SSC CGL EXAMS 2018 | Quantitative Aptitude Practice Questions (Day-19)
Dear Aspirants, Here we have given the Important SSC Exam 2018 Practice Test Papers. Candidates those who are preparing for SSC 2018 can practice these questions to get more confidence to Crack SSC 2018 Examination.
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Question 1 of 10
1. Question
Which one of the following numbers will completely divide (623Ã—824Ã—525Ã—226Ã—127)?
Correct
Answer:Â b)
623Ã—824Ã—525Ã—226Ã—127
By product of unit digit of the above numbers
3Ã—4Ã—5Ã—6Ã—7 = 20 (last two digits)
âˆ´ The product will be divisible by 10.
Incorrect
Answer:Â b)
623Ã—824Ã—525Ã—226Ã—127
By product of unit digit of the above numbers
3Ã—4Ã—5Ã—6Ã—7 = 20 (last two digits)
âˆ´ The product will be divisible by 10.
Question 2 of 10
2. Question
Two fifth of a number is greater than one third by 11. Find the number.
Correct
Answer: b)
2x/5 = x/3 + 11
2x/3 = x + 33/3
6x = 5x + 165
x = 165
Incorrect
Answer: b)
2x/5 = x/3 + 11
2x/3 = x + 33/3
6x = 5x + 165
x = 165
Question 3 of 10
3. Question
The sum of all odd numbers between 70 and 100 which are divisible by 9 isâ€”
Correct
Answer: d)
Odd numbers between 70 to 100 which are divisible by 9 is 81 & 99
Sum = 81 + 99 = 180
Incorrect
Answer: d)
Odd numbers between 70 to 100 which are divisible by 9 is 81 & 99
Sum = 81 + 99 = 180
Question 4 of 10
4. Question
Rahul bought a gift at 12.5% discount on the marked price and he sold it Raghu at a discount of 10% on the marked price. It Raghu paid Rs.900 for the gift they what percent of profit made by Rahul?
Correct
Answer: d)
Let x be the marked price of the gift.
Raghu buys the gift for is 900, which is the price after @ 10% discount on marked price.
0.9x = 900
x = Rs. 1000
As Rahul bought the cycle at 12.5% discount, then amount paid by Rahul
= 1000 Â 87.5% = 875 *1000/10 * 100 = Rs. 875
Profit made by Ragul = 900 â€“ 875 = 25
Profit % = 25* 100/875 = 2.85%
Incorrect
Answer: d)
Let x be the marked price of the gift.
Raghu buys the gift for is 900, which is the price after @ 10% discount on marked price.
0.9x = 900
x = Rs. 1000
As Rahul bought the cycle at 12.5% discount, then amount paid by Rahul
= 1000 Â 87.5% = 875 *1000/10 * 100 = Rs. 875
Profit made by Ragul = 900 â€“ 875 = 25
Profit % = 25* 100/875 = 2.85%
Question 5 of 10
5. Question
A 235 m long train running at a speed of 216km/h completely over takes a 115 m long train running in the same direction 10 seconds. The speed of the second train is:
Correct
Answer: a)
The speed of the first train = 216 kmph = 54 * 5/ 18 = 60 mps
Let x (in m/s) is the speed of the second train.
(60 â€“ x) * 10 = 235 + 115 = 350
600 â€“ 10x = 350
10x = 250
x = 25 mps
Incorrect
Answer: a)
The speed of the first train = 216 kmph = 54 * 5/ 18 = 60 mps
Let x (in m/s) is the speed of the second train.
(60 â€“ x) * 10 = 235 + 115 = 350
600 â€“ 10x = 350
10x = 250
x = 25 mps
Question 6 of 10
6. Question
Find the minimum value of y = sin^{12}x + cos^{12}x.
Correct
Answer: b)
For the value of y = sin^{12} x + cos^{12}x (Here x = 45^{o})
Find the maximum value of xyzw, where x + y + z + w = 27.
