MAT 2003 Question Paper

Let quantity of milk purchased be $$x$$ litres and water added be $$y$$ litres

Thus, cost price = Rs. $$6.4x$$

Selling price = Rs. $$8(x+y)$$

Also, profit % = 37.5 % = $$\frac{3}{8}$$

=> $$\frac{8(x+y)-6.4x}{6.4x}=\frac{3}{8}$$

=> $$1.6x+8y=2.4x$$

=> $$8y=0.8x$$

=> $$\frac{y}{x}=\frac{1}{10}$$

=> Ans - (B)

Distance covered by big wheel in 1 revolution = $$2\pi r$$

= $$2\times\frac{22}{7}\times7=44$$ cm

and by small wheel = $$22$$ cm

Since, distance $$\propto$$ speed

Let speed of small wheel = $$x$$ cm/s, => Speed of big wheel = $$2x$$ cm/s

Also, relative speed of both wheels = $$\frac{1980}{10}=198$$ cm

=> $$x+2x=198$$

=> $$x=\frac{198}{3}=66$$ cm/s

=> Ans - (C)

The given sequence is square of natural numbers, i.e $$(1)^2,(2)^2,(3)^2,(4)^2$$ and so on.

Thus, eighth term = $$(8)^2=64$$

=> Ans - (B)

Purchase price of cycle = $$\frac{90}{100}\times100=Rs.$$ $$90$$

After another discount of 15%, selling price = $$\frac{85}{100}\times90$$

= Rs. $$76.5$$

=> Ans - (C)

Expression : $$\left(\frac{1}{2}\right) \log_{10} 25 - 2\log_{10} 3 + \log_{10} 18$$

= $$\left(\frac{1}{2}\right) \log_{10} (5)^2 - 2\log_{10} 3 + \log_{10} (3^2\times2)$$

Using, $$\log (a\times b)=\log a+\log b$$ and $$\log a^b=b\log a$$

= $$\log_{10} 5-2\log_{10} 3+2\log_{10} 3+\log_{10} 2$$

= $$\log_{10} (5\times2)=\log_{10} 10=1$$

=> Ans - (B)

Let number of one rupee, 50 paise and 25 paise coins be $$2x,3x$$ and $$4x$$ respectively.

Total amount = $$(1\times2x)+(0.5\times3x)+(0.25\times4x)=216$$

=> $$2x+1.5x+x=4.5x=216$$

=> $$x=\frac{216}{4.5}=48$$

$$\therefore$$ Number of 50 paise coins = $$3\times48=144$$

=> Ans - (B)

Length of longest rod will be placed as a diagonal of length = $$\sqrt{l^2+b^2+h^2}$$

= $$\sqrt{(12)^2+(9)^2+(8)^2}$$

= $$\sqrt{144+81+64}=\sqrt{289}$$

= $$17$$ m

=> Ans - (C)

Let length of each train = $$x$$ m

Relative speed = $$(46-36)\times\frac{5}{18}=\frac{50}{18}$$ m/s

=> $$\frac{x+x}{\frac{50}{18}}=36$$

=> $$\frac{36x}{50}=36$$

=> $$x=50$$ m

=> Ans - (A)

342*2 = 684
784 - 684 = 100

Hence, remainder is 100. 

Let A be the top runner and B be the last runner

Both A and B starts running at same time.

After 5 mins A and B meet and A completes 400m and B completes 200m (Ratio of speed is 2:1)

After another 5 mins A completes 800m and B completes 400m. Thus A finishes the race

=> Time taken by A = 5+5=10mins

=> Ans - (C)