UPSC Insta–DART (Daily Aptitude and Reasoning Test) 17 Feb 2025
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 Ram and Sunita have some money. Product of the amounts of their money is Rs. 308700, and it’s known that Sunita has less money than Ram. If HCF of their money is Rs. 105, then how much money does Ram has? a) Rs. 420 b) Rs. 735 c) Rs. 815 d) Rs. 675 Correct Answer: B Explanation Let Ram and Sunita have Rs. 105x and Rs. 105y amounts respectively. ∴ 105x × 105y = 308700 Or 105x × 105y = (105 × 7) × (105 × 4) We know that, Sunita has less money than Ram. So, x > y. ∴ x = 7 and y = 4 Hence, Ram has 105x = 105 × 7 = Rs. 735. Incorrect Answer: B Explanation Let Ram and Sunita have Rs. 105x and Rs. 105y amounts respectively. ∴ 105x × 105y = 308700 Or 105x × 105y = (105 × 7) × (105 × 4) We know that, Sunita has less money than Ram. So, x > y. ∴ x = 7 and y = 4 Hence, Ram has 105x = 105 × 7 = Rs. 735.
#### 1. Question
Ram and Sunita have some money. Product of the amounts of their money is Rs. 308700, and it’s known that Sunita has less money than Ram. If HCF of their money is Rs. 105, then how much money does Ram has?
• a) Rs. 420
• b) Rs. 735
• c) Rs. 815
• d) Rs. 675
Explanation
Let Ram and Sunita have Rs. 105x and Rs. 105y amounts respectively.
∴ 105x × 105y = 308700
Or 105x × 105y = (105 × 7) × (105 × 4)
We know that, Sunita has less money than Ram. So, x > y.
∴ x = 7 and y = 4
Hence, Ram has 105x = 105 × 7 = Rs. 735.
Explanation
Let Ram and Sunita have Rs. 105x and Rs. 105y amounts respectively.
∴ 105x × 105y = 308700
Or 105x × 105y = (105 × 7) × (105 × 4)
We know that, Sunita has less money than Ram. So, x > y.
∴ x = 7 and y = 4
Hence, Ram has 105x = 105 × 7 = Rs. 735.
• Question 2 of 5 2. Question Which of the following is the least possible fraction that will get divided by each one of 5/18, 8/9, and 10/27? a) 25/9 b) 40/9 c) 25/3 d) 81/4 Correct Answer: B Explanation We basically have to find the L.C.M of the given fractions: 5/18, 8/9, 10/27 L.C.M of fractions = L.C.M. of numerators/H.C.F. of denominators = L.C.M of (5, 8, 10) / H.C.F of (18, 9, 27) = 40/9 Incorrect Answer: B Explanation We basically have to find the L.C.M of the given fractions: 5/18, 8/9, 10/27 L.C.M of fractions = L.C.M. of numerators/H.C.F. of denominators = L.C.M of (5, 8, 10) / H.C.F of (18, 9, 27) = 40/9
#### 2. Question
Which of the following is the least possible fraction that will get divided by each one of 5/18, 8/9, and 10/27?
Explanation
We basically have to find the L.C.M of the given fractions: 5/18, 8/9, 10/27
L.C.M of fractions = L.C.M. of numerators/H.C.F. of denominators
= L.C.M of (5, 8, 10) / H.C.F of (18, 9, 27) = 40/9
Explanation
We basically have to find the L.C.M of the given fractions: 5/18, 8/9, 10/27
L.C.M of fractions = L.C.M. of numerators/H.C.F. of denominators
= L.C.M of (5, 8, 10) / H.C.F of (18, 9, 27) = 40/9
• Question 3 of 5 3. Question Find the HCF of 0.24, 0.06 and 0.018. a) 0.6 b) 0.0006 c) 0.006 d) 0.06 Correct Answer: C Explanation Expressing each of the numbers without the decimal, we get: 0.240 × 1000 = 240 = 3×2×2×2×2×5 0.060 × 1000 = 60 = 2×2×3×5 0.018 × 1000 = 18 = 2×3×3 So, HCF of 18, 60 and 240 is 6. Hence, HCF of 0.018, 0.060 and 0.240 = 6/1000 = 0.006 Incorrect Answer: C Explanation Expressing each of the numbers without the decimal, we get: 0.240 × 1000 = 240 = 3×2×2×2×2×5 0.060 × 1000 = 60 = 2×2×3×5 0.018 × 1000 = 18 = 2×3×3 So, HCF of 18, 60 and 240 is 6. Hence, HCF of 0.018, 0.060 and 0.240 = 6/1000 = 0.006
#### 3. Question
Find the HCF of 0.24, 0.06 and 0.018.
Explanation
Expressing each of the numbers without the decimal, we get:
0.240 × 1000 = 240 = 3×2×2×2×2×5
0.060 × 1000 = 60 = 2×2×3×5
0.018 × 1000 = 18 = 2×3×3
So, HCF of 18, 60 and 240 is 6.
Hence, HCF of 0.018, 0.060 and 0.240 = 6/1000 = 0.006
Explanation
Expressing each of the numbers without the decimal, we get:
0.240 × 1000 = 240 = 3×2×2×2×2×5
0.060 × 1000 = 60 = 2×2×3×5
0.018 × 1000 = 18 = 2×3×3
So, HCF of 18, 60 and 240 is 6.
