Numerical Applications Test 44

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Numerical Applications Test 44

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Question 1
1 / -0

A train of 320 m cross a platform in 24 seconds at the speed of 120 km/h. while a man cross same platform in 4 minute.What is the speed of man in m/s?.

A
$$2.4$$

B
$$1.5$$

C
$$1.6$$

D
$$2.0$$

Solution

Speed of train $$=120\ kmph$$
$$=\dfrac{120\times5}{18}$$
$$=\dfrac{100}{3}\ m/sec$$
Let 'x' be the length of platform
Now,
$$\dfrac{Length\ of\ train\ and\ platform}{Time\ taken\ in\ crossing}=\dfrac{100}{3}$$
$$\Rightarrow\dfrac{320+x}{24}=\dfrac{100}{3}$$
$$\Rightarrow 320+x=\dfrac{24\times100}{3}$$
$$\Rightarrow 320+x=800$$
$$\Rightarrow x=480\ m$$
$$\therefore$$ Man's speed $$=\dfrac{480}{4\times60}=2\ m/sec$$
Question 2
1 / -0

A carpenter was hired to build 192 window frames. The first day he made five frames and each day, thereafter he made two more frames than he made the day before.How many days did it take him to finish the job?.

A
$$11$$

B
$$10$$

C
$$12$$

D
$$14$$

Solution

$$\textbf{Step 1 : Find the no. of days}$$
$$\text{Here, a=5 and d=2}$$
$$\text{Let the carpenter finish the job in n days}$$
$$\text{Then,}$$

$$S_n=192$$

$$\Rightarrow 192=\dfrac{n}{2}[2a+(n-1)d]$$

$$\Rightarrow 192=\dfrac{n}{2}[2\times5+(n-1)2]$$

$$\Rightarrow 192=n[5+n-1]$$

$$\Rightarrow n^2+4n-192=0$$

$$\Rightarrow n^2-12n+16n-192=0$$

$$\Rightarrow n(n-12)+16(n-12)=0$$

$$\Rightarrow (n-12)(n+16)=0$$

$$\therefore n=12$$ $$\quad \quad \textbf{[Rejecting n = -16 as number of days cannot be negative]}$$

$$\textbf{Hence, he will finish job in 12 days}$$
Question 3
1 / -0

A, B, and C together can finish a place of work in 12 days. A and C together work twice as much as B, A and B together work thrice as much as C. In what time ( day ) could each do it separately

A
144/5, 36, 48

B
36, 48, 144/5

C
48, 144/5, 36

D
none of these

Solution

Let A's 1 day work = a
B's 1 day work = b
C's 1 day work = c
$$ \therefore A+B+C = \frac{1}{12}$$ ...(1)
$$ A+C = 2B$$ ...(2)
$$ A+B = 3C$$ ...(3)
From(2)
$$A = 2B -C$$ ...(4)
from (3)
$$ A = 3C -B $$ ...(5)
comparing (4) and (5)
$$ 2B-C-3C-B$$
$$ 3B = 4C$$
$$ B = \frac{4}{3}C$$ ...(6)
putting in (5)
$$ A = BC -\frac{4C}{3}$$
$$ A = \frac{5C}{3}$$ ...(7)
putting (6) and (7) in (1)
$$ \frac{5C}{3}+\frac{4C}{3}+C = \frac{1}{12}$$
$$ \frac{12C}{3} = \frac{1}{12}$$
$$ C = \frac{1}{48}$$
$$ \therefore A = \frac{5}{3}\times \frac{1}{48} = \frac{5}{144}$$
$$ B = \frac{4}{3} \times \frac{1}{48} = \frac{1}{36}$$
$$ \therefore $$ A finishes in $$ \frac{144}{5}$$ days
B finishes in 36 days
C finishes in 48 days
Question 4
1 / -0

Pipe A can fill a tank in $$10h$$ and pipe $$B$$ can fill the same tank in $$12h$$. Both the pipes are opened to fill the tank and after $$3h$$ pipe $$A$$ is closed. Pipe $$B$$ will fill the remaining part of the tank in

