Numerical Applications Test 32

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Numerical Applications Test 32

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Question 1
1 / -0

Find the day of the week on $$\text{16 January, 1969}$$

A
Thursday

B
Monday

C
Tuesday

D
Friday

Solution

$$\Rightarrow\;1600$$ years have $$0$$ odd days .....(i)
$$300$$ years have $$1$$ odd day .....(ii)
$$68$$ years have $$17$$ leap years and $$51$$ ordinary years
$$=(17\times2+51\times1)$$
$$=85\,\text{odd days}=1\text{odd day}$$ ......(iii)
$$\text{16 January}$$ has $$2$$ odd days .........(iv)
Adding odd days,
$$0+1+1+2=4\,\text{odd days}=\text{Thursday}$$.
Question 2
1 / -0

At what time between $$2$$ and $$3$$ O'clock will the hands of a clock be together?

A
$$10 \cfrac{10}{11}$$ min. past $$2$$

B
$$11 \cfrac{10}{11}$$ min. past $$2$$

C
$$13 \cfrac{10}{11}$$ min. past $$2$$

D
$$7 \cfrac{10}{11}$$ min. past $$2$$

Solution

If the time is $$2$$ hours and $$x$$ minutes after $$2$$, when the two hands of the clock are same, then the angle formed by minute hand is: $$\cfrac{x}{60} \times 360 = 6x$$
The hour hand moves $$\cfrac{1}{2}$$ degree each minute, then angle traced by hour hand is = $$60 + \cfrac{x}{2}$$
Hence, $$60 + \cfrac{x}{2} = 6x $$
$$60 = \cfrac{11x}{2}$$
$$x = \cfrac{120}{11}$$
$$x = 10 \cfrac{10}{11}$$ min. past $$2$$
Question 3
1 / -0

The average maximum temperature for 7 days from the 12th September to 18th September is $$35^\circ C$$ and that for 7 days from 13th to 19th September is $$34^\circ C$$. From this we can conclude that

A
Maximum temperature on 19th is $$1^\circ C$$ less than that for 12th September

B
Maximum temperature on 12th is $$1^\circ C$$ less than that for 19th September

C
Maximum temperature on 19th is $$7^\circ C$$ less than that for 12th September

D
None of the above

Solution

The average maximum temp from 12th sept to 18th sept is 35 and The average maximum temp from 13th sept to 19th sept is 34
Then total temp from 12th sept to 18th sept =35$$\times$$7=245
And total temp from 13th sept to 19th sept =34$$\times$$7=238
Then diff=245-238=7
So Maximum temperature on 12th is7 $$^{0}$$ less than that for 19th September
7
77
Question 4
1 / -0

If the day before yesterday was Sunday, what day will fall on the day after tomorrow ?

A
Thursday.

B
Monday

C
Friday

D
Wednesday

Solution

$$\Rightarrow$$ Today it is Tuesday and day after tmorrow will be Thursday.
Question 5
1 / -0

$$3$$ persons weave $$168$$ shawls in $$14$$ days, how many shawls will $$ 8 $$ persons weave in $$5$$ days?

A
$$90$$

B
$$105$$

C
$$115$$

D
$$160$$

Solution

Here we are asked to find out the number of shawls we compare each item with the number of shawls as shown below:
More number of persons$$\displaystyle \Rightarrow $$ More shawls
Less number of persons$$\displaystyle \Rightarrow $$ Less shawls
i.e. the relationship is direct
Also More number of days$$\displaystyle \Rightarrow $$ More shawls
Less number of days$$\displaystyle \Rightarrow $$ Less shawls
i.e. the relationship is direct
Persons $$3:8::168:x$$
Days 14:5
$$\displaystyle \therefore \frac{3}{8}\times \frac{14}{5}=\frac{168}{x}$$
$$\displaystyle \Rightarrow 3\times 14\times x=168\times 8\times 5$$
$$\displaystyle \Rightarrow x=\frac{168\times 8\times 5}{3\times 14}=160$$
Question 6
1 / -0

$$15$$ men can build a wall $$108$$ m long in $$6 $$ days, what is the length of similar wall that can be built by $$25$$ men in $$3$$ days?

