Quiz

Looking at the number, one can guess certain attributes of the answer.

Since the question is squaring a number very close to 3, but less than that, the first digit of the answer should be a number close to 9 but less than that. Since option ‘5’ is 9.00098641, it is eliminated.

Since 9 x 9 = 81 i.e. the last digit is 1, when we multiply, 2.9999 with 2.9999, the last digit must be 1. Since option ‘4’ has the last digit as 2, it is also eliminated.

Since there are 4 digits after the decimal point, the answer should have 8 digits after the decimal. Since option ‘3’ has 7 digits, it is also eliminated

Since the squaring is of a number very close to 3, but less than that, the first digit should be the largest integer less than 9 i.e.8. Since option ‘2’ has the first digit as 7, it is also eliminated.

This leaves us with only one option i.e ‘1’, which is the answer.

Estimated Time to arrive at the answer = 10 seconds

Assume that the clock is 1 minute slow on Tuesday at 1.00 pm and 1 minute ahead on Thursday at 1.00 pm. Hence, the clock will show the correct time at 1.00 pm on Wednesday.

But the real gain of time is faster than assumed above. Thus, the clock will show the correct time earlier than 1.00 pm on Wednesday. The closest option is option ‘2’.

From the data in the question, it is clear that the descending order of speed is A>B>C, i.e. A being the fastest, followed by B and C is the slowest. Also, in a race of 970 meters, B can give a start of 50 meters (970 – 920) to C, Hence, it is obvious that in a 1000 meters race, the start for C must be proportionately greater than 50 as 1000 is greater than 970. Hence, the answer must be little over the difference of 970 and 920 meters or 50 meters; plus, something. Looking at the options, the only option matching this condition is option ‘3’ or 52 meters

105 x 107 = 11235 (using Mental Math technique discussed in Multiplication Shortcuts
Now, 12,000 x 1.1235 is approximately equal to a 12 % increase in 12,000
i.e. = 1440
Hence, the new Principle = 12,000 + 1440 = 13,440
However, this is an approximate answer. The real answer will be slightly greater than 13,440 (since, 1.1235 > 1.12). Thus, looking at the options the answer is Rs, 13,482 – option ‘5’.

Difference between and = 2 units

Difference between and = 6 units

Since, the lines PQ, and form a right-angled triangle with PQ as the hypotenuse. We already know that the hypotenuse of any right triangle should be more than the longest right side but less than the sum of the right sides. Thus, length of PQ should be lying between 6 < PQ < (6 + 2). The only value satisfying this range is 6.3 units.

Simply find the Arithmetic Mean of the numbers:
i.e. = 41
Now you already know that Arithmetic Mean > Geometric Mean > Harmonic Mean, hence, the answer should be less then 41 but close to it. From amongst the options, the only one satisfying this condition is option ‘3’.

Simply read the first part of the question and find those options that satisfy that condition as shown below:

=> = = 2 (Satisfies the condition)

=> = = (does not satisfy the condition)

=> = = 2 (Satisfies the condition)

=> = = (does not satisfy the condition)

=> = (does not satisfy the condition)

This leaves us with options ‘1’ and ‘3’ only. Now, look at the 2nd condition. Apply this to options ‘1’ and ‘3’ only as follows:

=> = = 5 (does not satisfy the condition)
Hence, answer is option ‘3’

Just to verify…c.
=> = = 7 (Satisfies the condition and hence is the answer)

Just solve the 1st level bracket i.e. (…), we get: 1 + 1/4 = 5/4.
5 > 4 or Numerator > Denominator

Also since, there are three division operation further left, in the final answer, the Numerator < Denominator. This eliminates options ‘2’, ‘3’ and ‘4’.

Also, note that if in the first level of operation, we get the numerator and denominator as 5 and 4 respectively, and that further operations involve addition, the final answer will have a value greater than 5 and 4. This eliminates option ‘1’. Hence ‘5’ is the final answer.

Observe that with a 25% reduction in speed, the person reaches his office later by 33% of an hour. Thus, by increasing speed by 33% of the present (not the usual speed), the person will take 25% of an hour less than what he took earlier (not the usual time).

Assuming t = hour. A 33% increase in speed on 3/4s, will give 1s. Thus 25% decrease of time on (t + 1/3) hours will give t hours which is the usual time. This satisfies the condition of traveling at usual speed s and also the arithmetical relationship that an increase of 33% can be brought to usual by a decrease of 25%. Thus t = 1 hour is true.

If you notice from the options that the option ‘4’ of 40 minutes is not possible If one pipe alone takes 36 minutes, then if other pipes are also running along with it, the time taken must be less than 36 minutes. Using the same logic, even 36 minutes of option ‘3’ also can’t be the answer.

Now, for simplicity sake, just ignore that the rate of flow of water is proportional to the square of diameter but take it as being directly proportional to the diameter itself. Adding up the diameters of the other pipes i.e. 1 + 2 + 3 = 6 cms This 6 cms is equal to the 6 cms diameter of the pipe whose time for filling the tank is given. This means that now there are two pipes each of 6 cms filling the tank simultaneously. Hence, the time taken by the two together will be equal to half of 36 minutes (half of the time taken by one 6 cms pipe to fill the tank) i.e. 18 minutes.

But in reality, the rate is proportional to the square of diameter of pipes. Hence, the answer should not be 18 minutes, but higher then that.
(Since, is > ).

Of the remaining options, only option ‘5’ satisfies the condition. Hence, the answer is 26 minutes.

Simply find the Arithmetic Mean (AM) of the three percentage values

i.e. = = 21

or 21.67%

We already know that AM > GM and that the difference is small when the individual values are ‘near’ each other. Thus, the actual answer must be less than 21.67% but close to it. Hence, the answer must be 21% or option ‘2’.

Since the y coordinate of point Q is negative with a value greater than that of point P (i.e. 3 > , the sum of these () will be negative. Thus, the y coordinate of the answer should be negative, and the x coordinate positive. This eliminates options ‘1’, ‘2’ and ‘4’.

Now looking at the coordinates of P and Q, we observe that the ‘x’ and ‘y’ coordinates of P are fractions with denominators 3 and 2 respectively, whereas the coordinates of Q are integers.

Hence, half of the sum of ‘x’ coordinates of P and Q would yield a fraction with denominator 3 x 2 = 6.

Similarly, half of the sum of y coordinates of P and Q would yield a fraction with denominator 2 x 2 = 4.

The only option that satisfies this condition is option ‘3’.

The difference between 50 lbs and 55 lbs = 5 lbs which is additional per boy in 30 boys. Hence, additional total weight = 30 x 5 = 150 lbs.

This 150 lbs should be reduced from the weight of girls. Since there are 50 girls, the average reduction per girl will be = 3 lbs.

Hence answer = 50 – 3 = 47 lbs.

Simply write the values of ‘x’ and ‘y’ such that x – y = 3 and x + y = the values in the options as shown below:
5 => 4, 1
7 => 5, 2
9 => 6, 3
11 => 7, 4
13 => 8, 5

Now, start cubing the values of ‘x’ and ‘y’ and find the option for which the difference is = 189.

This should be the answer. There is no need to solve further as there cannot be more than one answer.

= 9

Just rewrite the expression as:

=

Now observe the variable ‘y’ and its powers
LHS => and

RHS => and
(after opening the brackets)
This means that the variable ‘y’ cancels out and so the answer should have only the variable ‘x’. The only option that satisfies this condition is option ‘2’.

