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1.The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when

divided by 9 leaves no remainder, is:

A 1677 B 1683 C2523 D3363

Answer: Option B

L.C.M. of 5, 6, 7, 8 = 840. Required number is in the of the form 840k + 3

Least value of k for which (840k + 3) is divisible by 9 is k= 2.

Required number = (840*2+3) = 1683.

2.There is a certain four digit number whose fourth digit is twice the first digit. Third digit is

three more than second digit. Sum of the first and fourth digits is twice the third number.

What was that number?

A3036 B4368 C4148 D3146

Answer: Option B

From the options, only 4368 is satisfying the conditions

As per First condition; 4*2 = 8As per second condition 3+3 = 6

As per third condition 4+8 = 2*6

3.A worker is paid Rs.20/- for a full days work. He works 1,1/3,2/3,1/8,3/4 days in a week.

What is the total amount paid for that worker ?

A57.50 B57 C57.9 D58

Answer: Option A

4.If the number 42573 * is exactly divisible by 72, then the minimum value of * is:

A4 B5 C6 D7

Answer: Option C

72 = 9 x8, where 9 and 8 are co-prime.

The minimum value of x for which 73x for which 73x is divisible by 8 is, x = 6.

Sum of digits in 425736 = (4 + 2 + 5 + 7 + 3 + 6) = 27, which is divisible by 9.

Required value of * is 6.

5.Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder

in each case.

A4% of a B5% of a C20% of a D None of these

Answer: Option A

20% ofa = b =>20

a = b.

100b% of 20 = (

bx 20 ) = (

20a x

1x 20 ) =

4a = 4% ofa.

100 100 100 100

6.A number which when divided by 3, 4, 5, 6, 7, leaves the remainder 2, 3, 4, 5 and 6

respectively. Such largest 5 digit number is :

A99960 B999579 C99539 D99959

Answer: Option D

Observe that 3-2.4-37-6 leave 1 as the common differenceLCM (3, 4, 5,6, 7) = 3*4*5*7 = 420

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Largest 5 digits number = 99999, Which on being divided by 420 will leave 238 and 39 as

remainder

Hence, the largest such number = 99999-(remainder + common difference) = 99999-40

=99959.

7.Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the sameremainder in each case. Then sum of the digits in N is:

A4 B5 C6 D8

Answer: Option A

N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305) = H.C.F. of 3360, 2240 and

5600 = 1120.

Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

8.Sanal has 60 red marbles, 156 blue marbles and 204 green marbles. He distributes them

amongst a group of kids, such that each kid gets equal marbles and no kid has marbles of

more than one colour. Find the number of kids.

A12 B21 C35 D36Answer: Option C

HCF of 60, 156 and 204 is 12

60/12 = 5, 156/12 = 13, 204/12 = 17

Each kid gets 12 marbles, 5 get red, 13 get blue and 17 get green

:. Total number of kids = 5 +13+17 = 35

9.Two numbers are in the ratio of 11 : 13 . If the H.C.F of these numbers is 19, determine the

numbers

A304, 369 B209, 247 C182, 199 D182,122

Answer: Option B

Let the numbers be 11x and 13xSince the H.C.F of given numbers is 19 which indicates that 19 is the common factor of these

two numbers. Hence, it is obvious that value of x is 19. Therefore these numbers are 209 and

247 respectively.

10.I lost Rs.68 in two races. My second race loss is Rs.6 more than the first race. My friend

lost Rs.4 more than me in the second race. What is the amount lost by my friend in the

second race?

A40 B39 C41 D45

Answer: Option C

x + x+6 = rs 68

2x + 6 = 68

2x = 68-62x = 62

x=31

x is the amt lost in I race

x+ 6 = 31+6=37 is lost in second race

then my friend lost 37 + 4 = 41 Rs

answers for synonyms:

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1. c2. a3. b4. a5. c6. a7. a8. d9. c10.b