CC 1501 Solutions

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CC-1501 Solutions

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Section I

1.

Answer: 2. 16

Solution: As per triangle inequality, Shortest possible length of third side > 24 – 9 = 15.

Thus shortest possible integral length of the third side of the triangle is = 16. (2)

2.

Solution: x4 + 1/x4 = 47

Or x4 + 1/x4 + 2 = 49

(x2 + 1/x2)2 = 49 or x2 + 1/x2 = 7

Again adding 2 to both sides

We get (x+1/x)2 = 9

or x+1/x = 3

Cubing both sides we get

x3 + 1/x3 + 3*x*(1/x)(x+1/x) = 27

x3 + 1/x3 +3*3 = 27

Or x3+ 1/x3 = 18

3.

Answer: 4. 675 miles

Solution: Let total distance be „x‟ miles, then x = 100 180

80+

𝑥−180

110 .

Thus x = 675 miles. (4)

4.

Answer: 4. 61 ft

Solution: Let x be the distance of the bottom of the ladder from the wall, then (x+1) is

the length of the ladder which acts as hypotenuse of the right triangle. Thus (11, x, x+1)

is a Pythagorean triplet, which is simply (11, 60, 61). Thus required length of the ladder

is 61 ft. (4)

5.

Solution:

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As [x] and ]x[ are always integers ,from the given equations we can easily determine

that {a}=0.9,{b}=0.4,{c}=0.7

Now, [a] = a-{a} and ]a[= a+1-{a}

Adding all three equations we get,

2(a+b+c)+3-2({a}+{b}+{c})=30

a+b+c = 15.5

6.

Answer: 2. 504

Solution: Let the three sides of the right triangle be a - d, a, a + d.

Using Pythagoras Theorem, we get that (a + d)2 = a2 + (a - d)2.

Simplifying, we get that a = 4d i.e. the three sides are: 3d, 4d, 5d.

Now given is that 3d + 4d + 5d = 12d > 2015.

i.e. d > 2015/12

i.e. d > 167

i.e. dminimum = 168.

So shortest possible length of the shortest side is = 3 dminimum = 3(168) = 504. (2)

7.

Answer: 2. -1

Solution: Just check the options, for b = -2, polynomial becomes (x + 1)2 and both roots

are integers and satisfy other condition also.

For b = 2, polynomial becomes (x - 1)2 and both roots are integers and satisfy other

condition also.

But for b = -1, polynomial becomes x2 + x + 1 and roots are non-real. Thus b cannot be

equal to -1. (2)

8.

Answer: 1. 0

Solution: S19 = S22 i.e. a20 + a21 + a22 = 0

i.e. (a21 – d) + (a21) + (a21 + d) = 0

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i.e. a21 = a + 20d = 0

Now a7 + a35 = (a + 6d) + (a + 34d) = 2a + 40d = 2(a + 20d) = 2(0) = 0. (1)

9.

Answer: 4. not possible

Solution: Given that x2 + 4x – n – 1 < 0 for every real number x.

But this is not possible as the given quadratic polynomial is an upward parabola

graphically, and will achieve positive values certainly. (4)

10.

Answer: 1. 215 rupees

Solution: Given that 3D + 5P + 1S = 465 (I)

And 4D + 9P + 1S = 715 (II)

Now 2(II) – (I) gives, 2D + 1P + 1S = 215. (1)

11.

Answer: 2. 1

Solution: N4 + N2 + 1 = (N2 – N + 1)( N2 + N + 1)

Only for N = 1, the given number is a prime number and the number has exactly two

divisors. For all other values of N, the given number has more than 2 divisors. (2)

12.

Solution- T = A +B + C- Exactly 2-2 Exactly 3+none

Where T is Total number of days

A- Total number of days heard him meow

B- Total number of days heard him hiss

C- Total number of days heard him purr

Let x is the number of days I heard him meow and purr but not hiss 24= 16+ 7+ 11- (3+1+x) - 2×2 +0

After solving x= 2

Answer option 2

13.

Answer: 3. 2

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Solution: Given that PQRS(PQ + QR + RP) = 1

i.e. PQR(PQS + PRS + QRS) = 1

And we want to find the minimum value of sum of these two terms i.e. PQR & (PQS +

PRS + QRS).

