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44.08

44.08

45

45

None

NA

BED

BED

20

? = 148 X 3.75 x 4/111 = 20

197

Let 'N' is the smallest number which divided by 13 and 16 LEAVES RESPECTIVE REMAINDERS of 2 and 5.

Required number = (LCM of 13 and 16) - (common DIFFERENCE of divisors and remainders) = (208) - (11) = 197.

125

Let the number = X,

Then, x-(3/5)x = 50,

=> (2/5)x = 50 => 2x = 50*5,

=> x = 125

135

The correct number is 136.
3 x 1 + 1 = 4
4 x 2 + 2 = 10
10 x 3 + 3 = 33
33 x 4 + 4 = 136
136 x 5 + 5 = 685
685 x 6 + 6 = 4116

0

0

If the data in statement (I) alone is SUFFICIENT to answer the question while that in statement (II) alone is not sufficient.

Let the number be 10X + y
∴ 3x = 4y......(1)
From (I),
x + y = 14 .....(2)
Solving equations (1) and (2),
x = 8, y = 6
∴ Number = 86 From (II)
y = 75/100 x x 
∴ 3x = 4y.
Which is the same as eqn (1)

<P> 39/340

Total balls = 4 + 6 + 7 = 17
∴ N(S) = 17C1 = 680
Two red balls can be selected from four red balls in 4C2 = 6 ways.
and the third ball can be selected from the REMAINING 13 balls in 13C1 = 13 ways.
∴ P (E) = 13x6/680 = 39/340

36

∴ REQD . NUMBERS = 3 X 4P2 = 3 x 12 = 36

36

Out of the total six LETTERS, three are vowels (E, I, E) and three are consonants (SRS)
Here, E and S COME twice If we take all the vowels together the number of letters is 3 + 1 = 4
So, letters can be arranged in 41/21 ways and three vowels can be arranged in 31/21 ways
 ∴ Total number of ways = 41/21 X 31/21= 12 x 3 = 36

11! - 8! x 4!

without any restriction, 11 persons can sit in 11! ways, If we take all the GIRLS together as ONE, the total number of persons will be 7 + 1 = 8, who can sit in 8! ways and the FOUR girls can sit in 4! ways among themselves. So, when all the girlssit together the number of ways = 8!x 4!
When they don't sit together, the total number of ways = 11! - 8! x 4!

1953.6

? = 444 X 44 x 4/40 = 1953.6

140

4550 ÷ ? X 25/100 = 130
= 4550 ÷ ?/4 = 130
= 4550 x 4/? = 130
= ? x 130 = 4550 x 4
= ? = 4550 x 4/130 = 140

37

888 ÷ 1 = 888
888 ÷ 2 = 444
444 ÷ 3 = 148
148 ÷ 4 = 37
37 ÷ 5 = 7.4

12

The word TEST CONSISTS of 4 LETTERS in which T comes TWICE 
∴ Number of arrangements= 4!/2! = 4 x 3 x 2 x 1/2 x 1 = 12

47

? = 3√103823 = 3√47 X 47 x 47 = 47

7

? = (√6 + 1)² - 2√6
= (√6)² + 2 X √6 + 1 + 1² - 2√6
= 6 + 2√6 + 1 - 2√6 = 6 + 1 = 7

91/190

Total possible outcomes 
= 20C2 = 20 X 19/1 x 2 = 190
Total FAVOURABLE outcomes = 7C1 x 13C1 = 7 x 13 = 91
∴ Required probability = 91/190

19

? = √49 + 289 + 25 - 2 = √361 = 19

660

Let the consecutive even NUMBERS be
 A = 2x, B = 2x + 2, C = 2x + 4, D = 2x + 6 and
 E = 2x + 8
= A + B + C + D + E = 130
= 10X + 20 = 130
∴ x = 130 - 20/10 = 11
Therefore, product of A and E= 22 x 30 = 660

57.5

Given EXP. = 35 + 15 * 3/2 

= 35 + 45/2 = 35 + 22.5 = 57.5

2/5

A = 2B => B = 1/2 A, so, AB = 2/25 

1/2 A2 = 2/25

A2 = 4/25 

A = 2/5

24

Let the ten's digit be X. Then, unit's digit = x + 2. NUMBER = 10x + (x + 2) = 11x + 2 

Sum of digits = x + (x + 2) = 2x + 2 

(11x + 2)(2x + 2) = 144 

2x2 + 26x - 140 = 0 

(x - 2)(11x + 35) = 0 

x = 2

Hence, REQUIRED number = 11x + 2 = 24.

