Free aptitude test

A paddy processing unit purchases 10,000 kg of paddy. After subjecting it to milling operation, the husk and rice are separated. The husk forms 5% of the total paddy weight. The rice so extracted from the paddy contains 6% of brokens or defective material in it. Cost of purchase of paddy is Rs.3.50/kg and the selling price of rice is Rs.8.00/kg. The broken material is sold at Rs.2.00/kg. The husk is sold atRs.0.50/kg.

Questions

1. What is the amount of paddy to be processed, such that the sale of the rice extracted from it excluding brokens and husk fetches an amount of Rs.14, 288?
(1) 2000 kg (2) 1786kg (3) 2500 kg (4) 1886 kg

2. Assuming that all the rice produced by processing 10,000 kg of paddy is sold, what would be the decrease in profits if the brokens in the operations increased to 8%?
(1) Rs. 1,000 (2) Rs. 1,100 (3) Rs. 1, 140 (4) Rs. 1,200

3. Due to the evaporation of moisture, there is a 5% loss of weight before the processing of the. Paddy began. As a result the decrease in the weight of rice obtained (in kg) is
(1) 450 kg (2) 500 kg (3) 447 kg (4) 475 kg

4. What is the amount realised on the sale of by products i.e., broken and husk on processing 5,000 kg of paddy
(l) Rs.200 (2) Rs.180 (3) Rs.695 (4) Rs.190

Answers with Explanations

1. (1) Let the amount of paddy to be processed = q kg. the amount of rice it yields = q x 0.95 = 0.95 q
out of this if you exclude 6% broken
0.95q (1 - 6 / 100) = 0.893q

∴So the rice you get = 0.893q ∴0.893q x 8 = 14.288
0.893q = 1786 q = 2000 kg.

2. (3) The rice produced by processing 10.000 kg of paddy = 9,500 kg.
if the brokens increase by 2% of 9,500 ⇒ 190 kg
of additional brokens 190 kg of fine nee less ⇒190 x 8 = 1520 but this is offset by 190 x 2 = 380
profit over sale of additional brokens 1, 520 – 380 = Rs.1, 140

3. (4) 5% loss of weight would imply = 0.05 x 10,000 = 500kg.
the rice extracted will be = 9,500 x 95 / 100 = 9025
the decrease will be = 9, 500 - 9025 = 475 kg.

4. (3) Proceedings from sale of husk =
5000 x 0.05 x 1/2 = Rs.125
proceedings from the sale of brokens =
5000 x 0.95 x 6/100 x 2 = 570
hence amount realised = Rs.570 + Rs 125 = Rs.695

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Quantitative aptitude questions

Alord received a large order for stitching school uniforms from Mayflower school and Little Flower school. He has two cutters who will cut the fabric, five tailors who will do the stitching, and two assistants to stitch the buttons and button holes. Each of these nine persons will work for exactly 10 hours a day. Each of the May-flower uniforms requires 20 min. for cutting the fabric, one hour for stitching, and 15 min. for stitching buttons and button holes, whereas the Little Flower uniform requires 30 min., 1 hour, and 30 min. respectively for these activities.

                   Questions from this passage:     

1. What is the maximum number of little Flower uniforms that A lord can complete in a day?
(1) 50 (2) 20 (3) 40 (4) 30

2. On particular day, Alord decided to complete 20 Little Flower uniforms. How many Mayflower uniforms can he complete on that day?
(1) 30 (2) 40 (3) 20 (4) 0

3. If Alord decides to complete 30 Little Flower uniforms only and no other on a particular day, how
many total man-hours will be idle?
(1) 20 (2) 30 (3) 5 (4) 25

4. If he hires one more assistant, what is the maximum number of Mayflower uniforms that he can
complete in a day?
(1) 40 (2) 50 (3) 60 (4) 30

5. A lord has the option to hire one more employee of any category. Which category should he hire to get maximum increase in production capacity, assuming that he needs to stitch only Mayflower uniforms on that day?
(1) Tailor (2) Cutter (3) Assistant (4) Cannot be determined

                Answers with Explanations    

1. (3) No. of working hours for each person = 10 hour
∴Total time available for cutters in a day = 20 hours = 1200 mins.
Max. No. of uniform that 2 cutters can cut for Little Flower in a day = 1200/30 = 40
Time required for stitching 40 uniform = 40 x 60 = 2400 min and Alord has 5 tailors for stitching
that means the Total available time is (5 X 10 x 60) = 3000 mins.
Hence maximum number of Little Flower uniforms can be completed in a day = 40.

2. (1) To complete 20 Little Flower uniforms,
Time consumed by Cutters = 20 x 30 = 600 mins
Time consumed by Tailors = 20 x 60 = 1200 mins
Time consumed by Assistants = 20 x 30 = 600 mins
Time available for Cutters = (1200 – 600) = 600 mins
Time available for Tailors = (3000 – 1200) = 1800 mins
Time available for Assistant = (1200 – 600) = 600 mins
Max. No. of uniform that 2 cutters can cut for May Flowers uniform in a day = 600/20 = 30
Time required to stitch 30 uniforms = (30 x 60) = 1800 min and is exactly equal to the time
available for tailors. Hence No. of Mayflower uniform that can be completed on that day is 30.

3. (2) (1) To complete 30 Little Flower uniforms,
Time consumed by Cutters = (30 x 30) = 900 mins
Time consumed by tailors = (30 x 60) = 1800 mins
Time consumed by assistants = (30 x 30) = 900 mins
Time available for cutters = (1200 – 900) = 300 mins
Time available for tailors = (3000 – 1800) = 1200 mins
Time available for Assistant = (1200 – 900) = 300 mins
As they are not working in this available time it is the idle time.
Hence total man hours that are idle = (300 + 1200 + 300) = 1800 mins = 30 hours.

4. (2) Hiring one more assistance increases the time available for stitching buttons and button holes.
Max. No. of uniform those cutter can cut for Mayflower in a day = 1200/20 = 60
Time required by Tailors to stitch 60 uniforms = (60 x 60) = 3600 mins, whereas time available is
3000 mins only.
Hence maximum No. of uniform that can be stitched in a day = 3000/60 = 50
Time consumed by assistant to stitch buttons to 50 uniform = (50 x 50) = 750 mins
Time available for assistant = (3 x 10 x 60) = 1800 mins.
Hence 50 uniforms can be completed in a day.

5. (1) Max. number of uniforms that can be cut by those 2 cutters in a day for MayFlower is 60. And the
buttons and buttonholes can be easily stitched by the 2 assistant for these 60 uniforms in a day in
the time available for them. While 5 tailors can stitch only 50 uniforms in the available time.
Hence, A lord will hire 1 tailor to get maximum increase in production capacity in a day.

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