Quantitative Aptitude for Competitive Exam

It is given that;

The total amount $=\mathrm{Rs}. 172850$

The no. of people $=25$

The amount received by each person $=\mathrm{Rs}. \frac{172850}{25}$

$=\mathrm{Rs}. 6914$

Let smallest even number be $A$.

According to the question,

$B=A+2,C=A+4$ and $D=A+6$

Then,

$A+A+2+A+4+A+6=180$

$$\begin{gathered} \Rightarrow 4A+12=180 \\ \Rightarrow 4(A+3)=180 \end{gathered}$$

$\Rightarrow A+3=\frac{180}{4}$

$$\begin{gathered} \Rightarrow A=45-3 \\ \therefore A=42 \end{gathered}$$

Therefore, our numbers are $42, 44, 46$ and $48$.

So, next $4$ even numbers are $50, 52, 54$ and $56$.

The required sum of next $4$ even numbers$=50+52+54+56=212$

Hence, $212$ is the correct answer.

Let, the unit's digit $=y$

And, ten's digit $=x$

Then, the number $=10x+y$

When interchange the digits,

The new number $=10y+x$

According to the question,

$(10x+y)-(10y+x)=18$

$\Rightarrow 10x+y-10y-x=18$

$\Rightarrow 9x-9y=18$

$=x-y=2$ ... (i)

Also, $x+y=12$ ... (ii)

On adding Eqs. (i) and (ii), we get

$x-y=2$

$\Rightarrow x+y=12$

$\Rightarrow 2x=14$

$\Rightarrow x=7$

Therefore, $x=7$ and $y=5$

Product $=7\times 5=35$

It is given that;

Time taken by Monica $=48 \sec$

Time taken by Veronica $=64 \sec$

Time taken by Rachael $=72 \sec$

Now, the required time$=$ LCM $\left(48, 64, 72\right)$

Thus, LCM $\left(48, 64, 72\right)=576 \sec$

Let the two consecutive odd numbers are $x$ and $\left(x+2\right)$.

According to the question,

$x(x+2)=19043$

$\Rightarrow x^2+2x-19043=0$

$\Rightarrow x^2+139x-137x-19043=0$

$\Rightarrow x(x+139)-137(x+139)=0$

$\Rightarrow (x+139)(x-137)=0$

$\Rightarrow x=137, -139$

$\Rightarrow x=137$

Thus, the smallest one is $137$.

The given expression can be simplified as:

$?=56.6\times 16.6\times 6.6$

$?=939.56\times 6.6$

$?=6201.096$

Hence, none of these is the correct answer.

The given expression can be simplified as:

$?=\frac{1}{4}$ of $\frac{1}{2}$ of $\frac{3}{4}$ of $52000$

$?=\frac{1}{4}\times \frac{1}{2}\times \frac{3}{4}\times 52000$

$?=\frac{156000}{32}=4875$

The given expression can be simplified as:

$?=999.99+99.99+9.99$

$?=1099.98+9.99$

$?=1109.97$

Hence, none of these is the correct answer.

The given question is,

$11.6\times ?=899$

On simplifying we get,

$$\begin{gathered} \frac{116}{10}\times ?=899 \\ \Rightarrow ?=\frac{899\times 10}{116} \\ \Rightarrow ?=7.75\times 10 \\ \Rightarrow ?=77.5 \end{gathered}$$

Thus, the correct answer is $77.5$ 

The given expression can be simplified as:

$\Rightarrow 85332-11638-60994$

$\Rightarrow 85332-\left(11638+60994\right)$

$\Rightarrow 85332-72632$

$\Rightarrow 12700$

Thus, the correct answer is $12700$

Total number of employees $=1225$

$\%$ of employee got transferred to different places$=40\%$

Number of employees who got transferred$=40\% of$Total number of employees  in an organization

$$\begin{gathered} =40\% of 1225 \\ =\frac{40}{100}\times 1225 \\ =\frac{2}{5}\times 1225 \\ =2\times 245 \\ =490 \end{gathered}$$

Hence, total $490$ employees got transfered  to different places.

Let the original fraction be $\frac{x}{y}$.

