 Insights after the analysis of Group 4 Aptitude Questions of 2019

• Simple Interest(2)
• Simplification(5)
• L.C.M (2)
• Cube Root
• Mean & Average(2)
• A.P & G.P(2)
• Time and Work(2)
• Area & Volume (3)
• Ratio and Proportion(2)
• Ascending order of Fractional numbers & Expression (2)
• Sum of Natural Numbers
• Standard Deviation

105. Find the simple interest on Rs 7500 at 8% per annum per 1 year 6 months $SI = \frac{PNR}{100}$

P = 7500 , R=8%, $N= 1\frac{1}{2}=\frac{3}{2}$ $SI =\frac{7500\times8\times3}{100\times2}$

SI = 900

Syllabus – Simple Interest

1. #### Cricket player Dhoni’s average in first 30 matches was 72 runs. After 31st match, his average raised as 73 runs. How many runs did he make in 31st match? $Average = \frac{Sum of Runs}{Total Matches}$ $Sum of Runs = 72\times30$ $After 31st Match, Sum of runs = 31 \times 73 = 2263$

Difference between the two = 2263 – 2160 = 103

Syllabus – Simplification

#### 111. Find the LCM of $3(a-1),2(a^{2}-1),(a^{2}-1)$

By factorization method

3 (a-1) = 3 (a-1) $2(a^{2}-1) = 2(a-1)(a-1)$ $(a^{2}-1) = (a-1)(a+1)$

LCM = $6(a-1)^{2}(a+1)$

#### 112. Find the cube root of $1\frac{61}{64}$

Mixed fraction = $\frac{125}{64} = \frac{5}{4}$

Syllabus – Simplification

1. #### The maximum temperature in a city on 7 days of a certain week was 34.8* C, 38.5* C , 33.4* C, 34.7* C, 35.8* C, 32.8* C, 34.3* C. Find the mean temperature for the week.

Sum of maximum temperatures = 244.3

Mean temperature = $\frac{Sum}{Number of Days} = \frac{244.3}{7}$ = 34.9

Syllabus – Simplification

#### 135.  In a Geometric Progression $t_{2}=\frac{3}{5} , t_{3}=\frac{1}{5}$ then the common ration is $\frac{t_{3}}{t_{2}} = \frac{\frac{1}{5}}{\frac{3}{5}}$ –> $\frac{1}{5}\times\frac{5}{3} = \frac{1}{3}$

Syllabus – Simplification

#### 136. If $a_{1}$= -1,  then find $a_{2}, a_{3} in a_{n}= \frac{a_{n-1}}{n+2} , n > 1$ $a_{1}= -1$ –> $a_{2}= \frac{a_{1}}{4}$ –> $a_{3}= \frac{a_{2}}{3+2}= \frac{-\frac{1}{4}}{5} = -\frac{1}{20}$

Syllabus – Simplification

1. #### A can do a piece of work in 20 days and B can do it in 30 days. How long will they take to do the work together?

A’s one day work = $\frac{1}{20}$

B’s one day work = $\frac{1}{30}$

= $\frac{1}{20}+\frac{1}{30} = \frac{3+2}{60} = \frac{5}{60} = \frac{1}{12}$

So, 12 days

Syllabus – Time and work

1. #### If 14 compositors can compose 70 pages of a book in 5 hours, how many compositors will compose 100 pages of this book in 10 hours? $\frac{M_{1}D_{1}T_{1}}{W_{1}}= \frac{M_{2}D_{2}T_{2}}{W_{2}}$ $\frac{14\times5}{70}=\frac{M_{2}\times10}{100}$ $M_{2}=\frac{14\times5\times100}{70\times10}$  = 10

Syllabus – Time and work

1. #### The ratios of the respective heights and the respective radii of two cylinders are 1:2 and 2:1 respectively. Then their respective volumes are in the ratio.

Volume of a cylinder = $\pi\times r^{2}h$ $\frac{h_{1}}{h_{2}}=\frac{1}{2} --> h_{1}=\frac{h_{2}}{2}$ $\frac{r_{1}}{r_{2}}=\frac{2}{1} --> r_{1}=2r_{2}$

Volume = $\frac{r_{1}^{2}\times h_{1}}{r_{2}^{2}\times h_{2}}$

substituting the values $\frac{4r_{2}^{2}\times h_{2}}{2r_{2}^{2}\times h_{2}}$

= 2 : 1

Syllabus – Volume

1. #### At what rate of interest a sum of money doubles itself in 10 years in simple interest?

By fast track technique

If a sum of money becomes n times in T year at simple interest, then formula for calculating rate of interest will be

RT = 100 (n-1)

= $\frac{100(2-1)}{10}$ = 10

Syllabus – Simple Interest

#### 154. The average height of 12 students in a class was calculated as 152 cm. On verification it was found that one reading was wrongly recorded as 148 cm instead of 172 cm. Find the correct mean height.

Average = $\frac{Sum of height}{total students}$

Sum of height = 12 * 152 = 1824

Missed value in sum of height = 172 – 148 = 24

Adding and taking average = $\frac{1824 + 24}{12}$ = 154

Syllabus – Simplification

1. #### The difference of the squares of two positive numbers is 45. The square of the smaller number is 4 times the larger number. Find the numbers.

