A person gives $$47\frac{1}{2}\%$$ of his property to the elder son, $$32\frac{1}{2}\%$$ to the younger son, 5% to a temple and the balance of Rs. 6 lakhs to his daughter. The share of the younger son (in lakhs of rupees) is
TS Police SI Mains 2018 Arithmetic and Reasoning
For the following questions answer them individually
If 25% of a number is added to another number then the second number is increased by 10%. The ratio of the first number to the second is
A Person spends 80% of his monthly salary. His salary was increased by 20%. He increased his expenditure also by 20%. Then, the percentage increase in his savings is
A class has two sections A and B having 60 and 40 students respectively, If 45% of the section A and 55%of the section B pass in an examination then the pass percentage of the two sections together is
To pass an Examination a student has to secure 45% of marks.A boy getting 295 marks fails by 20 marks.The maximum marks for the examination is
In a fraction if the numerator is increased by 23% and the decreased by 32% we get new fraction $$\frac{13}{17}$$.The sum of the new fraction to the original is
The total number of students in a class with 44 girls and the rest 45% boys is
There are 60 students in a class.one of the student weighing 40 Kgs was replaced by another student.now the average weight of the student increased by 0.2 kg. The weight of the new student is (in kgs)
The average of 40 numbers is 35. If two numbers 44,45 are discarded, then the average of the remaining numbers is
There are 4 number. The average of first three numbers is 56 and the average of last three is 49,if the first number is 48 then the last number is
Two persons A and started a company with capitals in the ratio 7 : 8. At the end of the business A and B shared the profit in the ratio 5 : 4. If A kept his capital for 10 months, then the number of months that B kept his capital is
In a Partnership A,B,C invest capitals in the ratio 3:5:7.If at the end of the year the total profit is Rs.90,000 then difference between the profits of A and c in rupees is
A,B,c started a business with the capitals in the ratio 4:5:6 which are kept for 16 months,12 months,8 months respectively.At the end of the business if B shares in the profit is Rs.1,35,000 , then the total profit in rupees is
A,B enter a business with capitals Rs. 3 lakhs and Rs. 4 lakhs respectively, At the end of the year A receives monthly salary for looking after accounts. At the end of the year A and B receive their shares respectively equal to the amounts when the total profit of Rs. 1,96000 is divided in the ratio 4:3. Then A’s monthly salary in rupees,is
Two persons A and B enter in to a business with the capitals in the ratio 4:5.At the end of 5 months A with draw from the Business.At the end of business if they shared their profits in the ratio 1:6, then the number of months B's capital used in the Business is
A certain Amount is divided among three persons A,B,C. A gets $$\frac{1}{6}$$ of the amount and b gets $$\frac{1}{6}$$ of the amount and c gets remaining amount . If C gets 2800/- then A gets
$$\sqrt[3]{\sqrt{0.000729}} = $$
The Mess charge for 42 students for 28 days is Rs.47,040.Then the Mess charges for 35 Students for 25 days,in rupees, is equal to
$$y$$ ≠ 0, $$2x = 3y \Rightarrow \frac{x + y}{x - y} = $$
A person P started business with some capital and another person Q joined him four months later investing 60% more than the investment of P.If q gets 5000 more than P at the end of the year then the profit of P (in rupees) is
If $$a : b = 1 : 2 , b : c = 3 : 5 , c : d = 5 : 4 and e : d = 5 : 6$$ then $$a : b : c : d : e =$$
If the average of three persons is 20 years and the ages are in the ratio 2:3:5 then the ages (in years) of the youngest among them is
In a Joint business A invested thrice that of B and the period of Investment of A is twice that of B.If the share of B in the Profit of Rs.40,000/- and If A donates a sum of Rs.50000/- to a trust from his share then the amount left with A in the profit (in lakhs of rupees) is
A person A purchased a watch for Rs. 4800 and sold it to B for 8% profit. B sold it to C for 12% loss, The amount paid by C, corrected to nearest rupee, is
If 175 workers can dig a canal of 3150 m long in 36 days, then the number of workers require to dig a similar canal of 3900 m long in 24 days is
If a train run at speed of 40kmhr reaches destination late by 11 minutes but if it runs 50 kmph it is late by 5 min only, then the distance travlled by train in Kms is
The following table shows the marks of 20 students of class in Mathematics. Then the average marks of these 20 students is
An article has marked price of Rs.3600. By giving two successive discounts of 10% each instead of giving a discount of 20% on the marked price,then the gain of trader in rupees is
The average of least and greatest fractions among $$\frac{11}{13}, \frac{5}{6}, \frac{7}{8}, \frac{13}{15}$$ is
If $$a and b$$ are positive integers such that $$b < 100$$ and $$72.48 = a \left(24 + \frac{b}{100}\right)$$ then $$a + b =$$
The greatest five digit number which is a perfect square is
$$16^{2}+17^{2}+18^{2}+.....30^{2}$$
Four persons A, B, C, D can be around a track in 24, 48, 60 and 96 seconds respectively. All of them start at point p on the track simultaneously. When they meet at point p at second time the number of rounds made by C is
If n is any odd integer greater than 1 then the largest natural number among the following that certainly divides $$n(n^2 - 1)$$
The total number of divisors of $$2^53^45^3$$ is (including 1 and the number given)
Two positive integers x, y have their g.c.d =12 and l.c.m. = 5184. If x, y > 12, then x + y =
The traffic lights at three different road crossings respectively change after every45 seconds, 75 seconds and 100 seconds. [fall the traftic lights changesinultancously at 9:25:00 hours, then the timeat whichthe lights again change simultaneously
L.C.M. of two prime numbers a and b, (a > b) is 319. Then, a - 2b =
The least number which when divided by numbers 12, 16, 20, 25 leaves 4 as remainder but when divided by 7 leaves no remainder is
The sum of two numbers is 125.Their H.C.F and L.C.M are respectively 25 and 150. Then the sum of their reciprocals is
A Person P sells an article to Q at a profit of 5 % Qsells it to another person at profit R And R sells it to another person at profit of s For Rs.46305 by making a profit of 5%.Then the cost price of an article(in Rs)
In a school. the ratio of the number of boysand girls is 3:2. If 30% of the boys and 20% of the girls are scholarship holders, then the percentage of students that do not get scholarship is
A bag contains m coins of Rs. 5/- denomination, n coins of Rs. 2/- denomination and p coins of Rs. 1/- denomination whose total value is Rs. 630/- . If m : n : p = 2 : 3 : 5 then the ratio of the total values of Rs. 5/-, Rs. 2/- and Rs. 1/- denomination coins in the bag is
If $$x : y : z = 7 : 8 : 9$$, then $$\frac{x}{y} : \frac{y}{z} : \frac{z}{x} = $$
$$x, y, z$$ are positive integers such that $$x^3 + y^3 + z^3 = 8072$$. If $$x : y = y : z = 3 : 2$$, then $$y =$$
Two positive integers whose sum is 143, cannot bein the ratio
If $$\frac{8^3. (27)^4. 6^5}{(36)^2. 9^4. (18)^2} = 2^a. 3^b$$ then $$a : b = $$
Given y is inversely proportionalto the cube of x andthat y =2 when x = 3, Then the value of y when x = $$\frac{1}{2}$$ is
If $$\frac{x}{a} = \frac{y}{b} = \frac{z}{c}$$ and l,m,n are any positive numbers then these ratios are also equal to:
If 55% of 880 is 80 more than $$\frac{4}{5}$$ of x, then x=
If the number 19585*2, is divisible by 3, then the smallest digit in the place of * is
Solution
Given, the number 19585*2 is divisible by 3
The sum of all the digits apart from * is 30
So * can be 0 and the number would still be divisible by 3
hence, 0 is the answer.