The numerator is increased by $500\%$ and the denominator is increased by $300\%$.

Increased numerator $=x+5x=6x$

Increased denominator $=y+3y=4y$

According to the question,

$\frac{6x}{4y}=2\frac{4}{7}$

$\Rightarrow \frac{6x}{4y}=\frac{18}{7}$

$$\begin{gathered} \Rightarrow \frac{x}{y}=\frac{72}{42} \\ \therefore \frac{x}{y}=\frac{12}{7} \end{gathered}$$

Hence, the original fraction is $\frac{12}{7}$.

We have to calculate $25\%$ of $50\%$ of $\frac{2}{3}$rds of $630$.

$25\%\times 50\%\times$($\frac{2}{3}$of $630$)

$=\frac{25}{100}\times \frac{50}{100}\times \left(\frac{2}{3}\times 630\right)$

$=\frac{1}{4}\times \frac{1}{2}\times 420$

$=\frac{420}{8}=52.5$

Let the income of Shilpa be $x$.

Amount spent on school fees $=\frac{8x}{100}$

Amount spent on rent $=\frac{25x}{100}$

Amount spent on furniture $=\frac{17x}{100}$

Total amount spent on school fees, rent, and furniture $=\left(\frac{8x}{100}+\frac{25x}{100}+\frac{17x}{100}\right)=\frac{50x}{100}=\frac{x}{2}$

Remaining amount $=\left(x-\frac{x}{2}\right)=\frac{x}{2}$

Amount spent on medical bills $=\left(\frac{25}{100}\times \frac{x}{2}\right)=\frac{x}{8}$

Amount saved for investment $=$ Rs. $6000$

Now, $\frac{x}{2}+\frac{x}{8}+6000=x$

$\Rightarrow \frac{5x}{8}+6000=x$

$\Rightarrow x-\frac{5x}{8}=6000$

$\Rightarrow \frac{3x}{8}=6000$

$\Rightarrow x=16000$

Therefore, the total income of Shilpa is Rs.$16000$.

Amount spent on rent $=25\%$ of $16000$

$=\frac{1}{4}\times 16000=4000$

Hence, the amount she spent on rent is Rs. $4000$.

The difference between $89\%$ and $73\%$ of a given number will be equal to $(89-73)\%=16\%$ of the number.

Let the number be $x$.

Then,

$16\%$ of $x=448$

$\Rightarrow \frac{16x}{100}=448$

$\Rightarrow x=\frac{448\times 100}{16}=2800$

Now,

$49\%$ of $2800=\frac{49}{100}\times 2800$

$=49\times 28=1372$

Let the principal amount be $= Rs.x$

Then, interest will be $=Rs.\left(19050-x\right)$

We know that,

$P=\frac{SI\times 100}{R\times T}$

$x=\frac{\left(19050-x\right)\times 100}{3\times 9}$

$27x=\left(19050-x\right)\times 100$

$\frac{27x}{100}=\left(19050-x\right)$

$x+0.27x=19050$

$1.27x=19050$

$x=\frac{19050}{1.27}= Rs.15000$

Hence, the principal value is $Rs.15000$.

Let us assume that principal amount is $₹x$

We know that,

$$$\mathrm{Compound} \mathrm{Interest}=\mathrm{Principal}\left[{\left(1+\frac{\mathrm{Rate}}{100}\right)}^{\mathrm{Time}}-1\right]$

Substitute the values in the above formula, we get,

$$\begin{gathered} 1414.4=x\left[{\left(1+\frac{8}{100}\right)}^2-1\right] \\ 1414.4=x\left[{\left(\frac{108}{100}\right)}^2-1\right] \\ 1414.4=x\left[\left(\frac{11664}{10000}\right)-1\right] \\ 1414.4=x\left[\frac{11664-10000}{10000}\right] \\ x=\frac{1414.4\times 10000}{1664} \\ x=₹8500 \end{gathered}$$

Now, we know that,

Amount$=$Principal$+$Compound Interest

$$\begin{gathered} =8500+1414.40 \\ =₹9914.4 \end{gathered}$$

Hence, the amount that Suresh got at the end of two years was $₹9914.4$.

