MAT 2002 Question Paper

Let number of women that have both be $$b$$

Then, according to ques, $$a+b=7$$ and $$b+c=8$$ and $$d=3$$

Also, $$a+b+c+d=15$$

=> $$a+b+c=12$$ -----------------(2)

Substituting value of $$(a+b)$$ in equation 2, => $$c=12-7=5$$

and thus $$b=3$$

$$\therefore$$ Number of women having both nose studs and ear rings = 3

=> Ans - (C)

Of the 500 seats, number of seats in executive class = 50 and chair cars = 450

Now, on the given journey, total seats booked = $$\frac{85}{100}\times500=425$$

Seats booked in executive class = $$\frac{96}{100}\times50=48$$

=> Seats occupied in chair car = $$425-48=377$$

$$\therefore$$Number of empty chair car seats = $$450-377=73$$

=> Ans - (B)

The order of installments in Rs. is 100, 95, 90, 85 and so on.

The above series forms an A.P. with first term = $$a=100$$ and common difference = $$d=-5$$ and total sum = Rs. 975. Let in $$n$$ months, total amount is paid.

Sum of an A.P. = $$\frac{n}{2}[2a+(n-1)d]$$

=> $$\frac{n}{2}[2\times100+(n-1)(-5)]=975$$

=> $$\frac{n}{2}\times[200-5n+5]=975$$

=> $$205n-5n^2=1950$$

=> $$n^2-41n+390=0$$

=> $$(n-15)(n-26)=0$$

=> $$n=15, 26$$

Now, if $$n=26$$, number of installments will become negative (after 20 installments), hence it is not possible.

$$\therefore$$ Total installments = 15 months

=> Ans - (C)

Let capacity of vessel be $$x$$ litres

Now, special liquid in the vessel = $$(x-4)$$ litres and water = $$4$$ litres

=> Ratio = $$\frac{x-4}{4}=\frac{1}{2}$$

=> $$x-4=2$$

=> $$x=6$$

$$\therefore$$ Capacity of vessel = 6 litres

Radius of cone = 7 cm and height of cone = 51 cm

=> Volume of cone = $$\frac{1}{3}\pi r^2h$$

= $$\frac{1}{3}\times\frac{22}{7}\times49\times51=2618$$ $$cm^3$$

$$\therefore$$ Weight of cone = $$2618\times10=26180$$ gm = 26.18 kg

=> Ans - (C)

Six balls can be selected in the following ways : (1 red ball and 5 blue balls) or (2 red balls and 4 blue balls)

Since, all the six balls cannot be blue.

=> Total number of ways = $$(C^3_1\times C^7_5)+(C^3_2\times C^7_4)$$

= $$(3\times21)+(3\times35)$$

= $$63+105=168$$

=> Ans - (B)

Let $$x$$ be the face value of debentures.

Then, 15% of $$x$$ = 1500

=> $$x=10,000$$

=> Available value = $$\frac{104}{100}\times10,000=10,400$$

Brokerage = 1% of 10400 = 104

$$\therefore$$ Total money invested = $$10400+104=Rs.$$ $$10504$$

=> Ans - (C)

Let selling price of 1st grade cycle be Rs. $$4x$$ and of 2nd grade cycle be Rs. $$3x$$

Total selling price = $$\frac{140}{100}\times3150=Rs.$$ $$4410$$

According to ques,

=> $$(8\times4x)+(22\times3x)=4410$$

=> $$32x+66x=98x=4410$$

=> $$x=\frac{4410}{98}=45$$

$$\therefore$$ Selling Price of 1st grade cycles = $$4\times45=Rs.$$ $$180$$

=> Ans - (C)

For three terms $$A,B,C$$ to be in GP, it must follow : $$B^2=AC$$

Now, $$\log_{x} a, a^{\frac{x}{2}}$$ and $$\log_b x$$ are in GP

=> $$(a^{\frac{x}{2}})^2=(\frac{\log a}{\log x})\times(\frac{\log x}{\log b})$$

=> $$a^x=\frac{\log a}{\log b}$$

=> $$a^x=\log_b a$$

=> $$x=\log_a(\log_b a)$$

=> Ans - (A)