 litre Total amount of alcoh

Total amount of alcoh ol out of 10 litres of mixture is 0.5 + 2.4 = 2.9 litreol out of 10 litres of mixture is 0.5 + 2.4 = 2.9 litre

Hence, the concentration of the mixture is 29%

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

HHGG

22 910109 100100

KKJJ

.. .. Ans.(4) Ans.(4) Ø Ø Ø Ø A l l o y A c o n t a i n s 4 0 % 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 % g o l d a n d 6 0 % s i l v es i l v er . A lr . A ll o y B c o n t a i n s 3 5 % l o y B c o n t a i n s 3 5 % g o l d a n d 4 0 % g o l d a n d 4 0 % s i l v es i l v er a n d 2 5 %r a n d 2 5 % c o p p e r . A l l o y s A a n d B a r e m i x e d i n t c o p p e r . A l l o y s A a n d B a r e m i x e d i n t h e r a t i o 1 : 4 . h e r a t i o 1 : 4 . W h a t i s tW h a t i s t h e r a t i o o f g o l d a n d s i l v e r i n th e r a t i o o f g o l d a n d s i l v e r i n t h eh e n e w l y f o r m e d a l l o y i s ? n e w l y f o r m e d a l l o y i s ? ((11)) 22 00 % % aann d d 33 00 %% ((22)) 33 66 % % aann d d 44 44 %% ((33)) 22 55 % % aann d d 3355 %% ((44)) 4499 % % aann d d 3366 %% Sol.

Sol. Assume the weight of Alloy A is 100 kg. Assume the weight of Alloy A is 100 kg.

∴

∴ The weight of alloy B is 400 kg.The weight of alloy B is 400 kg. ∴

∴ GGoolldd SSiillvveerr CCooppppeerr

A A 4400 kkgg 6600 kkgg 00 kkgg B B 141400 kkgg 116600 kkgg 110000 kkgg total total →→ 118800 kkgg 222200 kkgg 110000 kkgg ∴

∴ Ratio of Gold and Silver in new alloy =Ratio of Gold and Silver in new alloy = 180180

500 500 200 200 500 500 :: = 36%:44%.= 36%:44%. A n s . ( 2 ) A n s . ( 2 ) Ø Ø Ø Ø D i a a n d U r e a a r e t w o 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 sc 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 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 yi 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 a n d P . A m i x t u r e o f D i a a n d U r e a N a n d P . A m i x t u r e o f D i a a n d U r e a i s p r e p a r e d i s p r e p a r e d i n w h i c h t hi n w h i c h t h e r a t i o o f N , e r a t i o o f N , P a n d K i s 2 6 %P a n d K i s 2 6 % , , 6 8 % 6 8 % a n da n d 6 % 6 % r e s p er e s p ec t i v e lc t i v e ly . Ty . Th e r a t i o o f N , h e r a t i o o f N , P a n d K P a n d K i n D i a i s 2 0 %i n D i a i s 2 0 % , 7 0 % , 7 0 % a n d 1 0 % a n d 1 0 % r e s p er e s p ec t i v e lc t i v e ly . W h a t i s t h ey . W h a t i s t h e r a t i o o f N a n d P i n t h e U r e a ? r a t i o o f N a n d P i n t h e U r e a ? ((11)) 22 77 % % aann d d 66 33 %% ((22)) 33 33 % % aann d d 66 77 %% ((33)) 33 55 % % aann d d 6655 %% ((44)) 7700 % % aann d d 3300 %% Sol.

Sol. _Urea_Urea N N NN N N P P PP P P K K K K K K x x 2200%% 26% 26% y y 7700%% 68% 68% 0 0 1100%% 6% 6% Dia Dia Mixture Mixture

This 6% of K is obtained only from Dia. This 6% of K is obtained only from Dia.

N N NN N N P P PP P P K K K K K K x x 112200 260 260 y y 680 680 0 0 6060 60 60 Dia Dia Mixture Mixture Urea Urea ∴ ∴

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For PT Faculty use only

N N_U_U + N+ N_D_D = N= N_M_M ⇒⇒ NN U U + 120 = 260+ 120 = 260 ⇒⇒ NNUU = 140.= 140. a anndd PP_U_U + P+ P_D_D = P= P_M_M⇒⇒ PP U U + 420 = 680+ 420 = 680 ⇒⇒ PPUU = 260= 260 U, D, M

U, D, M →→Urea, Dia and mixtureUrea, Dia and mixture ∴

∴ Amount of N in UAmount of N in Urea = 140rea = 140 and amount of P and amount of P in Urea = 260in Urea = 260 ∴

∴ Ratio of N : P = 7:13Ratio of N : P = 7:13 ⇒⇒ 35 : 65.35 : 65. Ans.(3) Ans.(3)

Ø Ø Ø Ø L a s t y e a r i n C A T , 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 he 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 wa d a d i f f e r e n t w e i g h t a g e . Te i g h t a g e . Th e wh e w e i g h t a g ee i g h t a g e o f Q A , o f Q A , D I a n d V A /D I a n d V A / R C s e c t i o n s w a s 8 , 9 a n d 1 0 r e s p e c t i v e l y . R C s e c t i o n s w a s 8 , 9 a n d 1 0 r e s p e c t i v e l y . T h e m a x i m u m mT h e m a x i m u m m a r k s i n a l l t h e t h r e ea r k s i n a l l t h e t h r e e s e c t i o n s t o g e t h e r w e r e 8 1 0 . W r o n g a n s w e r d i d n o t c a r r y n e g a t i v e m a r k s a s a p e n a l t y . I f P a d m a s e c t i o n s t o g e t h e r w e r e 8 1 0 . W r o n g a n s w e r d i d n o t c a r r y n e g a t i v e m a r k s a s a p e n a l t y . I f P a d m a h a d g o t t e h a d g o t t en 2 0 % n 2 0 % m o r e m a r k s i n Q A am o r e m a r k s i n Q A an d 8 % n d 8 % m o r e m a r k s i n D I a n d 7 .m o r e m a r k s i n D I a n d 7 .1 4 % 1 4 % m o r e m a r k s i n V A / R C ,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 % 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 sm 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. T. Th e t o t a l m a r k s t h a t P a d m a h a dh e t o t a l m a r k s t h a t P a d m a h a d s c o r e d s c o r e d ((11)) 773300 ((22)) 770000 ((33 )) 775500 ((44 )) 777755 Sol. Sol. 8x + 9x + 10x = 8108x + 9x + 10x = 810 ⇒ ⇒ x = 30x = 30

Total marks in QA

Total marks in QA →→ 240240

DI DI →→ 270270 VA VA RC RC →→ 300300

Now her score in QA

Now her score in QA →→ 240240

1

1 2..2 ==200200 Her score in DI =Her score in DI = 270 270 1

1 08..08 ==250250

Her score in Her score in VA VA

RC RC ==

300 300 1