Straight Lines Test 13

Result

Straight Lines Test 13

Score
-
out of -
Rank
-
out of -

TIME Taken - -

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Question 1
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 2
1 / -0

$$A$$ is twice as fast as $$B$$ is thrice as fast as $$C$$. The journey covered by $$C$$ in $$42$$ minutes, what will be covered by $$A$$ is

A
$$21$$ Min

B
$$64$$ Min

C
$$17$$ Min

D
$$40$$ Min

Solution

time taken ny C to complete the journey $$=$$ 42 minutes
time taken by A to complete the journey $$=\dfrac{42}{2}$$ minutes ($$\because $$ A is twice as fast as C)
$$=21 $$ minutes.
Question 3
1 / -0

If the letter of the word $$LATE$$ be permuted and the words so formed be arranged as in a dictionary . Then the rank of $$LATE$$ is :

A
$$12$$

B
$$13$$

C
$$14$$

D
$$15$$

Solution

LATE
No. of words that start with 'A'
$$\Rightarrow 3!=6$$
No. of words that start with E
$$\Rightarrow 3!=6$$
$$6+6=12$$
$$13^{th}$$ word starts with L
$$13^{th}$$ word = LAET
$$14^{th}$$ word = LATE
$$\therefore $$ Rank of LATE is $$14$$
Question 4
1 / -0

Choose the most appropriate option which fits this pattern

A

B

C

D
Question 5
1 / -0

Solve:$$\dfrac{2}{2}+\dfrac{3}{3}+\dfrac{4}{4}+$$...... + upto $$1000$$ terms= ?

A
$$1000$$

B
twice of $$500$$

C
four times of $$250$$

D
All of these

Solution

All the terms simplify to $$1$$ which is being added $$1000$$ times.
Therefore, the final answer is $$1000$$ and it can be seen that all the options are correct.
Question 6
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 7
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 8
1 / -0

Choose the most appropriate option which follows the pattern

A

B

C

D
None of the above
Question 9
1 / -0

Find the number suitable for blank in given series.
1,2,3,5,8,13,21,___,55

A
35

B
33

C
36

D
34

Solution

The required number can be found by adding two numbers before it .
Examples : 3 can be found by adding two number before it.
i.e => (1+2) = 3
Similarly 5  can be found by adding two number before it.
i.e => (2+3) = 5 and so on.
In the same way the missing number can be found by adding  two number before it .
i.e => 13 + 21 = 34
The required number is 34. (Ans)
Question 10
1 / -0

958, 833, 733, 658, 608 ?

A
577

B
583

C
567

D
573

E
none of these

Solution

$$958, 833, 733, 658, 608, - $$
$$a_{0}= 958$$
$$a_{1}= 833 = a_{0}- 125$$
$$a_{2}= 733=a_{1}-100$$
$$a_{3}= 658=a_{2}-75$$
$$a_{4}= 608 = a_{3}-50$$
These numbers are in A.P.
Where common difference = +25
Answer = $$a_{4}-25= 608-25=583$$