If the number 45135*2, is divisible by 8, then the smallest whole number in the place of * is
Solution
The number formed by last 3 digits of any number should be divisible by 8 for the original entire number to be divisible by 8.
The number formed would be 5*2
502 is not divisible by 8.
512 is divisible by 8
Hence, smallest whole number in place of * is 1.
If $$4^{th}$$ October is 2008 is Saturday, then the day of the week on $$28^{th}$$ may 2014 is
Solution
Different methods are possible to solve this question.
One can work backwards and determine the day of 28 May in 2008
Number of days from 28 May to 4 October = 129
129 divided by 7 gives remainder 3.
So, if 4 October was a Saturday, then 28 May in same year has to be 3 days prior to Saturday i.e. Wednesday
Now, 28 May from 2008 to 2014 respectively would be on:
Wednesday, thursday, Friday, Saturday, Monday, Tuesday, Wednesday.
The time between 2 hours and 3 hours,when the minutes hand and hours hand coincide is 2 hours x minutes when x=:
Solution
Angular distance between hour hand and minute hand at sharp 2 PM would be 60 degrees.
Relative velocity of minute hand with respect to hour hand = 5.5 degrees per minute
Hence, the 2 hands coincide at = 60/5.5 = 120/11 minutes after 2 PM
Hence, option A
The minutes hand of an incorrect clock overtakes the hours hand at an interval of 63 minutes,Then the time it gains in a day, in minutes, is
Solution
The actual time taken by minute hand to overtake the hour hand every hour must be = 360/5.5 = 720/11 minutes.
However, the time getting taken in the incorrect clock = 63 minutes.
Thus the time gained every hour = (720/11) - 63 = 27/11 minutes
Hence, the time gained during the entire day = Just slightly over 56 minutes i.e. option A
Two trains of same length 80 meters, running in opposite direction with same speed cross each other in 3 seconds. The speed of each train is (km/hr)
Solution
Total relative distance = 160 meters
Let the speed of each train be s m/s
Hence, $$\frac{160}{2s}=3$$
Hence, s = (80/3) m/s
s converted into kmph = $$\frac{80}{3}\cdot\frac{18}{5}=96\ kmph$$
Two persons A and B are 45 km apart. If they travel in opposite direction to each other, they meetin 3 hrs. If they travel in the same direction they meet in 5 hrs. If A travels faster than B, then the speed of A, in kmph, is
Solution
Let the speed of A be a kmph and speed of B be b kmph
Hence, in first case: a + b = (45/3) = 15 kmph
In second case, a - b = 45/5 = 9 kmph
Addding the 2 equations, we get a = 12 kmph
A car reached a distance of 340 km in 4 hours .It travels some distance at a speed of 90 kmph and the remaining at speed of 60 kmph,then distance travelled with speed of 60 kmph , in kilometers, is
Solution
Let the distance travelled at 60 kmph be d km
Hence, distance travelled at 90 kmph be (340-d) km.
Hence, $$\frac{d}{60}+\frac{340-d}{90}=4$$
Solving this, we get d = 40 km
Read the data given in the table below and answer questions.
The total number of students appeared in the examinations is
Solution
Total number of students from P = 360*5 = 1800
Total number of students from Q = 520*(5/4) = 650
Total number of students from R = 540*(10/9) = 600
Total number of students from S = 420*(10/7) = 600
Hence, total number of students = 3650
The total number of students who failed in the examinations from the schools Q, R and S is
Solution
Number of students of Q failing the exam = 520/4 = 130
Number of students of R failing the exam = 540/9 = 60
Number of students of S failing the exam = (3/7)*420 = 180
Hence, total number = 370
For the following questions answer them individually
P lends Rs. 15,000 to Q for a simple interest 8% per annum and Q lends the same amount to R for a simple interest 11% per annum. Then the gain of Q in 3 years, in rupees, is
Solution
Simple Interest for Rs. 15000 at 3% per annum for 3 years = $$\frac{15000\times3\times3}{100}$$ = Rs.1350
$$\therefore\ $$Gain of Q in 3 years = Rs.1350
Hence, the correct answer is Option C
In 4 year Rs.6000 amount became Rs.8500 for some rate of simple interest.Then period required for Rs.1200 to become an amount of 1825 at same rate of interest is
Solution
Given, Principal amount = Rs.6000
Time = 4 years
Total amount = Rs.8500
Let the rate of interest = R
$$\Rightarrow$$ 8500 - 6000 = $$\frac{6000\times4\times R}{100}$$
$$\Rightarrow$$ 2500 = 240R
$$\Rightarrow$$ R = $$\frac{125}{12}$$%
Let the time required for Rs.1200 to become Rs.1825 at $$\frac{125}{12}$$% is T
$$\Rightarrow$$ 1825 - 1200 = $$\frac{1200\times T\times\frac{125}{12}}{100}$$
$$\Rightarrow$$ 625 = $$\frac{12\times T\times125}{12}$$
$$\Rightarrow$$ T = 5 years
Hence, the correct answer is Option A
A sum of money becomes Rs. 1,08,000/- after 3 years and becomes Rs. 1,92,000/- after 9 years on same rate of compound interest. Then, that sum of money is (in rupees)
Solution
Let the rate of interest be R and $$1+\frac{R}{100}=m$$
Hence, 108000 = P*$$m^3$$ and 192000 = P*$$m^9$$
Dividing the second by first, we get $$m^6=\frac{16}{9}$$
Hence, $$m^3=\frac{4}{3}$$ , substituting in the first equation, we get P = 81,000
Asum of moneyis doubled at some compoundinterest in 8 years. The numberofyears in which the amount becomes 4 times with samerate ofinterestis
Solution
Let the principal be P, rate of interest be R, $$1+\frac{R}{100}=m$$
We are given that 2P = P*$$m^8$$ which gives us $$m^8=2$$
Now, let N be the number of years when sum of money becomes 4 times itself, hence:
$$4=m^N$$
$$2^2=m^N$$
$$\left(m^8\right)^2=m^N$$
Hence, N = 16
An amount of Rs. 5,00,000/- becomes Rs. 5,51,250/- on some compound interest compoundedhalf-yearly in one year. Then the rate of compoundinterestis
Solution
Let the rate of interest be 2R% p.a. and let $$1+\frac{R}{100}=m$$
Hence, by the given data:
$$551250=5,00,000\cdot m^2$$
$$m^2=1.1025$$
$$m=1.05$$
$$R=5$$
Hence, rate of interest = 2R = 10% p.a.
