👆Ans. B | Explanation:
Since each stick must be cut into parts of equal length and each part must be as long as possible. So, we need to find out the HCF.
HCF of 85, 119, 136 and 204 = 17
Number of parts from 85 cm long stick = 85/17 = 5
Number of parts from 119 cm long stick = 119/17 = 7
Number of parts from 136 cm long stick = 136 /17 = 8
Number of parts from 204 cm long stick = 204 /17 = 12
∴ Maximum number of pieces = 5 + 7 + 8 + 12 = 32
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Since each stick must be cut into parts of equal length and each part must be as long as possible. So, we need to find out the HCF.
HCF of 85, 119, 136 and 204 = 17
Number of parts from 85 cm long stick = 85/17 = 5
Number of parts from 119 cm long stick = 119/17 = 7
Number of parts from 136 cm long stick = 136 /17 = 8
Number of parts from 204 cm long stick = 204 /17 = 12
∴ Maximum number of pieces = 5 + 7 + 8 + 12 = 32
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👍7
Dr. Bunti made a medicine by mixing two herbs X and Y. In 50 g of that medicine, the percentage of herb X is 40%. How much of herb X should Dr. Bunti add into it further so as to increase its percentage to 84%?
Anonymous Quiz
12%
(a) 142 g
23%
(b) 132.8 g
59%
(c) 137.5 g
7%
(d) 141.7 g
👍3
👆Ans. C | Explanation:
According to the question,
Quantity of herb X in 50 g of medicine = (40/100) × 50 = 20 g
Let x g of herb X is further mixed in the medicine.
Then, [(20 + x)/(50 + x)] × 100 = 84
Or 2000 + 100x = 4200 + 84x
Or 16x = 2200
Or x = 2200/16
Or x = 137.5 g
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According to the question,
Quantity of herb X in 50 g of medicine = (40/100) × 50 = 20 g
Let x g of herb X is further mixed in the medicine.
Then, [(20 + x)/(50 + x)] × 100 = 84
Or 2000 + 100x = 4200 + 84x
Or 16x = 2200
Or x = 2200/16
Or x = 137.5 g
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👍3❤1
A faulty clock was set correct at 6:00 a.m. on Thursday. This clock gains 5 seconds in every 5 minutes. What must be the approximate time displayed by this clock at 5:30 p.m. on the same day?
Anonymous Quiz
53%
(a) 5:41 p.m.
20%
(b) 5:13 p.m.
19%
(c) 5:23 p.m.
7%
(d) 5:56 p.m.
👍3
👆Ans. A | Explanation:
Total time from 6 a.m. to 5:30 p.m. = 11 hours 30 minutes = 690 minutes
Clock gains 5 seconds in every 5 minutes.
∴ Gain in 1 minute = 5/5 = 1 second
∴ Gain in 60 minutes = 60 × 1 = 60 seconds
∴ Gain in 690 minutes = 690 seconds = 690/60 minutes = 69/6 minutes = 23/2 minutes = 11.5 minutes OR 11 minutes 30 seconds
Hence, the correct time = 5:30 + 0:11 = 5:41 p.m. (approximately)
(as the clock is gaining time, it means it is running faster than it should and hence display time will be more than the real time)
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Total time from 6 a.m. to 5:30 p.m. = 11 hours 30 minutes = 690 minutes
Clock gains 5 seconds in every 5 minutes.
∴ Gain in 1 minute = 5/5 = 1 second
∴ Gain in 60 minutes = 60 × 1 = 60 seconds
∴ Gain in 690 minutes = 690 seconds = 690/60 minutes = 69/6 minutes = 23/2 minutes = 11.5 minutes OR 11 minutes 30 seconds
Hence, the correct time = 5:30 + 0:11 = 5:41 p.m. (approximately)
(as the clock is gaining time, it means it is running faster than it should and hence display time will be more than the real time)
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👍5
If α and β are two real numbers such that α + β = - q/p and αβ = r/p, where 0<p<q<r (p, q, r are natural numbers), then which one of the following alternatives represents the highest value?
Anonymous Quiz
6%
(a) 1 / (α + β)
37%
(b) (1/ α) + (1/ β)
47%
(c) – 1/ (αβ)
10%
(d) αβ/(α + β)
👍2
👆Ans. C | Explanation:
It is given that α + β = - q/p and αβ = r/p, where 0<p<q<r
So, let’s assume p = 1, q = 2 and r = 3.
Putting these values in answer options, we get
(A) 1 / (α + β) = - p/q = - (1/2)
(B) (1/ α) + (1/ β) = (α + β)/ αβ = -(q/p) × (p/r) = - (q/r) = - (2/3)
(C) – 1/ (αβ) = -(p/r) = - (1/3)
(D) αβ/(α + β) = r/p × (-p/q)= - r/q = - (3/2)
We can see that, - (1/3) is the greatest among the above values.
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It is given that α + β = - q/p and αβ = r/p, where 0<p<q<r
So, let’s assume p = 1, q = 2 and r = 3.
