How to Prepare for Quantitative Aptitude for CAT (2014)

Block I: Numbers

Chapter 2. PROGRESSIONS

The chapter on progressions essentially yields common-sense based questions in examinations.

Questions in the CAT and other aptitude exams mostly appear from either Arithmetic Progressions (more common) or from Geometric Progressions.

The chapter of progressions is a logical and natural extension of the chapter on Number Systems, since there is such a lot of commonality of logic between the problems associated with these two chapters.

ARITHMETIC PROGRESSIONS

Quantities are said to be in arithmetic progression when they increase or decrease by a common difference.

Thus each of the following series forms an arithmetic progression:

3, 7, 11, 15,…

8, 2, –4, –10,…

a, a + d, a + 2d, a + 3d,…

The common difference is found by subtracting any term of the series from the next term.

That is, common difference of an AP = (t_N – t_{N – 1}).

In the first of the above examples the common difference is 4; in the second it is –6; in the third it is d.

If we examine the series a, a + d, a + 2d, a + 3d,… we notice that in any term the coefficient of d is always less by one than the position of that term in the series.

Thus the rth term of an arithmetic progression is given by T_r = a + (r – 1)d.

If n be the number of terms, and if L denotes the last term or the nth term, we have

To Find the Sum of the given Number of Terms in an Arithmetic Progression

Let a denote the first term d, the common difference, and n the total number of terms. Also, let L denote the last term, and S the required sum; then

If any two terms of an arithmetical progression be given, the series can be completely determined; for this data results in two simultaneous equations, the solution of which will give the first term and the common difference.

When three quantities are in arithmetic progression, the middle one is said to be the arithmetic mean of the other two.

Thus a is the arithmetic mean between a – d and a + d. So, when it is required to arbitrarily consider three numbers in AP take a – d, a and a + d as the three numbers as this reduces one unknown thereby making the solution easier.

To Find the Arithmetic Mean between any Two given Quantities

Let a and b be two quantities and A be their arithmetic mean. Then since a, A, b, are in AP. We must have

b – A = A – a

Each being equal to the common difference;

This gives us A =

Between two given quantities it is always possible to insert any number of terms such that the whole series thus formed shall be in AP. The terms thus inserted are called the arithmetic means.

To Insert a given Number of Arithmetic Means between Two given Quantities

Let a and b be the given quantities and n be the number of means.

Including the extremes, the number of terms will then be n + 2 so that we have to find a series of n + 2 terms in AP, of which a is the first, and b is the last term.

Let d be the common difference;

then b = the (n + 2)th term

= a + (n + 1)d

Hence, d =

and the required means are

Till now we have studied APs in their mathematical context. This was important for you to understand the basic mathematical construct of A.Ps. However, you need to understand that questions on A.P. are seldom solved on a mathematical basis, (Especially under the time pressure that you are likely to face in the CAT and other aptitude exams). In such situations the mathematical processes for solving progressions based questions are likely to fail or at the very least, be very tedious. Hence, understanding the following logical aspects about Arithmetic Progressions is likely to help you solve questions based on APs in the context of an aptitude exam.

Let us look at these issues one by one:

1. Process for finding the nth term of an A.P.

Suppose you have to find the 17th term of the

A.P. 3, 7, 11…….

The conventional mathematical process for this question would involve using the formula.

T_n = a + (n – 1) d

Thus, for the 17th term we would do

T₁₇ = 3 + (17 –1) × 4 = 3 + 16 × 4 = 67

Most students would mechanically insert the values for a, n and d and get this answer.

However, if you replace the above process with a thought algorithm, you will get the answer much faster.

The algorithm goes like this:

In order to find the 17th term of the above sequence add the common difference to the first term, sixteen times. (Note: Sixteen, since it is one less than 17).

Similarly, in order to find the 37th term of the A.P. 3, 11 …, All you need to do is add the common difference (8 in this case), 36 times.

Thus, the answer is 288 + 3 = 291.

(Note: You ultimately end up doing the same thing, but you are at an advantage since the entire solution process is reactionary.)

2. Average of an A.P. and Corresponding terms of the A.P.

Consider the A.P., 2, 6, 10, 14, 18, 22. If you try to find the average of these six numbers you will get: Average = (2 + 6 + 10 + 14 + 18 + 22)/6 = 12

Notice that 12 is also the average of the first and the last terms of the A.P. In fact, it is also the average of 6 and 18 (which correspond to the second and 5th terms of the A.P.). Further, 12 is also the average of the 3rd and 4th terms of the A.P.

(Note: In this A.P. of six terms, the average was the same as the average of the 1st and 6th terms. It was also given by the average of the 2nd and the 5th terms, as well as that of the 3rd and 4th terms. )

We can call each of these pairs as “CORRESPONDING TERMS” in an A.P.

What you need to understand is that every A.P. has an average.

And for any A.P., the average of any pair of corresponding terms will also be the average of the A.P.

If you try to notice the sum of the term numbers of the pair of corresponding terms given above:

1st and 6th (so that 1 + 6 = 7)

2nd and 5th(hence, 2 + 5 =7)

3rd and 4th (hence, 3 + 4 = 7)

This rule will hold true for all A.P.s.

For example, if an A.P. has 23 terms then for instance, you can predict that the 7th term will have the 17th term as its corresponding term, or for that matter the 9th term will have the 15th term as its corresponding term. (Since 24 is one more than 23 and 7 + 17 = 9 + 15 = 24.)

3. Process for finding the sum of an A.P.

Once you can find a pair of corresponding terms for any A.P., you can easily find the sum of the A.P. by using the property of averages:

i.e. Sum = Number of terms × Average.

In fact, this is the best process for finding the sum of an A.P. It is much more superior than the process of finding the sum of an A.P. using the expression (2a+(n–1)d).

4. Finding the common difference of an A.P., given 2 terms of an A.P.

Suppose you were given that an A.P. had its 3rd term as 8 and its 8th term as 28. You should visualize this A.P. as – , – , 8 , – , – , –, – , 28.

From the above figure, you can easily visualize that to move from the third term to the eighth term, (8 to 28) you need to add the common difference five times. The net addition being 20, the common difference should be 4.

Illustration: Find the sum of an A.P. of 17 terms, whose 3rd term is 8 and 8th term is 28.

Solution: Since we know the third term and the eighth term, we can find the common difference as 4 by the process illustrated above.

The total = 17 × Average of the A.P.

Our objective now shifts into the finding of the average of the A.P. In order to do so, we need to identify either the 10th term (which will be the corresponding term for the 8th term) or the 15th term (which will be the corresponding term for the 3rd term.)

Again: Since the 8th term is 28 and d = 4, the 10th term becomes 28 + 4 + 4 = 36.

Thus, the average of the A.P.

= Average of 8th and 10th terms

= (28 + 36)/2 = 32.

Hence, the required answer is sum of the A.P. = 17 × 32 = 544.

The logic that has applied here is that the difference in the term numbers will give you the number of times the common difference is used to get from one to the other term.

For instance, if you know that the difference between the 7th term and 12th term of an AP is –30, you should realize that 5 times the common difference will be equal to –30. (Since 12 – 7 = 5).

Hence, d = – 6.

5. Types of APs:Increasing and Decreasing A.P.s.

Depending on whether ‘d’ is positive or negative, an A.P. can be increasing or decreasing.

Let us explore these two types of A.P.s further:

(A) Increasing A.P.s:

Every term of an increasing AP is greater than the previous term.

Depending on the value of the first term, we can construct two graphs for sum of an increasing A.P.

Case 1: When the first term of the increasing A.P. is positive. In such a case the sum of the A.P. will show a continuously increasing graph which will look like the one shown in the figure below:

Case 2: When the first term of the increasing A.P. is negative. In such a case, the Sum of the A.P. plotted against the number of terms will give the following figure:

The specific case of the sum to n₁ terms being equal to the sum to n₂ terms.

In the series case 2 above, there is a possibility of the sum to ‘n’ terms being repeated for 2 values of ‘n’. However, this will not necessarily occur.

This issue will get clear through the following example:

Consider the following series:

Series 1: –12, –8, –4, 0, 4, 8, 12

As is evident the sum to 2 terms and the sum to 5 terms in this case is the same. Similarly, the sum to 3 terms is the same as the sum to 4 terms. This can be written as:

S₂ = S₅ and S₃ = S₄.

In other words the sum to n₁ terms is the same as the sum to n₂ terms.

Such situations arise for increasing A.P.’s where the first term is negative. But as we have already stated that this does not happen for all such cases.

