MAT 2005 Question Paper

Shekhar invested Rs. 25,000 for 1st year, Rs. 35,000 for 2nd year and Rs. 45,000 for 3rd year. Rajeev invested Rs. 35,000 for 2 years and Jatin invested Rs. 35,000 fr 1 year

=> Ratio of profits earned by Shekhar, Rajeev and Jatin respectively = $$(25000\times1+35000\times1+45000\times1):(35000\times2):(35000\times1)$$

= $$(25+35+45):(70):(35)$$

= $$105:70:35=3:2:1$$

$$\therefore$$ Rajeev's share in total profit of Rs. 1,50,000 = $$\frac{2}{(3+2+1)}\times1,50,000$$

= $$2\times25000=Rs.$$ $$50,000$$

=> Ans - (B)

Let principal sum be Rs. $$P$$ and rate of interest = $$r\%$$

Difference between S.I. and C.I. for 2 years = $$\frac{Pr^2}{100^2}$$ and for 3 years = $$\frac{Pr^2}{100^2}(3+\frac{r}{100})$$

=> Ratio = $$[\frac{Pr^2}{100^2}(3+\frac{r}{100})]:[\frac{Pr^2}{100^2}]=25:8$$

= $$\frac{3+\frac{r}{100}}{1}=\frac{25}{8}$$

=> $$\frac{r}{100}=\frac{25}{8}-3$$

=> $$\frac{r}{100}=\frac{25-24}{8}=\frac{1}{8}$$

=> $$r=\frac{100}{8}=12\frac{1}{2}\%$$

=> Ans - (D)

Let the capacity of the tank be 100 litres
Initially: A type petrol = 100 litres

After first operation:
A type petrol = $$\frac{100}{2}=50$$ litres
B type petrol = $$50$$ litres

After second operation:
A type petrol = $$50+\frac{50}{2}=75$$ litres
B type petrol = $$\frac{50}{2}=25$$ litres

After third operation:
A type petrol = $$\frac{75}{2}=37.5$$ litres
B type petrol = $$50+\frac{25}{2}=62.5$$ litres

$$\therefore$$ Required percentage = 37.5%

=> Ans - (B)

Let quantity of each be L.C.M.(9,18) = 18 units

=> Gold in alloy A = 14 units and copper in alloy A = 4 units

Similarly, gold in alloy B = 7 units and copper in alloy B = 11 units

$$\therefore$$ Ratio of gold and copper in alloy C = $$\frac{(14+7)}{(4+11)}$$

= $$\frac{21}{15}=7:5$$

=> Ans - (C)

Let number of boys be $$x$$ and number of girls be $$y$$

According to ques,

=> $$16.4x+15.4y=15.8(x+y)$$

=> $$16.4x+15.4y=15.8x+15.8y$$

=> $$0.6x=0.4y$$

=> $$\frac{x}{y}=\frac{0.4}{0.6}=2:3$$

=> Ans - (B)

Distance travelled by Bombay Express in 2 hours = $$60\times2=120$$ km

Thus, at 16.30 hrs, the two trains are at a distance of 120 km from each other.

Travelling in same direction, relative speed = $$80-60=20$$ km/hr

=> Time taken by the trains to meet = $$\frac{120}{20}=6$$ hours

$$\therefore$$ Distance from Delhi when the trains meet = $$80\times6=480$$ km

[Note :- If we multiply the speed of Bombay Express, i.e. 60 km/hr, we need to keep in mind that it has already travelled 120 km (for 2 hours) in the beginning.]

=> Ans - (C)

Let AB be the tower and BC = 60 m. It takes 5 seconds for the boat to go from C to point D.

In right $$\triangle$$ ABC,

=> $$tan(45^\circ)=\frac{AB}{BC}$$

=> $$1=\frac{AB}{60}$$

=> $$AB=60$$ m ----------------(i)

Similarly, in right $$\triangle$$ ABD,

=> $$tan(30^\circ)=\frac{AB}{BD}$$

=> $$\frac{1}{\sqrt3}=\frac{60}{BD}$$

=> $$BD=60\sqrt3$$ m ----------------(ii)

Thus, CD = $$60\sqrt3-60=60(\sqrt3-1)$$ m

Using speed = distance/time, => Speed of boat = $$\frac{60(\sqrt3-1)}{5}$$

= $$12(\sqrt3-1)=12\times0.732=8.784$$ m/s

$$\therefore$$ Speed of boat (in km/hr) = $$8.784\times3.6\approx32$$ km/hr

=> Ans - (A)

Expression : $$\left[\frac{1}{\log_{xy} (xyz)} + \frac{1}{\log_{yz} (xyz)} + \frac{1}{\log_{zx} (xyz)}\right]$$

Using, $$\log_m(n)=\frac{1}{\log_n(m)}$$ and $$\log(mn)=\log(m)+\log(n)$$

= $$\log_{xyz}(xy)+\log_{xyz}(yz)+\log_{xyz}(zx)$$

= $$\log_{xyz}(xy\times yz\times zx)$$

= $$\log_{xyz}(xyz)^2$$

= $$2\log_{xyz}(xyz)$$      [$$log_m(m)=1$$]

= $$2\times1=2$$

=> Ans - (D)

Sides of rectangle are $$l=26$$ cm and $$b=18$$ cm

=> Perimeter of rectangle = $$2\times(26+18)=2\times44=88$$ cm

Let radius of circle be $$r$$ cm

=> Circumference of circle = $$2\pi r=88$$

= $$2\times\frac{22}{7}\times r=88$$

=> $$r=88\times\frac{7}{44}=14$$ cm

$$\therefore$$ Area of circle = $$\pi r^2$$

= $$\frac{22}{7}\times14\times14=616$$ $$cm^2$$

=> Ans - (D)

Let true discount T.D. be Re 1

=> B.D. = Rs. $$\frac{11}{10}=1.10$$

$$\therefore$$ Sum = $$\frac{(1.10\times1)}{(1.10-1)}=Rs.$$ $$11$$

=> S.I. on Rs. 11 for 2 years is Rs. 1.10

Let rate of interest be $$r\%$$

=> $$\frac{11\times r\times2}{100}=1.1$$

=> $$r=\frac{110}{22}=5\%$$

=> Ans - (C)