Straight Lines Test 42

Result

Straight Lines Test 42

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Question 1
1 / -0

Find $$x$$, if $$\dfrac {1}{4!}-\dfrac {1}{x}=\dfrac {1}{5!}$$.

A
$$5$$

B
$$4$$

C
$$30$$

D
$$None$$

Solution
Question 2
1 / -0

Find the sum of first $$15$$ terms of the sequence whose $${n}^{th}$$ term is $$3+4n$$.

A
$$525$$

B
$$425$$

C
$$495$$

D
$$None\,of\,thes$$

Solution

$$n^{th}$$ term = 3+4 n
$$1^{st}$$ term = 3+4(1)=7
$$2^{nd}$$ term = 3+4(2)=11
$$3^{rd}$$ term 3+4(3)=15
It is an AP with a=7, d=4
$$S_{15}=\frac{15}{2}[2(7)+14(4)]$$
$$= 15[7+28]$$
$$= 525$$
Question 3
1 / -0

solve

A
$$225$$

B
$$175$$

C
$$335$$

D
$$350$$

Solution

Pattens is as follows:-

$$ (20 - 9)^{2} = 121 , $$ $$(24 - 11)^{2} = 169 $$

So $$ (28 - 13)^{2} = 15^{2} = 225$$

Hence missing place = 225
Question 4
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 5
1 / -0

insert the missing number: 5,8,12,17,23,___,38

A
$$29$$

B
$$30$$

C
$$32$$

D
$$25$$

Solution

$$5,8,12,17,23,......38$$
$$5+3=8;8+4=12,12+5=17,17+6=23$$
$$23+7=30,30+8=30$$
$$\therefore 30$$
Question 6
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 7
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 8
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 9
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 10
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