1. NHAI employs 100 men to build a highway of 2km in 50 days working 8 hours a day. If in 25 days they completed 1/3 part of work .than how many more employees should NHAI hire to finish it in time working 10 hours a day?

Explanation:

Here 2km is immaterial. The given problem can be written in a tabular form like below.

2. A mixture of 125 gallons of wine and water contains 20% of water. How much water must be added to the mixture in order to increase the percentage of water to 25% of the new mixture?

a) 10 gallons

b) 8.5gallons

c) 8gallons

d) 8.33gallons

Explanation:

Initially water in the mixture = 20%(125) = 25

Let x gallons of water be added to change to water concentration to 25% or 1/4

$ \Rightarrow \dfrac{{25 + x}}{{125 + x}} = \dfrac{1}{4}$

$ \Rightarrow x = \dfrac{{25}}{3}$ = 8.33 gallons.

3. Side of a square is 10 cm. after joining the mid point of all sides makes a another inner square and this process goes to infinite.Find the sum of perimeter of all squares.

Explanation:

So if you observe carefully, the side of the small square is $\dfrac{1}{{\sqrt 2 }}$ part of the side of the bigger square.

So the side of the square inside the small square = $\dfrac{{10}}{{\sqrt 2 }} \times \dfrac{1}{{\sqrt 2 }} = \dfrac{{10}}{2} = 5$ and so on...

So areas of the perimeters = $4\left( {10 + \dfrac{{10}}{{\sqrt 2 }} + \dfrac{{10}}{2} + ...} \right)$

= $40\left( {1 + \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{2} + ...} \right)$

The terms in the bracket are in GP withe common ratio of ${\dfrac{1}{{\sqrt 2 }}}$

So Sum of the perimeters = $40\left( \vcenter{\dfrac{1}{{1 - \dfrac{1}{{\sqrt 2 }}}}} \right)$ = $40\left( {\dfrac{{\sqrt 2 }}{{\sqrt 2 - 1}}} \right)$

4. Data sufficiency question:

What will be the percentage profit of selling one liter milk.?

1) 16 liter of milk is sold at cost price after adding 20% water to it.

2) the cost price of one liter milk is Rs.16.

Explanation:

Let us assume one liter costs Rs.1. So C.P = Rs.16

When 20% water is added, then total volume = 20 liters. So SP = 20. Profit can be calculated.

Statement 1 is sufficient.

Statement 2 is not required.

5. y=x/(x-k),where k is a constant,and x is real number.show that.

1.y increase with increase in x.

2.y decreases first and then increase with the value of x.

3.y increase then decrease with the value of x.

4.it remains constant.

Explanation:

Typical question. Taking k =5 and we draw the graph,

6. What is the maximum value of v

a) 9

b) 12

c) 15

d) none of these

Explanation:

To maximize the value of v

${v^x} - yz = {( - 3)^2} - ( - 3 \times 2)$ = 15

7. Given a Number 123456, from this number put any three values in numerator and remaining three are in denominator. So how many values you can make from this number less than 1/5.

1. 2

2. 4

3. 8

4.27

Explanation:

If the given value is 120, then denominator should be slightly greater than 600. If for 130, it is 650. So if we take numerator as 132, then denominator should be greater than 660 which is not possible as we have only 5 and 4 available. So numerator is less than 130. The following numbers are possible.

123/654, 123/645, 124/635, 124/653, 125/634, 125/643.

8. A square was given. Inside the square there are white tiles and black tiles. Black tiles was among the diagonal of the square and dimensions of both white as well as black tiles is 1cm x 1cm. If there are 81 black tiles in the square. Then find the no of white tiles in it.

In a square, number of squares on the diagonal is equal to the tiles on a single row. If there are even number of square on a side, then total squares on the diagonal is 2n - 1, otherwise 2n. As the total tiles on the diagonal are given as 81, then number of tiles on a side = 2n - 1 = 81 so n = 41.

So number of white tiles = ${41^2} - 81 = 1681 - 81$ = 1600

Explanation:

Here 2km is immaterial. The given problem can be written in a tabular form like below.

We can apply chain rule now.

Total men required to complete the remaining work = $100 \times \dfrac{{25}}{{25}} \times \dfrac{8}{{10}} \times \dfrac{{{\textstyle{2 \over 3}}}}{{{\textstyle{1 \over 3}}}}$ = 160

So additional men required = 160 - 100 = 60

a) 10 gallons

b) 8.5gallons

c) 8gallons

d) 8.33gallons

Explanation:

Initially water in the mixture = 20%(125) = 25

Let x gallons of water be added to change to water concentration to 25% or 1/4

$ \Rightarrow \dfrac{{25 + x}}{{125 + x}} = \dfrac{1}{4}$

$ \Rightarrow x = \dfrac{{25}}{3}$ = 8.33 gallons.

