# Daily Quantitative Aptitude Quiz-02 February 2022

Safalta Experts Published by: Shashank Srivastava Updated Mon, 07 Feb 2022 09:33 PM IST

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Q.1 What will come in the place of the question mark ‘?’ in the following question?
(3/13) of [325/(3)ˉ³] × ?² = 25 × 10⁴ × (1.5)⁴
A. 675
B. 75
C. 575
D. 625
E. Source: Safalta.com

25
Solution-
(3/13) of [325/(3)ˉ³] × ?² = 25 × 10⁴ × (1.5)⁴
=> 3 × 25 × 3³ × ?² = 25 × 10⁴ × (1.5)⁴
=> 3(¹⁺³) × 25 × ?² = 25 × 10000 × (3)⁴/(2)⁴
=> ?² = 10000/16
=> ?² = 625
=> ? = 25
Hence, the value of (?) is 25.

Q.2 What should come in the place of the question mark ‘?’ in the following question?
12 × 87 + 12 × 114 + 93 × 12 – 44 × 12 = ?
A. 2000
B. 2500
C. 3000
D. 3500
E. None of these
Solution-
12 × 87 + 12 × 114 + 93 × 12 – 44 × 12 = ?
=> 1044 + 1368 + 1116 – 528 = ?
=> 3528 – 528 = ?
=> ? = 3000
Hence, the value of ‘?’ is 3000.

Q.3 A milkman, in 50 litres mixture of milk and water keeps the ratio of milk and water in 3 : 2. One day he decided this ratio is to be 2 : 3, then find quantity of water to be further added in the mixture.
A. 25 litres
B. 30 litres
C. 40 litres
D. 41 litres
E. None of these
Solution-
In actual ratio
Quantity of milk in the mixture,
=> (50 × 3)/5 litres = 30 litres,
Quantity of water in the mixture,
=> (50 – 30)litres = 20 litres
According of question new ratio = 2: 3
Let the quantity of water added in the mixture be x liters
Then,
Milk: Water = 30: (20 + x)
=> 2/3  = 30/(20 + x)
=> 90 = 40 + 2x
=> 50 = 2x
=> x = 25 liters
Hence, the quantity of water to be added in the mixture is 25 litres.

Q.4 An amount of Rs. 400 becomes Rs. 424 in 3 years at a certain rate of simple interest, If the rate of interest increased by 8%, then find what amount will Rs. 400 becomes in 2 years?
A. Rs. 450
B. Rs. 425
C. Rs. 480
E. None of these
Solution-
S.I. = Amount – Principle = 424 – 400 = Rs. 24
R = (24 × 100)/(400 × 3) = 2%
It is given in the question that new rate is 8% more than previous rate of interest.
New rate = 2% + 8% = 10%
New S.I. = (400 × 10 × 2)/100 = Rs. 80
New Amount = 400 + 80 = Rs. 480
Hence, the new amount is Rs. 480

Q.5 A train 300 metre long crosses a lamppost in 25 seconds. Calculate the speed of the man running in opposite direction who takes 20 seconds to cross the train.
A. 15 m/s
B. 5 m/s
C. 10 m/s
D. 2 m/s
E. 3 m/s
Solution-
Speed of train = 300/25 = 12 m/s
Let the speed of man be x m/s
Time taken to cross the train = 20s = distance of train/relative speed
Relative speed = 12 + x
20 = 300/(12 + x)
12 + x = 15
Speed of man = 15 – 12 = 3 m/s
Hence, Speed of the running man is 3 m/s.

Q.6 A vegetable vendor claims to sell his vegetables at a cost price but uses a weight of 800 gm instead of kg weight. Thus he makes a profit of:
A. 10%
B. 20%
C. 30%
D. 22%
E. 25%
Solution-
According to the formula,
Profit% = True weight – false weight/false weight × 100
=> 1000 – 800/800 × 100
=> 100/4
=> 25%
Hence, the vegetable vendor made a profit of 25%.

Direction (7-8): Study the bar graph and answer the following questions
In a bar graph, shows the number of students passed in XII class from 6 schools.

Q.7 if the fail percentage of school P is 60% then, find number of students who failed from school P is what percentage of the number of students passes from school T.
A. 80%
B. 133.33%
C. 90%
D. 105%
E. 110%
Solution-
From the given graph,
Students who passed in school P = 80
=> 40% = 80
=> 1% = 2
=> 60% = 60 × 2 = 120 (Failed students of school P)
Now, the number of students passed from school T = 90
Then, the required percentage = (120/90) × 100 = (4/3) × 100
=> 133.33%
Hence, the required percentage is 133.33%.

Q.8 If the ratio between the total number of students who passed to the total number of students who failed from all school is 7: 3, then find the total number of failed students from all schools together.
A. 210
B. 250
C. 140
D. 220
E. None of these
Solution-
Total number of passed students = 80 + 90 + 70 + 50 + 90 + 110 = 490
=> 7 units = 490
=> 1 unit = 70
Failed students = 3 units = 70 × 3 = 210
Hence, the total number of failed student from all school is 210.

Q.9 In the given question, two equations numbered I and II are given. Solve both the equations and mark the appropriate answer.
I. 5x² - 18x + 9 = 0
II. 3y² + 5y – 2 = 0
A. x > y
B. x < y
C. x  y
D. x  y
E. x = y or relationship between x and y cannot be established
Solution-
I. 5x² - 18x + 9 = 0
=> 5x² - 15x – 3x + 9 = 0
=> 5x(x – 3) -3(x – 3) = 0
=> (5x – 3)(x – 3) = 0
=> x = 3/5 or 3
II. 3y² + 5y – 2 = 0
=> 3y² + 6y - y – 2 = 0
=> 3y(y + 2) -1(y + 2)
=> (3y – 1)(y + 2) = 0
=> y = 1/3 or -2
Clearly x > y.

Q.10 In the given question, two equations numbered I and II are given. Solve both the equations and mark the appropriate answer.
I. 3x² - 7x + 2 = 0
II. 2y² - 11y + 15 = 0
A. x > y
B. x < y
C. x  y
D. x  y
E. x = y or relationship between x and y cannot be established
Solution-
I. 3x² - 7x + 2 = 0
=>  3x² - 6x - x + 2 = 0
=> 3x(x – 2) -1(x -2) = 0
=> (x – 2)(3x – 1) = 0
=> x = 2 or 1/3

II. 2y² - 11y + 15 = 0
=> 2y² - 6y – 5y + 15 = 0
=> 2y(y – 3) – 5(y – 3) = 0
=> (y – 3)(2y – 5) = 0
=> y = 3 or 5/2
Clearly x < y.

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