MAT 2009 Question Paper

Let sum invested at 5% be Rs. $$100x$$ and sum invested at 6% be Rs. $$(12000-100x)$$ and time period = 4 years

Simple interest = $$\frac{P\times R\times T}{100}$$

According to ques,

=> $$(\frac{100x\times5\times4}{100})+(\frac{(12000-100x)\times6\times4}{100})=2580$$

=> $$20x+24(120-x)=2580$$

=> $$20x-24x+2880=2580$$

=> $$4x=300$$

=> $$x=\frac{300}{4}=75$$

$$\therefore$$ Sum invested at 5% per annum = $$100\times75=Rs.$$ $$7500$$

=> Ans - (C)

Under compound interest, money will become 8 times in 3 years. Let it takes $$t$$ years to become 16 times.

Substituting values in the formula = $$(x)^{\frac{1}{a}}=(y)^{\frac{1}{b}}$$

=> $$(8)^{\frac{1}{3}}=(16)^{\frac{1}{t}}$$

=> $$(2)^1=(2)^{\frac{4}{t}}$$

Comparing the exponents, we get : $$t=4$$ years

=> Ans - (B)

Let number of papers be $$x$$ and total marks in all papers = $$63x$$

According to ques,

=> $$\frac{63x+20+2}{x}=65$$

=> $$63x+22=65x$$

=> $$2x=22$$

=> $$x=\frac{22}{2}=11$$

=> Ans - (D)

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

According to ques, time taken in 12 km downstream = Time taken 4 km upstream

=> $$\frac{12}{x+y}=\frac{4}{x-y}$$

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

=> $$2x=4y$$

=> $$x=2y$$ --------------(i)

Also, $$\frac{45}{x+y}+\frac{45}{x-y}=20$$

=> $$\frac{15}{y}+\frac{45}{y}=20$$

=> $$20y=60$$

=> $$y=\frac{60}{20}=3$$

$$\therefore$$ Speed of stream = 3 km/hr

=> Ans - (A)

Speed of A = 7 m/s and speed of B = 8 m/s

Time taken by A to complete 1 round = $$\frac{300}{7}$$ sec

Time taken by B to complete 1 round = $$\frac{300}{8}$$ sec

=> Meeting at STARTING POINT = L.C.M. $$(\frac{300}{7},\frac{300}{8})$$

= $$\frac{300}{1}=300$$ seconds

=> Ans - (C)

Let number of questions attempted wrong be $$x$$, and number of questions attempted correct be $$(90-x)$$

According to ques,

=> $$5(90-x)-2x=387$$

=> $$450-5x-2x=387$$

=> $$7x=63$$

=> $$x=\frac{63}{7}=9$$

=> Ans - (C)

Let quantity of rice bought at Rs 4.50 per kg be $$x$$ kg and at Rs 5 per kg be $$(50-x)$$ kg

According to ques,

=> $$4.5x+5(50-x)=240$$

=> $$4.5x+250-5x=240$$

=> $$0.5x=10$$

=> $$x=\frac{10}{0.5}=20$$ kg

=> Ans - (C)

Principal sum = Rs. 16000 and time period = 2 years. Let rate of interest = $$r\%$$

Amount under Compound interest = $$P(1+\frac{R}{100})^T$$

=> $$16,000(1+\frac{r}{100})^2=17,640$$

=> $$(1+\frac{r}{100})^2=\frac{17640}{16000}$$

=> $$1+\frac{r}{100}=\sqrt{1.1025}$$

=> $$\frac{r}{100}=1.05-1=0.05$$

=> $$r=0.05\times100=5\%$$

=> Ans - (A)

Let $$x$$ ml of kerosene be there in 1 litre mixture, => Quantity of petrol in 1 litre mixture = $$(1000 - x)$$ ml

=> Required average = $$\frac{40}{1000}(1000-x)+\frac{18}{900}x=38$$

=> $$\frac{x}{25}-\frac{x}{50}=2$$

=> $$x=100$$

So, 1 litre mixture has 900 ml petrol and 100 ml kerosene.

Also, cost of 1 litre petrol = Rs. 30 and 1 litre kerosene = $$30-\frac{2}{3}\times30=Rs.$$ $$10$$

=>Actual Cost of 1 litre mixture = $$(\frac{30}{1000}\times900)+(\frac{10}{1000}\times100)$$

= $$27+1=Rs.$$ $$28$$

$$\therefore$$ Additional amount earned by pump-owner = Rs. (30-28) = Rs. 2

=> Ans - (D)

Let initial speed of train be $$5x$$ km/hr and total distance travelled be $$d$$ km

According to ques, => $$\frac{150}{5x}+\frac{(d-150)}{3x}=\frac{d}{5x}+8$$

=> $$\frac{30}{x}-\frac{50}{x}+\frac{d}{3x}-\frac{d}{5x}=8$$

=> $$\frac{2d}{15}-8x=20$$ --------------(i)

Similarly, $$\frac{510}{5x}+\frac{(d-510)}{3x}=\frac{d}{5x}+4$$

=> $$\frac{102}{x}-\frac{170}{x}+\frac{d}{3x}-\frac{d}{5x}=4$$

=> $$\frac{2d}{15}-4x=68$$ --------------(ii)

Subtracting equation (i) from (ii), we get : $$4x=48$$

=> $$x=12$$

Substituting above value in equation (i), => $$\frac{2d}{15}=20+96$$

=> $$d=58\times15=870$$ km

=> Ans - (B)

SHORTCUT METHOD (to find original speed) = $$\frac{360}{3x}-\frac{360}{5x}=4$$ => $$x=12$$