Straight Lines Test 29

Result

Straight Lines Test 29

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

13, 74, 290, 650,.......

A
$$1248$$

B
$$1370$$

C
$$1346$$

D
$$1452$$

E
$$1625$$

Solution

Here we observe the first term 13 can be written as: $$2^2+3^2$$
Similarly other numbers can be written as:
$$74: 5^2+7^2$$
$$290: 11^2+13^2$$
$$650: 17^2+19^2$$
Here, we observe that all above term of series can be written as sum of squares of two prime numbers which comes next to it's previous pair.
Like: $$(2,3);(5,7);(11,13);(17,19)$$ So, other pair of prime number will be $$ (23,29).$$
Hence next term of given series will be:
$$\Rightarrow 23^2+29^2=1370$$
So, correct answer is $$1370$$.
Question 2
1 / -0

Figures $$1$$ and $$2$$ are related in a particular manner. Establish the same relationship between figures $$3$$ and $$4$$ by choosing a figure from amongst the options.

A

B

C

D

Solution

Hexagon is rotating in anti-clockwise direction two times to get the next figure. (1 to 2).
Same for getting the figure (3 to 4) rotate the figure in an anti-clockwise direction (two times) we will get the desired figure.
Question 3
1 / -0

Select the INCORRECT match

A
$$3249-MMMCCXLIX$$

B
$$1667-MDCLXVII$$

C
$$207-CCXVII$$

D
$$499-CDXCIX$$

Solution

(a) $$MMMCCXLIX=3000+200+40+9=3249$$
(b) $$MDCLXVII=1000+500+100+60+7=1667$$
(c) $$CCXVII=217=200+17$$
(d) $$CDXCIX=400+90+9=499$$
Hence the correct match is option C.
Question 4
1 / -0

If D (3, -1), E (2, 6) and F (5, 7) are the vettices of the sides of $$\Delta DEF$$, the area of triangle DEF is sq. units.

A
$$11$$

B
$$22$$

C
$$48$$

D
$$50$$

Solution

Area of triangle = $$\dfrac{1}{2} {x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}$$

Area of triangle $$DEF=|\dfrac{1}{2}(3(6-7)+2(7+1)+5(-1-6))|$$

$$=|\dfrac{1}{2}(-3+16-35)|$$

$$=11$$ sq. Units.
Question 5
1 / -0

A telephone number $$d_1d_2d_3d_4d_5d_6d_7$$ is called memorable if the prefix sequence $$d_1d_2d_3$$ is exactly the same as either of the sequence $$d_4d_5d_6$$ or $$d_5d_6d_7$$(or possibly both). If each $$d_1\epsilon\{x|0\leq x\leq 9, x\epsilon W\}$$, then number of distinct memorable telephone number is(are).

A
$$19810$$

B
$$19,910$$

C
$$19,990$$

D
$$20,000$$

Solution
Question 6
1 / -0

The number of ways in which $$6$$ rings can be worn on the four fingers of one hand is

A
$$4^{6}$$

B
$$^{6}C_{4}$$

C
$$6^{4}$$

D
None of these

Solution

Each ring can be worn on $$4$$ fingers, means each have $$4$$ possibilities.
$$\therefore$$ Total ways$$=4\times4\times4\times4\times4\times4=4^{6}$$
Hence, $$(A)$$
Question 7
1 / -0

If $$15! =2^{\alpha}\cdot 3^{\beta}\cdot 5^{\gamma}\cdot 7^{\delta}\cdot 11^{\theta}\cdot 13^{\Phi}$$, then the value of expression $$\alpha -\beta +\gamma -\delta +\theta -\Phi$$ is

A
$$4$$

B
$$6$$

C
$$8$$

D
$$10$$

Solution

$$15!=1\times 2\times 3\times 4 \times 5\times 6\times 7\times 8\times 9\times 10\times 11\times 12\times 13\times 14\times 15$$

$$15!=2\times 3\times 2^2 \times 5\times 3\times 2 \times 7\times 2^3\times 3^2\times 2\times 5\times 11\times 3\times 2^2 \times 13\times 2\times 7 \times 3\times 5$$

$$15!=2^{11}\times 3^6\times 5^3\times 7^ 2\times 11\times 13 $$

$$\implies \alpha =11$$

$$\implies \beta =6$$

$$\implies \gamma=3$$

$$\implies \delta =2$$

$$\implies \theta =1$$

$$\implies \Phi=1$$

$$\alpha-\beta+\gamma-\delta+\theta-\Phi=11-6+3-2+1-1=6$$
Question 8
1 / -0

Geometric mean of $$7,\ { 7 }^{ 2 },\ { 7 }^{ 3 }....{ 7 }^{ n }$$ is

A
$$\displaystyle { 7 }^{ \dfrac { \left( n+1 \right) }{ 2 } }$$

B
$${ 7 }^{ { n }/{ 2 } }$$

C
$${ 7 }^{ \dfrac { \left( n-1 \right) }{ 2 } }$$

D
$${ 7 }^{ \left( \dfrac { n+1 }{ 3 } \right) }$$

Solution

$$Gm$$ of two no $$a$$ & $$b=\sqrt {ab}$$
For $$aGP$$
$$GM=\sqrt {a_1. an}$$
$$\Rightarrow \ GM $$ of $$7, 7^2, ..... 7^n$$
$$=\sqrt {7.7^n}$$
$$=7\dfrac {(n+1)}{2}$$
Question 9
1 / -0

Sum of the three digit numbers (no digit being zero) having the property that all digits are perfect squares, is

A
$$3108$$

B
$$6216$$

C
$$13986$$

D
$$None\ of\ these$$

Solution

Perfect square digits from $$1 $$ to $$9 $$ are - $$1,4,9$$

The 3 digit nos. having the property that all digits are perfect squares are,

$$149, 194, 491, 419, 914, 941$$

The sum is,

$$149+194+491+419+914+941=3108$$
Question 10
1 / -0

Choose the correct answer from the alternatives given.
A loss of $$20%$$ is incurred when $$6$$ articles are sold for a rupee. To gain $$20% $$ how many articles should be sold for a rupee?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

: Given that, SP = $$1$$. Loss$$ % = 20%$$ We know that, $$SP = 0.8$$ $$\times$$ $$CP$$
$$CP \, = \, \dfrac{5}{4}$$
To gain$$ 20%$$,SP=1.2 $$\displaystyle \times \, CP \, = \, \dfrac{120}{100} \, \times \, \dfrac{5}{4} \, = \, \dfrac{3}{2} \, \Rightarrow \, = \, \dfrac{3}{2}$$
For$$ Rs. 3/2$$ number of articles sold
For $$Rs. 1$$ number of articles to be sold $$= 6 $$ $$\times$$ $$ \dfrac{2}{3}$$ = $$ 4 $$ articles.