(CIVIL SERVICES APTITUDE TEST)

CLASSROOM
HANDOUT (I-V)

SOLUTIONS

DELHI CENTRE: DELHI CENTRE: DELHI CENTRE: PRAYAGRAJ CENTRE: JAIPUR CENTRE:
Vivekananda House Tagore House Mukherjee Nagar 31/31 Sardar Patel Marg, Plot No. 6 & 7, 3rd Floor,
6-B, Pusa Road, 27-B, Pusa Road, 637, Banda Bahadur Civil Lines, Prayagraj Sree Gopal Nagar,
Metro Pillar No. 111, Metro Pillar No. 118, Marg, Mukherjee Nagar, Uttar Pradesh-211001 Gopalpura Bypass,
Near Karol Bagh Metro Near Karol Bagh Metro Delhi-110009 Phone: 9958857757 Jaipur-302015
New Delhi-110060 New Delhi-110060 Phone: 9311667076 Phone: 9358200511
Phone: 8081300200 Phone: 8081300200

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2 | CSAT HANDOUT : SOLUTIONS

CLASS - I : SOLUTIONS
1. (c) Number = 2k, 3k Statement-II:
LCM = 2 × 3 × k = 48 LCM (4, 5, 6, 15) = 60
k = 8 Number = 2520 = 2497 + 23
Numbers = 16, 24 Hence statement (II) is incorrect.
Sum = 40 Hence option (a) is correct.
Shortcut  2 + 3 = 5 only 40 is divisible by 5.
6. (d)
Number = 13k + 3
2. (d) Number = LCM(5, 8, 12, 30)k + 2 When divided by 5, R = 2
= 120k + 2 13k + 3 = 5y + 2
For k = 2 Number 13k + 1 = 5y
= 242 – divisible by 11
13 k  1
y=
3. (a) 5
For k = 3, 13(3) + 3 = 42
Remainder: 4 5

2 3 7. (a) LCM(36, 48, 72 108) = 432 sec

(4-2) (5-3) = 7 min 12 sec

Remainder which is equal to that number will be: 8. (c) Number = LCM(15, 20, 25, 35) – 10

n*LCM of [4, 5] – 2. = 2100 – 10 = 2090

We get 18, 38, 58, 78, 98, 118 when value of n is 1, 2, 9. (b) Number = LCM(3, 4, 5, 6)k + 2
3, 4, 5 and 6 respectively. Highest double digit such
= 60k + 2
number will be
For k = 17, Number = 1022
4. (c)
10. (a) Number = LCM(5, 6, 8, 12)k + 1
Statement-I: Number = LCM(18, 24, 36) + 7
= 120k + 1  divisible by 13
= 72 + 7 = 79
For k = 4, Number = 481
Hence statement (I) is correct.
11. (a) Distance = LCM(40, 42, 45)
Statement-II: Number = LCM(8, 12, 16) – 3
= 2520 cm = 25 m 20 cm
= 48 – 3 = 45
Hence statement (II) is correct. 12. (a) LCM(2, 3, 4, 5, 6) = 60
Hence option (c) is correct. Number of times they meet in 180 days = 3
(On 1st, 61st and 121st day)
5. (a)
Statement-I: Number = LCM(56, 98) + 2 13. (c) LCM(6, 7, 8, 9, 12) = 504 sec
= 392 + 2 = 394 1 hour = 3600 sec

Hence statement (I) is correct. Total times = 8 (including the time when they started
firing together)

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 CSAT HANDOUT : SOLUTIONS | 3

14. (b) 19. (b)

For cricket and football, student id must be divisible It is given that
by 4 and 6 i.e., by 12 GCD(a, b) × LCM(a, b) = a × b = 1680
No. divisible by 12 are 12, 24, 36, 48, 60, 72, 84  8 × LCM(a, b) = 1680
and 96. 1680
 LCM(a, b) = = 210
i.e., Students = 8 8

15. (c) 20. (b)