We know that if product of two positive real numbers is constant then their sum is

minimised when the two numbers are equal to each other.

So in this case, the two terms should be equal and equal to 1 each. Thus required

minimum value is = 1 + 1 = 2. (3)

14.

Solution: Volume of a pyramid = 1/3 * Area of base * Height

4 = 1/3 * Area of base * √3

4√3 = (√3/4)a2

Or a = 4 where is the side length of the base of the pyramid.

Using cosine rule we can find the length of side AE.

So, AE = √13

15.

Solution:

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In the second race, as A and B reach simultaneously, A has covered a distance of 750

and by that time B has covered a distance of 900 m

So Va/2Vb = 750/900 = 5/6

Or Va: Vb = 5:3

In the first case

Va/Vb = 1000/(1000-50-50Vb) = 5/3

Or Vb = 7m/s

16.

Solution (1): Total no of letters in the word „APPLICATIONS‟ is 12.

Total no of vowels = 5

Total no of consonants = 7

Number of words where all vowels are together is 8!*5!/(2!*2!2!)

So total number of ways where vowels are not together,

=12!/(2!*2!*2!) – 8!*5!/(2!*2!*2!) =3

2* 11! – 7!*5!

17.

Solution:

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As by the time A reaches P, B reaches Q. So ratio of speeds of A and B is 2:1

Now the distance between A and B remains √3a { Distance between P and Q)

When they meet each other, this distance PQ must be divided in the ratio of 2:1. So,

RQ = 1/3 * (√3a) = a/√3

SO RO = 2a/√3

18.

Solution:

Let‟s suppose that common ratio of the given infinite G.P. is „r‟.

It‟s given that a1 + a2 + a3 + -------+ a = 4

And (a1)2 + (a2)

2 + (a3)2 + -------+ (a)

2 = 8

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So , 𝑎

1−𝑟 = 4 and

a^2

1−𝑟^2 = 8

Solving these equations we get, a = 8

3 and r =

1

3

Hence 27a3 = 27 * 8

3 *

1

3 *

1

3 = 8

19.

Solution:

(x2-9x+19)^(𝑥2+8x-33= 1 has two different meaning,

Either x2+8x-33 = 0, then x=-11 or 3

Or, x2-9x+19 = 1 or -1.

When x2-9x+19 = 1, then x=6 or 3

When x2-9x+19 = -1, then x=5 or 4

But if x2-9x+19 = -1 then value of (x2+8x-33) must be even.

So, x 4.

Hence sum of all the possible values of x = -11+3+6+5 = 3

20.

Solution:

Let‟s suppose that degree of the given polynomial f(x) = n then degree of the polynomial

f(f(x)) will be n2 and degree of xf(x) is n+1.

f(f(x)) = xf(x)+x

Equating degrees of polynomials on both sides, we get that for any positive value of x,

there is no such polynomial is possible.

So, x < 0.

Now, if x = 1 , f(f(1) = f(1) – 1 ,

Or, f(x) = (x+1)

Or, f(-91) = -92

21.

Solution: 𝑙𝑜𝑔10 𝑥2 + 4𝑥 − 21 + 𝑙𝑜𝑔10

(𝑥+1)

(𝑥+7)= 2 + 𝑙𝑜𝑔10 (2𝑥 + 2)

𝑙𝑜𝑔10 { 𝑥2 + 4𝑥 − 21 ∗ 𝑥 + 1

𝑥 + 7 } = 𝑙𝑜𝑔10 100 + 𝑙𝑜𝑔10 (2𝑥 + 2)

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𝑙𝑜𝑔10 {(𝑥 + 1)(𝑥 − 3)} = 𝑙𝑜𝑔10 200(𝑥 + 1)

Comparing both sides we get

(x+1)(x-3) = 200(x+1)

Or x = 203

Now number of zeroes in 203! = 203/5 + 203/25 + 203/125 = 49

22.

Solution:

As we can see that the altitude of the given triangle is the y coordinate of the point of

intersection of the lines 2x+3y =6 and y-x =1

Or y = 8/5

23.

Answer: 3. 1512

Solution: First select 5 distinct digits in C(10, 5) = 252 ways.