Consider the following A.P.s.

Series 2: –8, –3, +2, + 7, + 12…

Series 3: –13, –7, –1, + 5, + 11…

Series 4: –12, –6, 0, 6, 12 …

Series 5: –15, –9, –3, + 3, 9, 15 …

Series 6: –20, –12, –4, 4 , 12, …

If you check the series listed above, you will realize that this occurrence happens in the case of Series 1, Series 4, Series 5 and Series 6 while in the case of Series 2 and Series 3 the same value is not repeated for the sum of the Series.

A clear look at the two series will reveal that this phenomenon occurs in series which have what can be called a balance about the number zero.

Another issue to notice is that in Series 4,

S₂ = S₃ and S₁ = S₄

While in series 5

S₁ = S₅ and S₂ = S₄.

In the first case (where ‘0’ is part of the series) the sum is equal for two terms such that one of them is odd and the other is even.

In the second case on the other hand (when ‘0’ is not part of the series) the sum is equal for two terms such that both are odd or both are even.

Also notice that the sum of the term numbers which exhibit equal sums is constant for a given A.P.

Consider the following question which appeared in CAT 2004 and is based on this logic:

The sum to 12 terms of an A.P. is equal to the sum to 18 terms. What will be the sum to 30 terms for this series?

Solution: If S₁₂ = S₁₈, S₁₁ = S_19… and S₀ = S₃₀

But Sum to zero terms for any series will always be 0. Hence S₃₀ = 0.

(B) Decreasing A.P.s.

Similar to the cases of the increasing A.Ps, we can have two cases for decreasing APs –

Case 1– Decreasing A.P. with first term negative.

Case 2– Decreasing A.P. with first term positive.

I leave it to the reader to understand these cases and deduce that whatever was true for increasing A.Ps with first term negative will also be true for decreasing APs with first term positive.

GEOMETRIC PROGRESSION

Quantities are said to be in Geometric Progression when they increase or decrease by a constant factor.

The constant factor is also called the common ratio and it is found by dividing any term by the term immediately preceding it.

If we examine the series a, ar, ar², ar³, ar⁴,…

we notice that in any term the index of r is always less by one than the number of the term in the series.

If n be the number of terms and if l denote the last, or nth term, we have

l = ar^{n – 1}

When three quantities are in geometrical progression, the middle one is called the geometric mean between the other two. While arbitrarily choosing three numbers in GP, we take a/r, a and a/r. This makes it easier since we come down to two variables for the three terms.

To Find the Geometric Mean between two given Quantities

Let a and b be the two quantities; G the geometric mean. Then since a, G, b are in GP,

b/G = G/a

Each being equal to the common ratio

G² = ab

Hence G =

To Insert a given Number of Geometric Means between two given Quantities

Let a and b be the given quantities and n the required number of means to be inserted. In all there will be n + 2 terms so that we have to find a series of n + 2 terms in GP of which a is the first and b the last.

Let r be the common ratio;

Then b = the (n + 2)th term = ar^{n + 1};

\ r^{(n + 1)} =

\ r = (1)

Hence the required number of means are ar, ar², … arⁿ, where r has the value found in (1).

To Find the Sum of a Number of Terms in a Geometric Progression

Let a be the first term, r the common ratio, n the number of terms, and S_n be the sum to n terms.

If r > 1, then

S_n = (1)

If r < 1, then

S_n = (2)

Sum of an infinite geometric progression when r < 1

S_• =

Obviously, this formula is used only when the common ratio of the GP is less than one.

Similar to APs, GPs can also be logically viewed. Based on the value of the common ratio and its first term a G.P. might have one of the following structures:

(1) Increasing GPs type 1:

A G.P. with first term positive and common ratio greater than 1. This is the most common type of G.P.

e.g: 3, 6, 12, 24…(A G.P. with first term 3 and common ratio 2)

The plot of the sum of the series with respect to the number of terms in such a case will appear as follows:

(2) Increasing GPs type 2:

A G.P. with first term negative and common ratio less than 1.

e.g: –8, –4,–2, –1, – ……

As you can see in this GP all terms are greater than their previous terms.

[The following figure will illustrate the relationship between the number of terms and the sum to ‘n’ terms in this case]

(3) Decreasing G.Ps type 1:

These GPs have their first term positive and common ratio less than 1.

e.g: 12, 6, 3, 1.5, 0.75 …..

(4) Decreasing GPs type 2:

First term negative and common ratio greater than 1.

e.g: –2, –6, –18 ….

In this case the relationship looks like.

HARMONIC PROGRESSION

Three quantities a, b, c are said to be in Harmonic Progression when a/c = .

In general, if a, b, c, d are in AP then 1/a, 1/b, 1/c and 1/d are all in HP.

Any number of quantities are said to be in harmonic progression when every three consecutive terms are in harmonic progression.

The reciprocals of quantities in harmonic progression are in arithmetic progression. This can be proved as:

By definition, if a, b, c are in harmonic progression,

=

\ a(b – c) = c(a – b),

dividing every term by abc, we get

which proves the proposition.

To Find the Harmonic Mean between two given Quantities

Let a, b be the two quantities, H their harmonic mean; then 1/a, 1/H and 1/b are in A.P.;

\ =

=

i.e. H =

THEOREMS RELATED WITH PROGRESSIONS

If A, G, H are the arithmetic, geometric, and harmonic means between a and b, we have

A = (1)

G = (2)

H = (3)

Therefore, A × H = = ab = G²

that is, G is the geometric mean between A and H.

From these results we see that

A – G =

=

which is positive if a and b are positive. Therefore, the arithmetic mean of any two positive quantities is greater than their geometric mean.

Also from the equation G² = AH, we see that G is intermediate in value between A and H; and it has been proved that A > G , therefore G > H and A > G > H.

The arithmetic, geometric, and harmonic means between any two positive quantities are in descending order of magnitude.

As we have already seen in the Back to school section of this block there are some number series which have a continuously decreasing value from one term to the next – and such series have the property that they have what can be defined as the sum of infinite terms. Questions on such series are very common in most aptitude exams. Even though they cannot be strictly said to be under the domain of progressions, we choose to deal with them here.

Consider the following question which appeared in CAT 2003.

Find the infinite sum of the series:

1 + + + + + . . . . . . .

(a) 27/14    (b) 21/13

(c) 49/27    (d) 256/147

Solution: Such questions have two alternative widely divergent processes to solve them.

The first relies on mathematics using algebraic solving. Unfortunately this process being overly mathematical requires a lot of writing and hence is not advisable to be used in an aptitude exam.

The other process is one where we try to predict the approximate value of the sum by taking into account the first few significant terms. (This approach is possible to use because of the fact that in such series we invariably reach the point where the value of the next term becomes insignificant and does not add substantially to the sum). After adding the significant terms we are in a position to guess the approximate value of the sum of the series.

Let us look at the above question in order to understand the process.

In the given series the values of the terms are:

First term = 1

Second term = 4/7 = 0.57

Third term = 9/63 = 0.14

Fourth term = 16/343 = 0.04

Fifth term = 25/2401 = 0.01

Addition upto the fifth term is approximately 1.76

Options (b) and (d) are smaller than 1.76 in value and hence cannot be correct.

That leaves us with options 1 and 3

Option 1 has a value of 1.92 approximately while option 3 has a value of 1.81 approximately.

At this point you need to make a decision about how much value the remaining terms of the series would add to 1.76 (sum of the first 5 terms)

Looking at the pattern we can predict that the sixth term will be

36/7⁵ = 36/16807 = 0.002 (approx.)

And the seventh term would be 49/7⁶ = 49/117649 = 0.0004 (approx.).

The eighth term will obviously become much smaller.

It can be clearly visualized that the residual terms in the series are highly insignificant. Based on this judgement you realize that the answer will not reach 1.92 and will be restricted to 1.81. Hence the answer will be option 3.

Try using this process to solve other questions of this nature whenever you come across them. (There are a few such questions inserted in the LOD exercises of this chapter)

WORKED-OUT PROBLEMS

Problem 2.1 Two persons—Ramu Dhobi and Kalu Mochi have joined Donkey-work Associates. Ramu Dhobi and Kalu Mochi started with an initial salary of ` 500 and ` 640 respectively with annual increments of ` 25 and ` 20 each respectively. In which year will Ramu Dhobi start earning more salary than Kalu Mochi?

Solution The current difference between the salaries of the two is ` 140. The annual rate of reduction of this dfference is ` 5 per year. At this rate, it will take Ramu Dhobi 28 years to equalise his salary with Kalu Dhobi’s salary.