Let the numbers be x and y.
Remainder : 4 5 6
According to the question,
2 3 4
x + y = 45 ...(i)
Which is equal to (4 - 2), (5 - 3), (6 - 4), that Difference of two numbers
number will be: n LCM of [4, 5, 6] - 2 = n * 60 – 2
When n = 1 we get 58. 1
=  Sum of two numbers
Highest possible three digits such number will 9
be 958.  x–y = 5 ...(ii)
16. (c) On adding eqs. (i) and (ii), we get
Statement-I:
x  y  45
 That number will be n bullet LCM of [5, 6, 7] + 2
xy  5
= 210 + 2 = 212.
2 x  50  x  25
Statement-II: LCM of fraction = LCM of
numerators/H.C.F. of Denominators.
From eq. (i)
LCM = LCM of (5, 8, 11)/HCF of (2, 9, 14)
x + y = 45
= 440/1 = 440.
 y = 45 – x
Hence, both statements are correct.
 y = 45 – 25 = 20
So, option (c) is correct.
Now, LCM of 25 and 20
17. (b)
5 25, 20
LCM of 5, 8, 12, 20 will not be a multiple of 9.
5, 4
18. (b)
(x + 3)(6x 2 + 5x - 4) = (x + 3)(2x - 1)(3x + 4)  Required LCM = 4 × 5 × 5 = 100
(2x 2 + 7x + 3)(x + 3) = (2x + 1)(x + 3)(x + 3)
LCM = (2x+1) (2x - 1)(x + 3)² (3x + 4) 
= (4x2 - 1) * (x + 3)2 * (3x + 4)

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4 | CSAT HANDOUT : SOLUTIONS

CLASS - II : SOLUTIONS
1. (b) LCM of (2, 3, 4, 5, 6)

Number = HCF(66 – 38, 80 – 66) or a factor of this 60 – 1 = 59

HCF.
Hence, there is only one integer between 0 and 100
HCF(28, 14) = 14 which satisfies the given condition.

So, number = 1, 2, 7, 14 6. (c)

Putting and checking the numbers. I. If a = bc with HCF (b, c) = 1

b and c are co-prime numbers.
Largest possible number = 14
HCF (c, bd) = HCF (c, d) which is the correct.
Method II:
II. If a = bc with HCF (b, c) = 1  b and c are
Put the options and check.
coprime numbers.
For (a), R = 3 in each case. LCM (b, c) = bc
For (b), R = 10 in each case. Now, LCM(a, d)= LCM(bc, d)
For (c), R = 17, 3, 3  LCM (a, d) = LCM (c, bd) which is the correct.
For (d), R = 10, 10, 24 7. (a) HCF(480, 720, 2640) = 240
2. (d) 786 – 4 = 782 8. (c)

991 – 5 = 986 HCF(5, 2, 4) 1

Statement-I: HCF = LCM(6, 9, 27) =
54
1196 – 6 = 1190
LCM (6, 5, 15) 30
Number = HCF(782, 986, 1190) = 34 Statement-II: LCM = HCF(11, 22, 44) =
11
3. (d)
Hence, both the statements are correct.
Number = HCF(1879 – 1709, 2399 – 1879)
Hence, option (c) is correct.
= HCF(170, 520) or a factor of HCF
 7 35  HCF(7, 35)
= 10 or a factor of 10. 9. (d) HCF  ,  =
2 4  LCM (2, 4)
Dividing by 10, R = 9, 9, 9
7 3
= m =1 m
4. (b) HCF(117, 130, 143) = 13 4 4
117  130  143 10. (a) LCM + HCF = 310
Total containers = = 30
13 LCM = 30 HCF
5. (b)
31 HCF = 310
Let p = 2, q = 3, r = 4 , s = 5 and t = 6 and remainders
HCF = 10
a = 1, b = 2, c = 5 , d = 4 , e = 5
LCM = 300
Now, 2 - 1 = 1; 3 - 2 = 1; 4 - 3 = 1 5 - 4 = 1; 6 – 5 = 1
LCM × HCF = Product of numbers
k=1
300 × 10 = 20 × Number2
Number that is divisible by 2, 3, 4, 5 and 6 leaving
the remainders 1, 2, 3, 4 and 5 Number2 = 150

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 CSAT HANDOUT : SOLUTIONS | 5

11. (a) 21 × 4641 = N1 × N2 Since numbers are co-prime,

21 × 3 × 7 × 221 = N1 × N2 c = 37 a = 19
= 441 × 221 = N1 × N2 551
b = = 29
= 21 × 3 × 7 × 13 × 17 = N1 × N2 19

= (21 × 13) × (21 × 17) = 273 × 357 Sum = 85

[Shortcut : only in option (a), unit place = 1] 16. (d)

