UPSC Insta–DART (Daily Aptitude and Reasoning Test) 21 Aug 2024
Kartavya Desk Staff
Considering the alarming importance of CSAT in UPSC CSE Prelims exam and with enormous requests we received recently, InsightsIAS has started Daily CSAT Test to ensure students practice CSAT Questions on a daily basis. Regular Practice would help one overcome the fear of CSAT too.We are naming this initiative as Insta– DART – Daily Aptitude and Reasoning Test. We hope you will be able to use DART to hit bull’s eye in CSAT paper and comfortably score 100+ even in the most difficult question paper that UPSC can give you in CSP-2021. Your peace of mind after every step of this exam is very important for us.
Looking forward to your enthusiastic participation (both in sending us questions and solving them on daily basis on this portal).
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• Question 1 of 5 1. Question A tube contains a mixture of two liquid P and Q in the ratio of 4:1. When 20 liters of the mixture is taken out and 20 liters of liquid Q is poured in the jar, then the ratio becomes 2:3. How many liters of liquid P was contained in the jar initially? a) 32 b) 16 c) 8 d) 12 Correct Answer: (a) 32 Explanation: Let initially the amount of liquid P in the tube=4x Initially the amount of liquid Q in the tube =x (4x-4/5 20 )/ x-1/5 20 +20 =2/3 (4x-16)/x+16=2/3 12x-48=2x+32 10x=80 X=8liters Quantity of liquid P=48=32liters. Incorrect Answer: (a) 32 Explanation: Let initially the amount of liquid P in the tube=4x Initially the amount of liquid Q in the tube =x (4x-4/5 20 )/ x-1/5 20 +20 =2/3 (4x-16)/x+16=2/3 12x-48=2x+32 10x=80 X=8liters Quantity of liquid P=48=32liters.
#### 1. Question
A tube contains a mixture of two liquid P and Q in the ratio of 4:1. When 20 liters of the mixture is taken out and 20 liters of liquid Q is poured in the jar, then the ratio becomes 2:3. How many liters of liquid P was contained in the jar initially?
Answer: (a) 32
Explanation:
Let initially the amount of liquid P in the tube=4x
Initially the amount of liquid Q in the tube =x
(4x-4/5 20 )/ x-1/5 20 +20 =2/3
(4x-16)/x+16=2/3
12x-48=2x+32
Quantity of liquid P=4*8=32liters.
Answer: (a) 32
Explanation:
Let initially the amount of liquid P in the tube=4x
Initially the amount of liquid Q in the tube =x
(4x-4/5 20 )/ x-1/5 20 +20 =2/3
(4x-16)/x+16=2/3
12x-48=2x+32
Quantity of liquid P=4*8=32liters.
• Question 2 of 5 2. Question In an alloy A, copper and zinc present in the ratio 4:3 respectively. And in alloy B, the same elements are in the ratio 3:5 respectively. If these two alloys be mixed to form a new alloy in which same elements are un the ratio 1:1 respectively, then find the ratio of alloy A and alloy B in the new alloy? a) 7:4 b) 7:3 c) 4:5 d) 2:7 Correct Answer: (a) 7:4 Explanation: 1/8:1/14 Required ratio=7:4 Incorrect Answer: (a) 7:4 Explanation: 1/8:1/14 Required ratio=7:4
#### 2. Question
In an alloy A, copper and zinc present in the ratio 4:3 respectively. And in alloy B, the same elements are in the ratio 3:5 respectively. If these two alloys be mixed to form a new alloy in which same elements are un the ratio 1:1 respectively, then find the ratio of alloy A and alloy B in the new alloy?