Correct
Answer: a)
By using AM Â GM property;
[TheÂ AMâ€“GMÂ inequality, states that the arithmetic mean of a list of non-negative real numbers isÂ greater than or equal toÂ the geometric mean of the same list]
So, xyzw is Maximum for (27/4)^{4}
Incorrect
Answer: a)
By using AM Â GM property;
[TheÂ AMâ€“GMÂ inequality, states that the arithmetic mean of a list of non-negative real numbers isÂ greater than or equal toÂ the geometric mean of the same list]
So, xyzw is Maximum for (27/4)^{4}
Question 8 of 10
8. Question
The curved surface area of a cylindrical pillar is 704 cm2 and its volume is 2816 cm3. Find the ratio of the diameter of its base to its height.
Correct
Answer: b)
Let r and h (in cm) be the radius of the base and the height of the pillar then,
Radius = 8 cm
Diameter = 2r = 2 * 8 = 16 cm
Ï€rh = 22/7 * 8 * h = 704
Height = 28cm
Radius : Diameter = 16 : 14 = 8 : 7
Incorrect
Answer: b)
Let r and h (in cm) be the radius of the base and the height of the pillar then,
Radius = 8 cm
Diameter = 2r = 2 * 8 = 16 cm
Ï€rh = 22/7 * 8 * h = 704
Height = 28cm
Radius : Diameter = 16 : 14 = 8 : 7
Question 9 of 10
9. Question
P, Q, R and S complete a work/ job is 10 days, 20 days, 25 days and 50 days respectively. P & Q start the work and working for one day only, and then Q left the Job, and in place of him R & S coin for the same job but after 2 days also let the job. Remaining work will be done by R & S only on the conditions of alternate day starting from R. In how many days work will complete?
Correct
Answer: b)
A/c to Q. 1day work by P & Q = 10U+ 5U = 15U.
2 daysâ€™ work by P, R&S = 2 (10 + 4 + 2) units
Total work in 3 days = 15 units + 32 units = 47 units
Remaining work = 100 units â€“ 47 units = 5 3 units
According to the questions remaining work will be done by R & S on alternate day,
So, 1 day work for R = 4 units.
& 1 day work for S = 2 units
In 2 days R & S complete the 6 unit works.
2 days ->6 units
2 days ->Â 6 units….. And so on…
16 days total work by R & S = 6 Â 8 = 48 units
Remaining = 53 units â€“ 48 units = 5 units => 4 units by R in one day + 1 units by S in Â½ day
Total days = 1+ 2 + 16 + 1 + Â½ = 20 days
Incorrect
Answer: b)
A/c to Q. 1day work by P & Q = 10U+ 5U = 15U.
2 daysâ€™ work by P, R&S = 2 (10 + 4 + 2) units
Total work in 3 days = 15 units + 32 units = 47 units
Remaining work = 100 units â€“ 47 units = 5 3 units
According to the questions remaining work will be done by R & S on alternate day,
So, 1 day work for R = 4 units.
& 1 day work for S = 2 units
In 2 days R & S complete the 6 unit works.
2 days ->6 units
2 days ->Â 6 units….. And so on…
16 days total work by R & S = 6 Â 8 = 48 units
Remaining = 53 units â€“ 48 units = 5 units => 4 units by R in one day + 1 units by S in Â½ day
Total days = 1+ 2 + 16 + 1 + Â½ = 20 days
Question 10 of 10
10. Question
How many liters of a solution containing milk and water in the ratio 4: 5 should be added to 90 liters of 60% milk solution in order to get 50% milk solution?
Correct
Answer: c)
Let x (in liters) of 4 : 5 milk solution is added to 90 liters of 60% milk solution.
Volume of the milk before = Volume of the milk after
Incorrect
Answer: c)
Let x (in liters) of 4 : 5 milk solution is added to 90 liters of 60% milk solution.
Volume of the milk before = Volume of the milk after
1) Which one of the following numbers will completely divide (623Ã—824Ã—525Ã—226Ã—127)?
a) 7
b) 10
c) 9
d) 11
2) Two fifth of a number is greater than one third by 11. Find the number.
a) 164
b) 165
c) 200
d) None of these
3) The sum of all odd numbers between 70 and 100 which are divisible by 9 isâ€”
a) 160
b) 170
c) 252
d) None of these
4) Rahul bought a gift at 12.5% discount on the marked price and he sold it Raghu at a discount of 10% on the marked price. It Raghu paid Rs.900 for the gift they what percent of profit made by Rahul?