Note that speed of B follows Arithmetic mean and that of A follows Harmonic Mean.

We know that for a given set of numbers, Harmonic Mean is always < Arithmetic Mean

Hence speed of A < speed of B, so B reaches Y before A.

The first step is to eliminate those options whose sum 47. Thus option ‘5’ is eliminated as sum of 10 + 20 + 18 = 48.

The next step is to identify the metal with maximum or minimum quantity. Because A = 4/5 B and B = 3/4 C; the metal C is of maximum quantity and A is of minimum quantity. Thus options ‘1’ and ‘4’ are eliminated. This leaves us with options ‘2’ and ‘3’ only. Since, A = 4/5 B; 5A = 4B.

For option ‘2’, 5 x 12 4 x 13, hence ‘2’ is also eliminated. Thus, the answer is option ‘3’.

Simply find the simple interest over 4 years as follows:

0.05 x 4 x 2000 = 400

Hence the new principal becomes 2000 + 400 = 2400.

We also know that Compound Interest in this case will be > Simple Interest and also since the rate of interest is just 5% for only 4 years, the interest difference between Compound and Simple will not be ‘large’. Hence, the answer should be greater than $2400 but close to it. The only option that satisfies this condition is $2431. Hence the answer is option ‘3’.

If a line is cut by a point in a certain ratio, then its length along the X-axis and that along the Y-axis is also cut in the same ratio.

In this case, the x coordinate of A (–3) and that of C (2) are –3 –2 = –5 or 5 units apart along the X-axis. When this must be divided in the ratio of 2 : 3; 2 parts towards point A and 3 parts towards point C, we get the coordinate of point B.

As can be seen in the above figure; the value of x coordinate of point B will be –1 (2 units on the right of –3 or 3 units on the left of 2). Once we know that the x coordinate is –1, we can look at the options and see that only option ‘2’ satisfies the requirement and hence the answer should be option ‘2’.

While reading the question itself, you can figure out that Algebra is a common subject in the two groups. Hence, when we add up the marks of the two groups, the marks of Algebra are repeated.

Also, the sum of marks of the two groups will end in a ‘0’

For 50; 0 x 4 = 0

For 70; 0 x 4 = 0

And 0 + 0 = 0, so last digit = 0.

The last digit of the sum of marks for all 7 subjects = 6

For 58; 8 x 7 = 56, so last digit = 6.

Hence, the last digit of marks of Algebra = 0 – 6 = 4.

From amongst the options, the only option with last digit = 4 is option ‘4’.

Simply look at the options and start substituting the values in the expressions and just check which one satisfies the equation. Also, remember to start substituting with the value that is easiest of all to substitute. In this case, for example, the easier values are ‘0’ and ‘1’. You would notice that ‘0’ satisfies the equation as follows:

Hence option ‘5’.

Visually you can find the HCF as as ‘3’, ‘’ and ‘’ are the only terms common in all the three expressions. This eliminates options ‘2’ and ‘3’.

Also note that the LCM must have ‘’, as it is the only term appearing in the third expression and not in the other two expressions. Hence, the answer is option ‘4’.

To cover 1 mile @ 3 mph, it takes 20 minutes

To cover 1 mile @ 4 mph, it takes 15 minutes

Hence, time gap = 20 – 15 = 5 minutes

Thus, for a distance of 1 mile, time gap = 5 minutes

Hence, a time gap of 4 minutes (2 + 2), distance is correspondingly mile.

If 3 men and 6 women take 4 days, hypothetically, 1.5 men and 3 women will take 8 days. Hence, 1 man and 8 women will obviously take less than 8 days. Also, 1 man and 8 women will also take more than 4 days; hence the answer lies between 8 days and 4 days i.e. 6 days.

(Note: there is another shortcut here which is shorter than the one shown above.
If you look at the x coordinate of X; it is a fraction with denominator 3. Now, look at the options and find that option whose sum of Numerator and Denominator is 3. In this case, the option is ‘4’ which is the answer.)

Last digit of total age of 10 boys = 0

Last digit of total age of 12 boys = 0 (Note: 12.5 x 12 is same as 25 x 6 and its last digit is 5 x 6 => 0)

Last digit of total age of 18 boys = 4

Hence, last digit of the sum of age of the class = 0 + 0 + 4 = 4.

Also, since there are more students with age of 13 than with the age of 12 or 12.5, the average should definitely be more than 12.5. From the options, we can eliminate all but options ‘4’ and ‘5’.

To choose between options ‘4’ and ‘5’, simply check the last digits of as follows:

For option ‘4’ =>12.6 x 40 = 126 x 4, last digit = 4

For option ‘5’ => 12.7 x 40 = 127 x 4, last digit = 8.

Since only option ‘4’ satisfies the condition of last digit = 4, option ‘4’ is the answer.

Just check the last digit of 785 x 1005 i.e 5 x 5 = 25, so last digit = 5. Adding 108 will make the last digit = 5 + 8 = 13 i.e 3.

This eliminates options ‘1’, ‘4’ and ‘5’. From between options ‘2’ and ‘3’, option ‘3’ can also be eliminated since, the first three digits in it are 785…The logic here is when 785 is multiplied by a number greater than 1000 (1005 in this case) the answer will surely be greater than 785000 and also when 108 is added to it, the answer will be greater than 785108. Since 785033 < 785108, we eliminate option ‘3’. Hence answer is option ‘2’.

Just get to the ‘’ level. Note that the product of these two numbers should be 2+3 = 5 digits after the decimal point. This eliminates option ‘5’.

Also, the product of the last digits is ‘6’ (9 x 4 = 36). Hence, ‘6’ should be the last digit of the answer. This eliminates option ‘1’.

Also, note that 13 x 17 > 200. Hence, the answer should be a number starting with 2.something…

Hence, the answer is option’4’.

Since Distance = speed x time,

The length of the train should be a multiple of both the speed and the time.

Time taken is known as 15 seconds. Hence, length of the slower train should be a multiple of 15. The only option which is a multiple of 15 is option ‘1’.

Original ratio = 50 : 1

New ratio = 75 : 1

The additional 75 – 50 = 25 is attributable to the additional 500 agents. Hence, each unit or 1 corresponds to 500 / 25 = 20. Hence, 20 mangers and so option ‘3’.

Remember that the difference of $180 is nothing but the interest on interest earned in the 2^nd year.

Interest earned in 1^st year by Compound as well as Simple is the same = 80r

Interest earned on this interest in the next year =

% i.e option ‘2’.

Since, the median includes point A, the coordinates of point A should satisfy the equation of the median. Hence, just check which amongst the equations is satisfied by the coordinates of point A (4, 8).

x + y = 4 + 8 = 12 7

x + y = 4 + 8 = 12 10

3x + y = 3 x 4 + 8 = 20 10

2x + y = 2 x 4 + 8 = 16 =16

6x + y = 6 x 4 + 8 = 32 =32

As can be seen, that only options‘d and ‘5’ are satisfied by the coordinates of point A, thus we can eliminate options ‘1’, ‘2’ and ‘3’.

Since we are left with two options, we need to do further elimination. This can be done by substituting the coordinates of point D.

Coordinates of point D can be found mentally as it is the average value of the corresponding coordinates of point B and C (D being the midpoint of B and C).

Now, checking between the options’4’ and ‘5’ which one is getting satisfied with the coordinates of point D (5, 2).