We know that,

$Rate=\frac{SI\times 100}{P\times T}$

So,

$Rate=\frac{0.40\times 100}{1\times 4}$

$Rate=0.10\times 100=10\%$

Now interest on $Rs.450$,

$SI=\frac{P\times R\times T}{100}$

$=\frac{450\times 10\times 2}{100}$

$=45\times 2= Rs.90$

We know that

$SI=\frac{P\times R\times T}{100}$

$SI=\frac{31400\times 8\times 12}{100}$

$=314\times 96$

$=Rs.30144$

Now,

$Amount=Principal+SI$

$=Rs.\left(31400+30144\right)$

$=Rs.61544$

We know that

$CI=P\left[{\left(1+\frac{r}{100}\right)}^n-1\right]$

$CI=15800\left[{\left(1+\frac{6}{100}\right)}^2-1\right]$

$=15800\left[{\left(\frac{106}{100}\right)}^2-1\right]$

$=15800\left[\left(1.06\times 1.06\right)-1\right]$

$=15800\left[1.1236-1\right]$

$=15800\times 0.1236= Rs. 1952.88$

A man bought a horse and a carriage for $\mathrm{Rs}. 3000$.

Let us assume the cost price of horse $=x$

Then, the cost price of the carriage $=\mathrm{Rs}. \left(3000-x\right)$

According to the question,

$20\%$ of $x-10\%$ of $\left(3000-x\right)=2\%$ of $3000$

$\frac{20}{100}\times x-\frac{10}{100}\times \left(3000-x\right)=\frac{2}{100}\times 3000$

$\frac{x}{5}-\frac{\left(3000-x\right)}{10}=60$

$\frac{2x-3000+x}{10}=60$

$3x-3000=60\times 10$

$3x=600+3000$

$3x=3600$

$x=1200$

Therefore, the cost of horse is $\mathrm{Rs}. 1200$.

The selling price of DVD $=Rs. 10980$

Profit percent $=22\%$

We know that,

$CP=SP\times \left(\frac{100}{100+P\%}\right)$

Therefore, the required $CP$

$=10980\times \frac{100}{100+22}$

$=10980\times \frac{100}{122}$

$=\frac{1098000}{122}=Rs. 9000$

Selling price for school books $=₹4544$

The owner of the shop charges $28\%$ more than $CP$ means it is the profit per cent he earned.

$CP=SP\left(\frac{100}{100+P\%}\right)$

Then, the cost price of books

$=4544\left(\frac{100}{100+28}\right)$

$=4544\times \frac{100}{128}$

$=₹3550$

Thus, $₹3550$ is the correct answer.

The selling price of $12$ toys $=Rs. 300$

And, profit percent on $12$ toys $=25\%$

We know that,

$CP=SP\left(\frac{100}{100+P\%}\right)$

So, the cost price of $12$ toys is

$=300\left(\frac{100}{100+25}\right)$

$=300\times \frac{100}{125}$

$=Rs. 240$

Hence, the cost of one toy $=\frac{240}{12}=Rs. 20$

Selling price of television set $=₹11500$

The owner of shop charges $25\%$ more, then it is profit for the owner of shop.

Therefore, the cost price of television is-

$CP=SP\left(\frac{100}{100+P\%}\right)$

$CP=11500\left(\frac{100}{100+25}\right)$

$CP=11500\times \frac{100}{125}$

$CP=₹9200$

Let the third number be $=x$

we know that

Average $=\frac{Sum of terms}{Number of terms}$

$281=\frac{2\times 280+x+2\times 178.5}{5}$

$281\times 5=560+x+357$

$x=1405-917$

$x=488$

Let the common ratio be $=x$

Then son and mother age will be $2x$ and $7x$ respectively.

