General Test

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General Test - 26

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Weekly Quiz Competition

Question 1
4 / -1

A train covers a distance of 20 km in 15 minutes. If its speed is increased by 10 km/hr, the time taken by it to cover the same distance will be:

A
13 minutes 20 seconds

B
12 minutes

C
11 minutes 10 seconds

D
10 minutes

Solution

Distance = 20km

Time = 15 minutes = $\frac{1}{4}$ hr

∴ Speed of the train = $\frac{Distance}{Time}$

= $\frac{20}{(\frac{1}{4})}$ = 80 km/h

After increasing its speed by 10 km/hr,

New speed = 90 km/hr

Distance = 20 km (given)

∴ Time = $\frac{Distance}{Speed}$

∴ Time = $\frac{20}{90}$ = $\frac{2}{9}$ th of an hour = 60 × $\frac{2}{9}$

Thus, the time taken by the train will be 13 minutes and 20 seconds.
Question 2
4 / -1

The speed of Ram and Shaym are in the ratio 7 ∶ 8. They both cover a certain distance but Ram takes 45 minutes more than Shyam. Find the time taken by Ram to cover the distance?

A
6 hours

B
7 hours

C
8 hours

D
9 hours

Solution

Given:

The speed of Ram and Shaym are in the ratio 7 ∶ 8.

Ram takes 45 minutes more than Shyam to cover the certain distance.

Speed = $\frac{Distance}{Time}$

Time = $\frac{Distance}{S p e e d}$

Let the distance be D.

Let the speed of Ram be 7x and the speed of Shaym be 8x.

Let Ram takes t₁ time to cover the distance and Shyam takes t₂ time to cover the distance.

Time taken by Ram = t₁ = $\frac{D}{7 x}$

Time taken by Shyam = t₂ = $\frac{D}{8 x}$

Now, Ram takes 45 min more than Shyam.

∴ t₁ – t₂ = $(\frac{D}{7 x}) - (\frac{D}{8 x}) = \frac{45}{60}$

⇒ D = 42x

Now, Time taken by Ram = $\frac{42 x}{7 x}$ = 6 hours.

∴ Ram will take 6 hours to cover the dame distance.
Question 3
4 / -1

Two trains of lengths 160 m and 200 m travel at speeds of 48 m/sec and 52 m/sec respectively in opposite directions to each other. What is the total time taken by them to cross each other?

A
3.6 sec

B
4 sec

C
5.2 sec

D
6.8 sec

Solution

Relative speed $=48+52=100 $ m/sec

Total distance covered by both the trains $=160+200=360 $ m

Speed $=\frac{\text { Distance }}{\text { Time }}$

$\therefore 100=\frac{360}{\text { Time }}$

$\therefore$ Time $=3.6$ sec
Question 4
4 / -1

If the roots of $ax^2+ bx + c =0$ are in the ratio $m : n$, then:

A
$m n a^{2}=(m+n) c^{2}$

B
$m n b^{2}=(m+n) a c$

C
$m n b^{2}=(m+n)^{2} a c$

D
$m n b^{2}=(m-n)^{2} a c$

Solution

We have$\frac{\alpha}{\beta}=\frac{m}{n}$

$\Rightarrow \frac{\alpha}{m}=\frac{\beta}{n}$

$ \Rightarrow \frac{\alpha+\beta}{m+n}=\sqrt{\frac{\alpha \beta}{m n}}$ by ratio proportion

$\therefore m n(\alpha+\beta)^{2}=\alpha \beta(m+n)^{2}$

$\Rightarrow m n\left(\frac{-b}{a}\right)^{2}=(m+n)^{2} \frac{c}{a}$

$\therefore m n b^{2}=(m+n)^{2} a c$
Question 5
4 / -1

If $24$ is $\frac{2}{3}$ of $\frac{3}{4}$ of a number, then what will be $\frac{1}{4}$ of that number?

A
$8$

B
$12$

C
$10$

D
$32$

Solution

Let the number be $x$.

Accordimng to the question,
$\frac{2}{3} \times \frac{3}{4} \times x=24$
$\Rightarrow \frac{1}{2} \times x=24$
$\Rightarrow x=48$
So,$\frac{1}{4}$ of that number $=\frac{1}{4} \times 48=12$
Question 6
4 / -1

A man bought $13$ articles at Rs. $70$ each,$15$ at Rs.$60$ each and$12$ at Rs.$65$ each. The average price per article is:

A
Rs. $60.24$

B
Rs. $64.24$

C
Rs. $64.75$

D
Rs. $62.25$

Solution

Cost of $13$ articles $=13 \times 70=$ Rs. $910$

Cost of $15$ articles $=15 \times 60=$ Rs. $900$

Cost of $12$ articles $12 \times 65=780$

Average of each article $= \frac{910+900+780}{40}= \frac{2590}{40}=$ Rs. $64.75$
Question 7
4 / -1

What is not the value of 5 $\div$ ( -1) between?