2x + y = 2x5 + 2 = 12 16

6x + y = 6x5 + 2 = 32 =32

Hence the answer is option ‘5’.

The average age has increased by 0.1 years with 50 boys. Hence, total increase = 0.1 x 50 = 5 years. This 5 years increase is due to 10 boys, so additional age over the normal average age of 12 years is = 0.5 years.

Hence, average age of new boys = 12 + 0.5 = 12.5 years

Since the product of two numbers = product of LCM and HCF of two numbers, just multiply the options to see which is = = 4500 as follows:

15 x 30 = 450 4500

60 x 75 = 4500 = 4500 hence this should be the answer. There is no need to check further as there cannot be more than one correct answer.

Applying the test of divisibility, observe from options that option ‘1’ and option ‘3’ are both multiples of 3 and hence will not leave any remainder. Hence both options ‘1’ and ‘3’ can be eliminated.

Also, the remainders when 949 and 929 when divided by 4 will give remainder of 1 and not 3 as required. Hence options’4’ and ‘5’ are also eliminated.

Thus we are left out with only option ‘2’, which actually satisfies the conditions and hence is the answer.

The Arithmetic Mean (AM) of the values 50 and 40 is 45.

We already know that the Harmonic Mean (HM) of a set of numbers is always less than the AM of the same set. (i.e. HM < AM).

Hence, the answer should be < 45. The only option < 45 is 44.4 km/hr

Should B take 12 days working alone, then A and B working together would take 6 days. But A and B together actually take 8 days; hence B would take more than 12 days to finish the work working alone. This eliminates options ‘1’ and ‘2’

Also note that the difference between 6 days and actual 8 days is that of 2 days. Simply multiplying 2 with 12 we get 24 days which is the answer.

% marks of A = %.

From the data, marks of C > B, also marks of A > C, hence marks of A > B. This eliminates options ‘1’ and ‘5’. Also, the difference between marks of A and C is 10% and that between B and C is also 10%. Thus % marks B 84 – 10 – 10 = 64%. Thus, the answer has to be close to 64%. The closest to it is options ‘4’ = 68%.

We already know the slope intercept form as y = mx + c.

The slope here is given as –4.

Thus, y = –4x + c or 4x + y = c

Thus, the coefficient of x should be 4.

Only option ‘3’ satisfies this condition and hence it is the answer.

For 9 players, if the average age goes below by 1 year, the total reduction in age = 9 years. Since, the average age of the 9 players gets reduced due to not including the ages of captain and vice-captain. Hence, the captain and vice captain’s age contribute to the reduction of 9 years. Total age of captain and vice-captain = 25 + 28 = 53 years.

53 – 9 = 44 years will be the age of 2 players of the team. Hence, average age of the team = years

Just look at the options and see which numbers are in the ratio of 5 : 7. Thus all options other than ‘1’ and ‘3’ are eliminated.

Also, for option ‘3’:

Thus, answer is option ‘1’:

Option ‘1’ and ‘4’ can be eliminated since, the answer cannot be an even number nor can it end in 5, else it will not leave any remainder when divided by 2 or 5 respectively.

Option ‘5’ is also eliminated since when it is divided by 3, it leaves a remainder 1 instead of 2.
( and 16/3 = 5 + 1 as remainder) Finally subtract 6 from the remaining options and check for divisibility of 7. You will find that 30029 – 6 = 30023 is divisible by 7 and hence option ‘2’ is the answer.

At 6 km/hr it takes th of an hour to travel 1 km

At 4 km/hr it takes th of an hour to travel 1 km

Hence, the time difference is hour = 5 minutes.

Actual difference in time = 10 minutes.

If difference in time = 5 minutes for 1 km,

For difference in time = 10 minutes, distance = 2 kms

Simply find those options that are multiples of 4. This is because the answer is x some integer. In this case, the multiples of 4 are options ‘2’, ‘4’ and ‘5’. Of these, option ‘2’ can be eliminated as x some integer < x some integer, but in case of option ‘2’, 4500 > 1125.

Also, option ‘5’ can also be eliminated as 1124 is very close to 1125. Hence the answerer is option ‘4’.

If A’s earnings = $100

B will earn = $120 (20% more than A)

C will earn = $150 (25% more than B)

D will earn $200 (30% more than C)

Since, the ratio of A’s earnings to that of D is almost 1 : 2, A’s earning will be very close to half of D’s earning.

Thus, approximate earnings of A = 975 / 2 = 487.5.

The only answer closest to this is option’4’ = $500

The perimeter of parallelogram ABCD = sum of length of all sides

= + + +

=

Now, from the options, just subtract 14. This will give us the value of , which is a multiple of 5 .

Hence, that option which is a multiple of 5 after subtracting 14 from it should be the answer.

22 – 14 = 8      (Not a multiple of 5 and so it is not the correct answer)

48 – 14 = 34    (Not a multiple of 5 and so it is not the correct answer)

53 – 14 = 39    (Not a multiple of 5 and so it is not the correct answer)

59 – 14 = 45    (Is a multiple of 5 and so it is the correct answer)

32 – 14 = 18    (Not a multiple of 5 and so it is not the correct answer)

The average increases by 0.1 years for 25 + 1 = 26 students, so total increase = 2.6 years. Thus, the age of new student = 2.6 + original average age of the 25 students = 2.6 + 10 = 12.6 years.

Note that if the age 20 years ago was 1/3^rd of the present age, then 20 years represents 2/3^rd of the present age. Hence, present age = 30 years.

Since C’s present age is 50% more than the average age and A’s present age is 50% less than the average age, the average age of the three will be same as the average age of C and A.

Since,

Also, since B’s age is equal to 10 years, the present average age of all three is 10 years. So, the average age of all three two years ago is 10 – 2 = 8 years.

Note that in a kilometer race A beats B by 100 meters and B beats C by 50 meters. Since, B runs 100 meters less than 1000 meters of A (1000 – 100 = 900 meters), C will lag little less than 50 meters behind B in this race where B runs 900 meters. Hence the gap between A and C will be sum of gap between A and B (100 meters) and gap between B and C (little less than 50 meters). Hence, gap between A and C is little less than 150 meters. The only option close to 150 and less than 150 is option ‘2’ i.e 145 meters.

Since ‘2’ is common in both the relationships, simply find the product of coefficients of ‘1’ and ‘3’ i.e. 5 x 11 = 55 which will be the corresponding term for ‘2’. This eliminates all options other than ‘3’ and ‘5’.

If 3b = 11c, it means that ‘3’ is smaller than ‘2’, Hence in a : b : c, if the corresponding figure for ‘2’ is 55, ‘3’ should be less than 55. This is true only for option ‘5’.

Assume that the cost price to A = $100

Hence, cost price to B = $110

Hence, cost price to C = $99

Note that cost price to C < cost price to A. Thus, 980,100 < cost price to A. This eliminates options ‘1’, ‘2’ and ‘3’. Also note that, if the cost to A = 100; cost to C = 99, thus if cost to A = $1,000,000; cost to C = $990,000, but actual cost to C = $980,100. Thus option ‘4’ is the answer.

Using some fundamental concepts of geometry as shown below:

The diagonal cannot be larger than the sum of all the three sides: 11 + 8 + 6 = 25 < 25.3. This eliminates option ‘5’. Find the pairs of sums of length of sides as follows: 11 + 8 =19, 11 + 6 = 17 and 8 + 6 = 14. The answer has to lie between the largest sum and the smallest sum. i.e. between 19 and 14. This leaves us only with option ‘2’ as the answer.