Let two numbers be x(greater) and y(smaller) $x^{2}-y^{2}=45$ –> 1 $y^{2}=4x$ –> 2

Substituting 2 in 1 $x^{2}-4x=45 --> x^{2}-4x-45 --> x^{2}-9x+4x-45$ –>

x(x – 9) + 5(x – 9) –> (x+5)(x-9) –> Then x = -5,9

x = -5 –> substitute in 2 –> $y = \sqrt{-20}$ (negative so neglected)

x = 9 –> substitute in 2 –> y = 6

Hence the numbers are 9 and 6

Syllabus – Simplification

1. #### The radius of a cart wheel is 35 cm. How many revolutions does it make in travelling a distance of 154 m.

Circumference of wheel = $\frac{Total Distance}{Total Revolutions}$ $2\pi r = \frac{154 \times 100}{Total Revolutions}$ $Total Revolutions = \frac{154 \times 100 \times 7}{2 \times 22}$ = 70

Syllabus – Area

#### 165. Saran is 6 times as old as his son sankar. After 4 years, he will be 4 times as old as his son. What are their present ages?

If t year after the age of one person is n times that of another person and at present, the age of the first person is m times of that of another person, then

Age of first person = $\frac{tm(n-1)}{(m-n)}$

Age of second person = $\frac{t(n-1)}{(m-n)}$

Age of first person = $\frac{4\times6\times(4-1)}{(6-4)} = 36$

Age of Second person = $\frac{4\times(4-1)}{(6-4)} = 6$

Syllabus – Ratio and Proportion

#### 166. Arrange in ascending order $\frac{3}{4}, \frac{1}{2}, \frac{5}{8}$ $\frac{3}{4} = 0.75$ –> $\frac{1}{2} = 0.5$ –> $\frac{5}{8} = 0.625$

so, $\frac{1}{2}, \frac{5}{8} , \frac{3}{4}$

Syllabus – Number series

1. #### In a T-20 cricket match, Raju hit a “Six” 10 times out of 50 balls he played. If a ball was selected at random. Find the probability that he would not have hit a “Six”

Favourable cases = 10

Total outcomes = 50

Probability happening of an event = Number of favourable cases / Total no of possible outcomes

Probability of hitting a six = $\frac{10}{50}= \frac{1}{5}$

Probability of not hitting a six = $1-\frac{1}{5} =\frac{4}{5}$

#### 170. If 1+2+…….+n = K then $1^{3}+2^{3}+----+n^{3}$ is equal to

Formula for sum of n numbers

1+2+3+ …….+ n = $\frac{n(n+1)}{2}$ $1^{3}+2^{3}+----+n^{3} = (\frac{n(n+1)}{2})^{2}$

So 1+2+…….+n = K means $1^{3}+2^{3}+----+n^{3} = k^{2}$

Syllabus – Number series

#### 171. The common ratio of the G.P $a^{m-n}, a^{m}, a^{m+n}$ is

Ratio = $\frac{t_{2}}{t_{1}}$ $\frac{a^{m}}{a^{m-n}} =a^{n}$

Syllabus – Number series

1. #### Express 0.35 into fraction

Pure Recurring Decimal fraction : When all the digits in a decimal fraction are repeated after the decimal point, then the decimal fraction is called pure recurring

To convert pure recurring decimal fractions into simple fractions, write down the repeated digits only once in numerator and place as many nines in the denominator as the number of digits repeating

So, 35 / 99

Syllabus – Number series

1. #### How many numbers are there between 200 and 300 which are exactly divisible by 6, 8 and 9

Take L.C.M of 6,8,9 = 72

So the multiples of 72 between 200 and 300 is 216 and 288. Answer will be 2

Syllabus – LCM

1. #### A group of 100 candidates have their average height 163.8 cm with coefficient of variation 3.2. What is the standard deviation of their heights?

Mean = 163.8 cm

C.V = $\frac{S.D}{Mean}\times 100$ $3.2 = \frac{S.D}{163.8}\times100$

S.D = 5.24

Syllabus – Simplification

1. #### The present ages of Reena and Usha are 24 years and 36 years respectively what was the ratio between the ages of Usha and Reena, 8 years ago? $\frac{Reena}{Usha}=\frac{(24-8)}{(36-8)}=\frac{4}{7}$

Since the ratio asked in Usha and reena, ans is 7/4

Syllabus – Ratio & Proportion

#### 197.  Simplify $\frac{(3^{3})^{-2}\times(2^{3})^{-3}}{(2^{4})^{-2}\times 3^{-4}\times 4^{-2}}$ $\frac{27^{-2}\times 4^{-3}}{16^{-2}\times 3^{-4} \times 4^{-2}}$ –> $\frac{16\times16\times3\times3\times3\times3}{27\times4}$ –> $7\frac{1}{9}$

Syllabus – Simplification

#### 198. The total surface area of cube is 384 $m^{2}$. Find the side of the cube.

TSA = $6a^{2}$ = 384 $a^{2}=64$

a = 8m

Syllabus – Area