Numerical Applications Test 43

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Numerical Applications Test 43

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Question 1
1 / -0

Number of five-digit numbers divisible by 5 that can be formed from the digits $$0,1, 2, 3, 4, 5$$ without repetition of digits are

A
$$240$$

B
$$360$$

C
$$148$$

D
$$216$$

Solution
Question 2
1 / -0

$$S, T$$ and $$U$$ can complete a work in $$40, 48$$ and $$60$$ days respectively. They received$$Rs. 10800$$ to complete the work. They begin the work together but $$T$$ left $$2$$ days before the completion of the work and $$U$$ left $$5$$ days before the completion of work. $$S$$ has completed the remaining work alone. What is the share of $$S$$ (in Rs) from total money.

A
$$4000$$

B
$$4320$$

C
$$4500$$

D
$$4860$$

Solution

$$S=40\times 6=240$$
$$T=48\times 5=240$$
$$U=60\times 4=240$$
Let time taken by them to complete the work $$=x$$ days
$$6x+5(x-2)+4(x-5)=240$$
$$\Rightarrow 15x=240+30=270$$
$$x=18$$
Workdone by $$S=6\times 18=108$$
Share of $$S=\dfrac{103}{240}\times 10800=4860\ Rs$$
Question 3
1 / -0

Let $$5 < n_1 < n_2 < n_3 < n_4$$ be integers such that $$n_1+n_2+n_3+n_4=35$$. The number of such distinct arrangements $$(n_1, n_2, n_3, n_4)$$.

A
$$^{38}C_3$$

B
$$^8C_3$$

C
$$5$$

D
$$6$$

Solution

$${ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }+.....+{ n }_{ k }=R$$
for this arrangement,
No. of arrangement or distinct arrangements are $${ R+K-1 }_{ { C }_{ K-1 } }$$
So, for $${ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }+{ n }_{ 4 }=35$$
No. of arrangements $$={ 35+4-1 }_{ { C }_{ 4-1 } }={ 38 }_{ { C }_{ 3 } }$$
Question 4
1 / -0

Let $$p=\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+\dfrac{5}{1\times6}+.......+\dfrac{1}{2013\times2014}$$ and $$Q=\dfrac{1}{1008\times2014}+\dfrac{1}{1009\times2013}+.........+\dfrac{1}{2014\times1008}$$
then $$\dfrac{P}{Q}=$$

A
$$2013$$

B
$$2014$$

C
$$1511$$

D
$$2$$
Question 5
1 / -0

Let the eleven letters, $$A, B, ....K$$ denote an artbitrary permutation of the integers $$(1,2,....11)$$, then $$(A-1)(B-2)(C-3)...(K-11)$$ is

A
Necessarily zero

B
Always odd

C
Always evem

D
None of these

Solution

Given set of numbers is {1,2,3.....11} in which 5 are even and six are odd, which demands that in the given product it is not possible to arrange to subtract only even number from odd numbers. There must be at least one factor involving subtraction of an odd number from another odd number. So at least one of the factors is even. Hence product is always even.
Question 6
1 / -0

Ten persons, amongst whom are $$A$$,$$B$$ and $$C$$ to speak at a function. The number of ways in which it can be done if $$A$$ wants to speak before $$B$$ AND $$B$$ wants to speak before $$C$$ is

A
$$\dfrac{{10!}}{6}$$

B
$$3!$$ $$7!$$

C
$$^{10}{P_3}.7!$$

D
none of these

Solution
Question 7
1 / -0

$$24$$ men working at $$8$$ hours per day can do a piece of work in $$15$$ days. In how many days can $$20$$ men working at $$9$$ hours per day do the same work?

