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QUANT PRACTICE SET 30 FOR SBI PO PRELIM 2020

QUANT PRACTICE SET 30 FOR SBI PO PRELIM 2020


SBI Clerk and SBI PO Prelims exam are going to be held in the upcoming months. We have already provided you with the PDFs of many topics of Quantitative Aptitude like Simplification/Approximation, Number Series,datainterpretation.important Arithmetic questions to prepare for SBI Clerk and PO Pre exam. Practicing these questions will help you to know about the level of the questions. To increase your speed and accuracy, enhance your calculations.

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                           START TEST

1.The average age of employees working in the company is 35 years. If there are total 40 employees and the retirement age is 60 years, find the average age of these employees in next 2 yearswhen 10 employees will retire.
A) 87/13 B) 83/7
C) 86/13 D) 88/3
E) None of these

2. A dishonest shopkeeper marked the price of 1 kg of rice at 50% above its cost price. While buying he got 1250 grams instead of one kg of rice and while selling he used false weight and gave 1176 gramsi of one kg by allowing 20% discount. What
is his actual profit/loss percent?
A) Profit 27.55% B) Loss 22.3%
C) Loss 21.6% D) Profit 25.45%
E) None of these

3. A man borrowed Rs. X at a simple interest of 20% per annum for 6 years and Rs. Y at a simple interest of 10% per annum for 6 years. If he had borrowedRs. (X + Y) at a simple interest of 16% for 6 years,
he would have paid the same interest. Find the ratio ofX and Y.
A) 1 ∶ 1 B) 1 ∶ 2
C) 2 ∶ 1 D) 2 ∶ 3
E) 3 ∶ 2

4. The speed of boat in still water is 18 kmph and speed of stream is 2 kmph. Boat travels total of 320 km upstream and downstream. It take 3 hour more to travel the same distance in upstream compare to downstream. Time taken to travel the upstream is ....... min.
A) 559.23 B) 602.5
C) 633.6 D) 643.75
E) 661.96

5. If the diagonal of a rectangle is 10 cm long, find the relation between its area ‘A’ and perimeter ‘P’.
A) A = P²/2 – 200
 B) A = P²/4 – 100
C) A = P²/8 – 50
 D) A = P²/8 - 505
E) A = P²/4 - 200

6. In the given question, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answer
I. 7x2- 168x = 0
II. 11y2- 891 = 0
A) x < y B) x > y
C) x ≤ y D) x ≥ y
E) x = y or the relationship cannot be
determined

7. Directions: In the following questions twoequations numbered I and II are given. You have tosolve both the equations and give the answer:
I. 35x2 + 26x + 3 = 0
II. 7y2- 12y - 4 = 0
A) x > y B) x ≥ y
C) x < y D) x ≤ y
E) x = y or the relationship cannot be 
established

8. Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations and give the answer:
I. x3 + 5x2 = 6x
II. y3 = 5y2 – 6y
A) x > y B) x ≥ y
C) x < y D) x ≤ y
E) x = y or the relationship cannot be 
established

9. Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations and give the answer:
I. 25x2- 30x + 8 = 0
II. y2- 2y + 1 = 0
A) x > y B) x ≥ y
C) x < y D) x ≤ y
E) x = y or the relationship cannot be 
established

10. Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations and give the answer:
(i) 2x2 + 7x + 3 = 0
(ii) 4y2 + 16y + 15 = 0
A) x > y B) x > y
C) x < y D) x < y
E) x = y or no relation can be established 
between x and y.





SOLUTIONS:

1.Total employees in the company = 40
Average age of employees = 35
Total age of employees = (40 × 35) = 1400
In next 2 years,
Total remaining employees = 40 – 10 = 30
Retirement age = 60 years
Total age of 30 employees after two years = 1400 + (40 × 2) – (60 × 10)
= 1400 + 80 – 600 = 880
∴ Average age after two years = 880/30 = 88/3 years.

2.Let the actual cost of 1 kg of rice be 100x
So, the cost of per gram of purchased rice is = (100x/1250) = 2x/25
Marked price = (100 + 50) % of 100x = 150x
Selling price per gram of rice = (100 - 20) % of MP × (1/1176) = 150x × 0.8 × (1/1176) = 5x/49
∴ His profit% is = [(5x/49 - 2x/25) /(2x/25) ] × 100 = (27/98) × 100 = 27.55%

3.Simple interest = (Principal × Rate × Time)/100
Simple interest on Rs. X = (X × 20 × 6)/100 = 6X/5
Simple interest on Rs. Y = (Y × 10 × 6)/100 = 3Y/5
Simple interest on Rs. (X + Y) = [(X + Y) × 16 × 6]/100 = (24/25)(X + Y)
But, the simple interest paid is equal,
⇒ 6X/5 + 3Y/5 = 24X/25 + 24Y/25
⇒ X/Y = 9/6 = 3/2
∴ X ∶ Y = 3 ∶ 2

