AP ICET 2013 Question Paper

Solution:

Let Original dimensions be l and b. 

Area = l x b

New Dimension be L and B

% Increase in l = 20

L = (100+20)% Of l = 1.2l

% Increase in b = 20

B = (100+20)% Of b = 1.2b

New Area = L x B = (1.2l)(1.2b) = 1.44 lb

% change increase in area = $$\frac{Difference In Area}{Old Area}$$ x 100

                                       = $$\frac{1.44lb-lb}{lb}$$ x 100 =44% Answer

Solution:

Gain % = 5

Loss % = 5

Difference in profit percentages = 5-(-5) = 10%

This 10%(of CP) corresponds to monetary value of Rs 15

1% of CP corresponds to Rs. 15/10 = Rs.1.5

100% of CP = Rs. 1.5* 100 = Rs. 150 Answer

Solution

A/q 

20 C.P. = 15 S.P.

15 C.P. + 5 C.P. = 15 S.P.

5 C.P. = 15 S.P. - 15 C.P.

C.P =  $$\frac{15}{5}$$ (S.P. - C.P.)  = 3 (S.P. - C.P.)

C.P =  3 (S.P. - C.P.)

$$\frac{1}{3}$$ =  $$\frac{(S.P. - C.P.)}{CP}$$ = Fractional Profit

Profit percentage = Fractional Profit x 100 = $$\frac{100}{3}$$ %

=33$$\frac{1}{3}$$ % Answer

Solution 

Let C's investment be C

So 

B = C+ 5000

 A = B+ 4000 = (C+ 5000)+ 4000 = C+ 9000 =A

So A +B +C = 50000

    C+ 9000 +C+ 5000 + C = 50000

    14000 + 3C = 50000

     3C = 36000 

       C = 12000    

Hence A =  C+ 9000 = 21000

         B = C+ 5000 = 17000

Ratio of A:B:C = 21:17:12

Profit = 35000

As share = $$\frac{21}{21+17+12}$$ x 35000 = $$\frac{21}{50}$$ x 35000 = 21 x 700 

            = 14700 Answer              

Solution:

A, B and C 's investments are in the ratio 2 : 4 : 5.

So profits will be shared in the same ratio.

For every Rs.(2+4+5=)11 profit, C gets Rs 5 and A get Rs 2

So if  C gets Rs 5...A gets Rs 2

     if    C gets Rs 1...A gets Rs $$\frac{2}{5}$$

     if    C gets Rs 2200...A gets Rs $$\frac{2}{5}$$ x 2200 = Rs $$\frac{4400}{5}$$ 

     = Rs.880 Answer

Solution:

Let total work be 60 units [LCM of 20,30]

A does 60 units in 20 minutes. Hence in 1 minute A does (60/20)units = 3 units.

B does 60 units in 30 minutes. Hence in 1 minute B does (60/30)units =2 units.

Total work in 1 minute = 5 units

Now, for the given scenario,

First,

A and B work for 6 minutes together @ 5 units/min

Work done = (5 x 6) units = 30

Work remaining =Total units of work - work done = 60-30 =30 UNITS .

Then,

A work alone @ 3 units/min.

Time taken = (Work remaining / speed) = 30/3 mins. = 10 minutes

Total time = 6  +10 = 16  minutes

Solution:

Work done by A in 1 hour = 1 unit

Work done by B in 1 hour = - 1/6 unit

Total work done by A and B in 1 hr. = 1- $$\frac{1}{6}$$ = $$\frac{5}{6}$$

Time taken to fill the tank = $$\frac{1}{WorkIn1hour}$$ = 1/$$\frac{5}{6}$$       $$\frac{1}\frac{5}{6}$$

= $$\frac{6}{5}$$ Answer

Solution

Distance = 60km

Time = 40 minutes = $$\frac{40}{60}$$ hour = $$\frac{2}{3}$$ Hour

Speed = $$\frac{Distance}{Time}$$

Speed = $$\frac{60}{\frac{2}{3}}$$ = $$\frac{(60)(3)}{2}$$ = 90 kmph

Reduced speed = 90-15 = 75 kmph

Distance = 60 km

Speed = $$\frac{Distance}{Speed}$$

Time = $$\frac{60}{75}$$ = $$\frac{4}{5}$$ hr.=  48 minutes.

Solution:

Let,The distance between A-B be $$d$$.

While going from A to B Speed = 75 kmph

Time = $$\ \ \frac{\ Dis\tan ce}{Speed}$$ = $$\ \ \frac{\ d}{75}$$ hours

While going from B to A Speed = 75 kmph

Time = $$\ \ \frac{\ Dis\tan ce}{Speed}$$ = $$\ \ \frac{\ d}{75 - 15}$$ = $$\ \ \frac{\ d}{60}$$ hours

Total time = 3 hours

=> $$\ \ \frac{\ d}{75}$$ + $$\ \ \frac{\ d}{60}$$ = 3

$$\ \frac{\ d}{75}+\frac{d}{60\ }\ =\ 3$$

$$\ \frac{\ 4d\ +5d}{300}\ =\ 3$$

$$\ \frac{\ 9d}{300}\ =\ 3$$

$$\ \ d=\ 100\ km$$ Answer

Solution

A, B can together complete the work in 12 days

Hence efficiency of (A+B) = $$\left(\frac{100}{12}\right)\%$$ = $$8\frac{1}{3\ }\%$$

B and C together complete the same work in 8 days.

Hence efficiency of (B+C) = $$\left(\frac{100}{8}\right)\%$$ = $$12\frac{1}{2\ }\%$$

A, B, C work together, then that work can be finished in 6 days

Hence efficiency of (A+B+C) = $$\left(\frac{100}{6}\right)\%$$ = $$16\frac{2}{3\ }\%$$

Effeciencies

A+B = $$8\frac{1}{3\ }\%$$         .......(i)     

B+C = $$12\frac{1}{2\ }\%$$        ......(ii)

A+B+C =$$16\frac{2}{3\ }\%$$    .....(iii)

Solving for B, 

iii - ii

(A+B+C) - (B+C) = A= $$16\frac{2}{3\ }\%$$ - $$12\frac{1}{2\ }\%$$ = $$4\frac{1}{6\ }\%\ =\ \frac{25}{6\ }\%$$

Putting this in (i)

$$\ \frac{25}{6\ }\%$$ + B = $$8\frac{1}{3\ }\%$$

B =$$8\frac{1}{3\ }\%$$ -  $$\ \frac{25}{6\ }\%$$ = $$\frac{25}{6\ }\%$$ %

Time taken by B = $$\left(\frac{1}{\ efficicency}\right)$$ = $$\frac{100}{\frac{25}{6\ }}\ =\ 24$$ 

24 days Answer