TS ICET 2019 Question Paper Shift-2 (23rd May)

Solution 

Let original salary be Rs.100

After increment of 20% on original salary (Rs.100)

Salary = (100+20)/100 x Rs.100 = 1.2 x Rs. 100 = Rs. 120.

After decrement of 10% on increased salary (Rs.120)

Salary = (100-10)/100 x Rs.120 = 0.9 x Rs. 120 = Rs. 108.

So the final salary stands at Rs. 108

Net change w.r.t. original salary =  Rs. (108-100) = Rs. 7

% change = (Rs.8/100) x Rs.100 = 8% Answer

Solution:

Old loss % = 15

New loss % = 9

Difference in loss percentages = 15-9 = 6%

This 6%(of CP)loss corresponds to monetary loss of Rs 420.

1% of CP corresponds to Rs. 420/6 = Rs.70

100% of CP = Rs. 70 * 100 =Rs. 7000 Answer

Solution

Let Original price of the car be Rs. 100

After 20% discount car's price = (100-20)/100 x Rs.100 = 0.8 x Rs. 100      

                                            = Rs. 80.

So CP for Man = Rs.80

Car was sold at 20% higher price,

So SP = (100+20)/100 x Rs. 100

        = 1.2 x 100 = Rs. 120

SP for man = Rs. 120

Profit % = (SP-CP)/CP

             =(120 - 80) / 80

              = 50 % Answer

Solution:

The ratio of investments of P,O and R is 7200:12600:9600.    

The standard ratio will be 12:21:16.

So for every profit of Rs.(12+21+16) , R will get Rs. 16.

i.e. So for every profit of Rs.49,R will get Rs. 16.

for Profit of Re 1, R will get Rs.(16/49)

For Profit of Rs 6860, R will get Rs. (16 /49 ) x 6860                                                  

                                             = Rs. 2240. Answer

Solution

Let Ps and Os initial investment be 3x and 4x respectively.

P's investment was for 10 months so it will be proportional to (3x)(10) = 30x

let Q's investment be for 'm' months, so its proportional investment will be (4x)(m).

So according to investment profits will be divided in the ratio 30x : 4mx = 30:4m  

A/q 

=>  30:4m =  5:6

=> 30/4 =

=>  m= 9    

Hence O invested for 9 months.                                                  

Solution:

Let total capacity of the tank be 60 units [LCM of 15,20]

First pipe fills 60 units (full tank) in 15 hrs. Hence in 1 hr. it fills (60/15) =4 units

Second pipe empties 60 units (full tank) in 20 hrs. Hence in 1 hr. it empties (60/20)units = 3 units or fills (-)3 units.

If both the pipes are opened simultaneously, total tank filled in 1 hr = 4 + (-)3 units = 1 unit.

So to fill full capacity (60 units), time taken = (60/1) = 60 hrs.

Solution:

Let total work be 60 units [LCM of 12,15]

P does 60 units in 12 minutes. Hence in 1 minute P does (60/12)units = 5 units.

O does 60 units in 15 minutes. Hence in 1 minute O does (60/15)units = 4 units. 

Total work in 1 minute = 9 units

Now, for the given scenario,

First, 

P and O work for 3minutes together @ 9 units/min

Work done = (9 x 3) units = 27

Work remaining =Total units of work - work done = 60-27 =33 UNITS .

Then,

O work alone @ 4 units/min.

Time taken = (Work remaining / speed) = 33/4 mins. = 8.25 minutes

Total time = 3+8.25 = 11.25 minutes or 11$$\frac{1}{4}$$ minutes

Solution 

Method 1

Distance = 200 m

Time = 20s

Speed in m/s = (200/20) m/s

                    =      10m/s

Now Important pointer to remember 

5m/s = 18 km/h

So 10m/s = 36kmph Answer

Method 2

20 s for 200m.

1 Hr. (20 x 3 x 60 s) for (200 x 3 x60) m

1 Hr for 36000m

1 Hr for 36 km

Hence speed 36kmph Answer

Solution:

Let old speed and time be S1 and T1.

Let new speed and time be S2 and T2.

A/q,

S2=(5/4)S1 = 1.25 S1 .....Eqn (i)

T1 - T2 = 30minutes = 0.5 hr.....Eqn (ii)

As Distance remains constant ; S x T = constant;

S1 x T1 = S2 x T2

S1 x T1 = (1.25 S1) x T2......from Eqn.(i)

T1/T2 = 1.25

T1= 1.25 T2

and T1- T2 = 0.5 hr .........from Eqn.(ii)

1.25T2 - T2 = 0.5Hr

0.25 T2 = 0.5 Hr.

T2 = 2 Hr.

T1 = old time = T2 + 0.5 Hr. = 2.5 Hr.Answer

Solution:

A/q

Let Q's efficiency be x%.

Then P's efficiency is 2x%

Total efficiency (if they work together) = (x + 2x)%= 3x %

With 3x% efficiency, time taken = 12 days.

So, with x% efficiency time taken = 13 x 3 = 36 days..... (Efficiency is indirectly proportional to Time taken)

So time taken if P works alone @ 2x % is = 36/2 days    

= 18 days. Answer