Thus, in the 29th year he will earn more.

This problem should be solved while reading and the thought process should be 140/5 = 28. Hence, answer is 29th year.

Problem 2.2 Find the value of the expression

1 – 6 + 2 – 7 + 3 – 8 +……… to 100 terms

(a) –250    (b) –500

(c)–450    (d) –300

Solution The series (1 – 6 + 2 – 7 + 3 – 8 +……… to 100 terms) can be rewritten as:

fi (1 + 2 + 3 + … to 50 terms) – (6 + 7 + 8 + … to 50 terms)

Both these are AP’s with values of a and d as Æ

a = 1, n = 50 and d = 1 and a = 6, n = 50 and d = 1 respectively.

Using the formula for sum of an AP we get:

Æ 25(2 + 49) – 25(12 + 49)

Æ 25(51 – 61) = –250

Alternatively, we can do this faster by considering (1 – 6), (2 – 7), and so on as one unit or one term.

1 – 6 = 2 – 7 = … = –5. Thus the above series is equivalent to a series of fifty –5’s added to each other.

So, (1 – 6) + (2 – 7) + (3 – 8) + … 50 terms = –5 × 50 = –250

Problem 2.3 Find the sum of all numbers divisible by 6 in between 100 to 400.

Solution Here 1st term = a = 102 (which is the 1st term greater than 100 that is divisible by 6.)

The last term less than 400, which is divisible by 6 is 396.

The number of terms in the AP; 102, 108, 114…396 is given by [(396 – 102)/6] +1= 50 numbers.

Common difference = d = 6

So, S = 25 (204 + 294) = 12450

Problem 2.4 If x, y, z are in GP, then 1/(1 + log₁₀x), 1/ (1 + log₁₀y) and 1/(1 + log₁₀z) will be in:

(a) AP    (b) GP

(c) HP    (d) Cannot be said

Solution Go through the options.

Checking option (a), the three will be in AP if the 2nd expression is the average of the 1st and 3rd expressions. This can be mathematically written as

2/(1 + log₁₀y) = [1/(1 + log₁₀x)] + [1/(1 + log₁₀z)]

=

=

Applying our judgement, there seems to be no indication that we are going to get a solution.

Checking option (b)

[1/(1 + log₁₀y)]² = [1/(1 + log₁₀x)] [1/(1 + log₁₀z)]

= [1/(1 + log₁₀(x + z) + log₁₀xz)]

Again we are trapped and any solution is not in sight.

Checking option (c).

1/(1 + log₁₀x ), 1/(1 + log₁₀y) and 1/(1 + log₁₀z) are in HP then 1 + log₁₀x, 1 + log₁₀y and 1 + log₁₀z will be in AP.

So, log₁₀x, log₁₀y and log₁₀z will also be in AP.

Hence, 2 log₁₀y = log₁₀x + log₁₀z

fiy² = xz which is given.

So, (c) is the correct option.

Alternatively, you could have solved through the following process.

x, y and z are given as logarithmic functions.

Assumex = 1, y = 10 and z = 100 as x, y, z are in GP

So,1 + log₁₀x = 1, 1 + log₁₀y = 2 and 1 + log₁₀z = 3

fi Thus we find that since 1, 2 and 3 are in AP, we can assume that

1 + log₁₀x , 1 + log₁₀y and 1 + log₁₀z are in AP

fi Hence, by definition of an HP we have that 1/(1 + log₁₀x), 1/(1 + log₁₀y ) and 1/(1 + log₁₀z) are in HP. Hence, option (c) is the required answer.

Problem 2.5 Find t₁₀ and S₁₀ for the following series:

1, 8, 15,…

Solution This is an AP with first term 1 and common difference 7.

t₁₀ = a + (n – 1) d = 1 + 9 × 7 = 64

S₁₀ =

= = 325

Alternatively, if the number of terms is small, you can count it directly.

Problem 2.6 Find t₁₈ and S₁₈ for the following series:

2, 8, 32, …

Solution This is a GP with first term 2 and common ratio 4.

t₁₈ = ar^{n – 1} = 2.4¹⁷

S₁₈ =

Problem 2.7 Is the series 1, 4,… to n terms an AP, or a GP, or an HP, or a series which cannot be determined?

Solution To determine any progression, we should have at least three terms.

If the series is an AP then the next term of this series will be 7

Again, if the next term is 16, then this will be a GP series (1, 4, 16 …)

So, we cannot determine the nature of the progression of this series.

Problem 2.8 Find the sum to 200 terms of the series

1 + 4 + 6 + 5 + 11 + 6 +…

(a) 30,200   (b) 29,800

(c) 30,200   (d) None of these

Solution Spot that the above series is a combination of two APs.

The 1st AP is (1 + 6 + 11 +…) and the 2nd AP is (4 + 5 + 6 +…)

Since the terms of the two series alternate, S = (1 + 6 + 11 +… to 100 terms) + (4 + 5 + 6 + … to 100 terms)

= Æ (Using the formula for the sum of an AP)

= 50[497 + 107] = 50[604] = 30200

Alternatively, we can treat every two consecutive terms as one.

So we will have a total of 100 terms of the nature:

(1 + 4) + (6 + 5) + (11 + 6) … Æ 5, 11, 17…

Now, a = 5, d = 6 and n = 100

Hence the sum of the given series is

S = × [2 × 5 + 99 × 6]

= 50[604] = 30200

Problem 2.9 How many terms of the series –12, –9, –6,… must be taken that the sum may be 54?

Solution Here S = 54, a = –12, d = 3, n is unknown and has to be calculated. To do so we use the formula for the sum of an AP and get.

54 =

or 108 = –24n – 3n + 3n² or 3n² – 27n – 108 = 0

or n² – 9n – 36 = 0, or n² – 12n + 3n – 36 = 0

n(n – 12) + 3(n – 12) = 0 fi (n + 3) (n – 12) = 0

The value of n (the number of terms) cannot be negative. Hence –3 is rejected.

So we have n = 12

Alternatively, we can directly add up individual terms and keep adding manually till we get a sum of 54. We will observe that this will occur after adding 12 terms. (In this case, as also in all cases where the number of terms is mentally manageable, mentally adding the terms till we get the required sum will turn out to be much faster than the equation based process.

Problem 2.10 Find the sum of n terms of the series 1.2.4 + 2.3.5 + 3.4.6 + …

(a)n(n + 1)(n + 2)

(b)(n(n + 1)/12)(3n² + 19n + 26)

(c)((n + 1)(n + 2)(n + 3))/4

(d)(n² (n + 1)(n + 2)(n + 3))/3

Solution In order to solve such problems in the examination, the option-based approach is the best. Even if you can find out the required expression mathematically, it is advisable to solve through the options as this will end up saving a lot of time for you. Use the options as follows:

If we put n = 1, we should get the sum as 1.2.4 = 8. By substituting n = 1 in each of the four options we will get the following values for the sum to 1 term:

Option (a) gives a value of: 6

Option (b) gives a value of: 8

Option (c) gives a value of: 6

Option (d) gives a value of: 8

From this check we can reject the options (a) and (c).

Now put n = 2. You can see that up to 2 terms, the expression is 1.2.4 + 2.3.5 = 38

The correct option should also give 38 if we put n = 2 in the expression. Since, (a) and (c) have already been rejected, we only need to check for options (b) and (d).

Option (b) gives a value of 38

Option (d) gives a value of 80.

Hence, we can reject option (d) and get (b) as the answer.

LEVEL OF DIFFICULTY (I)

LEVEL OF DIFFICULTY (II)

Directions for Questions 37 to 39: Answer these questions based on the following information.

There are 250 integers a₁, a₂…….a₂₅₀, not all of them necessarily different. Let the greatest integer of these 250 integers be referred to as Max, and the smallest integer be referred to as Min. The integers a₁ through a₁₂₄ form sequence A, and the rest form sequence B. Each member of A is less than or equal to each member of B.

Directions for Questions 42 and 43: These questions are based on the following data.

At Burger King—a famous fast food centre on Main Street in Pune, burgers are made only on an automatic burger making machine. The machine continuously makes different sorts of burgers by adding different sorts of fillings on a common bread. The machine makes the burgers at the rate of 1 burger per half a minute. The various fillings are added to the burgers in the following manner. The 1st, 5th, 9th, ...... burgers are filled with a chicken patty; the 2nd,9th, 16th, ........ burgers with vegetable patty; the 1st, 5th, 9th, ...... burgers with mushroom patty; and the rest with plain cheese and tomato fillings.

