Straight Lines Test 11

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Straight Lines Test 11

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Weekly Quiz Competition

Question 1
1 / -0

The average IQ of $$4$$ people is $$110$$. If three of these people each have an IQ of $$105$$, what is the IQ of the fourth person?

A
110

B
115

C
120

D
125

Solution

Average IQ of $$4$$ people $$=110$$
$$\therefore$$ total IQ of $$4$$ people $$=$$ $$110\times 4=440$$
Average IQ of $$3$$ people $$=105$$
$$\therefore $$ total IQ of $$3$$ people $$=$$ $$105\times 3=315$$
Then IQ of forth person $$=$$ $$440-315=125$$
Question 2
1 / -0

A garrison of '$$n$$' men had enough food to last for $$30$$ days. After $$10$$ days, $$50$$ more men joined them. If the food now lasted for $$16$$ days, what is the value of $$n$$?

A
$$200$$

B
$$240$$

C
$$280$$

D
$$320$$

Solution

After $$10$$ days, the food for n men is there for $$20$$ days.
This food can be eaten by $$(n + 50)$$ men in $$16$$ days.
$$\therefore 20n = 16(n + 50)$$
$$\therefore n = 200$$
Question 3
1 / -0

A bag contains Rs. $$112$$ in the form of $$1$$-rupee, $$50$$-paise and $$10$$-paise coins in the ratio $$3 : 8 : 10$$. What is the number of $$50$$-paise coins?

A
$$112$$

B
$$128$$

C
$$96$$

D
$$24$$

Solution

Here bag contains Rs. $$112$$ in the form of $$1$$-rupee, $$50$$-paise and $$10$$-paise.
Coin ratio is $$ 3 : 8 : 10$$.
Value ratio $$= 3 \times 1: 8\times \dfrac {1}{2} : 10 \times \dfrac {1}{10}$$
$$= 3 : 4 : 1$$
Number of $$50$$ paise coins $$= \dfrac {4}{8} \times$$ Rs. $$112 =$$ Rs. $$56$$
Therefore, number of $$50$$ paise coins $$= 2\times 56 = 112$$.
Question 4
1 / -0

The sum of two numbers is $$80$$. If the larger number exceeds four times the smaller by $$5$$, what is the smaller number?

A
$$5$$

B
$$15$$

C
$$20$$

D
$$25$$

Solution

Given, sum of two numbers is $$80$$.
Let the smaller number be $$x$$
Thus the larger will be $$80 - x$$.
Also given, larger number exceeds four times the smaller number by $$5$$.
Therefore, $$ 80 - x = 4x + 5$$
$$\Rightarrow 5x = 75$$
$$\Rightarrow x = 15$$
Thus the smaller number is $$15$$.
Question 5
1 / -0

A man has 9 friends, 4 boys and 5 girls. In how many ways can be invite them, if there have to be exactly three girls in the invites?

A
$$320$$

B
$$160$$

C
$$80$$

D
$$200$$

Solution

He can select three girls in $$^5C_3$$ ways = 10 ways
He can select boys in $$ [^4C_0+^4C_1 +^4C_2 +^4C_3 + ^4C_4]$$ ways = 16 ways
Total number of ways $$= 10\times16$$ $$ = 160$$
Question 6
1 / -0

Complete the addition square by finding the missing numbers.
Find $$(A-B)-(C-D)$$
$$+$$ $$10,923$$ $$8,473$$
$$18,732$$ $$A$$ $$B$$
$$9,018$$ $$C$$ $$D$$

A
$$0$$

B
$$450$$

C
$$1000$$

D
$$300$$

Solution

From the given table, we have
$$A=18,732+10,923=29,655$$
$$B=18,732+8,473=27,205$$
$$C=10,923+9,018=19,941$$
and $$D=8473+9,018=17,491$$
Now $$A-B=29,655-27,205=2,450$$
$$C-D=19,941-17,491=2,450$$
$$\therefore$$ $$(A-B)-(C-D)=2,450-2,450=0$$
Question 7
1 / -0

The models are shaded to show which of the following?