A person invested in three different schemes A, B and C atthe rates of simple interests-8%, 10% and 12% respectively. The amount invested in the schemeC is equal to 225%- of the amountinvested in scheme A and is also equal to 90% of the amount invested inschemeB. If the total interest earned by him in one yearis Rs. 48,000/-, then the amount “ the’ personinvested in schemeC is (in rupees)
Solution
C = 2.25A = 0.9B
Hence, A = 4C/9, B = 10C/9
Total interest earned in one year = 0.08A + 0.1B + 0.12C = 0.0355C + 0.1111C + 0.12C = 0.2666C
48,000 = 0.2666C
C = 1,80,000
A Bank lends Rs, 30,00,000/- to a person P at 12.5% simple interest and Rs. 10,00,000/- to another person Q at 10.5% simple interest, for t years. If the Bank received an interest of Rs. 16,80,000/- at the end of the period t, then t =
Solution
Interest received from first person = 3000*125*t = 3,75,000*t rupees
Interest received from second person = 1000*105*t = 1,05,000*t rupees
3,75,000t*t + 1,05,000*t = 16,80,000
4,80,000*t = 16,80,000
t = 3.5
Two persons A and B undertake to paint a house for Rs. 54,000/-, A alone can paint the house in 40 days and Balone canpaint the house in 45 days, A and B along with a third person C complete painting in 20 days. Then, the share of C is (in rupees)
Solution
Let the total work be = 360 units
Hence, A can complete 360/40 = 9 units/day and B can complete 360/45 = 8 units/day
It is given that A,B,C together finish the work in 20 days i.e. all 3 complete 18 units/day
Hence, C alone does 18-9-8 = 1 unit/day
Hence, share of C = (1/18)*54,000 = 3000 rupees
A person A takes $$7\frac{1}{2}$$ days to finish $$\frac{5}{8}^{th}$$ of a work then the number of days required for A to finish the complete work is
Solution
Given, A takes $$7\frac{1}{2}$$ days to finish $$\frac{5}{8}^{th}$$ of the work
$$\Rightarrow$$ Number of days required to complete the total work = $$\frac{7.5}{\frac{5}{8}}$$ = $$\frac{7.5\times8}{5}$$ = 12 days
Hence, total work is done in 12 days
A and B can complete a workin 20 days .B and C can complete the same work in 24 days.The work was started by A and he work for 8 days,then after B wors for 11 days and the C completes the remaining work in 22 days,then the number of days required for c alone to complete the work is
Solution
Let the total work be 120 units and let the per day capacities of A,B,C be a, b, c units
Hence,
a + b = 6 ..... (A)
b + c = 5 ..... (B)
Hence, a + 2b + c = 11 ..... (C) & a - c = 1 ..... (D)
Also, given that:
8a + 11b + 22c = 120 ..... (E)
Multiplying (A) by 6 and (B) by 5, adding both, we get 6a + 11b + 5c = 61 ..... (F)
Subtracting (F) from (E), we get: 2a + 17c = 59 ..... (G)
Multiplying (D) by 2, subtracting from (G), we get
19c = 57
c = 3
Hence, C alone takes 120/3 = 40 days to complete work
A person P can complete a work in 40 days . A second person Q can complete the same work in 35 days. Both P and Q work together for 14 days and then P goes away. The number of days that Q takes to complete the remaining work is
Solution
Let total work be 280 units. Hence, P completes 7 units/day and Q completes 8 units/day
Total work done per day working together = 15 units
Total work done in 14 days together = 15*14 = 210 units
Remianing work = 70 units
Time taken by Q to complete work = 70/8 days
Twenty men complete a work in 16 days. Twenty women complete the same work in18 days. If all these men and women worktogether, then the numberof days required to complete that work
Solution
Let the capacity of each man and woman be m and w units/day respectively
320 m*d = 360 w*d
m = (9/8)*w
All men and women work together , then work done in a day = 20m + 20w = 20 * (17/8) * w = 42.5*w
Total days needed to complete work = 360 w*d/42.5*w = 8.47 days
A person A can complete a work in 15 days working 8 hours daily. A second person B can complete the same workin 20 days working 7 hours a day. If both A and B work together, working 7 hours a day, then the number of days required to complete that workis
Solution
Total hours required by A = 120 hours
Total hours required by B = 140 hours
Total hours needed by both working together = (120*140)/260 = 840/13 hours
But, A and B work 7 hours each day
Hence, total days required together = 840/(13*7) = 120/13 days
Twopersons A and B can do a workin 24 and 30 daysrespectively. After working together for some days, left. B finished the remaining workin 12 days. Then, the number of days that A workedis
Solution
Let the total work be 120 units
Hence, A works a = 5 units/day and B works b = 4 units/day
B completes remaining work in 12 days. Hence, B completes 48 units alone.
This means A and B together completed = 120 - 48 = 72 units
Total days A and B worked together = 72/(5+4) = 8 days
A merchant marked his goods 25% above the cost price and then allows a discount of 5%. Then the profit percent in this transaction is
Solution
Effective selling price = 0.95 x MP = 0.95 x 1.25 x CP = 1.1875 CP
$$\therefore\ $$Profit percentage = $$\frac{1.1875CP-CP}{CP}\times100$$ = 18.75% = $$18\frac{3}{4}$$%
Hence, the correct answer is Option A
The Profit Earned by selling an article for Rs.2500 is equal to 7 times the loss if the same is sold for 2100.Then the selling price of an article in (Rs) to make a profit of 20 percent is
Solution
Let the cost price be CP
2500 - CP = 7*(CP - 2100)
8*CP = 17200
CP = 2150
Now, to earn 20% profit = 1.2*2150 = 2580
A shopkeeper sold anarticle at a certain price. If it ts sold at 80%of that price, then his loss would be 8%. Then, the profit percent at the original selling price is
Solution
Let the price at which item is originally sold = S
Given that 0.8*S = 0.92*CP (CP is cost price)
S = 1.15*CP
This means profit earned by selling at S is 15% over CP
A person A purchased some pens for Rs. 350 and sold them for Rs. 450. Another person B purchases some pens for Rs. 540 and sold them for Rs. 660. Then the difference between their profit percents is
Solution
Profit percentage of A = $$\frac{100}{350}\times100$$ = $$\frac{2}{7}\times100$$%
Profit percentage of B = $$\frac{120}{540}\times100$$ = $$\frac{2}{9}\times100$$%
Difference = $$\frac{2}{7}\times100-\frac{2}{9}\times100$$ = $$\frac{400}{63}$$
Hence, the correct answer is Option A
A person travels $$\frac{2}{3}$$ of distance x km with a speed of 4 kmph and the remaining distance with a speed of 5 kmph.If the total time taken for the journey is 56 minutes, then x =
Solution
Given, Total time = 56 min = $$\frac{56}{60}$$ hr
According to the problem,
$$\frac{\frac{2}{3}x}{4}+\frac{\frac{x}{3}}{5}=\frac{56}{60}$$
$$\frac{2x}{4}+\frac{x}{5}=\frac{56}{20}$$
$$\frac{10x+4x}{20}=\frac{56}{20}$$
$$14x=56$$
$$x=4$$ km
Hence, the correct answer is Option C
The nearest year after 2000 having same calendar as that of 2000 is
Solution
Assume Jan 1, 2000 was a Sunday. Also, 2000 was a leap year.