Simply adding 32 + 5 = 37 gives you the answer.

The reasoning here is that if the average of the first 5 days is 32 then the attendance of the 5^th day would be 32. Since the average is 33, 1 day more than 32; then for 5 days it is an extra 5, the additional 1 being true for 5 days. So, it is 32 + 5 = 37.

Use the alligation technique here. The average value of coin = or 45 cents
Thus,

Thus, for every one 25 cents coins, there are four 50 cents coins. Dividing 100 coins in this ratio, we get 80 coins of 50 cents.

Observe from the equation has ‘8’ in its denominator. This 8 cannot be reduced in further operations as 5, 3 and 10 are relatively prime to 8 and hence have an HCF of 1. Should the HCF of these numbers be 2 or any multiple of 2, then the denominator ‘8’ will get to 4, 2 or 1 accordingly. But in this particular case, it is not so. Hence, the answer should have 8 or a multiple of 8 in its denominator. The only option satisfying this criterion is options ‘3’.

Distance traveled by A and B will be in the ratio of their speeds since time taken to meet is same for both. Thus, ratio of distance traveled by A and B = ratio of speeds of A and B = 18 : 14 = 9 : 7.

Total distance traveled by both A and B together = twice the distance between Mumbai and Pune = 2 x 192 = 384 kms. Dividing 384 kms between A and B in the ratio of 9 : 7, we get distance traveled by A = 216 kms.

Hence the meeting point is 192 – 216 = 24 kms from Pune.

Of 5x and 6x, the greater is 6x, which means that it is a multiple of 6. Thus, all options other than ‘3’ and ‘4’ are eliminated as they are not multiples of 6. Also, when that number is to be subtracted by 4, it should yield a number which is a multiple of 5 since the ratio after subtraction changes to 4 : 5.

24 – 4 = 20 is a multiple of 5 but 30 – 4 = 26 is not a multiple of 5 and hence 24 is the number and option ‘3’ is the answer.

If earnings increase by 25%, to get the original earnings, the new earning should be reduced by 20%

Thus, old earnings = $800 less 20% = 800 – 160 = 640

Hence option ‘3’.

Observe that both 98 and 14 are multiples of 7 and when multiplied by , will have in the numerator a number which is a multiple of 11 (because 22 is in the numerator of the value of ). Only option ‘3’ is a multiple of 11 and hence it is the answer.

The average age of Allen and Beth = 40 years. Note from options, that only options ‘2’ and ‘5’ satisfies this condition ( from option ‘2’ and from option ‘5’). This eliminates all other options.

We also know from the question that the average age of Beth and Chris is 38 years. Now between the options ‘2’ and ‘5’, only option ‘5’ satisfies this condition ( ).

Hence the answer is option ‘5’.

If the worker remains absent, he not only loses $5 but also loses the $10 daily wage. Thus, the net loss for the worker is $15 (10 + 5).

Had the worker attended work on all 30 days, his earnings would have been 30 x 10 = $300. But his actual earnings = $240. Hence, the difference = 300 – 240 = $60 is attributable to the number of days he was absent. Since, loss per day of absence = $ 15, so for a loss of $60, the worker was absent for 4 days.

Looking at the information in the problem, it is evident that the LCM is a multiple of 7 and 5. The only option that is a multiple of 7 and 5 amongst the options is option ‘2’.

Note that when B travels 1000 meters, C travels 900 meters and A 850 meters. Thus, when C travels 900 meters, A travels 850 meters. Thus, in a 900 meters race C can give a lead of a little over 50 meters to A. Hence answer must be closer to 50 meters but more than that. From the options, only option ‘3’ is closest to 50 and more than that (500/9 > 50).

If x is the total score, then A’s score = and B’s score = . Since B’s score is a fractional multiple of x, with denominator = 49, the value of x, should be a multiple of 49 as B’s score has to be an integer. The only option that is a multiple of 49 from the options is option ‘4’.

Instead of taking the initial figure as ‘x’, to avoid working with fractions, assume the figure as $300.

Thus 1/3^rd of $300 = $100 @ 3% gives interest of $3

And 1/6^th of $300 = $50 @ 6% gives interest of $3

And the balance $300 – $100 – $50 = $150 @ 8% gives interest of $12

Thus, total interest earned = 3 + 3 + 12 = $18.

For a total amount of $300, the interest earned = $18,

For what total amount would the interest = $600

Doing cross multiplication, we get $10,000

Ratio of height to length = 2 : 5 = 4 : 10

Ratio of height to breadth = 4 : 5

Thus, ratio of length to breadth = 10 : 5

Hence, area of ceiling =

Hence, the area of the ceiling should be a multiple of 50. The only option satisfying this condition is option ‘2’.

Difference between the average weight of the original sacs and the new sacs = 50 – 47.4 = 2.6lbs

Total difference = 2.6 x 5 = 13 lbs. Now this 13 lbs is reduced from the total weight of (x + 5) bags and the reduction is that of 0.2 lb each. Hence

therefore x+5=65

therefore x=60 lbs

2880 can be split as a product of 288 and 10. 288 is a product of 2 and a perfect square 144. 10 is a product of 2 and 5. Hence, the only number that has a power of 1 is 5, so it has to be divided to make 2880 a perfect square.

Option ‘3’ can be eliminated as 4 itself is a perfect square. Hence, multiplying it by any number will not make that number a perfect square. Option ‘5’ can also be eliminated as 18 = 9 x 2 and 9 being a perfect square will not make any number a perfect square and the 2 does not make sense here as it already is in option ‘1’.

Now, simply divide 2352 with 6. The number 392 is not divisible by 3 hence, you will have to multiply 3 with 2352 to make 2352 a perfect square.

Since the time taken by the train to cover the platform is 3 times the time it takes to cover the pole, the total length that the train traverses while crossing the platform is three times of what it does while traversing the pole.

When the train traverses the pole, its traverses its own length and when it traverses the platform, it traverses its length and the length of the platform, hence if ‘l’ is the length of the train, then:

meters

Since A works on the 1^st day and B on the 2^nd and so on alternately, A will be working only on odd numbered days. Thus, option ‘1’ can be eliminated since A cannot work on the 26^th day. Similarly, options ‘3’, ‘4’ and ‘5’ can also be eliminated since B works only on even numbered days. This leaves us only with the option ‘2’.

Interest on $400 for 2 years is equivalent to that for $200 for 4 years. Thus effectively $200 can be said to be lent for 4 + 3 = 7 years

therefore, Interest earned =

r=5%

Add squares of 3, 4 and 5 to get 9 + 16 + 25 = 50.

50 is a factor of the total surface areas of the three smaller cubes and hence 50 or its factor should be one of the values in the proportion. Only option ‘3’ has the factor of 50; 25, so option ‘3' must be the answer.

Since the average temperatures for the first 5 days and the last 5 days is ‘equidistant’ from the average temperature for the 9 days, we can simply take the arithmetic mean of 25 and 15 as 20°C as the temperature of the 5^th day.

In = 20, ‘ ’ are mixed fractions which when converted to improper fractions will become .

20 being an integer, should yield an integer, which is possible only if is a multiple of 4 or ‘1’. Since we do not know the value of ‘1’, we look at the option check for that value of ‘1’, which makes a multiple of 4. The only value possible from the options to make a multiple of 4 is 3. Hence, a = b = 3.