As per question

$2x+7x=2\times 27$

$9x=54$

$x=\frac{54}{9}=6$

The present age of mother $=7x=7\times 6=42$ years

The present age of son $=2x=2\times 6=12$ years

Mother age after $7$ years

$=7x+7$

$=7\times 6+7$

$=42+7=49$

We know that

Average $=\frac{Sum of terms}{Number of terms}$

Required average

$=\frac{965+362+189+248+461+825+524+234}{8}$

$=\frac{3808}{8}=476$

Given,

$21a+21b=1134$

$21(a+b)=1134$

$a+b=\frac{1134}{21}$

$a+b=54$

Required average $=\frac{Sum of terms}{number of terms}$

$=\frac{54}{2}=27$

Let the first number be $=6x$

then the Second number will be $=3x$

and the third number will be $=2x$

As per the question,

$6x+3x+2x=154\times 3$

$11x=154\times 3$

$x=\frac{154\times 3}{11}$

$x=42$

Required difference $=6x-2x=4x$

$=4\times 42=168$

It is given that;

Total number of students in the school $=819$

Number of girls $=364$

Therefore, number of boys $=819-364=455$

Thus, the required ratio $=\frac{455}{364}=5:4$

The dividend to be divided $=₹57834$ and their ratio is $3:2:1$

Therefore,

Meena's share $=\frac{3}{3+2+1}\times 57834=₹28917$

Urmila's share $=\frac{2}{3+2+1}\times 57834=₹19278$

Vaishali's share $=\frac{1}{3+2+1}\times 57834=₹9639$

Hence, $₹19278$ is the correct answer.

We assume that,

The another sum of money that is to be divided between $A$ and $B$ be $\mathrm{Rs}. Q$.

According to the question,

$A+B=Q$ ... $\left(1\right)$

It is given that,

$A=B$ ... $\left(2\right)$

$Z=A-2000$ ... $\left(3\right)$

Putting the value of $A$ in equation $\left(1\right)$,

$A+B=Q$

$B+B=Q$

$B=\frac{Q}{2}, A=\frac{Q}{2}$

Therefore,

$Z=\frac{Q}{2}-2000$

Again from the question;

$Z : X : Y$ is in proportion of $4 : 5 : 6$.

Therefore, the part of $Z=4p$

$\Rightarrow 4p=\frac{Q}{2}-2000$

$\Rightarrow p=\frac{Q}{8}-500$

Hence, the amount received by $X=5p$

$5p=5\times \left(\frac{Q}{8}-500\right)$

The amount of $Q$ is unknown, so we can't determine the answer.

We assume that;

The present age of Sushma$=6\mathrm{x}$

The present age of Karishma$=7\mathrm{x}$

According to the question,

$\frac{6\mathrm{x}+8}{7\mathrm{x}+8}=\frac{8}{9}$

$\Rightarrow 9\left(6\mathrm{x}+8\right)=8\left(7\mathrm{x}+8\right)$

$\Rightarrow 54\mathrm{x}+72=56\mathrm{x}+64$

$\Rightarrow 2\mathrm{x}=8$

$\Rightarrow \mathrm{x}=4$

The required ratio$=\frac{\left(6\times 4\right)+12}{\left(7\times 4\right)+12}$

$\Rightarrow \frac{36}{40}=9:10$

Let the number of girls be $\mathrm{x}$.

According to the question,

No. of boys$=\mathrm{x}+16\%$ of$x$

$\Rightarrow \mathrm{x}+\left(\mathrm{x}\times \frac{16}{100}\right)=\frac{29}{25}\mathrm{x}$

Now, the required ratio$=\frac{\left({\displaystyle \frac{29}{25}}\mathrm{x}\right)}{\mathrm{x}}=29:25$

Work $=$ Time $\times$ Efficiency

Let us assume, work$=1$

A alone can complete the work $=6$ days 

A's efficiency $=\frac{1}{6}$

B alone can complete the work$=12$ days

B's efficiency$=\frac{1}{12}$

When both A and B work together, then the efficiency of (A and B)$=\frac{1}{6}+\frac{1}{12}=\frac{2+1}{12}=\frac{1}{4}$

Time is taken to complete the work$=\frac{1}{{\displaystyle \frac{1}{4}}}=4$ days

Hence, A and B together completed the work in $4$ days.

Use formula: $M_1\times D_1=M_2\times D_2$

It is given that $16$ men finish a work in $8$ days.