3. Side of a square is 10 cm. after joining the mid point of all sides makes a another inner square and this process goes to infinite.Find the sum of perimeter of all squares.

Explanation:

You can calculate the side of the small square in two ways. Take half of the side which is equal to 5 cm. Using Pythagorean theorem, ${x^2} = {5^2} + {5^2} $ $ \Rightarrow {x^2} = 50$ $ \Rightarrow x = \sqrt {50} = 5\sqrt 2 = \dfrac{{10}}{{\sqrt 2 }}$

In another way, we can equate the squares of the sides of the small square to square of the side which is equal to 10. $ \Rightarrow {x^2} + {x^2} = {10^2}$

$ \Rightarrow 2{x^2} = 100$

$ \Rightarrow x = \sqrt {50} = 5\sqrt 2 = \dfrac{{10}}{{\sqrt 2 }}$So if you observe carefully, the side of the small square is $\dfrac{1}{{\sqrt 2 }}$ part of the side of the bigger square.

So the side of the square inside the small square = $\dfrac{{10}}{{\sqrt 2 }} \times \dfrac{1}{{\sqrt 2 }} = \dfrac{{10}}{2} = 5$ and so on...

So areas of the perimeters = $4\left( {10 + \dfrac{{10}}{{\sqrt 2 }} + \dfrac{{10}}{2} + ...} \right)$

= $40\left( {1 + \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{2} + ...} \right)$

The terms in the bracket are in GP withe common ratio of ${\dfrac{1}{{\sqrt 2 }}}$

So Sum of the perimeters = $40\left( \vcenter{\dfrac{1}{{1 - \dfrac{1}{{\sqrt 2 }}}}} \right)$ = $40\left( {\dfrac{{\sqrt 2 }}{{\sqrt 2 - 1}}} \right)$

4. Data sufficiency question:

What will be the percentage profit of selling one liter milk.?

1) 16 liter of milk is sold at cost price after adding 20% water to it.

2) the cost price of one liter milk is Rs.16.

Explanation:

Let us assume one liter costs Rs.1. So C.P = Rs.16

When 20% water is added, then total volume = 20 liters. So SP = 20. Profit can be calculated.

Statement 1 is sufficient.

Statement 2 is not required.

5. y=x/(x-k),where k is a constant,and x is real number.show that.

1.y increase with increase in x.

2.y decreases first and then increase with the value of x.

3.y increase then decrease with the value of x.

4.it remains constant.

Explanation:

Typical question. Taking k =5 and we draw the graph,

If x increases y decreases but when x equal to k, y value becomes infinite. But when x is greater than k, y value slowly reaches to 1. So it decrease from infinite to 1.

^{x }- yz. If the value of v,x,y,z have to be chosen from the set A where A(-3,-2,-1,0,1,2,3)a) 9

b) 12

c) 15

d) none of these

Explanation:

To maximize the value of v

^{x }- yz, we make yz negative and v^{x }as maximum as possible using given value.${v^x} - yz = {( - 3)^2} - ( - 3 \times 2)$ = 15

7. Given a Number 123456, from this number put any three values in numerator and remaining three are in denominator. So how many values you can make from this number less than 1/5.

1. 2

2. 4

3. 8

4.27

Explanation:

If the given value is 120, then denominator should be slightly greater than 600. If for 130, it is 650. So if we take numerator as 132, then denominator should be greater than 660 which is not possible as we have only 5 and 4 available. So numerator is less than 130. The following numbers are possible.

123/654, 123/645, 124/635, 124/653, 125/634, 125/643.

8. A square was given. Inside the square there are white tiles and black tiles. Black tiles was among the diagonal of the square and dimensions of both white as well as black tiles is 1cm x 1cm. If there are 81 black tiles in the square. Then find the no of white tiles in it.

In a square, number of squares on the diagonal is equal to the tiles on a single row. If there are even number of square on a side, then total squares on the diagonal is 2n - 1, otherwise 2n. As the total tiles on the diagonal are given as 81, then number of tiles on a side = 2n - 1 = 81 so n = 41.

So number of white tiles = ${41^2} - 81 = 1681 - 81$ = 1600

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