You can simply approximate the expression (Note, that the actual value will be less than 5 and hence the ‘–’ sign in brackets)

Hence,

Since the actual denominator is less than 5, the final answer will be greater than 82. The only option coming close to this figure is 88, i.e option ‘2’.

20 being an integer, should yield an integer, which is possible only if is a multiple of 4 or ‘1’. Since we do not know the value of ‘1’, we look at the option check for that value of ‘1’, which makes a multiple of 4. The only value possible from the options to make a multiple of 4 is 3. Hence, a = b = 3.

Note that it takes 90 secs for the train traveling at the speed of 10 mt/sec (36 km/hr) to traverse a length equal to its own plus that of the platform.

i.e. length of train + length of platform = 90

aaaaaaaaaaaaaaa10

length of train + length of platform = 90 x 10 = 900 meters.

Hence, the length of platform has to be < 900 meters. Hence 700 meters (option ‘1’).

Assuming A and C together as a new person D, then we can say that B and D together finish the work in 10 days. Also, B is half as efficient as D. In other words, D has double the efficiency of B and so can do double the work that B can do in the same time. Hence, if the work is divided in 3 parts, D will do 2 parts of the work and 1 part would be done by B.

Hence, the ratio of total work, to the amount of work done by B alone is 3 : 1.Hence, the ratio of the number of days, it takes to finish the work by working together to that taken by B working alone will be 1 : 3. Since, if efficiency decreases, the number of days took increase by the same proportion. Hence, the number of days taken by B alone will be 10 x 3 = 30 days.

We know that the selling price is 1.4 times the cost price. Hence, = the cost price.

We know that = 0.1428…
is a division by 7 (Since, 14 is a multiple of 7). Hence, the numbers after the decimal point will be of the type .14… The only option matching this criterion is option ‘5’.

Hence, the ratio of total work to the amount of work done by B alone is 3 : 1. Hence, the ratio of the number of days it takes to finish the work by working together to that taken by B working alone will be 1 : 3. Since, if efficiency decreases, the number of days took increase by the same proportion. Hence, the number of days taken by B alone will be 10 x 3 = 30 days.

Since there are three equal parts of a circle in the equilateral diagram, the above diagram is same as the one shown below as we can take three parts of the circle as one complete circle.

Hence area of the shaded region = Area of triangle – Area of circle.

= sq. units

Note that the reduction of 10 lbs in the total weight of 10 sacks is attributable to the two rice sacks. Hence, each rice sack should have 5 lbs less weight than the average weight of the two wheat sacks of 26 and 30 lbs.

Average of 26 and 30 = lbs

Hence average weight of a sac of rice =28 – 5 = 23 lbs.

We know that R’s age 10 years ago is three times S’s age.

i.e. where ‘r’ and ‘s’ are present ages of R and S respectively.

Hence, (r – 10) = multiple of 3.

From the options, subtract 10 to get the age of R 10 years ago and see which amongst these is a multiple of 3.

30 => 30 – 10 = 20 (Not a multiple of 3)

50 => 50 – 10 = 40 (Not a multiple of 3)

60 => 60 – 10 = 50 (Not a multiple of 3)

70 => 70 – 10 = 60 (is a multiple of 3)

80 => 80 – 10 = 70 (Not a multiple of 3)

Since, only option ‘4’ satisfies the conditions, 70 years is the present age of R.

We already know the value of the reciprocals of 7, 11, 15, 17 and 9 as:

= 0.1482…

= 0.09090…

= 0.0666…

= 0.0588…

= 0.1111…

From above it is clear that,

< < < <

During the course of a complete day, the monkey climbs 30 – 10 = 20 meters, so to climb 120 meters, it will take days. We know that since the monkey has slipped during the 6^th night, the total complete days needed to reach 120 meters has to be less than 6 days. This eliminates all options from ‘3’ to ‘5’. This means that the answer can be or days. You can simply check that in 5 complete days, the monkey climbs 5 x 20 = 100 meters. Since during day time, the monkey climbs 30 meters, so to climb the balance (120 – 100) 20 meters, of the 6^th day time will be used. Since the complete day has equal duration of both daytime and nighttime, the 2/3 of daytime is equivalent to 1/3 of the whole day (Day time and nighttime together). Hence total duration needed to reach 120 meters = days.

(Note: Students commit mistakes in considering daytime as whole day and get the answer as days instead of days).

Assuming option ‘1’ to be the answer,

A’s share = 700

B’s share = 1400 and

C’s share = 2800

But, 700 + 1400 + 2800 7000 and hence options ‘1’ is not the answer.

Assuming option ‘2’ to be the answer,

A’s share = 900

B’s share = 1800 and

C’s share = 3600

But, 900 + 1800 + 3600 7000 and hence options ‘2’ is not the answer.

Assuming option ‘3’ to be the answer,

A’s share = 1000

B’s share = 2000 and

C’s share = 4000

But, 1000 + 2000 + 4000 = 7000 and hence options ‘3’ is the answer.

No need to check further as there can be only one single answer.

Once you reach to a the level of identifying the dimensions of the triangle (2, 2 and 3), put the values as , to get the answer.

Now, do not solve this. Just by observing it, you will see that there are four 2s in the denominator with the sign. Hence, the denominator will be = 4. Also 7 is a prime number so … will have (or one of its form) as the numerator. Hence the answer should have . This format is only in option ‘4’.

Should the two parts invested in be equal in size i.e. $500 each, the earnings will be $45 (average of 40 and 50). But since the actual earning is more than $45 (46.50 > 45.00), the amount invested at 5% is more than the amount invested at 4%. (Using the concept of weighted average, where average tends to shift towards the higher weight). Hence, the amount invested @ 4% is less than $500 and that invested @ 5% is more than $500. The only options satisfying this are ‘1’ and ‘3’.

Option ‘3’ is eliminated by working backwards:

Hence, option ‘1’ is the right option.

Now, do not solve this. Just by observing it, you will see that there are four 2s in the denominator with the sign. Hence, the denominator will be = 4. Also 7 is a prime number so … will have (or one of its form) as the numerator. Hence the answer should have . This format is only in option ‘4’.

After admission of 10 new students, the total number of students becomes 20 + 10 = 30. If 10 students have to increase the average age by 1 year of 30 students, total increase needed = 1 x 30 = 30 years. This extra 30 years is from 10 students only. Hence extra age per new student = years. Hence average age of the 10 new students = 12 + 3 = 15 years.

The last digit of the square of 1884 will be 6 ( = 16, Last digit = 6)

Similarly, the last digit of the square of 1020 will be 0.

Hence, last digit of difference between the squares of the numbers = 6 – 0 = 6

This eliminates options ‘2’ and ‘3’ as both when squares will have their last digit not equal to 6.

From amongst the remaining options, the option ‘4’ is most likely to be the answer as the number whose square is equal to the difference of squares of 1884 and 1020, will be also a 4-digit number lying between 1884 and 1020. The only 4-digit number satisfying the conditions is 1584 i.e. option ‘4’.

Simply subtract each option from 10420 starting with the least number and check for perfect square as follows:

e. 10420 – 16 = 10404 = (Perfect Square)

(Note: We are taking option ‘5’ first as it is the one with the lowest number.)

No need to check any further.

Hence, option ‘5’.

The monkey climbs 5 – 3 = 2 meters every 2 minutes. Hence to climb 21 meters, he should take 21 minutes (purely looking at this relationship itself). Hence the answer should be 21 minutes. Of the options, it can be either 17 or 18 minutes.