Total work $= 16 \times 8$

Number of days it will take for $12$ men to finish the work

$=\frac{16\times 8}{12}$

$=\frac{32}{3}=10\frac{2}{3}$ days

A alone completes the work $=12$ days 

A's one day work $=\frac{1}{12}$

(A $+$ B) can complete the work $=8$ days

(A $+$ B)'s one day work $=\frac{1}{8}$

Let us suppose, B can complete the work in $x$ days.

Then, B's one day work $=\frac{1}{x}$

Thus, we get,

$\frac{1}{8}=\frac{1}{12}+\frac{1}{x}$

$\Rightarrow \frac{{\displaystyle 1}}{{\displaystyle x}}=\frac{{\displaystyle 1}}{{\displaystyle 8}}-\frac{1}{12}$

$\Rightarrow \frac{{\displaystyle 1}}{{\displaystyle x}}=\frac{3-2}{24}$

$\Rightarrow \frac{{\displaystyle 1}}{{\displaystyle x}}=\frac{1}{24}$

Thus, B alone can complete the work in $\frac{1}{{\displaystyle \frac{1}{24}}}=24$ days

Hence, $24$ days is the correct answer.

Use formula: $\frac{M_1\times D_1}{W_1}=\frac{M_2\times D_2}{W_2}$

It is given that $12$ men finish one-third of work in $8$ days.

Thus, they can complete the total work in $8\times 3=24$ days.

Number of days $16$ men take to finish the work

$=\frac{12\times 24}{16}$

$=18$

Let us assume, the work $=1$

A alone can complete the work $=6$ days 

A's one day work $=\frac{1}{6}$

B alone can complete the work $=12$ days

B's one day work $=\frac{1}{12}$

When both A and B work together, then efficiency of (A $+$ B)

$=\frac{1}{6}+\frac{1}{12}$

$=\frac{2+1}{12}$

$=\frac{1}{4}$

Work is completed $=\frac{1}{{\displaystyle \frac{1}{4}}}=4$ days

Therefore, 'none of these' is the correct answer.

We know that

$Speed=\frac{dis\tan ce}{time}$

Speed of car$=\frac{520}{8}=65$ $kmph$

Then the speed of the train in $kmph$

$=3\times speed of car$

$=3\times 65=195$

Distance cover by train in $13$ hour in $km$$=13\times 195$

$=2535$

We know that

$Speed=\frac{dis\tan ce}{time}$

Then,

$$Speed of the train in $kmph$

$=\frac{1560}{26}=60$

Time taken by A to complete round$=252$ seconds

Time taken by B to complete round$=308$ seconds

Time taken by C to complete round$=198$ seconds

They meet again at the starting point after time

$=$ LCM of $252, 308, 198=2772$ seconds

Time in minute$=\frac{2772}{60}$$=46$ minutes $12$ seconds

We know that,

$Time=\frac{dis\tan ce}{speed}$

Time taken by car to travels $45$ $kms$ by $15$ $kmph$$=\frac{45}{15}=3$ hours

Time taken by car to travels $50$ $kms$ by $25$ $kmph$$=\frac{50}{25}=2$ hours

Time taken by car to travels $25$ $kms$ by$10$ $kmph$$=\frac{25}{10}=2.5$ hours

Average speed$=\frac{Total dis\tan ce}{Total time}$

Average speed in $kmph$

$=\frac{(45+50+25)}{(3+2+2.5)}$

$=\frac{120}{7.5}=16$

We know that,

$Time=\frac{Dis\tan ce}{Speed}$

Let the distance between the village and the school be $=x$ $km$

As per the question,

$\frac{x}{4}+\frac{x}{4}=6$

$\frac{2x}{4}=6$

$x=6\times 2=12$

The distance between the village and the school is $12$ $km$.

Let the breadth of rectangular plot be$=x$ $m$

then the length will be$=3x$ $m$

$Area of the rec\tan gle=Length\times Breadth$

Area of a rectangular plot

$=x\times 3x$

$7803=3x^2$

$x^2=\frac{7803}{3}=2601$

$x=\sqrt{2601}=\sqrt{51\times 51}$

$x=51$

Thus, the breadth is $51$ $m$.