Putting these values in answer options, we get
(A) 1 / (α + β) = - p/q = - (1/2)
(B) (1/ α) + (1/ β) = (α + β)/ αβ = -(q/p) × (p/r) = - (q/r) = - (2/3)
(C) – 1/ (αβ) = -(p/r) = - (1/3)
(D) αβ/(α + β) = r/p × (-p/q)= - r/q = - (3/2)
We can see that, - (1/3) is the greatest among the above values.
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👍2
What is the ratio of expected police-to-population ratio of West Bengal, Bihar and Andhra Pradesh taken together to the existing police-to-population ratio of Uttar Pradesh, Andhra Pradesh and Maharashtra taken together?
Anonymous Quiz
47%
(a) 2962 : 2955
22%
(b) 331 : 297
21%
(c) 93 : 97
11%
(d) 5920 : 6131
👍5
👆Ans. A | Explanation:
Expected police-to-population ratio of West Bengal = 1187
Expected police-to-population ratio of Bihar = 1133
Expected police-to-population ratio of Andhra Pradesh = 642
Existing police-to-population ratio of Uttar Pradesh = 1173
Existing police-to-population ratio of Andhra Pradesh = 953
Existing police-to-population ratio of Maharashtra = 829
Required ratio = [(1187 + 1133 + 642)/ (1173 + 953 + 829)] = 2962/2955 = 2962 : 2955
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Expected police-to-population ratio of West Bengal = 1187
Expected police-to-population ratio of Bihar = 1133
Expected police-to-population ratio of Andhra Pradesh = 642
Existing police-to-population ratio of Uttar Pradesh = 1173
Existing police-to-population ratio of Andhra Pradesh = 953
Existing police-to-population ratio of Maharashtra = 829
Required ratio = [(1187 + 1133 + 642)/ (1173 + 953 + 829)] = 2962/2955 = 2962 : 2955
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👍3❤1
एक फोन का अंकित मूल्य 16,000 रुपये है, जो फोन के लागत मूल्य से 60% अधिक है। यदि इसे अंकित मूल्य पर 12% की छूट पर बेचा जाए, तो लाभ प्रतिशत ज्ञात कीजिए।
Anonymous Quiz
6%
(a) 38.6%
21%
(b) 42.7%
36%
(c) 48.6%
36%
(d) 40.8%
👍3
👆Ans. D | Explanation:
दिया गया है कि अंकित मूल्य है = 16000 रुपये
अत:, लागत मूल्य = (16000/160) × 100 = 10000 रुपये
यदि इसे 12% की छूट पर बेचा जाता है, तो विक्रय मूल्य = 16000 × (88/100) = 14080 रुपये
लागत मूल्य पर लाभ = 14080 - 10000 = 4080 रुपये
इस प्रकार, लाभ % = (4080/10000) × 100 = 40.8%
इसलिए विकल्प (d) सही उत्तर है।
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दिया गया है कि अंकित मूल्य है = 16000 रुपये
अत:, लागत मूल्य = (16000/160) × 100 = 10000 रुपये
यदि इसे 12% की छूट पर बेचा जाता है, तो विक्रय मूल्य = 16000 × (88/100) = 14080 रुपये
लागत मूल्य पर लाभ = 14080 - 10000 = 4080 रुपये
इस प्रकार, लाभ % = (4080/10000) × 100 = 40.8%
इसलिए विकल्प (d) सही उत्तर है।
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👍1🔥1
Animesh borrowed Rs.1625 from Anurag. He repaid Rs.10 in the first month and promised to increase the repayment amount by Rs.7.5 every month. In how many months will Animesh be able to return the whole amount (including the first month)?
Anonymous Quiz
8%
(a) 10 months
23%
(b) 15 months
25%
(c) 18 months
44%
(d) 20 months
👍1
👆Ans. D | Explanation:
Payment done by Animesh in the first month = Rs.10
Payment done in the second month = Rs. 10 + 7.5 = Rs. 17.5, and so on.
It’s an arithmetic series, wherein a = 10, d = 7.5, Sn = 1625
We know that, Sn = (n/2) [2a + (n – 1) × d]
Or 1625 = (n/2) [2 × 10 + (n – 1) × 7.5]
Or 1625 = (n/2) [20 + (n – 1) × 15/2]
Or 1625 = (n/4) [40 + (n – 1) × 15]
Or 6500 = 40n + 15n2 – 15n
Or 15n2 + 25n – 6500 = 0
Or 3n2 + 5n – 1300 = 0
On solving it we get, n = 20 months
Hence, option (d) is correct.
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Payment done by Animesh in the first month = Rs.10
Payment done in the second month = Rs. 10 + 7.5 = Rs. 17.5, and so on.
It’s an arithmetic series, wherein a = 10, d = 7.5, Sn = 1625
We know that, Sn = (n/2) [2a + (n – 1) × d]
Or 1625 = (n/2) [2 × 10 + (n – 1) × 7.5]
Or 1625 = (n/2) [20 + (n – 1) × 15/2]
Or 1625 = (n/4) [40 + (n – 1) × 15]
Or 6500 = 40n + 15n2 – 15n
Or 15n2 + 25n – 6500 = 0
Or 3n2 + 5n – 1300 = 0
On solving it we get, n = 20 months
Hence, option (d) is correct.
