Straight Lines Test 48

Result

Straight Lines Test 48

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Question 1
1 / -0

Find the missing number, if same rule is followed in all the three figures.

A
12

B
16

C
14

D
9

Solution

REF.Image.
Given

If we consider down two number of fig(i)

then their product is $$ \Rightarrow 4\times 16 = 64 $$

Now if we take square root then =$$ \sqrt{64}$$

$$ = 8 \to$$upper are we are getting.

similarly from fig (iii)

down number product $$ \Rightarrow 12 \times27 = 324$$

take square roots $$ = \sqrt{324} = 18 \to$$upper number.

So, similarly from fig(ii)

Down product = $$ 18 \times 8 = 144 $$

so square root = $$ \sqrt{144} = 12 \rightarrow $$ option A
Question 2
1 / -0

The number of the three digits numbers having only two consecutive digits identical is:

A
153

B
160

C
180

D
161

Solution
Question 3
1 / -0

Find the missing number in the given figure?

A
$$25$$

B
$$361$$

C
$$196$$

D
$$144$$

Solution
Question 4
1 / -0

A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is

A
140

B
196

C
280

D
346

Solution

There would be two cases,

(1) When 4 is selected from first five and rest 6 from remaining 8.
This is done in $$^{5}C_{4}\times ^{8}C_{6}$$ ways.

(2) When all 5 are selected from first five and rest five from remaining 8.
This can be done in $$^{5}C_{5}\times ^{8}C_{5}$$ ways.

$$\therefore $$ A total of $$^{5}C_{4}\times ^{8}C_{6}+^{5}C_{5}\times ^{8}C_{5}$$

$$=5\times \cfrac{8!}{6!2!}+\cfrac{8!}{3!5!}$$

$$=\cfrac{5\times 8\times 7}{2}+\cfrac{8\times 7\times 8}{3\times 2}$$

$$=140+56=196\,ways$$
Question 5
1 / -0

If $$\displaystyle\sum^m_{k=1}(k^2+1)k!=1999(2000!)$$, then m is?

A
$$1999$$

B
$$2000$$

C
$$2001$$

D
$$2002$$

Solution
Question 6
1 / -0

In the following number series,one number is wrong.find out the ?

1,2,8,33,149,765,4626

A
$$33$$

B
$$8$$

C
$$149$$

D
$$0$$

Solution

1,2,8,33,149,765,4626
Carefully analyzing the series, we can see the following pattern
$$1\times 1+(1)^{2}=2$$
$$2\times 2+(2)^{2}=8$$
$$8\times 3+(3)^{2}=33$$
$$33\times 4+(4)^{2}=148$$
$$148\times 5+(5)^{2}=765$$
$$765\times 6+(6)^{2}=4626$$
So, we can clearly see that 149 is the wrong entry in the series.
Question 7
1 / -0

If $$ '-'$$ denotes $$'\div ',\ '\div '$$ denotes $$'\times ',\ '+'$$ denotes' - 'and $$'\times '$$ denotes $$'+'$$, then find the value of $$116+9\div 52-4\times 5$$.

A
$$16$$

B
$$8$$

C
$$9$$

D
$$4$$

Solution

$$-$$ denoted $$\div$$
$$\div$$ denoted $$\times $$
$$+$$ denoted $$-$$
$$\times $$ denoted $$+$$ $$\boxed{BODMAS}$$
$$116+9\div 52-4\times 5$$
$$116-9\times 52\div 4+5$$
$$116-9\times 13+5$$
$$116-117+5$$
$$116-112=4$$
Question 8
1 / -0

If $$n\ \in\ N$$ & $$n$$ is even, then $$\dfrac {1}{1\ .\ (n-1)\ !}+\dfrac {1}{3\ !(n-3)\ !}+\dfrac {1}{5\ !\ (n-5)\ !}+....+\dfrac {1}{(n-1)\ !\ 1\ !}=$$

A
$$2^{n}$$

B
$$\dfrac {2^{n-1}}{n\ !}$$

C
$$2^{n}n\ !$$

D
$$none\ of\ these$$

Solution
Question 9
1 / -0

$$435, 354, 282, 219, 165, ?$$

A
$$103$$

B
$$112$$

C
$$120$$

D
$$130$$

E
$$none\ of\ these$$

Solution

$$\therefore 165-45=x$$
$$x=120$$.
Question 10
1 / -0

$$10$$ IIT and $$2$$ NIT students sit at random in a row, and then number of ways in which exactly $$3$$ IIT students sit between $$2$$ NIT students is

A
$$16\times 10!$$

B
$$15\times 10!$$

C
$$10\times 16!$$

D
$$None\ of\ these$$

Solution