21

Let the NUMBER of CHILDREN be 'X' Therefore, each child will receive
X x 60/100 = 3x/5 CHOCOLATES
∴ X x 3x/5 = 735
∴ x² x 735x5/3 = 1225
∴ x = 1225 = 35
∴ Number of chocolates = 3x/5 = 21

20160

No. of WORDS = l2l4l7/ l3l2= 20160

NONE of these

None of these

31

31

21

21

64

64

343

343

83

Let the first odd NUMBER be X.
Then, the other odd number would be x + 2.
∴ x(x+2) = 6723
= x2 + 2x = 6723
= x2 + 2x = 6723
= x2 + 2x - 6723 = 0
= x2 + 83 - 81 x - 6723 = 0
= x(x+83) - 81 (x + 83) = 0
= (x + 83) (x - 81) = 0
= (x + 83) (x - 81) = 0
= x - 81 = 0
∴ x = 81
Hence, GREATER number = x + 2 = 81 + 2 = 83

18

4th No. = (Total of 7 Nos) - (Total of FIRST THREE + total of last 3)

2444

? = √2704 X 2209
= 52 x 47 = 2444

10

(?)² = 3² X 4² x 5/36 + 80
= 20 + 80 = 100
∴ ? = √100 = 10

10

Clearly, (2272 - 875) = 1397, is exactly DIVISIBLE by N.

Now. 1397 = 11 * 127

The REQUIRED 3-DIGIT number is 127, the sum of whose digit is 10.

15

In this simple ADDITION with REPETITION series, each number in the series repeats itself, and then INCREASES by 12 to ARRIVE at the NEXT number.

53

This is a simple subtraction series in which a random number, 93, is interpolated as EVERY third number. In the subtraction series, 10 is SUBTRACTED from each number to ARRIVE at the NEXT.

3

This is a MULTIPLICATION SERIES; each number is 3 times the PREVIOUS number.

IV

This is an ALTERNATING addition and SUBTRACTION series. ROMAN NUMBERS alternate with Arabic numbers. In the Roman numeral PATTERN, each number decreases by 1. In the Arabic numeral pattern, each number increases by 1.

78

This SERIES ALTERNATES the addition of 4 with the subtraction of 3.

60

The world LEASE CONSISTS of 5 LETTERS in which E COMES TWICE
 ∴ 51/21 = 5x4x3x2x1/2x1 = 60

17 years

Let RADHA's PRESENT age = x years.
∴ x = 2 (x = 12) - 3
= x = 2X - 24 - 3
= x = 27
∴ Raja's present age = 4/9 x 27= 12 years 
∴ Raja's age after 5 years = 12 + 5 = 17 years

29 years

Age of the fourth BOY 
= (26 x 7 - 3 x 19 - 3 x 32) years 
= (182 - 57 - 96) years 
= 29 years

239

Let the TEN's digit be X.
∴ Unit's digit = 3X
Hundered's digit = 2X/3
∴ x + 3x + 2x/3 = 14
= 3x + 9X + 2x = 14 x 3
= 14x = 14 x 3
= x = 3
∴ Number = 239

17

? = √360 - 225 x 2 + 379
 = √360 - 450 + 379 
= √289
= 17

NONE of these

Let the greater and the SMALLER NUMBER be g and s respectively.

gs = 2560

g + s EXCEEDS g - s by 64 i.e., g + s - (g - s) = 64

i.e., 2s = 64 => s = 32.

g = 2560/s = 80.

670

Let the five consecutive ODD NUMBERS of set p be 2n - 3, 2n - 1, 2n + 1, 2n + 3, 2n + 5.

Sum of these five numbers

= 2n - 3 + 2n - 1 + 2n + 1 + 2n + 3 + 2n + 5

= 10N + 5 = 435 => n = 43

Largest number of set p = 2(43) + 5 = 91

The largest number of set q = 91 + 45 = 136

=> The five numbers of set q are 132, 133, 134, 135, 136.

Sum of above numbers = 132 + 133 + 134 + 135 + 136 = 670.