1) If the square root of 55625 is 75, then √5625 + √56.25 + √0.5625 is equal to

1) 82.25

2) 83.25

3) 80.25

4) 79.25

Sol: 2) √5625 = 75; √56.25 = 7.5; √. 5625 = .75

-- > 75+7.5+0.75 = 83.25

2) Which of the following integers has most number of divisors?

1) 176

2) 182

3) 99

4) 101

Sol: 2) 176 = 2,4,8,11,16,22,44,88

182 = 2,7,13,14,26,91

99 = 3, 9, 11, 33

101= 101

3) A boy was asked to find the value of 3/8 of a sum of money. Instead of multiplying The sum by

3/8 he divided it by 3/8 and then his answer exceeded by Rs. 55. Find the Correct be x.

Sol: Let amount be x

8/3* - 3/8 * = 55

-- > 64x – 9x/24 = 55 -- > 55x/24 = 55

-- > x = 24*55/55 = 24

: . 3/8 of x = 3/8 * 24 = Rs.9

4) A boy was asked to find the value of 7/12 of a sum of money. Instead of multiplying The sum by

7/12 he divided it by 7/12 and thus his answer exceeded the correct Answer By Rs.95. Find the correct

answer.

Sol: Let sum = Rs. K

: . 12/7 k – 7k/12 = 95

-- > 144k – 49k/84 = 95 -- > k = 84

:. 7/12 k -- > 7/12 * 84 = Rs. 49

5) In a boat 25 persons were sitting. Their average weight increased one kilogram when One man

goes and a new man comes in. The weight of the new man is 70kgs. Find the Weight of the man who is

going.

Sol: Weight increased per person is 1 kg.

Total increase in weight = 25 kgs

Weight of new man is 70 kgs,

(Which means his weight is 25 kgs heavier)

The weight of the old man was 70 – 25 = 45 kgs

6) What is the greatest possible length that can be used to measure exactly the following Lengths 7m,

3m 85cm, 12m 95cm?

The length to be measured is

700cm, 385cm, 1295cm.

The required length in cm is the H.C.F of 700, 385, and 1295, which is 35 cm.

7) The product of two-digit number is 2160 and their H.C.F is 12. The numbers are

Let the number are 12a and 12b

Then 12a * 12b = 2160

ab = 15

Value of co-primes a and b are (1, 15) (3,5)

: . The two digit numbers are (3*12, 5*12)

= (36, 60)

8) The least number of 6 digits which it exactly divisible by 12, 15 and 18 is

Least number of 6 digits is 100000

L.C.M of 12, 15, 18, is 180.

On dividing 100000 by 180, the remainder is 100

. Required number = 100000 + (180-100) = 100080

9) Two third of three fifth of one fourth of a number is 24. What is 40% of that number?

1) 96

2) 72

3) 120

4) 156

Sol: 1) let the number be x, then

X of 2/3 of 3/5 of . = 24

X * 2/3 * 3/5 * . = 24, x = 240.

Hence 40% of 240 = 40/100 * 240 = 96

10) It was calculated that 75 men could complete a piece of work in 20 days.

When work was scheduled to commence, it was found necessary to send 25

men to another project. How much longer will it take to complete the work?

Answer:

30 days.

Explanation:

Before:

One day work = 1 / 20

One man’s one day work = 1 / ( 20 * 75)

Now:

No. Of workers = 50

One day work = 50 * 1 / ( 20 * 75)

The total no. of days required to complete the work = (75 * 20) / 50 =30

11) A student divided a number by 2/3 when he required to multiply by 3/2. Calculate the percentage of error in his result.

Answer:

0 %

Explanation:

Since 3x / 2 = x / (2 / 3)

12) A dishonest shopkeeper professes to sell pulses at the cost price, but he uses a false weight of 950gm. for a kg. His gain is …%.

Answer:

5.3 %

Explanation:

He sells 950 grams of pulses and gains 50 grams.

If he sells 100 grams of pulses then he will gain (50 / 950) *100 =5.26

13) A software engineer has the capability of thinking 100 lines of code in five minutes and can type 100 lines of code in 10 minutes. He takes a break for five minutes after every ten minutes. How many lines of codes will he complete typing after an hour?