A
$$75$$

B
$$80$$

C
$$85$$

D
$$90$$

Solution

As we are asked to find out the length of the wall we compare each item with the length as shown below:
More number of men$$\displaystyle \Rightarrow $$ More length of wall
Less number of men$$\displaystyle \Rightarrow $$ Less length of wall
i.e. This forms a direct relationship Moreover,
Less number of days$$\displaystyle \Rightarrow $$ Less length built
More number of days$$\displaystyle \Rightarrow $$ More length built
i.e. this relationship is also direct
Let x be the required length of the wall
Now we have two equations
Men $$15:25$$
Days$$ 6:3::108:x$$
Now to solve we multiply the given ratios together equaling the required ratio
i.e. $$\displaystyle \frac{15}{25}\times \frac{6}{3}=\frac{108}{x}$$
$$\displaystyle \Rightarrow 15\times 6x=108\times 25\times 3$$
$$\displaystyle \Rightarrow x=\frac{108\times 25\times 3}{15\times 6}=18\times 5=90$$
Question 7
1 / -0

$$15$$ labourers can fill $$35$$ boxes in $$7 $$ days. how many labourers can fill $$65$$ boxes om $$5$$ days ?

A
$$13$$

B
$$39$$

C
$$45$$

D
$$55$$

Solution

We are asked to find out the number of labourers So we compare each item with the number of labourers as below
More number of boxes$$\displaystyle \Rightarrow $$ More labourers
i.e. the relationship is direct
Less number of days$$\displaystyle \Rightarrow $$ More number of labour
i.e. the relationship is inverse
$$\displaystyle \therefore $$ Boxes $$35:65$$
Days $$5:7::15:x$$
$$\displaystyle \therefore $$ $$\displaystyle \frac{35}{65}\times \frac{5}{7}=\frac{15}{x}$$
$$\displaystyle \therefore x=\frac{15\times 65\times 7}{35\times 5}=13\times 3=39$$
Question 8
1 / -0

The weights of 9 apples are 50, 60, 65, 62, 67, 70, 64, 45, 48 grams. Their mean weight is

A
60.5 gram

B
60 gram

C
59 gram

D
62 gram

Solution

Given weights of 9 apples are 50, 60, 65, 62, 67, 70, 64, 45, 48 grams.

Then total weight of 9 apples =50+60+65+62+67+70+64+45+48=531
Then mean=$$\frac{539}{9}$$=59
Question 9
1 / -0

How many times in a day the two hands of a clock are at $$ \displaystyle 90^{\circ} $$

A
4

B
12

C
22

D
44

Solution

In 12 hours, they are at right angles 22 times.
$$\therefore $$In 24 hours, they are at right angles 44 times.
Option D is correct.
Question 10
1 / -0

What day of the week will 1st January 2008 be given that 1st January 2000 is a Saturday?

A
Sunday

B
Wednesday

C
Tuesday

D
None of these

Solution

Since there will be $$2$$ leap years between January $$1^{st},$$ $$2000$$ and January $$1^{st},2008$$ and $$6$$ simple years
Hence, the date $$1^{st}$$ January, $$2008$$ have been jumped $$\dfrac{(4+6)}{7}=3(Remainder)$$ days since January $$1^{st},2000$$.
So, It will be $$Tuesday$$ on January $$1^{st},2000$$.

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Numerical Applications Test 32

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Numerical Applications Test 32

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SHARING IS CARING

Question 1

Thursday

Monday

Tuesday

Friday

Solution

Question 2

$$10 \cfrac{10}{11}$$ min. past $$2$$

$$11 \cfrac{10}{11}$$ min. past $$2$$

$$13 \cfrac{10}{11}$$ min. past $$2$$

$$7 \cfrac{10}{11}$$ min. past $$2$$

Solution

Question 3

Maximum temperature on 19th is $$1^\circ C$$ less than that for 12th September

Maximum temperature on 12th is $$1^\circ C$$ less than that for 19th September

Maximum temperature on 19th is $$7^\circ C$$ less than that for 12th September

None of the above

Solution

Question 4

Thursday.

Monday

Friday

Wednesday

Solution

Question 5

$$90$$

$$105$$

$$115$$

$$160$$

Solution

Question 6

$$75$$

$$80$$

$$85$$

$$90$$

Solution

Question 7

$$13$$

$$39$$

$$45$$

$$55$$

Solution

Question 8

60.5 gram

60 gram

59 gram

62 gram

Solution

Question 9

4

12

22

44

Solution

Question 10

Sunday

Wednesday

Tuesday

None of these

Solution

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