DPM 27 (Solutions)

C to E

C. RC Passage (1 to 4)

Answer Key: 1-C, 2-A, 3-C, 4-B

Explanation:
1) Options A, B, D can be easily ruled out. The politicians are indeed vying for power
but ‘under the carpet’ refers to concealed from outside world. This meaning is given
in C.
2) The meaning of the word can be inferred from the 3rd & 4th sentences of par1- ‘blue-
print for chaos’. Hence option A.
3) Para 1, sentence 3 mentions about the constitution being the reason for conflict.
Sentence 4 tells us who had written it & that it was amended so many times that
‘illegibility ‘had crept in. Sentence 5 tells us that the constitution could have created
a confusion or crisis even in good times…hence A, B, D are true. Option C is false.
4) Para 2, sentences 1 & 2 point to option B as the answer.
D. Quantitative Aptitude (1 to 10)

Keys

1. A
2. D
3. 20
4. A
5. C
6. C
7. C
8. 3
9. D
10. C

Solutions –

1. Consider the second statement first.

The time now is 10:00 + 3T min.
From the first statement, 10:00 + (3T +24) = 9:00 + 5T
 9:00 + 2T = 10:24
 2T = 84
 T = 42
The time 30 min. from now is 10:00 + 126 min + 30 min = 12:36 pm.
Option(A)

2. Let the mixture of each mixture drawn be x.

11 3 17
Quantity of milk in the first, second and third mixture drawn are 20 𝑥, 4 𝑥 and 20 𝑥
Concentration of milk in the resultant mixture

11 3 17
𝑥+ 𝑥+ 𝑥 𝟐
=( 20 4 20
) . (100%) = 𝟕𝟏 𝟑 %
3𝑥

Option (D)
3. Let the number of chocolates distributed among the children in A be T.
Hence number of chocolates distributed among the children in B = (T – 600).
Let the number of children in A be x.
Hence the number of children in B = x – 3 0

𝑇 𝑇−600
Given 𝑥 = 𝑥−30
𝑇. 𝑥 − 30. 𝑇 = 𝑇. 𝑥 − 600. 𝑥
 30T = 600x
𝑇
 𝑥 = 20
∴ Each Child in A must have received 20 chocolates.
Ans.: (20)

4. Let the CP be 100. In the first case, the profit is 36%.

Hence the SP is 136. This was at a discount of 20%. Therefore, the marked price has to be =
136
= 170
0.8
If he offered a discount of only 15%, the discount would be 25.5, or the SP would be 144.5.
The profit (on the CP) would have been 44.5%.
Option(A)

5. Let a and d be first term and common difference respectively of the AP.
∴ 6(a+8d) = 3(a+11d)
 a + 5d = 0
 The 6th term is zero.
Consider the options.
a. t1+t2+t14 = (a)+ (a+d) + (a + 13d) = 3a + 14d = - d
b. t3+t5+t9 = (a+ 2d) + (a+ 4d) + (a + 8d) = 3a + 14d = - d
c. t2+t4+t12 = (a+ d) + (a+ 3d) + (a + 11d) = 3a + 15d = 0
d. t4+t5+t10 = (a+ 3d) + (a+ 4d) + (a + 9d) = 3a + 16d = d
Among the options, only option (C) can be uniquely determined.
Option (C)

6. Let the four two-digit odd numbers be ab, ab+2, ab+4 and ab+6.
S = ab + (ab+2) + (ab+4) + (ab+6) = 4ab + 12
𝑆
√10 is an integer.
𝑆
∴ is a perfect square.
10
𝑆
As 10 is a perfect square, it is an integer.
∴ S is divisible by 10 i.e., units’ digit of S is 0.
∴ 4(ab) has a unit digit of 8.
∴ ab has a units digit of 7. (ab is odd)
S = 4(10a+ 7) + 12 = 40a + 40 = 4. (10). (a+1)
𝑆
= 4(a+1). This is a perfect square.
10
∴ a+1 is a perfect square.

𝑎 ≥ 1 ∴ 𝑎 + 1 ≥ 2, 𝐴𝑙𝑠𝑜 𝑎 + 1 ≤ 10
the only (perfect square) values of a+1 are 4 and 9.
The only values of a are 3 and 8.
ab = 37 or 87
Both I and II are true.
Option(C)

7. X = {1, 2, 3, …., 30}

There are 10 multiples of 3 in X (and 20 non multiples). To construct a subset such that the
product of all the elements of the subset is a multiple of 3. We should ensure that we do
include at-least one of these 10 multiples.

The number of ways of selecting these 10 multiples is 210 – 1 (We have to exclude the case
where all 10 multiples are omitted). The number of ways of selecting the other 20 numbers
is 220. Therefore, the number of subsets in which the product of all the elements is a multiple
of 3 is 220(210 – 1)
Option(C)

8. 𝑆𝑖𝑛3 𝜃 + 𝐶𝑜𝑠 3 𝜃 = 1 ( 𝐴𝑙𝑠𝑜 𝑆𝑖𝑛2 𝜃 + 𝐶𝑜𝑠 2 𝜃 = 1)………..(1)