Checking for 18 minutes, we find that after 18 minutes, the monkey will be at 18 meters (Note: At even numbered minutes, the monkey will have climbed its numerical equivalent in height).

Checking for 17 minutes, we should first check for 16 minutes (refer to the above note). At 16 minutes, the monkey will climb 16 meters and in the next minute, it would climb 5 meters reaching 16 + 5 = 21 meters. Hence 17 meters is the answer.

To have equal number of boys and girls, the ratio of boys to girls should change from 5 : 4 to 1 : 1. Also, 1 : 1 is equivalent to 5 : 5. The number of boys = 5x and the number of girls = 5x. Hence total number of students = 10x.

This means that the total number of students after new girls join is a multiple of 10 and hence the last digit of the sum should be = 0.

The existing total number of students = 432 with last digit = 2. Hence, the last digit of the number of girls who join should have last digit = 8, to get the last digit of the final sum = 0. The only option satisfying this condition is option ‘2’.

As can be seen that by a reduction in cost price, there is an increase in profit %. The increase in profit % = %

So, for a 50% increase, there should be a 33.33% reduction in cost price (as explained in the Percentages section of Stuff to Remember)

Hence, the original cost price should be such that a reduction of $20 is equivalent to a reduction of 33.33%.

Hence, original cost price =

Since height of pole and its shadow are in proportion, we notice that for a pole of 3 feet height, the shadow is 4 feet long. Thus, by reducing ¼th (25%) of the length of the shadow, we can get the length of the pole. Thus, by doing a 25% reduction on the length of 24 feet long shadow, we can get the height of the pole that casts this shadow.

25% reduction on 24 feet is 24 – 6 = 18 feet.

The average age of the couple 4 years ago = 25 years, then present average age = 25 + 4 = 29 years. Present average age of family = 20 years. Reduction of (29 – 20) 9 years is attributable to the child. Total reduction = 9 x 3 = 27 years (Since there are 3 members). Hence the age of child = 29 – 27 = 2 years.

Let the cost of 1 mango be ‘m’ cents and one apple be ‘n’ cents.

Hence  8m + 6n = 960                         ..…. Eq. 1.

And     6m + 8n = 1000                       ..…. Eq. 2.

Subtraction Eq. 1 from Eq. 2, we get:

2m + 2n = 40

– m + n = 20

This means that the price of the apple is 20 p more than that of a mango.

Assuming that m = n (i.e. the price of mango = price of apple)

therefore 6m + 8n = 1000 becomes 6n + 8n = 1000

therefore 14n = 1000

therefore p

Now since n > m by 20p, n > average of n and m by 10p.

therefore p

Just by observing you can make out that the numerator has a higher value of power of base 3 and the denominator has a higher value of power of base 2. Hence, the answer has to have 3 in the numerator and 2 in the denominator. The only option satisfying these conditions is option ‘3’.

Since the upstream travel speed = 4.5 – 1.5 = 3 km/hr and downstream travel speed = 4.5 + 1.5 = 6 km/hr (both being integers), but total travel time = hours (being a fraction with denominator as 3), the answer should also be a fraction with denominator as 3. The only answer satisfying this condition is 5.33 kms.

Since B’s investment remains constant at $4000 for 12 months, the proportion that B should get of the profit will be a multiple of 4. This eliminates all options other than ‘2’ and ‘4’. Also note that B’s total investment is less than that of A’s. Hence, A’s share of profit should be higher than B’s share. Thus option ‘4’ is also eliminated. Thus option ‘2’ is the answer.

A discount of 10% means 90% of the price or 0.9 times the price. The answer can be found by solving the expression 400 x 0.85 x 0.90. The answer should be a multiple of 9 due to the multiplication of 0.90. The only answer that is a multiple of 9 from amongst the options is $306 i.e. option ‘4’.

Since the size of bins is given in feet and that of the bricks is given in inches, we convert feet to inches by multiplying by 12. Thus, the answer should be a multiple of 12. Only option ‘3’ is a multiple of 12.

(Note: to find the multiple of 12, find the multiple of 3 as well as 4).

If 8 men out of 9 men spent $ 6 each, assume the average to be $6.

We also know that the 9^th man spent $ 4 more than the rest. This $4 has to be apportioned to the 8 men who spent $6 each. Hence, has to be added to the original spent of $6. i.e 60 + 0.50 = $6.50 which is the new average.

Observe that in the equation .

Hence x must be a negative number. Also, because the denominators are 3 and 5, the denominator of x will be LCM of 3 and 5, i.e. 15. Since only option ‘4’ satisfies the conditions of the denominator being negative as well as =15, it should be the answer.

Just by observing we see that the numerator has 6.5, which is a multiple of 5 and there is no multiple of 5 in the denominator, so the answer must be a multiple of 5. This eliminates the options ‘1’, ‘3’ and ‘4’. Also, the numerator must be significantly higher than the denominator since addition is carried out in numerator and subtraction in denominator. Hence the answer should be a higher multiple of 5. Hence option ‘5’.

Presently A works for a total of 10 + 5 + 4 days, i.e 19 days. A alone can finish the work in 20 days. This means that contribution of B is equivalent to saving 1 day's work of A or of the total work. B contributes this of the work by working for 5 days, so in 1 day, B will do of the work. Thus, B will take 100 days to finish the work alone.

Simply add 3% and 5%, we get 8%(-)

Approximately 8% of the rice is reduced from the original quantity. Note that the (-) sign indicates that the actual value will be slightly less than 8%.

8% of 11058 or 11000 (rounded) = 880. Add this to 11058, we get:

11058 + 880 = 11938. Thus the actual answer should be close to 11938. The only option satisfying this condition is option ‘3’.

Note that capacities of vessel A to that of B = 3 : 1. Hence in one turn of each 3x + 1x = 4x tank will be emptied (where ‘x’ the constant of proportionality). The tank in turn is large.

Hence, turns of each vessel will be needed. Hence a total of 90 turns are needed.

If you would observe each component keenly, you would observe that except in the first component i.e. , you get 4 x 48 as denominator. In all other components, you get only 48 as the denominator. It is important to note that in order to know the denominator, you do not have to ‘solve’ each component rather you can simply get it from the LCM of 4, 6 and 48.

Thus, you will have a picture in your mind that would look as shown below:

Canceling all 48s, we get . This means that in the final answer, you should have 4 or a multiple of it in the denominator. The only option satisfying this condition is option ‘3’ i.e.

First get the ratio of earnings of A and B by the usual method, i.e. 7 : 6.

The investment period of A is 4 months and 4 is not a factor of 7. The amount of investment has to be a factor of 7 so as to get 7 in the ratio. The only option which has a multiple of 7 is option ‘3’ $3500.

5% of $4000 = $200

Remember that whenever in Compound interest, whenever if the interest rate is halved but compounded twice, the net effect will always be slightly more than if the rate was taken as it is for 1 year.

Hence, in this case, the actual interest earned will be slightly more than $200, i.e $202.50

Simply by knowing that the ratios are in the form 3 : 2 between the first & second and 3 : 2 between the second & third number, we can know that the second number is a multiple of 6. (Since a : b : c = 9 : 6 : 4)

This means that the answer from amongst the options is either 6 or 12. All other options are eliminated.