So a leap year which starts on a Sunday would have the same calendar as 2000.
2004 starts on a Friday. 2008 starts on a Wednesday. Following the same pattern, we can see that 2028 begins on Sunday.
Hence, the answer.
A person put one coin at 1’O clock, two coins at 2’O clock, 3 coins at 3°O clock and so on in a box.Then the number of coins he has put in the box between 6:30 am to 6:30pm in a day is
Solution
From 7AM to 6PM, all numbers from 1 to 12 would have been come across by.
So the total is = 1+2+3....+11+12 = 78
The following diagram shows the expenditure of a company on different heads A, B, C, D, E in the year 2018. Based on the information given in below figure answer the following questions
If the amount spent on head A is Rs. 90,000 more than the amount spent oC, then the amount spent on E in rupees is
Solution
Amount spent on is twice than on C.
Hence, amount spent on C = 90,000
Now, for amount spent on E, we use direct variation:
$$\frac{90,000}{60}=\frac{?}{70}$$
? = 1,05,000
If the amount spent on B is Rs. 1,96,000, then the amount spent on A and E together, in rupees, is
Solution
A and E together correspond to 190 degrees
Using direct Variation:
$$\frac{196000}{80}=\frac{?}{190}$$
? = 4,65,500
For the following questions answer them individually
A dealer gets 10% of profit afier giving 10% discount on the marked price of anarticle. If the cost price of the article is Rs. 9000, its marked price, in rupees, is
Solution
0.9*MP = 1.1*CP = 1.1*9000 = 9900
MP = 9900/0.9 = 11,000
By selling an article each for Rs.600/- a trader get 10% profit on one end and 10% loss on other . The result in the transaction is
Solution
CP of article on which profit was made = 600/1.1 = 6000/11 rupees
CP of article on which loss was made = 600/0.9 = 6000/9 rupees
Total cost price = 40,000/33 rupees
Total selling price = 1200 = 39600/33 rupees
Hence, it is a loss. The only option with a loss is B.
Consider the following statements.
A. If the radius ofa sphereis increased by 8%, then the Percentage increase in its surface area is 20.86.
B. Ifthe side of'a cubeis increased by 12%, then the percentageincreasein its surface area 1s 25.44.
C. Ifthe radius of acircle is increased by 16%, then the percentage increasein its area is 34.56.
The correct statements among these are
Solution
If the surface area of original sphere is A, then after increasing radius by 8%, new surface area = 1.08*1.08*A = 1.1664*A
16.64% increase in area
If the surface area of original cube is A, then after increasing side by 12%, new surface area = 1.12*1.12*A = 1.2544*A
25.44% increase in area.
If the area of original circle is A, then after increasing radius by 16%, new area = 1.16*1.16*A = 1.3456*A
34.56% increase in area.
The area of the regular hexagon of side 12 cm in square centimeters, is
Solution
Area of hexagon = $$\frac{3\sqrt{\ 3}}{2}\cdot\left(side\right)^2$$
Putting side = 12 cm, we get area = 216*(root 3)
The area of a rectangle is 6 times that of a square. If the length of the rectangle is 16 cm more and the breadth of the rectangle is 8 cm morethan the side of the square, then the perimeter of the rectangle, in cm, is
Solution
Let side of square = S
Hence, (S+16)*(S+8) = 6*S*S
$$S^2+24S+128=6S^2$$
$$5S^2-24S-128=0$$
Solving for S, we get S = 8
Hence, length of rectangle = 24 cm and breadth = 16
Perimeter of rectancgle = 80 cm
A rectangular wooden piece of dimensions $$8 cm \times 6 cm \times 5 cm$$ cut in to pieces of dimensions $$\frac{5}{4} cm \times \frac{3}{8} cm \times \frac{1}{4} cm$$.then the number of pieces are
Solution
Number of pieces
= Volume of original piece/Volume of each cut piece
= $$\frac{\left(8\cdot6\cdot5\right)}{\frac{5}{4}\cdot\frac{3}{8}\cdot\frac{1}{4}}$$
= $$8\cdot2\cdot4\cdot8\cdot4$$
= $$2^{11}$$
If a right circular cone has height 56 cm and base radius 42 cm, then its curved surface area, in square centimeters, is (Take $$\pi = \frac{22}{7}$$)
Solution
Given, height of the right circular cone h = 56 cm
Radius of the right circular cone r = 42 cm
The slant height of the cone l = $$\sqrt{56^2+42^2}$$ = $$\sqrt{3136+1764}$$ = $$\sqrt{3136+1764}$$ = $$\sqrt{4900}$$ = 70 cm
Curved surface area of the cone = $$\pi rl$$
$$=\frac{22}{7}\left(42\right)\left(56\right)$$
$$=$$ 9240 sq.cm
The cost of a fencing around a circular field at the rate of Rs. 45 per meter is Rs. 15,840. Then the area of the field, in square meters, is (Take $$\pi = \frac{22}{7}$$)
Solution
Circumference of the field = 15840/45 = 352 meters
Also, 2*(22/7)*R = 352
R = 56 meters
Area of the field = (22/7)*56*56 = 9856 sq. meters
A wheel makes 60 revolutions in covering a distance of 990 meters. The radius of the wheel, in meters, is (Take $$\pi = \frac{22}{7}$$)
Solution
60*2*(22/7)*R = 990
R = 2.625 meters
The volume of a cube in cubic feet whose total surface area is 486 square feet, is
Solution
6*Side*Side = 486
Side = 9 feet
Volume = 9*9*9 = 729 cubic feet
By what percentage should the radius of a sphere be increased so that the volume increases by 200%?
Solution
Volume increases by 200% means the volume must becomes 3 times the original one
Hence, V2 = 3*V1
(4/3)*(22/7)*(R2)*(R2)*(R2) = 3*(4/3)*(22/7)*(R1)*(R1)*(R1)
R2 = $$\sqrt[\ 3]{3}$$ * R1
Hence, percentage increase = ({[$$\sqrt[\ 3]{3}$$ * R1] - R1}/R1)*100 = $$\left(\sqrt[\ 3]{3}-\ 1\right)\cdot100$$
The length and breadth of a rectangle are 100 cms. and 60 cms. If the length increases by 5% and the breadth decreases by 4%, then, the error percentage in the area calculated from these measurements is
Solution
Length tabulated would be 105 cm and breadth considered = 0.96*60 = 57.6 cm
Area calculated = 105*57.6 = 6048 sq. cm
Actual area = 6000 sq. cm
error = (48/6000)*100% = 0.8%
The sum of all two digit numbers which leave remainder 4 when divided by 11 is
Solution
It would be sum of numbers from (11*1)+4 to (11*8)+4
Adding them up, we get: (4*8)+11*(1+2+3+4+5+6+7+8) = 32 + 11*36 = 428
If in a leap year, January $$8^{th}$$ is a Sunday, the day on August $$15^{th}$$ in the same year is
Solution
January 1 will also be a Sunday.
