MAT 2001Question Paper

Let initial speed be $$x$$ km/hr, => New speed = $$(x-100)$$ km/hr

Let original duration of flight = $$t$$ hours and actual time taken = $$(t+1)$$ hours

Total distance = $$xt=3000$$ ------------(i)

Speed is inversely proportional to time,

=> $$\frac{x}{x-100}=\frac{t+1}{t}$$

=> $$xt=xt+x-100t-100$$

=> $$x-100t=100$$

=> $$\frac{3000}{t}-100t=100$$

=> $$-100t^2-100t+3000=0$$

=> $$t^2+t-30=0$$

=> $$(t+6)(t-5)=0$$

=> $$t=5,-6$$

$$\because$$ Time cannot be negative, => Original duration = 5 hours

=> Ans - (A)

Let number of students in each row be $$x$$ and number of rows = $$y$$

=> Total number of students = $$xy$$ 

Case 1 : $$(x+4)\times(y-2)=xy$$

=> $$xy-2x+4y-8=xy$$

=> $$2y-x=4$$ ----------(i)

Case 2 : $$(x-4)\times(y+4)=xy$$

=> $$xy+4x-4y-16=xy$$

=> $$x-y=4$$ ------------(ii)

Adding equations (i) and (ii), => $$y=8$$

and similarly, $$x=12$$

$$\therefore$$ Total number of students = $$12\times8=96$$

=> Ans - (C)

A derangement is a permutation of objects that leave no object in its original position.Number of derangement's of set with n elements is

$$D_n=n![1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-.................+(-1)^n\frac{1}{n!}]$$

The probability of derangement = $$\frac{D_n}{n!}$$

Now, in the given case $$n=4$$

$$\therefore$$ Probability that none of the objects occupy the place corresponding to their number

= $$1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}$$

= $$1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}$$

= $$\frac{12-4+1}{24}=\frac{9}{24}=\frac{3}{8}$$

=> Ans - (B)

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

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

According to ques,

=> $$\frac{24}{x-y}+\frac{28}{x+y}=6$$ ----------(i)

and $$\frac{30}{x-y}+\frac{21}{x+y}=6.5$$ -------(ii)

Using the operation : 4 x (ii)-3 x (i)

=> $$\frac{48}{x-y}=26-18$$

=> $$x-y=\frac{48}{8}=6$$ --------(iii)

Similarly, $$x+y=14$$ -------------(iv)

Now, subtracting equation (iii) from (iv), => $$2y=8$$

=> $$y=\frac{8}{2}=4$$

$$\therefore$$ Speed of stream = 4 km/hr

=> Ans - (C)

Distance covered at 22 km/hr in 30 hours = $$22\times30=660$$ km = $$66\times10^6$$ cm

Radius of cycle = 35 cm

In one revolution, distance covered by the wheel = Circumference of cycle = $$2\pi r$$

= $$2\times\frac{22}{7}\times35=220$$ cm

$$\therefore$$ Number of revolutions in the total journey = $$\frac{66\times10^6}{220}=3\times10^5$$

= 3 lakh revolutions

=> Ans - (B)

Let number of notes of Rs. 50 denomination = $$x$$ and number of notes of Rs. 100 denomination = $$(85-x)$$

According to ques,

=> $$50x+100(85-x)=5000$$

=> $$50x+8500-100x=5000$$

=> $$50x=3500$$

=> $$x=70$$

$$\therefore$$ Amount in only Rs. 50 denomination = $$70\times50=Rs.$$ $$3500$$

=> Ans - (A)

Since amount spend each year is constant, thus average cost is the harmonic mean of each price

= $$3\div(\frac{1}{7.5}+\frac{1}{8}+\frac{1}{8.5})$$

= $$3\div(\frac{2}{15}+\frac{1}{8}+\frac{2}{17})$$

= $$3\div(\frac{272+255+240}{2040})$$

= $$3\times\frac{2040}{767}=7.98$$

=> Ans - (C)

Sunder's age = 14 years and Shyam's age = $$14-9=5$$ years

Let present age of Ram = $$x$$ years

According to ques, => $$\frac{(x-10)}{14}=5$$

=> $$x-10=70$$

=> $$x=70+10=80$$

$$\therefore$$ Ram's age = 80 years

=> Ans - (A)

Marked price = Rs. 600

Relative discount after successive discounts of 25 and 10% = $$25+10-(\frac{25\times10}{100})=32.5\%$$

$$\therefore$$ Sale price = $$600-(\frac{32.5}{100}\times600)$$

= $$600-195=Rs.$$ $$405$$

=> Ans - (C)

Mean of 30 values = 150

After replacing 135 by 165, correct mean = $$\frac{(30\times150)+(-135+165)}{30}$$

= $$150+\frac{30}{30}=151$$

=> Ans - (A)