RPF 10 Jan 2019 Slot 1

If total salary is 100%, then % salary remaining after the donations = $$100-(13+12+14+16)=45\%$$

According to ques, $$45\%\equiv Rs.$$ $$40,950$$

Thus, amount donated to orphanage = $$12\%\equiv\frac{40950}{45}\times12$$

= $$910\times12=Rs.$$ $$10,920$$

=> Ans - (A)

Selling price = Rs. 2736 and profit % = 12.5%

=> Cost price = $$\frac{2736}{(100+12.5)}\times100$$

= $$24.32\times100=Rs.$$ $$2432$$

$$\therefore$$ Amount which is equal to half of the cost price of the product = $$\frac{2432}{2}=Rs.$$ $$1216$$

=> Ans - (C)

Let share of each kid be $$a,b,c,d$$ respectively.

Then according to ques, => $$a=2b=3c=4d=k$$

=> $$a=k$$ , $$b=\frac{k}{2}$$ , $$c=\frac{k}{3}$$ , $$d=\frac{k}{4}$$

Thus, total = $$a+b+c+d=500$$

=> $$k+\frac{k}{2}+\frac{k}{3}+\frac{k}{4}=775$$

=> $$k\times(\frac{12+6+4+3}{12})=775$$

=> $$k=775\times\frac{12}{25}=372$$

$$\therefore$$ Sum of gifts received by 1st kid and 2nd kid = $$a+b=k+\frac{k}{2}=3\frac{k}{2}$$

= $$3\times\frac{372}{2}=558$$

=> Ans - (A)

Let length of train = $$l$$ m and speed = $$s$$ m/s

Speed of train which takes 52 seconds to cross a bridge of length 204 m = $$s=\frac{204+l}{52}$$ -----------(i)

Also, $$s=\frac{l}{18}$$ --------------(ii)

Substituting value of $$s$$ from equation (ii) in (i), we get :

=> $$\frac{l}{18}=\frac{204+l}{52}$$

=> $$52l=(204\times18)+18l$$

=> $$52l-18l=34l=204\times18$$

=> $$l=\frac{204\times18}{34}=108$$

$$\therefore$$ Length of train = 108 m

=> Ans - (D)

Total sales of branch C2 for both years (in thousands) = 79 + 91 = 170

Total sales of branch C4 for both years (in thousands) = 66 + 86 = 152

=> Required ratio = 170:152

=> Ans - (D)

Since, the distance travelled is equal on the three sides, hence average speed will be the harmonic mean of the speeds.

= $$\frac{3}{\frac{1}{12}+\frac{1}{24}+\frac{1}{8}}$$

= $$\frac{3}{\frac{2+1+3}{24}}$$

= $$\frac{3}{6}\times24=12$$ km/hr

=> Ans - (C)

Rate of interest = 4% and time period = 2 years

Let principal sum = Rs. $$P$$

Also, difference between compound interest and simple interest for 2 years = $$P(\frac{r}{100})^2$$

=> $$P(\frac{4}{100})^2=60$$

=> $$P=60\times\frac{10000}{16}$$

=> $$P=60\times625Rs.$$ $$37,500$$

=> Ans - (A)

Expression : $$196^2 \times 56 \div 14^5 \times 1021 = ?$$

= $$(\frac{14^4}{14^5})\times56\times1021$$

= $$\frac{1}{14}\times56\times1021$$

= $$4\times1021=4084$$

=> Ans - (B)

Let number of 1 rupee, 5 rupees and 10 rupees respectively be $$3x,5x$$ and $$7x$$

Total amount = $$(1\times3x)+(5\times5x)+(10\times7x)=2352$$

=> $$3x+25x+70x=294$$

=> $$x=\frac{294}{98}=3$$

$$\therefore$$ Number of 10 rupees coins = $$7\times3=21$$

=> Ans - (D)

Let side of cube be $$a$$ cm

=> Diagonal = $$\sqrt{a^2+a^2+a^2}=22\sqrt3$$

=> $$\sqrt3a=22\sqrt3$$

=> $$a=22$$ cm

Let height of cuboid = $$h$$ cm, length, $$l=22$$ cm and breadth, $$b=11$$ cm

According to ques, Volume of cuboid = Volume of cube

=> $$lbh=a^3$$

=> $$22\times11\times h=(22)^3$$

=> $$h=\frac{22\times22}{11}=44$$ cm

=> Ans - (B)