DECIMAL FRACTIONS

1.) Convert 0.78787878787878.......... into fraction.

         (a)45/90            (b)78/99                        (c)78/90             (d)78/990

Solution: option (b)

Explnation: 0.7878787878.........=78/99

2.) what is a rational no for reoccuring decimal 0.890890890........?

     (a)890890/1000             (b)67/421                 (c)890/999             (d)none of these

Solution: option (c)

Explanation: 0.890890890....=890/999

3.) 926 + 9.026 + 0.926 + 9.0026 =

    (a) 944.9546             (b)944.1246      (c)944.9456              (d)954

Solution: option (a)

Explanation: 926+ 9.026 + 0.926 + 9.0026 = 944.9546

4.) Add 3/4 x 5/8 + 8/2 +79.34

      (a) 83.43875           (b)834              (c)84              (d) 83.45678

Solution: option (a)

Explanation: 3/4 x 5/8 +8/4 +79.345 = 83.43875

PERCENTAGE

1.) Two numbers A and B are such that the sum of 5% of A and 4% of B is two-third of the sum of 6% of A and 8% of B. Find the ratio of A : B.

A. 2 : 1                   B. 1:2                       C. 1:1                      D. 4:3 SOLUTION: (D) EXPLANATION: 5% OF A +4% OF B= 2/3(6%OF A+  8%OF B) 5A/100 +4B/100= 2/3(6A/100 +8B/100) 5A+4B =2/3(6A+8B) 15A+12B=12A+ 16B 15A-12A=16B-12B 3A=4B A/B=4/3 A:B=4:3 2.) The population of a town increased from 1,75,000 to 2,62,500 in a decade. What is the average percent increase of population per year?   A. 4%                       B.6%                      C.5%                           D.50% SOLUTION: (C) EXPLANATION: INCREASE IN POPULATION AFTER 10 YEARS= 262500-175000                                                                                  =87500 %INCREASE IN POPULATION AFTER 10 YEARS =(87500/175000)*100 =50% AVERAGE %INCREASE IN POPULATIONPER YEAR= 50%/10 =5%

MIXTURE AND ALLIGATION

1.)A container contains 40 litres of milk.From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container.

a) 27.36 liters            b)29.16 liters             c)  28 liters                 d) none of these

solution: b) 29.16 liters

explanation: Amount of milk left after 3 operations

 litres

2.) 8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the water is 16 : 65. How much wine the cask hold originally?

a) 24 liters              b) 32 liters                 c) 48 liters              d) none of these

solution:(a)

explanation:    Let the quantity of the wine in the cask originally be x litres

Then, quantity of wine left in cask after 4 operations =litres

AVERAGE 1.)  In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?
(A) 6.25                        (B)5.5                               (C)7.4                                      (D) 5
SOL.: (A) EXPLANATION: Runs scored in the first 10 overs = 10 × 3.2 = 32 Total runs = 282 Remaining runs to be scored = 282 - 32 = 250 Remaining overs = 40 Run rate needed = $\frac{250}{40}=6.25$ x$$ \frac{\text{Sum of ages of all the 11 members of the team}}{11}=x$$ 11x$$ =11x-26-29=11x-55$$ =\frac{11x-55}{9}$$ \frac{11x-55}{9}=(x-1)$$ \Rightarrow 11x-55=9(x-1)\Rightarrow 11x-55=9x-9\Rightarrow 2x=46\Rightarrow x=\frac{46}{2}=23\text{years}$$

CHAIN RULE

1.) If 7 spiders make 7 webs in 7 days, then 1 spider will make 1 web in how many days?

A) 1	B) 3
C)  7	D) 14

SOLUTION: (C)

Let the required number days be x.

Less spiders, More days (Indirect Proportion)

Less webs, Less days (Direct Proportion)

 1 x 7 x x = 7 x 1 x 7

=> x= 7

2.) 3 pumps working 8 hours a day, can empty a tank in 2 days. How many hours a day must 4 pumps work to empty the tank in 1 day?

A) 9	B) 10
C) 11	D) 12

SOLUTION: (D)

Let the required no of working hours per day be x.

More pumps , Less working hours per day  (Indirect Proportion)

Less days, More working hours per day      (Indirect Proportion)

       X=      x=12

PIPES ANS CISTERN

Two pipes A and B can fill a cistern in 371⁄2 minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if pipe B is turned off after:
A. 5 min	B. 9 min
C. 10 min	D. 15 min

SOLUTION: (B)

EXPLANATION: Pipe A can fill a tank in 37¹⁄₂ minutes = ⁷⁵⁄₂ minutes

=>Part filled by pipe A in 1 minute = ²⁄₇₅

Pipe B can fill a tank in 45 minutes

=> Part filled by pipe B in 1 minute = ¹⁄₄₅

=> Part filled by Pipe A and B in 1 minute = =275+145=6+5225=11225

Assume that B is turned off after x minutes. i.e., for x minutes, both pipe A and B were open.

Part filled in x minutes by Pipe A and B = x×11225=11x225

Now, the cistern must be filled in (30-x) minutes by pipe A alone

Part filled in (30-x) minutes by pipe A = (30−x)×275=2(30−x)75

11x225+2(30−x)75=111x225+2(30−x)75=1⇒11x+6(30−x)=225⇒11x+180−6x=225⇒5x=45x=9