Numerical Applications Test 17

Result

Numerical Applications Test 17

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A and B together can do a job in $$2$$ days; B and C can do it in four days; and A and C in $$2\dfrac{2}{5}$$ days. The number of days required for A to do the job alone is:

A
$$1$$

B
$$3$$

C
$$6$$

D
$$12$$

Solution

Let A, B, C can do a job alone in a, b, c days.
A and B together can do in 2 days.
$$\cfrac { 1 }{ a } +\cfrac { 1 }{ b } =\cfrac { 1 }{ 2 } $$ -------(1)
B and C together can do in 4 days.
$$\cfrac { 1 }{ b } +\cfrac { 1 }{ c } =\cfrac { 1 }{ 4 } $$ ------(2)
C and A together can do in $$\cfrac{12}{5}$$ days.
$$\cfrac { 1 }{ c } +\cfrac { 1 }{ a } =\cfrac { 5 }{ 12 } $$ -------(3)
(1) - (2), $$\Longrightarrow \cfrac { 1 }{ a } -\cfrac { 1 }{ c } =\cfrac { 1 }{ 4 } $$ -------(4)
(3) + (4), $$\Longrightarrow \cfrac { 2 }{ a } =\cfrac { 8 }{ 12 } $$
$$\therefore a=3$$
$$\therefore $$ A can do job in 3 days.
Question 2
1 / -0

$$A$$ and $$B$$ together can finish a work in $$30$$ days. They worked on it for $$20$$ days and then $$B$$ left the work. The remaining work was done by $$A$$ alone in $$20$$ days more. In how many days can $$A$$ alone finish the work?

A
$$48\ days$$

B
$$50\ days$$

C
$$54\ days$$

D
$$60\ days$$

Solution

Let the total work be $$60$$ units. [LCM of $$30$$ and $$20$$]
$$A$$ and $$B$$ do $$\dfrac {60}{30} = 2\ units/ day$$
As they worked for $$20$$ days, they did $$20\times 2 = 40\ units$$ of work.
Work left is $$60 - 40 = 20\ units$$
Thus, $$20$$ units is finished by $$A$$ in $$20\ days$$.
Hence, $$A$$ does $$\dfrac {20}{20} = 1\ unit/ day$$.
So, $$A$$ alone can finish the work in $$\dfrac {60}{1} = 60\ days$$.
Question 3
1 / -0

$$A$$ 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 $$2$$ more days, $$B$$ alone could do the work in

A
$$4$$ days

B
$$6$$ days

C
$$8$$ days

D
$$10$$ days

Solution

Let the total unit of work be $$6$$ units (LCM of $$2$$ and $$3$$)
$$(A + B)do = \dfrac {6}{3} = 2\ units/ day$$
$$(A + B)'s 2\ day$$ work $$= 2\times 2 = 4\ units$$
Work left $$= 6 - 4 = 2\ units$$
This remaining work is done by $$A$$ in $$2$$ days.
$$\therefore A's$$ per day work $$= \dfrac {2}{2} = 1\ unit/ day$$
$$\therefore B's$$ per unit work $$= (A + B)'s\ work - A's\ work$$
$$= 2 - 1 = 1\ unit/ day$$
$$B$$ alone can complete the work in $$= \dfrac {6}{1} = 6\ days$$.
Question 4
1 / -0

$$8$$ taps of the same size fill a tank in $$27$$ minutes. If two taps go out of order, how long would the remaining taps take to fill the tank?

A
$$86$$ minutes

B
$$56$$ minutes

C
$$66$$ minutes

D
$$36$$ minutes

Solution

Given that,
$$8$$ taps of the same size fill a tank in $$27$$ minutes.

So, $$1$$ tap can fill the tank in $$8 \times 27 = 216$$ minutes

If two taps go out of order, then the remaining taps $$= 8 - 2 = 6$$

Therefore, Time taken by $$6$$ taps to fill the tank $$= \dfrac{216}{6} = 36$$ minutes.
Question 5
1 / -0

$$12$$ workers can complete a piece of work in $$10$$ days. If the number of workers are reduced to $$\dfrac {1}{3}rd$$ of the original number, then how many more days would be required to complete the same work?