Sin 𝜃 as well as Cos 𝜃 range from – 1 to 1.
Equation (1) is satisfied when (sin 𝜃 , cos 𝜃) = (0, 1) 𝑜𝑟 (1, 0)
If (sin 𝜃 , cos 𝜃) = (0, 1), 𝜃 = 0° 𝑜𝑟 360°
If (sin 𝜃 , cos 𝜃) = (1, 0), 𝜃 = 90°
If any of sin 𝜃 𝑎𝑛𝑑 cos 𝜃 is negative, the other exceeds 1. This is not possible.
If any of sin 𝜃 𝑎𝑛𝑑 cos 𝜃 lies between 0 and 1, the other lies between 0 and 1.
∴ 𝑆𝑖𝑛3 𝜃 < 𝑆𝑖𝑛2 𝜃 and 𝐶𝑜𝑠 3 𝜃 < 𝐶𝑜𝑠 2 𝜃
∴ 𝑆𝑖𝑛3 𝜃 + 𝐶𝑜𝑠 3 𝜃 < 𝑆𝑖𝑛2 𝜃 + 𝐶𝑜𝑠 2 𝜃 = 1

This is not possible.

(sin 𝜃 , cos 𝜃) 𝑐𝑎𝑛 𝑏𝑒 (0, 1) 𝑜𝑟 (1,0) 𝑜𝑛𝑙𝑦.
𝜃 = 0° , 90° 𝑜𝑟 360°
Ans. (3)
9. 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐴 = 6. (8𝑥 2 + 1) 𝑖. 𝑒. 48𝑥 2 + 6
A, B, C are distinct positive integers.
∴ ( 𝐴 − 𝐵 )2 + ( 𝐵 − 𝐶 )2 + (𝐶 − 𝐴 )2 > 0
(∵ 𝐴 − 𝐵, 𝐵 − 𝐶, 𝐶 − 𝐴 are all non-zero)
(∴ 𝑇ℎ𝑒 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒𝑖𝑟 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝑖𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒)
2(𝐴2 + 𝐵2 + 𝐶 2 − 𝐴𝐵 − 𝐵𝐶 − 𝐶𝐴) > 0
𝐴2 + 𝐵2 + 𝐶 2 − 𝐴𝐵 − 𝐵𝐶 − 𝐶𝐴 > 0
Adding 2(AB + BC + CA) both sides,
(𝐴 + 𝐵 + 𝐶 )2 > 3(𝐴𝐵 + 𝐵𝐶 + 𝐶𝐴)
(𝐴 + 𝐵 + 𝐶 )2 > 3(𝐴𝐵 + 𝐵𝐶 + 𝐶𝐴) ≥ 3(48𝑥 2 + 6)
(𝐴 + 𝐵 + 𝐶 )2 > 144𝑥 2 + 18
A+B+C is greater than 12x.(∵ A, B, C are positive)
Option(D)

10. The number of children is n. the number of chocolates is 14n + 13. The is equal to (say) =
11n + R.
14𝑛 + 13 = 11𝑛 + 𝑅
∴ 𝑅 = 3𝑛 + 13 or (R – 13) is a multiple of 3.
Among the choices, only 37 is a possible value.
Option(C)
E. DILR SET-1(1 to 4)

Keys

1. 8
2. 2
3. 2
4. 1

Solution for the questions 1 to 4:

Given,
P Q R S T
Mechanics 2
Philosophy 3
Criminology 1
Total b z 9

Given that in any subject P did not get the 1st rank, Q did not get the second rank. R did not
get the 3rd rank, S did not get the 4th rank and T did not get the 5th rank.
S's Ranks must be 1, 3 and 5
As b ≤ z, if Q's ranks are 3, 4. 5 then R's ranks must be 3, 4, 5, which is not possible as R
cannot get the 3rd rank.
Q got the 1st rank in one of the subjects.
- As it cannot be philosophy or criminology, Q got the 1st rank in mechanics.
S get the 1st rank in philosophy.
 R's ranks are 2, 4 and 5 and he got the second rank in criminology. Since P, Q, S and T
cannot get the second rank in criminology.
Now, R or T got the fourth rank in mechanics.
Case (i)
Let us assume that R got the 4th rank in mechanics.
 T got the third rank and S got the 5th ranks in mechanics.
- S got the third rank in criminology.
 T got the second rank in philosophy.
 P got the fourth rank in philosophy.
 In criminology, P got the 5th rank and Q got the 4th rank.
The distribution will be as follows: -
P Q R S T
Mechanics 2 1 4 5 3
Philosophy 4 3 5 1 2
Criminology 5 4 2 3 1
Total 11 8 11 9 6

Case (ii)
Let us assume that T got the 4h rank in mechanics.
So, R got the fifth rank in mechanics and the fourth rank in philosophy. S got the third rank
in mechanics and the fifth rank in criminology.
In philosophy, P got the fifth rank and T got the second rank.
 In criminology, P got the third rank and Q got the fourth rank.

.: The final distribution will be as follows.

P Q R S T
Mechanics 2 1 5 3 4
Philosophy 5 3 4 1 2
Criminology 3 4 2 5 1
Total 10 8 11 9 7

1. The sum of ranks obtained by Q in both the cases is 1 + 3 + 4 = 8. Ans: (8)

2. R got the second rank in criminology. Ans: (2)
3. In case (a) three persons and in case (b) two persons got the ranks between ranks of R
and S in philosophy. At least two persons. Ans: (2)
4. Difference in ranks of Q and S in criminology is 1 in either case. Ans: (1)