August 15 is 227 days from January 1.
227/7 ==> R = 3
Move 3 days from Sunday: Monday, Tuesday, Wednesday.
If A earns 50% more than B and B earns x % less than A then x =
Solution
Let B earn Rs. 100. Hence, A earns Rs. 150
Hence, earning of B with respect to A = (100/150)*100% = (200/3)% i.e. 66.67%
Hence, B earns 33.33% less than A.
The largest positive integer k such that $$12^k$$ divides $$(109)!$$ is
The number of positive integral divisors of 37800 is
Solution
$$37800=2^3\cdot3^3\cdot5^2\cdot7$$
Hence, number of factors of 37800 = (3+1)*(3+1)*(2+1)*(1+1) = 96
In questions pick out the odd thing among the given items.
15, 35, 99, 143
56, 34, 14, 7
Solution
It should be a geometric progression with common ratio of 0.5
Every successive term should be half of the term just before it.
Hence, 34 is the odd one out. It should be 28 in its place.
A, D, I, P, T, Y
Solution
Number the alphabets as A = 1 to Z = 26
A, D, I, P, Y have numbers which are perfect squares.
9, 49, 81, 121, 169, 289
Solution
All are squares of prime numbers except 81 which is a square of 9.
Hence, odd one out.
ZA, YE, XI, WO, TU
Solution
The second alphabet in every pair is a vowel.
And the first letter in each pair is reverse chronology from Z.
Instead of T, there should be V.
343, 1331, 2197, 4913, 6859, 9261
Solution
All are cubes of prime numbers except 9261 which is a cube of 21 i.e. not a prime number
Hence, odd one out.
1156, 1444, 1764, 2430
Solution
1156 = 34*34
1444 = 38*38
1764 = 42*42
But 2430 is not 46*46
Hence, odd one out.
Beetroot, Brinjal, Carrot, Potato
Solution
The edible plant of the brinjal plant grows above the ground.
The edible parts of all the other plants grows below the ground.
Reservoir, River, Tank, Watershed
Solution
River is the only body in which water is not stationary. Hence, the odd one out.
21V23X, 13N15P, 9J11L, 5F7H
Solution
21-13 = 8
13-9 = 4
9-? = 2
? should be 7 i.e. the first number at the start of fourth term
Similarly
23-15 = 8
15-11 = 4
11-? = 2
? should be 9 i.e. the term at the center of fourth term
hence, fourth term in question is odd one out
LO12, UF06, VE22, ZA26
Solution
The last 2 digits stand for the number of the first alphabet in the pair.
L is 12th, Z is 26th and so on.
U is not 06th. Hence, odd one.
In below questions you find the left two terms/group of letters which have a relation between them. Fill the blank on the right whith one of the options given so that the terms/group of letters on the right have the same relation as the first two.
FILM : 40 :: MILK : ..............
Solution
Number the alphabets from A to Z as 1 to 26
F+I+L+M = 6 + 9 + 12 + 13 = 40
Hence, MILK = 13 + 9 + 12 + 11 = 45
MAIL : PIUN :: TOUR : .........
Solution
M --> (N,O skip) --> P
A --> (E skip) --> I
I --> (O skip) --> U
L --> (M skip) --> N
Applying same pattern to vowels and consonants in TOUR, we get WAET
CAST : 43 :: RIPE : .........
Solution
Number the alphabets from 1 to 26.
C+A+S+T = 3+1+19+20 = 43
R+I+P+E = 18+9+16+5 = 48
Choose a suitable option to fill in the blank given below:
School : Head Master :: College : Principal :: University : ..........
Solution
The institute should be followed by its head of operations.
A university is headed by a Vice Chancellor.
Amritsar : Golden Temple :: Ajmer : Darga :: Hyderabad : .........
Solution
The name of city should be followed by a popular tourist spot or a landmark in the city.
Charminar is a famous monument in the city of Hyderabad.
For the following questions answer them individually
Four figures are given below. The next figure is
Solution
The sequence from first to next is:
45 degrees rotate in anti clockwise
90 degrees rotate in clockwise
Repeat
Hence, the fifth figure will be what we get after rotating figure 4 90 degrees clockwise.
Option C
In problems there is a relation between the left two figures. Choose a correct figure from the given options such that the same relation exist with the two figures on the right.
Answer the following questions below based on the information given below.
In the figure given below the triangular region denotes the persons who read newspaper A. The circular region denotes the persons who read newspaper B and the rectangular region denotes the persons who read newspaper C.
Number of persons who read only newspaper A is
Solution
Number of people who read only A = 10 + 20 = 30
Number of persons who read both newspapers A and B is
Solution
Intersection areas of triangle and circle = 15+7 = 22
Number of persons who read only one of the newspapers A, B and C is
Solution
People who read only A = 10 + 20 = 30
People who read only B = 25 + 20 = 45
People who read only C = 20 + 40 = 60
Pople who read only one newspaper = 30+45+60 = 135
Number of persons who read all three newspapers is
Solution
Intersection area of triangle, rectangle and circle = 15
Number of persons who read either newspaper B or newspaper C is
Solution
People who read B or C (avoiding overlap or repeat count) = 25+15+30+7+20+20+17+40 = 174
For the following questions answer them individually
If $$\triangle$$ denotes addition. $$\Box$$ denotes multiplication, * denotes substraction and $$\triangledown$$ denotes division then $$3 * 4 \Box 5 \triangle 6 \triangledown 2 =$$
Solution
The given expression in standard mathematical operations can be termed as:
3 - 4 x 5 + 6/2
= 3 - 4 x 5 + 3
= 3 - 20 + 3
= -14
From point P. a man rides a cycle 9 km towards East. Then, he rides the cycle 2 km towards North. From there, he rides the cycle 3 km towards East. Finally he rides the cycle 3 km towards North to a point Q. Then the distance between the points P and Q is (in km)
Solution
Consider P at the origin (0,0)
Then Q is located at (12,6)
Hence, distance between P and Q = $$\sqrt{\ \left(12-0\right)^2+\left(6-0\right)^2}$$ = 13
A person walks 3 km towards East, then walks 3 km towards South, then walks 7 km towards West and stops. The distance from the starting point and the end pointis (in km)
Solution
Let starting point have co ordinates (0,0)
Then finishing point = (-4, 3)
Distance between starting point and finish point = $$\sqrt{\ \left(-4-0\right)^2+\left(3-0\right)^2}$$ = 5
P + Q means P is the daughter of Q, P * Q means P is the husband of Q. If P + Q * R then which of the following is true?
Solution
P+Q*R meaans that Q and R is a couple where Q is the husband, R is the wife and P is their daughter.
This condition is satisfied only by option A.
Which of the three figures from the following five figures A, B, C, D and E form a square?
If the paper in the following shape is folded to form a die then
(a) 1 lies opposite 6.
(b) 2 lies opposite 4.
(c) 3 lies opposite 5.
Then the true statement(s) among (a), (b), (c) is(are)
Solution
After folding, we get :
2 and 4 opposite each other
1 and 5 opposite each other
3 and 6 opposite each other
Among the given four options, which one when placed in the blank space of given figure would complete the pattern symmetrically?