Now the middle digit is the smallest one, we need to select 2 out of remaining 4 in C(4,

2) = 6 ways for first two places and all the digits can be arranged in only one possible

arrangement.

So total number of such numbers is = 252*6 = 1512. (3)

24.

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Solution: Total cost = Fixed cost + variable cost

68*5 = Fixed + k*5 (1)

50*8 = Fixed + k*8 (2)

Solving these we get k = 20 and F = 240

Total cost for 15 guests = 240 + 15*20 = 540

So average cost for 15 guests = 540/15 = Rs 36

25.

Solution: For the same work if a man takes 1 day to finish it, then a woman takes 2

days to finish it while a child takes 4 days to finish it.

So ratio of work done by them is 4:2:1 ( Man: Women: Children)

As total number of men, women and children employed is 60

So 6x+5x+4x = 60 or 15x =60

X = 4

Number of men = 24, women = 20 and children = 16

Amount of work done by them = 24*4 : 2*20 :1*16 = 12:5:2

So total wages = 12a + 5a + 2a { As wages are proportional to work done}

19a = 1900 or a = 100

Thus total wages earned by 16 children = Rs 200

So, the daily wage of a child is Rs 200/16 = 12.5

26.

Solution:

According to question, if R is the radius of the circles then,

πR∗R

2𝜋𝑅

= 2 * 2πR

𝜋𝑅∗𝑅

R = 2 2

Area of all the squares = 4* π*R*R = 32π

Side of the square is = 4R =8 2

Area of the square = 8 2* 8 2 = 128

So, required area = 128 - 32π = 32(4 - π)

27.

Solution: Kerosene and petrol in the 16 L solution is 12 and 4 L respectively.

Let the quantity of 30% solution be x litres.

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So Kerosene added = 0.3x litres and petrol added = 0.7x

Now the concentration becomes 50% i.e. quantity of kerosene = quantity of petrol

12 + 0.3x = 4 + 0.7x

8 = 0.4x or x = 20 L

28.

Solution: Given N = 24335374

Number of factors of N divisible by 20 will be the number of factors of 22335274

i.e.(2+1)(3+1)(2+1)(4+1) = 180

Thus p = 180

For q we have to find the factors of N that are perfect squares. Now any factor of N

would be of the form 2a3b5c7d and for it to be a perfect square a,b,c and d must take

even values.

q = 3*2*2*3 = 36

Thus, p/q = 5

29.

Solution:

Solving given equation we get C2 = 1 , c=1 or -1

When c=1, number of ordered pairs (a,b) = 20*20 = 400

When c=1, number of ordered pairs (a,b) = 20*20 = 400

When a=0, number of ordered pairs (b,c) = 20*20 = 400

Hence total number of ordered triplets (a,b,c) = 400+400+400 = 1200

30.

Solution: 6! is divisible by 48 and so every other factorials bigger than 6.

K mod 48= (1! +2! +3! +4! +5!) Mod 48 = 9

31.

Solution: Amount received in Scheme A is

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𝐴 = 𝑃(1 +18

2 ∗ 100)2∗2

Similarly amount received in Scheme B is

𝐴 = 𝑃(1 +𝑟

100)2

(+9

100)4 = 1 +

𝑟

100

2

Solving we get r = 18.81%

32.

Answer: 2. 1

Solution: Equating the sum of numbers in middle column and forward diagonal, we get

3 + x = 0 + 2

i.e. x = -1.

So required sum of three numbers in first row is = 2(-1) + 3 + 0 = 1. (2)

33.

Solution- 29 can be written as (31-2), 30 can be written as (37-7), 32 can be written as

(37-5). Only 31 can not be equal to (a-b) where a and b are prime numbers.

Answer option 3

34.

Solution- let six consecutive numbers are x, x+1, x+2, x+3, x+4, x+5.

Sum= 6x+15= n

Therefore, x+5 = 𝒏+𝟏𝟓

𝟔

Answer option 1

SECTION II

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1. Solution: „4‟. 4 is the only case where no data or information of the incident is

preserved, something that has been described as the characteristic feature of

quantum immortality in the passage.

2. Solution: „2‟. 2 is what we can infer about Hume‟s thought on the matter stated in

the second paragraph. 1 and 3 have also been mentioned to be Hume‟s thoughts

but only to explain his stand which has been expressed in 2. 1 and 3 do not sum

up his stand on the central idea of immortality of humams. 4 runs contrary to

what Hume believes.

