MAT 2009 Question Paper

Let number of Electronic graduates be $$x$$ and number of mechanical graduates be $$y$$

According to ques,

=> $$3.56x+2.45y=3.12(x+y)$$

=> $$356x+245y=312x+312y$$

=> $$44x=67y$$

=> $$\frac{x}{y}=\frac{67}{44}$$

$$\because$$ 67 and 44 are co-primes, hence minimum value of $$x$$ and $$y$$ are 67 and 44 respectively. (Both must be integers)

$$\therefore$$ Minimum number of Electronic Graduates = 67

=> Ans - (C)

Let number of students in class be $$x$$ and average weight of class be $$y$$ kg

=> Total weight of class = $$xy$$ kg

According to ques, => $$\frac{xy+50}{x+1}=y+1$$

=> $$xy+50=xy+x+y+1$$

=> $$x+y=49$$ ------------(ii)

Similarly, $$\frac{xy+100}{x+2}=y+1.5$$

=> $$xy+100=xy+1.5x+2y+3$$

=> $$1.5x+2y=97$$ ------------(ii)

Solving equations (i) and (ii), we get : $$x=2$$ and $$y=47$$

$$\therefore$$ Original average weight = 47 kg

=> Ans - (D)

Let number of officers be $$x$$ and number of workers be $$(400-x)$$

According to ques,

=> $$(10,000\times x)+2,000\times(400-x)=3,000\times400$$

=> $$10,000x+8,00,000-2000x=12,00,000$$

=> $$8000x=4,00,000$$

=> $$x=50$$

$$\therefore$$ Number of officers = 50

=> Ans - (B)

Let the mixture drawn from 1st vessel be $$9x$$ gallons and from 2nd vessel be $$6y$$ gallons

Now, in the third vessel (26 gallons), quantity of milk = 13 gallons and water = 13 gallons

According to ques, milk from first two vessels = $$8x+y=13$$ --------------(i)

and water = $$x+5y=13$$ -------------(ii)

Solving the two equations, we get : $$x=\frac{4}{3}$$ and $$y=\frac{7}{3}$$

$$\therefore$$ Mixture taken from second vessel = $$6\times\frac{7}{3}=14$$ gallons

=> Ans - (B)

Let speed of stream be $$x$$ km/hr

=> Downstream speed = $$(4.5+x)$$ km/hr and Upwnstream speed = $$(4.5-x)$$ km/hr

According to ques,

=> $$2\times(4.5-x)=1\times(4.5+x)$$

=> $$9-2x=4.5+x$$

=> $$3x=4.5$$

=> $$x=\frac{4.5}{3}=1.5$$

$$\therefore$$ Speed of stream = 1.5 km/hr

=> Ans - (A)

Ratio of time taken by a woman, man and a boy = 8 : 6 : 12

= 4 : 3 : 6

=> 4 women $$\equiv$$ 3 men $$\equiv$$ 6 boys

Similarly, 12 men + 12 women + 12 boys $$\equiv(12)+(9)+(6)$$ men $$=27$$ men

Let time taken by 12 men, 12 women and 12 boys (or 27 men) be $$x$$ days

=> $$9\times6\times6=27\times8\times x$$

=> $$x=\frac{54\times6}{27\times8}=1.5$$

$$\therefore$$ Number of days required = $$1\frac{1}{2}$$ days

=> Ans - (D)

Let initial number of days planned be $$x$$, then number of pieces done per day = $$\frac{360}{x}$$

According to ques,

=> $$x\times\frac{360}{x}=(x-1)\times(4+\frac{360}{x})$$

=> $$4x-4+360-\frac{360}{x}=360$$

=> $$4x^2-4x-360=0$$

=> $$x^2-x-90=0$$

=> $$(x-10)(x+9)=0$$

=> $$x=10, -9$$

$$\because$$ $$x$$ cannot be negative, => Number of days taken = 10 days

=> Ans - (A)

Speed of Rahul = 12 km/hr and let speed of stream = $$x$$ km/hr

=> Downstream speed = $$(12+x)$$ km/hr and Upstream speed = $$(12-x)$$ km/hr

According to ques, since distance in both cases is same and distance = time x speed

=> $$6\times(12+x)=9\times(12-x)$$

=> $$72+6x=108-9x$$

=> $$15x=36$$

=> $$x=\frac{36}{15}=2.4$$

$$\therefore$$ Speed of stream = 2.4 km/hr

=> Ans - (B)

Let $$L$$ = Large ships, $$M$$ = Medium ships and $$S$$ = Small ships

According to ques, => $$4L=7S$$

=> $$\frac{L}{S}=\frac{7}{4}$$

=> $$L=7x$$ and $$S=4x$$

Also, $$3M=2L+S$$

=> $$M=\frac{2\times7x+4x}{3}=6x$$

Thus, ratio of large : medium : small = 7:6:4

$$\therefore$$ Number of journeys required = $$\frac{(15\times7)+(7\times6)+(14\times4)}{(12\times7)+(14\times6)+(21\times4)}\times36$$

= $$\frac{7308}{252}=29$$

=> Ans - (C)