The machine makes exactly 660 burgers per day.

LEVEL OF DIFFICULTY (III)

Level of Difficulty (I)

Hints

Level of Difficulty (II)

2.a + (a + d) + (a + 2d) + (a + 3d) = 20

and a(a + 3d) = (a + d) (a + 2d)

4.Calculate the sum of an AP with first term 1, common difference 1 and last term 12. Multiply this sum by 4 for 2 days.

5.The maximum sum will occur when the last term is either 2 or 0.

6.Visualise the AP as 7, 14…196.

9.The AP is 105, 112 …994.

10.The common difference is = 29.2.

11.See the terms of the series in 33 blocks of 3 each. This will give the AP – 4, –5, –6…–33. Further, the hundredth term will be 34.

12.Solve through options.

14.The first drop is 120 metres. After this the ball will rise by 96 metres and fall by 96 meters. This process will continue in the form of an infinite GP with common ratio 0.8 and first term 96.

The required answer will be got by

120 + 96 * 1.25 * 2

15.Take any GP and solve by using values.

18.Solve by using values to check options.

22.The difference between the seventh and third term is given by

(a + 6d) – (a + d)

23. = 7.

27.The required answer will be by adding 20 terms of the GP starting with the first term as 1000 and the common ratio as 1.05.

30.Visualise it as an infinite GP with common ratio 0.9.

Level of Difficulty (III)

1.Difference between the tenth and the sixth term = –16

or (a + 9d) – (a + 5d) = –16

Æ d = –4

2.Sum of the first term and the fifth term = 10

or a + a + 4d = 10

or a + 2d = 5(1)

and, the sum of all terms of the AP except for the 1st term = 99

or 9a + 45d = 99

a + 5d = 11(2)

Solve (1) and (2) to get the answer.

3.The second statement gives the equation as a + 8d = 2(a + 3d) + 6

or a – 2d = 6

Now, use the options to find the value of d, and put these values to check the equation obtained from the first statement.

i.e. (a + 2d) (a + 5d) = 406

4.To plant the 1st sapling, Mithilesh will cover 20 m; to plant the 2nd sapling he will cover 40 m and so on. But for the last sapling, he will cover only the distance from the starting point to the place where the sapling has to be planted.

5.Assume a series having a few number of terms e.g.

1, 2, 4, 8, 16, 32…

Now sum of all the terms at the even places = 42 (P₂)

and sum of all the terms at the odd places = 21 (P₁)

common ratio of this series = = 2 = P₂/P₁.

6.Use the same process as illustrated above.

7.Check the options by putting n = 1, 2, … and then equate it with the original equation given in question.

9.For 1 term, the value should be:

6² + 8 = 44

Only option (b) gives 44 for n = 1

10.Sum of the AP for n sides = Sum of interior angles of a polygon of n sides.

× (2a + (n – 1)d) = (2n – 4) × 90

where a = 120° and d = 5°.

11.Solve using options to check for the correct answer.

12.a + (a + 4d) = 26 and (a + d) (a + 3d) = 160

Alternatively, you can try to look at the factors of 160 and create an AP such that it meets the criteria.

Thus, 160 can be written as

2 × 80

4 × 40

8 × 20

10 × 16

and so on.

If we consider 8 × 20, then one possibility is that d = 6 and the first and fifth terms are 2 and 26. But 2 + 26 π 28. Hence, this cannot be the correct factors.

Try 10 × 16. This will give you,

a = 7, d = 3 and 5th term = 19

and 7 + 19 = 26 satisfies the condition

13.A possible AP satisfying this condition is

0, 1, 2, 3, 4, 5, 6, 7, 8, …

14.Assume the fifth term as (a + 4d), the eleventh term as (a + 10d), the second term as (a + d), the fourteenth term as (a + 13d), the first term as (a) and the fifteenth term as (a + 14d).

Then, First term × Fifteenth term

= a × (a + 14d) = a² + 14ad

Also (a + d) (a + 13d) = P

and (a + 4d)² + (a + 10d)² = 3

16.The solution (from the options) has got something to do with either 2¹⁰⁰ or 2¹⁰¹ for 100 terms. Hence, for 3 terms recreate the options and crosscheck with the actual sum.

For 3 terms: Sum = 2 + 8 + 24 = 34.

(a)100 × 2¹⁰¹ + 2 for 100 terms becomes 3 × 2⁴ + 2 for 3 terms.

= 48 + 2 = 50 π 34. Hence is not correct.

(b)99 × 2¹⁰⁰ + 2 for 100 terms becomes 2 × 2³ + 2 for 3 terms.

But this does not give 34. Hence is not correct.

(c)99 × 2¹⁰¹ + 2 Æ 2 × 2⁴ + 2 = 34

(d)100 × 2¹⁰⁰ + 2 Æ 3 × 2³ + 2 π 34.

Hence, option (c) is correct.

17.r = 2 and a + ar³ = 108.

20.Solve by checking options and using principles of logarithms.

21.The sum of the first 20 terms will be

a + (a + d) + (a + 2d) … + (a + 19d)

i.e.20a + 190d = 400

Similarly, use the sum to fifty terms.

23.For a product a × b × c to be maximum, given a + b + c = constant, the condition is a = b = c.

27.The length of sides of successive triangles form a GP with common ratio 1/3.

28.The area of successive triangles form an infinite GP with a common ratio 1/4.

29.Common ration = 1/.

31.Ration of sum of the first 2n terms to the first n terms is equal to 3.

Thus,

= 3

Solve to get, 2a = (n + 1)d.

Put values for a and n to get a value for d and check for the conditions given in the question.

33.Visualize the AP and solve using standard formulae.

34.Take any values for the numbers.

Say, the two positive numbers are 1 and 27.

Then, a = 14, b = 3 and c = 9.

Solutions and Shortcuts

Level of Difficulty (I)

1.In order to count the number of terms in the AP, use the short cut:

[(last term – first term)/ common difference] + 1. In this case it would become:

[(130 – 20)/5] +1 = 23. Option (b) is correct.

2.7000 – 500 – 12500 means that the starting scale is 7000 and there is an increment of 500 every year. Since, the total increment required to reach the top of his scale is 5500, the number of years required would be 5500/500 = 11. Option (a) is correct.

3.Since the 8th and the 12th terms of the AP are given as 39 and 59 respectively, the difference between the two terms would equal 4 times the common difference. Thus we get 4d = 59 – 39 = 20. This gives us d = 5. Also, the 8^th term in the AP is represented by a + 7d, we get:

a + 7d = 39 Æ a + 7 × 5 = 39 Æ a = 4. Option (c) is correct.

4.If we take the sum of the sides we get the perimeters of the squares. Thus, if the side of the respective squares are a₁, a₂, a₃, a₄… their perimeters would be 4a₁, 4a₂, 4a₃, 4a₄. Since the perimeters are in GP, the sides would also be in GP.

5.The number of terms in a series are found by:

6.The first common term is 3, the next will be 9 (Notice that the second common term is exactly 6 away from the first common term. 6 is also the LCM of 2 and 3 which are the respective common differences of the two series.)

Thus, the common terms will be given by the A.P 3, 9, 15 ….., last term. To find the answer you need to find the last term that will be common to the two series.

The first series is 3, 5, 7 … 239

While the second series is 3, 6, 9 ….. 240.

Hence, the last common term is 237.

Thus our answer becomes + 1 = 40

7.Trying Option (a),

We get least term 5 and largest term 30 (since the largest term is 6 times the least term).

The average of the A.P becomes (5 + 30)/2 = 17.5

Thus, 17.5 × n = 105 gives us:

to get a total of 105 we need n = 6 i.e. 6 terms in this A.P. That means the A.P. should look like:

5, _ , _ , _, _, 30.

It can be easily seen that the common difference should be 5. The A.P, 5, 10, 15, 20, 25, 30 fits the situation.

The same process used for option (b) gives us the A.P. 10, 35, 60. (10 + 35 + 60 = 105) and in the third option 15, 90 (15 + 90 = 105).

Hence, all the three options are correct.

8.The first term is 20 and the common difference is –5, thus the 15th term is:

20 + 14 × (–5) = –50. Option (c) is correct.

9.The difference between the amounts at the end of 4 years and 10 years will be the simple interest on the initial capital for 6 years.

Hence, 360/6 = 60 =(simple interest.)

Also, the Simple Interest for 4 years when added to the sum gives 1240 as the amount.

