MAT 2004 Question Paper

Let length of rectangular plot be 4 m and width be 10 m

=> Area = $$lb=40$$ $$m^2$$

After length is increased by 25%, new length = $$5$$ m

=> New width if area is same = $$\frac{40}{5}=8$$ m

$$\therefore$$ Reduction in width = $$\frac{(10-8)}{10}\times100=20\%$$

=> Ans - (D)

Rate of interest = 5% and income tax rate on interest = 20%

=> 80% of interest is added for next year principal

Now effective rate of interest = 80% of 5% = $$4\%$$ Principal sum = Rs. 5000 and time = 3 years

$$\therefore$$ Amount under compound interest = $$P(1+\frac{r}{100})^t$$

= $$5000\times(1+\frac{4}{100})^3$$

= $$5000\times\frac{26}{25}\times\frac{26}{25}\times\frac{26}{25}$$

= $$17576\times\frac{8}{25}=Rs.$$ $$5624.32$$

=> Ans - (A)

Let number of 10 paise coins be $$x$$ and number of 25 paise coins be $$(180-x)$$

According to ques,

=> $$\frac{1}{10}\times(x)+\frac{1}{4}\times(180-x)=36.90$$

=> $$\frac{x}{10}-\frac{x}{4}+45=36.90$$

=> $$\frac{3x}{20}=8.10$$

=> $$x=8.10\times\frac{20}{3}=54$$

$$\therefore$$ Number of 10 paise coins = 54 and number of 25 paise coins = 126

=> Ans - (B)

1 candle is made from 9 stubs, => 116 candles will be made from 1044 stubs

From 116 candles, 12 candles can be made with 8 stubs left.

Now total stubs left = $$12+8=20$$, out of which two candles can be made with two stubs left.

Thus, Maximum number of candles that can be made = $$(116+12+2)=130$$

(or) Number of candles made = $$(\frac{1044}{9^1})+(\frac{1044}{9^2})+(\frac{1044}{9^3})\approx130$$

=> Ans - (C)

Since, there are only two types of socks in the bag. So, if Badri pick up 3 socks then certainly two of them are of same type. Thus this is a certain event.

=> Required probability = 1

=> Ans - (C)

The two boys in corner can be arranged in = $$2!$$ ways

Now, the remaining boys can be arranged in = $$18!$$ ways

There are 6 girls who can be arranged in = $$6!$$ ways

$$\therefore$$ Total number of ways in which all the students can be arranged

= $$2!\times18!\times6!$$

= $$18!\times1440$$

=> Ans - (A)

Let the two numbers be $$7x$$ and $$8x$$

Now, L.C.M. $$(7x,8x)=7\times8\times x$$

According to ques,

=> $$56x=280$$

=> $$x=\frac{280}{56}=5$$

$$\therefore$$ The two numbers are : 35, 40

=> Ans - (C)

Using $$M_1D_1=M_2D_2$$

Let $$x$$ people were supposed to work initially.

=> $$x\times24=(x-9)\times32$$

=> $$3x=4(x-9)$$

=> $$3x=4x-36$$

=> $$x=36$$

=> Ans - (C)

Two circles with centres A and A' and having radius = 2 cm

$$\angle$$ BAC = $$120^\circ$$ [since ABA' forms a equilateral triangle] and length of an arc = $$\frac{\theta}{360^\circ}\times2\pi r$$

Perimeter of shaded portion = $$2\times($$ length of arc BC$$)$$

= $$2\times(\frac{120^\circ}{360^\circ}\times2\pi r)$$

= $$2\times(\frac{1}{3}\times2\pi\times2)$$

= $$2\times\frac{4\pi}{3}=\frac{8\pi}{3}$$

=> Ans - (A)

Time taken to make $$(pq)$$ metres of cloth = $$r$$ min

=> Time taken to make 20,000 m of cloth = $$\frac{r}{pq}\times20,000$$ minutes

$$\therefore$$ Required time (in hour) = $$20,000\times\frac{r}{60(pq)}$$

=> Ans - (D)