**SJVN-Ltd Placement Paper : 42402**

# SJVN-Ltd Question-Paper placement paper Question and answers

**Executive aptitude questions for learn and practice**

1. 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.

One day work = 1 / 20

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

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

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

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

3. 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 %

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

4. 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?

250 lines of codes

5. 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.

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.

6. 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.

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.

8. 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.

et 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

9. 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 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.

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.

10. 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-5

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