A
$$3$$

B
$$5$$

C
$$15$$

D
$$20$$

Solution

12 workers can complete a piece of work in 10 days
I/3 rd of original No's of workers = 12*1/3 = 4 workers
No of workers reduced then more no days are required
12*10÷ 4 = 30
Ans = 30 days
hence 20 days more
Question 6
1 / -0

If $$^{n + 1}{C_3} = 4\,{\,^n}{C_2}$$ then $$n=$$

A
$$12$$

B
$$10$$

C
$$16$$

D
$$11$$

Solution

Given that,$$^{n+1}{{C}_{3}}={{4.}^{n}}{{C}_{2}}$$

Then $$n=?$$

$$ ^{n+1}{{C}_{3}}={{4.}^{n}}{{C}_{2}} $$

$$ \Rightarrow \dfrac{\left( n+1 \right)!}{3!\left( n+1-3 \right)!}=4.\dfrac{n!}{2!\left( n-2 \right)!} $$

$$ \Rightarrow \dfrac{\left( n+1 \right)n!}{3\times 2!\left( n-2 \right)!}=4.\dfrac{n!}{2!\left( n-2 \right)!} $$

$$ \Rightarrow \dfrac{\left( n+1 \right)}{3}=4 $$

$$ \Rightarrow n+1=12 $$

$$ \Rightarrow n=12-1=11\, $$

Hence, this is the answer.
Option (D) is correct.
Question 7
1 / -0

A can do a piece of work in $$80 days $$.He work at it for $$10$$ days and then $$B$$ alone finishes the work in $$42$$ day.In how much time will A and B working together,finish the work?

A
$$30$$ days

B
$$25$$ days

C
$$20$$ days

D
$$35$$ days

Solution

A can complete a work in $$80$$ days
A can work in one day$$=\dfrac{1}{80}$$
then,
A leave the work after $$10$$ days
$$=\dfrac{1}{80}\times 10$$
$$=\dfrac{1}{8}$$
Remaining work$$=1-\dfrac{1}{8}$$
$$=\dfrac{7}{8}$$
B complete remaining work$$=42$$ days
$$\dfrac{7}{8}=42$$
B can complete whole work in
$$=\dfrac{42\times 3}{7}$$
$$=48$$
B can work in one day$$=\dfrac{1}{48}$$
A and B can complete the work together$$=\dfrac{1}{80}+\dfrac{1}{48}$$
$$=\dfrac{3+5}{240}$$
$$=\dfrac{8}{240}$$
$$=\dfrac{1}{30}$$
$$30$$ days.
Question 8
1 / -0

$$3$$ letters are posted in $$5$$ letters boxes. If all the letters are not posted in the same box, then number of ways of posting is

A
$$120$$

B
$$125$$

C
$$130$$

D
$$124$$

Solution

According to problem,
$$3$$ letters are posted $$5$$ boxes
implies that,
we have a letter and can be posted in any of the $$5$$ boxes.
Similarly next letter can be posted in 5 ways, and the all other follow same method
implies that
$${\left(5\right)}^3$$
$$ = 5 \times 5 \times 5$$
Hence, $$ 125 $$ possible ways for posting in $$5$$ boxes.
Question 9
1 / -0

A _____ is an arrangement of all or part of set of object in a definite order.

A
permutation

B
function

C
combination

D
factorial

Solution
Question 10
1 / -0

$$8$$ men can dig a pit in $$20$$ days. If a man works one and a half as much again as a boy, then $$4$$ men and $$8$$ boys can dig it in

A
$$10\ days$$

B
$$12\ days$$

C
$$15\ days$$

D
$$16\ days$$

Solution

$$\begin{array}{l} from\, the\, given\, data \\ work\, done\, by\, men=1+\frac { 1 }{ 2 } \\ work\, done\, by\, boy=1 \\ So,\, ratio\, of\, work\, done\, by\, man\, and\, boy=3:2 \\ Now, \\ Total\, work\, =8\times 3\times 20 \\ Now, \\ 4men+9Boy=4\times 3+9\times 2 \\ =12+18 \\ =30 \\ Hence,\, \\ No.\, of\, days=\dfrac { { 8\times 3\times 20 } }{ { 30 } } \\ =16days \\ Hence,\, option\, \, D\, \, is\, the\, correct\, answer. \end{array}$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Numerical Applications Test 17

Result

Numerical Applications Test 17

Score

-

Rank

-

SHARING IS CARING

Question 1

$$1$$

$$3$$

$$6$$

$$12$$

Solution

Question 2

$$48\ days$$

$$50\ days$$

$$54\ days$$

$$60\ days$$

Solution

Question 3

$$4$$ days

$$6$$ days

$$8$$ days

$$10$$ days

Solution

Question 4

$$86$$ minutes

$$56$$ minutes

$$66$$ minutes

$$36$$ minutes

Solution

Question 5

$$3$$

$$5$$

$$15$$

$$20$$

Solution

Question 6

$$12$$

$$10$$

$$16$$

$$11$$

Solution

Question 7

$$30$$ days

$$25$$ days

$$20$$ days

$$35$$ days

Solution

Question 8

$$120$$

$$125$$

$$130$$

$$124$$

Solution

Question 9

permutation

function

combination

factorial

Solution

Question 10

$$10\ days$$

$$12\ days$$

$$15\ days$$

$$16\ days$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.