12. (c) From option (d), we can say that the HCF of 36 and
24 is 12 and it also satisfies the given condition
Let number be 32a, 32b
a > b > 12 ... a = 36 and b = 24
HCF = 32
17. (d)
and Sum = 32(a + b) = 256
We know that, LCM of two numbers must be the
a+b=8 multiple of their HCF. In the given options, 60 is
not a multiple of 8 and hence 60 cannot be the LCM
(1) Numbers not multiple of 32
of the numbers.
(2) 32a = 32 a = 1
18. (a)
32b = 224 b = 7 a+b=8
Given numbers are 12,18,21, 28 and remainder,
(3) 32a = 96 a = 3 k = 3.
32b = 160 b = 5 a+b=8 LCM of 12, 18, 21 and 28 = 252
and greatest 4-digit number = 9999
[Shortcut : (1) is there in all options except (c)]
On dividing 9999 by 252,
13. (a) LCM = 12 HCF
39
252 9999
LCM + HCF = 403
756
13 HCF = 403
2439
HCF = 31 2268
LCM = 372 171  R

 Required number = [Greatest 4-digit

31  372
= 124 = Number2 number – R] + k
93
= (9999 - R) + k
14. (a) Numbers = 3x, 4x, 5x
= (9999-171) + 3
LCM × HCF2 = 3x × 4x × 5x
= 9828 + 3 = 9831
2400 × x2 = 60 × x3
19. (c)
40 = x = HCF Let P(x) = 36{x4 + 5x3 – 2x2}
15. (c) Co-prime  HCF = 1 = 36x2 {3x2 + 5x – 2}

a × b = 551 = 36x2 {3x2 + 6x – x – 2}

= 36x2 (3x(x + 2) – 1(x + 2)}
b × c = 1073
= 2 × 2 × 3 × 3 × x + x × (x + 2) (3x – 1)
c 1073 29  37 37 Q(x) = 9(6x3 + 4x2 – 2x) = 9x(6x2 + 4x – 2)
= = =
a 551 19  29 19
= 18x (3x2 + 2x – 1)

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6 | CSAT HANDOUT : SOLUTIONS

= 18x {3x2 + 3x – x – 1} 20. (a)

= 18x {3x(x + 1) – 1(x + 1)} HCF = 12
= 2 × 3 × 3 × x × (3x – 1)(x + 1) LCM = 72
and R(x) = 54(27x4 – x) = 54x(27x3 – 1) 12  72
Second number =  36
= 2 × 3 × 3 × 3 × x × (3x – 1)(9x2 + 3x + 1) 24
[ a3 – b3 = (a – b)(a2 + b2 + ab)] Difference between two numbers will be 36 – 24

So, HCF of [P(x), Q(x), R(x)] = 2 × 3 × 3 × x(3x – 1) = 12

= 18x(3x – 1)


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 CSAT HANDOUT : SOLUTIONS | 7

CLASS - III : SOLUTIONS

1. (c) 5. (a) by 5 – unit digit should be 0 or 5
71640,321970
803642 + x
(b) by 3 — 1 1 4 3, 4 7 2 0 5, 2 3 1 6
(8 + 3 + 4) – (0 + 6 + 2 + x) = 11 k
(c) by 6 — 1 4 2 8, 9 2 5 2, 7 2 1 2
15 – (8 + x) = 11k
(d) by 8 — 7 1 6 4 0, 3 2 1 9 7 6
7 – x = 11k
7 – x = 0 or 11 6. (c)
x = 7 or – 4
5172355 = Sum = 28 – not divisible by 9
 x=7
7. (c)
2. (c) A381 is divisible by 11 if and only if (A + 8) – (3 + 1)
24 — divisible by 8 and 3. i.e. (A + 4) is divisible by 11.
(c) — 1236480 – divisible by 8.
So, A = 7 satisfies the condition.
Sum = 24 – divisible by 3.
8. (b)
3. (b)
381 A is divisible by 9.

(x + 7 + 7) – (3 + 1) = 11k or 0 So, 3 + 8 + 1 + A = 12 + A is divisible by 9.

14 + x – 4 = 11k or 0 So, A = 6
x + 10 = 11
9. (d)
x = 1 (For x = 1 number will be divisible by 11)
15 B is divisible by 6.
4. (a&c) So B should be even number
Divisibility rule for 2, 3, 6
also 1 + 5 + B should be divisible by 3
Rule-2 : unit place even
Rule-3 : sum divisible by 3 So only 6 is the value for B.

Rule-6 : divisible by both 2 and 3. 10. (d)

3 + 4 + 0 + 9 + 1 + 2 + 2 = 21 – divisible by 3.
7921782
even — divisible by 2.
From options, it is divisible by 11 and either of 5, 6,
(b) 634 — not divisible by 8. 8, 9.

