Numerical Applications Test 33

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Numerical Applications Test 33

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Weekly Quiz Competition

Question 1
1 / -0

Ranjana correctly rememers that Kiran's birthday was after Tuesday but before Friday Rajan correctly remembers but before Sunday On which day of the week does kiran's birthday definitely fad?

A
Monday

B
Thursday

C
Saturday

D
Cannot be determined

Solution

According to Birth Day
Ranjana Wednesday, Thursday
Rajan Thursday, Friday, Saturday
Clearly Thursday is common to both Hence confirm day is the Thursday
Question 2
1 / -0

If the first day of the year 1990 was a Monday what day of the week was 1st January 1998?

A
Thursday

B
Tuesday

C
Wednesday

D
Friday

Solution

IN the leap year odd no. of day in the year $$=2$$

Leap year has $$29$$ day in feb.
The Leap in that divided by $$4$$
So,
$$1990=1$$
$$1991=1$$
$$1992=2$$
$$1993=1$$
$$1994=1$$
$$1995=1$$
$$1996=2$$
$$1997=1$$
$$1998=1$$
$$(1+1+2+1+1+1+2+1+1)=11$$

The odd no. of the year $$=$$ No. of day in year $$7$$
$$11$$/$$7$$=$$4$$
Remainder=$$4$$
So, $$4$$=TUESDAY

Hence, this is the answer.
Question 3
1 / -0

If $$20$$ people can build a wall in $$7$$ days, then the number of person required to complete the work in $$35$$ days is:

A
$$4$$

B
$$8$$

C
$$12$$

D
$$20$$

Solution

We first find out the proportionality The more
the number of people to build the wall the less is the time taken and the less the number of people to build the wall the more is the time taken
Hence the relationship is inverse
$$\displaystyle \therefore 20\times 7=35x$$
where x is the number of people required
$$\displaystyle \therefore x=\frac{20\times 7}{35}=\frac{20}{5}=4$$ people
Question 4
1 / -0

If $$34$$ men completed $$\dfrac{2}{5}$$th of a work in $$8$$ days working $$9$$ hours a day. How many more man should be engaged to finish the rest of the work in $$6$$ days working $$9$$ hours a day?

A
$$68$$ men

B
$$45$$ men

C
$$34$$ men

D
$$28$$ men

E
None of these

Solution

Work efficiency of $$34$$ men in $$8$$ days working $$9$$ hours a day is $$\dfrac{2}{5}$$th
Hence, work efficiency of $$x$$ men to finish rest of the work in $$6$$ days working $$9$$ hours a day = $$1 - \dfrac{2}{5} = \dfrac{3}{5}$$
$$\therefore$$ By direct proportionality,
$$\dfrac{34\times8\times9}{\dfrac{2}{5}} =\dfrac{x\times6\times9}{ \dfrac{3}{5}}$$
$$\therefore 34\times72\times\dfrac{3}{5} = x\times54\times\dfrac{2}{5}$$
$$\therefore x = \dfrac{34\times72\times3\times5}{54\times2\times5}$$
$$\therefore x = 68$$
Hence extra men required = $$68 - 34 = 34$$men
So, the answer is option C
Question 5
1 / -0

The arithinetic mean of $$5, 6, 8, 9, 12, 13, 17$$ is

A
$$20$$

B
$$15$$

C
$$10$$

D
$$25$$

Solution

Mean $$ = \dfrac { \text{Sum of observations}}{\text{Total number of observations}} = \dfrac {5 + 6 + 8 + 9 + 12 + 13 + 17}{7} = \dfrac {70}{7} = 10 $$
Question 6
1 / -0

If p persons working p hours a day for each of p days produce p units of work, then the unit of work produced by q persons working q hours a day for each of q days is

A
$$\displaystyle\dfrac{q^3}{p^2}$$

B
$$\displaystyle\dfrac{q^2}{p^3}$$

C
$$\displaystyle\dfrac{p^2}{q^3}$$

D
$$\displaystyle\dfrac{p^3}{q^2}$$

Solution

We know that
$$\dfrac{M_1\times T_1\times D_1}{W_1}=\dfrac{M_2\times T_2\times D_2}{W_2}$$

Therefore,
$$\dfrac{p\times p\times p}{p}=\dfrac{q\times q\times q}{W_2}$$
$$W_2=\dfrac{q^3}{p^2}$$

Hence, this is the answer.
Question 7
1 / -0

Sujal covers a distance in $$40$$ min, if she drives at a speed of $$60$$ km/h on an average . Find the speed at which she must drive at to reduce the time of the journey by $$25\%$$.

