Numerical Applications Test 41

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Numerical Applications Test 41

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Weekly Quiz Competition

Question 1
1 / -0

The length of the base of a square pyramid is $$2\ cm$$ and the height is $$6\ cm$$. Calculate the volume.

A
$$8\ cm^3$$

B
$$6\ cm^3$$

C
$$4\ cm^3$$

D
$$2\ cm^3$$

Solution

Volume of square pyramid $$=\dfrac { 1 }{ 3 } \times { a }^{ 2 }\times h=\dfrac { 1 }{ 3 } \times 2\times 2\times 6=8{ cm }^{ 3 }$$
Question 2
1 / -0

If $$3$$ men with $$4$$ boys can earn $$Rs. 210$$ in $$7$$ days and $$11$$ men with $$13$$ boys can earn $$Rs. 830$$ in $$8$$ days, in what time will $$7$$ men with $$9$$ boys earn $$Rs. 1100$$?

A
$$6\ days$$

B
$$16\ days$$

C
$$10\ days$$

D
$$12\ days$$

Solution

$$(3\ men + 4\ boys)$$ can earn $$Rs. (210 + 7)$$, i.e., $$Rs. 30.00$$ in one day. And $$(11\ men + 13\ boys)$$ can earn $$Rs. (830\div 8)$$, i.e., $$Rs. 103.75$$ in one day. Find out the earning of $$1$$ man and $$1$$ boy and then proceed.
$$\left.\begin{matrix}3m + 4b & = 30.00\\ 11m + 13b & = 103.75\end{matrix}\right\}\Rightarrow 1m = 5, 1b = 3.75$$
Wages for $$7m + 9b = 7\times 5 + 9\times 3.75 = Rs. 68.75$$ per day
$$\therefore$$ No. of required days $$= Rs. 1100 + Rs. 68.75 = 16$$
Question 3
1 / -0

16 men can do a piece of work in 16 days. 4 days after they started the work 8 more men joined them. How many days will they now take to complete the remaining work?

A
$$10\ days$$

B
$$6\ days$$

C
$$8\ days$$

D
$$12\ days$$

Solution

16 men can do work in 16 days
$$16 \times 16=16 \times 4 +24 \times x$$
$$x=16\dfrac{[16-4]}{24}=8 days$$
Question 4
1 / -0

A contractor employed 210 men to build a house in 60 days. After 12 days, he was joined by 70 more men. In how many days will the remaining work be finished?

A
35

B
36

C
38

D
40

Solution

So, $$210$$ men employed for $$60$$ days to complete work.
So, total expected work $$=(210\times 60)$$
So, According to the question-
$$\underbrace {(210\times 12)}_{First\quad 12\quad days}+\underbrace {x}_{Next\quad no.\quad of\quad days\quad}\underbrace {(210+70)}_{When\quad 70 \quad more\quad men\quad join\quad}=\underbrace {210\times 60}_{Total\quad work}$$
$$\implies 2520+230x=12600$$
$$\implies 230x=10080$$
$$\implies x=36$$
So, $$36$$ more days are required.
Question 5
1 / -0

Choose the correct answer from the alternatives given :
Two taps can separately fill a cistern in $$20$$ and $$30$$ minutes respectively. Each of the taps is opened for a minute in turn, but due to a hole at the bottom of the cistern and discharge of the water through that, it takes $$3$$ minutes more to fill the cistern. If the cistern is full of water, find the time taken by the hole to empty it.

A
$$2\frac {4}{7}$$ hours

B
$$5\frac {1}{3}$$ hours

C
$$3\frac {3}{5}$$ hours

D
$$4\frac {1}{3}$$ hours

Solution

Let the total unit of work be $$60$$ units (LCM of $$20$$ & $$30$$).
Tap- $$1$$ per minute work = $$\dfrac{60}{20}$$ = $$3$$ units / min.
Tap - $$2$$ per minute work = $$\dfrac{60}{30}$$ = $$2$$ units /min.
Time taken by Tap - $$1$$ & Tap - $$2$$
$$5$$ unit of work = $$2$$ minute
$$60$$ unit of work = $$\dfrac{2}{5} \times 60$$ = $$24$$ minute.
Due to the leakage. it takes $$$3 more minutes I.e. $$27$$ minutes to fill the tank.
Let the total unit of work be $$216$$ units. [LCM of $$24$$ and $$27$$]
Tap($$1 + 2$$) per minute work = $$\dfrac{216}{24}$$ =$$9$$ units / min.
Tap + Holes per minute work = $$\dfrac{216}{27}$$ =$$8$$ units / min.
Holes per minute work = $$9 - 8= 1$$ unit/min.
Time taken by hole to empty the tank = $$\dfrac{216}{1}$$
= $$3$$ hours + $$36$$ minute
=$$3$$+ $$\dfrac{3}{5}$$ = 3$$\dfrac{3}{5}$$ hour
Question 6
1 / -0