Hence, the original sum must be 1000.

10.The three parts are 3, 5 and 7 since 3² + 5² + 7² = 83. Since, we want the smallest number, the answer would be 3. Option (b) is correct.

11.a = 5, a + 2d = 15 means d = 5. The 16th term would be a + 15d = 5 + 75 = 80. The sum of the series would be given by:

[16/2] × [5 + 80] = 16 × 42.5 = 680. Option (d) is correct.

12.Use trial and error by using various values from the options.

If you find the sum of the series till 18 terms the value is 513. So also for 19 terms the value of the sum would be 513. Option (c) is correct.

13.Solve this question through trial and error by using values of n from the options:

For 19 terms, the series would be 5 + 8 + 11 + …. + 59 which would give us a sum for the series as 19 × 32 = 608. The next term (20th term of the series) would be 62. Thus, 608 + 62 = 670 would be the sum to 20 terms. It can thus be concluded that for 20 terms the value of the sum of the series is not less than 670. Option (a) is correct.

14.His total earnings would be 60 + 63 + 66 + … + 117 = ` 1770. Option (a) is correct.

15.The series would be 5, 20, 80, 320, 1280, 5120, 20480. Thus, there are a total of 7 terms in the series. Option (c) is correct.

16.Sum of a G.P. with first term 1 and common ratio 2 and no. of terms 20.

= 2²⁰ – 1

17.16r⁴ = 81 Æ r⁴ = 81/16 Æ r = 3/2. Thus, 4th term = ar³ = 16 × (3/2)³ = 54. Option (c) is correct.

18.In the case of a G.P. the 7th term is derived by multiplying the fourth term thrice by the common ratio. (Note: this is very similar to what we had seen in the case of an A.P.)

Since, the seventh term is derived by multiplying the fourth term by 8, the relationship.

r³ = 8 must be true.

Hence, r = 2

If the fifth term is 48, the series in reverse from the fifth to the first term will look like:

48, 24, 12, 6, 3. Hence, option (b) is correct.

19.Visualising the squares below 84, we can see that the only way to get the sum of 3 squares as 84 is: 2² + 4² + 8² = 4 + 16 + 64. The largest number is 8. Option (a) is correct.

20.The series would be given by: 1, 5, 9… which essentially means that all the numbers in the series are of the form 4n + 1. Only the value in option (c) is a 4n + 1 number and is hence the correct answer.

21.The series will be 301, 308, …….. 497

Hence, Answer = + 1 = 29

22.The answer to this question can be seen from the options. Both 2, 6, 18 and 18, 6, 2 satisfy the required conditions- viz: GP with sum of first and third terms as 20. Thus, option (c) is correct.

23.We need the sum of the series 20 + 24 + 28 to cross 1000. Trying out the options, we can see that in 20 years the sum of his savings would be: 20 + 24 + 28 + … + 96. The sum of this series would be 20 × 58 =1160. If we remove the 20th year we will get the series for savings for 19 years. The series would be 20 + 24 + 28 + …. 92. Sum of the series would be 1160 – 96 = 1064. If we remove the 19th year’s savings the savings would be 1064 – 92 which would go below 1000. Thus, after 19 years his savings would cross 1000. Option (a) is correct.

24.The answer to this question cannot be determined because the question is talking about income and asking about savings. You cannot solve this unless you know the value of the expenditure the man incurs over the years. Thus, “Cannot be Determined” is the correct answer.

25.Similar to what we saw in question 18,

4th term × r⁶ = 10th term.

The 4th term here is 3^{– 1}and the tenth term is 3⁵.

Hence 3^{– 1}× r⁶ = 3⁵

Gives us: r = 3.

Hence, the second term will be given by (fourth term/r²)

¹/₃ × ¹/₃ × ¹/₃ = ¹/₂₇

[Note: To go forward in a G.P. you multiply by the common ratio, to go backward in a G.P. you divide by the common ratio.]

26.a + 6d = 6 and a + 20d = –22. Solving we get 14d = –28 Æ d = –2. 26th term = 21^st term + 5d = –22 + 5 × (–2) = –32. Option (b) is correct.

27.Since the sum of 5 numbers in AP is 30, their average would be 6. The average of 5 terms in AP is also equal to the value of the 3^rd term (logic of the middle term of an AP). Hence, the third term’s value would be 6. Option (b) is correct.

28.The answer will be given by:

[10 + 11 + 12 + …….. + 50] – [16 + 24 + … + 48]

= 41 × 30 – 32 × 5

= 1230 – 160 = 1070.

29.Think like this:

The average of the first 4 terms is 7, while the average of the first 8 terms must be 11.

Now visualize this :

Hence, d = 4/2 = 2 [Note: understand this as a property of an A.P.]

Hence, the average of the 6th and 7th terms = 15 and the average of the 8th and 9th term = 19

But this (19) also represents the average of the 16 term A.P.

Hence, required answer = 16 × 19 = 304.

30.Go through the options. The correct option should give value as 1, when n = 3 and as 8 when n = 8.

Only option (a) satisfies both conditions.

31.The series is: 1/81, –1/27, 1/9, –1/3, 1, –3, 9, –27, 81, –243, 729. There are 11 terms in the series. Option (a) is correct.

32.1/8 × r⁵ = 128 Æ r⁵ = 128 × 8 = 1024 Æ r = 4. Thus, the series would be 1/8, 1/2, 2, 8, 32, 128. The third geometric mean would be 8. Option (d) is correct.

33.AM = 25 means that their sum is 50 and GM = 7 means their product is 49. The numbers can only be 49 and 1. Option (c) is correct.

34.Trial and error gives us that for option (b):

With the ratio 4:1, the numbers can be taken as 4x and 1x. Their AM would be 2.5x and their GM would be 2x. The GM can be seen to be 20% lower than the AM. Option (b) is thus the correct answer.

35.The total savings would be given by the sum of the series:

100 + 150 + 200 + 650 = 12 × 375 = ` 4500. Option (b) is correct.

36.In order to find how many times the alarm rings we need to find the number of numbers below 100 which are not divisible by 2,3, 5 or 7. This can be found by:

100 – (numbers divisible by 2) – (numbers divisible by 3 but not by 2) – (numbers divisible by 5 but not by 2 or 3) – (numbers divisible by 7 but not by 2 or 3 or 5).

Numbers divisible by 2 up to 100 would be represented by the series 2, 4, 6, 8, 10..100 Æ A total of 50 numbers.

Numbers divisible by 3 but not by 2 up to 100 would be represented by the series 3, 9, 15, 21…99 Æ A total of 17 numbers. Note use short cut for finding the number of number in this series :

[(last term – first term)/ common difference] + 1 = [(99 – 3)/6] + 1 = 16 + 1 = 17.

Numbers divisible by 5 but not by 2 or 3: Numbers divisible by 5 but not by 2 up to 100 would be represented by the series 5, 15, 25, 35…95 Æ A total of 10 numbers. But from these numbers, the numbers 15, 45 and 75 are also divisible by 3. Thus, we are left with 10 – 3 = 7 new numbers which are divisible by 5 but not by 2 and 3.

Numbers divisible by 7, but not by 2, 3 or 5:

Numbers divisible by 7 but not by 2 upto 100 would be represented by the series 7, 21, 35, 49, 63, 77, 91 Æ A total of 7 numbers. But from these numbers we should not count 21, 35 and 63 as they are divisible by either 3 or 5. Thus a total of 7 – 3 = 4 numbers are divisible by 7 but not by 2, 3 or 5.

37.For looking at the zeroes in the expression we should be able to see that the number of zeroes in the third term onwards is going to be very high. Thus, the number of zeroes in the expression would be given by the number of zeroes in:

4 + 24²⁴. 24²⁴ has a unit digit 6. Hence, the number of zeroes in the expression would be 1. Option (d) is correct.

38.Since the sum of 22 terms of the AP is 385, the average of the numbers in the AP would be 385/22 = 17.5. This means that the sum of the first and last terms of the AP would be 2 × 17.5 = 35. Trial and error gives us the terms of the required GP as 7, 14, 28. Thus, the common ratio of the GP can be 2.

39.The sum of the interior angles of a polygon are multiples of 180 and are given by (n – 1) × 180 where n is the number of sides of the polygon. Thus, the sum of interior angles of a polygon would be a member of the series: 180, 360, 540, 720, 900, 1080, 1260

The sum of the series with first term 100 and common difference 10 would keep increasing when we take more and more terms of the series. In order to see the number of sides of the polygon, we should get a situation where the sum of the series represented by 100 + 110 + 120… should become a multiple of 180. The number of sides in the polygon would then be the number of terms in the series 100, 110, 120 at that point.