A
$$5h$$

B
$$4h$$

C
$$5h$$ $$24min$$

D
$$3h$$

Solution

Let capacity of tank = V liters
Speed of A = V/10 lit/hour
B = V/12 lit/hour
In 3hrs both fill $$ = 3(\frac{V}{10})+3(\frac{V}{12}) = 3V(\frac{11}{60}) $$
$$ = 33V/60 $$ Remaining $$ 27V/60 $$
B fills remaining $$ = \frac{\frac{27V}{60}}{\frac{V}{12}} = \frac{27}{5} = 5hr.24\, min $$
$$ \Rightarrow (c) $$
Question 5
1 / -0

$$A,B$$ and $$c$$ can do a piece of work in $$24,30$$ and $$40$$ days respectively. The began the work together but $$C$$ left the work four days before the completion of the work. The work was completed in

A
$$9\ days$$

B
$$10\ days$$

C
$$11\ days$$

D
$$14\ days$$

Solution
Question 6
1 / -0

$$150$$ workers were engaged to finish piece of work in a certain number of day. Four workers dropped the second day, four more workers dropped the third day and so on, it take $$8$$ more days to finish the work now. Then the number of days in which the work was completed is

A
$$29$$ days

B
$$24$$ day

C
$$25$$ days

D
None of these

Solution
Question 7
1 / -0

Find the average of the following set of scores $$253,124,255,534,836,375,101,443,760$$

A
$$427$$

B
$$413$$

C
$$141$$

D
$$409$$

Solution
Question 8
1 / -0

The number of permutations of letters of the word "PARALLAL" atken four at a time must be,

A
$$216$$

B
$$244$$

C
$$286$$

D
$$1680$$

Solution

Permutation of word PARALLAL taken four at a time
Four letter word might be $$LLL$$_
So total of $$16$$ words [As $$4$$ways, $$4$$spots]
For $$AALL$$, $$6$$ different arrangements
$$AA$$_ _, can be $${ 4 }_{ { C }_{ 2 } }$$ ways $$=6$$ ways
Arranged in $$12$$ ways is total of $$72$$ words
And extra words can be formed in $$120$$ ways
$$16+6+144+120=286$$ ways
Question 9
1 / -0

$$A$$ can do a piece of work in $$15$$ days. $$B$$ is $$50\%$$ more efficient than $$A$$. $$B$$ can finish it in

A
$$10$$ days

B
$$7 \dfrac { 1 } { 2 }$$ days

C
$$12$$ days

D
$$12 \frac { 1 } { 2 }$$ days

Solution

$$A\rightarrow 15$$ days

capacity of $$A$$ is $$\dfrac{W}{15}$$ where $$W$$ is worke

$$B$$ is $$50\%$$ more efficient than $$A$$

$$\dfrac{W}{15}+\dfrac{50}{100}(\dfrac{W}{15})=\dfrac{150}{100}(\dfrac{W}{15})$$

$$=\dfrac{3}{2}(\dfrac{W}{15}$$)

Let $$B$$ take $$t$$ days to complete $$W$$ work

$$\dfrac{3}{2}(\dfrac{W}{15})\times t=W$$

$$t=10\,days$$
Question 10
1 / -0

Six people are going to sit in a row on a bench. $$A$$ and $$B$$ are adjacent, $$C$$ does not want to sit adjacent to $$D.E$$ and $$F$$ can sit anywhere. Number of ways in which these six people can be seated is

A
$$200$$

B
$$144$$

C
$$120$$

D
$$56$$

Solution

A, B, C, D, E, F
Consider AB as group so we have AB, C, D, E, F.
We have totally $$5$$
No. of ways$$(w_1)=5!\times 2$$
$$=240$$
Let CD are adjacent now AB, CD, E, F
No. of ways$$(w_2)=4!2!2!$$
$$=96$$
Total no. of ways
$$W=w_1-w_2$$
$$=240-96$$
$$=144$$.