MAT 2005 Question Paper

Letters of the word : 

M A T   H E I C S
M A T

Case 1 : 4 are different (all four words chosen are different, no repition)

= $$C^8_4\times4!=70\times24=1680$$

Case 2 : 2 alike + 2 different (out of the chosen four letters, two are same, rest two are chosen of the remaining seven)

= $$C^3_1\times C^7_2\times\frac{4!}{2!}$$

= $$3\times21\times12=756$$

Case 3 : 2 alike + 2 alike (four letters are chosen from M,A,T) 

= $$C^3_2\times\frac{4!}{2!2!}=18$$

$$\therefore$$ Total number of ways = $$1680+756+18=2454$$

=> Ans - (B)

Let height of cylindrical vessel = $$h=2$$ cm => Radius of cylinder = $$3$$ cm and radius of hemisphere bowl = $$3$$ cm

=> Volume of hemispherical bowl = $$\frac{2}{3}\pi r^3$$

= $$\frac{2}{3}\pi (3)^3=18\pi$$ $$cm^3$$

Now, volume of cylindrical vessel = $$\pi r^2h$$

= $$\pi \times(3)^2\times2=18\pi$$ $$cm^3$$

$$\because$$ Volume of both are same, thus volume of beverage in cylindrical vessel is $$100\%$$ of hemispherical bowl.

=> Ans - (C)

Effective percentage change = $$(+25)+(-20)+(\frac{25\times-20}{100})$$

= $$5-5=0$$

$$\because$$ The effective % change is zero, the final price will be same as original price.

=> Ans - (C)

Let radius of both be $$r=5$$ and height be $$h=12$$ units

=> Slant height of cone = $$l=\sqrt{(5)^2+(12)^2}=\sqrt{25+144}$$

=> $$l=\sqrt{169}=13$$ units

=> Total surface area of cylinder : Total surface area of cone = $$2\pi r(r+h):\pi r(r+l)$$

= $$2r+2h:r+l$$

= $$\frac{10+24}{5+13}=\frac{34}{18}$$

= $$17:9$$

=> Ans - (C)

Tom's speed = 30 pages/hr and Jan's speed = 40 pages/hr

Now, in 4:30 to 5:20 = 50 minutes, Tom has already read 25 pages.

Let $$t$$ minutes after 5:20, they will be reading the same page. Now, Tom is 25 pages ahead of Jan, and they are reading the same novel, thus effective speed = $$40-30=10$$ pages/hr

=> Time taken to catch up Tom = $$\frac{25}{10}=2.5$$ hours

$$\therefore$$ At 5:20 + 2:30 = 7:50, they will be reading the same page.

=> Ans - (D)

Using, $$\frac{M_1D_1H_1}{W_1}$$ $$=\frac{M_2D_2H_2}{W_2}$$

Initially, 117 men completed $$\frac{4}{7}$$ work in 33 days working 8 hours per day. Let $$x$$ men complete the remaining $$\frac{3}{7}$$ work in 13 days working 9 hours per day.

=> $$\frac{117\times33\times8}{4}=\frac{x\times13\times9}{3}$$

=> $$x=33\times2\times3=198$$

$$\therefore$$ Additional men required = $$198-117=81$$

=> Ans - (B)

Let lengths of longer parallel side and shorter side be $$5x$$ and $$3x$$ m respectively.

=> Area of trapezium = $$\frac{1}{2}\times$$ (height) $$\times$$ (sum of parallel sides)

=> $$\frac{1}{2}\times24\times(5x+3x)=1440$$

=> $$12\times8x=1440$$

=> $$x=\frac{1440}{96}=15$$

$$\therefore$$ Length of longer parallel side = $$5\times15=75$$ m

=> Ans - (C)

Number of words in which all the 5 letters are repeated = $$10^5=100000$$

Number of words in which no letter is repeated = $$P^{10}_5=30240$$

=> Number of words which have at least one letter repeated = $$100000-30240=69760$$

=> Ans - (A)

Let speed of man = $$x$$ km/hr and speed of stream = $$y$$ km/hr

=> Downstream speed = $$(x+y)$$ km/hr and upstream speed = $$(x-y)$$ km/hr

According to ques, => $$\frac{4}{x+y}=\frac{3}{x-y}$$

=> $$4x-4y=3x+3y$$

=> $$x=7y$$ --------------(i)

Also, $$\frac{48}{x+y}+\frac{48}{x-y}=14$$

=> $$\frac{48}{8y}+\frac{48}{6y}=14$$

=> $$6+8=14y$$

=> $$y=1$$

$$\therefore$$ Speed of stream = 1 km/hr

=> Ans - (A)

Total cost price = $$(150\times250)+2500=Rs.$$ $$40,000$$

Total marked price = $$150\times320=Rs.$$ $$48,000$$

Total Selling price after 5% discount = $$\frac{95}{100}\times48,000=Rs.$$ $$45,600$$

$$\therefore$$ Profit % = $$\frac{(45600-40000)}{40000}\times100$$

= $$\frac{5600}{400}=14\%$$

=> Ans - (A)