Now assuming that b = 6, we will have a = 9 and c = 4 (Since a : b : c = 9 : 6 : 4)

Now simply add the last digits of the cubes 9, 6 and 4 i.e. 9 + 6 + 4 = 19 (Last digit of cubes of 9, 6 and 4 are 9, 6 and 4 respectively). Here the last digit is 9 (from 19). But the last digit of the sum of cubes of these number is actually 2 (from 8 072). Hence the middle number is not 6 but 12.

5% of $4000 = $200

Remember that whenever in Compound interest, whenever if the interest rate is halved but compounded twice, the net effect will always be slightly more than if the rate was taken as it is for 1 year.

Hence, in this case, the actual interest earned will be slightly more than $200, i.e $202.50

Consider each option. Assume A’s share = 410, C’s will be = 510 as A’s share is $100 less than C’s. B’s share will be 510 + 70 = 580.

Now check 410 + 580 + 510 = 1500

This satisfies the condition that the total amount = $1500 and hence, option ‘1’ is the correct option.

Adding 20% on 100 is equivalent to removing 16.67% from 120. (as explained in the Percentages section of Stuff to Remember)

16.67% = of the number

of 1890 = 315.

Reducing 16.67% from 1890, we get 1890 – 315 = 1575

However, this is inclusive of transport costs of 5%

Hence, the actual answer is less by approximately 5%. The only option less than $1575 is $1500. Hence, option ‘2’. Note that $1417.50 is not the correct value since it is more than 5% less than 1575.

Assume each option to be the right answer and reverse calculate to check as follows:

Option Age of Dick Age of Tom Age of Tom 20 years back

a.                     44                    50                    30                    (multiple of 15)

b.                     48                    54                    34                    (not a multiple of 15)

c.                     50                    56                    36                    (not a multiple of 15)

d.                     52                    58                    38                    (not a multiple of 15)

e.                     54                    60                    40                    (not a multiple of 15)

(Note: If you go sequentially taking one option at a time and start with option ‘1’, you do not have to go further as option ‘1’ will give the age of Tom 20 years back as a multiple of 15. No other option leads to a multiple of 15.)

Observe that the term (0.4 – 4.04) will yield a negative value. Hence the term will also be negative number slightly more than 1. Hence, 4 + (some negative number slightly more than 1) = a number slightly less than 4 = x

Hence, will yield a value slightly more than 1. Hence option ‘1’.

The two 5% increases can be looked at approximately 10% increase. Reducing 10% from 70560, we get: 70560 – 7056 = 63560 which is closest to 64000. Hence option ‘1’.

Note that the age of A three years earlier should have been as even number because . From the options, the various values of (a – 3) that we get are:

Option # Value of (a – 3)

a                      24 – 3 = 21 (Odd)

b                      35 – 3 = 32 (Even)

c                      45 – 3 = 42 (Even)

d                      48 – 3 = 45 (Odd)

e                      54 – 3 = 51 (Odd)

Options ‘1’, ‘4’ and ‘5’ are eliminated as they yield odd values.

Note that half of 83 = 41.5. From the given conditions, we can also deduce that a > b. Hence, a > 41.5. This leaves us only with option ‘3’.

Since the denominator is , its conjugate is which is multiplied to it and the numerator . Since both the terms ( and ) are positive terms, their product too would be positive and there would not be any negative sign. The only option that does not have a negative sign is option ‘1’.

It takes 1 boy 12 days to finish the work, so it will take 12 boys to finish the work in 1 day. Also, 3 men = 4 women = 12 boys.
Thus 1 man = 4 boys and 1 woman = 3 boys.
Thus if 12 boys are needed to finish the work in 1 day, if 1 man and 1 woman are working together, they will do the work of 4 + 3 = 7 boys. Hence, balance number of boys i.e. 12 – 7 = 5 will be needed to finish the work in 1 day.

By reducing $10 from the selling price the profit reduces by 5%, hence $10 is equivalent to 5%, so 100% will be equivalent to $200.

If the student had answered 30 out of 60 questions correctly, his score would have been 30 x 2 – 30 x 1 = 60 – 30 = 30. Since the actual score is > 30, the attempts are also > 30. Hence answer is 33 questions.

Since the sum and product of two numbers are integers, the two numbers are also integers. Reciprocals of integers will result in fractions and so will the sum of the reciprocals. This eliminates options ‘3’ and ‘5’.

Since both the sum and product of the two numbers is even, the two numbers individually are also even. Hence the sum of their reciprocals will have their denominator a multiple of 2 x 2 = 4. The only option satisfying these conditions is option ‘4’.

Since, the original ratio of ladies to gents = 1 : 2, the total number of guests = 1x + 2x = 3x.

Also, the new ratio of ladies to gents = 1 : 3, the total number of guests = 1y + 3y = 4y

Hence the total number of guests should be a multiple of both 3 and should become a multiple of 4 after 4 is subtracted from it. This means that the original number of guest should be a multiple of 3 as well as 4 or a multiple of 12. The only option satisfying this condition is option ‘5’.

Interest earned in 2^nd year has to be slightly larger than $375 (5% of $7500), since we have to add the interest accrued on the interest of the previous year. However, this additional amount being small (due to 5%), we can approximate the actual answer to a value slightly greater than $375, but close to it. The only options satisfying these conditions are ‘2’ and ‘3’. Option ‘2’ is eliminated as it is ‘too close to’ $375.

Difference between the selling prices = 75 – 60 = 15.

15 when broken up in the ratio of 1 : 4 (1 being loss and 4 being gain), we get loss = . So, cost price = 60 + 3 = $63

Similarly, from the gain side, gain = . Hence cost price = 75 – 12 = $63

Since HCF of any set with a prime number in it is always 1, denominator will be = 1. This eliminates options ‘1’ and ‘5’.

LCM of numerators = product of numerators if they are relatively prime

Hence LCM of the numerators = 4 x 3 x 5 = 60

Hence answer = 60

(Note: The example here is seemingly simple but the important part is the approach in the technique. This approach is universal in its application and so can be applied to other more complex problems also).

The average age has increased by 0.1 years with 50 boys. Hence, total increase = 0.1 x 50 = 5 years. This 5 years increase is due to 10 boys, so additional age over the normal average age of 12 years is = 0.5 years.

Hence, average age of new boys = 12 + 0.5 = 12.5 years

Simple observing 5 : 4 : 4 : 3 : 4 we notice that the average of 5, 4, 4, 3 and 4, is 4.

Hence, 4 corresponds to the average of 60% and so 5 will correspond to the average of 75% (using cross multiplication).

A’s age 7 years back = of B’s present age. All options being natural numbers, B’s present age is also a natural number and so is A’s present age. Hence A’s age 7 years back is also a natural number.

Now if (A’s age 7 years back) a natural number is equated to (B’s present age) another natural number, then B’s present age should be evenly / completely divisible by 6 to have A’s age 7 years back as a natural number.

Since the options are half of B’s present age, just double them up and check which of these numbers is divisible by 6. Better still, just check which from amongst the options are divisible by 3 (half of 6).

Only option ‘2’ is divisible by 3.

Between and , is the smaller one. This eliminates option ‘4’.

Also, =

Similarly, =

Since, 0.35 < 0.43, eliminate option ‘1’.

Similarly, =

Since, 0.35 < 0.52, eliminate option ‘3’.

Similarly, =

Since, 0.35 < 0.65, eliminate option ‘5’.

Hence the answer will be the left out option ‘2’.

