Quantitative Aptitude Practice Question and Answer
8Q: एक रेलगाड़ी A की लम्बाई अपनी विपरित दिशा में आती हुई रेलगाड़ी B की लम्बाई से 50 प्रतिशत अधिक है। रेलगाड़ी A की चाल 40 किमी प्रति घंटा तथा B की चाल 32 किमी प्रति घंटा है तथा दोनों रेलगाड़ियों ने एक दूसरे को पार करने में 15 सैंकण्ड लिये तो रेलगाड़ी B की लम्बाई ज्ञात कीजिए। 2952 05d15fa5c5fdb2c68495d9df9
5d15fa5c5fdb2c68495d9df9- 1150 मीटरfalse
- 2120 मीटरfalse
- 3130 मीटरfalse
- 4180 मीटरtrue
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Answer : 4. "180 मीटर"
Q: A merchant buys two articles for Rs.600. He sells one of them at a profit of 22% and the other at a loss of 8% and makes no profit or loss in the end. What is the selling price of the article that he sold at a loss ? 2944 05b5cc71be4d2b4197774f4fe
5b5cc71be4d2b4197774f4fe- 1Rs. 404.80true
- 2Rs. 536.80false
- 3Rs.440false
- 4Rs. 160false
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Answer : 1. "Rs. 404.80"
Explanation :
Answer: A) Rs. 404.80 Explanation: Let C1 be the cost price of the first article and C2 be the cost price of the second article.Let the first article be sold at a profit of 22%, while the second one be sold at a loss of 8%. We know, C1 + C2 = 600.The first article was sold at a profit of 22%. Therefore, the selling price of the first article = C1 + (22/100)C1 = 1.22C1The second article was sold at a loss of 8%. Therefore, the selling price of the second article = C2 - (8/100)C2 = 0.92C2. The total selling price of the first and second article = 1.22C1 + 0.92C2. As the merchant did not make any profit or loss in the entire transaction, his combined selling price of article 1 and 2 is the same as the cost price of article 1 and 2. Therefore, 1.22C1 + 0.92C2 = C1+C2 = 600As C1 + C2 = 600, C2 = 600 - C1. Substituting this in 1.22C1 + 0.92C2 = 600, we get1.22C1 + 0.92(600 - C1) = 600or 1.22C1 - 0.92C1 = 600 - 0.92*600or 0.3C1 = 0.08*600 = 48or C1 = 48/(0.3) = 160.If C1 = 160, then C2 = 600 - 160 = 440.The item that is sold at loss is article 2. The selling price of article 2 = 0.92*C2 = 0.92*440 = 404.80.
Q: The value of a machine depreciates at the rate of 10% every year. It was purchased 3 years ago. If its present value is Rs. 8,748, its purchase price was ? 2941 05b5cc75ce4d2b4197774fc2f
5b5cc75ce4d2b4197774fc2f- 1Rs.14,000false
- 2Rs.12,000true
- 3Rs. 12,800false
- 4Rs. 15,000false
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Answer : 2. "Rs.12,000"
Explanation :
Answer: B) Rs.12,000 Explanation: We have P = 8748, R = -10 and n = 3Therefore, the purchase price of the machine= P1+R100n= 87481-101003= 8748 x (100/90) x (100/90) x (100/90)= 12,000. Therefore, the purchase price of the machine was Rs. 12,000.
Q: In a regular week, there are 5 working days and for each day, the working hours are 8. A man gets Rs. 2.40 per hour for regular work and Rs. 3.20 per hours for overtime. If he earns Rs. 432 in 4 weeks, then how many hours does he work for ? 2930 05b5cc74fe4d2b4197774fb35
5b5cc74fe4d2b4197774fb35- 1145false
- 2165false
- 3175true
- 4135false
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Answer : 3. "175"
Explanation :
Answer: C) 175 Explanation: Given that, working days = 5 working hours = 8 A man get rupess per hour is Rs.2.40 So in one day the man get total rupees is 2.40 x 8 = 19.2 So in 5 days week the man get total rupees is 19.2 x 5 = 96 So in 4 week the man get total rupees is 96 x 4 = 384 So the man worked for = 160hours in 4 weeks But given that the man earned Rs.432 Hence remaning money is (432-384 = 48)which is earn by doing overtime work Overtime hours = 48/3.20 = 15 So total worked hours is = 15 + 160 = 175.
Q: If 6 engines consume 24 metric tonnes of coal, when each is working 8 hours day, how much coal will be required for 9 engines, each running 13hours a day, it being given that 2 engines of former type consume as much as 3 engines of latter type ? 2925 05b5cc723e4d2b4197774f672
5b5cc723e4d2b4197774f672- 145 metric tonnestrue
- 247 metric tonnesfalse
- 355 metric tonnesfalse
- 434 metric tonnesfalse
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Answer : 1. "45 metric tonnes"
Explanation :
Answer: A) 45 metric tonnes Explanation: 2 engines of former type for one hour consumes (2 x 24)/(6 x8) = 1 metric ton i.e. 3 engines of latter type consumes 1 ton for one hour Hence 9 engines consumes 3 tons for one hour For 15 hours it is 15 x 3 = 45 metric tonnes.
Q: A does half as much work as Band C does half as much work as A and B together. If C alone can finish the work in 40 days, then together ,all will finish the work in ? 2923 05b5cc728e4d2b4197774f714
5b5cc728e4d2b4197774f714- 117 + 4/7 daysfalse
- 213 + 1/3 daystrue
- 315 + 3/2 daysfalse
- 416 daysfalse
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Answer : 2. "13 + 1/3 days"
Explanation :
Answer: B) 13 + 1/3 days Explanation: C alone can finish the work in 40 days.As given C does half as much work as A and B together => (A + B) can do it in 20 days(A + B)s 1 days wok = 1/20.A's 1 days work : B's 1 days Work = 1/2 : 1 = 1:2(given)A's 1 day’s work = (1/20) x (1/3) = (1/60) [Divide 1/20 in the raio 1:2]B's 1 days work = (1/20) x (2/3) = 1/30(A+B+C)'s 1 day's work = (1/60) + (1/30) + (1/40) = 9/120 = 3/40 All the three together will finish it in 40/3 = 13 and 1/3 days.
Q: Divide Rs.118000 among three persons A, B and C such that the ratio of the shares of A and B is 3:4 and that of B:C is 5:6 ? 2915 05b5cc6ebe4d2b4197774eeaa
5b5cc6ebe4d2b4197774eeaa- 1A = 48000 B = 40000 C = 30000false
- 2A = 30000 B = 40000 C = 48000true
- 3A = 30000 B = 48000 C = 40000false
- 4A = 40000 B = 48000 C = 30000false
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Answer : 2. "A = 30000 B = 40000 C = 48000"
Explanation :
Answer: B) A = 30000 B = 40000 C = 48000 Explanation: Compound ratio of A:B:CA:B = 3:4B:C = 5:6----------A:B:C = 15:20:24 We can divide Rs.118000 in this ratio. Share of A = 15/59 x 118000 = 30000 Share of B = 20/59 x 118000 = 40000 Share of C = 24/59 x 118000 = 48000
Q: If (89)2 is added to the square of a number, the answer so obtained is 16202. What is the (1/26) of that number? 2913 05b5cc6bde4d2b4197774d947
5b5cc6bde4d2b4197774d947- 15.65false
- 22.7false
- 33.5true
- 46.66false
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Answer : 3. "3.5"
Explanation :
Answer: C) 3.5 Explanation: Let the number is = x (89)2 + x2 = 16202 x2 = 8281 x = 91 => (1/26) of 91 = 3.5