A
$$14$$ days

B
$$16$$ days

C
$$13$$ days

D
$$17$$ days

Solution

$${\textbf{Step -1: Finding number of work units for the first case}}{\text{.}}$$

$${\text{Total 24 men work for 8 hours a day and takes a total of 15 days to complete it}}{\text{.}}$$

$${\text{So total work units}} = 24 \times 8 \times 15 = 2880$$

$${\text{Thus, for the first case, it's a total of 2880 work units}}{\text{.}}$$

$${\textbf{Step -2: Finding number of days for the second case}}{\text{.}}$$

$${\text{Let it takes }}x{\text{ days for the worker to complete the task, so we can write}}$$

$${\text{Number of work units}} = 20 \times 9 \times x = 180x$$

$${\text{Also, the work is same so the work units must be same}}$$

$${\text{So, equating the work units for the two cases, we get}}$$

$$\Rightarrow 2880 = 180x$$

$$\Rightarrow x = \dfrac{{2880}}{{180}} = 16$$

$${\textbf{Thus, it will take 16 days for the 20 workers to complete the work}}{\textbf{. Thus, option B }}$$$$\textbf{is correct.}$$
Question 8
1 / -0

$$10$$ Men begin to work together on a job, but after some days, $$4$$ if them left the job. As a result the job which could have been completed in $$40$$ days is completed in $$50$$ days. How many days after the commencement of the work did the $$4$$ men level?

A
$$25$$

B
$$30$$

C
$$10$$

D
$$15$$

Solution

$$10$$ Men begin work
It could been done in $$40$$ days.
Let $$4$$ men left after x days.
Left days are $$40-x$$ if they do not left and $$50-x$$ if they left.
$$(40-x) 10= (50-x)6$$
$$400-10 x=300-6x$$
$$1000=4x$$
$$x=25$$ days.
Question 9
1 / -0

Consider all permutations of the letters of the word MORADABAD.
The number of permutations which contain the word BAD is:

A
$$21 \times 5!$$

B
$$7 \times 5!$$

C
$$6 \times 5!$$

D
$$2 \times 5!$$

Solution

Total number of letters = $$9$$

Consider BAD as a group, then the number of total letters = $$6+1=7$$

$$\therefore$$ Permutation of 7 things = $$7!$$

but letters remaining after removing BAD will have two A's

$$\therefore$$ Total permutation = $$\dfrac{7!}{2}$$ = $$21\cdot 5!$$
Question 10
1 / -0

The number of permutation of the letters of the word $$HINDUSTAN$$ such that neither the pattern $$'HIN'$$ nor $$'DUS'$$ nor $$'TAN'$$ appears, are :

A
$$166674$$

B
$$169194$$

C
$$166680$$

D
$$181434$$

Solution

Total number of letters $$=9$$ in which N is repeated twice

Therefore total number of permutation $$=\dfrac{9!}{2}$$

The number of permutation in which HIN comes as block $$=9-3+1=7!$$

The number of permutation in which DUS comes as block $$=7!$$

The number of permutation in which TAN comes as block $$=\dfrac{7!}{2}$$

The number of permutation in which HIN and DUS comes as block $$=5!$$

The number of permutation in which DUS and TAN comes as block $$=5!$$

The number of permutation in which HIN and TAN comes as block $$=5!$$

Therefore the required number of permutations $$=\dfrac{9!}{2}-\left(7!+7!+\dfrac{7!}{2}-3\times5!+3!\right)$$

$$=\dfrac{362880}{2}-\left(5040+5040+\dfrac{5040}{2}-360+6\right)$$

$$=181440-\left(10080+2520-360+6\right)$$

$$=181440-12246$$

$$=169194$$

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Re-Attempt Weekly Quiz Competition

Numerical Applications Test 43

Result

Numerical Applications Test 43

Score

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Rank

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SHARING IS CARING

Question 1

$$240$$

$$360$$

$$148$$

$$216$$

Solution

Question 2

$$4000$$

$$4320$$

$$4500$$

$$4860$$

Solution

Question 3

$$^{38}C_3$$

$$^8C_3$$

$$5$$

$$6$$

Solution

Question 4

$$2013$$

$$2014$$

$$1511$$

$$2$$

Question 5

Necessarily zero

Always odd

Always evem

None of these

Solution

Question 6

$$\dfrac{{10!}}{6}$$

$$3!$$ $$7!$$

$$^{10}{P_3}.7!$$

none of these

Solution

Question 7

$$14$$ days

$$16$$ days

$$13$$ days

$$17$$ days

Solution

Question 8

$$25$$

$$30$$

$$10$$

$$15$$

Solution

Question 9

$$21 \times 5!$$

$$7 \times 5!$$

$$6 \times 5!$$

$$2 \times 5!$$

Solution

Question 10

$$166674$$

$$169194$$

$$166680$$

$$181434$$

Solution

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