If we explore the sums of the series represented by

100 + 110 + 120…

We realize that the sum of the series becomes a multiple of 180 for 8 terms as well as for 9 terms.

It can be seen in: 100 + 110 + 120 + 130 + 140 + 150 + 160 + 170 = 1080

Or 100 + 110 + 120 + 130 + 140 + 150 + 160 + 170 + 180 = 1260.

40.The sum of the total distance it travels would be given by the infinite sum of the series:

420 × 8/1 + 367.5 × 8/1 = 3360 + 2940 = 6300. Option (b) is correct.

41.The two series till their hundredth terms are 13, 15, 17….211 and 14, 17, 20…311. The common terms of the series would be given by the series 17, 23, 29….209. The number of terms in this series of common terms would be 192/6 + 1 = 33. Option (d) is correct.

42.The area of the first square would be 64 sq cm. The second square would give 32, the third one 16 and so on. The infinite sum of the geometric progression 64 + 32 + 16 + 8… = 128. Option (a) is correct.

43.It can be seen that for the series the average of two terms is 2, for 3 terms the average is 3 and so on. Thus, the sum to 2 terms is 2², for 3 terms it is 3² and so on. For 11111 terms it would be 11111² = 123454321. Option (d) is correct.

44.The maximum score would be the sum of the series 9 + 13 + …. + 389 + 393 + 397 = 98 × 406/2 = 19894. Option (d) is correct.

45.The series would be 8, 8/3, 8/9 and so on. The sum of the infinite series would be 8/(1 – 1/3) = 8 × 3/2 = 12. Option (a) is correct.

46.The maximum sum would occur when we take the sum of all the positive terms of the series.

The series 25, 24.5, 24, 23.5, 23, ….…. 1, 0.5, 0 has 51 terms. The sum of the series would be given by:

n × average = 51 × 12.5 = 637.5

Option (a) is correct.

47.The side of the first equilateral triangle being 24 units, the first perimeter is 72 units. The second perimeter would be half of that and so on.

72, 36, 18 …

The infinite sum of this series = 72(1 – 1/2) = 144. Option (d) is correct.

48.Solve using options. Option (a) fits the situation as 16 + 2 + 2/8 + 2/64 meets the conditions of the question. Option (a) is correct.

49.The 33^rd term of the sequence would be the 17^th term of the sequence 3, 9, 15, 21 ….

The 17^th term of the sequence would be 3 + 6 × 16 = 99.

50.The sum to 33 terms of the sequence would be:

The sum to 17 terms of the sequence 3, 9, 15, 21, …99 + The sum to 16 terms of the sequence 8, 13, 18, 83.

The required sum would be 17 × 51 + 16 × 45.5 = 867 + 728 = 1595.

Level of Difficulty (II)

1.Identify an A.P. which satisfies the given condition.

Suppose we are talking about the second and third terms of the A.P.

Then an A.P. with second term 3 and third term 2 satisfies the condition.

a times the ath term = b times the bth term.

In this case the value of a = 2 and b = 3.

Hence, for the (a + b)^th term, we have to find the fifth term.

It is clear that the fifth term of this A.P. must be zero.

Check the other three options to see whether any option gives 0 when a = 2 and b = 3.

Since none of the options b, c or d gives zero for this particular value, option (a) is correct.

2.Since the four parts of the number are in AP and their sum is 20, the average of the four parts must be 5. Looking at the options for the largest part, only the value of 8 fits in, as it leads us to think of the AP 2, 4, 6, 8. In this case, the ratio of the product of the first and fourth (2 × 8) to the product of the first and second (4 × 6) are equal. The ratio becomes 2:3.

3.View: 1 – 4 + 5 – 8 + 9 – 12 . . . . . 50 terms as

(1 – 4) + (5 – 8) + (9 – 12) …… 25 terms.

Hence, –3 + –3 + –3 . . . . 25 terms

= 25 × – 3 = –75.

4.In the course of 2 days the clock would strike 1 four times, 2 four times, 3 four times and so on. Thus, the total number of times the clock would strike would be:

4 + 8 + 12 + ..48 = 26 × 12 = 312. Option (b) is correct.

5.Since this is a decreasing A.P. with first term positive, the maximum sum will occur upto the point where the progression remains non – negative.

44, 42, 40 ….. 0

Hence, 23 terms × 22 = 506.

6.The sum of the required series of integers would be given by 7 + 14 + 21 + ….196 = 28 × 101.5 = 2842. Option (b) is correct.

7.A little number juggling would give you 2nd term is 1/3 and 3rd term is ½ is a possible situation that satisfies the condition.

The A.P. will become:

1/6, 1/3, 1/2, 2/3, 5/6, 1

or in decimal terms, 0.166, 0.333, 0.5, 0.666, 0.833, 1

Sum to 6 terms = 3.5

Check the option with m = 2 and n = 3. Only option (c) gives 3.5. Hence, must be the answer.

8.101 + 103 + 105 + … 199 = 150 × 50 = 7500

9.The required sum would be given by the sum of the series 105, 112, 119, …. 994. The number of terms in this series = (994 – 105)/7 + 1 = 127 + 1 = 128. The sum of the series = 128 × (average of 105 and 994) = 70336. Option (c) is correct.

10.5 × average of 107 and 253 = 5 × 180 = 900. Option (c) is correct.

11.The first 100 terms of this series can be viewed as:

(1 – 2 –3) + (2 – 3 –4) + . . . . + (33 – 34 – 35) + 34

The first 33 terms of the above series (indicated inside the brackets) will give an A.P.: –4, –5, –6 .... –36

Sum of this A.P. = 33 × –20 = –660

Answer = –660 + 34 = –626

12.Solve this one through trial and error. For n = 2 terms the sum upto 2 terms is equal to 96. Putting n = 2, in the options it can be seen that for option (b) the sum to two terms would be given by

8 × (1000 – 10 – 18)/81 = 8 × 972/81 = 8 × 12 = 96.

13.If we take the values of a, b, and c as 10,100 and 1000 respectively, we get log a, log b and log c as 1, 2 and 3 respectively. This clearly shows that the values of log a, log b and log c are in AP.

14.The path of the rubber ball is:

In the figure above, every bounce is 4/5th of the previous drop.

In the above movement, there are two infinite G.Ps (The GP representing the falling distances and the GP representing the Rising distances.)

The required answer: (Using a/(1–r) formula)

= 1080

15.Solve this for a sample GP. Let us say we take the GP as 2, 6, 18, 54. x, the first term is 2, let n = 3 then the 3^rd term y = 18 and the product of 3 terms p = 2 × 6 × 18 = 216 = 6³. The value of p² = 216 × 216 = 6⁶.

Putting these values in the options we have:

Option (a) gives us (xy)^n–1 = 36² which is not equal to the value of p² we have from the options

Option (b) gives us (xy)ⁿ = 36³ = 6⁶ which is equal to the value of p² we have from the options.

It can be experimentally verified that the other options yield values of p² which are different from 6⁶ and hence we can conclude that option (b) is correct.

16.Trying to plug in values we can see that the infinite sum of the GP 16, 8, 4, 2… is 32 and hence the third term is 4.

17.The expression can be written as x^{(1/2 + ¼ + 1/8 +1/16…..)} = x^{INFINTE SUM OF THE GP} = x¹. Option (b) is correct.

18.For n = 1, the sum should be 6. Option (b), (c) and (d) all give 6 as the answer.

For n = 2, the sum should be 30.

Only option d gives this value. Hence must be the answer.

19.From the facts given in the question it is self evident that the common ratio of the GP must be 2 (as the sum of the 2^nd and 4^th term is twice the sum of the first and third term). After realizing this, the question is about trying to match the correct sums by taking values from the options.

The GP formed from option (c) with a common ratio of 2 is: 8,16,32,64 and this GP satisfies the conditions of the problem- sum of 1^st and 3^rd term is 40 and sum of 2^nd and 4^th term is 80.

20.Since the sum of the first five even terms is 15, we have that the 2^nd, 4^th, 6^th, 8^th and 10^th term of the AP should add up to 15. We also need to understand that these 5 terms of the AP would also be in an AP by themselves and hence, the value of the 6^th term (being the middle term of the AP) would be the average of 15 over 5 terms. Thus, the value of the 6^th term is 3. Also, since the sum of the first three terms of the AP is –3, we get that the 2^ndterm would have a value of –1. Thus, the AP can be visualized as:

_, -1, _,_,_,3,….