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👍2
In a class, 40% of the boys is same as half of the girls, and there are 20 girls in total. Total number of students in the class is:
Anonymous Quiz
11%
(a) 70
62%
(b) 45
14%
(c) 35
13%
(d) 25
👍1
👆Ans. B | Explanation:
40% of the boys is same as half of the girls, and there are 20 girls.
So, 40% of boys = 20/2 = 10
Therefore, 100% of boys = (10/40%) × 100% = 25
Therefore, total number of boys = 25
Thus, total number of students = 25 + 20 = 45
Hence, option (b) is correct.
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40% of the boys is same as half of the girls, and there are 20 girls.
So, 40% of boys = 20/2 = 10
Therefore, 100% of boys = (10/40%) × 100% = 25
Therefore, total number of boys = 25
Thus, total number of students = 25 + 20 = 45
Hence, option (b) is correct.
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👍3
A person deposited Rs. 13,200 in a bank, which pays 14% simple interest. If he rather had invested in Rs. 110 stocks which pay a dividend of 15% on the face value of Rs. 100, how much would he had lost or gained?
Anonymous Quiz
14%
(a) Loses Rs. 48
60%
(b) Gains Rs. 48
17%
(c) Loses Rs. 132
9%
(d) Gains Rs. 132
👍4
👆Ans. B | Explanation:
Interest earned on Rs. 13200 at a rate of 14% = Rs. 1848
Number of shares purchased = 13200/110 = 120
Dividend earned by him on 120 shares which pays a dividend of 15% per share = 120 × [(15/100) × 100] = Rs. 1800
Therefore, net profit = 1848 – 1800 = Rs. 48
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Interest earned on Rs. 13200 at a rate of 14% = Rs. 1848
Number of shares purchased = 13200/110 = 120
Dividend earned by him on 120 shares which pays a dividend of 15% per share = 120 × [(15/100) × 100] = Rs. 1800
Therefore, net profit = 1848 – 1800 = Rs. 48
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👍3
A machine costs m rupees per day to maintain and n paise for each unit it produces. If the machine produces r units in a week, then which of the following is the total cost of operating the machine for a week?
Anonymous Quiz
12%
(a) 7m + 100nr
58%
(b) (700m + nr) /100
21%
(c) m + nr
9%
(d) 700mnr
👍2
👆Ans. B | Explanation:
Cost of 1 day to maintain the machine = Rs. m
Therefore 7 day’s cost for maintenance = Rs. 7m
Similarly, cost of 1 unit = n paise (for 1 day) = Rs. n/100
Therefore, cost of r units = Rs. rn/100
Hence, Total cost = 7m +( rn/100) = (700m + nr) /100
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Cost of 1 day to maintain the machine = Rs. m
Therefore 7 day’s cost for maintenance = Rs. 7m
Similarly, cost of 1 unit = n paise (for 1 day) = Rs. n/100
Therefore, cost of r units = Rs. rn/100
Hence, Total cost = 7m +( rn/100) = (700m + nr) /100
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👍2
Sonali leaves Delhi at 4 pm by a train and reaches Pune at 10 pm. Priyanka leaves Pune at 6 pm by a train and reaches Delhi at 11 pm on the same day. What is the time when they crossed each other?
Anonymous Quiz
40%
(a) 7:49 pm
24%
(b) 7:20 pm
22%
(c) 6:50 pm
14%
(d) 8:20 pm
👍4
👆Ans. (a) | Explanation:
Let distance between Delhi and Pune be A km.
Time taken by Sonali to reach Pune = 10 – 4 = 6 hours
Time taken by Priyanka to reach Delhi = 11 – 6 = 5 hours
Speed of train which leaves Delhi = A/6 km/hr
And Speed of train which leaves Pune = A/5 km/hr
Relative speed = A/6 + A/5 = 11A/30 km/hr
Distance covered by train which leaves Delhi in 2 hours = 2 × A/6 = A/3 km
Distance left to be covered = A – A/3 = 2A/3 km
Time taken by both the trains to cross each other = Distance left to be covered/ Relative speed = (2A/3)/(11A/30) = 20/11 hours = 1 hour 49 minutes
Required time = 6 pm + 1 hour 49 minutes = 7: 49 pm
Hence, option (a) is the correct answer.
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Let distance between Delhi and Pune be A km.
Time taken by Sonali to reach Pune = 10 – 4 = 6 hours
Time taken by Priyanka to reach Delhi = 11 – 6 = 5 hours
Speed of train which leaves Delhi = A/6 km/hr
And Speed of train which leaves Pune = A/5 km/hr
Relative speed = A/6 + A/5 = 11A/30 km/hr
Distance covered by train which leaves Delhi in 2 hours = 2 × A/6 = A/3 km
Distance left to be covered = A – A/3 = 2A/3 km
Time taken by both the trains to cross each other = Distance left to be covered/ Relative speed = (2A/3)/(11A/30) = 20/11 hours = 1 hour 49 minutes
Required time = 6 pm + 1 hour 49 minutes = 7: 49 pm
Hence, option (a) is the correct answer.
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👍5❤2