Answer:

250 lines of codes

14) A man was engaged on a job for 30 days on the condition that he would get a wage of Rs. 10 for the day he works, but he have to pay a fine of Rs. 2 for each day of his absence. If he gets Rs. 216 at the end, he was absent for work for ... days.

Answer:

7 days

Explanation:

The equation portraying the given problem is:

10 * x – 2 * (30 – x) = 216 where x is the number of working days.

Solving this we get x = 23

Number of days he was absent was 7 (30-23) days.

15) A contractor agreeing to finish a work in 150 days, employed 75 men each working 8 hours daily. After 90 days, only 2/7 of the work was completed. Increasing the number of men by ________ each working now for 10 hours daily, the work can be completed in time.

Answer:

150 men.

Explanation:

One day’s work = 2 / (7 * 90)

One hour’s work = 2 / (7 * 90 * 8)

One man’s work = 2 / (7 * 90 * 8 * 75)

The remaining work (5/7) has to be completed within 60 days, because

the total number of days allotted for the project is 150 days.

So we get the equation

(2 * 10 * x * 60) / (7 * 90 * 8 * 75) = 5/7 where x is the number of

men working after the 90th day.

We get x = 225

Since we have 75 men already, it is enough to add only 150 men.

16) What is a percent of b divided by b percent of a?

(a) a

(b) b

(c) 1

(d) 10

(e) 100

Answer:

(c) 1

Explanation:

a percent of b : (a/100) * b

b percent of a : (b/100) * a

a percent of b divided by b percent of a : ((a / 100 )*b) / (b/100) * a )) = 1

17) A man bought a horse and a cart. If he sold the horse at 10 % loss and the cart at 20 % gain, he would not lose anything; but if he sold the horse at 5% loss and the cart at 5% gain, he would lose Rs. 10 in the bargain. The amount paid by him was Rs._______ for the horse and Rs.________ for the cart.

Answer:

Cost price of horse = Rs. 400 & the cost price of cart = 200.

Explanation:-

Let x be the cost price of the horse and y be the cost price of the cart.

In the first sale there is no loss or profit. (i.e.) The loss obtained is equal to the

gain.

Therefore (10/100) * x = (20/100) * y

X = 2 * y -----------------(1)

In the second sale, he lost Rs. 10. (i.e.) The loss is greater than the profit by

Rs. 10.

Therefore (5 / 100) * x = (5 / 100) * y + 10 -------(2)

Substituting (1) in (2) we get

(10 / 100) * y = (5 / 100) * y + 10

(5 / 100) * y = 10

y = 200

From (1) 2 * 200 = x = 400

18) A tennis marker is trying to put together a team of four players for a tennis tournament out of seven available. males - a, b and c; females – m, n, o and p. All players are of equal ability and there must be at least two males in the 1 team. For a team of four, all players must be able to play with each other

under the following restrictions:

b should not play with m,

c should not play with p, and

a should not play with o.

Which of the following statements must be false?

1. b and p cannot be selected together

2. c and o cannot be selected together

3. c and n cannot be selected together.

Answer:

3.

Explanation:

Since inclusion of any male player will reject a female from the team.

Since there should be four member in the team and only three males are

available, the girl, n should included in the team always irrespective of others

selection.

19) Five farmers have 7, 9, 11, 13 & 14 apple trees, respectively in their orchards. Last year, each of them discovered that every tree in their own orchard bore exactly the same number of apples. Further, if the third farmer gives one apple to the first, and the fifth gives three to each of the second and the fourth, they would all have exactly the same number of apples. What were the yields per tree in the orchards of the third and fourth farmers?

Answer:

11 & 9 apples per tree.

Explanation:

Let a, b, c, d & e be the total number of apples bored per year in A, B,

C, D & E ‘s orchard. Given that a + 1 = b + 3 = c – 1 = d + 3 = e – 6

But the question is to find the number of apples bored per tree in C and D ‘s

orchard. If is enough to consider c – 1 = d + 3.

Since the number of trees in C’s orchard is 11 and that of D’s orchard

is 13. Let x and y be the number of apples bored per tree in C & d ‘s orchard

respectively.

Therefore 11 x – 1 = 13 y + 3

By trial and error method, we get the value for x and y as 11 and 9

20) Five boys were climbing a hill. J was following H. R was just ahead of G. K was between G & H. They were climbing up in a column. Who was the second?

Answer:

G.

Explanation:

The order in which they are climbing is R – G – K – H – J

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