Thus, it is obvious that the AP would be –2, –1, 0, 1, 2, 3. The second term is –1. Thus, option (c) is correct.

21.Second term = a + d, Fifth term = a + 4d; third term = a + 2d, seventh term = a + 6d.

Thus, 2a + 5d = 8 and 2a + 8d = 14 Æ d = 2 and a = –1.

The eleventh term = a + 10d = –1 + 20 = 19. Thus, option (a) is correct.

22.If the difference between the seventh and the second term is 20, it means that the common difference is equal to 4. Since, the third term is 9, the AP would be 1, 5, 9, 13, 17, 21, 25, 29 and the sum to 8 terms for this AP would be 120. Thus, option (d) is correct.

23.5d = 35 Æ d = 7. Thus, the numbers are 4, 11, 18, 25, 32, 39. The largest number is 32. Option (c) is correct.

24.Find sum of the series:

104, 109, 114 . . . . 999

Average × n = 551.5 × 180 = 99270

25.Since the sum of the first three terms of the AP is 30, the average of the AP till 3 terms would be 30/3 = 10. The value of the second term would be equal to this average and hence the second term is 10. Using the information about the sum of squares of the first and second terms being 116, we have that the first term must be 4. Thus, the AP has a first term of 4 and a common difference of 6. The seventh term would be 40. Thus, option (b) is correct.

26.The combined travel would be 25 on the first day, 26 on the second day, 27 on the third day, 28 on the fourth day, 29 on the fifth day and 30 on the sixth day. They meet after 6 days. Option (c) is correct.

27.This is a calculation intensive problem and you are not supposed to know how to do the calculations in this question mentally. The problem has been put here to test your concepts about whether you recognize how this is a question of GPs. If you feel like, you can use a calculator/ computer spreadsheet to get the answer to this question.

The logic of the question would hinge on the fact that the value of the investment of the fifteenth year would be 1000. At the end of the 15^th year, the investment of the 14^th year would be equal to 1000 × 1.05, the 13^th year’s investment would amount to 1000 × 1.05² and so on till the first year’s investment which would amount to 1000 × 1.05¹⁴ after 15 years.

Thus, you need to calculate the sum of the GP:

1000, 1000 × 1.05, 1000 × 1.05², 1000 × 1.05³ for 15 terms.

28.Since, sum to n terms is given by (n + 8),

Sum to 1 terms = 9

Sum to 2 terms = 10

Thus, the 2nd term must be 1.

29.Solve this question by looking at hypothetical values for n and 2n terms. Suppose, we take the sum to 1 (n = 1) term of the first series and the sum to 2 terms (2n = 2) of the second series we would get A/B as 1/1.5 = 2/3.

For n = 2 and 2n = 4 we get A = 1.25 and B = 1.875 and A/B = 1.25/1.875 = 2/3.

Thus, we can conclude that the required ratio is always constant at 2/3 and hence the correct option is (c).

30.We need to find the infinite sum of the GP: 100, 90, 81…(first term = 100 and common ratio= 0.9)

We get: Infinite sum of the series as 1000. Thus, option (b) is correct.

31.Questions such as these have to be solved on the basis of a reading of the pattern of the question. The sum upto the first term is: 1/5. Upto the second term it is 2/9 and upto the third term it is 3/13. It can be easily seen that for the first term, second term and third term the numerators are 1, 2 and 3 respectively. Also, for 1/5 – the 5 is the second value in the denominator of 1/1 × 5 (the first term); for 2/9 also the same pattern is followed—as 9 comes out of the denominator of the second term of series and for 3/13 the 13 comes out of the denominator of the third term of the series and so on. The given series has 56 terms and hence the correct answer would be 56/225.

32.Solve this on the same pattern as Question 31 and you can easily see that for the first term sum of the series is , for 2 terms we have the sum as and so on. For the given series of 120 terms the sum would be = 10.

Option (a) is correct.

33.If you look for a few more terms in the series, the series is:

1, 1/3, 1/6, 1/10, 1/15, 1/21, 1/28, 1/36, 1/45, 1/55, 1/66, 1/78, 1/91, 1/105, 1/120, 1/136, 1/153 and so on. If you estimate the values of the individual terms it can be seen that the sum would tend to 2 and would not be good enough to reach even 2.25.

Thus, option (a) is correct.

34.Solve this using trial and error. For 1 term the sum should be 40 and we get 40 only from the first option when we put n = 1. Thus, option (a) is correct.

35.For this question too you would need to read the pattern of the values being followed. The given sum has 13 terms.

It can be seen that the sum to 1 term = ½

Sum to 2 terms = 2/3

Sum to 3 terms = ¾

Hence, the sum to 13 terms would be 13/14.

36.The sum to infinite terms would tend to 1 because we would get (infinity)/(infinity + 1).

37.All members of A are smaller than all members of B. In order to visualize the effect of the change in sign in A, assume that A is {1, 2, 3…124} and B is {126, 127…250}. It can be seen that for this assumption of values neither options (a), (b) or (c) is correct.

38.If elements of A are in ascending order a₁₂₄ would be the largest value in A. Also a₁₂₅ would be the largest value in B. On interchanging a₁₂₄ and a₁₂₅, A continues to be in ascending order, but B would lose its descending order arrangement since a₁₂₄ would be the least value in B. Hence, option a is correct.

39.Since the minimum is in A and the maximum is in B, the value of x cannot be less than Max-Min.

40.It is evident that the whole question is built around Arithmetic progressions. The 5^th row has an average of 55, while the 15^th row has an average of 65. Since even column wise each column is arranged in an AP we can conclude the following:

1^st row – average 51 – total = 23 × 51

2^nd row – average 52 – total 23 × 52…..

23^rd row – average 73 – total 23 × 73

The overall total can be got by using averages as:

23 × 23 × 62 = 32798

41.The numbers forming an AP would be:

123, 135, 147, 159, 210, 234, 246, 258, 321, 345, 357, 369, 432, 420, 456, 468, 543, 531, 567, 579, 654, 642, 630, 678, 765, 753, 741, 789, 876, 864, 852, 840, 987, 975, 963 and 951. A total of 36 numbers.

If we count the GPs we get:

124, 139, 248, 421, 931, 842—a total of 6 numbers.

Hence, we have a total of 42 3 digit numbers where the digits are either APs or GPs.

Thus, option (d) is correct.

42.Total burgers made = 660

Burgers with chicken and mushroom patty = 165 (Number of terms in the series 1, 5, 9…657)

Burgers with vegetable patty = 95 (Number of terms in the series 2, 9, 16, …660)

Burgers with chicken, mushroom and vegetable patty = 24 (Number of terms in the series 9, 37, 65….653)

Required answer = 660 – 165 – 95 + 24 = 424

43.From the above question, we have 24 such burgers.

44.The key to this question is what you understand from the statement- ‘for two progressions out of P₁, P₂ and P₃ the average is itself a term of the original progression P.’ For option (a) which tells us that the Progression P has 20 terms, we can see that P₁ would have 7 terms, P₂ would have 7 terms and P₃ would have 6 terms. Since, both P₁ and P₂ have an odd number of terms we can see that for P₁ and P₂ their 4^th terms (being the middle terms for an AP with 7 terms) would be equal to their average. Since, all the terms of P₁, P₂ and P₃ have been taken out of the original AP P, we can see that for P₁ and P₂ their average itself would be a term of the original progression P. This would not occur for P₃ as P₃ has an even number of terms. Thus, 20 is a correct value for n.

Similarly, if we go for n = 26 from the second option we get:

P₁, P₂ and P₃ would have 9, 9 and 8 terms respectively and the same condition would be met here too.

For n = 36 from the third option, the three progressions would have 12 terms each and none of them would have an odd number of terms.

Thus, option (d) is correct as both options (a) and (b) satisfy the conditions given in the problem.

45.Since, P₁ is formed out of every third term of P, the common difference of P₁ would be three times the common difference of P. Thus, the common difference of P would be 2.

TRAINING GROUND FOR BLOCK I

HOW TO THINK IN PROBLEMS ON BLOCK I

1. A number xis such that it can be expressed as a + b + c = x where a, b, and c are factors of x. How many numbers below 200 have this property?

(a)31   (b) 32

(c)33   (d) 5

Solution: In order to think about this question you need to think about whether the number can be divided by the initial numbers below the square root like 2, 3, 4, 5… and so on.