If the image of a clock in a mirror shows the time as 6:30, then the actual time shown by the clock is
Solution
While the minute hand will retains its image in both the actual object and mirror image, the hour hands will symmetrically change its position about minute hand.
Hence, time in actual clock = 5:30
If P is the brother of Q, Q is the daughter of R andS is the father of P, then how is R related to S?
Solution
P and Q are brother - sister pair respectively.
Hence, R is the mother of both and S is their father.
Hence, R is the wife of S.
If A is B’s brother, B is C’s sister, and C is the mother of the a D, then D is A’s
Solution
A, B, and C are of the same generation.
D is of the next generation and the child of C
Thus C would definitely be nephew/niece of A.
Only Nephew is mentioned, hence, the answer.
Showing a lady, a man said “The son of her only brother is the brother of my wife”, then the relation between that lady and the man is
Solution
Read the following:
Six students P, Q, R, S, T and U sit on the ground at the vertices of a regular hexagon. P is neither adjacent to Q nor R. S is neither adjacentto R nor T. Q, R are adjacent. U is in the middle of S and R. Onthe basis of the information given above answer the questions.
Who sits opposite to T ?
Solution
The possible 2 arrangements:
If one neighbour of P is T, then who is the other one?
Solution
The possible 2 arrangements:
Hence, the other neighbor of P is S.
Which of the following is not a pair of neighbours?
Solution
Possible 2 arrangements:
For the following questions answer them individually
There are 39 persons standing in a line. When counted from left to right. A and B are at $$9^{th}$$ and $$27^{th}$$ positions in the line. If C is between A andB and if there are five persons between B and C, then the position of C in the line from the left is
Solution
The 5 persons to the left of B would people who are 22nd to 26th when counted from left.
C is just beside the leftmost of these - hence, it would be at 21st place from the left.
To answer the following questions consider the following information.
Given that
U = {1, 2, 3, ......., 500}
A = the set of all multiples of 6 in U,
B = the set of all multiples of 15 in U and
C = the set of all multiples of 10 in U.
Let $$\mid S \mid$$, denotes the number of elements in a set S. Then
$$\mid A \cup B \mid = $$
Solution
n(A U B) = n(A) + n(B) - n(overlap of both A and B)
n(A) = 83 (as 1 to 500 contains 6*1 = 6 to 6*83 = 498)
n(B) = 33 (as 1 to 500 contains 15*1 = 15 to 15*33 = 495)
n(overlap of A and B) = Multiples of LCM of 6 and 15 i.e. 30 from 1 to 500
n(overlap of A and B) = 16 (30*1 = 30 to 30*16 = 480)
Hence, n(A U B) = 83+33-16 = 100
$$\mid (A \cup B \cup C) \mid = $$
The number of integers in U that are divisible by exactly any two of 2, 3, 5 is
Consider the following information.
In a hostel there are 220 students. Out of them 105 read newspaper A, 90 read newspaper B, 75 read newspaper C. 35 read both A and B. 20 read both A and C, 23 read bath B and C and 9 read all three newspapers. Then
Number of students who do not read any newspaper is
Solution
After plotting the 3-circle Venn diagram, we get :
People who read only A: 59
People who read only B: 41
People who read only C: 41
People who read only A&B: 26
People who read only B&C: 14
People who read only A&C: 11
People who read all 3: 9
People who read none of the 3: 19
Number of students who read only one newspaper Is
Solution
After plotting the 3-circle Venn diagram, we get :
People who read only A: 59
People who read only B: 41
People who read only C: 41
People who read only A&B: 26
People who read only B&C: 14
People who read only A&C: 11
People who read all 3: 9
People who read none of the 3: 27
Hence, number of people who read only one newspaper = 59+41+41 = 141
Number of students who read exactly two newspapers Is
Solution
After plotting the 3-circle Venn diagram, we get :
People who read only A: 59
People who read only B: 41
People who read only C: 41
People who read only A&B: 26
People who read only B&C: 14
People who read only A&C: 11
People who read all 3: 9
People who read none of the 3: 27
Hence, number of people who read exactly 2 newspapers = 26+11+14 = 51
Number of students who read newspapers B and C but not A is
Solution
After plotting the 3-circle Venn diagram, we get :
People who read only A: 59
People who read only B: 41
People who read only C: 41
People who read only A&B: 26
People who read only B&C: 14
People who read only A&C: 11
People who read all 3: 9
People who read none of the 3: 27
Hence, number of people who read B&C but not A = 14
Number of students who read two or less number of newspapers is
Solution
After plotting the 3-circle Venn diagram, we get :
People who read only A: 59
People who read only B: 41
People who read only C: 41
People who read only A&B: 26
People who read only B&C: 14
People who read only A&C: 11
People who read all 3: 9
People who read none of the 3: 27
Hence,
Number of people who read two or less number of newspapers = 211
For the following questions answer them individually
In a college having 600 science students, 450) students are doing Physics, 300 students are doing Mathematics and 400 students are doing Chemistory On These students 180 students are doing both Physics and Mathematics and 350 students are doing Physics and Chemistry 200 students are studying all three subjects. Then the number of students doing both Mathematics and Chenustry is
A misfit in the following sequence
1682, 1683, 1684, 1685, 1686, 1687, 1688, 1689, ...
In the following figure, the regions 1, 2, 3, 4, 5, 6, 7, 8 are shaded with the colours Voilet (V), Purple (P), Blue (B), Yellow (Y), Orange (O), Red (R), Green (G) and Maroon (M) respectively. If the figure is reflected with respect to $$l_1$$ and then reflected with respect to $$l_2$$ then the new order of the colours in the clockwise direction
Solution
The population of a town at present is 3,00,000. The growth rate of the population in Successive years is expected to be 6%, $$7\frac{1}{2}\%, 9\%, 10\frac{1}{2}\% ....$$ of the present population The expected population at the end of $$8^{th}$$ year from now is
Solution
Population today: 3,00,000
Addition in first year: 18,000
Addition in second year: 22,500
Addition in third year: 27,000
Addition in fourth year: 31,500
Addition in fifth year: 36,000
Addition in sixth year: 40,500
Addition in seventh year: 45,000
Addition in eighth year: 49,500
Hence, population at the end of eighth year = 2,70,000 + 3,00,000 = 5,70,000
Observe the statements given below:
I. The sum of the interior angles of a regular pentagon is $$540^\circ$$.
II. Each interior angle of a regular polygon of n sides is $$\left(l - \frac{2}{n}\right)\pi$$, Then
Solution
Both the statements are standard geometry concepts and are true.
The value of x in the figure given below is
Solution
The sum of all angles of a triangle is 180 degrees
Hence, X + 45 + 70 = 180
X = 65
When thirty persons stand in a row the positions of A and B from left are seventh and twenty seventh respectively. Another person C standing between them has exactly five persons between Band C. If P is the middle person among those standing between A and C then the position of p from the left is
Solution
Position of C from left : 21
P is exactly between 7th position and 21st position.
Hence, position of P from left is 14th.
