A and B together can complete a work in 3 days.

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A and B together can complete a work in 3 days. They start together but after 2 days, B left the work. If the work is completed after two more days, B alone could do the work in

A5 days

B6 days

C9 days

D10 days

Nithin K ? Apr 7 '2018 at 19:44

Answer is : 6 days

Explanation:

(A+B)'s one day's work \(= \frac{1}{3}\) part
(A+B) works 2 days together \(= \frac{2}{3}\) part;
Remaining work \(= 1-\frac{2}{3} \implies \frac{1}{3} \)part;
1/3 part of work is completed by A in two days;
Hence, one day's work of A = \(\frac{1}{6}\);
Then, one day's work of B\( = \frac{1}{3}-\frac{1}{6}\implies \frac{1}{6}\);
So, B alone can complete the whole work in 6 days.

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What part of a ditch, 48 m long 16.5 m broad and 4 m deep can be filled

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What part of a ditch, 48 m long 16.5 m broad and 4 m deep can be filled by the earth got by digging a cylindrical tunnel of diameter 4 m and length 56 m ?

A$$\frac{1}{9}$$

B$$\frac{2}{9}$$

C$$\frac{7}{9}$$

D$$\frac{8}{9}$$

Nithin K ? Apr 7 '2018 at 16:54

Answer is : b.) \(\frac{2}{9}\)

Explanation:

Volume of the earth dugout as a tunnel = πr2h

= \(\Big(\frac{22}{7}\Big) \times 2 \times 2 \times 56 \)

= \(\text{704 }m^3 \)
\(\text{Volume of the ditch = }\frac{(48 \times 33)}{(2)}\times 4 \implies 24 \times 33 \times 4 \implies \text{3168 }m^3 \)
∴ \(\text{Part required = }\frac{704}{3168} \implies \frac{2}{9}\)

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The area of circle is increased by 22 cm. its radius is increased by 1 cm. The original radius of the circle is

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The area of circle is increased by 22 cm. its radius is increased by 1 cm. The original radius of the circle is

A6 cm

B3.2 cm

C3 cm

D3.5 cm

Nithin K ? Apr 7 '2018 at 16:39

Answer is : c.) 3 cm

Explanation:

Let original radius be r.

Then, according to the question,
\(\Large \pi (r+1)^{2}-\pi r^{2} =22\)

\(\implies \Large \pi\times [(r+1)^{2}-r^{2}] =22\)

\(\implies \Large \frac{22}{7}\times (r+1+r)(r+1-r) =22\)

\(\implies 2r+1=7 \implies 2r=6\)

\(r= \Large \frac{6}{2} =\text{3 cm}\)

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The perimeter of the floor of a room is 18 metres.

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The perimeter of the floor of a room is 18 metres. What is the area of the walls of the room. If the height of the room is 3 metres?

A$$21 m^2$$

B$$42 m^2$$

C$$54 m^2$$

D$$108 m^2$$

Nithin K ? Apr 7 '2018 at 16:11

Answer is : c.) \(54\text{ m}^2\)

Explanation:

If the floor is in rectangular shape

Then your answer is, 

Perimeter of rectangle =2(l+b) 

18=2(l+b) 

Then, 

\(\text{Area of four walls =}2(l+b)\times h\)

\(\text{Area of four walls=}18\times3 \)
\(\text{Area of four walls = }54\text{ m²}\)

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A train travelling at a speed of 30 m/s crosses a platform, 600 m. long, in 30 sec

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A train travelling at a speed of 30 m/s crosses a platform, 600 m. long, in 30 sec.The length (in metres) of train is

A120

B150

C200

D300

Nithin K ? Apr 7 '2018 at 16:2

Answer is : B.) 150

Explanation:

Let the length of the train is x meter and Speed of the train is y meter/second
Then \(\frac{x}{y} = 15\) [because \(\frac{distance}{speed} = time\)] 
\(\implies y = \frac{15}{x}\)

\(\begin{aligned} => \frac{x+100}{25} = \frac{x}{15} \\ x = 150 \text{ meters} \end{aligned}\)

So length of the train is 150 meters

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A man completes 30 km of a journey at the speed of 6 km/h and the remaining

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A man completes 30 km of a journey at the speed of 6 km/h and the remaining 40 km of the journey in 5 hr. His average speed for the whole journey is

A7 km/h

B$$6\frac{4}{11}km/h$$

C8 km/h

D7.5 km/h

Nithin K ? Apr 7 '2018 at 15:36

Answer is : 7 km/h

Explanation:

Total distance = 30 + 40 = 70 km 

 

Total time taken =  \(\Large \frac{30}{6}+5=\frac{60}{6} = 10 h\)

Therefore, Required average speed =\(\Large \frac{70}{10} = 7 km/h\)

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Two trains 105 m and 90 m long, run at the speeds of 45 km/h and 72 km/h respectively

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Two trains 105 m and 90 m long, run at the speeds of 45 km/h and 72 km/h respectively, in opposite directions on parallel tracks. The time which they take to cross each other is

A8 sec.