A
0 and -10

B
0 and 10

C
-4 and -15

D
-6 and 6

Solution

5 $\div$ ( -1) = -5

It is placed on the number line as

Now, as we can see, -5 lies between (0 and -10), (-4 and -15) and (-6 and 6). But it does not lie between 0 and 10.
Question 8
4 / -1

A vessel of 80 litres is filled with milk and water. 70% of milk and 30% of water is taken out of the vessel. It is found that the vessel is vacated by 55%. Find the initial quantity of milk and water.

A
50, 20

B
50, 30

C
30, 10

D
30, 20

Solution

Let the initial quantity of milk be x L

Since The total mixture is 80 L so the initial quantity of water be (80 - x) L

⇒ 70% of x + 30% of (80 - x) = 55% of 80⇒0.7x + 0.3 × (80 - x) = 0.55 × 80

⇒ x = 50

Therefore, Initial quantity of milk = 50 L and the initial quantity of water = 30 L
Question 9
4 / -1

Two letters are chosen randomly from English alphabet. What is the probability that none of them is a vowel and they are consecutive letters?

A
$\frac{8}{325}$

B
$\frac{12}{325}$

C
$\frac{16}{325}$

D
$\frac{4}{325}$

Solution

Any two letters can be chosen in $=\frac{(26 \times 25)}{2}=325$ ways

If they are consecutive, they can be $A B, B C, C D, \ldots Y Z$, a total of $25$ cases

Out of these, $AB, DE, EF, HI, IJ, NO, OP, TU$ and $UV$ will have vowels

$\therefore$ Required probability $=\frac{(25-9)}{325}=\frac{16}{325}$
Question 10
4 / -1

The sides of a triangle are $m, n$ and $\sqrt{m^{2}+n^{2}+m n}$. What is the sum of the acute angles of the triangle?

A
$45^{\circ}$

B
$60^{\circ}$

C
$75^{\circ}$

D
$90^{\circ}$

Solution

Let $m=n=1$ unit

Then $\sqrt{ m ^{2}+ n ^{2}+ mn }=\sqrt{3}$ unit

Using cosine rule;

$\cos\theta=\frac{a^{2}+b^{2}-c^{2}}{2ab}$

After putting the value,

$\cos \theta=\frac{1^{2}+1^{2}-\sqrt{3}^{2}}{2 \times 1 \times 1}$

$\Rightarrow \cos\theta=\frac{1+1-3}{2\times1\times1}$

$\Rightarrow \cos \theta=-\frac{1} {2}$

$\therefore \theta=120^{\circ}$ $\left[\because \cos 120^{\circ}=-\frac{1}{2}\right]$

Now, the sum of the acute angles of the triangle $=180^{\circ}-120^{\circ}$

$=60^{\circ}$

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General Test - 26

Result

General Test - 26

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SHARING IS CARING

Question 1

13 minutes 20 seconds

12 minutes

11 minutes 10 seconds

10 minutes

Solution

Question 2

6 hours

7 hours

8 hours

9 hours

Solution

Question 3

3.6 sec

4 sec

5.2 sec

6.8 sec

Solution

Question 4

\(m n a^{2}=(m+n) c^{2}\)

\(m n b^{2}=(m+n) a c\)

\(m n b^{2}=(m+n)^{2} a c\)

\(m n b^{2}=(m-n)^{2} a c\)

Solution

Question 5

\(8\)

\(12\)

\(10\)

\(32\)

Solution

Question 6

Rs. \(60.24\)

Rs. \(64.24\)

Rs. \(64.75\)

Rs. \(62.25\)

Solution

Question 7

0 and -10

0 and 10

-4 and -15

-6 and 6

Solution

Question 8

50, 20

50, 30

30, 10

30, 20

Solution

Question 9

\(\frac{8}{325}\)

\(\frac{12}{325}\)

\(\frac{16}{325}\)

\(\frac{4}{325}\)

Solution

Question 10

\(45^{\circ}\)

\(60^{\circ}\)

\(75^{\circ}\)

\(90^{\circ}\)

Solution

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