Choose the correct answer from the alternatives given.
A contractor took a contract for building $$12$$ kilometre road in $$15$$ days and employed $$100$$ labours on the work. After $$9$$ days he found that only $$5$$ kilometre road had been constructed. How many more labours should be employed to ensure that the work may be completed with in the given time?

A
$$120$$

B
$$90$$

C
$$110$$

D
$$100$$

Solution

$$100$$ labours worked for $$9$$ days,
Hence $$900$$ man days = $$5$$ km
$$1$$ km = $$\dfrac{900}{5}$$ man days.
Now $$7$$ km of the road needs to be built
Then
$$7km$$=$$\dfrac{900}{5} \times 7 = 180 \times 7$$ = $$1260$$ man days.
Now thus $$1260$$ man days of work needs to he completed in $$6$$ days.
Let number of more workers required be x.
$$1260=(100 + x) $$ $$\times $$ $$6$$
$$210 = 100 + x \Rightarrow x= 110$$
Hence, number of more workers required are $$110$$.
Question 7
1 / -0

A telephone number $$d_1d_2d_3d_4d_5d_6d_7$$ is called memorable if the prefix sequence $$d_1d_2d_3$$ is exactly the same as either of the sequence $$d_4d_5d_6$$ or $$d_5d_6d_7$$(or possibly both). If each $$d_1\epsilon\{x|0\leq x\leq 9, x\epsilon W\}$$, then number of distinct memorable telephone number is(are).

A
$$19810$$

B
$$19,910$$

C
$$19,990$$

D
$$20,000$$

Solution
Question 8
1 / -0

An alphabet contains a $$A^{'s}$$ and b $$ B^{'s}$$ . (In all a+b letters ). The number of words each containing all the $$ A^{'s}$$ and any number of $$ B^{'s}$$, is

A
$$^{a+b}C_b$$

B
$$^{a+b+1}C_a$$

C
$$^{a+b+1}C_b$$

D
none of these

Solution
Question 9
1 / -0

The maximum number of intersection points of n circles and n straight lines , among themselves is 80.The value of n is

A
$$7$$

B
$$6$$

C
$$5$$

D
$$8$$

Solution
Question 10
1 / -0

The number of ways in which $$6$$ rings can be worn on the four fingers of one hand is

A
$$4^{6}$$

B
$$^{6}C_{4}$$

C
$$6^{4}$$

D
None of these

Solution

Each ring can be worn on $$4$$ fingers, means each have $$4$$ possibilities.
$$\therefore$$ Total ways$$=4\times4\times4\times4\times4\times4=4^{6}$$
Hence, $$(A)$$

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Re-Attempt Weekly Quiz Competition

Numerical Applications Test 41

Result

Numerical Applications Test 41

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SHARING IS CARING

Question 1

$$8\ cm^3$$

$$6\ cm^3$$

$$4\ cm^3$$

$$2\ cm^3$$

Solution

Question 2

$$6\ days$$

$$16\ days$$

$$10\ days$$

$$12\ days$$

Solution

Question 3

$$10\ days$$

$$6\ days$$

$$8\ days$$

$$12\ days$$

Solution

Question 4

35

36

38

40

Solution

Question 5

$$2\frac {4}{7}$$ hours

$$5\frac {1}{3}$$ hours

$$3\frac {3}{5}$$ hours

$$4\frac {1}{3}$$ hours

Solution

Question 6

$$120$$

$$90$$

$$110$$

$$100$$

Solution

Question 7

$$19810$$

$$19,910$$

$$19,990$$

$$20,000$$

Solution

Question 8

$$^{a+b}C_b$$

$$^{a+b+1}C_a$$

$$^{a+b+1}C_b$$

none of these

Solution

Question 9

$$7$$

$$6$$

$$5$$

$$8$$

Solution

Question 10

$$4^{6}$$

$$^{6}C_{4}$$

$$6^{4}$$

None of these

Solution

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