B6 sec.

C7 sec.

D5 sec.

Nithin K ? Apr 7 '2018 at 15:28

Answer is : B. 6 sec.

Explanation:

Length of the 1st train = 105 m;
Length of the 2nd train = 90 m.
Relative speed of the trains,
= 45+72 = 117kmph
\(= 117\times\frac{5}{18} = 32.5 m/sec\)
\(\text{Time taken to cross each other},\)
\(=\frac{\text{ (Length of 1st train + length of 2nd train)}}{\text{relative speed of the trains}}.\)
Time taken = \(\frac{195}{32.5}\) = 6 secs.

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A man sells his typewriter at 5% loss. If he sells it for Rs.80 more, he will gain 5%.

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A man sells his typewriter at 5% loss. If he sells it for Rs.80 more, he will gain 5%. The cost price of the type writer is

ARs.1600

BRs.1200

CRs.1000

DRs.800

Aryan bahuguna ? Jun 6 '2020 at 21:10

CP=X

LOSS =5%

GAIN=5%

SP=CP(100-LOSS)/100

95×/100=19×/20

SP=CP (100 +PROFIT)/100= X(100+5)/100= 105X/100 = 21X/100 

21X/20-19X/20=80

X=80×20=1600

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If a man were to sell his chair for Rs.720

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If a man were to sell his chair for Rs.720, he would loss 25%. To gain 25% he should sell it for

ARs.1200

BRs.1000

CRs.960

DRs.900

Nithin K ? Apr 7 '2018 at 10:17

Answer is : A. Rs.1200

Explanation:

If there is Loss then ,

\(\text{Cost Price} = \Big[ \frac{100}{(100 -Loss\%)}\Big] \times \text{Selling Price}\)

CP of chair = \(\frac{100}{(100-25)}\times720\)

= \(\frac{100}{75}\times 720\)\(\) 

= \(Rs. 960\)

To gain 25%,

\(\text{Selling Price} = \Big[\frac{(100 +Gain \%) }{ 100}\Big] \times \text{Cost Price}\)

\(\text{Selling Price} = \Big[\frac{(100 + 25)}{100}\Big]\times 960\)

\(=\frac{125}{100}\times 960\)

\(= Rs. 1200\)

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A book seller sells a book at a profit of 10%. If he had bought it at 4% less

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A book seller sells a book at a profit of 10%. If he had bought it at 4% less and sold it for Rs.6 more, he would have gained $$18\frac{3}{4}%$$. The cost price of the book is

ARs.130

BRs.140

CRs.150

DRs.160

pavan ? Apr 6 '2018 at 13:37

Correct Answer:

C) Rs.150

Explanation:

Let CP = x

The, SP =\(\Large\frac{110x}{100} = Rs.\frac{11x}{10}\)

Now, CP = 96% of x =\(\Large\frac{96x}{100} = Rs.\frac{24x}{25}\)

Now according to the question,

SP = Rs.\(\left( \frac{11x}{10} + 6 \right)\)

\(\left( \frac{11x}{10} + 6 \right) = \Large118\frac{3}{4}% of \frac{24x}{25}\)

\(\Large\frac{11x + 60}{10} = \frac{475}{400}\times\frac{24x}{25} = \frac{57x}{50}\)

\(\implies 550x + 3000 = 570x \)
\(20x = 3000\)
\(X = \Large\frac{3000}{20} = 150\)

CP = 150

 

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Krishnan bought a camera and paid 20% less than its original price.

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Krishnan bought a camera and paid 20% less than its original price. He sold it as 40% profit on the price he had paid. The percentage of profit earned by krishnan on the original price was

A22

B32

C12

D15

Nithin K ? Apr 3 '2018 at 20:1

Answer is : 12

Explanation :

Let the original price be x Rs.

\(=\frac{80}{100} \times x = \frac{4x}{5} Rs\)

\(SP = \frac{4x}{5} \times \frac{140}{100} = \frac{28x}{25} Rs\)

Gain on original price

\(\frac{28x}{25} - x =\frac{3x}{25}\)

\(\therefore \text{Gain % =}\frac{3x}{\frac{25}{x} \times 100} \text{= 12%}\)

=12%
 

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Due to 20% reduction in the price of sugar, a man is able to purchase 6 kg

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Due to 20% reduction in the price of sugar, a man is able to purchase 6 kg more sugar in Rs.120. The original rate of sugar was

ARs.5 per kg

BRs.6.15 per kg

CRs.5.65 per kg

DRs.6.20 per kg

Nithin K ? Apr 2 '2018 at 22:21

Answer is : Rs.5 per kg

Explanation:

20% of 120/- is 24 /- 

Therefore we get 6kg sugar by 24/- 
Then
24=6kg
? =1kg
Then 1kg price is 4/- ( reduction price )
Original price:
100% = ?
80% = 4/- ( because in question he gives 20% reduction)
There fore \(\frac{(100 \times 4)}{80}\) = 5/- 

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