A
$$\cfrac { 1 }{ 3 } =\cfrac { 2 }{ 4 } $$

B
$$\cfrac { 1 }{ 4 } >\cfrac { 1 }{ 3 } $$

C
$$\cfrac { 2 }{ 3 } <\cfrac { 2 }{ 4 } $$

D
$$\cfrac { 2 }{ 4 } <\cfrac { 2 }{ 3 } $$

Solution

Fraction of shaded part in upper rectangle $$=2\times\dfrac{1}{3}=\dfrac{2}{3}$$
Fraction of shaded part in lower rectangle $$=2\times\dfrac{1}{4}=\dfrac{2}{4}$$
In upper rectangle shaded part is greater
So, $$\dfrac { 2 }{ 4 } <\dfrac { 2 }{ 3 } $$
Option D is correct.
Question 8
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Fathom is a unit once used by sailors to measure the depth of water. If a sunken ship was located underwater at $$240$$ feet, which expression would describe the location of the ship in fathoms?
$$ { 1\ fathom=6\ feet } $$

A
$$240\times 6$$

B
$$240\div 6$$

C
$$240-6$$

D
$$240+6$$

Solution

We have $$1fathom=6feet$$
$$1feet=\cfrac{1}{6}fathom$$
Ship was located underwater at $$240$$ feet
$$\therefore$$ location of ship underwater in fathoms $$=240\div6$$
Question 9
1 / -0

In the Number series given below, one number is missing. Series is followed by five alternatives. One of them is the right answer. Identify and indicate it as per the "Instructions".
$$2,35,104,209,....$$

A
$$350$$

B
$$248$$

C
$$256$$

D
$$311$$

E
$$413$$

Solution

To solve this question, we use the method of difference.
let $${T_5}=x$$
Now, series is $${S}:\ 2,\ 35,\ 104,\ 209,\ x$$
Let us take the difference between any 2 adjacent terms be a series $$dS$$
$${dS}: \ 33,\ 69,\ 105,\ (x-209)$$
We again take the difference between 2 adjacent terms. Let the series be $$ddS: \ 36,\ 36,\ (x-209-105) $$
For the series to be consistent,
$${x-209-105}={36}$$
$$\implies x=350$$
Thus, $${(A)}$$ must be the answer.
Question 10
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Directions For Questions

In these questions, numbers are placed in the figures on the basis of some rules. One place is vacant which if indicated as'?' Find out the correct alternatives to replace the question mark '?'

...view full instructions

The figure is given.

A
$$7$$

B
$$4$$

C
$$10$$

D
$$8$$

E
$$6$$

Solution

The no. in the triangle is:
$$\dfrac { (\text{largest}+\text{smallest})-\text{middle} }{ 2 } =\text{ number}$$
Triangle 1: $$\quad \dfrac { (13+5)-12 }{ 2 } = 3$$
Triangle 2: $$\quad \dfrac { (13+4)-9 }{ 2 } = 4$$
Thus the answer is: $$ \dfrac { (16+5)-7}{ 2 } =7$$
Thus, the answer is $$7$$.

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$$+$$	$$10,923$$	$$8,473$$
$$18,732$$	$$A$$	$$B$$
$$9,018$$	$$C$$	$$D$$

Straight Lines Test 11

Result

Straight Lines Test 11

Score

-

Rank

-

SHARING IS CARING

Question 1

110

115

120

125

Solution

Question 2

$$200$$

$$240$$

$$280$$

$$320$$

Solution

Question 3

$$112$$

$$128$$

$$96$$

$$24$$

Solution

Question 4

$$5$$

$$15$$

$$20$$

$$25$$

Solution

Question 5

$$320$$

$$160$$

$$80$$

$$200$$

Solution

Question 6

$$0$$

$$450$$

$$1000$$

$$300$$

Solution

Question 7

$$\cfrac { 1 }{ 3 } =\cfrac { 2 }{ 4 } $$

$$\cfrac { 1 }{ 4 } >\cfrac { 1 }{ 3 } $$

$$\cfrac { 2 }{ 3 } <\cfrac { 2 }{ 4 } $$

$$\cfrac { 2 }{ 4 } <\cfrac { 2 }{ 3 } $$

Solution

Question 8

$$240\times 6$$

$$240\div 6$$

$$240-6$$

$$240+6$$

Solution

Question 9

$$350$$

$$248$$

$$256$$

$$311$$

$$413$$

Solution

Question 10

$$7$$

$$4$$

$$10$$

$$8$$

$$6$$

Solution

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