MAT 2002 Question Paper

Arithmetic mean of $$x$$ and $$y$$ = 40

=> $$x+y=80$$ ---------(i) and $$z=10$$

Average of $$x,y$$ and $$z$$ = $$\frac{(x+y+z)}{3}$$

= $$\frac{80+10}{3}=30$$

=> Ans - (B)

Let radius of each ball be $$r$$ cm and radius of solid sphere = $$R=10$$ cm

=> Volume of sphere = $$\frac{4}{3}\pi R^3$$

According to ques,

=> $$\frac{4}{3}\pi R^3=8\times\frac{4}{3}\pi r^3$$

=> $$(10)^3=8r^3$$

=> $$r^3=(\frac{10}{2})^3$$

=> $$r=5$$ cm

$$\therefore$$ Surface area of each ball = $$4\pi r^2$$

=> $$4\times \pi\times25=100\pi$$ $$cm^2$$

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

Let marked price be Rs. $$100y$$

It is given that 7% of MP = 8% of CP

=> $$7y=8x$$ -------------(i)

Also, $$9y-10x=1$$ -----------(ii)

Solving equations (i) and (ii), we get : $$y=4$$ and $$x=3.5$$

$$\therefore$$ Cost price of article = $$100\times3.5=Rs.$$ $$350$$

=> Ans - (B)

Area of garden = 100 sq m

Let length = l and width = b

l x b = 100

Option A : b = 10 => l = 10 . It is a square and not a rectangle.

Option C : b = 50 , l= 2 . Since, he fences only three sides, either 2l +b should be 30 or l+2b should be 30. Neither of these values satisy the condition.

Option D: b alone cannot be 100

hence option B

Let salary for full day = Rs. $$2x$$, half day salary = Rs. $$x$$ and absent penalty = Rs. $$15$$

The worker was absent for 4 days, fully paid for 18 days, => He was late for 8 days

According to ques,

=> $$(18\times2x)+(8\times x)-(4\times15)=1700$$

=> $$36x+8x-60=1700$$

=> $$x=\frac{1760}{44}=40$$

$$\therefore$$ Salary paid to the worker who came on time every day and was never absent = $$30\times2\times40=Rs.$$ $$2400$$

=> Ans - (A)

Total number of squares in a $$n\times n$$ chessboard = $$\sum_1^n$$ $$(n)^2$$

and total number of rectangles = $$C_2^{(n+1)}\times C_2^{(n+1)}$$

=> Number of rectangles = $$C_2^9\times C_2^9$$

= $$36\times36=1296$$

=> Ans - (D)

Time taken when he travels at 40 km/hr = $$\frac{200}{40}=5$$ hours

and time taken travelling at 20 km/hr = $$\frac{200}{20}=10$$ hours

=> Total time = 15 hours

$$\therefore$$ Average speed = $$\frac{400}{15}=26.67$$ km/hr

=> Ans - (C)

Equivalent increase in area = $$30+30+(\frac{30\times30}{100})$$

= $$60+9=69\%$$

=> Ans - (C)

Let their monthly incomes be Rs. $$4x$$ and $$5x$$ respectively, and their monthly expenditures be Rs. $$7y$$ and $$9y$$

According to ques,

=> $$4x-7y=50$$ -----------(i)
and $$5x-9y=50$$ -------------(ii)

Solving above equations, we get : $$y=50$$ and $$x=100$$

$$\therefore$$ Their monthly incomes = Rs. 400 and Rs. 500

=> Ans - (D)