No. Not divisible by 5(unit place = 2)

Not divisible by 8 (782 not divisible by 8)
(c) 04 — divisible by 4
Divisible by 9 (sum divisible by 9)
Yes. So, option is (d).

(d) (2 + 7 + 6 + 3 + 2)— (3 + 4 + 8 + 9 + 3) 11. (c)

= 20 – 27 = – 7 – not a factor of 11 3576ab

b =0
No.
6a0 – divisible by 8

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8 | CSAT HANDOUT : SOLUTIONS

So, a = 0, 8 17. (d)

a + b = 0, 8. 136
5B7
12. (d) 7A3
882 3 + B + 1 = A + 10
6a4
B=A+6
14 a 6
also, 7 + A + 3 = 3k
8 + a = must be a single digit number as no number
was carried forward (Since 8 + 6 = 14) A = 2, 5, 8
So, 8 + a = b = 8 or 9 B = 8, 11, 14
a = 0, 1 So, only possible values are A = 2
b = 8, 9 B=8
From options a + b = 10, a, b = 1, 9 18. (a)
Number = 1496 — divisible by 22. 58N
13. (b) N = 4 – divisible by 8.
(M + 9 + 4 + 4 + 8) – (3 + 0 + 8 + 5 + 4)
146
= then highest value of 5 will be M + 25 – 20  M + 5
5n
M=6
146 146 146
 2  3 ... = 29 + 5 + 1 = 35
5 5 5 19. (b)
R = 5a
14. (b)
S = 5b
A number to be divisible by 88 it should be divisible
by 8 and 11 because 8 and 11 are co-prime numbers R – S = 5(a – b) – divisible by 5
whose multiplication gives 88. R × S = 25 ab – divisible by 25
R2 + S2 = 25(a2 + b2) – divisible by 5
15. (d)
R + S = 5(a + b) – divisible by 5 but not 10.
n – divisible by 3 but not 6.
So, n = 9, 15, 21 ... 20. (b)
From options, 10x – 1
3n + 1 = 9 × 3 + 1 = 28 x19
x  2  99 .... divisible by 11
16. (c)
x  3  999
4 + 2 + 5 + 2 + 7 + 4 + 6 + B = 3k x  4  9999 .... divisible by 11
30 + B = 3k
B = 0, 3, 6, 9 

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 CSAT HANDOUT : SOLUTIONS | 9

CLASS - IV : SOLUTIONS
5. (a) 4% = 0.04 (b) 11% = 0.11
1 2 2
1. (a) = 50% (b)  66 %
2 3 3 2
(c) 17 % = 0.174 (d) 19% = 0.19
3 5 1 5
(c)  75% (d)  83 %
4 6 3
3
(e) 185 % = 1.8575
8 2 9 4
(e) = 114 % (f) = 180%
7 7 5
(f) 223% = 2.23
23 1 55 (g) 12650% = 126.50
(g) = 209 % (h) = 250%
11 11 22
7
1 3 (h) 876543 % = 8765.43875
2. (a) 20% = (b) 30% = 8
5 10
6. (a) 0.14 = 14% (b) 0.23 = 23%
13 17
(c) 65% = (d) 85% = (c) 0.768 = 76.8% (d) 0.815 = 81.5%
20 20
(e) 3.84 = 384% (f) 4.89 = 489%
3 1
(e) 115% = 1 (f)225% = 2
20 4 (g) 123.75 = 12375% (h) 678.54 = 67854%

3 3 7. Hostelers = 70% of 2800 = 1960

(g) 375% = 3 (h)960% = 9
4 5
510
8. Amit =  85%
3 1 600
3. (a) 15% = (b) 25% =
20 4
420
Ashish = = 84%
21 14 500
(c) 42% = (d) 56% =
50 25 Amit performed better.

