MAT 2004 Question Paper

To rectify the mistake, we need to do the operation on the total = $$(-18)+(-13+31)=0$$

$$\because$$ By adding zero, there will be no change in the sum of the numbers, thus the average will remain same = 40.2

=> Ans - (A)

The total students who speak English are = 23

According to the venn diagram = $$9+3+6+x=23$$

=> $$x=23-18=5$$

$$\therefore$$ 5 students speak all the three languages.

=> Ans - (D)

Effciency of machine A alone = $$\frac{1}{10}-\frac{1}{15}$$

= $$\frac{3-2}{30}=\frac{1}{30}$$

Thus, machine A alone produce 3,00,000 metres of cloth in 30 hours

$$\therefore$$ In 10 hours, machine A alone can produce = $$\frac{300000}{30}\times10=1,00,000$$ metres

=> Ans - (B)

Let $$x$$ and $$(35-x)$$ litres of each solution be mixed.

=> $$37.85\%$$ of $$x+$$ $$92\%$$ of $$(35-x)=89\%$$ of $$35$$

=> $$37.85x+92(35-x)=89\times35$$

=> $$37.85x-92x=(89\times35)-(92\times35)$$

=> $$54.15x=35\times3=105$$

=> $$x=\frac{105}{54.15}\approx1.94$$

$$\therefore$$ 1.94 litres of the first and 33.06 litres of the second

=> Ans - (A)

Time taken by first train = 4 hours and time taken by second train = 3.5 hours

Since, distance travelled for both is same, ratio of time = $$8:7$$

=> Ratio of speeds of both trains (speed is inversely proportional to time) = $$7:8$$ respectively.

Let speed of first train be $$7x$$ km/hr and speed of second train be $$8x$$ km/hr and thus total distance between two stations = $$28x$$ km

Thus, distance travelled by first train in first two hours (till second train starts) = $$14x$$ km

=> Remaining distance between the trains at 7 am = $$14x$$ km and relative speed coming in opposite directions = $$7x+8x=15x$$ km/hr

Time taken for them to meet each other = $$\frac{14x}{15x}=\frac{14}{15}\times60=56$$ minutes

$$\therefore$$ The two trains will meet at 7:56 am

=> Ans - (B)

Expression : M39048458N

For above number to be divided by 8, the last three digits, i.e. $$58N$$ must be divided by 8, => $$N=4$$

Now, for the number $$M390484584$$ to be divided by 11, difference between sum of even digits and sum of odd digits must be divided by 11.

=> Sum of even digits = $$(3+5+8+0+4)=20$$
=> Sum of odd digits = $$(8+4+4+9+M)=(25+M)$$

required difference = $$(25+M)-20=11$$

=> $$M=6$$

$$\therefore$$ $$(M,N)=(6,4)$$

=> Ans - (C)

Let maximum marks in the examination be $$100x$$ and passing marks be $$y$$

Peter got 30% of the maximum marks, => $$30x+10=y$$ ----------(i)

Similarly, for Paul = $$40x-15=y$$ -------(ii)

Subtracting equation (i) from (ii), => $$10x=25$$

=> $$x=2.5$$

Substituting it in equation (ii), we get : Passing marks = $$y=85$$

=> Ans - (D)

Let cost price of article be Rs. $$100x$$

=> Marked price = Rs. $$130x$$

After 20% discount, selling price = $$\frac{80}{100}\times130x=208$$

=> $$4\times26x=208$$

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

Thus, cost price of article = Rs. 200

$$\therefore$$ Net profit as percent of cost price = $$\frac{(SP-CP)}{CP}\times100$$

= $$\frac{(208-200)}{200}\times100=4\%$$

=> Ans - (B)

Let the breadth of the field be $$x$$ m and length = $$(3x+2)$$ m

=> Perimeter of rectangular field = $$2[(x)+(3x+2)]=52$$

=> $$4x+2=26$$

=> $$4x=26-2=24$$

=> $$x=\frac{24}{4}=6$$ m

=> Ans - (D)

To select first pair, we need to choose 2 students out of 6, this can be done in $$C^6_2$$ ways

Similarly, for the next 2 pairs = $$C^4_2$$ and $$C^2_2$$ ways

Now, to eliminate repetition, required number of ways = $$\frac{C^6_2\times C^4_2\times C^2_2}{3!}$$

= $$\frac{15\times6\times1}{6}=15$$

=> Ans - (C)