Data Caselets Practice Questions: Level 02

DIRECTIONS for questions 1to 5: Refer to the data below & answer the questions that follow.
A man goes on dividing a certain sum of money into half. He does this 9 times. He gives the amount in round 1 and round 4 and round 7 to A. To B he gives the amount in rounds 2, 5 and 8. To C he gives the amount in Rounds 3, 6 and 9. Assume that the balance sum at the end of Round 9 is Rs. 195.31.
1. What is the original sum of money?
1. Rs. 10 x 104
2. Rs. 104
3. Rs. 106
4. Rs. 2.5 x 105
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Answer Option: 1
If the original amount is x at the end of round 9, the amount left is x / 29 = x / 512
Since the amount is Rs. 195.31 the original sum, x = 195.31 x 512
Approximately 200 x 500 = 10 x 104.
2. What is the ratio of shares of A : B : C ?
1. 10 : 5 : 2
2. 12 : 6 : 2
3. 4 : 2 : 1
4. 1 : 1 : 1
Answer Option: 3
Ratio of shares cannot be equal.
The ratio has to be such that share of C is half of B which is half of A. So answer is 3.
3. If A gets amounts in Rounds 1, 2 and 3, ; B in rounds 4, 5 and 6 ; C in Rounds 7 , 8 and 9 what is the ratio of shares of A : B : C ?
1. 64 : 8 : 1
2. 256 : 128 : 4
3. 128 : 64 : 8
4. 12 : 4 : 1
Answer Option: 1
A:B:C =
(1/2 + 1/4 + 1/8) : (1/16 + 1/32 + 1/64) : (1/128 + 1/256 + 1/512)
= 64:8:1
4. If the balance at the end of Round 9 is Rs. 390.625, the original sum is
1. Rs. 2.5 x 104
2. Rs. 2 x 105
3. Rs. 2 x 106
4. Rs. 3 x 108
Answer Option: 2
If the sum at the end of round 9 is Rs. 195.31, original sum is 10. If amount is Rs. 390.625, then the original amount is 2 x 105.
5. How much does A get in all (nearest 100)?
1. Rs. 45100
2. Rs. 48100
3. Rs. 57000
4. Rs. 62000
Answer Option: 3
A gets x/2 + x/16 + x/128 = (64+ 8 + 1)x/128
= 73x/128 is approximately 57,000.
Hence Option (3).
DIRECTIONS for questions 6 to 8: Refer to the following data and answer the questions that follow.
Nizam tailor,the famous tailor of Naharpur, after facing losses for 3 consecutive years diversified his business.He started taking orders for school uniform. Initially he got orders from two schools:Moti Ram School and Lala Hardyal School.
Nizam hired two cutters,five tailors and two helpers for this. Cutters cut the cloth, tailors stitch the clothes and helpers do the finishing work like buttoning etc .All employees work for ten hours a day. Timings for making uniform for both schools are different. The details are given below:
  Cutting time Stitching time Finishing time (buttoning etc)
MOTI RAM SCHOOL 20 Min 1Hour 15 Min
LALA HARDYAL SCHOOL 30 Min 1 Hour 30Min
6. One day Nizam decided to make 25 uniforms of Moti Ram School only, the man hours that will remain idle is(approx)
1. 30
2. 50
3. 55
4. 26
Answer Option: 2
Time taken to make 25 uniforms of MOTI RAM SCHOOL Time taken by cutter=25 x 20 = 500 min. Time taken by tailors=25 x 60 = 1500 min. Time taken by helper=25 x 15 = 375 min. So man hours that are idle=(1200 – 500)+(3000 – 1500) + (1200 – 375) = 700+1500+825 = 3025= 3025/60 = 50 hours (approx)
7. If he hires two more cutters and one more tailor, then what is the maximum number of Moti Ram School uniform can be made by him in a day?
1. 120
2. 70
3. 80
4. 60
Answer Option: 4
For MOTI RAM SCHOOL
Total cutters now = 2+2=4
Number of uniforms cut by cutters = 2400/20 = 120
Total tailors now =5+1=6
Max Number of uniforms stitched by tailors = 3600/60 = 60
Max Number of uniforms finished by helpers = 1200/15 = 80
Hence limiting factor is uniforms stitched by tailors. So 60 is the answer.
8. If he hires two more cutters and two more tailors, then what is the maximum number of Lala Hardyal school uniform can be made by him in a day?
1. 120
2. 40
3. 80
4. 60
Answer Option: 2
For LALA HARDYAL SCHOOL
Total cutters now = 2+2 = 4
Number of uniforms cut by cutters = 2400/30 = 80
Total tailors now = 5+2 = 7
Max Number of uniforms stitched by tailors=4200/60 = 70
Max Number of uniforms finished by helpers = 1200/30 = 40
Hence limiting factor is for helpers.So 40 is the answer.
DIRECTIONS for questions 9 to 10: Refer to the data below and answer the questions that follow.
The "Product Tree" of a consumer good X is as follows:
1. Unit of X is made up of two units of A, three units of B and two units of C. 2. Each unit of A is made up of raw material M of 3 units. 3. Each unit of B is made up of 1 unit of P and Q each. One unit of P is made of 2 units of M. One unit of Q is made up of 2 units of A. 4. One unit of C is made up of three units of D, each of which consumes two units of M.
9. If market demand for product X is 500 units and each of the unit of M costs Rs. 2.5, what is the total raw material cost?
1. Rs. 58000
2. Rs. 62500
3. Rs. 52500
4. Rs. 35000
Answer Option: 3

i) amount of M required for A = 2 x 3 = 6
ii) amount of M required for B
= [2 x 1 + 3 x 2 x 1] x 3 = 24
iii) amount of M required for C = 2 x 3 x 2 = 12
∴ One unit of X consumes 42 units of M.
∴ Total raw material cost = 42 x 500 x 2.5
= Rs. 52,500.
Hence, [3].
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10. The raw material cost constitutes of 70% of the total variable cost and the manufacturer incurs a fixed cost of Rs. 1 lac for producing X. If he decides to sell X at Rs. 250 per unit, how many units of X should he produce to break even his costs?
1. 1000
2. 1500
3. 2000
4. 2500
Answer Option: 1

From answer to previous question, raw material cost
per unit of X = 52500 /500
= Rs. 105
Total variable cost = 105/0.7
= Rs. 150
Selling price = Rs. 250.
To break even, he should cover the total variable cost as well as the fixed cost.
Let n be the number of units produced at break-even.
Then 250n = 1,00,000 + 150n
∴ n = 1,00,000 /100 = 1000.
Hence,option 1.
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