3. Solution: „3‟. The author first talks about information immortality. He then uses

this concept to dwell on the idea of immortality of human beings and the various

aspects associated with it. Hence 3 is the answer.

4. Solution: „2‟. The author, while explaining the principle of unitarity, which states

that information is indestructible, takes the example of burning a book. For this

example to exemplify the above stated principle, 2 must be true. 1 has been

stated to be untrue in the second last line of the first paragraph. 3 too can be

counted by using the same sentence. The author‟s taking the example of a

burning book would make sense even if 4 were untrue, but same cannot be said

for 2.

5. Solution: „2‟. In the third paragraph, the author says that if death is a wall, then

life ends before you reach that wall, since the moment you touch that wall, you

are dead. Hence one can never feel death. 2 is what comes closest in meaning to

that thought.

6. Solution: „1‟. A can be inferred from the first paragraph where the author says

that death has no definite meaning and that one cannot know about it. C states

the second of the two ideas that the author talks about in the last paragraph. In

the last line the author says that “it is useful to think about death only to the

point…”. He has proposed an idea which is somewhat midway of the two

philosophical ideas he mentioned in the beginning of the paragraph. B is beyond

the scope of the paragraph whereas D simply puts the phrases used in the

passage in some arrangement which does not support anything said in the

paragraph.

7. Solution: „4‟. In a metaphor, we change the form of one object/person to the

form of the object/person we are comparing it to, and thus the object/person

being compared is transformed beyond its original form. Changing the form of

death to a wall is typical of a metaphor.

8. Solution: „3‟. In the entire passage, the author has tried to discuss the concept of

death. Fear of death and the inevitability of death have been mentioned but only

to expatiate on the concept called death.

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9. Solution: „1‟. A is what we can infer from the 6th and 7th lines of the first

paragraph. B can be inferred from the last line of the third paragraph. Given his

take on beliefs, the author would most likely disagree with C. D can be inferred

from what the author says in the last 3rd, 4th and 5th lines of the first paragraph.

10. Solution: „3‟. Read the first 4 lines of the second paragraph to know that the

author says that the difference between belief and knowledge is that knowledge is

always true. Hence one can infer that belief is not always true. Using this as a

basis, the author says that we should not confuse knowledge with belief.

11. Solution: „4‟. The author says in the beginning of the last paragraph that the

unquestioned beliefs are the shackles of the mind which hold us captive, and we

therefore need to cultivate skeptical attitude. Hence a skeptical attitude would

promote abstinence from unquestioned or assumed beliefs and therefore a

skeptic‟s mind would be freer.

12. Solution: „1‟. Questioning things would make people who do not want to think

much confused and shake their comprehension of things and ideas. 1 comes

closest to our contextual need.

13. Solution: „4‟. The answer is very clearly stated in the first few lines of the first

paragraph.

14. Solution: „1‟. B can be logically derived from the first two lines of the second

paragraph. C is true as per the 3rd line of the first paragraph. No statement or

data mentioned in the passage supports A or D.

15. Solution: „2‟. In the last paragraph the author has mooted the idea that the entire

state and its organs be run by the charity given by citizens. The question asks

why this idea is practically unviable. 1 is not a practical limitation at all to this

idea and 4 is untrue as the author himself has confessed in the last paragraph

that many people do donate large sums of their wealth. The effect of such a

system on the civil society is debatable but the possibility of not gathering

enough funds through willful charity, as against the present system of taxes, is

the most glaring practical flaw. 2 points out to this possibility.

16. Solution: „3‟. The author first raises the question of „making up for the difference

between the living wage and the minimum wage‟ in the first paragraph. He has

then, throughout the passage, tried to fix accountability for the same.

17. Solution: „2‟. „This spiritual land‟ in D refers to India mentioned in B. Placing D

after C without acknowledging the turn in idea would be inappropriate. E marks a

swerve in the idea and begins with a befitting connector. „Them‟ in C can refer to

either the people mentioned in A (they) or in E (few).

