RPF 10 Jan 2019

Number of trailing zeroes = $$\frac{64}{5}+\frac{64}{25}$$

= $$12+2=14$$     [Number will be round off to the lowest value]

=> Ans - (B)

Average weight of first 4 rings = 27 grams

=> Weight of first 4 rings = $$27\times4=108$$ grams

Similarly, weight of remaining 6 rings = $$28\times6=168$$ grams

$$\therefore$$ Average weight of the total rings = $$\frac{108+168}{10}=27.6$$ grams

=> Ans - (A)

Let the length and breadth of rectangle be 100 cm

=> Original area = $$100\times100=10000$$ $$cm^2$$

Increasing the length by 2%, => new length = $$100\times\frac{102}{100}=102$$ cm

Similarly, new breadth = $$98$$ cm

=> New area = $$102\times98=9996$$ $$cm^2$$

$$\therefore$$ Area is decreased by = $$\frac{10000-9996}{10000}\times100=0.04\%$$

=> Ans - (C)

Expression : $$\frac{1}{8} \div \frac{1}{8} \div \frac{1}{8} \div \frac{1}{8} \div \frac{1}{8}$$

= $$[\frac{1}{8} \div \frac{1}{8}] \div [\frac{1}{8} \div \frac{1}{8} \div \frac{1}{8}]$$

= $$(1 \div \frac{1}{8}) \div \frac{1}{8} \div \frac{1}{8}$$

= $$(8\div\frac{1}{8})\div\frac{1}{8}$$

= $$8\times8\div\frac{1}{8}$$

= $$64\times8=512$$

=> Ans - (A)

% of non defective bulbs = $$100-11=89\%\equiv2848$$

=> Number of defective products = $$11\%\equiv\frac{2848}{89}\times11$$

= $$32\times11=352$$

=> Ans - (C)

Expression : $$((3^4)^2 \times \frac{1}{27^2})^2 + 9$$ = ?

= $$[(3)^8\times(3)^{-6}]^2+9$$

= $$[(3)^{8-6}]^2+9$$

= $$3^4+9=90$$

=> Ans - (D)

Product of the 2 numbers = Product of L.C.M. and H.C.F.

= $$25\times8=200$$

=> Ans - (A)

Let usual speed = $$6$$ m/min and usual time taken = $$t$$ min

=> New speed = $$5$$ m/min and new time = $$(t+16)$$ min

Also, speed is inversely proportional to time.

=> $$\frac{6}{5}=\frac{t+16}{t}$$

=> $$6t=5t+80$$

=> $$6t-5t=t=80$$

$$\therefore$$ Usual time taken = 1 hr 20 mins

=> Ans - (D)

Principal sum = Rs. 4400

Rate of interest = 8% and time period = 5 years

Simple interest = $$\frac{P\times R\times T}{100}$$

= $$\frac{4400\times8\times5}{100}=Rs.$$ $$1760$$

Thus, total amount after 5 years = $$Rs.$$ $$(4400+1760)=Rs.$$ $$6160$$

$$\therefore$$ Remaining amount after half of the total amount is been invested in Share market = $$\frac{6160}{2}=Rs.$$ $$3080$$

=> Ans - (A)

Let side of cube be $$a$$ cm

=> Diagonal = $$\sqrt{a^2+a^2+a^2}=6\sqrt3$$

=> $$\sqrt3a=6\sqrt3$$

=> $$a=6$$ cm

Let height of cuboid = $$h$$ cm, length, $$l=6$$ cm and breadth, $$b=3$$ cm

According to ques, Volume of cuboid = Volume of cube

=> $$lbh=a^3$$

=> $$6\times3\times h=(6)^3$$

=> $$h=\frac{6\times6}{3}=12$$ cm

=> Ans - (B)