(Note: In this case we are trying to bring the denominators close to 100 to get the decimal value of the fraction, so that it can directly be compared)

Assuming that there are 100 men working for 100 days, hence total work is for 10,000 man-days. With a reduction in number of men to 90, days increase to 110. Hence, work is equivalent to 90 x 110 = 9900 man-days. This is slightly less than 10,000 man-days. Hence, the actual number of men should be slightly greater than 100. As per the options, the only option slightly greater than 100 is 110.

Population 2 years hence can be approximated using 81250 as 80000 and instead of 3% cumulative over 2 years take 3% additive over 2 years i.e. 6% of 80000 = 4800 (+)

Hence approximate population = 81250 + 4800(+) = 86050(+)

(Note: The (+) sign denotes that the actual answer will be slightly greater than 86050 since our assumption of taking 3% addition as 6% will yield a value lesser than 3% cumulative.

Also 80000 (Approximated) < 81250 (Actual). Hence actual increase will be slightly greater than 4800. Hence the (+) sign) As can be seen from the options, only option ‘1’ is closest to 86050 and slightly greater than that.

We know that 80% of the boys present out of the total 40% boys make 0.32x = 1152.

We also know that 1/3 = 0.33. Hence, we can approximate 0.32 as 0.33 and so 1152 is approximately 1/3 of the total enrollment. Hence total enrollment = (+).

(Note: The ‘+’ sign indicates that the answer is slightly greater than 3300).

From the options we see that 3600 is the only one closest to 3300 and greater than that. Hence, 3600 is the answer.

One can intuitively know that to have sum of 5 consecutive numbers to be 100, we need as the middle number. We also know that in Arithmetic Progression, the product of the two consecutive numbers equidistant from a central number is always less than the square of the central number. Since 20 is the central number, the product of the first and last number (equidistant from the central number) will be less than . Also, since the numbers are close to each other (being consecutive), the product will not be ‘too’ less than 400. Only 396 satisfies this condition.

Since A’s share = of B,

, where ‘1’ and ‘2’ are the shares of A and B respectively.

Since all options are multiples of 5, the share of B under each option will be:

A’s share                     B’s share

75                    336 – 75 = 261 (not a multiple of 16)

80                    336 – 80 = 256 (is a multiple of 16)

No need to check further since there can be only one answer. Hence, option ‘2’ is the answer.

Just make a table as follows:

Sale Value               Scheme I                        Scheme II           Difference (II – I)

4000                            200                              1000+0                        800

8000                            400                              1000+100                    700

12000                          600                              1000+200                    600

Hence the value of sale at which the difference is $600 is $12000.

If 10 lbs is reduced from a weight, then it becomes 75% of the original. Thus, the reduction of 10 lbs is equivalent to 25%. Hence 100% is equal to 40 lbs. Thus, Mira’s weight is 40 lbs and her mother’s weight is 90 lbs.

Observe that 6822 > 5142 per se. Hence number system of 5142 > that of 6822 i.e. 10.

Assume number system as 11 and check as follows:

. This satisfies the equation as hence our assumption of 11 as the base number system is correct.

Now the number 3333 has to be written in the number system of 11.

Note that the last digit of the answer will be ‘0’ as 3333 is completely divisible by 11 and so the remainder will be 0. (Note that to convert the a number in the decimal system to some other system, we have to divide the number and read the remainders bottom up as is done in the long explanation above.)

This leaves us with options ‘1’, ‘2’ and ‘3’ only. Amongst these options, option ‘2’ is eliminated since 4320 > 3333. As the base of a number system increases, the number per se decreases. However, in this option, which is incorrect and so option ‘2’ is ruled out.

Option ‘3’ is also eliminated since 1220 << 3333. Although, as explained in the above paragraph, as the base of a number system increases, the number per se decreases, but there is a certain relationship between the ‘size’ of a number and the base of the number system. Since in this case, the base system changes from 10 to 11, which is a relatively marginal increase, the decrease in the ‘size’ of the number should also be relatively small. However, in case of option ‘3’, the number is far smaller than

and so it is eliminated.

This leaves us with option ‘1’ which is the answer.

Visualize that the total profit is divided in 8 parts, of which 1 part goes to B as wages and the remaining 7 parts are divided between B and A in the ratio of 6000 : 8000 or 3 : 4. Thus B receives 3 more parts of the profit and A receives 4 parts. Thus B receives in total 3 + 1 = 4 parts of the profit which is same as 4 parts received by A. Thus A receives the same money as B receives i.e. $1000.

Observe that if A’s cost price would have been $100, his selling price with 10% loss would be $90 and his selling price with 20% gain would be $120. Hence gain = 120 – 90 = 30

Selling price of A with 10% loss = Cost price of B = $90.

Hence Profit %for B = %

We know that . Substitute the values of the option in this equation.

Note that we substitute options starting with the easiest and going towards the tougher ones. In this case, we use the following order: 0 , –1, 5 , –5 and then 9.

Thus, for y = 0, , so it is not the answer.

For y = –1, , so it is not the answer.

For y = 5, , so it is the answer.

No need to check further as there can be only one correct answer.

Simply observe that since all the terms in the numerator are positive, simplifying the numerator will also give all positive terms.

The denominator has a negative sign. However, the process of rationalization would require its conjugate to be multiplied to result in a rational number in the denominator. This conjugate multiplier will have positive terms. Hence, when it multiplies with the all positive term numerator, we will get a product in the numerator with all positive terms. The only option that has all positive terms in the numerator is option ‘5’.

Total investment = 1100 x 10 = 11000 = 11 x 1000

Let a, b and c be 1000^th of the capital invested by A, B and C respectively.

Now, 2a = b and b = 3c

i.e. 2a = 3c

Thus 2a should be a multiple of 3, but since, 2 is not a multiple of 3, ‘1’ should be the multiple of 3. The only value that ‘1’ can take from amongst the options is 3 and 6 (or 3000 and 6000).

A’s capital cannot be $6000, as B’s capital will be 2 x 6000 = 12000, which is more than the total capital of $11000, so the answer has to be $3000 or option ‘1’.

Approximating, we can guesstimate that if the discount is 10%, the gap between the marked price and the cost price is 20%. Now if the discount is only 5% the gap between the cost price and selling price = 15%. However, the discount being calculated on the marked price and profit % on the cost price (cost price < marked price in this case since profit is generated), the actual gap will be slightly more than 15%. Hence 16%.

In order to reduce concentration from 60 to 40 (i.e. by 33.33%) we need to increase the quantity of the solution by 50% (See concept explained in section of Percentages in Stuff to Remember.)

Thus if solution initially was 100 cc, it should now be 150 cc. Thus an additional 50 cc is to be added.

For , should be a perfect square of = 0.

Equating to 0, we get:

= 0

= 0

1or –5

Substituting y = 1 in the original equation, we get:

, but this cannot be true, as , hence, 1 is not the correct solution to the equation.

Substituting y = –5 in the original equation, we get:

, this is true, as the equation is satisfied. Hence, –5 is the correct solution to the equation. Thus there is only a single solution to the equation.

Note that we are not checking by equating with perfect squares, as:

= 4 (a perfect square)

= 16

3 or –7

Substituting y = 3 in the original equation, we get:

, but this cannot be true, as , hence, 3 is not the correct solution to the equation.

Substituting y = –7 in the original equation, we get:

, but this cannot be true, as , hence, –7 is not the correct solution to the equation.

Since, 100 men join to work only for 5 more days, had they not joined, the work would have taken 5 more days. Hence, option ‘1’.