a) 2%
b) 20%
c) 28.5%
d) 2.85%
5) A 235 m long train running at a speed of 216km/h completely over takes a 115 m long train running in the same direction 10 seconds. The speed of the second train is:
a) 25 m/s
b) 20m/s
c) 10 m/s
d) 15 m/s
6) Find the minimum value of y = sin^{12}x + cos^{12}x.
a) 1/64
b) 1/32
c) 1/16
d) 1
7) Find the maximum value of xyzw, where x + y + z + w = 27.
a) 9^{4}
b) 4^{9}
c) (27/4)^{4}
d) 9
8) The curved surface area of a cylindrical pillar is 704 cm2 and its volume is 2816 cm3. Find the ratio of the diameter of its base to its height.
a) 7 : 8
b) 8 : 7
c) 16 : 14
d) None of these
9) P, Q, R and S complete a work/ job is 10 days, 20 days, 25 days and 50 days respectively. P & Q start the work and working for one day only, and then Q left the Job, and in place of him R & S coin for the same job but after 2 days also let the job. Remaining work will be done by R & S only on the conditions of alternate day starting from R. In how many days work will complete?
a) 20 days
b) 20 Â½ days
c) 16 days
d) 11. 83 days
10) How many liters of a solution containing milk and water in the ratio 4: 5 should be added to 90 liters of 60% milk solution in order to get 50% milk solution?
a) 160
b) 140
c) 162
d) 100
Answers:
1) Answer:Â b)
623Ã—824Ã—525Ã—226Ã—127
By product of unit digit of the above numbers
3Ã—4Ã—5Ã—6Ã—7 = 20 (last two digits)
âˆ´ The product will be divisible by 10.
2) Answer: b)
2x/5 = x/3 + 11
2x/3 = x + 33/3
6x = 5x + 165
x = 165
3) Answer: d)
Odd numbers between 70 to 100 which are divisible by 9 is 81 & 99
Sum = 81 + 99 = 180
4) Answer: d)
Let x be the marked price of the gift.
Raghu buys the gift for is 900, which is the price after @ 10% discount on marked price.
0.9x = 900
x = Rs. 1000
As Rahul bought the cycle at 12.5% discount, then amount paid by Rahul
= 1000 Â 87.5% = 875 *1000/10 * 100 = Rs. 875
Profit made by Ragul = 900 â€“ 875 = 25
Profit % = 25* 100/875 = 2.85%
5) Answer: a)
The speed of the first train = 216 kmph = 54 * 5/ 18 = 60 mps
Let x (in m/s) is the speed of the second train.
(60 â€“ x) * 10 = 235 + 115 = 350
600 â€“ 10x = 350
10x = 250
x = 25 mps
6) Answer: b)
For the value of y = sin^{12} x + cos^{12}x (Here x = 45^{o})
[TheÂ AMâ€“GMÂ inequality, states that the arithmetic mean of a list of non-negative real numbers isÂ greater than or equal toÂ the geometric mean of the same list]
So, xyzw is Maximum for (27/4)^{4}
8) Answer: b)
Let r and h (in cm) be the radius of the base and the height of the pillar then,
Radius = 8 cm
Diameter = 2r = 2 * 8 = 16 cm
Ï€rh = 22/7 * 8 * h = 704
Height = 28cm
Radius : Diameter = 16 : 14 = 8 : 7
9) Answer: b)
A/c to Q. 1day work by P & Q = 10U+ 5U = 15U.
2 daysâ€™ work by P, R&S = 2 (10 + 4 + 2) units
Total work in 3 days = 15 units + 32 units = 47 units
Remaining work = 100 units â€“ 47 units = 5 3 units
According to the questions remaining work will be done by R & S on alternate day,
So, 1 day work for R = 4 units.
& 1 day work for S = 2 units
In 2 days R & S complete the 6 unit works.
2 days ->6 units
2 days ->Â 6 units….. And so on…
16 days total work by R & S = 6 Â 8 = 48 units
Remaining = 53 units â€“ 48 units = 5 units => 4 units by R in one day + 1 units by S in Â½ day
Total days = 1+ 2 + 16 + 1 + Â½ = 20 days
10) Answer: c)
Let x (in liters) of 4 : 5 milk solution is added to 90 liters of 60% milk solution.
Volume of the milk before = Volume of the milk after