Answer: (a) 7:4
Explanation:
Required ratio=7:4
Answer: (a) 7:4
Explanation:
Required ratio=7:4
• Question 3 of 5 3. Question A vessel contains 120 liters of petrol. 10 liters of petrol is taken out from it and completely replaced by kerosene and then again from the mixture 10 liters of mixture taken out and completely replaced by kerosene. Find the remaining quantity of petrol in the final mixture. a) 200.83 b) 100.83 c) 50.12 d) 250.3 Correct Answer: (b) 100.83 Explanation: Fraction of petrol taken out from the initial quantity =10/120=1/12 So, remaining quantity of petrol in the final mixture=12011/1211/12=1210/12=100.833 liters Incorrect Answer: (b) 100.83 Explanation: Fraction of petrol taken out from the initial quantity =10/120=1/12 So, remaining quantity of petrol in the final mixture=12011/1211/12=1210/12=100.833 liters
#### 3. Question
A vessel contains 120 liters of petrol. 10 liters of petrol is taken out from it and completely replaced by kerosene and then again from the mixture 10 liters of mixture taken out and completely replaced by kerosene. Find the remaining quantity of petrol in the final mixture.
Answer: (b) 100.83
Explanation:
Fraction of petrol taken out from the initial quantity =10/120=1/12
So, remaining quantity of petrol in the final mixture=12011/1211/12=1210/12=100.833 liters
Answer: (b) 100.83
Explanation:
Fraction of petrol taken out from the initial quantity =10/120=1/12
So, remaining quantity of petrol in the final mixture=12011/1211/12=1210/12=100.833 liters
• Question 4 of 5 4. Question Mixture of milk and water has 14 liters of water. When 4 liters of milk and 22 liters of water are added to the mixture , then concentration of milk in mixture becomes 80%. Find the total quantity of initial mixture. a) 70 b) 100 c) 140 d) 50 Correct Answer: (c) 140 Explanation: Mixture=milk+ water Milk=x Water=14 Now adding Milk=x+4 Water=14+22=36 ATQ, X+4 /36 :4/1 On solving we get x=140 liters. Incorrect Answer: (c) 140 Explanation: Mixture=milk+ water Milk=x Water=14 Now adding Milk=x+4 Water=14+22=36 ATQ, X+4 /36 :4/1 On solving we get x=140 liters.
#### 4. Question
Mixture of milk and water has 14 liters of water. When 4 liters of milk and 22 liters of water are added to the mixture , then concentration of milk in mixture becomes 80%. Find the total quantity of initial mixture.
Answer: (c) 140
Explanation:
Mixture=milk+ water
Now adding
Water=14+22=36
X+4 /36 :4/1
On solving we get x=140 liters.
Answer: (c) 140
Explanation:
Mixture=milk+ water
Now adding
Water=14+22=36
X+4 /36 :4/1
On solving we get x=140 liters.
• Question 5 of 5 5. Question Rice of two category that is Rs. 300 per kg and Rs. 400 per kg are mixed with third category in the ratio 1:2:2. If mixture of these was sold at profit on 20% for Rs. 504, find cost price of third category. a) 100 b) 200 c) 300 d) 500 Correct Answer: (d) 500 Explanation: Let cost of third category rice is x Rs/kg ATQ, ( 3001 +4002 +x2 ) /2+2+1 =504/120 120 On solving we get x=500 Rs/kg. Incorrect Answer: (d) 500 Explanation: Let cost of third category rice is x Rs/kg ATQ, ( 3001 +4002 +x2 ) /2+2+1 =504/120 120 On solving we get x=500 Rs/kg.
#### 5. Question
Rice of two category that is Rs. 300 per kg and Rs. 400 per kg are mixed with third category in the ratio 1:2:2. If mixture of these was sold at profit on 20% for Rs. 504, find cost price of third category.
Answer: (d) 500
Explanation:
Let cost of third category rice is x Rs/kg
( 3001 +4002 +x2 ) /2+2+1 =504/120 120
On solving we get x=500 Rs/kg.
Answer: (d) 500
Explanation:
Let cost of third category rice is x Rs/kg
( 3001 +4002 +x2 ) /2+2+1 =504/120 120
On solving we get x=500 Rs/kg.
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