Numerical Applications Test 2

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Numerical Applications Test 2

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Weekly Quiz Competition

Question 1
1 / -0

Find the mean of the data $$10, 15, 17, 19, 20$$ and $$21$$.

A
$$11$$

B
$$16$$

C
$$17$$

D
$$21$$

Solution

Data observations are: $$10, 15, 17, 19, 20\ and\ 21$$

mean = $$\dfrac {sum}{Number}$$

mean = $$\dfrac {10 + 15 + 17 + 19+ 20+ 21}{6}$$

mean = $$\dfrac {102}{6}$$ = $$17$$
Question 2
1 / -0

Factorial of negative numbers is always greater than 1.

A
True

B
False

C
Either

D
Neither

Solution

Factorial : product of an integer with integer less than it.
Factorial can be interpolated using gamma function and gamma function and
gamma function is not defined for negative integer.
Factorial is not defined for negative integer
Question 3
1 / -0

Choose the correct option for the following.
$$n!=n(n-1)(n-2).....3.2.1$$

A
True

B
False

C
Ambiguous

D
Data insufficient

Solution

Factorial : the product of an integer and all integer less than that
$$\therefore n!=n\times (n-1)\times (n-2)............3\times 2\times 1$$

$$\therefore $$ The given statement
$$n!=n\times (n-1)\times (n-2)............3\times 2\times 1$$ is True
Question 4
1 / -0

A contractor estimates that $$3$$ persons could rewire Jasminders house in $$4$$ days. If he uses $$4$$ persons instead of three, how long should they take to complete the job?

A
$$8$$ days

B
$$3$$ days

C
$$4$$ days

D
$$9$$ days

Solution

The given example is of inverse proportionality

$$I$$ $$II$$

No. of persons $$3$$ $$4$$

No. of days $$4$$ $$x$$

By inverse proportionality,

$$3\times 4 = 4\times x$$

$$x = 3$$ days

Answer is option B
Question 5
1 / -0

A school has $$8$$ periods a day each of $$45$$ minutes duration. How long would each period be, if the school has $$9$$ periods a day, assuming the number of school hours to be the same?

A
$$32$$ minutes

B
$$40$$ minutes

C
$$45$$ minutes

D
$$20$$ minutes

Solution

Let the duration of period be : $$t$$ (minutes) if the school has $$9$$ periods
No. of period : $$9$$ ...... (given)
To find $$t$$ ?
Since no. of school hours are same for both $$8$$ periods and $$9$$ periods,
So, we have $$8 \times 45 = 9\times t$$
$$\implies 360 = 9 t$$
$$\implies t = 360/9$$
$$\implies t= 40$$ Minutes
Hence, each period will be of $$40$$ minutes
Question 6
1 / -0

$$A$$ can do a piece of work in $$24$$ days. If $$B$$ is $$60\%$$ more efficient than $$A$$, then the number of days required by $$B$$ to do the twice as large as the $$A's$$ work is -

A
$$24$$

B
$$36$$

C
$$15$$

D
$$30$$

Solution

A can do a work say $$w$$ in $$24$$ days
$$A\rightarrow \dfrac { w }{ 24 } $$ (work done by A in one day)
$$B$$ is $$60$$% more efficient
$$B\rightarrow 1.6\times \dfrac { w }{ 24 } $$ (work done by B in one day)
Now B has to do work $$2w$$
For one day B can do $$B\rightarrow 1.6\times \dfrac { w }{ 24 } $$
Say B takes $$x$$ days to complete $$2w$$ amount of work.
$$1.6\times \dfrac { w }{ 24 } \times x=2w$$
$$x=\dfrac { 48 }{ 1.6 } =30$$
Question 7
1 / -0

The mean of $$8, 7, 9, 10, 12, x$$ and $$14$$ is $$12$$, then find the value of $$x$$.

A
$$21$$

B
$$22$$

C
$$23$$

D
$$24$$

Solution

Given mean is $$12$$.
Therefore, mean $$=\dfrac {8+7+9+10+12+x+14}{7}$$
$$\Rightarrow 12=\dfrac {60+x}{7}$$
$$\Rightarrow 84=60+x$$
$$\Rightarrow x=24$$
Question 8
1 / -0

Find the arithmetic mean of integers from $$-5$$ to $$5$$

A
$$-1$$

B
$$0$$

C
$$1$$

D
None of these

Solution

Integers from $$-5$$ to $$+5$$ are $$= -5,-4,-3,-2,-1,0,1,2,3,4,5$$
sum of all these integers $$=0$$
so mean $$= 0$$
Question 9
1 / -0

If 09/12/2001 happens to be Sunday,then 09/12/1971 would have been at

A
Wednesday

B
Tuesday

C
Saturdays

D
Thursday

Solution

09/12/2001 -----Sunday
No. of days between 9/12/71 & 9/12/2001
we know every year has 1 odd days
we know leap year has 2 odd days
Here, No. of normal years= 22
And no. of leap years =8
So odd days =22+16 =38 i.e 3 odd days
(remainder when 38 is divided by 7,i.e. 3)
Hence it was a Thursday.
Question 10
1 / -0

Find the mean of integers from $$-4$$ to $$5$$

A
$$-1$$

B
$$0$$

C
$$0.5$$

D
$$1$$

Solution

integers from $$-4$$ to $$+5$$ are $$= -4,-3,-2,-1,0,1,2,3,4,5$$
sum of all these integers $$=5$$
total elements $$= 10$$
so mean $$= \dfrac{5}{10} = 0.5$$

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Numerical Applications Test 2

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Numerical Applications Test 2

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SHARING IS CARING

Question 1

$$11$$

$$16$$

$$17$$

$$21$$

Solution

Question 2

True

False

Either

Neither

Solution

Question 3

True

False

Ambiguous

Data insufficient

Solution

Question 4

$$8$$ days

$$3$$ days

$$4$$ days

$$9$$ days

Solution

Question 5

$$32$$ minutes

$$40$$ minutes

$$45$$ minutes

$$20$$ minutes

Solution

Question 6

$$24$$

$$36$$

$$15$$

$$30$$

Solution

Question 7

$$21$$

$$22$$

$$23$$

$$24$$

Solution

Question 8

$$-1$$

$$0$$

$$1$$

None of these

Solution

Question 9

Wednesday

Tuesday

Saturdays

Thursday

Solution

Question 10

$$-1$$

$$0$$

$$0.5$$

$$1$$

Solution

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