Let us say, if we think of a number that is not divisible by 2, in such a case if we take the number to be divisible by 3, 5 and 7, then the largest factors of x that we will get would be .

Even in this situation, the percentage value of these factors as a percentage of x would only be: for = 33.33%, = 20% and = 14.28%

Hence, if we try to think of a situation where a = , b = , and c = the value of a + b + c would give us only (33.33% + 20% + 14.28%) = 67.61% of x, which is not equal to 100%

Since the problem states that a + b + c = x, the value of a + b + c should have added up to 100% of x.

The only situation, for this to occur would be if

a = = 50% of x  b = = 33.33% of x and

c = = 16.66% of x.

This means that 2, 3 and 6 should divide x necessarily. In other words x should be a multiple of 6.The multiples of 6 below 200 are 6, 12, 18 … 198, a total of 33 numbers.

Hence the correct answer is c.

2. Find the sum of all 3 digit numbers that leave a remainder of 3 when divided by 7.

(a)70821    (b) 60821

(c)50521    (d) 80821

Solution: In order to solve this question you need to visualise the series of numbers, which satisfy this condition of the remainder 3 when divided by 7.

The first number in 3 digits for this condition is 101 and the next will be 108, followed by 115, 122 …

This series would have its highest value in 3 digits as 997.

The number of terms in this series would be 129 (using the logic that for any AP, the number of terms is given by + 1 )

Also the average value of this series is the average of the first and the last term i.e., the average of 101 and 997. Hence the required sum = 549 × 129 = 70821.

Hence the correct Option is (a).

3. How many times would the digit 6 be used in numbering a book of 639 pages?

(a)100    (b) 124

(c)150    (d) 164

Solution: In order to solve this question you should count the digit 6 appearing in units digit, separately from the instances of the digit 6 appearing in the tens place and appearing in the hundreds place.

When you want to find out the number of times 6 appears in the unit digit, you will have to make a series as follows: 6, 16, 26, 36 … 636.

It should be evident to you that the above series has 64 terms because it starts from 06, 16, 26 … and continues till 636. The digit 6 will appear once in the unit digit for each of these 64 numbers.

Next you need to look at how many times the digit 6 appears in the ten’s place.

In order to do this we will need to look at instances when 6 appears in the tens place. These will be in 6 different ranges 60s, 160s, 260s, 360s, 460s, 560s and in each of these ranges there are 10 numbers each with exactly one instance of the digit 6 in the tens place, a total of 60 times.

Lastly, we need to look at the number of instances where 6 appears in the hundreds place. For this we need to form the series 600, 601, 602, 603 … 639. This series will have 40 numbers each with exactly one instance of the digit 6 appearing in the hundreds place.

Therefore the required answer would be 64 + 60 + 40 = 164.

Hence, Option (d) is correct.

4. A number written in base 3 is 100100100100100100. What will be the value of this number in base 27?

(a) 999999   (b) 900000

(c) 989999   (d) 888888

The number can be visualised as:

Now in the base 27, we can visualise this number as:

When you want to write this number in Base 27 the unit digit of the number will have to account for the value of 3² in the above number. Since the unit digit of the number in Base 27 will correspond to 27 ⁰ we would have to use 27 ⁰, 9 times to make a value 3². The number would now become:

Similarly, the number of times that we will have to use 27¹ in order to make the value of 3⁵ (or 243) will be = 9. Hence the second last digit of the number will also be 9.

Similarly, to make a value of 3⁸ using 27² the number of times we will have to use 27² will be given by = 9. The number would now look as:

It can be further predicted that each of subsequent 1’s in the original number will equal 9 in the number, which is being written in Base 27. Since the original number in Base 3 has 18 digits, the left most ‘1’ in that number will be covered by a number corresponding to 27⁵ in the new number (i.e. 3¹⁵). Hence the required number will be a 6-digit number 999999. Hence, Option (a) is correct.

5. Let x= 1640, y = 1728 and z = 448. How many natural numbers are there that divide at least one amongst x, y, z.

(a) 47    (b) 48

(c) 49    (d) 50

Solution: 1640 can be prime factorised as 2³ × 5¹ × 41¹. This number would have a total of 16 factors.

Similarly, 1728 can be prime factorised as 2⁶ × 3³. Hence, it would have 28 factors. While 448 = 2⁶ × 7¹ would have 14 factors. Thus there are a total of (16 + 28 + 14) = 58 factors amongst x, y and z.

However, some of these factors must be common between x, y, z. Hence, in order to find the number of natural numbers that would divide at least one amongst x, y, z, we will need to account for double and triple counted numbers amongst these 58 numbers (by reducing the count by 2 for each triple counted number and by reducing the count by 1 for each double counted number).

The number of cases of triple counting would be for all the common factors of (x, y, z). This number can be estimated by finding the HCF of x, y and z and counting the number of factors of the HCF. The HCF of 1640, 1728 and 448 is 8 and hence the factors of 8, i.e., 1, 2, 4 and 8 itself must have got counted in each of the 3 counts done above. Thus each of these 4 numbers should get subtracted twice to remove the triple count. This leaves us with 58 – 4 × 2 = 50 numbers.

We still need to eliminate numbers that have been counted twice, i.e., numbers, which belong to the factors of any two of these numbers (while counting this we need to ensure that we do not count the triple counted numbers again). This can be visualised in the following way:

Number of common factors that are common to only 1640 and 1728 and not to 448 (x and y but not z):

1640 = 2³ × 5¹ × 41¹

1728 = 2⁶ × 3³

It can be seen from these 2 standard forms of the numbers that the highest common factors of these 2 numbers is 8. Hence there is no new number to be subtracted for double counting in this case.

The case of 1640 and 448 is similar because 1640 = 2³ × 5¹ × 41¹ while 448 = 2⁶ × 7¹ and HCF = 2³ and hence they will not give any more numbers as common factors apart from 1, 2, 4 and 8.

Thus there is no need of adjustment for the pair 1640 and 448.

Finally when we look for 1728 and 448, we realise that the HCF = 2⁶ = 64 and hence the common factors between 448 and 1728 are 1, 2, 4, 8, 16, 32, 64. But we are looking for factors which are common for 1728 and 448 but not common to 1640 to estimate the double counting error for this case.

Hence we can eliminate the number 1, 2, 4, 8 from this list and conclude that there are only 3 numbers 16, 32 and 64 that divide both 1728 and 448 but do not divide 1640.

If we subtract these numbers once each, from the 50 numbers we will end up with 50 – 3 × 1 = 47.

The complete answer can be visualised as 16 + 28 + 14 – 4 × 2 – 3 × 1 = 47.

Hence, Option (a) is correct.

6. How many times will the digit 6 be used when we write all the six digit numbers?

(a) 5,50,000   (b) 5,00,000

(c) 4,50,000    (d) 4,00,000

Solution: When we write all 6-digit numbers we will have to write all the numbers from 100000 to 999999, a total of 9 lac numbers in 6 digits without omitting a single number. There will be a complete symmetry and balance in the use of all the digits. However the digit 0 is not going to be used in the leftmost place.

Using this logic we can visualise that when we write 9 lakh, 6-digit numbers, the units place, tens place, hundreds place, thousands place, ten thousands place and lakh place – Each of these places will be written 9 lakh times. Thinking about the units place, we can think as follows: In writing the units place 9 lakh times (once for every number) we will be using the digit 0, 1, 2, 3 … 9 an equal number of times. Hence any particular digit like 6 would get used in the units digit a total number of 90000 times (9 lakh/10). The same logic will continue for tens, hundreds, thousands and ten thousands, i.e., the digit 6 will be used (9 lakh/10) = 90000 times in each of these places. (Note here that we are dividing by 10 because we have to equally distribute 9 lakh digits amongst the ten digits 0, 1, 2, 3,4, 5, 6, 7, 8, 9.)

Finally for the lakh place since “0” is not be used in the lakh place as it is the leftmost digit of the number, the number of times the digit 6 will be used would be 9 lakh/9 = 1 lakh. Hence the next time you are solving the problem of this type, you should solve directly using = 5,50,000.

Hence, Option (a) is the correct answer.

BLOCK REVIEW TESTS

Review Test 1

Review Test 2

Directions for Questions 1 to 4: Four sisters Suvarna, Tara, Uma and Vibha playing a game such that the loser doubles the money of each of the other player. They played four games and each sister lost one game in alphabetical order. At the end of fourth game each sister had ` 32.

Review Test 3

Review Test 4

Review Test 5