18. Solution: „3‟. Placing B after E will lead to ambiguity as „it‟ in B could then refer to

both hair thinning and hair loss in B. B towards the end brings us back to the

general statement where the paragraph began from, whereas one would expect

the flow of the paragraph to be more towards treatment after DA. Placing B after

A would not make any sense.

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19. Solution: „1‟. B would follow E since B has the answer that the reader would most

likely give to the question asked in E. Between 1 and 4, 1 would be a better

choice since E would be a better opener and also CDA would follow B as B ends

with „hear me out‟, which tells us that the author will now explain what he has

introduced.

20. Solution: „2‟. B and D are same ideas, both talking about internal beauty/peace.

Hence BD would be placed together. Also, since B names the God being prayed

to, it is more likely to be the beginning of the prayer. Hence B would precede B. C

too is the continuation of that prayer.

21. Solution: „4‟. All other statements focus on trees.

22. Solution: '2'. All other options talk about the Dinkas whereas 2 talks about the

region.

23. Solution: „1‟. All other statements are somewhat positive in tone about

fermentation.

24. Solution: „4‟. All other sentences talk about the role/strengthening of tongue.

25. Solution: C. If using car pool lanes reduced driving time (commuting time) by

more than 25 percent, people who used car pool lanes would not spend more

time than people who used subways and trains. This would contradict the

passage. Thus, C has to be true.

A. There could be other factors apart from stopping at traffic lights that accounted

for 25 percent of the journey time.

B. “Majority” does not have to be true.

D. Covering the distance between a subway or train station and one's final

destination could take longer than covering the distance between one's car and

one's final destination and yet the statements could still stand.

26. Solution: C. Simple math. Fall in the number of students but a constant tuition

revenue would mean that tuition fees per student have increased.

A. No such comparison can be made based on the information given.

B. Had there been a positive correlation, fewer enrollments would have seen a

decrease in total tuition revenue.

D. What will happen in a few years is out of scope.

27. Solution: D. The author seeks to point out that the Greeks are in fact not

stubborn and their not being able to continue with the Euro, if it so happens, will

be because the “currency offers no respite for countries in trouble.” The use of

quotes for “insubordinate” (stubborn/obstinate) in the first sentence also

indicates that the author distances herself from the term.

A has a number of issues. We do not know about the current nature of the euro

and the Greek's opinion of it. Also, A is not the author's main focus. B is out of

scope and extreme. The author does not advocate leaving the euro. C is out of

scope.

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28. Solution: B. curb on govt. spending ---> raised value of currency ---> country's

exports not more competitive

A. The statements imply that curbing government spending can ultimately hurt

the economy. Any positives, however, such as A, have not been implied.

C. What economists recommend cannot be inferred. Also, economists could

simply recommend not putting a curb on spending.

D. Reduced govt. spending leads to an increase in the value of a currency and not

the other way round.

29. Solution: A. D leads to A. But A is all encompassing. The author is arguing that

public funding for medical care is justified as it is similar to public funding for

education. So, the author is arguing that healthcare is not a privilege but a right.

B. While B may be true, it is not the author's main point.

C. C acts as a supporting detail for the author's main point that healthcare is a

right.

D. D leads to A, but A is all encompassing.

30. Solution: A. A is evident from “It may be an open secret that India depends on

Israel for its defense preparedness but that does not mean that we should

sacrifice our self-esteem and shy away from doing our duty towards humanity.”

The author is basically saying that safeguarding our interests with Israel should

not keep us from doing our humanitarian duties.

B. Not the main point. Also, B misconstrues a possibility as a certainty.

C. While C is true, it is not the author's conclusion.

D. Out of scope.

31. Solution: „3‟. In „A‟, it should be “…when they are present…” or “…when you

are present…” In „E‟, it should be “…resort to…”

32. Solution: „5‟. Idiomatic error in „B‟- it should be „stood for’. In „C‟, it should be

“…except perhaps born to…”

33. 33.Solution: „5‟. Modifier error in „A‟. In „C‟, it should be „poetry’ instead of

„poetic‟.

34. Solution: „1‟. In D, it should be „its‟ instead of „it’s‟. In E, „genre‟ is misspelt.

Section III

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1. 3

2. 1

3. 3

4. 2

5. 3

6. 2

7. 1

8. 4

Solution 9-12

Solution: From (1) we get

From (2) we get Eva is to the left of Bharti and opposite to F

From (3) we get

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From (4) we get the final arrangement as :

9. Eva is wearing jersey number 4.

10. Honey is wearing jersey number 8 and Bharti is sitting two places to the left

of Honey.