The cost of 2 pens is the same as the cost of 5 pencils. If it costs Rs. 34 for buying 4 pens and 7 pencils then the cost of 7 pens and 4 pencils (in rupees) is
Solution
Let cost of one pen be Rs. x and cost of one pencil be Rs. y
Hence, 2x = 5y
And 4x+7y = 34 ==> 17y = 34 ==> y = 2 and x = 5
Hence, cost of 7 pens and 4 pencils = 35 + 8 = 43
A train starting at a station at 1:10 pm reaches its destination at 2:20 pm the same day. If the speed of the train is 48 kmph then the distance (in kms) travelled by the train is
Solution
The time for which the train was travelling is = 1 hour 10 mins = (7/6) hours
Hence, distance travelled = (7/6)*48 = 56 km
A water glass in the shape of a frustum of a cone of height 14 cm. The diameters at the ends are 4 cm and 2 cm. Then the volume of the glass, in cubic centimeters, is (Take $$\pi = \frac{22}{7}$$)
Solution
The radii would be 2 cm and 1 cm respectively.
Let R = 2 cm and r = 1 cm
Let H = 14 cm
Hence, volume of this frustum is given as
V = $$\frac{\pi}{3}\cdot H\cdot\left(R^2+R\cdot r+r^2\ \right)$$
Substituting the values as declared above, we get
V = 102.667 cubic cm
The students of a class are standing in some rows. All the rows have the same number of students. If 3 students are added in each row, then there will be one row less. If three students are removed in each row,then there wi]] be two rows more. Then the number of students in the class is
Solution
Let there be r rows and n students in each row.
Hence, according to given two cases:
r*n = (r-1)*(n+3)
3r - n = 3
r*n = (r+2)*(n-3)
-3r + 2n = 6
Adding both, we get n = 9, r = 4
Hence, total studentss = 36
GDP growth rate of some countries for two years 2015-16, 2016-17 are tabulated below, Observe the table and answer the following questions.
The country which witnessed the highest percentage of GDP growthrate nextto C is
The country that witnessed the lowest percentage of GDP growth rate is
Solution
The net percentage growth rate of all the countries at end of 2016-17 with respect to their starting values at the beginning of 2015-16 are:
A: 15.13%
B: 15.24%
C: 4.94%
D: 9.3%
E: 5.98%
F: 5.7%
Hence, we can see that country C had lowest GDP growth in term of percentage
For the following questions answer them individually
The number that does not suit in the following sequence is
5042, 5043, 5044, 5045, 5046, 5047, 5048, 5049, 5050, 5051, ...
If $$T_n = \frac{2(n + 2)}{n(n + 1)}$$, for n = 1, 2, 3, ..... Then $$T_6 =$$
Solution
$$T_6=\frac{2\cdot8}{6\cdot7}=\frac{8}{21}$$
Answer the questions using the following information.
Given that
1 USD = 70.99 INR, | AUD = 0.708 USD, | EURO = 1.137 USD, and | NZD = 9.679 USD, (Here USD (U) = US Dollar, AUD (A) = Australian Dollar, EURO (FE) = European Union Currency, NZD (N) = Newzealand Dollar, INR (1) = Indian rupee)
Increasing order of exchange values of INR (I), NZD (N), EURO (E), AUD (A) with USD is
Solution
We can see that the most expensive currency is NZD while INR is the cheapest.
Also, E is more expensive than AUD.
Hence, option C
The currency among the following which has the highest exchange value with EURO is
Solution
1 EUR = 1.137 USD
1 EUR = 1.137*(1/0.708) AUD = 1.605 AUD
1 EUR = 1.137*(1/9.679) NZD = 0.117 NZD
Hence, it can be seen that NZD has the highest value against EUR.
The currency which hasthe least exchange value with USD is
Solution
1 INR = 0.014 USD
1 AUD = 0.708 USD
1 EURO = 1.137 USD
1 NZD = 9.679 USD,
Hence, it is clear that INR has the least value against USD
The currency among the following whichhas the highest exchange value with AUD is
Solution
1 AUD = 0.708 USD
Hence,
1 AUD = 0.708*70.99INR = 50.26 INR
1 AUD = 0.708*(1/1.137)EUR = 0.622 EUR
1 AUD = 0.708*(1.9.679) NZD = 0.073 NZD
From the above, we can see that NZD has the highest value against AUD and INR has the least value.
For the following questions answer them individually
In a certain code if the word WRITE is Coded as 136 and the word STUDYis coded 157, then in the same code the word ‘READ? is Coded as
In a code language, the word ‘MISTER’ ig coded as SGWWNS andthe word ‘FOREST’ is coded us ‘UUHVTL’, then the code for the word ‘PERSON’, in the same code language is
In a code the letters of the English alphabet are coded by numbers as follows.
$$A\leftrightarrow1, Z\leftrightarrow2, B\leftrightarrow3, Y\leftrightarrow4, C\leftrightarrow5, X\leftrightarrow6 .....$$
Then the code for COME TODAY is
Solution
The entire alphabet can be coded as:
(A,Z) = (1,2) (B,Y) = (3,4) (C, X) = (5,6)
(D,W) = (7,8) (E,V) = (9,10) (F,U) = (11,12)
(G,T) = (13,14) (H,S) = (15,16) (I,R) = (17,18)
(J,Q) = (19,20) (K,P) = (21,22) (L,O) = (23,24)
(M,N) = (25,26)
Hence, COME TODAY gets coded as: 5,24,25,9 14,24,7,1,4
In a certain code language HYDERABBD is coded as KBGHUDEDG. Then WARANGAL will be coded in that language as
Solution
2 alphabets are skipped and the next one is substituted in the code.
Hence, WARANGAL gets coded as ZDUDQJDO
In a certain code language, if AGRA is coded as 1493241, then the code of BHADRA in that code language is
Solution
Number the alphabets as A = 1, B = 2, C = 3 and so on
Then the number standing for each alphabet is squared and every alphabet is replaced with this respective square in the given coding system
Hence, A = 1, G = 7 squared = 49, R = 18 squared = 324 and so on
BHADRA becomes 4 64 1 16 324 1
In a certain code language, if ADILABAD is coded as 76547276 and WARANGAL is coded as 37170874, then in the same code language BANGLA is coded as
Solution
From the first code, we can deduce that B = 2, A = 7, L = 4 and from second we can deduce that N = 0 and G = 8
Hence, BANGLA = 270847
If the code for A is 1, that of B is 3, that of C is 5,..., then the code for G is
Solution
Every alphabet's code is given by:
Numeric code = 1 +(n-1)*2 where n = chronological position of alphabet in the 26-alphabet system
G is 7th alphabet
Hence, its numeric code = 1 + 12 = 13
If PRASAD is coded as 123435 then match the words of List-I to the codes of List-II.
Solution
PRASAD 123435
A --> 3, D --> 5, P--> 1, R--> 2, S --> 4
Use the code to generate for other words given and we get
SARADA --> 432353
DASARA --> 534323 and so on.
If $$-$$ denotes addition, $$+$$ denotes multiplications and $$\times$$ denotes division, then $$\left\{(39 \times 13) + 6\right\} - 8 =$$
Solution
The standard conversion to mathematical expression would be:
[(39/13)*6] + 8 = 18 + 8 = 26
Read the following:
In a certain code language MONKEY is coded as LNMJDX then consider the following.
