MAT 2006 Question Paper

Given, speed of the boat when moving perpendicular to the direction of current = 16 km/hr and speed of current = 3 km/hr

Speed of the boat in still water = 16 - 3 = 13 km/hr 

Speed of the boat against current = 13 - 3 = 10 km/hr

Hence, option C is the correct answer.

Given, R and S are moving at the speeds of 3 kmph and 4 kmph respectively. And distance between them is 17.5 km

Relative speed = 3 + 4 = 7 kmph

 Time taken for R and S to meet = $$\frac{17.5}{7}$$ = $$2.5$$ hours

They meet at (10 AM + 2.5 hours) = 12:30 PM

Hence, option D is the correct answer.

Let, cost price of the goods be 100x

As the shopkeeper marks up by 35% and gives 10% discount for cash transactions,

S.P = 90% of 135% of 100x

$$\Rightarrow$$ 0.09 $$\times$$ 1.35 $$\times$$ 100x

$$\Rightarrow$$ 121.5x i.e profit percentage = 21.5%

Hence, option C is the correct answer.

Work done by B in hour is given by $$\frac{1}{20}$$

As A can do 50% more efficient than B,

$$\frac{1}{A} = 150 % of \frac{1}{B}$$

$$\frac{1}{A} = \frac{3}{2} \times \frac{1}{B}$$ 

$$\frac{1}{A} =  \frac{3}{2} \times \frac{1}{20}$$

$$A = \frac{3}{40}$$

Work done by A and B together in hour = $$\frac{1}{A} + \frac{1}{B}$$ 

$$\Rightarrow \frac{3}{40} + \frac{1}{20}$$ 

$$\Rightarrow \frac{3 + 2}{40} = \frac{1}{8}$$ 

Total work done by A and B together = 8 hours.

Hence, option B is the correct answer.

Given that 4 times the amount received by A is equal to 6 times the amount received by B and 11 times the amount received by C.

This can be written as,

4A = 6B or A : B = 3 : 2 ....(1) and

4A = 11C or A : C = 11 : 4 ....(2)

Multiply equation (1) with 11 and equation (2) with 3 we get,

A :B = 33 : 22 and A : C = 33 : 12 (or)

A : B : C = 33 : 22 : 12 (or) 66 : 44 : 24

Hence, option D is the correct answer.

Let the actual distance travelled by train be 'd'. Similarly, actual speed and time be 's' and 't'

Given, If the train had been 6 km/hr faster, then it would have taken 4 hours less than the scheduled time,

st = (6 + s)(t - 4) (or) 6t - 4s -24 = 0................(1)

If the train were slower by 6 km/hr, then the train would have taken 6 hours more than the scheduled time,

st = (s - 6)(t + 6) (or) 6t - 6s + 36 = 0...............(2)

Subtract equation (1) and (2)

2s = 60 (or) s = 30

Substitute the value of 's' in equation (1)

6t - 4(30) = 24 (or) 6t = 144 (or) t = 24

The length of the journey (or) total distance = 24 x 30 = 720

Hence, option C is the correct answer.

Let, the number of rows = 'x'  and the number of students in the row be 'y'

Given that, if 4 students are extra in each row, then there would be 2 rows less,

(x - 2)(y + 4) = xy

xy + 4x -2y -8 = xy (or) 4x - 2y = 8 ...........(1)

If 4 students are less in each row, then there would be 4 more rows,

(x + 4)(y - 4) = xy 

xy - 4x + 4y - 16 = xy (or) 4x - 4y = -16 .......(2)

Subtract equations (1) and (2)

2y = 24 (or) y =12

Substitute value of 'y' in equation (1)

4x - 2(12) = 8 $$\Rightarrow$$ 4x = 8 + 24 $$\Rightarrow$$ x = 8

Number of students in the class = xy $$\Rightarrow$$ 8 x 12 = 96

Hence, option D is the correct answer.

Let, the part of monthly expenses which is constant be 'x' and 

The part which varies with the price of wheat be 'y'

Then according to the question,

x + 250y = 1000 ......(1)

x + 240y = 980 ........(2)

Subtract equations (1) and (2)

250y - 240y = 1000 - 980 

10y = 20 $$\Rightarrow$$ y = 2

Substitute value of 'y' in equation (1)

x + 250(2) = 1000 $$\Rightarrow$$ x = 500

The total monthly expenses of the family when the cost of wheat is Rs 350 a quintal is,

500 + 350(2) = 1200

Hence, option C is the correct answer.

Let usual speed of the plane be 's' km/hr and new speed be 'x + 250' km/hr

As the plane left 30 minutes late, the difference between this time and usual time will be equal to 30 minutes. 

$$\frac{1500}{s} - \frac{1500}{x + 250} = \frac{30}{60}$$

$$\frac{1500s + 1500 \times 250 - 1500s}{s(s + 250)} = \frac{1}{2}$$

$$1500(s + 250 - s) \times 2 = s(s+250)$$

$$75000 = s^{2} + 250s$$

 $$s^{2} + 250s - 7,50,000 = 0$$

Roots for the above equation are $$750$$ and '$$-1000$$'

As '$$-1000$$' is a negative value, consider only $$750$$.

Hence, option D is the correct answer.

Length of the box after the sheet is cut off (l) = 48 - (8 + 8) = 32 cm

Breadth of the box after the sheet is cut off (b) = 36 - (8 + 8) = 20 cm

Height (h) = 8 cm

Volume of the box = l x b x h

$$\Rightarrow$$ 32 x 20 x 8 = 5120 cm$$^{3}$$

Hence, option C is the correct answer.