21 39 15
(e) 84% = (f) 156% = 9. % change =  100%  25%
25 25 60

18 23 30
(g) 360% = (h)1150% = 10. % change =  100%  20%
5 2 150

1 22000
4. (a) 1 : 10 = 10% (b)2 : 15 = 13 % 11. (i) % salary =  100%  88%
3 25000

4 20 25000
(c) 3 : 35 = 8 % (d)5 : 84 = 5 % (ii) % salary =  100% = 113.63%
7 21 22000

2 12. Let b = 3, then a = 4

(e) 12 : 5 = 240% (f)25 : 6 = 416 %
3
1
2 % value = = 25%
4
(g) 124 : 15 = 826 %
3
13. Let b = 3, a = 2
5
(h) 386 : 225 = 171 % 1
9 % value =  100%  50%
2

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10 | CSAT HANDOUT : SOLUTIONS

14. Final salary = 5000[1.2][1.25]  160  90 

% change =  1    100%
= Rs. 75000  200  100 
= 28%
15. b = 1.5a, a = 1
c = 0.9b = 0.9 × 1.5a = 1.35a Method-II:

c is 35% more than a New area = area(1 – 0.2)(1 – 0.1)

20  10
16. Let salary = 100x % change = 20  10  = –28%
100
Travel = 10x
i.e., 28% decrease.
Personal = 18x
Shopping = 18x 21. (c)
Left = 54x = 21600 Fit products produced per day = 950
x = 400 19000
Total days = = 20 days
Salary = 40000 950

17. a = 0.75c 22. (b)

b = 0.6c Total correct question solved = 7 + 12 + 18 = 37
Minimum correct questions needed = 60% of 70
b 5 = 42
a = 0.75   b
0.6 4
More correct questions needed by Anita = 42 – 37
a is 125% of b. =5

18. Method-I: 23. (b)

Let original cost = 100 x = 0.8z
Increased price = 115 x 9x
y = 0.72z =  0.72 =
Final price = 115 × 0.85 = 97.75 0.8 10

2.25 y = 90% of x
Loss% =  100%  2.25%
100 24. (b)
Method-II: B = 0.95 A and also
A = B + 20
15  15
Loss% = % = 2.25%  A = 0.95A + 20
100
20
19. Method-I: A= = 400
0.05
Let price be a B = 380
ATQ, a(1.2)(0.9) = 54
25. (b)
a = Rs. 50
80% = 12 km
Method-II:
12
(20)( 10) 100% =  100 = 15 km
% change = 20  10  = 8% increase 80
100
54 = 1.08 (original price) 26. (b)
Original price = Rs. 50 Let the number of boys and girls be x and y
respectively.
20. Method-I:
Total sum of ages of boys = 20 x and total sum of
Original area = 200 × 100 cm2 ages of girls = 16y
New area = 160 × 90 cm2 Given that, 20x + 16y/(x+y)=18

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 CSAT HANDOUT : SOLUTIONS | 11

20x + 16y = 18x + 18y

  11x   9 x  
x=y   10    10  
      100%
= 
Hence, percentage of boys in group is 50%.   9x  
   
27. (a)   10  
Let the original income bex, i.e. I = x Rs 200 2
= % = 22 %
Saving income (s1 ) = x × 10% = rs x/10 9 9
Initial expenditure = x - (x/10)% = 9x/10 28. (b)
2 yr later, when income has increased by 20%.
40
Then, new income = x × 120% = 12x/10 Quantity of alcohol in 5 L of solution = 5
100
And new saving income (s2) = rs x/10
= 2L
New expenditure = New income -New saving
Quantity of alcohol in 6L of solution = 2L
income = [(12x)/10] - x/10 = Rs (11x)/10
 Strength of alcohol in new solution
Required percentage = (New expenditure - Initial
expenditure)/Initial expenditure × 100% 2  1
=   100  % = 33 %
6  3



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12 | CSAT HANDOUT : SOLUTIONS

CLASS - V : SOLUTIONS
1. (b) 15x = 60
Expense = Price × Quantity consumed
x = 4
(Price × Quantity)original = 1.25 × Price × Quantitynew
Passing marks = 25x + 40 = 140
4
Q original = Qnew = 80% of original 140
5 Pass% =  100% = 35%
400
i.e., consumption must decrease by 20%.

2. (a) 7. (b)
Milk = 35% of 200gm = 70gm Let total marks = 100x
Water = 130gm A.T.Q., 56x – 66 = 84x – 234
New water = 130 + 80 = 210 gm 28x = 168
70 x = 6
% milk =  100% = 25%
280 Passing marks = 56x – 66 = 270

3. (c) 8. (a)
Let original price = 100
45x
 27 = 36 Original expense = 100 × 30 = 3000
100
New expense = 3300
45x = 900
New price = 132
x = 20
3300
4. (b) New quantity = = 25
0.1Q = 0.2R
132
= 0.2(10000) = 2000 9. (d)
Q = 20000 Let maximum marks of each paper be 100
0.05P = 0.15Q = 3000 Total marks = 500
P = 60000 Marks scored = 300 = 6x + 7x + 8x + 9x + 10x
P + Q + R = 90000 40x = 300

