test: testing Latex 3 - Software Development MCQ

Explanation:

Ratio of their daily wages = 5 : 6 : 4

Total wage = Rs. 1800

6 × 5x + 4 × 6x + 4 × 9x = 1800 [As A, B, C works for 6, 4, 9 days respectively]

Or, 90x = 1800

$x=\frac{1800}{90}$

Total amount of A = $\frac{1800\times 30}{90}$

 = Rs. 600

Explanation:

Given:

$\frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{5}}}}$

$=\frac{1}{2+\frac{1}{3+\frac{1}{\frac{21}{5}}}}$

$=\frac{1}{2+\frac{1}{3+\frac{5}{21}}}$

$=\frac{1}{2+\frac{1}{\frac{68}{21}}}$

$=\frac{1}{2+\frac{21}{68}}$

$=\frac{1}{\frac{157}{68}}$

$=\frac{68}{157}$

Explanation:

Let C.P. be Rs. $x$.

Then, 

$\frac{1920-x}{x}\times 100=\frac{x-1280}{x}\times 100$

$\Rightarrow 1920-x=x-1280$

$\Rightarrow 2x=3200$

$\Rightarrow x=1600$

$\therefore$ Required S.P. $=125\mathrm{\%}$ of Rs. $1600$

$=$ Rs. $(\frac{125}{100}\times 1600)$

= Rs. 2000

Explanation:

Speed $=S=\frac{D}{T}$

Average Speed $=\frac{\text{ Total distance travelled }}{\text{ Total time taken }}$

Total distance $=96+124+105=325km$

Total Time $=t1+t2+t3$

$=\frac{d1}{s1}+\frac{d2}{s2}+\frac{d3}{s3}$

$=\frac{96}{16}+\frac{124}{31}+\frac{105}{7}$

$=25$ hours

$\therefore$ Average speed $=\frac{325}{25}$

= 13 km/hr

Explanation:

R cycles $\frac{1}{3}$rd as fast as Q.

$\therefore$ Speed of R = $\frac{1}{3}$rd of Speed of Q.

$\therefore$ R takes 3 times more time than Q.

R needs 45 minutes to finish the race so Q takes $\frac{45}{3}=15$ minutes to finish the race.

P is twice as fast as Q.

So, P takes half the time as Q =$\frac{15}{2}=7.5$ minutes.

 This means that when P reaches the finish line, Q is only half way.

So, Q is at half distance from finish line i.e., $\frac{4500}{2}$

= 2250 m

Explanation:

Let income of A and B be 3x and 2x respectively. 

Also, their expenditure is 5y and 3y.

Now, according to question,

3x - 5y = 1000 ------- (i) × 3

2x - 3y = 1000 ---------- (ii) × 5

9x - 15y - 10x + 15y = 3000 - 5000

Or, -x = -2000

x = 2000

Then, 

Income of A = 3x 

= 3 × 2000 

= Rs. 6000

Explanation:

LCM of 36, 42 and 72,

36 = 2 × 2 × 3 × 3

42 = 2 × 3 × 7

72 = 2 × 2 × 2 × 3 × 3

LCM = 2 ×2 × 2 × 3 × 3 × 7 = 504 seconds.

LCM of 36, 42 and 72 is 504

So, the lights will change simultaneously after 8 minutes and 24 seconds.

Explanation:

Let S be the sample space.

Then,

$n(S)={}^{52}{C}_2=\frac{(52\times 51)}{(2\times 1)}=1326$

Let E = event of getting 2 kings out of 4.

$\therefore n(E)={}^4{C}_2=\frac{(4\times 3)}{(2\times 1)}=6$

$\therefore P(E)=\frac{n(E)}{n(S)}=\frac{6}{1326}$

$=\frac{1}{221}$

Explanation:

Vimal's 1 day work $=\frac{1}{20}$

Since, Vimal and Kamal can together complete in 12 days.

i.e. (Vimal + Kamal)'s 1 day work $=\frac{1}{12}$

Then,

Kamal's 1 day work,

$=\frac{1}{12}-\frac{1}{20}$

$\Rightarrow \frac{2}{60}$ 

$\Rightarrow \frac{1}{30}$

If Kamal Works only for half a day daily, then his 1 day work becomes $\frac{1}{2}\times \frac{1}{30}=\frac{1}{60}$

Therefore, 1 day work of both Vimal and Kamal,

$=\frac{1}{20}+\frac{1}{60}$ 

$\Rightarrow \frac{4}{60}$ 

$\Rightarrow \frac{1}{15}$

So, the work will be completed in 15 days.

Explanation:

The possible cases can be:

1. 4 boys and 4 girls
2. 3 boys and 3 girls
3. 2 boys and 2 girls

Ways of selecting of 4 boys out of $4={}^4{C}_4=1$

Ways of selecting of 3 boys out of $4={}^4{C}_3=4$

Ways of selecting of 2 boys out of $4={}^4{C}_2=6$

Ways of selecting of 4 girls out of $6={}^6{C}_4=15$

Ways of selecting of 3 girls out of $6={}^6{C}_3=20$

Ways of selecting of 2 girls out of $6={}^6{C}_2=15$

$\therefore$ Total possible ways $N=(1\times 15)+(4\times 20)+(6\times 15)$

$N=15+80+90=185$

Explanation:

Let P, Q and R represent their respective monthly incomes. 

Then, we have:

P + Q = (5050 x 2) = 10100 .... (i)

Q + R = (6250 x 2) = 12500 .... (ii)

P + R = (5200 x 2) = 10400 .... (iii)

Adding (i), (ii) and (iii), we get:  

2(P + Q + R) = 33000  or   P + Q + R = 16500 .... (iv)

Subtracting (ii) from (iv), we get 

P = 4000.

$\therefore$ P's monthly income = Rs. 4000.

Explanation:

Cost Price of 1 toy = Rs. $\left(\frac{375}{12}\right)$

= Rs. 31.25

Selling Price of 1 toy = Rs. 33

So, Gain = Rs. (33 - 31.25) 

= Rs. 1.75

$\therefore$ Profit $\mathrm{\%}=(\frac{1.75}{31.25}\times 100)\mathrm{\%}$

$=\frac{28}{5}\mathrm{\%}$

= 5.6%

Explanation:

Let the side of cube = 10 cm

∴ Original volume = 10 × 10 × 10 = 1000 cm³

Now, side of new cube: 

= 10 - 25% of 10 = 7.5

∴ New volume = 7.5 × 7.5 × 7.5 = 421.875 cm³

∴ Required ratio:

$=\frac{1000}{421.875}$

$=\frac{1000000}{421875}$

$=\frac{64}{27}$

$=64:27$

Explanation:

$BO=\mathrm{OC}=15cm$

$OD=9cm$

$BD=\sqrt{{15}^2-9^2}$

$=\sqrt{24\times 6}$

$=12cm$

$BC=2\times 12$

= 24 cm

Explanation:

External radius = 3 cm

Internal radius = (3 - 0.5) cm 

= 2.5 cm

Volume of the metal:

$=[\frac{4}{3}\times \frac{22}{7}\times \{(3{)}^3-(2.5{)}^3\}]{cm}^3$

$=(\frac{4}{3}\times \frac{22}{7}\times \frac{91}{8}){cm}^3$

$=\left(\frac{143}{3}\right){\mathrm{cm}}^3$

$=47\frac{2}{3}{cm}^3$