Sequences and Series Test 40

Result

Sequences and Series Test 40

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Question 1
1 / -0

If $$^{8}C_{r}=^{8}C_{3}$$, then $$r$$ is equal to

A
$$5$$

B
$$4$$

C
$$8$$

D
$$6$$

Solution

$$^8C_r=^8C_3$$
$$\frac {8!}{r!(8-r)!}=\frac{8!}{3!\,5!}$$
$$\Rightarrow r!(8-r)!=3 !5!$$
$$\Rightarrow either \, r=3 \,\,\, or r=5 \Rightarrow (A)$$
Question 2
1 / -0

There is a certain relationship between the pair of figure on either side of ::. Identify the relationship on the left side and find the missing figure.

A

B

C

D

Solution

As per the first relation, in the first figure, there is a square on which there is 3 circle and in the second figure, there is a circle on which there is $$3$$ square.

Following the same in second relation, the correct option is $$B$$.
Question 3
1 / -0

Rahul told Anand, "Yesterday I defeated the only brother of the daughter of my grandmother." Whom did Rahul defeat ?

A
Son

B
Father

C
Brother

D
Cousin

Solution

Daughter of Grandmother $$\rightarrow$$ Aunt Aunt's only brother $$\rightarrow$$ Father
Question 4
1 / -0

Which number replaces the question mark ?

A
10

B
12

C
14

D
16

Solution

In each triangle, the central value equals the average of the 3 values around the outside.

Hence, the required value = $$\dfrac{15+17+4}{3}=12$$

Therefore, option B is correct.
Question 5
1 / -0

When $$n!+1$$ is divided by any natural number between $$2$$ and $$n$$ then remainder obtained is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$
Question 6
1 / -0

The number of ways in which $$9$$ persons can be divided into three equal groups, is

A
$$1680$$

B
$$840$$

C
$$560$$

D
$$280$$

Solution
Question 7
1 / -0

How many integers are there such that $$2 \le n \le 100$$ and the highest common factor of $$n$$ and $$36$$ is $$1$$?

A
$$166$$

B
$$332$$

C
$$331$$

D
$$416$$

Solution

$$36.2^{2}.3^{2}$$
Since HCF (36,n) =1
Therefore, n sholud not be a multiple of 2 or 3.
$$ 2 \leq n\leq 100$$
Total numbers = T= 1000.
Number of numbers divisible by 2= $$N_{2}$$
Number of numbers divisible by 3= $$N_{3}$$
Number of numbers divisible by 6= $$N_{6}$$
Therefore, there are $$T-N_{2}-N_{3}+N_{6}$$ total integers
a=2
l=1000
d=2
We know, $$l=a+(N_{2}-1)d.$$
$$N_{2}=\frac{1000-2}{2}+1$$
$$N_{2}=\frac{998}{2}+1$$
$$N_{2}=499+1=500$$
Similarly,
$$N_{3}=\frac{999-3}{3}+1=333$$
$$N_{6}=\frac{999-6}{3}+1=166$$
Answer
= 999-500-333+166
= 332
Question 8
1 / -0

If area of a triangle is $$35$$ square units with vertices $$\left( {2, - 6} \right),\,\,\left( {5,\,\,4} \right)$$ and $$({k},\,\,4)$$ then $${k}$$ is :

A
$$ -2 \space \text{or} \space 12$$

B
$$12$$

C
$$-2$$

D
$$ 2 \space \text{or} \space 12$$

Solution
Question 9
1 / -0

What is the next number in the series $$2,12,36,80,150?$$

A
$$194$$

B
$$210$$

C
$$252$$

D
$$258$$

Solution

$$12 - 2 = 10$$.
$$36 - 12 = 24$$.
$$80 - 36 = 44$$.
$$150 - 80 = 70$$.
Now,
$$24 - 10 = 14$$.
$$44 - 24 = 20$$.
$$70 - 44 = 26$$.
Then next difference of the difference will be $$26 + 6 = 32$$.
Hence answer is $$150 + 70 + 32 = 252$$.
Question 10
1 / -0

The number of permutations which can be formed out of the letters of the word "SERIES" three letters together, is:

A
120

B
60

C
42

D
none

Solution