11. Either Farida or Chhavi is wearing jersey number 1. So, it cannot be

determined.

12. Only second statement is true.

Solutions 13 to 16:

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13. 1

14. 3

15. 2

16. 4

Solutions 17 to 20:

17. 3

18. 4

19. 3

20. 2

Solution:

As D and F boarded at Dwarka and D got out at Karol Bagh, this can be shown as:

From (4) and (5), A boarded before B and E, so he must have boarded the metro at Tilak

Nagar.

As none of them got out at Rajouri Garden and only D got out at Karol Bagh. So A must

have got out at Rajiv Chowk and C must have boarded at that station and got down at

Barakhamba Road where G must have boarded.

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As B and E covered maximum number of intermediate stations, so we can say that B and

E got down at Mandi House while F must have got down at Tilak Nagar. The final

arrangement becomes:

Solutions 21 to 24

Solution:

As there are 6 teams, so number of matches = 6C2 = 15

Out of which there were 14 matches that had a definite outcome. So in all there are 14

wins. Also number of wins of South Korea, Germany, Pakistan, Netherlands, and Japan

won equal number of matches.

So 5x + y = 14 { Where x is the number of matches won by South Korea, Japan,

Germany, Netherlands and Pakistan while y represents the number of matches won by

Australia.

By this we get x = 2 and y = 4

Also Germany, Pakistan and Japan scored equal number of points. This further means

that the match that was drawn could not have been played by Germany, Pakistan, Japan

and Australia. So, the match between South Korea and Netherlands was a draw. The

table can be constructed as:

South

Korea

Japan Australia Germany Pakistan Netherlands Wins

South

Korea

X Lost Won Draw 2

Japan X Lost Lost Won 2

Australia Won Won X Lost Won Won 4

Germany Lost Won Won X Lost Lost 2

Pakistan Lost Won X Lost 2

Netherlands Draw Lost Lost Won Won X 2

21. Definitely the match between Netherlands and South Korea resulted in a draw.

22. South Korea lost 2 matches.

23. If Pakistan won the match against Japan, then the table finally becomes:

South

Korea

Japan Australia Germany Pakistan Netherlands Wins

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South

Korea

X Lost Lost Won Won Draw 2

Japan Won X Lost Lost Lost Won 2

Australia Won Won X Lost Won Won 4

Germany Lost Won Won X Lost Lost 2

Pakistan Lost Won Lost Won X Lost 2

Netherlands Draw Lost Lost Won Won X 2

Thus, South Korea lost to Japan.

24. Both the statement A and D are correct.

Solutions 25 to 28: Solution: From (1) we get

From (2) we get

From (3) we get

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From (4) we get

As only two of them are facing away from the centre, so case 1 gets ruled out and using

(5) we get the final arrangement as:

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25. E is from Indonesia.

26. F is from New Zealand.

27. E is to the right of the one from New Zealand.

28. The student from Brazil is sitting two places to the left of E.

Solutions 29 to 32:

29. 2

Solution:

As total profit was 30% after first year, company will invest 25% more,

So, total Investment for the second year = 2048 + 25% of 2048 = 2560

Hence total profit after two years,

= 30% of 2048 + 18.75% of 2560 = 614.40+480 = 1094.40 million

30. 4

Solution:

As all three years returns on investment were less than 37.5%, so company must have

increased its investment by 25% every year.

Total investment before 4th year = 2048(1.25) (1.25) (1.25) = 4000 million.

31. 4

Solution:

As It says nothing about next year investment that what will happen if company gains

more than 37.5% on a particular year ,So it cannot be determined.

32.1

Solution:

Total returns on investment for the first two years,

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= 31.25% of 2048+25% of (2048 x 1.25)

= 640 + 640 = 1280 million

So minimum Return on investment for the third year must be,

2048 – 1280 = 768 million

As for the first two years total investment was increased by 25%, so total investment

before third year = 2048(1.25) (1.25) =3200 million

Hence minimum required % Return on Investment for the third year,

= 768

3200 x 100 = 24%

CC 1501 Solutions

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