A
$$60$$ km/h

B
$$7$$0 km/h

C
$$75$$ km/h

D
$$80$$ km/h

Solution

Given speed $$= 60$$ kmph
It means Sujal covered $$60$$ km in $$60$$ min ($$1$$ hour). So, in $$40$$ min she will cover $$40$$ km.
$$25\%$$ time reduced $$= 40 - 25\%$$ of $$40 = 30$$ min
Thus, Sujal need to cover $$40$$ km in $$30$$ min ($$0.5$$ hour)
Speed $$\times$$ Time $$ =$$ Distance
Speed $$\times 0.5 = 40$$
Speed $$= 80$$ kmph.
Question 8
1 / -0

The arithmetic mean of $$2+\sqrt {(2)}$$ and $$2-\sqrt {(2)}$$ is

A
$$2$$

B
$$\sqrt {(2)}$$

C
$$0$$

D
$$4$$

Solution

The arithmetic mean of $$2+\sqrt 2$$ and $$2-\sqrt 2$$ $$=\displaystyle \frac {2+\sqrt 2+2-\sqrt 2}{2}$$
$$=\dfrac {4}{2}$$

$$=2$$
Option A is correct.
Question 9
1 / -0

A man can row $$\displaystyle 9\frac{1}{3}$$ kmph in still water and finds that it takes him thrice as much time to row up than as to row down the same distance in the river. Find the speed of the current in kmph.

A
$$2\dfrac{12}{7}$$

B
$$4\dfrac{2}{3}$$

C
$$2\dfrac{6}{5}$$

D
$$4\dfrac{4}{9}$$

Solution

Let the speed upstream be x kmph Then speed downstream = 3x kmph
Speed in still water = $$\displaystyle \frac{1}{2}$$ (3x + x) kmph = 2x kmph
$$\displaystyle \therefore $$ 2x = $$\displaystyle \frac{28}{3}$$ $$\displaystyle \Rightarrow x=\frac{14}{3}$$
So Speed upstream = $$\displaystyle \frac{14}{3}$$ km/hr;
Speed downstream = 14 km/hr
Hence speed of the current = $$\displaystyle \frac{1}{2}\left ( 14-\frac{14}{3} \right )$$ km/hr
=$$\displaystyle \frac{14}{3}$$ km/hr = $$\displaystyle 4\frac{2}{3}$$ km/hr
Question 10
1 / -0

An empty tank is connected with pipes A, B and C. A and B inlet pipes and they fill the tank in $$6$$ hours and $$8$$ hours respectively. While C is an outlet pipe and it empties the completely filled tank in $$5$$ hours. Find the time in which the tank will be completely filled if all the pipes are opened together.

A
$$\dfrac{119}{11}$$ hours

B
$$\dfrac{108}{11}$$ hours

C
$$\dfrac{120}{11}$$ hours

D
$$\dfrac{109}{11}$$ hours

Solution

Work rate of Pipe A = 100/6 = 16.66 % per hour
Work rate of Pipe B = 100/8 = 12.5 % per hour
Work rate (Negative) of Pipe C = 100/5 = 20% per hour
So, Combined work rate of A, B and C
= 16.66 + 12.50 - 20 = 9.16 % per hour
Time taken to fill the tank = 100/9.16 = 10.9 hours = 120/11 hours.
Alternatively you can solve it through fractional method as well,
A fills = 1/6 part per hour
B fills = 1/8 part per hour
C empties = 1/5 part per hour
Combined work
= 1/6 + 1/8 - 1/5
=(20 + 15 - 24) / 120 = 11/120 part per hour
Time taken to completed the work = 120 / 11 hours.

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Re-Attempt Weekly Quiz Competition

Numerical Applications Test 33

Result

Numerical Applications Test 33

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SHARING IS CARING

Question 1

Monday

Thursday

Saturday

Cannot be determined

Solution

Question 2

Thursday

Tuesday

Wednesday

Friday

Solution

Question 3

$$4$$

$$8$$

$$12$$

$$20$$

Solution

Question 4

$$68$$ men

$$45$$ men

$$34$$ men

$$28$$ men

None of these

Solution

Question 5

$$20$$

$$15$$

$$10$$

$$25$$

Solution

Question 6

$$\displaystyle\dfrac{q^3}{p^2}$$

$$\displaystyle\dfrac{q^2}{p^3}$$

$$\displaystyle\dfrac{p^2}{q^3}$$

$$\displaystyle\dfrac{p^3}{q^2}$$

Solution

Question 7

$$60$$ km/h

$$7$$0 km/h

$$75$$ km/h

$$80$$ km/h

Solution

Question 8

$$2$$

$$\sqrt {(2)}$$

$$0$$

$$4$$

Solution

Question 9

$$2\dfrac{12}{7}$$

$$4\dfrac{2}{3}$$

$$2\dfrac{6}{5}$$

$$4\dfrac{4}{9}$$

Solution

Question 10

$$\dfrac{119}{11}$$ hours

$$\dfrac{108}{11}$$ hours

$$\dfrac{120}{11}$$ hours

$$\dfrac{109}{11}$$ hours

Solution

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