Assertion(I): The code of FRANCE is EQZMBD.
Reason(II): The pattern of the given code is $$A \rightarrow Z, B \rightarrow A, C \rightarrow B, D \rightarrow C,...$$
Solution
We can see that the given reason works perfectly for the word and code given in the question.
Applying the same to FRANCE also gives the same answer as given.
Hence, both are correct and II is the right reason too.
Ina code language BREAK is written as XHUYO. Then, FLUTE is coded in that code language as
Solution
Number every alphabet as A = 1, B = 2, C = 3 and so on
Every alphabet's code thus becomes: (26 - Number of the alphabet)th alphabet
Thus B bcomes coded as (26-2) = 24th alphabet = X
Now, F => 26-6 = 20th alphabet = T
L => 26-12 = 14th alphabet = N and so on
Thus, FLUTE = TNEFU
If in a code language DREAM is coded as FTGCO, then CURSE is coded in that code language as
Solution
Every alphabet gets replaced with its +2 alphabet
Hence, CURSE becomes EWTUG
In a certain coding every consonant English alphabet is coded as the immediate next consonant and each vowel is codedas its previous vowel. Based on this information answer the following questions.
The code for the word ‘MANGO’ is
Solution
M --> N
A --> U
N --> (O) --> P
G -- > H
O --> I
The string of letters that is coded as ‘CDEFG’ is
Solution
The code is CDEFG
Vowel I gets coded as E
B gets written as C, C as D, D as F, F as G.
Hence, original word was BCIDF
The code for the word ‘LITMUS’ is
Solution
L --> M
I --> E
T --> V
M --> N
U --> O
S --> T
Data Sufficiency: In each of the following Questions you find a question followed by two statements labelled (I) and (II). You have to decide whether these statements answer the given questions. Mark your option as
Who is tallest among A, B, C, D, E?
I) D is taller than B, but shorter than E
II) C is taller than both B and E but shorter than A
Solution
Statement 1 alone tells us that both B and D are not the tallest.
However, we cannot say for sure whether E is the tallest or not.
From statement 2, we can see that even E is not taller than A.
A is the tallest of all when we combine the information given in the 2 sentences.
Hence, both statements are needed to determine answer.
What is the volume of the right circular cone?
I) Base radius of the cone is 12 cm,
II) Semi vertical angle of the cone is $$45^\circ$$
Solution
Statement 1 alone gives us no idea what the height of the cone might be.
Only when we combine statement 1 and 2 we can see that height = base radius because semi-vertical angle = 45 degrees.
Hence, both the statements are needed.
What is the total number of divisors of the positive integer n?
I) n is product of distinct numbers.
II) n is the cube of a prime number.
Solution
Statement 1 does not give us any idea of how many distinct numbers and what their powers are which will help us arrive at the number of factors of number n.
However, if statement 2 is true, then $$n=x^3$$ where x is a prime number, then number of factors of n = 3+1 = 4
Hence, statement 2 alone is sufficient.
Who is tallest among A, B, C?
I) A is taller than B.
II) B is shorter than C.
Solution
Even when we combine both the statements, the only certain conclusion we can make is that B is the shortest.
height in descending order can A,C,B or C,A,B
Hence, both the statements cannot answer the question.
What is the rate of simple interest?
I) Principal is Rs. 3,000
II) Interest is Rs. 450
Solution
We know that $$I\ =\ \frac{P\cdot N\cdot R}{100}$$
Knowing P and I is not sufficient to arrive at the value of R as number of years can be anything and thus, affect the rate of interest.
Let m and n be positive integers. Is n even?
I) $$m \times n$$ is an odd integer.
II) $$m + n$$ is an odd integer.
Solution
m*n = odd integer means that both m and n are odd integers.
m+n = Odd can be because of 2 cases:
a) m is odd, n is even
b) m is even, n is odd.
Hence, statement 1 alone can answer the question.
What is the area of the right angled isosceles triangle?
I) The angles of the triangle are $$90^\circ, 30^\circ, 60^\circ$$.
II) Length of the hypotenuse is $$15\sqrt2$$ cm.
Solution
The question makes it clear that the angles of the triangle would be 90-45-45.
Hence, statement 1 itself cannot be true in this case at any cost.
Knowing any side of the 90-45-45 triangle is enough to know the area of the triangle.
Hence, statement 2 alone can answer the question.
What is the sum of the ten terms of the arithmetic progression?
(I) The first term is 10.
(II) The tenth term is 2.
Solution
Sum of "n" terms in any arithmetic progression requires knowing either/both the first term and last term of the series And/or the common difference.
Here, the first and last term of the AP are given and hence, both are needed to compute the sum.
Is A, a sister of B?
I) B is asister of A
II) A is father of two children P and Q
Solution
Statement 1 alone leaves it quite ambiguous whether A is a male or a female.
However, statement 2 makes it clear that A is a male and thus, no chance of him being a sister of B.
Hence, statement 2 alone makes it possible to answer the question.
Did the person A get profit?
I) The person A purchased 9 bananas for Rs. 30.
II) The person A sold 8 bananas out of 9 bananas for Rs. 32.
Solution
Both the statements are needed to know how much money A spent and then earned by making sales.
Even if the last banana gets unsold, A still makes a profit, and hence =, we can answer the question.
For the following questions answer them individually
The missing terms in the following sequence in the order they appear are
S, 1, 9, W, 2, 3, E, 0, 5, ...... 1, .......
Solution
S is the 19th alphabet.
W is the 23rd alphabet.
N is the 14 alphabet. Hence, option A.
The missing term of the sequence 1, 3, 6, 10, 15,....., 28, 36 is
Solution
The difference between any 2 consecutive increases by 1 every time.
15-10 = 5
Hence, next number = 15+6 = 21
The next term in the sequence CF, FI, IL, LO is
Solution
C --> (D,E) --> F -->(G,H) --> I --> (J,K) --> L --> (M,N) --> O
F -->(G,H) --> I --> (J,K) --> L --> (M,N) --> O --> (P,Q) --> R
The next term in the sequence $$\frac{Z}{A},\frac{X}{C},\frac{V}{E},\frac{T}{G}$$ is
Solution
Z-->X-->V-->T-->R
A-->C-->E-->G-->I
The next term in the sequence B, C, E, G, K, M is
Solution
Number the alphabets as A = 1, B = 2 and so on.
B C E G K M are all prime numbered. Hence, the next prime number = 17 i.e. Q
Based on the below information Answer the questions. In the figure given a below the square region represents post-graduates, the rectangle represents graduates, circle represents private employees while the triangle represents Government employees.
The region/regions representing non-graduates employed either in Government or private is/are
Solution
We need areas in the triangle or the circle but outside the rectangle.
Hence, 2 and 9 are those areas.
The region 4 represents
The regions representing Post-graduate unemployed are
1, 5, 7 and 10 together represent
Solution
None of the 1,5,7,10 fall either in the private employees group or Govt employees group.
Hence, A.