Hence, HCF of 0.018, 0.060 and 0.240 = 6/1000 = 0.006
• Question 4 of 5 4. Question Ananya has some colour pencils in two boxes. If sum of the pencils in the two boxes is 112 and their HCF is 16, then what must be the difference between the number of pencils in the two boxes? a) 42 b) 38 c) 24 d) 16 Correct Answer: D Explanation Let the number of pencils in the two boxes be 16x and 16y. Then according to the question, 16x + 16y = 112 Or 16 (x + y) = 112 Or x + y = 7 ——— (1) (x, y) = (1, 6), (2, 5) or (3, 4) If x = 1, y = 6 Then, difference between the number of pencils in the two boxes = |16 (x – y)| = |16 (1 – 6)| = 16 × 5 = 80 If x = 2, y = 5 Then, difference between the number of pencils in the two boxes = |16 (x – y)| = |16 (2 – 5)| = 16 × 3 = 48 If x = 3 and y = 4 Then, difference between the number of pencils in the two boxes = |16 (x – y)| = |16 (3 – 4)| = 16 × 1 = 16 Incorrect Answer: D Explanation Let the number of pencils in the two boxes be 16x and 16y. Then according to the question, 16x + 16y = 112 Or 16 (x + y) = 112 Or x + y = 7 ——— (1) (x, y) = (1, 6), (2, 5) or (3, 4) If x = 1, y = 6 Then, difference between the number of pencils in the two boxes = |16 (x – y)| = |16 (1 – 6)| = 16 × 5 = 80 If x = 2, y = 5 Then, difference between the number of pencils in the two boxes = |16 (x – y)| = |16 (2 – 5)| = 16 × 3 = 48 If x = 3 and y = 4 Then, difference between the number of pencils in the two boxes = |16 (x – y)| = |16 (3 – 4)| = 16 × 1 = 16
#### 4. Question
Ananya has some colour pencils in two boxes. If sum of the pencils in the two boxes is 112 and their HCF is 16, then what must be the difference between the number of pencils in the two boxes?
Explanation
Let the number of pencils in the two boxes be 16x and 16y.
Then according to the question,
16x + 16y = 112
Or 16 (x + y) = 112
Or x + y = 7 ——— (1)
(x, y) = (1, 6), (2, 5) or (3, 4)
If x = 1, y = 6
Then, difference between the number of pencils in the two boxes = |16 (x – y)| = |16 (1 – 6)| = 16 × 5 = 80
If x = 2, y = 5
Then, difference between the number of pencils in the two boxes = |16 (x – y)| = |16 (2 – 5)| = 16 × 3 = 48
If x = 3 and y = 4
Then, difference between the number of pencils in the two boxes = |16 (x – y)| = |16 (3 – 4)| = 16 × 1 = 16
Explanation
Let the number of pencils in the two boxes be 16x and 16y.
Then according to the question,
16x + 16y = 112
Or 16 (x + y) = 112
Or x + y = 7 ——— (1)
(x, y) = (1, 6), (2, 5) or (3, 4)
If x = 1, y = 6
Then, difference between the number of pencils in the two boxes = |16 (x – y)| = |16 (1 – 6)| = 16 × 5 = 80
If x = 2, y = 5
Then, difference between the number of pencils in the two boxes = |16 (x – y)| = |16 (2 – 5)| = 16 × 3 = 48
If x = 3 and y = 4
Then, difference between the number of pencils in the two boxes = |16 (x – y)| = |16 (3 – 4)| = 16 × 1 = 16
• Question 5 of 5 5. Question Three numbers are in ratio of 41:47:53. If their HCF is 4, then these numbers must be: a) 172, 198, 224 b) 160, 180, 200 c) 152, 178, 204 d) 164, 188, 212 Correct Answer: D Explanation Numbers are in the ratio of 41:47:53. Let these three numbers be 41x, 47x, 53x. Given, HCF = 4 So, x = 4 So, required numbers are: 1st number = 41x = 41 × 4 = 164 2nd number = 47x = 47 × 4 = 188 3rd number = 53x = 53 × 4 = 212 Incorrect Answer: D Explanation Numbers are in the ratio of 41:47:53. Let these three numbers be 41x, 47x, 53x. Given, HCF = 4 So, x = 4 So, required numbers are: 1st number = 41x = 41 × 4 = 164 2nd number = 47x = 47 × 4 = 188 3rd number = 53x = 53 × 4 = 212
#### 5. Question
Three numbers are in ratio of 41:47:53. If their HCF is 4, then these numbers must be:
• a) 172, 198, 224
• b) 160, 180, 200
• c) 152, 178, 204
• d) 164, 188, 212
Explanation
Numbers are in the ratio of 41:47:53.
Let these three numbers be 41x, 47x, 53x.
Given, HCF = 4
So, required numbers are:
1st number = 41x = 41 × 4 = 164
2nd number = 47x = 47 × 4 = 188
3rd number = 53x = 53 × 4 = 212
Explanation
Numbers are in the ratio of 41:47:53.
Let these three numbers be 41x, 47x, 53x.
Given, HCF = 4
So, required numbers are:
1st number = 41x = 41 × 4 = 164
2nd number = 47x = 47 × 4 = 188
3rd number = 53x = 53 × 4 = 212
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