5. (d) x = 7.5
Original petrol price = 100 Marks = 45, 52.5, 60, 67.5, 75
New petrol price = 140 10. (c)
Original usage = 100 Let total employees = 100
New usage = x
100
Original expense = 100 × 100 = 10,000
New expense = 140x = 1.05(10000) 40 60
Males Females
x = 75
25 30 10
So, % decrease =  100% = 25% Salary Salary
100  30000  30000
6. (d)
Let total marks = 100x 45 employees = salary  30000

A.T.Q., 25x + 40 = 40x – 20  15 females salary  30000

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 CSAT HANDOUT : SOLUTIONS | 13

60 16. (d)
Females
Let A’s weight = 100
15 45 B’s weight = 60
 30000 < 30000
60
Salary Salary % decrease in weight =  100% = 37.5%
75% of female employees earn < 30000. 160
17. (a)
11. (a)
Amount he has = 75% = 9000
Water = 4950 gm
 25% = 3000
Rest = 50 gm
So he needs 1,000 more.
water
New mass = x 18. (b)
rest = 50gm = 2%
(1.25 price) (Quantity – 6) = 1200
100% = 2500 gm
and Price × Quantity = 1200
water = 2450 gm
1.25(1200) – (1.25 × 6) Price = 1200
12. (c)
5
Wage bill = 300 × w 300 =  6  price
4
New wage bill = 345 × 0.8w Rs. 40 = Price
 345  0.8 
% change =  1   100% 19. (c)
 300 
= 8% decrease 60 marks = 40%
Maximum marks = 100% = 150
13. (a) Milk = 324l
Water = 36l 20. (c)
(324 + 36 + x) × 19% = 36 + x 40% men = 30% women
(360 + x)19 = 100(36 + x) 4 × men = 3 × women
Let women = 100x
81x = 3240
men = 75x
x = 40l Married people = 30x + 30x = 60x
14. (a) 60x
% married population =  100%
Let total voters = 100x 175x
Voters voted = 75x 2
= 34 %
Valid votes = 70x 7
C = 1.4B = B + 84000 21. (b)
 0.4B = 84000 Pricefinal = Priceoriginal × (1.25)(0.8)(1.1)
= 1.1 Priceoriginal
B = 210000
= 110% of original
C = 294000
A = 40x 22. (a)
B + C = 70x – 40x = 30x 0.7x + y = 1.65y
7x = 6.5y
So A got maximum votes.
z + 0.6x = 1.65z
15. (b) 6x = 6.5z
Households between 30000 – 100000 = 66% x < y, x > z
66% = 990 households y>x>z
1% = 15 households 23. (d)
Between 1L – 2L = 22% – 6% = 16% Let A = 100, B = 50, then
= 240
P = 40% of 100 + 65% of 50

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14 | CSAT HANDOUT : SOLUTIONS

40 65 x
=  100   50 Proportion of females =
100 100 xy
= 40 + 32.5 = 72.5 No. of males in office B = x
Q = 50% of 100 + 50% of 50 No. of employees in office B = 1.5(x + y)
50 50 x
=  100   50 Proportion of females = 1 
100 100 1.5( x  y )
= 50 + 25 = 75
Hence, percentage of males in the overall
Here, P < Q
(x  y )
Again, let A = 101, B = 100, then employee group =
( x  y )  1.5( x  y )
P = 40% of 101 + 65% of 100
40 65 1
=  101   100 = = 40%
100 100 2.5

= 40.4 + 65 = 105.4 25. (d)

Q = 50% of 101 + 50% of 100 Let weight of total dry fruit be 'W' kg
50 50 % of pulp in fresh fruit = 30%
=  101   100
100 100
Weight of pulp in fresh fruit = 30% of 80
= 50.5 + 50 = 100.5
Here, P > Q 30  80
= = 24 kg.
100
Therefore, none of the relation can be
concluded with certainty. % of pulp in dry fruit = 75%
Weight of fruit pulp in dry fruit = 75% of W.
24. (b)
 weight of fruit pulp in constant in fresh
Let the no. of females in office A = x fruit and dry fruit
and no. of males in office A = y  24 kg = 75% of W
then total number of employees in office
A = (x + y) 24  100
 W=
75

 W = 32 kg

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