4.Let assume time taken to travel downstream as x hr.,
Speed of boat in upstream = Speed of boat - speed of stream = 18 - 2 = 16 kmph
Speed of boat in downstream = speed of boat + speed of stream = 18 + 2 = 20 kmph
Time for upstream = x + 3 hr.
By the formula,
Distance = speed × time,
Distance in upstream + Distance in downstream = 320
⇒ 16 × (x + 3) + 20 × x = 320
⇒ 16x + 48 + 20x = 320
⇒ 36x = 320 - 48
⇒ 36x = 272
⇒ x = 7.56 hr
Time taken for upstream = x + 3 = 7.56 + 3 = 10.56 hr
∴ Upstream time in min = 10.56 × 60 = 633.6 min.

5.Now, for a rectangle with length ‘l’, breadth ‘b’ and diagonal ‘d’, we can write,
⇒ d2 = l2 + b2
Area of rectangle = A = lb
Perimeter of rectangle = P = 2(l + b)
∵ (a2 + b2) = (a + b)2- 2ab
⇒ (l2 + b2) = (l + b)2- 2lb
⇒ d2 = (P/2)2- 2A
⇒ (10)2 = P2/4 - 2A
⇒ 100 = P2/4 - 2A
⇒ 2A = P2/4 - 100
⇒ A = P2/8 - 50

6.I. 7x2- 168x = 0
⇒ 7x(x - 24) = 0
Then, x = 0 or x = 24
II. 11y2- 891 = 0
⇒ 11(y2- 81) = 0
⇒ (y2- 92) = 0
Use: (x2- y2) = (x - y) (x + y)
⇒ (y - 9) (y + 9) = 0
Then, y = 9 or y = -9
So, when x = 0, x < y for y = 9 and x > y for y
= (-9)
And when x = 24, x > y for y = 9 and x > y for y = (-9)
∴ The relationship cannot be determined.

7.
I. 35x2 + 26x + 3 = 0
⇒ 35x2 + 21x + 5x + 3 = 0
⇒ 7x(5x + 3) + 1(5x + 3) = 0
⇒ (5x + 3) (7x + 1) = 0
⇒ x = -3/5 or x = -1/7
II. 7y2- 12y - 4 = 0
⇒ 7y2- 14y + 2y - 4 = 0
⇒ 7y(y - 2) + 2(y - 2) = 0
⇒ (y - 2) (7y + 2) = 0
⇒ y = 2 or y = -2/7
When x = -3/5, x < y for y = 2 and x < y for y = -2/7
And when x = -1/7, x < y for y = 2 and x > y for y = -2/7
∴ The relationship cannot be established.

8.
I. x3 + 5x2 = 6x
⇒ x(x2 + 5x – 6) = 0
⇒ x2 + 6x – x – 6 = 0
⇒ x(x + 6) – (x + 6) = 0
⇒ (x + 6) (x - 1) = 0
⇒ x = -6, 1 and 0
II. y3 = 5y2 – 6y
⇒ y(y2 – 5y + 6) = 0
⇒ y2- 3y - 2y + 6 = 0
⇒ y(y - 3) – 2(y - 3) = 0
⇒ (y – 3) (y –2) = 0
⇒ y = 3, 2 and 0
When x = -6, x < y for y = 3, x < y for y = 2 and x < y for y = 0
When x = 1, x < y for y = 3, x < y for y = 2 and x > y for y = 0
When x = 0, x < y for y = 3, x < y for y = 2 and x = y for y = 0
∴ The relationship cannot be established

9.
I. 25x2-30x + 8 = 0
⇒ 25x2- 20x – 10x + 8 = 0
⇒ 5x(5x - 4) - 2(5x - 4) = 0
⇒ (5x - 4) (5x - 2) = 0
⇒ x = 4/5 or x = 2/5
II. y2- 2y + 1 = 0
⇒ y2- y - y + 1 = 0
⇒ y(y - 1) - 1(y - 1) = 0
⇒ (y - 1) (y - 1) = 0
⇒ y = 1
When x = 4/5, x < y for y = 1
And when x = 2/5, x < y for y = 1
∴ x < y.

10.
I. 2x2 + 7x + 3 = 0
⇒ 2x2 + x + 6x + 3 = 0
⇒ x(2x + 1) + 3(2x + 1) = 0
⇒ (x + 3)(2x + 1) = 0
Then, x = (-3) or x = (-1/2)
II. 4y2 + 16y + 15 = 0
⇒ 4y2 + 6y + 10y + 15 = 0
⇒ 2y(2y + 3) + 5(2y + 3) = 0
⇒ (2y + 5)(2y + 3) = 0
Then, y = (-5/2) or y = (-3/2)
So, when x = (-1/2), x > y for y = (-5/2) and x > y for y = (-3/2)
And when x = (-3), x < y for y = (-5/2) and x < y for y = (-3/2)
∴ So, the relationship can not be determined.